MULTIVARIABLE
CALCULUS
EARLY TRANSCENDENTALS
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MULTIVARIABLE CALCULUS
EARLY TRANSCENDENTALS SIXTH EDITION
JAMES STEWART McMASTER UNIVERSITY
AUSTRALIA N BRAZIL N CANADA N MEXICO N SINGAPORE N SPAIN N UNITED KINGDOM N UNITED STATES
Publisher N Bob Pirtle
Assistant Editor N Stacy Green
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Multivariable Calculus: Early Transcendentals, Sixth Edition
James Stewart
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CONTENTS
Preface xi
To the Student xxi
10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 620
10.1 Curves Defined by Parametric Equations 621 Laboratory Project N Running Circles around Circles 629
10.2 Calculus with Parametric Curves 630 Laboratory Project N Bezier Curves 639
10.3 Polar Coordinates 639
10.4 Areas and Lengths in Polar Coordinates
10.5 Conic Sections 654
10.6 Conic Sections in Polar Coordinates
Review 669
Problems Plus 672
11 INFINITE SEQUENCES AND SERIES
11.1 Sequences 675
Laboratory Project N Logistic Sequences
650 662
11.2 Series 687
11.3 The Integral Test and Estimates of Sums 697
11.4 The Comparison Tests 705
11.5 Alternating Series 710
11.6 Absolute Convergence and the Ratio and Root Tests 714
11.7 Strategy for Testing Series 721
11.8 Power Series 723
11.9 Representations of Functions as Power Series 728
674
687
vii
viii
CONTENTS
LONDON
PARIS
O
12.1 ThreeDimensional Coordinate Systems
12.2 Vectors 770
12.3 The Dot Product 779
12.4 The Cross Product 786
Discovery Project N The Geometry of a Tetrahedron
12.5 Equations of Lines and Planes 794
Laboratory Project N Putting 3D in Perspective 804
12.6 Cylinders and Quadric Surfaces 804
Review 812
Problems Plus 815
11.10 Taylor and Maclaurin Series 734 Laboratory Project N An Elusive Limit 748
Writing Project N How Newton Discovered the Binomial Series 748
11.11 Applications of Taylor Polynomials 749
Applied Project N Radiation from the Stars 757 Review 758
Problems Plus 761
12 VECTORS AND THE GEOMETRY OF SPACE
764
765
794
13 VECTOR FUNCTIONS 816
13.1 Vector Functions and Space Curves 817
13.2 Derivatives and Integrals of Vector Functions 824
13.3 Arc Length and Curvature 830
13.4 Motion in Space: Velocity and Acceleration 838
Applied Project N Keplers Laws 848 Review 849
Problems Plus 852
14 PARTIAL DERIVATIVES 854
14.1 Functions of Several Variables 855
14.2 Limits and Continuity 870
14.3 Partial Derivatives 878
14.4 Tangent Planes and Linear Approximations 892
14.5 The Chain Rule 901
14.6 Directional Derivatives and the Gradient Vector 910
14.7 Maximum and Minimum Values 922
Applied Project N Designing a Dumpster 933
Discovery Project N Quadratic Approximations and Critical Points 933
14.8 Lagrange Multipliers 934
Applied Project N Rocket Science 941
Applied Project N HydroTurbine Optimization 943 Review 944
Problems Plus 948
15 MULTIPLE INTEGRALS 950
15.1 Double Integrals over Rectangles 951
15.2 Iterated Integrals 959
15.3 Double Integrals over General Regions 965
15.4 Double Integrals in Polar Coordinates 974
15.5 Applications of Double Integrals 980
15.6 Triple Integrals 990
Discovery Project N Volumes of Hyperspheres 1000
15.7 Triple Integrals in Cylindrical Coordinates 1000
Discovery Project N The Intersection of Three Cylinders 1005
15.8 Triple Integrals in Spherical Coordinates 1005
Applied Project N Roller Derby 1012
15.9 Change of Variables in Multiple Integrals 1012
Review 1021
Problems Plus 1024
Copyright 2008 Thomson Learning, Inc. All Rights Reserved.
CONTENTS
ix
x
CONTENTS
16 VECTOR CALCULUS
1026
16.1 Vector Fields
16.2 Line Integrals
16.3 The Fundamental Theorem for Line Integrals 1046
16.4 Greens Theorem 1055
16.5 Curl and Divergence 1061
16.6 Parametric Surfaces and Their Areas 1070
16.7 Surface Integrals 1081
16.8 Stokes Theorem 1092
Writing Project N Three Men and Two Theorems 1098
16.9 The Divergence Theorem 1099
16.10 Summary 1105
Review 1106
Problems Plus 1109
17 SECONDORDER DIFFERENTIAL EQUATIONS 1110
17.1 SecondOrder Linear Equations 1111
17.2 Nonhomogeneous Linear Equations 1117
17.3 Applications of SecondOrder Differential Equations 1125
17.4 Series Solutions 1133
Review 1137
APPENDIXES A1
F Proofs of Theorems A2
H Complex Numbers A5
I Answers to OddNumbered Exercises A13
INDEX A41
1027 1034
PREFACE
A great discovery solves a great problem but there is a grain of discovery in the solution of any problem.Your problem may be modest; but if it challenges your curiosity and brings into play your inventive faculties, and if you solve it by your own means, you may experience the tension and enjoy the triumph of discovery.
GEORGE POLYA
The art of teaching, Mark Van Doren said, is the art of assisting discovery. I have tried to write a book that assists students in discovering calculusboth for its practical power and its surprising beauty. In this edition, as in the first five editions, I aim to convey to the stu dent a sense of the utility of calculus and develop technical competence, but I also strive to give some appreciation for the intrinsic beauty of the subject. Newton undoubtedly experienced a sense of triumph when he made his great discoveries. I want students to share some of that excitement.
The emphasis is on understanding concepts. I think that nearly everybody agrees that this should be the primary goal of calculus instruction. In fact, the impetus for the current calculus reform movement came from the Tulane Conference in 1986, which formulated as their first recommendation:
Focus on conceptual understanding.
I have tried to implement this goal through the Rule of Three: Topics should be pre sented geometrically, numerically, and algebraically. Visualization, numerical and graph ical experimentation, and other approaches have changed how we teach conceptual reasoning in fundamental ways. More recently, the Rule of Three has been expanded to become the Rule of Four by emphasizing the verbal, or descriptive, point of view as well.
In writing the sixth edition my premise has been that it is possible to achieve concep tual understanding and still retain the best traditions of traditional calculus. The book con tains elements of reform, but within the context of a traditional curriculum.
ALTERNATIVE VERSIONS
I have written several other calculus textbooks that might be preferable for some instruc tors. Most of them also come in single variable and multivariable versions.
N Calculus, Sixth Edition, is similar to the present textbook except that the exponential, logarithmic, and inverse trigonometric functions are covered in the second semester.
N Essential Calculus is a much briefer book 800 pages, though it contains almost all of the topics in Calculus, Sixth Edition. The relative brevity is achieved through briefer exposition of some topics and putting some features on the website.
N Essential Calculus: Early Transcendentals resembles Essential Calculus, but the expo nential, logarithmic, and inverse trigonometric functions are covered in Chapter 3.
xi
xii
PREFACE
CONCEPTUAL EXERCISES
N Calculus: Concepts and Contexts, Third Edition, emphasizes conceptual understanding even more strongly than this book. The coverage of topics is not encyclopedic and the material on transcendental functions and on parametric equations is woven throughout the book instead of being treated in separate chapters.
N Calculus: Early Vectors introduces vectors and vector functions in the first semester and integrates them throughout the book. It is suitable for students taking Engineering and Physics courses concurrently with calculus.
WHATS NEW IN THE SIXTH EDITION?
Here are some of the changes for the sixth edition of Multivariable Calculus: Early Transcendentals.
N Sections 11.10 and 11.11 are merged into a single section. I had previously featured the binomial series in its own section to emphasize its importance. But I learned that some instructors were omitting that section, so I have decided to incorporate binomial series into 11.10.
N The material on cylindrical and spherical coordinates formerly Section 12.7 is moved to Chapter 15, where it is introduced in the context of evaluating triple integrals.
N New phrases and margin notes have been added to clarify the exposition. N A number of pieces of art have been redrawn.
N The data in examples and exercises have been updated to be more timely. N Extra steps have been provided in some of the existing examples.
N More than 25 of the exercises in each chapter are new. Here are a few of my favorites: 11.11.30, 14.5.44, and 14.8.2021.
N There are also some good new problems in the Problems Plus sections. See, for instance, Problem 24 on page 763.
N Tools for Enriching Calculus TEC has been completely redesigned and is accessible on the Internet at www.stewartcalculus.com. It now includes what we call Visuals, brief animations of various figures in the text. In addition, there are now Visual, Modules, and Homework Hints for the multivariable chapters. See the description on page xiii.
N The symbol V has been placed beside examples an average of three per section for which there are videos of instructors explaining the example in more detail. This material is also available on DVD. See the description on page xix.
FEATURES
The most important way to foster conceptual understanding is through the problems that we assign. To that end I have devised various types of problems. Some exercise sets begin with requests to explain the meanings of the basic concepts of the section. See, for instance, the first few exercises in Sections 14.2, and 14.3. Similarly, all the review sec tions begin with a Concept Check and a TrueFalse Quiz. Other exercises test conceptual understanding through graphs or tables see Exercises 10.1.2427, 11.10.2, 13.2.12, 13.3.3337, 14.1.12, 14.1.3038, 14.3.310, 14.6.12, 14.7.34, 15.1.510, 16.1.1118, 16.2.1718, and 16.3.12.
GRADED EXERCISE SETS
REALWORLD DATA
Each exercise set is carefully graded, progressing from basic conceptual exercises and skill development problems to more challenging problems involving applications and proofs.
My assistants and I spent a great deal of time looking in libraries, contacting companies and government agencies, and searching the Internet for interesting realworld data to intro duce, motivate, and illustrate the concepts of calculus. As a result, many of the examples and exercises deal with functions defined by such numerical data or graphs. For instance, functions of two variables are illustrated by a table of values of the windchill index as a function of air temperature and wind speed Example 2 in Section 14.1. Partial derivatives are introduced in Section 14.3 by examining a column in a table of values of the heat index perceived air temperature as a function of the actual temperature and the relative humid ity. This example is pursued further in connection with linear approximations Example 3 in Section 14.4. Directional derivatives are introduced in Section 14.6 by using a temper ature contour map to estimate the rate of change of temperature at Reno in the direction of Las Vegas. Double integrals are used to estimate the average snowfall in Colorado on December 2021, 2006 Example 4 in Section 15.1. Vector fields are introduced in Section 16.1 by depictions of actual velocity vector fields showing San Francisco Bay wind patterns.
One way of involving students and making them active learners is to have them work per haps in groups on extended projects that give a feeling of substantial accomplishment when completed. I have included four kinds of projects: Applied Projects involve applica tions that are designed to appeal to the imagination of students. The project after Section 14.8 uses Lagrange multipliers to determine the masses of the three stages of a rocket so as to minimize the total mass while enabling the rocket to reach a desired velocity. Laboratory Projects involve technology; the one following Section 10.2 shows how to use Bezier curves to design shapes that represent letters for a laser printer. Discovery Projects explore aspects of geometry: tetrahedra after Section 12.4, hyperspheres after Section 15.6, and intersections of three cylinders after Section 15.7. The Writing Project after Section 17.8 explores the historical and physical origins of Greens Theorem and Stokes Theorem and the interactions of the three men involved. Many additional projects can be found in the Instructors Guide.
The availability of technology makes it not less important but more important to clearly understand the concepts that underlie the images on the screen. But, when properly used, graphing calculators and computers are powerful tools for discovering and understanding those concepts. This textbook can be used either with or without technology and I use two special symbols to indicate clearly when a particular type of machine is required. The icon ; indicates an exercise that definitely requires the use of such technology, but that is not to say that it cant be used on the other exercises as well. The symbol CAS is reserved for problems in which the full resources of a computer algebra system like Derive, Maple, Mathematica, or the TI8992 are required. But technology doesnt make pencil and paper obsolete. Hand calculation and sketches are often preferable to technology for illustrating and reinforcing some concepts. Both instructors and students need to develop the ability to decide where the hand or the machine is appropriate.
TEC is a companion to the text and is intended to enrich and complement its contents. It is now accessible from the Internet at www.stewartcalculus.com. Developed by Har vey Keynes, Dan Clegg, Hubert Hohn, and myself, TEC uses a discovery and exploratory approach. In sections of the book where technology is particularly appropriate, marginal icons direct students to TEC modules that provide a laboratory environment in which they can explore the topic in different ways and at different levels. Visuals are animations of fig ures in text; Modules are more elaborate activities and include exercises. Instructors can
PROJECTS
TECHNOLOGY
PREFACExiii
TOOLS FOR ENRICHINGTM CALCULUS
xiv
PREFACE
ENHANCED WEBASSIGN
WEBSITE www.stewartcalculus.com
choose to become involved at several different levels, ranging from simply encouraging students to use the Visuals and Modules for independent exploration, to assigning spe cific exercises from those included with each Module, or to creating additional exercises, labs, and projects that make use of the Visuals and Modules.
TEC also includes Homework Hints for representative exercises usually odd numbered in every section of the text, indicated by printing the exercise number in red. These hints are usually presented in the form of questions and try to imitate an effective teaching assistant by functioning as a silent tutor. They are constructed so as not to reveal any more of the actual solution than is minimally necessary to make further progress.
Technology is having an impact on the way homework is assigned to students, particu larly in large classes. The use of online homework is growing and its appeal depends on ease of use, grading precision, and reliability. With the sixth edition we have been work ing with the calculus community and WebAssign to develop an online homework system. Up to 70 of the exercises in each section are assignable as online homework, including free response, multiple choice, and multipart formats.
The system also includes Active Examples, in which students are guided in stepbystep tutorials through text examples, with links to the textbook and to video solutions.
This site has been renovated and now includes the following.
N Algebra Review
N Lies My Calculator and Computer Told Me
N History of Mathematics, with links to the better historical websites
N Additional Topics complete with exercise sets: Fourier Series, Formulas for the Remainder Term in Taylor Series, Rotation of Axes
N Archived Problems Drill exercises that appeared in previous editions, together with their solutions
N Challenge Problems some from the Problems Plus sections from prior editions
N Links, for particular topics, to outside web resources
N The complete Tools for Enriching Calculus TEC Modules, Visuals, and Homework Hints
CONTENT
This chapter introduces parametric and polar curves and applies the methods of calculus to them. Parametric curves are well suited to laboratory projects; the two presented here involve families of curves and Bezier curves. A brief treatment of conic sections in polar coordinates prepares the way for Keplers Laws in Chapter 13.
The convergence tests have intuitive justifications see page 697 as well as formal proofs. Numerical estimates of sums of series are based on which test was used to prove conver gence. The emphasis is on Taylor series and polynomials and their applications to physics. Error estimates include those from graphing devices.
The material on threedimensional analytic geometry and vectors is divided into two chap ters. Chapter 12 deals with vectors, the dot and cross products, lines, planes, and surfaces.
11 N
Infinite Sequences and Series
12 N Vectors and The Geometry of Space
10 N
Parametric Equations and Polar Coordinates
13 N Vector Functions
This chapter covers vectorvalued functions, their derivatives and integrals, the length and curvature of space curves, and velocity and acceleration along space curves, culminating in Keplers laws.
Functions of two or more variables are studied from verbal, numerical, visual, and alge braic points of view. In particular, I introduce partial derivatives by looking at a specific column in a table of values of the heat index perceived air temperature as a function of the actual temperature and the relative humidity. Directional derivatives are estimated from contour maps of temperature, pressure, and snowfall.
Contour maps and the Midpoint Rule are used to estimate the average snowfall and average temperature in given regions. Double and triple integrals are used to compute probabilities, surface areas, and in projects volumes of hyperspheres and volumes of intersections of three cylinders. Cylindrical and spherical coordinates are introduced in the context of eval uating triple integrals.
Vector fields are introduced through pictures of velocity fields showing San Francisco Bay wind patterns. The similarities among the Fundamental Theorem for line integrals, Greens Theorem, Stokes Theorem, and the Divergence Theorem are emphasized.
Since firstorder differential equations are covered in Chapter 9, this final chapter deals with secondorder linear differential equations, their application to vibrating springs and electric circuits, and series solutions.
ANCILLARIES
Multivariable Calculus, Early Transcendentals, Sixth Edition, is supported by a complete set of ancillaries developed under my direction. Each piece has been designed to enhance student understanding and to facilitate creative instruction. The tables on pages xixxx describe each of these ancillaries.
ACKNOWLEDGMENTS
The preparation of this and previous editions has involved much time spent reading the reasoned but sometimes contradictory advice from a large number of astute reviewers. I greatly appreciate the time they spent to understand my motivation for the approach taken. I have learned something from each of them.
Marilyn Belkin, Villanova University
Philip L. Bowers, Florida State University
Amy Elizabeth Bowman, University of Alabama in Huntsville
M. Hilary Davies, University of Alaska Anchorage
Frederick Gass, Miami University
Paul Triantafilos Hadavas, Armstrong Atlantic State University
Nets Katz, Indiana University Bloomington
James McKinney, California State Polytechnic University, Pomona Martin Nakashima, California State Polytechnic University, Pomona Lila Roberts, Georgia College and State University
14 N
Partial Derivatives
15 N
Multiple Integrals
16 N Vector Calculus
17 N SecondOrder Differential Equations
PREFACExv
SIXTH EDITION REVIEWERS
xviPREFACE
TECHNOLOGY REVIEWERS
Maria Andersen, Muskegon Community College
Eric Aurand, Eastfield College
Joy Becker, University of WisconsinStout
Przemyslaw Bogacki, Old Dominion University
Amy Elizabeth Bowman, University of Alabama in Huntsville Monica Brown, University of MissouriSt. Louis
Roxanne Byrne, University of Colorado at Denver and Health Sciences Center
Teri Christiansen, University of MissouriColumbia Bobby Dale Daniel, Lamar University
Jennifer Daniel, Lamar University
Andras Domokos, California State University, Sacramento Timothy Flaherty, Carnegie Mellon University
Lee Gibson, University of Louisville
Jane Golden, Hillsborough Community College Semion Gutman, University of Oklahoma
Diane Hoffoss, University of San Diego
Lorraine Hughes, Mississippi State University
Jay Jahangiri, Kent State University
John Jernigan, Community College of Philadelphia
PREVIOUS EDITION REVIEWERS
B. D. Aggarwala, University of Calgary
John Alberghini, Manchester Community College Michael Albert, CarnegieMellon University
Daniel Anderson, University of Iowa
Donna J. Bailey, Northeast Missouri State University Wayne Barber, Chemeketa Community College
Neil Berger, University of Illinois, Chicago
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Bruce Gilligan, University of Regina Matthias K. Gobbert, University of Maryland,
Baltimore County
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Stuart Goldenberg, California Polytechnic State University John A. Graham, Buckingham BrowneNichols School Richard Grassl, University of New Mexico
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Prescott Campus
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Matt Kaufman
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at Geneseo
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Richard Nowakowski, Dalhousie University Hussain S. Nur, California State University, Fresno Wayne N. Palmer, Utica College
Vincent Panico, University of the Pacific
F. J. Papp, University of MichiganDearborn Mike Penna, Indiana UniversityPurdue University
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The George Washington University
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PREFACE
xvii
xviii
PREFACE
In addition, I would like to thank George Bergman, David Cusick, Stuart Goldenberg, Larry Peterson, Dan Silver, Norton Starr, Alan Weinstein, and Gail Wolkowicz for their suggestions; Dan Clegg for his research in libraries and on the Internet; Al Shenk and Den nis Zill for permission to use exercises from their calculus texts; John Ringland for his refinements of the multivariable Maple art; COMAP for permission to use project mate rial; George Bergman, David Bleecker, Dan Clegg, Victor Kaftal, Anthony Lam, Jamie Lawson, Ira Rosenholtz, Paul Sally, Lowell Smylie, and Larry Wallen for ideas for exer cises; Dan Drucker for the roller derby project; Thomas Banchoff, Tom Farmer, Fred Gass, John Ramsay, Larry Riddle, and Philip Straffin for ideas for projects; Dan Anderson, Dan Clegg, Jeff Cole, Dan Drucker, and Barbara Frank for solving the new exercises and sug gesting ways to improve them; Marv Riedesel and Mary Johnson for accuracy in proof reading; and Jeff Cole and Dan Clegg for their careful preparation and proofreading of the answer manuscript.
In addition, I thank those who have contributed to past editions: Ed Barbeau, Fred Brauer, Andy BulmanFleming, Bob Burton, Tom DiCiccio, Garret Etgen, Chris Fisher, Arnold Good, Gene Hecht, Harvey Keynes, Kevin Kreider, E.L. Koh, Zdislav Kovarik, Emile LeBlanc, David Leep, Gerald Leibowitz, Lothar Redlin, Carl Riehm, Peter Rosen thal, Doug Shaw, and Saleem Watson.
I also thank Kathi Townes, Stephanie Kuhns, and Brian Betsill of TECHarts for their production services and the following BrooksCole staff: Cheryll Linthicum, editorial pro duction project manager; Mark Santee, Melissa Wong, and Bryan Vann, marketing team; Stacy Green, assistant editor, and Elizabeth Rodio, editorial assistant; Sam Subity, technol ogy project manager; Rob Hugel, creative director, and Vernon Boes, art director; and Becky Cross, print buyer. They have all done an outstanding job.
I have been very fortunate to have worked with some of the best mathematics editors in the business over the past two decades: Ron Munro, Harry Campbell, Craig Barth, Jeremy Hayhurst, Gary Ostedt, and now Bob Pirtle. Bob continues in that tradition of editors who, while offering sound advice and ample assistance, trust my instincts and allow me to write the books that I want to write.
JAMES STEWART
ANCILLARIES
FOR INSTRUCTORS
Multimedia Manager Instructors Resource CDROM
ISBN 0495012416
Contains all art from the text in both jpeg and PowerPoint formats, key equations and tables from the text, complete prebuilt PowerPoint lectures, and an electronic version of the Instructors Guide.
TEC Tools for EnrichingTM Calculus
by James Stewart, Harvey Keynes, Dan Clegg, and developer Hu Hohn
TEC provides a laboratory environment in which students can explore selected topics. TEC also includes homework hints for representative exercises. Available online at www.stewartcalculus.com.
Instructors Guide
by Douglas Shaw and James Stewart ISBN 0495012548
Each section of the main text is discussed from several view points and contains suggested time to allot, points to stress, text discussion topics, core materials for lecture, workshopdiscus sion suggestions, group work exercises in a form suitable for handout, and suggested homework problems. An electronic version is available on the Multimedia Manager Instructors Resource CDROM.
Complete Solutions Manual
Multivariable
by Dan Clegg and Barbara Frank ISBN 0495012297
Includes workedout solutions to all exercises in the text.
Printed Test Bank
by William Steven Harmon ISBN 0495012424
Contains multiplechoice and shortanswer test items that key directly to the text.
ExamView
ISBN 049538240X
Create, deliver, and customize tests and study guides both print and online in minutes with this easytouse assessment and tutorial software on CD. Includes complete questions from the Printed Test Bank.
JoinIn on TurningPoint
ISBN 049511894X
Enhance how your students interact with you, your lecture, and each other. Thomson BrooksCole is now pleased to offer you bookspecific content for Response Systems tailored to Stewarts Calculus, allowing you to transform your classroom and assess your students progress with instant inclass quizzes and polls. Contact your local Thomson representative to learn more about JoinIn on TurningPoint and our exclusive infrared and radio frequency hardware solutions.
TextSpecific DVDs
ISBN 0495012432
Textspecific DVD set, available at no charge to adopters. Each disk features a 10 to 20minute problemsolving lesson for each section of the chapter. Covers both single and multi variable calculus.
Solution Builder
www.thomsonedu.comsolutionbuilder
The online Solution Builder lets instructors easily build and save personal solution sets either for printing or posting on password protected class websites. Contact your local sales representative for more information on obtaining an account for this instructor only resource.
ANCILLARIES FOR INSTRUCTORS AND STUDENTS
Stewart Specialty Website
www.stewartcalculus.com
Additional Topics N exercises N Challenge Problems N Web Links N
Contents: Algebra Review N
Mathematics N Tools for Enriching Calculus TEC
Enhanced WebAssign
ISBN 0495109630
Instant feedback, grading precision, and ease of use are just three reasons why WebAssign is the most widely used home work system in higher education. WebAssigns homework deliv ery system lets instructors deliver, collect, grade and record assignments via the web. And now, this proven system has been enhanced to include endofsection problems from Stewarts Calculusincorporating exercises, examples, video skill builders and quizzes to promote active learning and provide the immediate, relevant feedback students want.
Table continues on page xx.
Drill History of
Electronic itemsPrinted items
xix
The BrooksCole Mathematics Resource Center Website
www.thomsonedu.commath
When you adopt a Thomson BrooksCole mathematics text, you and your students will have access to a variety of teaching and learning resources. This website features everything from bookspecific resources to newsgroups. Its a great way to make teaching and learning an interactive and intriguing experience..
Maple CDROM
ISBN 0495012378 Maple 10 ISBN 0495390526 Maple 11
Maple provides an advanced, high performance mathematical computation engine with fully integrated numericssymbolics, all accessible from a WYSIWYG technical document environ ment. Available for bundling with your Stewart Calculus text at a special discount.
STUDENT RESOURCES
TEC Tools for EnrichingTM Calculus
by James Stewart, Harvey Keynes, Dan Clegg, and developer Hu Hohn
TEC provides a laboratory environment in which students can explore selected topics. TEC also includes homework hints for representative exercises. Available online at www.stewartcalculus.com.
Interactive Video SkillBuilder CDROM
ISBN 0495012378
Think of it as portable office hours! The Interactive Video Skillbuilder CDROM contains more than eight hours of video instruction. The problems worked during each video lesson are shown next to the viewing screen so that students can try work ing them before watching the solution. To help students evalu ate their progress, each section contains a tenquestion web quiz the results of which can be emailed to the instructor
and each chapter contains a chapter test, with answers to each problem.
Study Guide Multivariable
by Richard St. Andre ISBN 0495012270
Contains a short list of key concepts, a short list of skills to master, a brief introduction to the ideas of the section, an elaboration of the concepts and skills, including extra
workedout examples, and links in the margin to earlier and later material in the text and Study Guide.
Student Solutions Manual Multivariable
by Dan Clegg and Barbara Frank ISBN 0495012289
Provides completely workedout solutions to all oddnumbered exercises within the text, giving students a way to check their answers and ensure that they took the correct steps to arrive at an answer.
CalcLabs with Maple
Multivariable
by Philip Yasskin, Maurice Rahe, and Art Belmonte ISBN 0495012319
CalcLabs with Mathematica
Multivariable
by Selwyn Hollis ISBN 0495118907
Each of these comprehensive lab manuals will help students learn to effectively use the technology tools available to them. Each lab contains clearly explained exercises and a variety of labs and projects to accompany the text.
A Companion to Calculus
by Dennis Ebersole, Doris Schattschneider, Alicia Sevilla, and Kay Somers
ISBN 049501124X
Written to improve algebra and problemsolving skills of stu dents taking a calculus course, every chapter in this companion is keyed to a calculus topic, providing conceptual background and specific algebra techniques needed to understand and solve calculus problems related to that topic. It is designed for calcu lus courses that integrate the review of precalculus concepts or for individual use.
Linear Algebra for Calculus
by Konrad J. Heuvers, William P. Francis, John H. Kuisti, Deborah F. Lockhart, Daniel S. Moak, and Gene M. Ortner ISBN 0534252486
This comprehensive book, designed to supplement the calculus course, provides an introduction to and review of the basic ideas of linear algebra.
Electronic itemsPrinted items xx
TO THE STUDENT
Reading a calculus textbook is different from reading a newspaper or a novel, or even a physics book. Dont be discouraged if you have to read a passage more than once in order to understand it. You should have pencil and paper and calculator at hand to sketch a dia gram or make a calculation.
Some students start by trying their homework problems and read the text only if they get stuck on an exercise. I suggest that a far better plan is to read and understand a section of the text before attempting the exercises. In particular, you should look at the definitions to see the exact meanings of the terms. And before you read each example, I suggest that you cover up the solution and try solving the problem yourself. Youll get a lot more from looking at the solution if you do so.
Part of the aim of this course is to train you to think logically. Learn to write the solu tions of the exercises in a connected, stepbystep fashion with explanatory sentences not just a string of disconnected equations or formulas.
The answers to the oddnumbered exercises appear at the back of the book, in Appen dix I. Some exercises ask for a verbal explanation or interpretation or description. In such cases there is no single correct way of expressing the answer, so dont worry that you havent found the definitive answer. In addition, there are often several different forms in which to express a numerical or algebraic answer, so if your answer differs from mine, dont immediately assume youre wrong. For example, if the answer given in the back of the book is s21 and you obtain 11s2, then youre right and rationalizing the denominator will show that the answers are equivalent.
The icon ; indicates an exercise that definitely requires the use of either a graphing calculator or a computer with graphing software. Section 1.4 discusses the use of these graphing devices and some of the pitfalls that you may encounter. But that doesnt mean that graphing devices cant be used to check your work on the other exercises as well. The symbol CAS is reserved for problems in which the full resources of a computer algebra
xxi
system like Derive, Maple, Mathematica, or the TI8992 are required. You will also encounter the symbol , which warns you against committing an error. I have placed this symbol in the margin in situations where I have observed that a large proportion of my stu dents tend to make the same mistake.
Tools for Enriching Calculus, which is a companion to this text, is referred to by means of the symbol TEC and can be accessed from www.stewartcalculus.com. It directs you to modules in which you can explore aspects of calculus for which the computer is particu larly useful. TEC also provides Homework Hints for representative exercises that are indi cated by printing the exercise number in red: These homework hints ask you questions that allow you to make progress toward a solution without actually giving you the answer. You need to pursue each hint in an active manner with pencil and paper to work out the details. If a particular hint doesnt enable you to solve the problem, you can click to reveal the next hint.
An optional CDROM that your instructor may have asked you to purchase is the Inter active Video Skillbuilder, which contains videos of instructors explaining two or three of the examples in every section of the text.
I recommend that you keep this book for reference purposes after you finish the course. Because you will likely forget some of the specific details of calculus, the book will serve as a useful reminder when you need to use calculus in subsequent courses. And, because this book contains more material than can be covered in any one course, it can also serve as a valuable resource for a working scientist or engineer.
Calculus is an exciting subject, justly considered to be one of the greatest achievements of the human intellect. I hope you will discover that it is not only useful but also intrinsi cally beautiful.
JAMES STEWART
xxii
15.
MULTIVARIABLE
CALCULUS
EARLY TRANSCENDENTALS
620
10
PARAMETRIC EQUATIONS AND POLAR COORDINATES
Parametric equations and polar coordinates enable us to describe a great variety of new curvessome practical, some beautiful, some fanciful, some strange.
So far we have described plane curves by giving y as a function of x yfx or x
as a function of y xty or by giving a relation between x and y that defines y implicitly as a function of xf x, y0. In this chapter we discuss two new methods for describing curves.
Some curves, such as the cycloid, are best handled when both x and y are given in terms of a third variable t called a parameter xf t, ytt. Other curves, such as the cardioid, have their most convenient description when we use a new coordinate system, called the polar coordinate system.
yC
x, y ft, gt
0x FIGURE 1
10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS
Imagine that a particle moves along the curve C shown in Figure 1. It is impossible to describe C by an equation of the form yf x because C fails the Vertical Line Test. But the x and ycoordinates of the particle are functions of time and so we can write xf t and ytt. Such a pair of equations is often a convenient way of describing a curve and gives rise to the following definition.
Suppose that x and y are both given as functions of a third variable t called a param eter by the equations
xft ytt
called parametric equations. Each value of t determines a point x, y, which we can plot in a coordinate plane. As t varies, the point x, y f t, tt varies and traces out a curve C, which we call a parametric curve. The parameter t does not necessarily repre sent time and, in fact, we could use a letter other than t for the parameter. But in many applications of parametric curves, t does denote time and therefore we can interpret x, y f t, tt as the position of a particle at time t.
EXAMPLE 1 Sketch and identify the curve defined by the parametric equations xt2 2t yt1
SOLUTION Each value of t gives a point on the curve, as shown in the table. For instance, if t0, then x0, y1 and so the corresponding point is 0, 1. In Figure 2 we plot the points x, y determined by several values of the parameter and we join them to produce a curve.
N This equation in x and y describes where the particle has been, but it doesnt tell us when the particle was at a particular point. The parametric equations have an advantagethey tell us when the particle was at a point. They also indi cate the direction of the motion.
A particle whose position is given by the parametric equations moves along the curve in the direction of the arrows as t increases. Notice that the consecutive points marked on the curve appear at equal time intervals but not at equal distances. That is because the particle slows down and then speeds up as t increases.
It appears from Figure 2 that the curve traced out by the particle may be a parabola. This can be confirmed by eliminating the parameter t as follows. We obtain ty1 from the second equation and substitute into the first equation. This gives
xt2 2ty12 2y1y2 4y3
and so the curve represented by the given parametric equations is the parabola
xy24y3. M
y
t3 t2
t4
8
t2
t
x
y
2 1 0 1 2 3 4
8 3 0
1 0 3 8
1 0 1 2 3 4 5
t1 t0
0
FIGURE 2
0, 1 t1
x
621
622
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
y
No restriction was placed on the parameter t in Example 1, so we assumed that t could be any real number. But sometimes we restrict t to lie in a finite interval. For instance, the parametric curve
xt2 2t yt1 0t4
0,1
FIGURE 3
8, 5
shown in Figure 3 is the part of the parabola in Example 1 that starts at the point 0, 1 and 0x ends at the point 8, 5. The arrowhead indicates the direction in which the curve is traced
as t increases from 0 to 4.
In general, the curve with parametric equations
xft ytt atb has initial pointf a, ta and terminal pointf b, tb.
V EXAMPLE 2 What curve is represented by the following parametric equations? xcos t ysin t 0t2
SOLUTION If we plot points, it appears that the curve is a circle. We can confirm this impression by eliminating t. Observe that
x2 y2 cos2tsin2t1
Thus the point x, y moves on the unit circle x2y21. Notice that in this example the parameter t can be interpreted as the angle in radians shown in Figure 4. As t increases from 0 to 2 , the point x, ycos t, sin t moves once around the circle in the counterclockwise direction starting from the point 1, 0.
t y 2
cos t, sin t t0
t
1,0 x t2
EXAMPLE 3 What curve is represented by the given parametric equations? xsin 2t ycos 2t 0t2
SOLUTION Again we have
x2 y2 sin2 2tcos2 2t1
0x so the parametric equations again represent the unit circle x2y21. But as t increases from 0 to 2 , the point x, ysin 2t, cos 2t starts at 0, 1 and moves twice
around the circle in the clockwise direction as indicated in Figure 5. M
Examples 2 and 3 show that different sets of parametric equations can represent the same curve. Thus we distinguish between a curve, which is a set of points, and a parametric curve, in which the points are traced in a particular way.
t
0
FIGURE 4
t 3 2
M
t0, , 2
y
0, 1
FIGURE 5
SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS623
EXAMPLE 4 Find parametric equations for the circle with center h, k and radius r.
SOLUTION If we take the equations of the unit circle in Example 2 and multiply the expres sions for x and y by r, we get xr cos t, yr sin t. You can verify that these equations represent a circle with radius r and center the origin traced counterclockwise. We now shift h units in the xdirection and k units in the ydirection and obtain parametric equa tions of the circle Figure 6 with center h, k and radius r:
xhr cos t
ykr sin t
r h, k
0t2
y
FIGURE 6
xhr cos t, ykr sin t
y 1, 1
0x
0x
1, 1
FIGURE 7
V EXAMPLE 5 Sketch the curve with parametric equations xsin t, ysin2t.
SOLUTION Observe that ysin t2x2 and so the point x, y moves on the parabola
yx2. But note also that, since 1sin t1, we have 1×1, so the para metric equations represent only the part of the parabola for which 1×1. Since sin t is periodic, the point x, ysin t, sin2t moves back and forth infinitely often along the parabola from 1, 1 to 1, 1. See Figure 7. M
M
TEC Module 10.1A gives an animation of the relationship between motion along a parametric curve xf t, ytt and motion along the graphs of f and t as functions of t. Clicking on TRIG gives you the family of parametric curves
xa cos bt yc sin dt
Ifyouchooseabcd1andclick on animate, you will see how the graphs of
xcos t and ysin t relate to the circle in Example2.Ifyouchooseabc1,
d2, you will see graphs as in Figure 8. By clicking on animate or moving the tslider to the right, you can see from the color coding how motion along the graphs of xcos t and
ysin 2t corresponds to motion along the para metric curve, which is called a Lissajous figure.
yy
xt
FIGURE 8
xcost
ysin2t
ysin2t
x
xcos t
t
624
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES GRAPHING DEVICES
Most graphing calculators and computer graphing programs can be used to graph curves defined by parametric equations. In fact, its instructive to watch a parametric curve being drawn by a graphing calculator because the points are plotted in order as the correspon ding parameter values increase.
3 EXAMPLE 6 Use a graphing device to graph the curve xy43y2. SOLUTION If we let the parameter be ty, then we have the equations
3 3
xt4 3t2 yt
Using these parametric equations to graph the curve, we obtain Figure 9. It would be possible to solve the given equation xy43y2 for y as four functions of x and graph them individually, but the parametric equations provide a much easier method. M
In general, if we need to graph an equation of the form xty, we can use the para metric equations
xtt yt
Notice also that curves with equations yf x the ones we are most familiar with
graphs of functions can also be regarded as curves with parametric equations
xt yft
Graphing devices are particularly useful when sketching complicated curves. For instance, the curves shown in Figures 10, 11, and 12 would be virtually impossible to pro duce by hand.
FIGURE 9
6.5
FIGURE 10
xt2 sin 2t yt2 cos 5t
8
6.5 2.5
FIGURE 11
2.5
2.5 1 1
1
FIGURE 12
xsintcos 100t ycostsin 100t
3
TEC An animation in Module 10.1B shows how the cycloid is formed as the circle moves.
One of the most important uses of parametric curves is in computeraided design CAD. In the Laboratory Project after Section 10.2 we will investigate special parametric curves, called Bezier curves, that are used extensively in manufacturing, especially in the automotive industry. These curves are also employed in specifying the shapes of letters and other symbols in laser printers.
THE CYCLOID
EXAMPLE 7 The curve traced out by a point P on the circumference of a circle as the circle rolls along a straight line is called a cycloid see Figure 13. If the circle has radius r and rolls along the xaxis and if one position of P is the origin, find parametric equations for the cycloid.
8 2.5 1
x1.5 cos tcos 30t y1.5 sin tsin 30t
SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS625
P
P
y
SOLUTION We choose as parameter the angle of rotation of the circle 0 when P is at the origin. Suppose the circle has rotated through radians. Because the circle has been in contact with the line, we see from Figure 14 that the distance it has rolled from the origin is
OT arc PTr
Therefore the center of the circle is Cr , r. Let the coordinates of P be x, y. Then
from Figure 14 we see that
xOTPQr rsin r sin
yTCQCrrcos r1cosTherefore parametric equations of the cycloid are
xr sinyr1cos
One arch of the cycloid comes from one rotation of the circle and so is described by
0 2 . Although Equations 1 were derived from Figure 14, which illustrates the case where 0 2, it can be seen that these equations are still valid for other values of see Exercise 39.
Although it is possible to eliminate the parameter from Equations 1, the resulting Cartesian equation in x and y is very complicated and not as convenient to work with as the parametric equations. M
One of the first people to study the cycloid was Galileo, who proposed that bridges be built in the shape of cycloids and who tried to find the area under one arch of a cycloid. Later this curve arose in connection with the brachistochrone problem: Find the curve along which a particle will slide in the shortest time under the influence of gravity from a point A to a lower point B not directly beneath A. The Swiss mathematician John Bernoulli, who posed this problem in 1696, showed that among all possible curves that join A to B, as in Figure 15, the particle will take the least time sliding from A to B if the curve is part of an inverted arch of a cycloid.
The Dutch physicist Huygens had already shown that the cycloid is also the solution to the tautochrone problem; that is, no matter where a particle P is placed on an inverted cycloid, it takes the same time to slide to the bottom see Figure 16. Huygens proposed that pendulum clocks which he invented swing in cycloidal arcs because then the pendu lum takes the same time to make a complete oscillation whether it swings through a wide or a small arc.
FAMILIES OF PARAMETRIC CURVES
V EXAMPLE 8 Investigate the family of curves with parametric equations xacos t ya tan tsin t
Cr , r PQ
FIGURE 13
P
r
x OTx
r
FIGURE 14
y
A
cycloid
FIGURE 15
P
PP P
FIGURE 16
B
P
1
What do these curves have in common? How does the shape change as a increases?
626
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
a2
SOLUTION We use a graphing device to produce the graphs for the cases a2, 1, 0.5, 0.2, 0, 0.5, 1, and 2 shown in Figure 17. Notice that all of these curves except the case a0 have two branches, and both branches approach the vertical asymptote xa as x approaches a from the left or right.
a1 a0.5 a0.2
a0.5 a1 a2
When a1, both branches are smooth; but when a reaches 1, the right branch acquires a sharp point, called a cusp. For a between 1 and 0 the cusp turns into a loop, which becomes larger as a approaches 0. When a0, both branches come together and form a circle see Example 2. For a between 0 and 1, the left branch has a loop, which shrinks to become a cusp when a1. For a1, the branches become smooth again, and as a increases further, they become less curved. Notice that the curves with a posi tive are reflections about the yaxis of the corresponding curves with a negative.
These curves are called conchoids of Nicomedes after the ancient Greek scholar Nicomedes. He called them conchoids because the shape of their outer branches resembles that of a conch shell or mussel shell. M
a0
Members of the family xacos t, ya tan t sin t,
all graphed in the viewing rectangle 4, 4 by 4, 4
10.1 EXERCISES
14 Sketch the curve by using the parametric equations to plot points. Indicate with an arrow the direction in which the curve is traced as t increases.
1. x1st, yt2 4t, 0t5
2. x2cost, ytcost, 0t2
3. x5 sin t, yt2, t
xet t, yet t, 2t2
510
a Sketch the curve by using the parametric equations to plot points.Indicatewithanarrowthedirectioninwhichthecurve is traced as t increases.
b Eliminate the parameter to find a Cartesian equation of thecurve.
5. x3t5, y2t1
6. x1t, y52t, 2t3
7. xt2 2, y52t, 3t4
F I G U R E
1 7
8. x13t, y2t2
9.xst, y1t 10. xt2, yt3
1118
a Eliminate the parameter to find a Cartesian equation of the curve.
b Sketch the curve and indicate with an arrow the direction in which the curve is traced as the parameter increases.
11.xsin, ycos, 0
12. x4 cos , y5sin ,22 xsint, ycsct, 0t 2
14. xet 1, ye2t
15. xe2t, yt1
16. xlnt, yst, t1 17. xsinht, ycosht
4.
13.
18. x2 cosh t, y5 sinh t
1922 Describe the motion of a particle with position x, y as t varies in the given interval.
19. x32 cos t, y12 sin t, 2t3 2
20. x2 sin t, y4cos t, 0t3 2
x5 sin t, y2 cos t, t5
22. xsint, ycos2t, 2 t2
23. Suppose a curve is given by the parametric equations xf t, ytt, where the range of f is 1, 4 and the range of t is 2, 3. What can you say about the curve?
24. Match the graphs of the parametric equations xf t and ytt in ad with the parametric curves labeled IIV. Give reasons for your choices.
2527 Use the graphs of xft and ytt to sketch the parametric curve xft, ytt. Indicate with arrows the direction in which the curve is traced as t increases.
25.x y 1
1t 1t 1
26.x y
1
1t 1t
27. x y 1 1
SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS627
21.
1
a
I
t1t
xyy 21
2
1
1
b
1t
1t2x
II
bxt2 2t, yst
c xsin2t, ysintsin2t dxcos5t, ysin2t
e xtsin4t, yt2 cos3t fxsin2t, ycos2t
28.
Match the parametric equations with the graphs labeled IVI. Give reasons for your choices. Do not use a graphing device.
a xt4 t1, yt2
xyy 222
1t1t2x
c III
xyy 221
4t2 4t2
I II III
yyy
x
x
2t 2t12xIV V VI yyy
x
x
d IV
xy2y2 2
2 t 2 t
xx
;29. Graphthecurvexy3y3 y5.
;30. Graphthecurvesyx5 andxyy12 andfindtheir points of intersection correct to one decimal place.
2 x
628
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
31.
33.
a Show that the parametric equations
xx1 x2 x1t yy1 y2 y1t
where 0t1, describe the line segment that joins the points P1x1, y1 and P2x2, y2 .
b Find parametric equations to represent the line segment from 2, 7 to 3, 1.
Use a graphing device and the result of Exercise 31a to draw the triangle with vertices A1, 1, B4, 2, and C1, 5.
Find parametric equations for the path of a particle that moves along the circle x2y124 in the manner described.
a Once around clockwise, starting at 2, 1
b Three times around counterclockwise, starting at 2, 1
c Halfway around counterclockwise, starting at 0, 3
a Find parametric equations for the ellipse
x2a2y2b21. Hint: Modify the equations of the circle in Example 2.
b Use these parametric equations to graph the ellipse when a3 and b1, 2, 4, and 8.
If a and b are fixed numbers, find parametric equations for the curve that consists of all possible positions of the point P in the figure, using the angle as the parameter. Then elimi nate the parameter and identify the curve.
y
; 32.
;
a
b Ox
37. a xt3, yt2
c xe3t, ye2t
38. a xt, yt2 c xet, ye2t
b xt6, yt4
b xcost, ysec2t
x2a cot Sketch the curve.
y2a sin2 C
P
41.
34.
42.
Ifaandbarefixednumbers,findparametricequationsfor the curve that consists of all possible positions of the point P in the figure, using the angle as the parameter. The line seg ment AB is tangent to the larger circle.
y
A
P
c How does the shape of the ellipse change as b varies? ; 3536 Use a graphing calculator or computer to reproduce the
picture.
35. y 2
a
OBx
A curve, called a witch of Maria Agnesi, consists of all pos sible positions of the point P in the figure. Show that para metric equations for this curve can be written as
36. y 4
b
P
2 02x038x
3738 Compare the curves represented by the parametric equa tions. How do they differ?
43.
39. Derive Equations 1 for the case 2 .
40. Let P be a point at a distance d from the center of a circle of radius r. The curve traced out by P as the circle rolls along a straight line is called a trochoid. Think of the motion of a point on a spoke of a bicycle wheel. The cycloid is the spe cial case of a trochoid with dr. Using the same parameter
as for the cycloid and, assuming the line is the xaxis and
0 when P is at one of its lowest points, show that parametric equations of the trochoid are
xr dsin yrdcos Sketch the trochoid for the cases dr and dr.
Ox
y2a
y
a
A
44.
a Find parametric equations for the set of all points P as shown in the figure such that OPAB . This curve
is called the cissoid of Diocles after the Greek scholar Diocles, who introduced the cissoid as a graphical method for constructing the edge of a cube whose volume is twice that of a given cube.
b Use the geometric description of the curve to draw a rough sketch of the curve by hand. Check your work by using the parametric equations to graph the curve.
given by the parametric equations
12 xv0 cos t yv0 sin t2tt
where t is the acceleration due to gravity 9.8 ms2. aIfagunisfiredwith 30andv0 500ms,when
will the bullet hit the ground? How far from the gun will it hit the ground? What is the maximum height reached by the bullet?
b Use a graphing device to check your answers to part a. Then graph the path of the projectile for several other values of the angle to see where it hits the ground. Summarize your findings.
c Show that the path is parabolic by eliminating the parameter.
Investigate the family of curves defined by the parametric equations xt2, yt3ct. How does the shape change as c increases? Illustrate by graphing several members of the family.
LABORATORY PROJECT RUNNING CIRCLES AROUND CIRCLES629
y
P
B A
x2a
; 45.
Oax
Suppose that the position of one particle at time t is given by x13 sin t y12 cos t 0t2
and the position of a second particle is given by
x2 3cost y2 1sint 0t2
a Graph the paths of both particles. How many points of intersection are there?
;
;
b Are any of these points of intersection collision points?
In other words, are the particles ever at the same place at
the same time? If so, find the collision points. ;
c Describe what happens if the path of the second particle is given by
x2 3cost y2 1sint 0t2
46. If a projectile is fired with an initial velocity of v0 meters per second at an angle above the horizontal and air resistance is assumed to be negligible, then its position after t seconds is
;50. Investigatethefamilyofcurvesdefinedbytheparametric equations xcos t, ysin tsin ct, where c0. Start by letting c be a positive integer and see what happens to the shape as c increases. Then explore some of the possibilities that occur when c is a fraction.
; RUNNING CIRCLES AROUND CIRCLES
47.
;48. Theswallowtailcatastrophecurvesaredefinedbythepara metric equations x2ct4t3, yct23t4. Graph several of these curves. What features do the curves have
in common? How do they change when c increases?
Thecurveswithequationsxasinnt,ybcost are called Lissajous figures. Investigate how these curves vary when a, b, and n vary. Take n to be a positive integer.
49.
L A B O R AT O R Y PROJECT
y
C
In this project we investigate families of curves, called hypocycloids and epicycloids, that are generated by the motion of a point on a circle that rolls inside or outside another circle.
1. A hypocycloid is a curve traced out by a fixed point P on a circle C of radius b as C rolls on the inside of a circle with center O and radius a. Show that if the initial position of P is a, 0 and the parameter is chosen as in the figure, then parametric equations of the hypocycloid are
a
bb
O
b P a,0
A x xabcos bcos ab yabsin bsin ab
2.
Use a graphing device or the interactive graphic in TEC Module 10.1B to draw the graphs of hypocycloids with a a positive integer and b1. How does the value of a affect the graph? Show that if we take a4, then the parametric equations of the hypocycloid reduce to
x4cos3 y4sin3 This curve is called a hypocycloid of four cusps, or an astroid.
TEC LookatModule10.1Btoseehow hypocycloids and epicycloids are formed by the motion of rolling circles.
630
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
3. Nowtryb1andand,afractionwherenanddhavenocommonfactor.Firstletn1 and try to determine graphically the effect of the denominator d on the shape of the graph. Then let n vary while keeping d constant. What happens when nd1?
4. What happens if b1 and a is irrational? Experiment with an irrational number like
s2 or e2. Take larger and larger values for and speculate on what would happen if we were to graph the hypocycloid for all real values of .
5. If the circle C rolls on the outside of the fixed circle, the curve traced out by P is called an epicycloid. Find parametric equations for the epicycloid.
6. Investigate the possible shapes for epicycloids. Use methods similar to Problems 24.
10.2 CALCULUS WITH PARAMETRIC CURVES
Having seen how to represent curves by parametric equations, we now apply the methods of calculus to these parametric curves. In particular, we solve problems involving tangents, area, arc length, and surface area.
TANGENTS
In the preceding section we saw that some curves defined by parametric equations xf t and ytt can also be expressed, by eliminating the parameter, in the form yFx. See Exercise 67 for general conditions under which this is possible. If we substitute xft and ytt in the equation yFx, we get
ttF f t
and so, if t, F, and f are differentiable, the Chain Rule gives
ttF f tf tFx f t If f t0, we can solve for Fx:
Fxtt ft
Since the slope of the tangent to the curve yFx at x, Fx is Fx, Equation 1 enables us to find tangents to parametric curves without having to eliminate the parameter. Using Leibniz notation, we can rewrite Equation 1 in an easily remembered form:
It can be seen from Equation 2 that the curve has a horizontal tangent when dydt0 provided that dxdt0 and it has a vertical tangent when dxdt0 provided that dydt0. This information is useful for sketching parametric curves.
1
N If we think of a parametric curve as being traced out by a moving particle, then dydt and dxdt are the vertical and horizontal velocities of the particle and Formula 2 says that the slope of the tangent is the ratio of these velocities.
dy
dydt if dx0
dxdx dt dt
2
d2y 22
d dy d2yddy dtdx
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES631
As we know from Chapter 4, it is also useful to consider d 2 ydx 2. This can be found by replacing y by dydx in Equation 2:
Notethatdydt dx2 d2x
dt
dt2
dx2dx dx
dx
y
y3x3
3, 0
EXAMPLE 1 A curve C is defined by the parametric equations xt2, yt33t. a Show that C has two tangents at the point 3, 0 and find their equations.
b Find the points on C where the tangent is horizontal or vertical.
c Determine where the curve is concave upward or downward.
d Sketch the curve.
SOLUTION
a Noticethatyt3 3ttt2 30whent0orts3.Thereforethe point 3, 0 on C arises from two values of the parameter, ts3 and ts3 . This indicates that C crosses itself at 3, 0. Since
dydydt3t233 t1 dx dxdt 2t 2 t
the slope of the tangent when ts3 is dydx62s3 s3 , so the equa tions of the tangents at 3, 0 are
ys3 x3 and ys3 x3
b C has a horizontal tangent when dydx0, that is, when dydt0 and dxdt0. Since dydt3t23, this happens when t21, that is, t1. The corresponding points on C are 1, 2 and 1, 2. C has a vertical tangent when dxdt2t0, that is, t0. Note that dydt0 there. The corresponding point on C is 0, 0.
c To determine concavity we calculate the second derivative:
d dy 31 1
t2
V EXAMPLE 2
a Findthetangenttothecycloidxr sin ,yr1cos atthepoint where3. See Example 7 in Section 10.1.
b At what points is the tangent horizontal? When is it vertical?
SOLUTION
a The slope of the tangent line is
dydydrsin sin dx dxd r1cos1cos
t1
1, 2
0x t1
d2y dt dx 2 dx2 dx
3t2 14t3
1, 2
FIGURE 1
2t
Thus the curve is concave upward when t0 and concave downward when t0.
y 3×3
dt
d Using the information from parts b and c, we sketch C in Figure 1. M
632
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
When
3, we have
xr sin rs3 yr1cosr
3332 32 dysin 3s32 s3
and
Therefore the slope of the tangent is s3 and its equation is
yrs3 xrrs3or s3 xyr2 232 s3
The tangent is sketched in Figure 2.
dx 1cos 3 11 2
N The limits of integration for t are found as usual with the Substitution Rule. When xa, tiseither or .Whenxb,tisthe remaining value.
r, 2r
y
3
0 2r 4r x
r, 2r 3r, 2r 5r, 2r
FIGURE 2
b The tangent is horizontal when dydx0, which occurs when sin0 and
1cos0, that is,2n1 , n an integer. The corresponding point on the cycloid is 2n1 r, 2r.
When2n , both dxd and dyd are 0. It appears from the graph that there are vertical tangents at those points. We can verify this by using lHospitals Rule as follows:
limdylim sin limcos
l2ndx l2n1cos
A similar computation shows that dydx las
caltangentswhen 2n ,thatis,whenx2n r. AREAS
l2n sin
l 2n , so indeed there are verti
M
We know that the area under a curve yFx from a to b is Axb Fxdx, where a
Fx0. If the curve is traced out once by the parametric equations xf t and ytt,t, then we can calculate an area formula by using the Substitution Rule for
Definite Integrals as follows:
b
Ay ydxy ttftdt
a
V EXAMPLE 3 Find the area under one arch of the cycloid xr sinyr1cos
or ttftdt
y
See Figure 3.
y
0 2rx
FIGURE 3
N The result of Example 3 says that the area under one arch of the cycloid is three times the area of the rolling circle that generates the cycloid see Example 7 in Section 10.1. Galileo guessed this result but it was first proved by the French mathematician Roberval and the Italian mathematician Torricelli.
SOLUTION One arch of the cycloid is given by 0 2 . Using the Substitution Rule with yr1cosand dxr1cosd , we have
Ay2 r ydxy2 r1cos r1cos d 00
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES633
r2 y2 1cos 2 d r2 y2 12cos cos2 d 00
r2y2 12cos 11cos2d 02
2312232
r2 2sin4sin20r223r M
ARC LENGTH
We already know how to find the length L of a curve C given in the form yFx, axb. Formula 8.1.3 says that if F is continuous, then
L 1 dx ybdy2
Suppose that C can also be described by the parametric equations xft and ytt,t, where dxdtf t0. This means that C is traversed once, from left to right, as t increases from to and fa, fb. Putting Formula 2 into Formula
3 and using the Substitution Rule, we obtain
Since dxdt0, we have
P Ldt
i
y
3
ybdy2 a dx
a dx
ydydt2 dx L1dx1 dt
dxdt dt
C PTM
P
P
P i1
y dx2 dy2 dt dt
4
Pn
Even if C cant be expressed in the form yFx, Formula 4 is still valid but we obtain it by polygonal approximations. We divide the parameter interval,into n subintervals of equal width t. If t0, t1, t2, . . . , tn are the endpoints of these subintervals, then xif tiand yittiare the coordinates of points Pixi, yithat lie on C and the polygon with ver tices P0 , P1 , . . . , Pn approximates C. See Figure 4.
As in Section 8.1, we define the length L of C to be the limit of the lengths of these approximating polygons as n l :
n
Llim Pi1Pi
nl i1
The Mean Value Theorem, when applied to f on the interval ti1, ti , gives a number ti in
ti1, tisuch that
ftifti1ftiti ti1
If we let xixixi1 and yiyiyi1, this equation becomes
0x
FIGURE 4
xi ftit
634
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
Similarly, when applied to t, the Mean Value Theorem gives a number ti in ti1, tisuch that
Therefore
and so
yi ttit
Pi1Pi sxi 2yi 2s f tit2ttt2
i
sfti2 tt2 t i
n
Llims f ti2tt2 t
5
nl i1
i
The sum in 5 resembles a Riemann sum for the function s f t2 tt2 but it is not exactly a Riemann sum because titi in general. Nevertheless, if fand t are contin uous, it can be shown that the limit in 5 is the same as if ti and ti were equal, namely,
Lysft2tt2 dt
Thus, using Leibniz notation, we have the following result, which has the same form as
Formula 4.
THEOREM If a curve C is described by the parametric equations xf t, ytt, t ,where fandtarecontinuouson , andCistraversed exactly once as t increases from to , then the length of C is
L dtdtdt y dx 2 dy 2
6
Notice that the formula in Theorem 6 is consistent with the general formulas Lx ds and ds2dx2dy2 of Section 8.1.
EXAMPLE 4 If we use the representation of the unit circle given in Example 2 in Sec tion 10.1,
xcos t ysin t 0t2 then dxdtsin t and dydtcos t, so Theorem 6 gives
y2 dx2 dy2 2 2
L dtssin2tcos2t dtdt2
0 dt dt 0 0
as expected. If, on the other hand, we use the representation given in Example 3 in Sec
yy
tion 10.1,
then dxdt2 cos 2t, dydt2 sin 2t, and the integral in Theorem 6 gives
xsin 2t ycos 2t 0t2
y2 dx2 dy2 2 2
dts4 cos2 2t4 sin2 2t dt2 dt4
yy
0 dt dt 0 0
Notice that the integral gives twice the arc length of the circle because as t increases from 0 to 2 , the point sin 2t, cos 2t traverses the circle twice. In general, when finding the length of a curve C from a parametric representation, we have to be careful to ensure that C is traversed only once as t increases from to . M
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES635
V EXAMPLE 5 Find the length of one arch of the cycloid xrsin , yr1cos .
SOLUTION From Example 3 we see that one arch is described by the parameter interval 0 2 .Since
we have
dx r1cosand dy rsin dd
sr21cos 2r2sin2 d
Ly2 dx2
dy2 0dd0
d
y2 sr212 coscos2sin2
y
2
N The result of Example 5 says that the length of one arch of a cycloid is eight times the radius of the generating circle see Figure 5. This was first proved in 1658 by Sir Christopher Wren, who later became the architect of St. Pauls Cathedral in London.
dr y2 s21cos 00
y
0
FIGURE 5
To evaluate this integral we use the identity sin2x1 1cos 2x with 2
gives 1cos2 sin2 2. Since 0 2 , we have 02sin 20. Therefore
d
2x, which
and so
L8r
r
s21cos s4 sin2 22sin 2 2 sin 2 and so L2r y2 sin 2 d2r2 cos 220
0
2r228r
SURFACE AREA
In the same way as for arc length, we can adapt Formula 8.2.5 to obtain a formula for surface area. If the curve given by the parametric equations xf t, ytt,t, is rotated about the xaxis, where f, tare continuous and tt0, then the area of the resulting surface is given by
dx 2 dy 2 Sy2y dtdtdt
The general symbolic formulas Sx 2 y ds and Sx 2 x ds Formulas 8.2.7 and 8.2.8 are still valid, but for parametric curves we use
dx 2 dy 2 ds dtdt dt
EXAMPLE 6 Show that the surface area of a sphere of radius r is 4 r2. SOLUTION The sphere is obtained by rotating the semicircle
2r x
M
7
xr cos t yr sin t 0t
636
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
about the xaxis. Therefore, from Formula 7, we get Sy 2 rsintsrsint2 rcost2 dt
0
2 y rsintsr2sin2tcos2tdt2 y rsintrdt 00
2 r2 y sintdt2 r2cost0 4 r2 M 0
19. x2cos , ysin2
20. xcos3 , y2sin
10.2 EXERCISES 12 Find dydx.
1. xtsint, yt2 t
36 Find an equation of the tangent to the curve at the point corre
sponding to the given value of the parameter.
3. xt4 1, yt3 t; t1
4. xtt1, y1t2; t1
xest, ytlnt2; t1
6. xcos sin2 , ysin cos2 ; 0
78 Find an equation of the tangent to the curve at the given point by two methods: a without eliminating the parameter and b by first eliminating the parameter.
7. x1ln t, yt22; 1, 3 8. xtan , ysec ; 1,s2
; 910 Find an equation of the tangents to the curve at the given point. Then graph the curve and the tangents.
9. x6 sin t, yt2t; 0, 0
10. xcos tcos 2t, ysin tsin 2t ;
2. x1t, ystet
; 21.
; 22.
Use a graph to estimate the coordinates of the rightmost point on the curve xtt6, yet. Then use calculus to find the exact coordinates.
Use a graph to estimate the coordinates of the lowest point and theleftmostpointonthecurvext4 2t,ytt4.Then find the exact coordinates.
5.
; 2324 Graph the curve in a viewing rectangle that displays all the important aspects of the curve.
xt4 2t3 2t2, yt3 t 24. xt4 4t3 8t2, y2t2 t
1, 1
1116 Find dydx and d 2 ydx 2. For which values of t is the curve
27. a
Find the slope of the tangent line to the trochoid
xr dsin ,yrdcos intermsof .SeeExer cise 40 in Section 10.1.
; 26.
Showthatthecurvexcost,ysint costhastwotangents at 0, 0 and find their equations. Sketch the curve.
Graph the curve xcos t2 cos 2t, ysin t2 sin 2t to discover where it crosses itself. Then find equations of both tangents at that point.
concave upward?
x4t2, yt2 t3
b Show that if dr, then the trochoid does not have a vertical tangent.
11.
13. xtet, ytet
15. x2sint, y3 cost,
16. xcos2t, ycost, 0t
12. xt3 12t, yt2 1
28. a
Find the slope of the tangent to the astroid xa cos3 , ya sin3 in terms of . Astroids are explored in the Laboratory Project on page 629.
14. xtlnt, ytlnt 0t2
b At what points is the tangent horizontal or vertical?
c At what points does the tangent have slope 1 or 1?
29. Atwhatpointsonthecurvex2t3,y14tt2 doesthe tangent line have slope 1?
30. Find equations of the tangents to the curve x3t21, y2t31 that pass through the point 4, 3.
Use the parametric equations of an ellipse, xa cos , ybsin ,0 2 ,tofindtheareathatitencloses.
23.
25.
1720 Find the points on the curve where the tangent is horizontal or vertical. If you have a graphing device, graph the curve to check your work.
17. x10t2, yt3 12t
18. x2t3 3t2 12t, y2t3 3t2 1
31.
32. Findtheareaenclosedbythecurvext2 2t,yst and the yaxis.
33. Find the area enclosed by the xaxis and the curve x1et, ytt2.
34. Find the area of the region enclosed by the astroid
xa cos3 , ya sin3 . Astroids are explored in the Labo ratory Project on page 629.
y a
a 0 ax
a
35. Find the area under one arch of the trochoid of Exercise 40 in Section 10.1 for the case dr.
36. Letbe the region enclosed by the loop of the curve in Example 1.
a Find the area of .
b Ifis rotated about the xaxis, find the volume of the
resulting solid.
c Find the centroid of .
3740 Set up an integral that represents the length of the curve. Then use your calculator to find the length correct to four decimal places.
49. Use Simpsons Rule with n6 to estimate the length of the curvextet, ytet, 6t6.
50. In Exercise 43 in Section 10.1 you were asked to derive the parametric equations x2a cot , y2a sin2 for the curve called the witch of Maria Agnesi. Use Simpsons Rule with n4 to estimate the length of the arc of this curve given by 4 2.
5152 Find the distance traveled by a particle with position x, y as t varies in the given time interval. Compare with the length of the curve.
51. xsin2t, ycos2t, 0t3 52. xcos2t, ycost, 0t4
SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES637
53.
54. 55.
56.
Show that the total length of the ellipse xa sin , yb cos ,ab0,is
L4a y 2 s1e2 sin2 d 0
where e is the eccentricity of the ellipse eca, where csa2b2.
37. xtt2, y4t32, 1t2 3
38. x1et, yt2, 3t3
39. xtcost, ytsint, 0t2 40. xlnt, yst1, 1t5
4144 Find the exact length of the curve. x13t2, y42t3, 0t1
CAS
CAS
y11 sin t4 sin11t2
What parameter interval gives the complete curve?
b Use your CAS to find the approximate length of this curve.
AcurvecalledCornusspiralisdefinedbytheparametric equations
Find the total length of the astroid xacos3 where a0.
a Graph the epitrochoid with equations x11 cos t4 cos11t2
, yasin3 ,
xCtyt cos 0
yStyt sin 0
u22 du u22 du
41.
42. xet et, y52t, 0t3
43. x t , yln1t, 0t2
where C and S are the Fresnel functions that were introduced in Chapter 5.
a Graph this curve. What happens as t land as
t l ?
b Find the length of Cornus spiral from the origin to the
point with parameter value t.
1t
44. x3 cos tcos 3t,
y3 sin tsin 3t,
0t
; 4547 Graph the curve and find its length. xetcost, yetsint, 0t
5758 Set up an integral that represents the area of the surface obtained by rotating the given curve about the xaxis. Then use your calculator to find the surface area correct to four decimal places.
57. x1tet, yt2 1et, 0t1 58. xsin2t, ysin3t, 0t 3
45.
46. xcostlntan1t, ysint, 4t3 4 2
47. xet t, y4et2, 8t3
48. Find the length of the loop of the curve x3tt3, y3t2.
638CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
59. 60.
; 62.
63.
64.
xt3, yt2, 0t1
x3tt3, y3t2, 0t1 xacos3 , yasin3 , 02
Graph the curve
x2cos cos2 y2sin sin2
If this curve is rotated about the xaxis, find the area of the resulting surface. Use your graph to help find the correct parameter interval.
If the curve
5961 Find the exact area of the surface obtained by rotating the given curve about the xaxis.
b By regarding a curve yf x as the parametric curve
xx, yf x, with parameter x, show that the formula inpartabecomes
d2ydx2
1dydx2 32
y
P
0x
a Use the formula in Exercise 69b to find the curvature of the parabola yx 2 at the point 1, 1.
b At what point does this parabola have maximum curvature?
Use the formula in Exercise 69a to find the curvature of the cycloidx sin ,y1cos atthetopofoneofits arches.
a Show that the curvature at each point of a straight line is 0.
b Show that the curvature at each point of a circle of radius r is1r.
A string is wound around a circle and then unwound while being held taut. The curve traced by the point P at the end of the string is called the involute of the circle. If the circle has radius r and center O and the initial position of P is r, 0, and if the parameter is chosen as in the figure, show
that parametric equations of the involute are xrcossinyrsincos
y
r
74. A cow is tied to a silo with radius r by a rope just long enough to reach the opposite side of the silo. Find the area available for grazing by the cow.
61.
xtt3 yt 1 t2
70.
71.
72.
73.
1t2
is rotated about the xaxis, use your calculator to estimate the
area of the resulting surface to three decimal places.
If the arc of the curve in Exercise 50 is rotated about the xaxis, estimate the area of the resulting surface using Simp sons Rule with n4.
6566 Find the surface area generated by rotating the given curve about the yaxis.
x3t2, y2t3, 0t5
66. xet t, y4et2, 0t1
67. If fiscontinuousandft0foratb,showthatthe parametric curve xft, ytt, atb, can be put in the form yFx. Hint: Show that f 1 exists.
68. Use Formula 2 to derive Formula 7 from Formula 8.2.5 for the case in which the curve can be represented in the form yFx, axb.
69. The curvature at a point P of a curve is defined as d
T
65.
ds O
P
x
where
as shown in the figure. Thus the curvature is the absolute value of the rate of change of with respect to arc length. It can be regarded as a measure of the rate of change of direc tion of the curve at P and will be studied in greater detail in Chapter 13.
a For a parametric curve xxt, yyt, derive the
formula
xyxyx 2y 232
where the dots indicate derivatives with respect to t, so
xdxdt. Hint: Usetan1dydx and Formula 2 to find d dt. Then use the Chain Rule to find d ds.
is the angle of inclination of the tangent line at P,
; BE ZIER CURVES
SECTION 10.3 POLAR COORDINATES639
L A B O R AT O R Y PROJECT
The Bezier curves are used in computeraided design and are named after the French mathema tician Pierre Bezier 19101999, who worked in the automotive industry. A cubic Bezier curve is determined by four control points, P0x0, y0 , P1x1, y1, P2x2, y2 , and P3x3, y3 , and is defined by the parametric equations
xx01t3 3x1t1t2 3x2t21tx3t3 yy01t3 3y1t1t2 3y2t21ty3t3
where 0t1. Notice that when t0 we have x, yx0, y0and when t1 we have x, yx3, y3, so the curve starts at P0 and ends at P3.
1. Graph the Bezier curve with control points P04, 1, P128, 48, P250, 42, and P340, 5. Then, on the same screen, graph the line segments P0P1, P1P2, and P2P3. Exercise 31 in Section 10.1 shows how to do this. Notice that the middle control points P1 and P2 dont lie on the curve; the curve starts at P0, heads toward P1 and P2 without reaching them, and ends at P3 .
2. From the graph in Problem 1, it appears that the tangent at P0 passes through P1 and the tangent at P3 passes through P2. Prove it.
3. Try to produce a Bezier curve with a loop by changing the second control point in Problem 1.
4. Some laser printers use Bezier curves to represent letters and other symbols. Experiment with control points until you find a Bezier curve that gives a reasonable representation of the letter C.
5. More complicated shapes can be represented by piecing together two or more Bezier curves. Suppose the first Bezier curve has control points P0, P1, P2, P3 and the second one has con trol points P3, P4, P5, P6. If we want these two pieces to join together smoothly, then the tangents at P3 should match and so the points P2, P3, and P4 all have to lie on this common tangent line. Using this principle, find control points for a pair of Bezier curves that repre sent the letter S.
10.3 POLAR COORDINATES
Pr,
A coordinate system represents a point in the plane by an ordered pair of numbers called coordinates. Usually we use Cartesian coordinates, which are directed distances from two perpendicular axes. Here we describe a coordinate system introduced by Newton, called the polar coordinate system, which is more convenient for many purposes.
We choose a point in the plane that is called the pole or origin and is labeled O. Then we draw a ray halfline starting at O called the polar axis. This axis is usually drawn hor izontally to the right and corresponds to the positive xaxis in Cartesian coordinates.
If P is any other point in the plane, let r be the distance from O to P and let be the angle usually measured in radians between the polar axis and the line OP as in Figure 1. Then the point P is represented by the ordered pair r,and r, are called polar coordi nates of P. We use the convention that an angle is positive if measured in the counter clockwise direction from the polar axis and negative in the clockwise direction. If PO, then r0 and we agree that 0,represents the pole for any value of .
O
FIGURE 1
r
polar axis
x
640
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
r,
FIGURE 2
5 4
3 3 4
O
r,
We extend the meaning of polar coordinates r,to the case in which r is negative by agreeing that, as in Figure 2, the points r,and r,lie on the same line through O and at the same distance rfrom O, but on opposite sides of O. If r0, the point r,lies in the same quadrant as ; if r0, it lies in the quadrant on the opposite side of the pole. Notice that r,represents the same point as r,.
EXAMPLE 1 Plot the points whose polar coordinates are given.
a 1, 5 4 b 2, 3c 2, 2 3 d 3, 3 4
SOLUTION The points are plotted in Figure 3. In part d the point 3, 3 4 is located three units from the pole in the fourth quadrant because the angle 3 4 is in the second quadrant and r3 is negative.
O
O
2,3 O
O
2 1,5 3
4
FIGURE 3
2,23
3, 3 4
In the Cartesian coordinate system every point has only one representation, but in the polar coordinate system each point has many representations. For instance, the point 1, 5 4 in Example 1a could be written as 1, 3 4 or 1, 13 4 or 1, 4. See Figure 4.
4
M
5 O 13 4O4OO
3 4
1, 5 1,34444
FIGURE 4
y
1, 13 1,
In fact, since a complete counterclockwise rotation is given by an angle 2 , the point
Pr, Px,y
y
represented by polar coordinates r,is also represented by
r, 2nand r, 2n1
where n is any integer.
The connection between polar and Cartesian coordinates can be seen from Figure 5, in
which the pole corresponds to the origin and the polar axis coincides with the positive
xaxis. If the point P has Cartesian coordinates x, y and polar coordinates r, from the figure, we have
cos x sin y rr
and so
Although Equations 1 were deduced from Figure 5, which illustrates the case where r0and02,theseequationsarevalidforallvaluesofrand .Seethegen eral definition of sin and cos in Appendix D.
, then,
r
Oxx FIGURE 5
xr cos yr sin
1
SECTION 10.3 POLAR COORDINATES641
Equations 1 allow us to find the Cartesian coordinates of a point when the polar coor dinates are known. To find r and when x and y are known, we use the equations
which can be deduced from Equations 1 or simply read from Figure 5. EXAMPLE 2 Convert the point 2, 3 from polar to Cartesian coordinates.
SOLUTION Since r2 and3, Equations 1 give xrcos 2cos 2 1 1
32
yrsin 2sin 2s3s3 32
Therefore the point is 1, s3in Cartesian coordinates. M EXAMPLE 3 Represent the point with Cartesian coordinates 1, 1 in terms of polar
coordinates.
SOLUTION If we choose r to be positive, then Equations 2 give
rsx2 y2 s12 12 s2 tan y1
x
Since the point 1, 1 lies in the fourth quadrant, we can choose
7 4. Thus one possible answer is s2 ,4; another is s2 , 7 4. M
NOTE Equations 2 do not uniquely determine when x and y are given because, as increases through the interval 0 2 , each value of tan occurs twice. Therefore, in converting from Cartesian to polar coordinates, its not good enough just to find r and that satisfy Equations 2. As in Example 3, we must choose so that the point r,lies in the correct quadrant.
POLAR CURVES
The graph of a polar equation rf, or more generally Fr, 0, consists of all points P that have at least one polar representation r,whose coordinates satisfy the equation.
V EXAMPLE 4 What curve is represented by the polar equation r2?
SOLUTION The curve consists of all points r,with r2. Since r represents the distance
from the point to the pole, the curve r2 represents the circle with center O and radius
2. In general, the equation ra represents a circle with center O and radius a . See
Figure 6.
4 or
r21
r4
FIGURE 6
M
r2
r1
x
r2x2y2 tan y x
2
642
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
1
3, 1 2, 1
EXAMPLE 5 Sketch the polar curve1.
SOLUTION This curve consists of all points r,such that the polar angle is 1 radian. It
is the straight line that passes through O and makes an angle of 1 radian with the polar axis see Figure 7. Notice that the points r, 1 on the line with r0 are in the first quadrant, whereas those with r0 are in the third quadrant.
EXAMPLE 6
a Sketch the curve with polar equation r2 cos . b Find a Cartesian equation for this curve.
SOLUTION
1 O
1, 1 2,1
FIGURE 7
x
1, 1
a In Figure 8 we find the values of r for some convenient values of
corresponding points r, . Then we join these points to sketch the curve, which appears
to be a circle. We have used only values of
and plot the between 0 and , since if we let increase
beyond
, we obtain the same points again.
M
r2 cos
0
6 4 3 2 2 3 3 4 5 6
2
s3 s2 1 0 1 s2 s3 2
2 , 4
2
1, 3 2, 3
4
1 , 3
3 , 6
2, 0
3, 56
0 , 2
FIGURE 8
Table of values and graph of r2 cos
N Figure 9 shows a geometrical illustration that the circle in Example 6 has the equation r2 cos .TheangleOPQisarightangle Why?andsor2cos .
y
b To convert the given equation to a Cartesian equation we use Equations 1 and 2. Fromxrcos wehavecos xr,sotheequationr2cos becomesr2xr, which gives
2xr2 x2 y2 or x2 y2 2×0 Completing the square, we obtain
x12 y2 1
which is an equation of a circle with center 1, 0 and radius 1. M
r
O2Q
x
P
FIGURE 9
r 2
1
0 3 2
22
FIGURE 10
r1sin in Cartesian coordinates, 0 2
V EXAMPLE 7 Sketch the curve r1sin .
SOLUTION Instead of plotting points as in Example 6, we first sketch the graph of
r1sin in Cartesian coordinates in Figure 10 by shifting the sine curve up one unit. This enables us to read at a glance the values of r that correspond to increasing values of . For instance, we see that as increases from 0 to 2, r the distance from O increases from 1 to 2, so we sketch the corresponding part of the polar curve in Figure 11a. As increases from 2 to , Figure 10 shows that r decreases from 2 to 1, so we
SECTION 10.3 POLAR COORDINATES643
2
2
2
O
sketch the next part of the curve as in Figure 11b. As
r decreases from 1 to 0 as shown in part c. Finally, as
r increases from 0 to 1 as shown in part d. If we let
beyond 0, we would simply retrace our path. Putting together the parts of the curve from Figure 11ad, we sketch the complete curve in part e. It is called a cardioid, because its shaped like a heart.
increases from to 3 2, increases from 3 2 to 2 ,
increase beyond 2 or decrease
O 1 a
0
OOO2
b
F I G U R E 1 1 Stages in sketching the cardioid r1sin
3 3 22
c d
e
EXAMPLE 8 Sketch the curve rcos 2 .
M
TEC Module 10.3 helps you see how polar curves are traced out by showing animations similar to Figures 1013.
SOLUTION As in Example 7, we first sketch rcos 2 , 0 2 , in Cartesian coordi nates in Figure 12. As increases from 0 to 4, Figure 12 shows that r decreases from 1 to 0 and so we draw the corresponding portion of the polar curve in Figure 13 indi cated by !. As increases from 4 to 2, r goes from 0 to 1. This means that the distance from O increases from 0 to 1, but instead of being in the first quadrant this por tion of the polar curve indicated bylies on the opposite side of the pole in the third quadrant. The remainder of the curve is drawn in a similar fashion, with the arrows and numbers indicating the order in which the portions are traced out. The resulting curve has four loops and is called a fourleaved rose.
r 1
2
34
!
35 3 7 2 424424
4 !
FIGURE 13
Fourleaved rose rcos 2
0
8
FIGURE 12
rcos 2in Cartesian coordinates
M
644
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES SYMMETRY
When we sketch polar curves, it is sometimes helpful to take advantage of symmetry. The following three rules are explained by Figure 14.
a If a polar equation is unchanged when is replaced by, the curve is symmetric about the polar axis.
b If the equation is unchanged when r is replaced by r, or when is replaced by
, the curve is symmetric about the pole. This means that the curve remains
unchanged if we rotate it through 180 about the origin.
c If the equation is unchanged when about the vertical line2.
is replaced by
r,
, the curve is symmetric r,
r,
OO Or,
r,
r,
FIGURE 14
a
b c
The curves sketched in Examples 6 and 8 are symmetric about the polar axis, since cos cos . The curves in Examples 7 and 8 are symmetric about2 because sinsin and cos 2cos 2 . The fourleaved rose is also symmetric about the pole. These symmetry properties could have been used in sketching the curves. For instance, in Example 6 we need only have plotted points for 0 2 and then reflected about the polar axis to obtain the complete circle.
TANGENTS TO POLAR CURVES
To find a tangent line to a polar curve rf, we regard as a parameter and write its parametric equations as
xrcos f cos yrsin f sin
Then, using the method for finding slopes of parametric curves Equation 10.2.2 and the
Product Rule, we have
dy drsin rcos dyd d
dx dx drcos rsin dd
3
We locate horizontal tangents by finding the points where dyd0 provided that dxd0. Likewise, we locate vertical tangents at the points where dxd0 pro vided that dyd0.
Notice that if we are looking for tangent lines at the pole, then r0 and Equation 3 sim plifies to
dytan if dr0 dx d
drsin rcos dyd
dx dr cos rsin d
cos 12 sin12 sin2sin
cos sin 1sincos cos cos 1sin sin
cos 12 sin1sin 12 sin
SECTION 10.3 POLAR COORDINATES645
Forinstance,inExample8wefoundthatrcos2 0when4or3 4.This meansthatthelines4and 3 4oryxandyxaretangentlinesto rcos 2 at the origin.
EXAMPLE 9
a For the cardioid r1sin of Example 7, find the slope of the tangent line when3 .
b Find the points on the cardioid where the tangent line is horizontal or vertical.
SOLUTION Using Equation 3 with r1sin , we have
a The slope of the tangent at the point where3 is
dy cos 312sin 3 11s3 2
dx3 1sin 312sin 3 1s321s3
1s3
2s3 1s3
1s3 1 1s3
b Observe that
when,3 ,7 ,11 dx1sin 12 sin 0 when3 , , 5
d 266
Therefore there are horizontal tangents at the points 2, 2, 1 , 7 6, 1 , 11 6 and 22
dycos 12sin 0
d 2266
vertical tangents at 3 , 6 and 3 , 5 6. When3 2, both dyd 22
are 0, so we must be careful. Using lHospitals Rule, we have
and dxd
2 , 2m1
0,0
l3 2 12 sin l3 2 1sin
1 3 ,2 3
3 , 26
lim dy l3 2 dx
By symmetry,
lim 12 sin lim cos
3 , 52 6
1 lim cos 3l321sin
1 lim sin 3l32 cos
1,7 1,11 26 26
FIGURE 15
limdy l3 2 dx
Tangent lines for r1sin
Thus there is a vertical tangent line at the pole see Figure 15.
M
646
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
NOTE Instead of having to remember Equation 3, we could employ the method used to derive it. For instance, in Example 9 we could have written
xr cos1sincoscos1 sin 2 2
yrsin 1sin sin sin sin2 Then we would have
dydydcos 2sin coscos sin2 dx dxd sin cos2 sin cos2
which is equivalent to our previous expression.
GRAPHING POLAR CURVES WITH GRAPHING DEVICES
Although its useful to be able to sketch simple polar curves by hand, we need to use a graphing calculator or computer when we are faced with a curve as complicated as the ones shown in Figures 16 and 17.
1
FIGURE 16
1
rsin2.4 cos2.4
1
1.7
1.9 1.9
1.7
FIGURE 17
rsin1.2 cos6
1
Some graphing devices have commands that enable us to graph polar curves directly. With other machines we need to convert to parametric equations first. In this case we take the polar equation rf and write its parametric equations as
xrcos f cos yrsin f sin Some machines require that the parameter be called t rather than .
EXAMPLE 10 Graph the curve rsin8 5.
SOLUTION Lets assume that our graphing device doesnt have a builtin polar graphing command. In this case we need to work with the corresponding parametric equations, which are
xrcos sin8 5cos yrsin sin8 5sin
In any case, we need to determine the domain for . So we ask ourselves: How many complete rotations are required until the curve starts to repeat itself? If the answer is
n, then
sin 8 2nsin816n sin 8 5555
1
1 1
1
N In Exercise 55 you are asked to prove analyti cally what we have discovered from the graphs in Figure 19.
and so we require that 16n 5 be an even multiple of . This will first occur when n5. Therefore we will graph the entire curve if we specify that 0 10 . Switching from to t, we have the equations
xsin8t5cos t ysin8t5sin t 0t10 and Figure 18 shows the resulting curve. Notice that this rose has 16 loops.
V EXAMPLE 11 Investigate the family of polar curves given by r1c sin . How does the shape change as c changes? These curves are called limacons, after a French word for snail, because of the shape of the curves for certain values of c.
M
FIGURE 18
rsin8 5
SOLUTION Figure 19 shows computerdrawn graphs for various values of c. For c1 there
is a loop that decreases in size as c decreases. When c1 the loop disappears and the
curve becomes the cardioid that we sketched in Example 7. For c between 1 and 1 the 2
cardioids cusp is smoothed out and becomes a dimple. When c decreases from 1 to 0, 2
the limacon is shaped like an oval. This oval becomes more circular as c l 0, and when c0 the curve is just the circle r1.
c2.5
c0
FIGURE 19
Members of the family of limacons r1c sin
10.3 EXERCISES
c0.2
c0.5 c0.8 c1
12 Plot the point whose polar coordinates are given. Then find two other pairs of polar coordinates of this point, one with r0 and one with r0.
4. a s2 , 5 4
1. a 2, 3 2. a 1, 7 4
b 1, 3 4 b 3, 6
c 1, 2 c 1, 1
i Find polar coordinates r, 0 2 .
ii Find polar coordinates r, 02.
5. a 2, 2 6. a 3s3 , 3
of the point, where r0 andof the point, where r0 and
b 1, s3b 1, 2
34 Plot the point whose polar coordinates are given. Then find the Cartesian coordinates of the point.
3. a 1,b 2, 2 3 c 2, 3 4
c1.7
c1 c0.7 c0.5
c0.2
SECTION 10.3 POLAR COORDINATES647
The remaining parts of Figure 19 show that as c becomes negative, the shapes change in reverse order. In fact, these curves are reflections about the horizontal axis of the corre
sponding curves with positive c.
M
c 2, 7 56 The Cartesian coordinates of a point are given.
6
b 1, 5 2
c 2
648CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
712 Sketch the region in the plane consisting of points whose polar coordinates satisfy the given conditions.
7. 1r2
8.r0, 323 9.0r4, 26
10. 2r5, 3 4 5 4 2r3, 5 3 7 3
12.r1, 2
13. Find the distance between the points with polar coordinates
2, 3 and 4, 2 3.
14. Find a formula for the distance between the points with polar
coordinates r1, 1 and r2, 2 .
1520 Identify the curve by finding a Cartesian equation for the
curve.
15. r2 16.rcos1
43. r29 sin 2
45. r2 cos3 2 47. r12 cos 2
44. r2cos 4
46. r21
48. r12 cos 2
11.
r3 sin 19. rcsc
18. r2 sin
20. rtan sec
2126 Find a polar equation for the curve represented by the given Cartesian equation.
21. x3 22.x2y29
23. xy2
x2 y2 2cx
24. xy9 26. xy4
2 cos
4950 The figure shows the graph of r as a function of
sian coordinates. Use it to sketch the corresponding polar curve.
50. r r2
49.
2 1
02
51. Show that the polar curve r42 sec
has the line x2 as a vertical asymptote by showing that lim r lx2. Use this fact to help sketch the conchoid.
52. Show that the curve r2csc also a conchoid has the line y1 as a horizontal asymptote by showing that
lim r ly1. Use this fact to help sketch the conchoid.
53. Show that the curve rsin tan called a cissoid of Diocles has the line x1 as a vertical asymptote. Show also that the curve lies entirely within the vertical strip 0x1. Use these facts to help sketch the cissoid.
54. Sketch the curve x2y2 34×2 y2.
a In Example 11 the graphs suggest that the limacon
r1csin hasaninnerloopwhenc1.Prove that this is true, and find the values of that correspond to the inner loop.
b From Figure 19 it appears that the limacon loses its dimple when c1 . Prove this.
2
56. Match the polar equations with the graphs labeled IVI. Give reasons for your choices. Dont use a graphing device.
in Carte
022
called a conchoid
17.
55.
25.
2728 For each of the described curves, decide if the curve would be more easily given by a polar equation or a Cartesian equation. Then write an equation for the curve.
6 with
27. a
b A vertical line through the point 3, 3
A line through the origin that makes an angle of thepositivexaxis
a rs , 0 16 c rcos 3
e r2sin3
b r 2, 0 16 d r12 cos
f r12sin3
III
VI
A circle with radius 5 and center 2, 3
28. a
b A circle centered at the origin with radius 4
I
IV
II
V
2948 Sketch the curve with the given polar equation.
29. 6
31. rsin
33. r21sin ,
r, 0 37. r4 sin 3
r2 cos 4
41. r12sin
0
30.r23r20 32. r3 cos
34. r13cos 36.rln, 1 38. rcos 5
40. r3 cos 6 42. r2sin
35.
39.
5762 Find the slope of the tangent line to the given polar curve at the point specified by the value of .
;80.
; 81.
; 82.
Afamilyofcurvesisgivenbytheequationsr1csinn , where c is a real number and n is a positive integer. How does the graph change as n increases? How does it change as c changes? Illustrate by graphing enough members of the fam ily to support your conclusions.
A family of curves has polar equations
r1a cos 1a cos
Investigate how the graph changes as the number a changes. In particular, you should identify the transitional values of a for which the basic shape of the curve changes.
The astronomer Giovanni Cassini 1625 1712 studied the family of curves with polar equations
r4 2c2r2 cos2 c4 a4 0
where a and c are positive real numbers. These curves are called the ovals of Cassini even though they are oval shaped only for certain values of a and c. Cassini thought that these curves might represent planetary orbits better than Keplers ellipses. Investigate the variety of shapes that these curves may have. In particular, how are a and c related to each other when the curve splits into two parts?
Let P be any point except the origin on the curve rf. If is the angle between the tangent line at P and the radial line OP, show that
57. r2sin , r1 ,
61. rcos2 ,
64
58. r2sin , 60. rcos 3, 62. r12cos ,
3
3
SECTION 10.3 POLAR COORDINATES649
59.
6368 Find the points on the given curve where the tangent line is horizontal or vertical.
r3cos 64. r1sin
63.
65. r1cos 67. r2sin
66. re
68. r2 sin2
71. 72. 73. 74. 75. 76.
;77.
; 78.
; 79.
nephroid of Freeth hippopede
r
drd
in the figure.
69.
Show that the polar equation ra sinb cos
ab0, represents a circle, and find its center and radius.
70. Show that the curves ra sin right angles.
and ra cos
, where intersect at
; 7176 Use a graphing device to graph the polar curve. Choose the parameter interval to make sure that you produce the entire curve.
83.
r12 sin 2
rs10.8 sin 2
re sin2 cos4rsin24 cos4r25 sin 6 rcos2cos3
Howarethegraphsofr1sin6and r1sin3relatedtothegraphofr1sin ? In general, how is the graph of rf related to the graphofrf ?
Use a graph to estimate the ycoordinate of the highest points on the curve rsin 2 . Then use calculus to find the exact value.
a Investigate the family of curves defined by the polar equa tions rsin n , where n is a positive integer. How is the number of loops related to n?
b What happens if the equation in part a is replaced by
r sinn ?
;
gent line and the radial line is
thecurvere .
Illustrate part a by graphing the curve and the tangent lines at the points where0 and 2.
Prove that any polar curve rf with the property that the angle between the radial line and the tangent line is aconstantmustbeoftheformrCek ,whereCandk are constants.
84.
a
b c
Use Exercise 83 to show that the angle between the tan
Hint: Observe that
tan
butterfly curve
rf
P
y
O
4 at every point on
650
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.4 AREAS AND LENGTHS IN POLAR COORDINATES
In this section we develop the formula for the area of a region whose boundary is given by
r
a polar equation. We need to use the formula for the area of a sector of a circle
A1 r2 2
1
is the radian measure of the central angle.
FIGURE 1
b
O
rf
a
i
fi
i1
I a
where, as in Figure 1, r is the radius and
Formula 1 follows from the fact that the area of a sector is proportional to its central angle:
212
A 2r 2r .SeealsoExercise35inSection7.3.
Letbe the region, illustrated in Figure 2, bounded by the polar curve rf and by the raysa andb, where f is a positive continuous function and where 0ba2 . We divide the interval a, b into subintervals with endpoints 0 , 1 ,
2,…, n andequalwidth .Theraysi thendivideintonsmallerregionswith central angle ii1 . If we choose i in the ith subintervali1, i , then the area Ai of the ith region is approximated by the area of the sector of a circle with central angleand radius fi. See Figure 3.
Thus from Formula 1 we have
A i1fi 22
and so an approximation to the total area A ofis n
A 1fi 22
i1
It appears from Figure 3 that the approximation in 2 improves as n l . But the sums
in 2 are Riemann sums for the function t 1f 2, so 2
n
nl i1 a
It therefore appears plausible and can in fact be proved that the formula for the area A of
the polar regionis
Formula 3 is often written as
with the understanding that rf. Note the similarity between Formulas 1 and 4. When we apply Formula 3 or 4, it is helpful to think of the area as being swept out by
a rotating ray through O that starts with angle a and ends with angle b.
V EXAMPLE 1 Find the area enclosed by one loop of the fourleaved rose rcos 2 .
SOLUTION The curve rcos 2 was sketched in Example 8 in Section 10.3. Notice from Figure 4 that the region enclosed by the right loop is swept out by a ray that rotates from
FIGURE 2
b
O
FIGURE 3
lim1fi2 yb1f2d 22
b
a
2
Ayb 1f 2d
a
3
2
Ayb 1r2 d
a
4
2
rcos 2
rg a
O
FIGURE 6
aaa
4
4
r3 sin
6
r1sin
4 to
4. Therefore Formula 4 gives
Ay4 1r2d 1 y4 cos22 d y4 cos2 2 d
rf
Example 2 illustrates the procedure for finding the area of the region bounded by two polar curves. In general, letbe a region, as illustrated in Figure 6, that is bounded by curveswithpolarequationsrf ,rt , a,and b,where f t 0 and0ba2 .TheareaAofisfoundbysubtractingtheareainsidertfrom the area inside rf, so using Formula 3 we have
Ayb 1f 2d yb 1t 2d 1 yb f 2 t 2d 222
CAUTION The fact that a single point has many representations in polar coordinates
sometimes makes it difficult to find all the points of intersection of two polar curves.
For instance, it is obvious from Figure 5 that the circle and the cardioid have three
points of intersection; however, in Example 2 we solved the equations r3 sin and
r1sin and found only two such points, 3, 6 and 3, 5 6. The origin is also 22
a point of intersection, but we cant find it by solving the equations of the curves because the origin has no single representation in polar coordinates that satisfies both equations. Notice that, when represented as 0, 0 or 0, , the origin satisfies r3 sin and so it lies on the circle; when represented as 0, 3 2, it satisfies r1sin and so it lies on the cardioid. Think of two points moving along the curves as the parameter value increasesfrom0to2 .Ononecurvetheoriginisreachedat 0and;onthe
SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES651
22
44 0
y 4 1 1cos 4d1 1 sin 44 022408
M
FIGURE 4
5 6
FIGURE 5
V EXAMPLE 2 Find the area of the region that lies inside the circle r3 sin and outside the cardioid r1sin .
SOLUTION The cardioid see Example 7 in Section 10.3 and the circle are sketched in
Figure 5 and the desired region is shaded. The values of a and b in Formula 4 are deter
mined by finding the points of intersection of the two curves. They intersect when
3sin 1sin ,whichgivessin 1,so6,5 6.Thedesiredareacanbe 2
found by subtracting the area inside the cardioid between6 and the area inside the circle from 6 to 5 6. Thus
A1 y5 6 3sin 2d 1 y5 6 1sin 2d 2 6 2 6
Since the region is symmetric about the vertical axis2, we can write A21y29sin2 d 1y212sin sin2 d
5
6 from
O
22
6 6
y 2 8 sin212 sind 6
y 2 34 cos 22 sind 6
3 2sin2 2cos 26
because sin2
1 1cos 2 2
M
b
652
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
other curve it is reached at3 2. The points dont collide at the origin because they reach the origin at different times, but the curves intersect there nonetheless.
Thus, to find all points of intersection of two polar curves, it is recommended that you draw the graphs of both curves. It is especially convenient to use a graphing calculator or computer to help with this task.
EXAMPLE 3 Find all points of intersection of the curves rcos 2 and r1. 262
r1 2
FIGURE 7
21 , 3
1,
SOLUTION Ifwesolvetheequationsrcos2 andr1,wegetcos2 1 and,there 22
fore, 23, 5 3, 7 3, 11 3. Thus the values of between 0 and 2 that sat
isfy both equations are6, 5 6, 7 6, 11 6. We have found four points of
intersection: 1, 6, 1, 5 6, 1, 7 6, and 1, 11 6. 2222
rcos 2
However, you can see from Figure 7 that the curves have four other points of inter sectionnamely, 1, 3, 1, 2 3, 1, 4 3, and 1, 5 3. These can be found using
2222
symmetry or by noticing that another equation of the circle is r1 and then solving
theequationsrcos2 andr1. 2
ARC LENGTH
2
M
Tofindthelengthofapolarcurverf ,a b,weregard asaparameterand write the parametric equations of the curve as
xrcos f cos yrsin f sin Using the Product Rule and differentiating with respect to , we obtain
so, using cos 2 dx 2
d
dxdrcos rsin dydrsin rcos dd dd
sin 21 , we have
dy 2d
dr 2 dr
d
cos2 2rd cos sin r2sin2 dr2 dr
d sin2 2rd sin cos r2cos2 dr2
d r2
Assuming that fis continuous, we can use Theorem 10.2.6 to write the arc length as
yb dx2 dy2 Ld add
Therefore the length of a curve with polar equation rf, a b, is
V EXAMPLE 4 Find the length of the cardioid r1sin .
SOLUTION The cardioid is shown in Figure 8. We sketched it in Example 7 in Section 10.3. Its full length is given by the parameter interval 0 2 , so
yb 2 dr2 Lrd
ad
5
SECTION 10.4 AREAS AND LENGTHS IN POLAR COORDINATES
653
Formula 5 gives
y22dr22 22
Lrdy s1sin cos d
0d0 y2 s22sin d
O0
We could evaluate this integral by multiplying and dividing the integrand by
FIGURE 8 s22 sin , or we could use a computer algebra system. In any event, we find that the
r1sin length of the cardioid is L8. 10.4 EXERCISES
M
1 4 Find the area of the region that is bounded by the given curve and lies in the specified sector.
1.r 2, 04 2.re2,2 3. rsin , 3 2 3 4. rssin , 0
19.
22.
r3 cos 5
r12sin
20. r2 sin 6
7.
21.
inner loop
Find the area enclosed by the loop of the strophoid r2cos sec .
58 Find the area of the shaded region. 5. 6.
r
8.
r43 sin
914 Sketch the curve and find the area that it encloses.
2328 Find the area of the region that lies inside the first curve and outside the second curve.
r1cos
23. r2cos , r1 25. r28 cos 2 , r2
r3cos , r1cos 28.r3sin, r2sin
24. r1sin , r1
27.
26. r2sin ,
r3 sin
9. r3cos
r24 cos 2
10. r31cos12. r2sin
14. r2cos2
35. Find the area inside the larger loop and outside the smaller loop ofthelimaconr1 cos .
2
36. Find the area between a large loop and the enclosed small loop ofthecurver12 cos3 .
3742 Find all points of intersection of the given curves. 37. r1sin , r3 sin
38. r1cos , r1sin
39. r2 sin 2 , r1
rsin , rsin2
rsin 2
2934 Find the area of the region that lies inside both curves. 29.rs3cos, rsin
30. r1cos , r1cos
rsin2 , rcos2
32. r32 cos , r32 sin
33. r2sin2, r2cos2
34. rasin , rbcos , a0, b0
31.
11.
13. r2cos3
; 1516 Graph the curve and find the area that it encloses.
15. r12 sin 6 16. r2 sin3 sin 9 1721 Find the area of the region enclosed by one loop of
the curve.
17. rsin 2 18. r4 sin 3
40. rcos 3 , rsin 3 42. r2 sin2 , r2 cos2
41.
654CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
; 43.
44.
The points of intersection of the cardioid r1sin and thespiralloopr2 , 22,cantbefound exactly. Use a graphing device to find the approximate values of at which they intersect. Then use these values to estimate the area that lies inside both curves.
When recording live performances, sound engineers often use a microphone with a cardioid pickup pattern because it sup presses noise from the audience. Suppose the microphone is placed 4 m from the front of the stage as in the figure and the boundary of the optimal pickup region is given by the car dioid r88 sin , where r is measured in meters and the microphone is at the pole. The musicians want to know the area they will have on stage within the optimal pickup range of the microphone. Answer their question.
4952 Use a calculator to find the length of the curve correct to four decimal places.
stage
audience
Use Formula 10.2.7 to show that the area of the surface generated by rotating the polar curve
rfab
where fiscontinuousand0ab aboutthe
4548 Find the exact length of the polar curve.
56. a
Find a formula for the area of the surface generated by rotatingthepolarcurverf ,a bwhere fis continuousand0ab ,abouttheline2.
45. r3 sin , 0 47.r 2, 0 2
3 46. re2 ,
12 m 4m
b 2 dr2 ad
10.5
CONIC SECTIONS
FIGURE 1
Conics
microphone
048.r , 0 2
2
49. r3 sin 2 51. rsin 2
50. r4 sin 3
52. r1cos 3
; 5354 Graph the curve and find its length.
53. rcos4 4 54. rcos2 2
55. a
b Use the formula in part a to find the surface area gener ated by rotating the lemniscate r 2cos 2 about the polar axis.
polar axis is
Sy2 rsin rd
b Find the surface area generated by rotating the lemniscate r2 cos2 abouttheline2.
In this section we give geometric definitions of parabolas, ellipses, and hyperbolas and derive their standard equations. They are called conic sections, or conics, because they result from intersecting a cone with a plane as shown in Figure 1.
ellipse parabola hyperbola
PARABOLAS
SECTION 10.5 CONIC SECTIONS655
axis focus
parabola
A parabola is the set of points in a plane that are equidistant from a fixed point F called the focus and a fixed line called the directrix. This definition is illustrated by Figure 2. Notice that the point halfway between the focus and the directrix lies on the parabola; it is called the vertex. The line through the focus perpendicular to the directrix is called the axis of the parabola.
In the 16th century Galileo showed that the path of a projectile that is shot into the air at an angle to the ground is a parabola. Since then, parabolic shapes have been used in designing automobile headlights, reflecting telescopes, and suspension bridges. See Problem 18 on page 268 for the reflection property of parabolas that makes them so useful.
We obtain a particularly simple equation for a parabola if we place its vertex at the ori gin O and its directrix parallel to the xaxis as in Figure 3. If the focus is the point 0, p, then the directrix has the equation yp. If Px, y is any point on the parabola, then the distance from P to the focus is
PFsx2 yp2
and the distance from P to the directrix is yp . Figure 3 illustrates the case where
F
vertex
FIGURE 2
directrix
Px, y
y
F0, p
y
FIGURE 3
p0. The defining property of a parabola is that these distances are equal: sx2 yp2 yp
We get an equivalent equation by squaring and simplifying:
x2 yp2 yp2 yp2 x2 y2 2pyp2 y2 2pyp2
x2 4py
If we write a14p, then the standard equation of a parabola 1 becomes yax2. It opens upward if p0 and downward if p0 see Figure 4, parts a and b. The graph is symmetric with respect to the yaxis because 1 is unchanged when x is replaced by x.
Ox
p yp
An equation of the parabola with focus 0, p and directrix yp is x2 4py
1
yyyy
0, p
yp
0 0xx0x0x
yp 0, p xp xp
p, 0p, 0
a 4py, p0 b 4py, p0 c 4px, p0 d 4px, p0 FIGURE 4
656
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
FIGURE 5
The sketch is shown in Figure 5. M ELLIPSES
An ellipse is the set of points in a plane the sum of whose distances from two fixed points F1 and F2 is a constant see Figure 6. These two fixed points are called the foci plural of focus. One of Keplers laws is that the orbits of the planets in the solar system are ellipses with the sun at one focus.
y
52 , 0
If we interchange x and y in 1, we obtain
which is an equation of the parabola with focus p, 0 and directrix xp. Inter changing x and y amounts to reflecting about the diagonal line yx. The parabola opens to the right if p0 and to the left if p0 see Figure 4, parts c and d. In both cases the graph is symmetric with respect to the xaxis, which is the axis of the parabola.
EXAMPLE 1 Find the focus and directrix of the parabola y210x0 and sketch the graph.
SOLUTION If we write the equation as y210x and compare it with Equation 2, we see that 4p10, so p5. Thus the focus is p, 05, 0and the directrix is x5.
y2 4px
2
10×0
0x
x 52
222
P
F FTM
FIGURE 6
y
Fc, 0 0
FIGURE 7
Px, y
FTMc, 0 x
In order to obtain the simplest equation for an ellipse, we place the foci on the xaxis at the points c, 0 and c, 0 as in Figure 7 so that the origin is halfway between the foci. Let the sum of the distances from a point on the ellipse to the foci be 2a0. Then Px, y is a point on the ellipse when
PF1 PF2 2a
thatis, sxc2 y2 sxc2 y2 2a
or sxc2 y2 2asxc2 y2 Squaring both sides, we have
x2 2cxc2 y2 4a2 4asxc2 y2 x2 2cxc2 y2 which simplifies to asxc2y2a2cx
We square again:
which becomes a2c2x2a2y2a2a2c2
a2x2 2cxc2 y2a4 2a2cxc2x2
3
a, 0
y
0,b
b a a,0
From triangle F1F2P in Figure 7 we see that 2c2a, so ca and therefore a2c20. For convenience, let b2a2c2. Then the equation of the ellipse becomes b2x2a2y2a2b2 or, if both sides are divided by a2b2,
x2 y2
a2b2 1
Since b2a2c2a2, it follows that ba. The xintercepts are found by setting y0. Then x2a21, or x2a2, so xa. The corresponding points a, 0 and a, 0 are called the vertices of the ellipse and the line segment joining the vertices is called the major axis. To find the yintercepts we set x0 and obtain y2b2, so yb. Equation 3 is unchanged if x is replaced by x or y is replaced by y, so the ellipse is symmetric about both axes. Notice that if the foci coincide, then c0, so ab and the ellipse becomes a circle with radius rab.
We summarize this discussion as follows see also Figure 8.
If the foci of an ellipse are located on the yaxis at 0, c, then we can find its equa tion by interchanging x and y in 4. See Figure 9.
SECTION 10.5 CONIC SECTIONS657
c,0 0 c c,0 x
FIGURE 8
ab1, a b y
0, b
4
The ellipse
x2 y2
a2b2 1 ab0
has foci c, 0, where c2a2b2, and vertices a, 0.
0, a 0, c
b, 0
FIGURE 9
b, 0
5
The ellipse
x2 y2
b2a2 1 ab0
has foci 0, c, where c2a2b2, and vertices 0, a.
ba1, a b y
V EXAMPLE 2 Sketch the graph of 9×216y2144 and locate the foci. SOLUTION Divide both sides of the equation by 144:
x2 y2
1691
The equation is now in the standard form for an ellipse, so we have a216, b29, a4, and b3. The xintercepts are 4 and the yintercepts are 3. Also,
c2a2b27, so cs7 and the foci are s7, 0. The graph is sketched in Figure 10.
V EXAMPLE 3 Find an equation of the ellipse with foci 0, 2 and vertices 0, 3. SOLUTION Using the notation of 5, we have c2 and a3. Then we obtain
b2 a2 c2 945,soanequationoftheellipseis x2 y2
0x
0, c
0, a
4, 0
7 , 00
FIGURE 10
916144
0, 3
7 , 0
0, 3
4, 0
x
M
591
Another way of writing the equation is 9×25y245. M
658CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
y
Fc, 0 0
FIGURE 11
P is on the hyperbola when PFPFTM2a.
Px, y
FTMc, 0 x
Like parabolas, ellipses have an interesting reflection property that has practical conse quences. If a source of light or sound is placed at one focus of a surface with elliptical crosssections, then all the light or sound is reflected off the surface to the other focus see Exercise 63. This principle is used in lithotripsy, a treatment for kidney stones. A reflec tor with elliptical crosssection is placed in such a way that the kidney stone is at one focus. Highintensity sound waves generated at the other focus are reflected to the stone and destroy it without damaging surrounding tissue. The patient is spared the trauma of sur gery and recovers within a few days.
HYPERBOLAS
A hyperbola is the set of all points in a plane the difference of whose distances from two fixed points F1 and F2 the foci is a constant. This definition is illustrated in Figure 11.
Hyperbolas occur frequently as graphs of equations in chemistry, physics, biology, and economics Boyles Law, Ohms Law, supply and demand curves. A particularly signifi cant application of hyperbolas is found in the navigation systems developed in World Wars I and II see Exercise 51.
Notice that the definition of a hyperbola is similar to that of an ellipse; the only change is that the sum of distances has become a difference of distances. In fact, the derivation of the equation of a hyperbola is also similar to the one given earlier for an ellipse. It is left as Exercise 52 to show that when the foci are on the xaxis at c, 0 and the difference of distances is PF1 PF2 2a, then the equation of the hyperbola is
x2 y2
a2 b2 1
where c2a2b2. Notice that the xintercepts are again a and the points a, 0 and a, 0 are the vertices of the hyperbola. But if we put x0 in Equation 6 we get y2b2, which is impossible, so there is no yintercept. The hyperbola is symmetric with respect to both axes.
To analyze the hyperbola further, we look at Equation 6 and obtain
x2 y2
a2 1b2 1
Thisshowsthatx2 a2,soxsx2 a.Thereforewehavexaorxa.This means that the hyperbola consists of two parts, called its branches.
When we draw a hyperbola it is useful to first draw its asymptotes, which are the dashed lines ybax and ybax shown in Figure 12. Both branches of the hyperbola approach the asymptotes; that is, they come arbitrarily close to the asymptotes. See Exercise 69 in Section 4.5, where these lines are shown to be slant asymptotes.
byb yax yax
a, 0
c, 0 0
FIGURE 12
1 a b
a, 0
c, 0
x
6
7
The hyperbola
x2 y2
a2b2 1
has foci c, 0, where c2a2b2, vertices a, 0, and asymptotes ybax.
y
0, c
0, a 0, a
0, c
If the foci of a hyperbola are on the yaxis, then by reversing the roles of x and y we obtain the following information, which is illustrated in Figure 13.
x
SECTION 10.5 CONIC SECTIONS659
y ab x
FIGURE 13
1 a b
y 34 x y 4, 0
5,0 0
FIGURE 14
916144
y ab x
8
The hyperbola
y2 x2
a2 b2 1
has foci 0, c, where c2a2b2, vertices 0, a, and asymptotes yabx.
0
y34 x
4, 0
EXAMPLE 4 Find the foci and asymptotes of the hyperbola 9×216y2144 and sketch its graph.
SOLUTION If we divide both sides of the equation by 144, it becomes x2 y2
5,0 x
1691 whichisoftheformgivenin7witha4andb3.Sincec2 16925,the
foci are 5, 0. The asymptotes are the lines y3 x and y3 x. The graph is shown 44
in Figure 14. M EXAMPLE 5 Find the foci and equation of the hyperbola with vertices 0, 1 and
asymptote y2x.
SOLUTION From 8 and the given information, we see that a1 and ab2. Thus
ba21 andc2 a2 b2 5.Thefociare0,s52andtheequationofthe 24
hyperbola is
SHIFTED CONICS
y2 4×2 1 M
As discussed in Appendix C, we shift conics by taking the standard equations 1, 2, 4, 5, 7, and 8 and replacing x and y by xh and yk.
EXAMPLE 6 Find an equation of the ellipse with foci 2, 2, 4, 2 and vertices 1, 2, 5, 2.
SOLUTION The major axis is the line segment that joins the vertices 1, 2, 5, 2
and has length 4, so a2. The distance between the foci is 2, so c1. Thus
b2 a2 c2 3.Sincethecenteroftheellipseis3,2,wereplacexandyin4 by x3 and y2 to obtain
x32 y22 431
as the equation of the ellipse.
M
660CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
y
y1 32 x4
4, 4
4, 1
0x
4, 2
y1 32 x4
FIGURE 15
9472x8y1760
V EXAMPLE 7 Sketch the conic
9×2 4y2 72x8y1760
and find its foci.
SOLUTION We complete the squares as follows:
4y2 2y9x2 8×176
4y2 2y19x2 8×161764144
4y12 9×42 36 y12 x42
941
This is in the form 8 except that x and y are replaced by x4 and y1. Thus
a29, b24, and c213. The hyperbola is shifted four units to the right and one
unit upward. The foci are 4, 1s13 and 4, 1s13 and the vertices are 4, 4 and
10.5 EXERCISES
18 Find the vertex, focus, and directrix of the parabola and sketch
1718 Find an equation of the ellipse. Then find its foci. 17. 18.
its graph.
1. x2y2 3. 4×2 y
x22 8y3
7. y2 2y12x250
2. 4yx2 0
4. y2 12x
6. x1y52
8. y12x2x2 16
4, 2. The asymptotes are y13 x4. The hyperbola is sketched in 2
Figure 15.
M
y
1
0
1
x
y
1
2
x
5.
910 Find an equation of the parabola. Then find the focus and directrix.
9. 10.
1116 Find the vertices and foci of the ellipse and sketch its graph.
1924 Find the vertices, foci, and asymptotes of the hyperbola and sketch its graph.
y
1
0
2
x
y
1
2
x
x2 y2 144251
y2 x2
20. 16361
19.
21. y2 x2 4
24. y2 4×2 2y16x31
2530 Identify the type of conic section whose equation is given and find the vertices and foci.
25. x2y1 26.x2y21
x2 4y2y2 28. y2 8y6x16 29. y2 2y4x2 3 30. 4×2 4xy2 0
22. 9×2 4y2 36 23. 4×2 y2 24x4y280
11. 13.
16.
x2 y2
951
x2 y2
12. 641001
4×2 y2 16
9×2 18x4y2 27
x2 3y2 2x12y100
14. 4×2 25y2 25
15.
27.
31 48 Find an equation for the conic that satisfies the given conditions.
51.
In the LORAN LOng RAnge Navigation radio navigation
system, two radio stations located at A and B transmit simul
taneous signals to a ship or an aircraft located at P. The
onboard computer converts the time difference in receiving
these signals into a distance difference PAPB , and
this, according to the definition of a hyperbola, locates the ship or aircraft on one branch of a hyperbola see the figure. Suppose that station B is located 400 mi due east of station A on a coastline. A ship received the signal from B 1200 micro seconds s before it received the signal from A.
a Assuming that radio signals travel at a speed of 980 fts,
find an equation of the hyperbola on which the ship lies.
b If the ship is due north of B, how far off the coastline is
focus 0, 2 directrix x5
directrix x2 vertex 3, 2
vertical axis, 36. Parabola, horizontal axis,
31. Parabola,
32. Parabola,
Parabola,
34. Parabola,
33.
passing
Ellipse,
38. Ellipse,
39. Ellipse,
40. Ellipse,
41. Ellipse,
42. Ellipse,
the ship?
vertex 0, 0, vertex 1, 0, focus 4, 0, focus 3, 6, vertex 2, 3,
through 1, 0, 1, 1, and 3, 1
foci 2, 0, vertices 5, 0
foci 0, 5, vertices 0, 13
foci 0, 2, 0, 6, vertices 0, 0, 0, 8
foci 0, 1, 8, 1, vertex 9, 1
center 1, 4, vertex 1, 0, focus 1, 6 foci 4, 0, passing through 4, 1.8
35. Parabola,
passing through 1, 5
SECTION 10.5 CONIC SECTIONS661
37.
43. Hyperbola,
44. Hyperbola,
vertices 3, 0, foci 5, 0 vertices 0, 2, foci 0, 5 vertices 3, 4, 3, 6,
400 mi
transmitting stations
45. Hyperbola,
foci 3, 7, 3, 9
52. 53. 54. 55.
56.
57. 58.
59. 60.
Use the definition of a hyperbola to derive Equation 6 for a hyperbola with foci c, 0 and vertices a, 0.
Show that the function defined by the upper branch of the hyperbola y 2a 2x 2b 21 is concave upward.
Find an equation for the ellipse with foci 1, 1 and 1, 1 and major axis of length 4.
Determine the type of curve represented by the equation x2 y2
in each of the following cases: a k16, b 0k16, and c k0.
d Show that all the curves in parts a and b have the same
foci, no matter what the value of k is.
a Show that the equation of the tangent line to the parabola
y24px at the point x0, y0 can be written as y0y2pxx0
b What is the xintercept of this tangent line? Use this fact to draw the tangent line.
Show that the tangent lines to the parabola x24py drawn from any point on the directrix are perpendicular.
Show that if an ellipse and a hyperbola have the same foci, then their tangent lines at each point of intersection are perpendicular.
Use Simpsons Rule with n10 to estimate the length of the ellipse x24y24.
The planet Pluto travels in an elliptical orbit around the sun at one focus. The length of the major axis is 1.181010 km
46. Hyperbola, vertices 1, 2, 7, 2, foci 2, 2, 8, 2
Hyperbola, vertices 3, 0, 48. Hyperbola, foci 2, 0, 2, 8,
asymptotes y2x asymptotes y31 x and y51 x
22
A coastline
B
47.
49. The point in a lunar orbit nearest the surface of the moon is called perilune and the point farthest from the surface is called apolune. The Apollo 11 spacecraft was placed in an elliptical lunar orbit with perilune altitude 110 km and apolune altitude 314 km above the moon. Find an equation of this ellipse if the radius of the moon is 1728 km and the center of the moon is at one focus.
50. A crosssection of a parabolic reflector is shown in the figure. The bulb is located at the focus and the opening at the focus is 10 cm.
a Find an equation of the parabola.
b Find the diameter of the opening CD , 11 cm from the vertex.
kk161
V
C A
5 cm 11 cm
F
5 cm
D
B
P
662CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
and the length of the minor axis is 1.141010 km. Use Simp sons Rule with n10 to estimate the distance traveled by the planet during one complete orbit around the sun.
61. Find the area of the region enclosed by the hyperbola x2a2y2b21 and the vertical line through a focus.
62. a If an ellipse is rotated about its major axis, find the volume of the resulting solid.
b If it is rotated about its minor axis, find the resulting volume.
63. Let P1x1, y1 be a point on the ellipse x2a2y2b21 with foci F1 and F2 and let and be the angles between the lines PF1, PF2 and the ellipse as shown in the figure. Prove that
. This explains how whispering galleries and lithotripsy work. Sound coming from one focus is reflected and passes through the other focus. Hint: Use the formula in Problem 17
64. Let Px1, y1 be a point on the hyperbola x2a2y2b21 with foci F1 and F2 and let and be the angles between the lines PF1, PF2 and the hyperbola as shown in the figure. Prove that. This is the reflection property of the hyperbola. It shows that light aimed at a focus F2 of a hyperbolic mirror is reflected toward the other focus F1.
y
P
F 0 FTMx
P
a
on page 268 to show that tan y
tan . P,
a
F0FTMx FFTM a b 1
10.6 CONIC SECTIONS IN POLAR COORDINATES
In the preceding section we defined the parabola in terms of a focus and directrix, but we defined the ellipse and hyperbola in terms of two foci. In this section we give a more uni fied treatment of all three types of conic sections in terms of a focus and directrix. Further more, if we place the focus at the origin, then a conic section has a simple polar equation, which provides a convenient description of the motion of planets, satellites, and comets.
THEOREM Let F be a fixed point called the focus and l be a fixed line called the directrix in a plane. Let e be a fixed positive number called the eccentricity. The set of all points P in the plane such that
PFe Pl
that is, the ratio of the distance from F to the distance from l is the constant e is a conic section. The conic is
a an ellipse if e1 b a parabola if e1 c a hyperbola if e1
1
y
F
r cos
d
C
FIGURE 1
l directrix
xd
x
PROOF Notice that if the eccentricity is e1, then PF Pland so the given condi tion simply becomes the definition of a parabola as given in Section 10.5.
Let us place the focus F at the origin and the directrix parallel to the yaxis and
d units to the right. Thus the directrix has equation xd and is perpendicular to the polar axis. If the point P has polar coordinates r, , we see from Figure 1 that
P
r
PF r Pl drcos Thusthecondition PFPl e,or PF e Pl ,becomes
SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES663
redrcos
2
If we square both sides of this polar equation and convert to rectangular coordinates, we get
x2 y2 e2dx2 e2d2 2dxx2 or 1e2x2 2de2xy2 e2d2
After completing the square, we have
e2d 2 y2 e2d2 x 1e21e21e22
If e1, we recognize Equation 3 as the equation of an ellipse. In fact, it is of the form
3
y2 b2 1
e2d
h1e2 a21e22 b21e2
xh2 a2
where
e2d2 e2d2
In Section 10.5 we found that the foci of an ellipse are at a distance c from the center,
4
where
c2 a2 b2e4d2
1e22
5
This shows that ce2dh 1e2
and confirms that the focus as defined in Theorem 1 means the same as the focus defined in Section 10.5. It also follows from Equations 4 and 5 that the eccentricity is given by
ec a
If e1, then 1e20 and we see that Equation 3 represents a hyperbola. Just as we did before, we could rewrite Equation 3 in the form
and see that
xh2 y2
a2 b2 1
ec where c2a2b2 M a
664
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
By solving Equation 2 for r, we see that the polar equation of the conic shown in Fig ure 1 can be written as
r ed
1e cos
If the directrix is chosen to be to the left of the focus as xd, or if the directrix is cho sen to be parallel to the polar axis as yd, then the polar equation of the conic is given by the following theorem, which is illustrated by Figure 2. See Exercises 2123.
yy xd x d
directrix directrix
FxFx
yy
yd
directrix
Fx
Fx
directrix
ed 1e sin
a r ed 1e cos
FIGURE 2
Polar equations of conics
b r
ed 1e cos
c r
ed 1e sin
y d
d r
THEOREM A polar equation of the form
r ed or r ed
represents a conic section with eccentricity e. The conic is an ellipse if e1, a parabola if e1, or a hyperbola if e1.
6
1e cos 1e sin
V EXAMPLE 1 Find a polar equation for a parabola that has its focus at the origin and whose directrix is the line y6.
SOLUTION Using Theorem 6 with e1 and d6, and using part d of Figure 2, we see that the equation of the parabola is
6
r1sin M
V EXAMPLE 2 A conic is given by the polar equation r 10
32 cos
Find the eccentricity, identify the conic, locate the directrix, and sketch the conic. SOLUTION Dividing numerator and denominator by 3, we write the equation as
10
r3
12 cos 3
y
0
2,
10 32 cos
From Theorem 6 we see that this represents an ellipse with e2 . Since ed10 , 33
11
r
10
32 cos 4
SOLUTION We get the equation of the rotated ellipse by replacing equation given in Example 2. So the new equation is
with
4 in the
r
FIGURE 4
12 24 sin
6, 0
SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES665
x 5
directrix
FIGURE 3
r
we have
focus
10, 0
x
10 10
d3 3 5 e2
3
so the directrix has Cartesian equation x5. When
r2. So the vertices have polar coordinates 10, 0 and 2, in Figure 3.
,
EXAMPLE 3 Sketch the conic r12 . 24 sin
SOLUTION Writing the equation in the form r6
12 sin
we see that the eccentricity is e2 and the equation therefore represents a hyperbola. Since ed6, d3 and the directrix has equation y3. The vertices occur when
2 and 3 2, so they are 2, 2 and 6, 3 26, 2. It is also useful to plot the xintercepts. These occur when0, ; in both cases r6. For additional accuracy we could draw the asymptotes. Note that r lwhen 12 sin l 0 or 0 and 12 sin0 when sin1. Thus the asymptotes are parallel to the rays
0, r10; when
. The ellipse is sketched
M
7
6 and
11
2
6. The hyperbola is sketched in Figure 4. y
2, 2
6 , 2
y3 directrix 6,0 x
focus
When rotating conic sections, we find it much more convenient to use polar equations than Cartesian equations. We just use the fact see Exercise 77 in Section 10.3 that the graph of rfis the graph of rf rotated counterclockwise about the origin through an angle .
V EXAMPLE 4 If the ellipse of Example 2 is rotated through an angle origin, find a polar equation and graph the resulting ellipse.
4 about the
M
5 15 r10
r 10 32 cos
32cos4
6
FIGURE 5
We use this equation to graph the rotated ellipse in Figure 5. Notice that the ellipse has been rotated about its left focus. M
666
CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
In Figure 6 we use a computer to sketch a number of conics to demonstrate the effect of varying the eccentricity e. Notice that when e is close to 0 the ellipse is nearly circular, whereas it becomes more elongated as e l 1. When e1, of course, the conic is a parabola.
e0.1 e0.5
e0.68
e0.86 e0.96
e1
FIGURE 6
e1.1
KEPLERS LAWS
e1.4 e4
In 1609 the German mathematician and astronomer Johannes Kepler, on the basis of huge amounts of astronomical data, published the following three laws of planetary motion.
Although Kepler formulated his laws in terms of the motion of planets around the sun, they apply equally well to the motion of moons, comets, satellites, and other bodies that orbit subject to a single gravitational force. In Section 13.4 we will show how to deduce Keplers Laws from Newtons Laws. Here we use Keplers First Law, together with the polar equation of an ellipse, to calculate quantities of interest in astronomy.
For purposes of astronomical calculations, its useful to express the equation of an ellipse in terms of its eccentricity e and its semimajor axis a. We can write the distance d from the focus to the directrix in terms of a if we use 4:
KEPLERS LAWS
1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
2. The line joining the sun to a planet sweeps out equal areas in equal times.
3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
e2d2 a21e22
a21e22 ? d2e2 ? d
So eda1e2. If the directrix is xd, then the polar equation is
r eda1e2 1e cos 1e cos
a1e2 e
aphelion
FIGURE 7
perihelion
planet
r
The positions of a planet that are closest to and farthest from the sun are called its peri helion and aphelion, respectively, and correspond to the vertices of the ellipse. See Figure 7. The distances from the sun to the perihelion and aphelion are called the peri helion distance and aphelion distance, respectively. In Figure 1 the sun is at the focus F, so at perihelion we have0 and, from Equation 7,
r a1e2a1e1e a1e 1e cos 0 1e
Similarly, at aphelionand ra1e.
EXAMPLE 5
a Find an approximate polar equation for the elliptical orbit of the earth around the sun at one focus given that the eccentricity is about 0.017 and the length of the major axis is about 2.99108 km.
b Find the distance from the earth to the sun at perihelion and at aphelion.
SOLUTION
a Thelengthofthemajoraxisis2a2.99108,soa1.495108.Wearegiven that e0.017 and so, from Equation 7, an equation of the earths orbit around the sun is
sun
SECTION 10.6 CONIC SECTIONS IN POLAR COORDINATES667
The polar equation of an ellipse with focus at the origin, semimajor axis a, eccentricity e, and directrix xd can be written in the form
r a1e2 1e cos
7
The perihelion distance from a planet to the sun is a1e and the aphelion distance is a1e.
8
or, approximately,
ra1e21.49510810.0172 1e cos 10.017 cos
r1.49108 10.017 cos
b From 8, the perihelion distance from the earth to the sun is
a1e1.49510810.0171.47108 km and the aphelion distance is
a1e1.49510810.0171.52108 km M
668CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
10.6 EXERCISES
18 Write a polar equation of a conic with the focus at the origin
and the given data.
1. Hyperbola, eccentricity 7,
4
Show that a conic with focus at the origin, eccentricity e, and directrix xd has polar equation
r ed
1e cos
22. Show that a conic with focus at the origin, eccentricity e, and directrix yd has polar equation
r ed
1e sin
23. Show that a conic with focus at the origin, eccentricity e, and directrix yd has polar equation
r ed
1e sin
directrix y6 Ellipse, eccentricity 3, directrix x5
2. Parabola, directrix x4 4
9. 11.
15.
; 17.
; 18.
;19.
;20.
1
r1sin 10.r
12 310cos
3
22 cos 8
45 sin 10
56sin
tricity 0.093 and semimajor axis 2.28108 km. Find a polar equation for the orbit.
26. Jupiters orbit has eccentricity 0.048 and the length of the major axis is 1.56109 km. Find a polar equation for the orbit.
The orbit of Halleys comet, last seen in 1986 and due to return in 2062, is an ellipse with eccentricity 0.97 and one focus at the sun. The length of its major axis is 36.18 AU. An astronomical unit AU is the mean distance between the earth and the sun, about 93 million miles. Find a polar equa tion for the orbit of Halleys comet. What is the maximum distance from the comet to the sun?
28. The HaleBopp comet, discovered in 1995, has an elliptical orbit with eccentricity 0.9951 and the length of the major axis is 356.5 AU. Find a polar equation for the orbit of this comet. How close to the sun does it come?
29. The planet Mercury travels in an elliptical orbit with eccen tricity 0.206. Its minimum distance from the sun is
4.6107 km. Find its maximum distance from the sun.
30. The distance from the planet Pluto to the sun is
4.43109 km at perihelion and 7.37109 km at aphelion. Find the eccentricity of Plutos orbit.
31. Using the data from Exercise 29, find the distance traveled by the planet Mercury during one complete orbit around the sun. If your calculator or computer algebra system evaluates defi nite integrals, use it. Otherwise, use Simpsons Rule.
3.
4. Hyperbola,
5. Parabola,
916 a Find the eccentricity, b identify the conic, c give an equation of the directrix, and d sketch the conic.
eccentricity 2, vertex 4 , 32
directrix y2 vertex 1, 2
6. Ellipse,
7. Ellipse,
8. Hyperbola, eccentricity 3, directrix r6 csc
eccentricity 0.8, eccentricity 1,
directrix r4 sec
2
and
25. The orbit of Mars around the sun is an ellipse with eccen
24. Show that the parabolas rc1cos
rd1cosintersect at right angles.
r 12 4sin
r 9
62 cos
r 3
48 cos
12.r 14.r 16.r
13.
27.
a Find the eccentricity and directrix of the conic
r112 sinand graph the conic and its directrix.
b If this conic is rotated counterclockwise about the origin through an angle 3 4, write the resulting equation and graph its curve.
Graph the conic r456 cosand its directrix. Also graph the conic obtained by rotating this curve about the ori gin through an angle 3.
Graphtheconicsre1ecos withe0.4,0.6, 0.8, and 1.0 on a common screen. How does the value of e affect the shape of the curve?
aGraphtheconicsred1esin fore1andvar ious values of d. How does the value of d affect the shape of the conic?
b Graph these conics for d1 and various values of e. How does the value of e affect the shape of the conic?
21.
CHAPTER 10 REVIEW669
10 REVIEW
CONCEPT CHECK
1. a
b How do you sketch a parametric curve?
What is a parametric curve?
6. a Give a geometric definition of a parabola.
b Write an equation of a parabola with focus 0, p and direc
trix yp. What if the focus is p, 0 and the directrix is xp?
7. a Give a definition of an ellipse in terms of foci.
b Write an equation for the ellipse with foci c, 0 and
vertices a, 0.
8. a Give a definition of a hyperbola in terms of foci.
b Write an equation for the hyperbola with foci c, 0 and
vertices a, 0.
c Write equations for the asymptotes of the hyperbola in
part b.
9. a What is the eccentricity of a conic section?
b What can you say about the eccentricity if the conic section
is an ellipse? A hyperbola? A parabola?
c Write a polar equation for a conic section with eccentricity
e and directrix xd. What if the directrix is xd? yd? yd?
5. Thepolarcurvesr1sin2 andrsin2 1havethe same graph.
6. Theequationsr2,x2 y2 4,andx2sin3t, y2 cos3t0t2 allhavethesamegraph.
7. The parametric equations xt2, yt4 have the same graph asxt3,yt6.
8. Thegraphofy22y3xisaparabola.
9. A tangent line to a parabola intersects the parabola only once. 10. A hyperbola never intersects its directrix.
2. a
b How do you find the area under a parametric curve?
5. a
of a point if you knew the Cartesian coordinates?
How do you find the slope of a tangent line to a polar
How do you find the slope of a tangent to a parametric curve?
3. Write an expression for each of the following:
a The length of a parametric curve
b The area of the surface obtained by rotating a parametric
4. a
curve about the xaxis
Use a diagram to explain the meaning of the polar coordi
nates r,of a point.
b Write equations that express the Cartesian coordinates
x, y of a point in terms of the polar coordinates.
c What equations would you use to find the polar coordinates
curve?
b How do you find the area of a region bounded by a polar
curve?
c How do you find the length of a polar curve?
TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. If the parametric curve xf t, ytt satisfies t10, then it has a horizontal tangent when t1.
2. If xf t and ytt are twice differentiable, then d2yd2ydt2
dx2 d2xdt2
3. Thelengthofthecurvexft,ytt,atb,is
xabsft2tt2 dt.
4. If a point is represented by x, y in Cartesian coordinates where x0 and r,in polar coordinates, then
tan 1 yx.
670CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES
EXERCISES
14 Sketch the parametric curve and eliminate the parameter to find the Cartesian equation of the curve.
1. xt2 4t, y2t, 4t1 2. x1e2t, yet
3.xcos , ysec , 02
4. x2cos , y1sin
5. Write three different sets of parametric equations for the curve ysx.
6. Use the graphs of xft and ytt to sketch the para metric curve xf t, ytt. Indicate with arrows the direction in which the curve is traced as t increases.
xy 1
1t 1t 1
7. a Plot the point with polar coordinates 4, 2 3. Then find its Cartesian coordinates.
b The Cartesian coordinates of a point are 3, 3. Find two sets of polar coordinates for the point.
8. Sketch the region consisting of points whose polar coor
;
2124 Find the slope of the tangent line to the given curve at the point corresponding to the specified value of the parameter.
21. xlnt, y1t2; t1
22. xt3 6t1, y2tt2; t1 23.re;
24. r3cos 3 ;2
2526 Find dydx and d 2 ydx2. 25. xtsint, ytcost 26. x1t2, ytt3
27. Use a graph to estimate the coordinates of the lowest point on thecurvext3 3t,yt2 t1. Thenusecalculusto find the exact coordinates.
28. Find the area enclosed by the loop of the curve in Exercise 27. 29. At what points does the curve
x2a cos ta cos 2t y2a sin ta sin 2t have vertical or horizontal tangents? Use this information to
help sketch the curve.
30. Find the area enclosed by the curve in Exercise 29.
31. Find the area enclosed by the curve r29 cos 5 .
32. Find the area enclosed by the inner loop of the curve r13sin .
33. Find the points of intersection of the curves r2 and r4cos .
34. Find the points of intersection of the curves rcot and r2cos .
35. Find the area of the region that lies inside both of the circles r2sin andrsin cos .
36. Find the area of the region that lies inside the curve r2cos2 butoutsidethecurver2sin .
3740 Find the length of the curve.
37. x3t2, y2t3, 0t2
38. x23t, ycosh 3t, 0t1 39.r1, 2
40. rsin3 3, 0
dinates satisfy 1r2 and
916 Sketch the polar curve. 9. r1cos
11. rcos3
13. r1cos2
given Cartesian equation.
17. xy2 18. x2 y2 2
3
12 sin
6 5 6.
10. rsin4
12. r3cos3 14. r2cos 2
3
16. r
1718 Find a polar equation for the curve represented by the
15. r
22 cos
; 19.
; 20.
The curve with polar equation rsinis called a cochleoid. Use a graph of r as a function of in Cartesian coordinates to sketch the cochleoid by hand. Then graph it with a machine to check your sketch.
Graph the ellipse r243 cosand its directrix. Also graph the ellipse obtained by rotation about the origin through an angle 2 3.
41 42 Find the area of the surface obtained by rotating the given curve about the xaxis.
52. Find an equation of the ellipse with foci 3, 2 and major axis with length 8.
53. Find an equation for the ellipse that shares a vertex and a focus with the parabola x2y100 and that has its other focus at the origin.
54. Show that if m is any real number, then there are exactly two lines of slope m that are tangent to the ellipse
x2a2y2b21 and their equations are ymxsa2m2b2.
CHAPTER 10 REVIEW671
41. 42.
; 43.
;44.
t3 1
x4st, y32t2, 1t4
x23t, ycosh 3t, 0t1
The curves defined by the parametric equations
x t2 c y tt2 c t2 1 t2 1
are called strophoids from a Greek word meaning to turn or twist. Investigate how these curves vary as c varies.
A family of curves has polar equations r a sin 2where a is a positive number. Investigate how the curves change as a changes.
55. Find a polar equation for the ellipse with focus at the origin, 1
eccentricity 3 , and directrix with equation r4 sec . 56. Show that the angles between the polar axis and the
4548 Find the foci and vertices and sketch the graph.
asymptotes of the hyperbola red1e cos , e1, are given by cos11e.
57. In the figure the circle of radius a is stationary, and for every , the point P is the midpoint of the segment QR. The curve traced out by P for 0 is called the longbow curve.
45.
47. 48.
49. 50. 51.
x2 y2
98 1
6y2 x36y550
25×2 4y2 50x16y59
46. 4×2 y2 16
Find parametric equations for this curve.
y 2a
a
R
y2 a
Find an equation of the ellipse with foci 4, 0 and vertices 5, 0.
Find an equation of the parabola with focus 2, 1 and direc trix x4.
Find an equation of the hyperbola with foci 0, 4 and asymptotes y3x.
P Q
0x
PROBLEMS PLUS
CAS
1.
2.
A curve is defined by the parametric equations
xyt cosu du yyt sinu du 1u 1u
Find the length of the arc of the curve from the origin to the nearest point where there is a vertical tangent line.
a Find the highest and lowest points on the curve x4y4x2y2.
b Sketch the curve. Notice that it is symmetric with respect to both axes and both of the
lines yx, so it suffices to consider yx0 initially.
c Use polar coordinates and a computer algebra system to find the area enclosed by the
curve.
; 3. Whatisthesmallestviewingrectanglethatcontainseverymemberofthefamilyofpolar curves r1c sin , where 0c1? Illustrate your answer by graphing several mem bers of the family in this viewing rectangle.
4. Fourbugsareplacedatthefourcornersofasquarewithsidelengtha.Thebugscrawl counterclockwise at the same speed and each bug crawls directly toward the next bug at all times. They approach the center of the square along spiral paths.
a Find the polar equation of a bugs path assuming the pole is at the center of the square.
5.
Use the fact that the line joining one bug to the next is tangent to the bugs path. b Find the distance traveled by a bug by the time it meets the other bugs at the center.
a
aa
a
A curve called the folium of Descartes is defined by the parametric equations x 3t y 3t2
1t3 1t3
a Show that if a, b lies on the curve, then so does b, a; that is, the curve is symmetric with respect to the line yx. Where does the curve intersect this line?
b Find the points on the curve where the tangent lines are horizontal or vertical.
c Show that the line yx1 is a slant asymptote.
d Sketch the curve.
e Show that a Cartesian equation of this curve is x3y33xy.
f Show that the polar equation can be written in the form
r3sec tan 1tan3
g Find the area enclosed by the loop of this curve.
h Show that the area of the loop is the same as the area that lies between the asymptote
and the infinite branches of the curve. Use a computer algebra system to evaluate the integral.
CAS
672
PROBLEMS PLUS
;
6. A circle C of radius 2r has its center at the origin. A circle of radius r rolls without slipping in the counterclockwise direction around C. A point P is located on a fixed radius of the rolling circle at a distance b from its center, 0br. See parts i and ii of the figure. Let L be the line from the center of C to the center of the rolling circle and let be the angle that L makes with the positive xaxis.
a Using as a parameter, show that parametric equations of the path traced out by P are xb cos 33r cos yb sin 33r sin
Note: If b0, the path is a circle of radius 3r; if br, the path is an epicycloid. The path traced out by P for 0br is called an epitrochoid.
b Graph the curve for various values of b between 0 and r.
c Show that an equilateral triangle can be inscribed in the epitrochoid and that its centroid is on the circle of radius b centered at the origin.
Note: This is the principle of the Wankel rotary engine. When the equilateral triangle rotates with its vertices on the epitrochoid, its centroid sweeps out a circle whose center is at the center of the curve.
d In most rotary engines the sides of the equilateral triangles are replaced by arcs of circles
centered at the opposite vertices as in part iii of the figure. Then the diameter of the
rotor is constant. Show that the rotor will fit in the epitrochoid if b3 2s3 r. 2
yy PP
P
Px
2r r
i
b x
ii
iii
673
11
INFINITE SEQUENCES AND SERIES
y
T
T
T T
x
ysin x
The partial sums Tn of a Taylor series provide better and better approximations to a function as n increases.
Infinite sequences and series were introduced briefly in A Preview of Calculus in connection with Zenos paradoxes and the decimal representation of numbers. Their importance in calculus stems from Newtons idea of representing functions as sums
of infinite series. For instance, in finding areas he often integrated a function by first expressing it as a series and then integrating each term of the series. We will pursue his idea in Section 11.10 in order to integrate such functions as ex 2. Recall that we have previously been unable to do this. Many of the functions that arise in mathematical physics and chemistry, such as Bessel functions, are defined as sums of series, so it is important to be familiar with the basic concepts of convergence of infinite sequences and series.
Physicists also use series in another way, as we will see in Section 11.11. In studying fields as diverse as optics, special relativity, and electromagnetism, they analyze phe nomena by replacing a function with the first few terms in the series that represents it.
674
11.1 SEQUENCES
A sequence can be thought of as a list of numbers written in a definite order:
a1, a2, a3, a4, …, an,…
The number a1 is called the first term, a2 is the second term, and in general an is the nth term. We will deal exclusively with infinite sequences and so each term an will have a successor an1.
Notice that for every positive integer n there is a corresponding number an and so a sequence can be defined as a function whose domain is the set of positive integers. But we usually write an instead of the function notation fn for the value of the function at the number n.
NOTATION The sequence a1, a2, a3, . . . is also denoted by
an or ann1
EXAMPLE 1 Some sequences can be defined by giving a formula for the nth term. In the following examples we give three descriptions of the sequence: one by using the preced ing notation, another by using the defining formula, and a third by writing out the terms of the sequence. Notice that n doesnt have to start at 1.
n n 1234n
a n1 ann1 n1
b 1nn1 an1nn1
2, 3, 4, 5,…, n1,…
2, 3, 4 , 5 ,…, 1nn1,…
0, 1, s2 , s3 , . . . , sn3 , . . .
3n sn3
n n0
3n
ansn3 , n3
39 2781 3n
c d
n3
cos 6
n s31 n
ancos 6 , n0 1, 2 , 2 , 0, . . . , cos 6 , . . . M
V EXAMPLE 2 Find a formula for the general term an of the sequence 3, 4 , 5 , 6 , 7 ,…
5 25 125 625 3125
assuming that the pattern of the first few terms continues. SOLUTION We are given that
a13 a24 a3 5 a46 5 25 125 625
a5 7 3125
Notice that the numerators of these fractions start with 3 and increase by 1 whenever we go to the next term. The second term has numerator 4, the third term has numerator 5; in general, the nth term will have numerator n2. The denominators are the powers of 5, so an has denominator 5n. The signs of the terms are alternately positive and negative, so
675
676
CHAPTER 11 INFINITE SEQUENCES AND SERIES
a aTM a a
011 2
we need to multiply by a power of 1. In Example 1b the factor 1n meant we started with a negative term. Here we want to start with a positive term and so we use 1n1 or 1n1. Therefore
n1 n2
an 1 5n M
EXAMPLE 3 Here are some sequences that dont have a simple defining equation.
a The sequencepn , where pn is the population of the world as of January 1 in the year n.
b If we let an be the digit in the nth decimal place of the number e, then anis a well defined sequence whose first few terms are
7, 1, 8, 2, 8, 1, 8, 2, 8, 4, 5, . . .
c The Fibonacci sequencefnis defined recursively by the conditions
f1 1 f2 1 fn fn1 fn2 n3 Each term is the sum of the two preceding terms. The first few terms are
1, 1, 2, 3, 5, 8, 13, 21, . . .
This sequence arose when the 13thcentury Italian mathematician known as Fibonacci
solved a problem concerning the breeding of rabbits see Exercise 71. M
A sequence such as the one in Example 1a, annn1, can be pictured either by plotting its terms on a number line as in Figure 1 or by plotting its graph as in Figure 2. Note that, since a sequence is a function whose domain is the set of positive integers, its graph consists of isolated points with coordinates
1, a1 2, a23, a3. . . n, an. . .
From Figure 1 or 2 it appears that the terms of the sequence annn1 are
approaching 1 as n becomes large. In fact, the difference 1n1
FIGURE 1
an
1
0 1234567 FIGURE 2
a 78
n1 n1
can be made as small as we like by taking n sufficiently large. We indicate this by writing
n
In general, the notation
lim n 1 nl n1
lim anL nl
means that the terms of the sequence anapproach L as n becomes large. Notice that the following definition of the limit of a sequence is very similar to the definition of a limit of a function at infinity given in Section 2.6.
an
Figure 3 illustrates Definition 1 by showing the graphs of two sequences that have the limit L.
an
SECTION 11.1 SEQUENCES677
DEFINITION A sequence anhas the limit L and we write liman L or an lL as nl
nl
if we can make the terms an as close to L as we like by taking n sufficiently large. If limn lan exists, we say the sequence converges or is convergent. Otherwise, we say the sequence diverges or is divergent.
1
FIGURE 3
Graphs of two sequences with
lim anL n
LL
0n0n
A more precise version of Definition 1 is as follows.
DEFINITION A sequence anhas the limit L and we write liman L or an lL as nl
nl
if for every 0 there is a corresponding integer N such that
if nN then an L
2
N Compare this definition with Definition 2.6.7.
FIGURE 4 0
Definition 2 is illustrated by Figure 4, in which the terms a1, a2, a3, . . . are plotted on a number line. No matter how small an interval L, L is chosen, there exists an N such that all terms of the sequence from aN1 onward must lie in that interval.
a a aTM a aN1aN2 aa a a a
L L L
Another illustration of Definition 2 is given in Figure 5. The points on the graph of anmust lie between the horizontal lines yL and yL if nN. This picture must be valid no matter how smallis chosen, but usually a smallerrequires a larger N.
y
L
yL
yL
FIGURE 5
01234 N n
678
CHAPTER 11 INFINITE SEQUENCES AND SERIES
If you compare Definition 2 with Definition 2.6.7, you will see that the only difference between limnl anL and limxl fxL is that n is required to be an integer. Thus we have the following theorem, which is illustrated by Figure 6.
y
L
y
LIMIT LAWS FOR SEQUENCES
FIGURE 6
01234 x
In particular, since we know that limx l1x r 0 when r0 Theorem 2.6.5, we have
lim 10 if r0 nl nr
If an becomes large as n becomes large, we use the notation limnl an. The fol lowing precise definition is similar to Definition 2.6.9.
If lim n lan, then the sequence anis divergent but in a special way. We say that andiverges to .
The Limit Laws given in Section 2.3 also hold for the limits of sequences and their proofs are similar.
THEOREM If limxl fxL and fnan when n is an integer, then limnl anL.
3
4
DEFINITION limn lan means that for every positive number M there is an integer N such that
if nN then an M
5
If anand bnare convergent sequences and c is a constant, then liman bnliman limbn
nl nl nl liman bnliman limbn
nl nl nl
limcan climan limcc
nl nl nl limanbnliman limbn
nl nl nl
an nl bn
lim ann l
limbn nl
if lim bn0 nl
p
if p0 and an0
lim
lim anplim an nl nl
SQUEEZE THEOREM FOR SEQUENCES
The Squeeze Theorem can also be adapted for sequences as follows see Figure 7.
Another useful fact about limits of sequences is given by the following theorem, whose proof is left as Exercise 75.
EXAMPLE 4 Find lim n . nl n1
SOLUTION The method is similar to the one we used in Section 2.6: Divide numerator and denominator by the highest power of n and then use the Limit Laws.
SECTION 11.1 SEQUENCES679
cn
bn
an
0
FIGURE 7
n
The sequence bnis squeezed n
between the sequences an n
and cn.
an
1
01234 n 1
FIGURE 8
EXAMPLE 6 Determine whether the sequence an1n is convergent or divergent. SOLUTION If we write out the terms of the sequence, we obtain
1, 1, 1, 1, 1, 1, 1, . . .
The graph of this sequence is shown in Figure 8. Since the terms oscillate between 1 and 1 infinitely often, an does not approach any number. Thus limn l1n does not exist; that is, the sequence 1nis divergent. M
Ifan bn cn fornn0 andliman limcn L,thenlimbn L. nl nl nl
THEOREM If lim an0, then lim an0. nl nl
6
lim 1
nl lim1lim1
Here we used Equation 4 with r1. M EXAMPLE 5 Calculate lim ln n .
nl n
SOLUTION Notice that both numerator and denominator approach infinity as n l . We cant apply lHospitals Rule directly because it applies not to sequences but to func tions of a real variable. However, we can apply lHospitals Rule to the related function
lim n lim 1
nl n1
nl 11
n nl nln
N This shows that the guess we made earlier from Figures 1 and 2 was correct.
1 101
fxln xx and obtain
Therefore, by Theorem 3, we have
lim ln xlim 1×0 xlxxl1
lim ln n0 M nl n
680CHAPTER 11 INFINITE SEQUENCES AND SERIES
N The graph of the sequence in Example 7 is shown in Figure 9 and supports our answer.
EXAMPLE 7 Evaluate lim SOLUTION
Therefore, by Theorem 6,
if it exists.
1n nl n
an
1
0
1
1
n
lim1n lim10 nl n nln
1n
lim 0 M nl n
THEOREM If lim anL and the function f is continuous at L, then nl
lim fanfL nl
7
FIGURE 9
The following theorem says that if we apply a continuous function to the terms of a con vergent sequence, the result is also convergent. The proof is left as Exercise 76.
EXAMPLE 8 Find lim sin n. nl
SOLUTION Because the sine function is continuous at 0, Theorem 7 enables us to write limsin nsinlim nsin00 M
nl nl
V EXAMPLE 9 Discuss the convergence of the sequence ann!nn, where n!123n.
SOLUTION Both numerator and denominator approach infinity as n l , but here we have no corresponding function for use with lHospitals Rule x! is not defined when x is not an integer. Lets write out a few terms to get a feeling for what happens to an as n gets large:
N CREATING GRAPHS OF SEQUENCES
Some computer algebra systems have special
commands that enable us to create sequences and graph them directly. With most graphing calculators, however, sequences can be graphed by using parametric equations. For instance, the sequence in Example 9 can be graphed by enter ing the parametric equations
xt yt!tt
andgraphingindotmode,startingwitht1 andsettingthetstepequalto1.Theresultis shown in Figure 10.
1
a11
12 22
123 333
a2
an123n
a3nnnn
8
It appears from these expressions and the graph in Figure 10 that the terms are decreas ing and perhaps approach 0. To confirm this, observe from Equation 8 that
an1 23n n nnn
Notice that the expression in parentheses is at most 1 because the numerator is less than
or equal to the denominator. So 010 1
FIGURE 10
0ann
Weknowthat1nl0asnl.Thereforean l0asnlbytheSqueeze
Theorem. M
SECTION 11.1 SEQUENCES681 V EXAMPLE 10 For what values of r is the sequence r nconvergent?
SOLUTION We know from Section 2.6 and the graphs of the exponential functions in Section1.5thatlimxl ax fora1andlimxl ax 0for0a1.Therefore, putting ar and using Theorem 3, we have
It is obvious that
limrn ifr1
nl 0 if0r1
lim 1n1 and lim 0n0 nl nl
If 1r0, then 0r 1, so limrnlimrn 0
nl nl
and therefore limnl rn0 by Theorem 6. If r1, then rn diverges as in Example 6. Figure 11 shows the graphs for various values of r. The case r1 is shown in Figure 8.
an
1
0 1
r1
r1
an
1
1
1r0
0n
n r1
The results of Example 10 are summarized for future use as follows.
FIGURE 11
The sequence anrn
0r1
M
The sequence r nis convergent if 1r1 and divergent for all other values of r.
limrn 0 if 1r1 nl 1 ifr1
9
DEFINITION A sequence an is called increasing if anan1 for all n1, thatis,a1 a2 a3 .Itiscalleddecreasingifan an1 foralln1.It is called monotonic if it is either increasing or decreasing.
10
N The right side is smaller because it has a larger denominator.
EXAMPLE 11 The sequence3 is decreasing because n5
333
n5 n15 n6
andsoan an1 foralln1.
M
682
CHAPTER 11 INFINITE SEQUENCES AND SERIES
EXAMPLE 12 Show that the sequence ann is decreasing. n2 1
SOLUTION 1 We must show that an1an, that is, n1n
n12 1 n2 1
This inequality is equivalent to the one we get by crossmultiplication:
n1n ? n1n2 1nn12 1 n12 1 n2 1
? n3 n2 n1n3 2n2 2n ? 1n2n
Since n1, we know that the inequality n2n1 is true. Therefore an1an and so anis decreasing.
SOLUTION 2 Consider the function f xx : x2 1
x2 12×2 1×2
fx x2 12×2 12 0 whenever x2 1
Thus f is decreasing on 1,and so f nf n1. Therefore anis decreasing. M
DEFINITION A sequence anis bounded above if there is a number M such that
an M foralln1 It is bounded below if there is a number m such that
man for all n1
If it is bounded above and below, then anis a bounded sequence.
11
an M
L
0 123
FIGURE 12
n
For instance, the sequence ann is bounded below an0 but not above. The sequencean nn1isboundedbecause0an 1foralln.
We know that not every bounded sequence is convergent for instance, the sequence an1n satisfies 1an1 but is divergent from Example 6 and not every mono tonic sequence is convergent ann l . But if a sequence is both bounded and monotonic, then it must be convergent. This fact is proved as Theorem 12, but intuitively you can understand why it is true by looking at Figure 12. If anis increasing and anM for all n, then the terms are forced to crowd together and approach some number L.
The proof of Theorem 12 is based on the Completeness Axiom for the setof real numbers, which says that if S is a nonempty set of real numbers that has an upper bound M xM for all x in S, then S has a least upper bound b. This means that b is an upper bound for S, but if M is any other upper bound, then bM. The Completeness Axiom is an expression of the fact that there is no gap or hole in the real number line.
MONOTONIC SEQUENCE THEOREM Every bounded, monotonic sequence is convergent.
12
PROOF Suppose anis an increasing sequence. Since anis bounded, the set
San n1 has an upper bound. By the Completeness Axiom it has a least upper bound L . Given 0, L is not an upper bound for S since L is the least upper bound. Therefore
aN L forsomeintegerN Butthesequenceisincreasingsoan aN foreverynN.ThusifnN,wehave
anL so 0Lan
since a nL . Thus
Lan whenever nN
A similar proof using the greatest lower bound works if anis decreasing. M
The proof of Theorem 12 shows that a sequence that is increasing and bounded above is convergent. Likewise, a decreasing sequence that is bounded below is convergent. This fact is used many times in dealing with infinite series.
EXAMPLE 13 Investigate the sequence andefined by the recurrence relation a1 2 an1 1an 6 for n1,2,3,…
SOLUTION We begin by computing the first several terms:
a2 1264 a3 1465 22
a5 5.75 a6 5.875
a85.96875 a95.984375
These initial terms suggest that the sequence is increasing and the terms are approaching 6. To confirm that the sequence is increasing, we use mathematical induction to show thatan1 an foralln1.Thisistrueforn1becausea2 4a1.Ifweassume that it is true for nk, then we have
solimnl an L.
SECTION 11.1 SEQUENCES683
a1 2
a4 1565.5
2
N Mathematical induction is often used in dealing with recursive sequences. See page 77 for a discussion of the Principle of Mathematical Induction.
so
and Thus
ak1ak
ak1 6ak 6
1ak161ak6 22
ak2ak1
2
a75.9375
684
CHAPTER 11 INFINITE SEQUENCES AND SERIES
We have deduced that an1an is true for nk1. Therefore the inequality is true for all n by induction.
Next we verify that anis bounded by showing that an6 for all n. Since the sequence is increasing, we already know that it has a lower bound: ana12 for all n. We know that a16, so the assertion is true for n1. Suppose it is true for nk. Then
so
Thus
ak6 ak612
1ak 61126 22
ak16
This shows, by mathematical induction, that an6 for all n.
Since the sequence anis increasing and bounded, Theorem 12 guarantees that it has
a limit. The theorem doesnt tell us what the value of the limit is. But now that we know Llimn lan exists, we can use the recurrence relation to write
liman1 lim1an 61liman 61L6 222
Sincean lL,itfollowsthatan1 lL,tooasnl,n1ltoo.Sowehave L1 L6
Solving this equation for L, we get L6, as predicted. M
N A proof of this fact is requested in Exercise 58.
11.1 EXERCISES
1. a
b What does it mean to say that limnl an8?
c What does it mean to say that limnl an?
2
nl nl nl
What is a sequence?
12.1 , 2 ,3 , 4 , . . . 49 1625
14. 5, 1, 5, 1, 5, 1, . . . 15. List the first six terms of the sequence defined by
an n 2n1
Does the sequence appear to have a limit? If so, find it.
16. List the first nine terms of the sequence cosn 3. Does this sequence appear to have a limit? If so, find it. If not, explain why.
1746 Determine whether the sequence converges or diverges. If it converges, find the limit.
17. an 10.2n
11. 2, 7, 12, 17, . . . 248
13.
1,3 , 9 ,27 , . . .
2. a
b What is a divergent sequence? Give two examples.
What is a convergent sequence? Give two examples.
38 List the first five terms of the sequence.
4. ann1 3n1
6. 2462n an
assuming that the pattern of the first few terms continues.
9. 1,1,1,1,1,… 10. 1,1,1, 1 , 1 ,… 3 5 7 9 3 9 27 81
3. an 10.2n 31n
5. ann!
7. a13, an12an1
8. a14, an1an1 914 Find a formula for the general term an of the sequence,
18. an3
n3 n1
35n2 annn2
n3 20. ann1
22. an3n2 5n
24. ann1 9n1
1n n3 26. ann32n21
28. ancos2n 30. arctan 2 n
32. lnn ln 2n
34. ncosn
an lnn1lnn
38. ansn213n sin 2n
a Determine whether the sequence defined as follows is convergent or divergent:
a1 1 an1 4an forn1
b What happens if the first term is a12?
If 1000 is invested at 6 interest, compounded annually, then after n years the investment is worth a n10001.06 n dollars.
a Find the first five terms of the sequence an .
b Is the sequence convergent or divergent? Explain. Find the first 40 terms of the sequence defined by
1 an if an is an even number an1 2
3an1 if an is an odd number
and a111. Do the same if a125. Make a conjecture about this type of sequence.
For what values of r is the sequence nr nconvergent? a If anis convergent, show that
lim an1lim an nl nl
b A sequence anis defined by a11 and
an111an for n1. Assuming that an is convergent, find its limit.
Suppose you know that anis a decreasing sequence and all its terms lie between the numbers 5 and 8. Explain why the sequence has a limit. What can you say about the value of the limit?
25. ann21 27. ancosn2
SECTION 11.1 SEQUENCES685
21. ane1n
23. antan2n
55.
56.
57. 58.
18n 1 n1n
2n1! 29. 2n1!
31. enen e2n 1
33. n2en
ancos2n
2n
37. annsin1n
39.an 1n
2n
41. an ln2n2 1lnn2 1 0, 1, 0, 0, 1, 0, 0, 0, 1, . . .
45. ann! 2n
40.an1sn
35.
43.
42. 44. 46.
anlnn2 n
1,1,1,1,1,1,1,1,… 13243546
an3n n!
; 4753 Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is conver gent, guess the value of the limit from the graph and then prove your guess. See the margin note on page 680 for advice on graphing sequences.
60. an2n1
1 2n3
62. an2n3 3n4
64. annen 66. ann1
n
6066 Determine whether the sequence is increasing, decreasing, or not monotonic. Is the sequence bounded?
an
63. ann1n
47. an 12en
48. an sn sin sn 50. ansn 3n5n
n
n2 1
32n2 2
65. an
49. an
51. ann2 cosn
8n n 1n2
67. Find the limit of the sequence
s2 , s2s2 , s2s2s2 , . . .
52. an1352n1 n!
53. an1352n1 2nn
68. Asequenceanisgivenbya1 s2,an1 s2an .
a By induction or otherwise, show that anis increasing
and bounded above by 3. Apply the Monotonic Sequence
Theorem to show that lim n la n exists.
b Find limnl an.
36.
54.
19.
59.
61.
686CHAPTER 11 INFINITE SEQUENCES AND SERIES
69.
72. a
Leta1 a,a2 fa,a3 fa2ffa,…, an1f an , where f is a continuous function. If limnl anL, show that fLL.
; 73. a
Show that the sequence defined by
a11 an131
an
is increasing and an3 for all n. Deduce that anis conver
79.
e Usepartscanddtoshowthatan4foralln.
f Use Theorem 12 to show that limn l11nn exists.
The limit is e. See Equation 3.6.6.
Let a and b be positive numbers with ab. Let a1 be their
70.
gent and find its limit.
Show that the sequence defined by
a12 an1
arithmetic mean and b1 their geometric mean: ab
satisfies 0an2 and is decreasing. Deduce that the sequence is convergent and find its limit.
an1anbn 2
1
3an
a12 b1sab Repeat this process so that, in general,
71. a
Fibonacci posed the following problem: Suppose that rabbits live forever and that every month each pair produces a new pair which becomes productive at age
2 months. If we start with one newborn pair, how many pairs of rabbits will we have in the nth month? Show that the answer is fn , wherefnis the Fibonacci sequence defined in Example 3c.
bn1sanbn a Use mathematical induction to show that
an an1 bn1 bn
b Deduce that both anand bnare convergent.
c Show that limnl anlimnl bn. Gauss called the
common value of these limits the arithmeticgeometric mean of the numbers a and b.
a Show that if limnl a2nL and limnl a2n1L, then anis convergent and limn lanL.
bIfa1 1and
an111
1an
findthefirsteighttermsofthesequencean.Thenuse part a to show that limnl ans2. This gives the continued fraction expansion
b Let anfn1fn and show that an111an2 . Assuming that anis convergent, find its limit.
b Illustrate part a by taking f xcos x, a1, and estimating the value of L to five decimal places.
80.
74. 75.
76. 77.
78.
Use a graph to guess the value of the limit
n5 lim
nl n!
b Use a graph of the sequence in part a to find the smallest values of N that correspond to 0.1 and 0.001 in Definition 2.
Use Definition 2 directly to prove that limn lr n0 when r 1.
Prove Theorem 6.
Hint: Use either Definition 2 or the Squeeze Theorem.
Prove Theorem 7.
Prove that if limn lan0 and bn is bounded, then
s21
2
1
1
limnlanbn0. Letan1n.
1n
a Showthatif0ab,then
81.
2
The size of an undisturbed fish population has been modeled
by the formula
pn1bpn
apn
where pn is the fish population after n years and a and b are positive constants that depend on the species and its environ ment. Suppose that the population in year 0 is p00.
a Show that ifpnis convergent, then the only possible
values for its limit are 0 and ba.
b Show that pn1bapn.
c Usepartbtoshowthatifab,thenlimnl pn 0;
in other words, the population dies out. dNowassumethatab.Showthatifp0 ba,then
pn is increasing and 0pnba. Show also that ifp0 ba,thenpnisdecreasingandpn ba. Deducethatifab,thenlimnl pn ba.
bn1 an1
ban1bn
b Deduce that bnn1anban1. cUsea11n1andb11ninpartbto
show that anis increasing. dUsea1andb112ninpartbtoshowthat
a2n 4.
SECTION 11.2 SERIES687
CAS LOGISTIC SEQUENCES
L A B O R AT O R Y PROJECT
A sequence that arises in ecology as a model for population growth is defined by the logistic difference equation
pn1 kpn1pn
where pn measures the size of the population of the nth generation of a single species. To keep the numbers manageable, pn is a fraction of the maximal size of the population, so 0pn1. Notice that the form of this equation is similar to the logistic differential equation in Section 9.4. The discrete modelwith sequences instead of continuous functionsis preferable for modeling insect populations, where mating and death occur in a periodic fashion.
An ecologist is interested in predicting the size of the population as time goes on, and asks these questions: Will it stabilize at a limiting value? Will it change in a cyclical fashion? Or will it exhibit random behavior?
Write a program to compute the first n terms of this sequence starting with an initial population p0, where 0p01. Use this program to do the following.
1. Calculate 20 or 30 terms of the sequence for p01 and for two values of k such that 2
1k3. Graph the sequences. Do they appear to converge? Repeat for a different value of p0 between 0 and 1. Does the limit depend on the choice of p0? Does it depend on the choice of k?
2. Calculate terms of the sequence for a value of k between 3 and 3.4 and plot them. What do you notice about the behavior of the terms?
3. Experiment with values of k between 3.4 and 3.5. What happens to the terms?
4. For values of k between 3.6 and 4, compute and plot at least 100 terms and comment on the behavior of the sequence. What happens if you change p0 by 0.001? This type of behavior is called chaotic and is exhibited by insect populations under certain conditions.
11.2 SERIES
If we try to add the terms of an infinite sequence anwe get an expression of the form
n1
a1 a2 a3 an
which is called an infinite series or just a series and is denoted, for short, by the symbol
an or an n1
But does it make sense to talk about the sum of infinitely many terms? It would be impossible to find a finite sum for the series
12345n
because if we start adding the terms we get the cumulative sums 1, 3, 6, 10, 15, 21, . . . and, after the nth term, we get nn12, which becomes very large as n increases.
However, if we start to add the terms of the series
1111111 2 4 8 16 32 64 2n
1
688
CHAPTER 11 INFINITE SEQUENCES AND SERIES
n
Sum of first n terms
1 2 3 4 5 6 7
10 15 20 25
0.50000000 0.75000000 0.87500000 0.93750000 0.96875000 0.98437500 0.99218750 0.99902344 0.99996948 0.99999905 0.99999997
we get 1 , 3 , 7 , 15, 31, 63, . . . , 112n, . . . . The table shows that as we add more and more 2 4 8 16 32 64
terms, these partial sums become closer and closer to 1. See also Figure 11 in A Preview of Calculus, page 7. In fact, by adding sufficiently many terms of the series we can make the partial sums as close as we like to 1. So it seems reasonable to say that the sum of this infinite series is 1 and to write
1 111 11 1 n12n 2 4 8 16 2n
We use a similar idea to determine whether or not a general series 1 has a sum. We consider the partial sums
and, in general,
s1a1
s2 a1 a2
s3 a1 a2 a3
s4 a1 a2 a3 a4
n
sn a1 a2 a3 anai
i1
These partial sums form a new sequence sn, which may or may not have a limit. If lim n lsns exists as a finite number, then, as in the preceding example, we call it the sum of the infinite seriesan.
DEFINITION Given a series n1 ana1a2a3, let sn denote its nth partial sum:
n
snai a1 a2 an
i1
If the sequence snis convergent and lim n lsns exists as a real number, then
the seriesan is called convergent and we write
a1 a2 an s or an s
n1
The number s is called the sum of the series. Otherwise, the series is called divergent.
2
N Compare with the improper integral yfxdxlimyt fxdx
1 tl1
To find this integral, we integrate from 1 to t and then let tl. For a series, we sum from 1 to n and then let n l .
Thus the sum of a series is the limit of the sequence of partial sums. So when we write n1 ans, we mean that by adding sufficiently many terms of the series we can get as close as we like to the number s. Notice that
n anlim ai
n1 nl i1
EXAMPLE 1 An important example of an infinite series is the geometric series
aarar2 ar3 arn1arn1 a0 n1
N Figure 1 provides a geometric demonstration of the result in Example 1. If the triangles are constructed as shown and s is the sum of the series, then, by similar triangles,
sa sosa
Each term is obtained from the preceding one by multiplying it by the common ratio r. We have already considered the special case where a1 and r1 on page 687.
SECTION 11.2 SERIES689
22
a aar
1r
s
Ifr1,thensn aaanal.Sincelimnl sn doesntexist,the geometric series diverges in this case.
If r1, we have
sn aarar2 arn1
and rsnarar2 arn1 arn
Subtracting these equations, we get
sn rsn aarn
sna1r n
1r If1r1,weknowfrom11.1.9thatrn l0asnl,so
lim snlim a1r n aa lim r na nl nl 1r 1r 1rnl 1r
Thus when r 1 the geometric series is convergent and its sum is a1r.
If r1 or r1, the sequence rn is divergent by 11.1.9 and so, by Equation 3,
lim n lsn does not exist. Therefore the geometric series diverges in those cases. M We summarize the results of Example 1 as follows.
aar
ar
ar
ar ar
ar
3
aa
a
FIGURE 1
4
The geometric series
arn1 aarar2
n1
is convergent if r 1 and its sum is
n1 1r If r 1, the geometric series is divergent.
a arn1
r1
N In words: The sum of a convergent geometric series is
first term
1common ratio
V EXAMPLE 2 Find the sum of the geometric series 510 20 40
3 9 27
SOLUTION Thefirsttermisa5andthecommonratioisr2.Sincer2 1,the 33
series is convergent by 4 and its sum is
5102040553 M
3 9 27 12 5 33
690CHAPTER 11 INFINITE SEQUENCES AND SERIES
n
sn
1 2 3 4 5 6 7 8 9
10
5.000000 1.666667 3.888889 2.407407 3.395062 2.736626 3.175583 2.882945 3.078037 2.947975
N What do we really mean when we say that the sum of the series in Example 2 is 3? Of course, we cant literally add an infinite number of terms, one by one. But, according to Definition 2, the total sum is the limit of the sequence of partial sums. So, by taking the sum of sufficiently many terms, we can get as close as we like to the number 3. The table shows the first ten par tial sums sn and the graph in Figure 2 shows how the sequence of partial sums approaches 3.
sn
3
0 20 n FIGURE 2
EXAMPLE 3 Is the series22n31n convergent or divergent?
n1
SOLUTION Lets rewrite the nth term of the series in the form arn1:
4nn1
N Another way to identify a and r is to write out the first few terms:
416 6439
22n31n22n3n1 n1 n1 n1
44 3
3n1
We recognize this series as a geometric series with a4 and r4 . Since r1, the
series diverges by 4. M V EXAMPLE 4 Write the number 2.3172.3171717. . . as a ratio of integers.
n1
3
SOLUTION
2.3171717…2.3 171717103 105 107
After the first term we have a geometric series with a17103 and r1102. Therefore
17 103
1102 100 17 1147
2.3172.3 23
17 1000
1 2.3 99
10990 495
M
TEC Module 11.2 explores a series that depends on an angle in a triangle and enables you to see how rapidly the series
EXAMPLE 5 Find the sum of the seriesxn, where x 1. n0
SOLUTION Notice that this series starts with n0 and so the first term is x01. With series, we adopt the convention that x01 even when x0. Thus
xn 1xx2 x3 x4
n0
This is a geometric series with a1 and rx. Since r x 1, it converges and 4 gives
1
converges when
varies.
xnM n0 1x
5
EXAMPLE 6 Show that the series1 is convergent, and find its sum. n1 nn1
SOLUTION This is not a geometric series, so we go back to the definition of a convergent series and compute the partial sums.
snn 11111
i1 ii1 12 23 34 nn1
We can simplify this expression if we use the partial fraction decomposition
111 ii1 i i1
see Section 7.4. Thus we have
snn 1 n11 i1 ii1 i1 i i1
SECTION 11.2 SERIES691
N Notice that the terms cancel in pairs. This is an example of a telescoping sum: Because of all the cancellations, the sum collapses like a pirates collapsing telescope into just two terms.
N Figure 3 illustrates Example 6 by show
ing the graphs of the sequence of terms
an1nn1 and the sequence snof partial sums. Notice that an l 0 and sn l 1. See Exercises 62 and 63 for two geometric inter pretations of Example 6.
1 1111111 1
122334nn1
n1
lim snlim 11101
and so
Therefore the given series is convergent and
nl nl n1
1
11 M n1 nn1
V EXAMPLE 7 Show that the harmonic series
11111
sn
an
n1n 234
SOLUTION For this particular series its convenient to consider the partial sums s2, s4, s8, s16,
is divergent.
s32, . . . and show that they become large.
s11
s211
s4 11 1 111 1 112 234 244 2
s8 11 1 11 1 1 1 2345678
11 1 11 1 1 1 2448888
11 1 1 13 222 2
s16 11 1 11 1112345 89 16
11 1 11 1112 4 4 8 8 16 16
11 1 1 1 14 2222 2
0n
FIGURE 3
2
692CHAPTER 11 INFINITE SEQUENCES AND SERIES
N The method used in Example 7 for showing that the harmonic series diverges is due to the French scholar Nicole Oresme 13231382.
Similarly,s32 15,s64 16,andingeneral 22
s2n1n 2
This shows that s2 n las n land so snis divergent. Therefore the harmonic series diverges. M
PROOF Let sna1a2an. Then ansnsn1. Sincean is convergent, the sequencesnisconvergent.Letlimnl sn s.Sincen1lasnl,wealso have lim n lsn1s. Therefore
liman limsn sn1limsn limsn1 nl nl nl nl
ss0 M
NOTE 1 With any seriesan we associate two sequences: the sequence snof its par tial sums and the sequence anof its terms. Ifan is convergent, then the limit of the sequence sn is s the sum of the series and, as Theorem 6 asserts, the limit of the sequence anis 0.
NOTE 2 The converse of Theorem 6 is not true in general. If limn lan0, we cannot conclude that an is convergent. Observe that for the harmonic series 1n we have an 1nl0asnl,butweshowedinExample7that1nisdivergent.
The Test for Divergence follows from Theorem 6 because, if the series is not divergent, then it is convergent, and so limnl an0.
THEOREM If the seriesan is convergent, then lim an0.
6
n1
nl
THE TEST FOR DIVERGENCE If lim an does not exist or if lim an0, then the
7
seriesan is divergent. n1
nl nl
n2
EXAMPLE 8 Show that the series 2 diverges.
SOLUTION
n1 5n 4
n2 11
limanlim 2 lim
nl nl 5n 4 nl 54n
2 0 5
So the series diverges by the Test for Divergence.
M
NOTE3 If we find that limnl an0, we know that an is divergent. If we find that limnl an0, we know nothing about the convergence or divergence of an. Remem ber the warning in Note 2: If limnl an0, the series an might converge or it might diverge.
These properties of convergent series follow from the corresponding Limit Laws for Sequences in Section 11.1. For instance, here is how part ii of Theorem 8 is proved:
Let
nn
sn ai s an tn bi
t bn n1
i1 n1
The nth partial sum for the seriesanbnis
n
un ai bi
i1
and, using Equation 5.2.10, we have
nl nl i1 nl i1 i1 nn
limun lim ai bilim ai bi n nn
limai limbi limsnlimtn st nl i1 nl i1 nl nl
Thereforeanbnis convergent and its sum is
an bnst anbn M
n1 n1
EXAMPLE 9 Findthesumoftheseries 3
n1 nn1
n1
1. 2n
i1
SECTION 11.2 SERIES693
THEOREM Ifan andbn are convergent series, then so are the seriescan where c is a constant, anbn, and anbn, and
i can c an ii an bn anbn n1 n1 n1 n1 n1
iii anbnanbn n1 n1 n1
8
SOLUTION The series12n is a geometric series with a1 and r1 , so 22
In Example 6 we found that
11 21
n1 n1 2 12
11 n1 nn1
So, by Theorem 8, the given series is convergent and
3 13 113114 M n1 nn1 2n n1 nn1 n1 2n
NOTE 4 A finite number of terms doesnt affect the convergence or divergence of a series. For instance, suppose that we were able to show that the series
4 n 31 n
n
694
CHAPTER 11 INFINITE SEQUENCES AND SERIES
is also convergent.
11.2 EXERCISES
1. a What is the difference between a sequence and a series?
b What is a convergent series? What is a divergent series?
2. Explain what it means to say that n1 an5.
38 Find at least 10 partial sums of the series. Graph both the sequence of terms and the sequence of partial sums on the same screen. Does it appear that the series is convergent or divergent? If it is convergent, find the sum. If it is divergent, explain why.
3 n1
n 18. n1 4 n0
n 19.3n1 20.
n0 n1
is convergent. Since
n123 n n1 n31 2 9 28 n4 n31
it follows that the entire series n1 nn31 is convergent. Similarly, if it is known that the series nN1 an converges, then the full series
N anan an
n1 n1 nN1
1 s2
en 3 n1
17.
n
;
12 4. n1 5n
tann 6. n1
2n21
n1n21
0.6n1
n1
2134 Determine whether the series is convergent or divergent. If it is convergent, find its sum.
3.
1 21.
n1
22. 23. 2 24.
n1 2n3kk2
5.
7.
10.
n1 2nk2
2 k1 k3
1 1 1
k2k112n
8. n1 sn sn1
Letan2n. 3n1
n2 nn2
25. n 26. n1 3
13n n
n1 2
27.sn 2 28.0.8n10.3n n1 n1
nn32
9.
a Determine whether anis convergent. b Determine whether n1 an is convergent.
n2129. ln 2
30. cos1k k1
n1 2n 1
a Explain the difference between
ai and aj i1 j1
b Explain the difference between
n n ai and aj
arctann n1
32. n
31.
i1 i1
1120 Determine whether the geometric series is convergent or
divergent. If it is convergent, find its sum.
11. 324 812. 1 1 1 1 39 842
3540 Determine whether the series is convergent or divergent by expressing sn as a telescoping sum as in Example 6. If it is convergent, find its sum.
1
33.n
n1 e
1 nn1
n1 5 nen
34.2 n1 n
13. 3416 642 2 39 36.
35.
14. 10.40.160.064 n2 n2 1 n1 n2 4n310n 3 n
15.60.9n1 16.n1 37.38. ln
n1 n1 9 n1 nn3 n1 n1
1n 1n1 39. e e
56.
If the nth partial sum of a series n1 an is sn3n2n,
findanandan.
40.coscos 57. When money is spent on goods and services, those who
n1
1 1 n1
n1 n2 n12
41 46 Express the number as a ratio of integers. 0.20.2222…
receive the money also spend some of it. The people receiv ing some of the twicespent money will spend some of that, and so on. Economists call this chain reaction the multiplier effect. In a hypothetical isolated community, the local govern ment begins the process by spending D dollars. Suppose that each recipient of spent money spends 100c and saves 100s of the money that he or she receives. The values c and s are called the marginal propensity to consume and the mar ginal propensity to save and, of course, cs1.
a Let Sn be the total spending that has been generated after n transactions. Find an equation for Sn .
b Show that limn lSnkD, where k1s. The number k is called the multiplier. What is the multiplier if the marginal propensity to consume is 80?
Note: The federal government uses this principle to justify deficit spending. Banks use this principle to justify lending a large percentage of the money that they receive in deposits.
A certain ball has the property that each time it falls from a height h onto a hard, level surface, it rebounds to a height rh, where 0r1. Suppose that the ball is dropped from an initial height of H meters.
a Assuming that the ball continues to bounce indefinitely,
find the total distance that it travels. Use the fact that the
ball falls 1 t t 2 meters in t seconds. 2
b Calculate the total time that the ball travels.
c Suppose that each time the ball strikes the surface
with velocity v it rebounds with velocity kv, where 0k1. How long will it take for the ball to come to rest?
Find the value of c if
1cn 2 n2
Find the value of c such that
enc 10 n0
In Example 7 we showed that the harmonic series is diver gent. Here we outline another method, making use of the fact that ex1x for any x0. See Exercise 4.3.76.
If sn is the nth partial sum of the harmonic series, show that e snn1. Why does this imply that the harmonic series is divergent?
Graph the curves yxn, 0x1, for n0, 1, 2, 3, 4, . . . on a common screen. By finding the areas between successive curves, give a geometric demonstration of the fact, shown in Example 6, that
SECTION 11.2 SERIES695
41.
42. 0.730.73737373 . . .
43. 3.4173.417417417 . . .
44. 6.2546.2545454 . . .
45. 1.5342
46. 7.12345
4751 Find the values of x for which the series converges. Find the sum of the series for those values of x.
xn
n
49. 4nxn n0
48. x4n
n1
58.
47.
n1 3
x3n 50.
cosnx 51.
n0 2n
52. We have seen that the harmonic series is a divergent series
whose terms approach 0. Show that
n0 2n
ln 1 1
n1 n is another series with this property.
60.
61.
62.
CAS 5354 Use the partial fraction command on your CAS to find a convenient expression for the partial sum, and then use this expression to find the sum of the series. Check your answer by using the CAS to sum the series directly.
3n23n1 53. 2 3 n1 n n
If the nth partial sum of a series n1 an is
sn
n1
n1 1
54. 3
1 n2 n n
;
59.
55.
findanand
n1
an.
1 n1 nn1
696CHAPTER 11 INFINITE SEQUENCES AND SERIES
63. The figure shows two circles C and D of radius 1 that touch atP.Tisacommontangentline;C1 isthecirclethattouches C, D, and T; C2 is the circle that touches C, D, and C1; C3 is the circle that touches C, D, and C2. This procedure can be continued indefinitely and produces an infinite sequence of circles Cn . Find an expression for the diameter of Cn and thus provide another geometric demonstration of Example 6.
69. Ifan is convergent andbn is divergent, show that theseriesan bnisdivergent. Hint:Argueby contradiction.
70. Ifan andbn are both divergent, isanbn neces sarily divergent?
Suppose that a seriesan has positive terms and its partial sums sn satisfy the inequality sn1000 for all n. Explain whyan must be convergent.
72. The Fibonacci sequence was defined in Section 11.1 by the equations
f1 1, f2 1, fn fn1 fn2 n3
Show that each of the following statements is true.
P
CD
64. A right triangle ABC is given with Aand AC b. CD is drawn perpendicular to AB, DE is drawn perpen dicular to BC, EFAB, and this process is continued indefinitely, as shown in the figure. Find the total length of all the perpendiculars
CD DE EF FGin terms of b and .
111 fn1 fn1 fn1 fn fn fn1
T
a
b 1
1 n2 fn1 fn1
fn
c 2
n2 fn1 fn1
The Cantor set, named after the German mathematician
Georg Cantor 18451918, is constructed as follows. We
start with the closed interval 0, 1 and remove the open inter
val 1 , 2 . That leaves the two intervals 0, 1and 2, 1 and A3333
65.
F H
b
BGEC
What is wrong with the following calculation? 0000
111111 111111
1111111 10001
Guido Ubaldus thought that this proved the existence of God because something has been created out of nothing.
66. Suppose that n1 an an0 is known to be a convergent series. Prove that n1 1an is a divergent series.
67. Prove part i of Theorem 8.
68. Ifan is divergent and c0, show thatcan is divergent.
D
we remove the open middle third of each. Four intervals remain and again we remove the open middle third of each of them. We continue this procedure indefinitely, at each step removing the open middle third of every interval that remains from the preceding step. The Cantor set consists of the num bers that remain in 0, 1 after all those intervals have been removed.
a Show that the total length of all the intervals that are
removed is 1. Despite that, the Cantor set contains infi nitely many numbers. Give examples of some numbers in the Cantor set.
b The Sierpinski carpet is a twodimensional counterpart of the Cantor set. It is constructed by removing the center oneninth of a square of side 1, then removing the centers of the eight smaller remaining squares, and so on. The figure shows the first three steps of the construction. Show that the sum of the areas of the removed squares
is 1. This implies that the Sierpinski carpet has area 0.
71.
C
1 CTM 1
C
73.
74. a A sequence anis defined recursively by the equation
an1 an1an2for n3, where a1 and a2 can be any
2
real numbers. Experiment with various values of a1 and a2
and use your calculator to guess the limit of the sequence. b Find limn lan in terms of a1 and a2 by expressing
an1an in terms of a2a1 and summing a series.
75. Consider the series
a Find the partial sums s1, s2, s3, and s4. Do you recognize the denominators? Use the pattern to guess a formula for sn.
b Use mathematical induction to prove your guess.
c Show that the given infinite series is convergent, and find
its sum.
76. In the figure there are infinitely many circles approaching the vertices of an equilateral triangle, each circle touching other
circles and sides of the triangle. If the triangle has sides of length 1, find the total area occupied by the circles.
SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS697
n
n1 n1!
11.3
THE INTEGRAL TEST AND ESTIMATES OF SUMS
In general, it is difficult to find the exact sum of a series. We were able to accomplish this for geometric series and the series1nn1 because in each of those cases we could find a simple formula for the nth partial sum sn. But usually it isnt easy to compute lim n lsn. Therefore, in the next few sections, we develop several tests that enable us to determine whether a series is convergent or divergent without explicitly finding its sum. In some cases, however, our methods will enable us to find good estimates of the sum. Our first test involves improper integrals.
We begin by investigating the series whose terms are the reciprocals of the squares of the positive integers:
111111 n1n2 12 22 32 42 52
Theres no simple formula for the sum sn of the first n terms, but the computergenerated table of values given in the margin suggests that the partial sums are approaching a num ber near 1.64 as n land so it looks as if the series is convergent.
We can confirm this impression with a geometric argument. Figure 1 shows the curve y1×2 and rectangles that lie below the curve. The base of each rectangle is an interval of length 1; the height is equal to the value of the function y1×2 at the right endpoint of the interval. So the sum of the areas of the rectangles is
n
snn 1 i1 i2
5
10
50
100
500
1000
5000
1.4636 1.5498 1.6251 1.6350 1.6429 1.6439 1.6447
11111 1 12 22 32 42 52 n1 n2
y1
012345x
area 1 area 1 area 1 area 1 2 3 4 5
y
area 1 1
FIGURE 1
698
CHAPTER 11 INFINITE SEQUENCES AND SERIES
If we exclude the first rectangle, the total area of the remaining rectangles is smaller
than the area under the curve y1×2 for x1, which is the value of the integral
x 1x2dx.InSection7.8wediscoveredthatthisimproperintegralisconvergentandhas 1
value 1. So the picture shows that all the partial sums are less than
1y 1dx2 12 1×2
Thus the partial sums are bounded. We also know that the partial sums are increasing because all the terms are positive. Therefore the partial sums converge by the Monotonic Sequence Theorem and so the series is convergent. The sum of the series the limit of the partial sums is also less than 2:
111112 n1 n2 12 22 32 42
The exact sum of this series was found by the Swiss mathematician Leonhard Euler 17071783 to be 26, but the proof of this fact is quite difficult. See Problem 6 in the Problems Plus following Chapter 15.
Now lets look at the series
111111 n1sn s1 s2 s3 s4 s5
The table of values of sn suggests that the partial sums arent approaching a finite number, so we suspect that the given series may be divergent. Again we use a picture for confirma tion. Figure 2 shows the curve y1sx , but this time we use rectangles whose tops lie above the curve.
y
012345x area 1 area 1 area 1 area 1
1 2 3 4
The base of each rectangle is an interval of length 1. The height is equal to the value of the function y1sx at the left endpoint of the interval. So the sum of the areas of all the rectangles is
11111 1 s1 s2 s3 s4 s5 n1 sn
n
snn 1 i1 si
5
10
50
100
500
1000
5000
3.2317
5.0210 12.7524 18.5896 43.2834 61.8010
139.9681
y 1 x
FIGURE 2
This total area is greater than the area under the curve y1sx for x1, which is equal to the integral x 1sxdx. But we know from Section 7.8 that this improper integral is
1
divergent. In other words, the area under the curve is infinite. So the sum of the series must
be infinite; that is, the series is divergent.
The same sort of geometric reasoning that we used for these two series can be used to
prove the following test. The proof is given at the end of this section.
SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS699
THE INTEGRAL TEST Suppose f is a continuous, positive, decreasing function on
1,and let anf n. Then the series n1 an is convergent if and only if the
improper integral x f x dx is convergent. In other words: 1
i If y f x dx is convergent, thenan is convergent.
1 n1
ii If y f x dx is divergent, thenan is divergent. 1 n1
NOTE When we use the Integral Test, it is not necessary to start the series or the inte gral at n1. For instance, in testing the series
1 weuse y 1 dx n4 n32 4 x32
Also, it is not necessary that f be always decreasing. What is important is that f be ulti mately decreasing, that is, decreasing for x larger than some number N. Then nN an is convergent, so n1 an is convergent by Note 4 of Section 11.2.
EXAMPLE 1 Test the series1 n1 n21
for convergence or divergence.
SOLUTION The function f x1x 21 is continuous, positive, and decreasing on
1,so we use the Integral Test:
y1 d xl i m y t 1
d xl i m t a n1 x1t tl
1 x21 tl 1 x21
lim tan1t
N In order to use the Integral Test we need to be able to evaluate x1 f x dx and therefore we have to be able to find an antiderivative of f. Frequently this is difficult or impossible, so we need other tests for convergence too.
V EXAMPLE 2 For what values of p is the series1 convergent? n1 np
SOLUTION If p0, then limnl 1np. If p0, then limnl 1np1. In either case, lim n l1n p 0, so the given series diverges by the Test for Divergence 11.2.7. If p0, then the function f x1x p is clearly continuous, positive, and decreasing
on 1, . We found in Chapter 7 see 7.8.2 that
y 1 dxconvergesifp1anddivergesifp1
1xp
It follows from the Integral Test that the series1n p converges if p1 and diverges if 0p1. For p1, this series is the harmonic series discussed in Example 7 in Section 11.2. M
The series in Example 2 is called the pseries. It is important in the rest of this chapter, so we summarize the results of Example 2 for future reference as follows.
tl 4244
Thus x 1×21 dx is a convergent integral and so, by the Integral Test, the series
1
1n21 is convergent. M
700
CHAPTER 11 INFINITE SEQUENCES AND SERIES
EXAMPLE 3
a The series
The pseries1 is convergent if p1 and divergent if p1. n1 np
1
11111 n1 n3 13 23 33 43
is convergent because it is a pseries with p31. b The series
1 11111 n1 n 13 n1 s3 n s3 2 s3 3 s3 4
is divergent because it is a pseries with p11. 3
M
NOTE We should not infer from the Integral Test that the sum of the series is equal to the value of the integral. In fact,
12 1
2n1n 6
whereas y 2 dx1 1x
an y fxdx
n1 1
Therefore, in general,
V EXAMPLE 4 Determine whether the seriesln n converges or diverges. n1 n
SOLUTION The function f xln xx is positive and continuous for x1 because the logarithm function is continuous. But it is not obvious whether or not f is decreasing, so we compute its derivative:
fx1xxln x1ln x x2 x2
Thus fx0whenlnx1,thatis,xe.Itfollowsthat f isdecreasingwhenxe and so we can apply the Integral Test:
lnx2t
Since this improper integral is divergent, the seriesln nn is also divergent by the
ESTIMATING THE SUM OF A SERIES
Suppose we have been able to use the Integral Test to show that a seriesan is conver gent and we now want to find an approximation to the sum s of the series. Of course, any partial sum sn is an approximation to s because lim n lsns. But how good is such an approximation? To find out, we need to estimate the size of the remainder
t lnx
1x tl1x tl21tl2
lnx
y dxlim y
dxlim
Integral Test. M
lnt2
lim
Rn ssn an1 an2 an3
y
The remainder Rn is the error made when sn, the sum of the first n terms, is used as an approximation to the total sum.
We use the same notation and ideas as in the Integral Test, assuming that f is decreas ing on n, . Comparing the areas of the rectangles with the area under yf x for xn in Figure 3, we see that
Rn an1 an2 y fxdx n
an1
an2
0nx FIGURE 3
Similarly, we see from Figure 4 that
Rn an1 an2 y n1
So we have proved the following error estimate.
y
fxdx
SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS701
y
…
y
REMAINDER ESTIMATE FOR THE INTEGRAL TEST Suppose f kak , where f is a continuous, positive, decreasing function for xn andan is convergent. If Rn ssn,then
y fxdxRn y fxdx n1 n
2
an1
0 n1 x
FIGURE 4
an2
…
V EXAMPLE 5
a Approximate the sum of the series1n 3 by using the sum of the first 10 terms. Estimate the error involved in this approximation.
b How many terms are required to ensure that the sum is accurate to within 0.0005?
SOLUTION In both parts a and b we need to know x fx dx. With fx1×3, n
which satisfies the conditions of the Integral Test, we have
a
11 t1 11 y 3 dxlim2 lim222 n x tl 2x n tl 2t 2n 2n
1s10111 1 1.1975 n1 n3 13 23 33 103
According to the remainder estimate in 2, we have
R10y1dx 1 1 10 x3 2102 200
So the size of the error is at most 0.005.
b Accuracy to within 0.0005 means that we have to find a value of n such that
Rn0.0005. Since
Rny 1dx 1 n x3 2n2
we want 10.0005 2n2
702
CHAPTER 11 INFINITE SEQUENCES AND SERIES
Solving this inequality, we get
n21 1000 or ns1000 31.6 0.001
We need 32 terms to ensure accuracy to within 0.0005. M If we add sn to each side of the inequalities in 2, we get
because snRns. The inequalities in 3 give a lower bound and an upper bound for s. They provide a more accurate approximation to the sum of the series than the partial sum sn does.
EXAMPLE 6 Use 3 with n10 to estimate the sum of the series1 . n1 n3
SOLUTION The inequalities in 3 become
s10 y 1 dxss10 y 1 dx
sn y fxdxssn y fxdx n1 n
3
From Example 5 we know that
so s10Using s101.197532, we get
11 x3
y 1 dx 1 n x3 2n2
1ss102112
10 x3
1 2102
1.201664s1.202532
If we approximate s by the midpoint of this interval, then the error is at most half the
length of the interval. So
11.2021 with error0.0005 M n1 n3
If we compare Example 6 with Example 5, we see that the improved estimate in 3 can be much better than the estimate ssn. To make the error smaller than 0.0005 we had to use 32 terms in Example 5 but only 10 terms in Example 6.
PROOF OF THE INTEGRAL TEST
We have already seen the basic idea behind the proof of the Integral Test in Figures 1 and 2 for the series1n2 and1sn . For the general seriesan, look at Figures 5 and 6. The area of the first shaded rectangle in Figure 5 is the value of f at the right endpoint of 1, 2,
y
that is, f2a2. So, comparing the areas of the shaded rectangles with the area under yfx from 1 to n, we see that
n
a2 a3 an y fxdx
1
Notice that this inequality depends on the fact that f is decreasing. Likewise, Figure 6 shows that
y
012345…nx FIGURE 5
aTM
a
a
012345…nx FIGURE 6
since f x0. Therefore
sn a1ai a1 y fxdxM,say
a
an
an1
4
SECTION 11.3 THE INTEGRAL TEST AND ESTIMATES OF SUMS703
y
y
n
y fxdxa1 a2 an1
1
i If y f x dx is convergent, then 4 gives 1
n
ai yn fxdxy fxdx
i2 1 1
5
a
aTM
a
a
n
i2 1
11.3 EXERCISES
1. Draw a picture to show that
Since snM for all n, the sequence snis bounded above. Also sn1 sn an1 sn
since an1f n10. Thus snis an increasing bounded sequence and so it is convergent by the Monotonic Sequence Theorem 11.1.12. This means that an is convergent.
ii Ifx fxdxisdivergent,thenxn fxdxlasnlbecause fx0.But5 11
gives
n1
yn fxdx ai sn1
1 i1
and so sn1 l . This implies that sn land soan diverges. M
38 Use the Integral Test to determine whether the series is convergent or divergent.
1
n2 n1.31 x1.3 dx
y 1 What can you conclude about the series?
2. Suppose f is a continuous positive decreasing function for x1 and anfn. By drawing a picture, rank the following three quantities in increasing order:
y6 fxdx ai ai 1 i1 i2
1 3.5
1 4.5
5 6
n2 nen 8.
n1 sn 5.
n1 n
6.1
7.
1
n1 2n13
n1 sn4
n1 n1 n1
704CHAPTER 11 INFINITE SEQUENCES AND SERIES
926 Determine whether the series is convergent or divergent.
a Use the sum of the first 10 terms to estimate the sum of the series n1 1n2. How good is this estimate?
b Improve this estimate using 3 with n10.
c Find a value of n that will ensure that the error in the
approximation ssn is less than 0.001.
Find the sum of the series n1 1n5 correct to three decimal
places.
Estimate n1 2n16 correct to five decimal places.
How many terms of the series n2 1nln n2 would you need to add to find its sum to within 0.01?
Show that if we want to approximate the sum of the series n1 n1.001 so that the error is less than 5 in the ninth decimal place, then we need to add more than 1011,301 terms!
a Show that the series n1 ln n2n 2 is convergent.
b Find an upper bound for the error in the approximation
ssn.
c What is the smallest value of n such that this upper bound
is less than 0.05?
d Find sn for this value of n.
a Use 4 to show that if sn is the nth partial sum of the harmonic series, then
s 1lnn n
b The harmonic series diverges, but very slowly. Use part a to show that the sum of the first million terms is less than 15 and the sum of the first billion terms is less than 22.
Use the following steps to show that the sequence
tn1111ln n 23n
has a limit. The value of the limit is denoted by
Eulers constant.
a Draw a picture like Figure 6 with f x1x and interpret
tn asanareaoruse5toshowthattn 0foralln. b Interpret
tn tn1 lnn1lnn 1 n1
as a difference of areas to show that tntn10. There
fore, tnis a decreasing sequence.
c Use the Monotonic Sequence Theorem to show that tnis
convergent.
Find all positive values of b for which the series n1 bln n converges.
Find all values of c for which the following series converges.
9.
12. 13. 14. 15.
19.n1
82764125 1111
34.
35. 36.
37.
38.
2
0.85 10. n1.4 3n1.2
n1 n n1 1 1111
1 2s2 3s3 4s4 5s51 1111
3579
11 1115 8 11 14 17
52sn
3
n2 16.3
CAS
n1
1
18.
n
n1 n13n2
n1
2 n4
lnn n3
n1 nn1 20.1
23. 25.
1 n2 nlnn
e1n
2
22.
24. 26.
n1
n1 n24n51
n2 nln n 2n2
n n3 e
1 28. 1
33.
11.
17.
39.
21.
n1 n1
n n1 n41
40.
n1 n3n
2730 Find the values of p for which the series is convergent.
27.n2
and is called
nln np 29. n1n2p
n3 n ln n lnln np
lnn 30.p
n1 n is defined by
31. The Riemann zetafunction
xx
1 n1 n
and is used in number theory to study the distribution of prime numbers. What is the domain of ?
32. a Find the partial sum s10 of the series n1 1n4. Estimate the error in using s10 as an approximation to the sum of the series.
b Use 3 with n10 to give an improved estimate of the sum.
c Findavalueofnsothatsn iswithin0.00001ofthesum.
41. 42.
c 1
n1 n n1
SECTION 11.4 THE COMPARISON TESTS705
11.4
THE COMPARISON TESTS
In the comparison tests the idea is to compare a given series with a series that is known to be convergent or divergent. For instance, the series
1 n1 2n1
reminds us of the series n1 12n, which is a geometric series with a1 and r1 and 22
is therefore convergent. Because the series 1 is so similar to a convergent series, we have the feeling that it too must be convergent. Indeed, it is. The inequality
11 2n1 2n
shows that our given series 1 has smaller terms than those of the geometric series and therefore all its partial sums are also smaller than 1 the sum of the geometric series. This means that its partial sums form a bounded increasing sequence, which is convergent. It also follows that the sum of the series is less than the sum of the geometric series:
11 n1 2n1
Similar reasoning can be used to prove the following test, which applies only to series whose terms are positive. The first part says that if we have a series whose terms are smaller than those of a known convergent series, then our series is also convergent. The second part says that if we start with a series whose terms are larger than those of a known divergent series, then it too is divergent.
1
THE COMPARISON TEST Suppose thatan andbn are series with positive terms. i Ifbn is convergent and anbn for all n, thenan is also convergent.
ii Ifbn is divergent and anbn for all n, thenan is also divergent.
N It is important to keep in mind the distinction between a sequence and a series. A sequence is a list of numbers, whereas a series is a sum. With every seriesan there are associated two sequences: the sequence anof terms and the sequence snof partial sums.
Standard Series for Use with the Comparison Test
PROOF
i Let
nn
sn ai tn bi t bn
i1 i1 n1
Since both series have positive terms, the sequences snand tnare increasing
sn1 sn an1 sn.Alsotn lt,sotn tforalln.Sinceai bi,wehavesn tn. Thus snt for all n. This means that snis increasing and bounded above and therefore converges by the Monotonic Sequence Theorem. Thusan converges.
ii Ifbn is divergent, then tn lsince tnis increasing. But aibi so sntn. Thus sn l . Thereforean diverges. M
In using the Comparison Test we must, of course, have some known seriesbn for the purpose of comparison. Most of the time we use one of these series:
N A p series1n p converges if p1 and diverges if p1; see 11.3.1
N A geometric seriesar n1 converges if r 1 and diverges if r 1; see 11.2.4
706
CHAPTER 11 INFINITE SEQUENCES AND SERIES
V EXAMPLE 1 Determine whether the series5
n1 2n24n3
converges or diverges. SOLUTION For large n the dominant term in the denominator is 2n2 so we compare the
given series with the series52n2 . Observe that 55
2n24n3 2n2
because the left side has a bigger denominator. In the notation of the Comparison Test,
an is the left side and bn is the right side. We know that551
n1 2n2 2 n1 n2
is convergent because its a constant times a pseries with p21. Therefore
5
n1 2n24n3
is convergent by part i of the Comparison Test. M
NOTE 1 Although the condition anbn or anbn in the Comparison Test is given for all n, we need verify only that it holds for nN, where N is some fixed integer, because the convergence of a series is not affected by a finite number of terms. This is illustrated in the next example.
V EXAMPLE 2 Test the seriesln n for convergence or divergence. n1 n
SOLUTION This series was tested using the Integral Test in Example 4 in Section 11.3, but it is also possible to test it by comparing it with the harmonic series. Observe that ln n1 for n3 and so
ln n1 n3 nn
We know that1n is divergentpseries with p1. Thus the given series is divergent by the Comparison Test. M
NOTE 2 The terms of the series being tested must be smaller than those of a convergent series or larger than those of a divergent series. If the terms are larger than the terms of a convergent series or smaller than those of a divergent series, then the Comparison Test doesnt apply. Consider, for instance, the series
The inequality
1 n1 2n1
11 2n1 2n
is useless as far as the Comparison Test is concerned becausebn 1 n is convergent 2
and anbn . Nonetheless, we have the feeling that12n1 ought to be convergent
because it is very similar to the convergent geometric series1 n. In such cases the fol 2
lowing test can be used.
N Exercises 40 and 41 deal with the cases c0 and c.
for convergence or divergence. SOLUTION We use the Limit Comparison Test with
SECTION 11.4 THE COMPARISON TESTS707
THE LIMIT COMPARISON TEST Suppose thatan andbn are series with positive terms. If
liman c nl bn
where c is a finite number and c0, then either both series converge or both diverge.
PROOF Let m and M be positive numbers such that mcM. Because anbn is close to c for large n, there is an integer N such that
manM when nN bn
and so mbnanMbn when nN
Ifbn converges, so doesMbn. Thusan converges by part i of the Comparison Test. Ifbn diverges, so doesmbn and part ii of the Comparison Test shows thatan diverges. M
EXAMPLE 3 Test the series1 n1 2n1
an 1
2n1
and obtain
Since this limit exists and12n is a convergent geometric series, the given series con verges by the Limit Comparison Test. M
EXAMPLE 4 Determine whether the series2n23n converges or diverges. n1 s5n5
SOLUTION The dominant part of the numerator is 2n2 and the dominant part of the denomi nator is sn5n52. This suggests taking
bn1 2n
an 12n1
nl bn nl 12 nl 2 1 nl 112
2n
limlim n limn lim n10
1
an nl bn
2n23nlim
n12 2n523n32lim
lim
nl s5n5 23
2 nl 2s5n5
2n23n ans5n5
2n2 2 bnn52n12
limn 201 nl 5 2s01
2n1 5
708
CHAPTER 11 INFINITE SEQUENCES AND SERIES
Sincebn21n12 is divergent pseries with p11, the given series diverges 2
by the Limit Comparison Test.
M
Notice that in testing many series we find a suitable comparison seriesbn by keeping only the highest powers in the numerator and denominator.
ESTIMATING SUMS
If we have used the Comparison Test to show that a seriesan converges by comparison with a seriesbn, then we may be able to estimate the suman by comparing remainders. As in Section 11.3, we consider the remainder
Rn ssn an1 an2
For the comparison seriesbn we consider the corresponding remainder
Tn ttn bn1 bn2
Since anbn for all n, we have RnTn. Ifbn is a pseries, we can estimate its remain der Tn as in Section 11.3. Ifbn is a geometric series, then Tn is the sum of a geometric series and we can sum it exactly see Exercises 35 and 36. In either case we know that Rn is smaller than Tn.
V EXAMPLE 5 Use the sum of the first 100 terms to approximate the sum of the series1n31. Estimate the error involved in this approximation.
SOLUTION Since
11 n31 n3
the given series is convergent by the Comparison Test. The remainder Tn for the compari son series1n3 was estimated in Example 5 in Section 11.3 using the Remainder Esti mate for the Integral Test. There we found that
Tny 1dx 1 n x3 2n2
Therefore the remainder Rn for the given series satisfies RnTn 1
With n100 we have
2n2
R10010.00005 21002
Using a programmable calculator or a computer, we find that1 100 1
3
n1 n 1
3 0.6864538 n1 n 1
with error less than 0.00005.
M
SECTION 11.4 THE COMPARISON TESTS709
11.4 EXERCISES
Supposean andbn are series with positive terms andbn is known to be convergent.
a Ifan bn foralln,whatcanyousayaboutan?Why?
b If anbn for all n, what can you say about an? Why?
2. Supposean andbn are series with positive terms andbn is known to be divergent.
a If anbn for all n, what can you say about an? Why?
b If anbn for all n, what can you say about an? Why?
332 Determine whether the series converges or diverges.
3336 Use the sum of the first 10 terms to approximate the sum of the series. Estimate the error.
33.
35.
1.
14
sin2n 34.3
n1 sn 1
1 36.n
3.
5.
7.
9.
11.
13.
15.
19.
21.
23.
25.
27.
29.
n1
n
34. 2n1
n3
4
n2 n1
n1
n1 n
n1 12n n1 n13n
The meaning of the decimal representation of a number 0.d1d2d3 . . . where the digit di is one of the numbers 0, 1, 2,…,9isthat
0.d1d2d3d4… d1d2d3d410 102 103 104
Show that this series always converges.
p
For what values of p does the series n2 1n ln n converge?
Prove that if an0 andan converges, thenan2 also converges.
a Suppose thatan andbn are series with positive terms andbn is convergent. Prove that if
n1 nsn
6.
12.
14.
16.
18.
20.n1
n1
9n
n1 n 2sn43n
n n1 310
cos2n n1 n21
n1 n1 n4n
arctan n
1.2
n
38. 39.
40.
1 en n1 n
10.
n1
2
n21
37.
8.
3n411sinn
n1
n0
n2
n1
n1
10n sn
n1 1
sn31 1
2n3 n4n
n
an
lim 0 nl bn
is also convergent.
b Use part a to show that the series converges.
n1 n
21n
n1 nsn1
2
n1 sn1
14n
n
n1 2n2 n152n
22 n1 1n
thena n
lnn i
lnn ii
n1 n3
a Suppose thatan andbn are series with positive terms
andb n
lim an nl bn
thenan is also divergent.
b Use part a to show that the series diverges.
n1 snen is divergent. Prove that if
41.
17.
n1 13
sn2
22.
24.
n6n2
1nn2
26.n1
n3 n13n25n
n5
3 sn7n2
e 1n n
1 lnn iii
n2lnn n1 n
Give an example of a pair of seriesan andbn with positive
3
n1 n n1
42.
43. 44. 45. 46.
terms where lim n lanbn0 andbn diverges, butan converges. Compare with Exercise 40.
Showthatifan 0andlimnl nan 0, thenan is divergent.
Show that if an0 and an is convergent, then ln1an is convergent.
Ifan is a convergent series with positive terms, is it true thatsinanis also convergent?
Ifan andbn are both convergent series with positive terms, is it true thatan bn is also convergent?
n1 s1n2 n6 12
28.n1
1 n1 n!
30.
n! n1 nn
sin 32.
11
31.
n1 n n1 n
11n
710
CHAPTER 11 INFINITE SEQUENCES AND SERIES
11.5
ALTERNATING SERIES
The convergence tests that we have looked at so far apply only to series with positive terms. In this section and the next we learn how to deal with series whose terms are not necessarily positive. Of particular importance are alternating series, whose terms alternate in sign.
An alternating series is a series whose terms are alternately positive and negative. Here are two examples:
111111 1n1 2 3 4 5 6 n1 n
123456n1n
234567 n1 n1
We see from these examples that the nth term of an alternating series is of the form
an1n1bn or an1nbn where bn is a positive number. In fact, bnan .
The following test says that if the terms of an alternating series decrease toward 0 in absolute value, then the series converges.
THE ALTERNATING SERIES TEST If the alternating series
1n1bn b1 b2 b3 b4 b5 b6n1
bn 0
satisfies
then the series is convergent.
i ii
bn1bn
lim bn0 nl
for all n
Before giving the proof lets look at Figure 1, which gives a picture of the idea behind the proof. We first plot s1b1 on a number line. To find s2 we subtract b2, so s2 is to the leftofs1.Thentofinds3 weaddb3,sos3 istotherightofs2.But,sinceb3 b2,s3 isto the left of s1. Continuing in this manner, we see that the partial sums oscillate back and forth. Since bn l 0, the successive steps are becoming smaller and smaller. The even par tial sums s2, s4, s6, . . . are increasing and the odd partial sums s1, s3, s5, . . . are decreasing. Thus it seems plausible that both are converging to some number s, which is the sum of the series. Therefore we consider the even and odd partial sums separately in the follow ing proof.
b
FIGURE1
0 sTM s s s s s s
b
bTM
b
b
b
N Figure 2 illustrates Example 1 by showing the graphs of the terms an1n1n and the partial sums sn. Notice how the values of sn zigzag across the limiting value, which appears to be about 0.7. In fact, it can be proved that the exact sum of the series is ln 20.693 see Exercise 36.
V EXAMPLE 1 The alternating harmonic series
PROOF OF THE ALTERNATING SERIES TEST We first consider the even partial sums:
s2 b1 b2 0
s4 s2 b3 b4s2
In general s2ns2n2b2n1b2ns2n2
Thus
But we can also write
s2n b1 b2 b3b4 b5b2n2 b2n1b2n
Every term in brackets is positive, so s2 nb1 for all n. Therefore the sequence s2 nof even partial sums is increasing and bounded above. It is therefore convergent by the Monotonic Sequence Theorem. Lets call its limit s, that is,
lim s2ns nl
Now we compute the limit of the odd partial sums:
lim s2n1lim s2nb2n1 nl nl
lim s2nlim b2n1 nl nl
s0 s
by condition ii
SECTION 11.5 ALTERNATING SERIES
711
since b2nb2n1 0s2 s4 s6 s2n
Since both the even and odd partial sums converge to s, we have lim n lsns
see Exercise 80a in Section 11.1 and so the series is convergent. M
since b2 b1 sinceb4 b3
1 1
1n1
1
2 3 4 n1 n
1
i bn1bn because ii lim bnlim 10
1
satisfies
11 n1 n
sn
an
nl nl n
so the series is convergent by the Alternating Series Test.
V EXAMPLE 2 The series1n 3n is alternating but n1 4n1
M
0n
FIGURE 2
lim bnlim 3n nl nl 4n1
lim 3 nl 41
3 4
n
712
CHAPTER 11 INFINITE SEQUENCES AND SERIES
so condition ii is not satisfied. Instead, we look at the limit of the nth term of the series: lim anlim 1n 3n
nl nl 4n1
This limit does not exist, so the series diverges by the Test for Divergence. M
n2
EXAMPLE 3 Test the series1n1 for convergence or divergence.
SOLUTION The given series is alternating so we try to verify conditions i and ii of the Alternating Series Test.
Unlike the situation in Example 1, it is not obvious that the sequence given by bnn2n31 is decreasing. However, if we consider the related function
fxx2x31, we find that
fx x2x3
x3 12
Since we are considering only positive x, we see that fx0 if 2×30, that is,
xs3 2. Thus f is decreasing on the interval s3 2, . This means that fn1fn and therefore bn1bn when n2. The inequality b2b1 can be verified directly but all that really matters is that the sequence bnis eventually decreasing.
n1 n31
N Instead of verifying condition i of the Alter nating Series Test by computing a derivative, we could verify that bn1bn directly by using the technique of Solution 1 of Example 12 in Section 11.1.
Condition ii is readily verified:
1
n2 n lim
n3
Thus the given series is convergent by the Alternating Series Test. M
ESTIMATING SUMS
A partial sum sn of any convergent series can be used as an approximation to the total sum s, but this is not of much use unless we can estimate the accuracy of the approximation. The error involved in using ssn is the remainder Rnssn. The next theorem says that for series that satisfy the conditions of the Alternating Series Test, the size of the error is smaller than bn1, which is the absolute value of the first neglected term.
limbn lim 3
nl nln1 nl11
0
ALTERNATING SERIES ESTIMATION THEOREM If s 1n1bn is the sum of an alternating series that satisfies
i 0bn1 bn and ii limbn 0 nl
then Rnssnbn1
N You can see geometrically why the
Alternating Series Estimation Theorem is true
by looking at Figure 1 on page 710. Notice that ss4b5,ss5b6,andsoon.Notice also that s lies between any two consecutive partial sums.
PROOF We know from the proof of the Alternating Series Test that s lies between any two consecutive partial sums sn and sn1. It follows that
ssnsn1 snbn1 M
1n n0 n!
SOLUTION We first observe that the series is convergent by the Alternating Series Test because
i 11 1 n1! n!n1 n!
ii 011l0 so 1l0asnl n!n n!
To get a feel for how many terms we need to use in our approximation, lets write out the first few terms of the series:
V EXAMPLE 4 Find the sum of the series By definition, 0!1.
correct to three decimal places.
SECTION 11.5 ALTERNATING SERIES713
Noticethat
s11111111 0! 1! 2! 3! 4! 5! 6! 7!
11111111 2 6 24 120 720 5040
b711 0.0002 5040 5000
s611111 1 1 0.368056 2 6 24 120 720
and
By the Alternating Series Estimation Theorem we know that
N In Section 11.10 we will prove that
exn0 xnn! for all x, so what we have obtained in Example 4 is actually an approxi mation to the number e1.
11.5 EXERCISES
1. a What is an alternating series?
b Under what conditions does an alternating series converge? c If these conditions are satisfied, what can you say about the
remainder after n terms?
220 Test the series for convergence or divergence.
2. 1 2 3 4 534567
4444 47 8 9 10 11
4. 11111s2 s3 s4 s5 s6
9.
15.
nsn 1 n 10. 1 n
ss6b7 0.0002
This error of less than 0.0002 does not affect the third decimal place, so we have
s0.368 correct to three decimal places. M
NOTE The rule that the error in using sn to approximate s is smaller than the first neglected term is, in general, valid only for alternating series that satisfy the conditions of the Alternating Series Estimation Theorem. The rule does not apply to other types of series.
n1 10n
n2
1n1
n1 n34
n1 12sn
11.
1n
n2 lnn
12.
e1n n1 n
1n1
nlnn
14.1n1
n1 n
sinn 2 16.
n1 n!
1nsin n 18. 1ncos n
13.
3.
cosn34
n1 n
17.
1n1 5.
1 n1 n1 lnn4
n1
n1 2n1
6.
n1
nn nn
3n1n19.
1n 20.
1n 8.1n
n1 2n1 n1 sn 2
n1 n! n1 5
7.
3
714CHAPTER 11 INFINITE SEQUENCES AND SERIES
; 2122 Calculate the first 10 partial sums of the series and graph both the sequence of terms and the sequence of partial sums on the same screen. Estimate the error in using the 10th partial sum to approximate the total sum.
1n1 22.3
31. Is the 50th partial sum s50 of the alternating series
n1 1n1n an overestimate or an underestimate of the total sum? Explain.
3234 For what values of p is each series convergent?n1
1n1 21. 32
1 n1 np
n1 n
2326 Show that the series is convergent. How many terms of
the series do we need to add in order to find the sum to the indi cated accuracy?
n1 n
n1
29.
n1
n 1n1 n2
n
h2n ln2nl
Use these facts together with part a to show that
N We have convergence tests for series with positive terms and for alternating series. But what if the signs of the terms switch back and forth irregularly? We will see in Example 3 that the idea of absolute convergence sometimes helps in such cases.
1n n 28.n
n1 8
1n 30.n
n1 3n!
hn lnnl and therefore
as nl as nl
1n 33.
34.
1n1
ln np n
n1 np
Showthattheseries1n1bn,wherebn 1nifnisodd
and bn1n2 if n is even, is divergent. Why does the Alter nating Series Test not apply?
Use the following steps to show that
n1
nating harmonic series.
a Show that s2nh2nhn.
b From Exercise 40 in Section 11.3 we have
n2
n1
n1
1n1
6
n
error0.00005 error 0.0001 error 0.000005
35.
36.
24. 25. 26.
decimal places.
1n1 27. 5
ln 2
Let hn and sn be the partial sums of the harmonic and alter
n
1n
n 5
1n
1 n1 n
n0 10nn!
1n1nen n1
error 0.01
2730 Approximate the sum of the series correct to four
10
s2n l ln 2 as n l .
11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS
Given any seriesan, we can consider the corresponding series
an a1 a2 a3n1
whose terms are the absolute values of the terms of the original series.
Notice that ifan is a series with positive terms, then an an and so absolute con vergence is the same as convergence in this case.
EXAMPLE 1 The series
32.
23.
DEFINITION A seriesan is called absolutely convergent if the series of absolute valuesanis convergent.
1
1n1 1 1 1
n1 n2 122 32 42
SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS715
is absolutely convergent because
1n1 11111 n1 n2 n1 n2 22 32 42
is a convergent pseriesp2.
EXAMPLE 2 We know that the alternating harmonic series
1n1 1 1 1
n1 n 1234
is convergent see Example 1 in Section 11.5, but it is not absolutely convergent because the corresponding series of absolute values is
1n1 11111 n1 n n1n 234
M
N Figure 1 shows the graphs of the terms an and partial sums sn of the series in Example 3. Notice that the series is not alternating but has positive and negative terms.
which is the harmonic seriespseries with p1 and is therefore divergent. M
Example 2 shows that the alternating harmonic series is conditionally convergent. Thus it is possible for a series to be convergent but not absolutely convergent. However, the next theorem shows that absolute convergence implies convergence.
PROOF Observe that the inequality
0an an2an
is true because anis either an or an. Ifan is absolutely convergent, thenanis convergent, so2anis convergent. Therefore, by the Comparison Test,ananis convergent. Then
an an anan
is the difference of two convergent series and is therefore convergent. M
V EXAMPLE 3 Determine whether the series
cosncos1cos2cos3
DEFINITION A seriesan is called conditionally convergent if it is conver gent but not absolutely convergent.
2
THEOREM If a seriesan is absolutely convergent, then it is convergent.
3
0.5
n1 is convergent or divergent.
n2 12 22 32
sn
an
SOLUTION This series has both positive and negative terms, but it is not alternating.
The first term is positive, the next three are negative, and the following three are posi tive: The signs change irregularly. We can apply the Comparison Test to the series of absolute values
cosn cosn n1 n2 n1 n2
0n
FIGURE 1
716
CHAPTER 11 INFINITE SEQUENCES AND SERIES
Since cos n 1 for all n, we have
cosn 1
n2 n2
We know that1n2 is convergentpseries with p2 and thereforecos n n2 is convergent by the Comparison Test. Thus the given seriescos nn2 is absolutely convergent and therefore convergent by Theorem 3. M
The following test is very useful in determining whether a given series is absolutely convergent.
THE RATIO TEST
an1nl an
L1, then the series an is absolutely convergent
i If lim
and therefore convergent.
n1
,thentheseries
an1nl an
an1nl an
n1
ii If lim
is divergent.
L1or lim
an
iii If lim an11, the Ratio Test is inconclusive; that is, no conclusion can be nl an
drawn about the convergence or divergence ofan.
PROOF
i The idea is to compare the given series with a convergent geometric series. Since L1, we can choose a number r such that Lr1. Since
liman1 L and Lr nl an
the ratio an1anwill eventually be less than r; that is, there exists an integer N such that
or, equivalently,
an1r whenever nN an
an1an r whenever nN
4
Putting n successively equal to N, N1, N2, . . . in 4, we obtain aN1 aN r
and, in general,
aN2aN1 raN r2 aN3aN2 raN r3
aNkaN rk for all k1
5
SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS717
Now the series
aNrk aNraNr2 aNr3
k1
is convergent because it is a geometric series with 0r1. So the inequality 5, together with the Comparison Test, shows that the series
an aNkaN1aN2aN3
nN1 k1
is also convergent. It follows that the series n1 anis convergent. Recall that a finite number of terms doesnt affect convergence. Thereforean is absolutely convergent.
ii If an1an l L1 or an1an l , then the ratio an1anwill eventually be greater than 1; that is, there exists an integer N such that
an11 whenever nN an
This means that an1 anwhenever nN and so lim an0
nl
Thereforean diverges by the Test for Divergence. M
NOTE Part iii of the Ratio Test says that if limn lan1an 1, the test gives no information. For instance, for the convergent series1n2 we have
as n l
as nl
1
an1 n12 n2 1
2 2 l 1 an 1 n1 11
n2 n whereas for the divergent series1n we have
1
an1n1n1 l1 an 1n111 nn
Therefore, if limn lan1an 1, the seriesan might converge or it might diverge. In this case the Ratio Test fails and we must use some other test.
EXAMPLE 4 Test the series
SOLUTION We use the Ratio Test with an1n n 33n:
n1
1n
n 3 3n
for absolute convergence.
N ESTIMATING SUMS
In the last three sections we used various meth
ods for estimating the sum of a seriesthe
method depended on which test was used to
prove convergence. What about series for which
the Ratio Test works? There are two possibilities:
If the series happens to be an alternating series,
as in Example 4, then it is best to use the meth
ods of Section 11.5. If the terms are all positive,
then use the special methods explained inExercise 34.
1n1n13
3n1 n13 3n
an1
an 1n 3 n
n 3n13 3n
1 n13 3 n
113 1
3 n
l
1 3
1
718
CHAPTER 11 INFINITE SEQUENCES AND SERIES
Thus, by the Ratio Test, the given series is absolutely convergent and therefore convergent.
M
nn
n1 n!
SOLUTION Since the terms annnn! are positive, we dont need the absolute value signs.
V EXAMPLE 5 Test the convergence of the series
.
an1 n1n1 n! n1n1n n! nn
an n1! n n1n! n n1n1n
n 1nle asnl
See Equation 3.6.6. Since e1, the given series is divergent by the Ratio Test. M
NOTE Although the Ratio Test works in Example 5, an easier method is to use the Test for Divergence. Since
nn nnnn
ann!123nn
it follows that an does not approach 0 as n l . Therefore the given series is divergent by the Test for Divergence.
The following test is convenient to apply when nth powers occur. Its proof is similar to the proof of the Ratio Test and is left as Exercise 37.
THE ROOT TEST
i If lim sn an L1, then the seriesan is absolutely convergent
nl
and therefore convergent.
n1
iiIflimsn anL1orlimsn an,thentheseriesanisdivergent.
nl nl
iii If lim sn an 1, the Root Test is inconclusive. nl
n1
If limn lsn an 1, then part iii of the Root Test says that the test gives no infor mation. The seriesan could converge or diverge. If L1 in the Ratio Test, dont try the Root Test because L will again be 1. And if L1 in the Root Test, dont try the Ratio Test because it will fail too.
2n3n V EXAMPLE 6 Test the convergence of the series . n1 3n2
SOLUTION
an 2n3n 3n2
23
sna n 2 n3n l 21
3n2 32 3 n
Thus the given series converges by the Root Test.
M
6
N Adding these zeros does not affect the sum of the series; each term in the sequence of partial sums is repeated, but the limit is the same.
then any rearrangement ofan has the same sum s.
However, any conditionally convergent series can be rearranged to give a different sum. To
illustrate this fact lets consider the alternating harmonic series
11111111ln 2 2345678
See Exercise 36 in Section 11.5. If we multiply this series by 1 , we get 2
1 1 1 1 1ln2 2468 2
Inserting zeros between the terms of this series, we have
01 01 01 01 1 ln2 24682
Now we add the series in Equations 6 and 7 using Theorem 11.2.8: 11 1 1 1 1 3 ln2
SECTION 11.6 ABSOLUTE CONVERGENCE AND THE RATIO AND ROOT TESTS719 REARRANGEMENTS
The question of whether a given convergent series is absolutely convergent or condi tionally convergent has a bearing on the question of whether infinite sums behave like finite sums.
If we rearrange the order of the terms in a finite sum, then of course the value of the sum remains unchanged. But this is not always the case for an infinite series. By a rearrangement of an infinite seriesan we mean a series obtained by simply changing the order of the terms. For instance, a rearrangement ofan could start as follows:
It turns out that
a1 a2 a5 a3 a4 a15 a6 a7 a20ifan is an absolutely convergent series with sum s,
7
8
32574 2
Notice that the series in 8 contains the same terms as in 6, but rearranged so that one negative term occurs after each pair of positive terms. The sums of these series, however, are different. In fact, Riemann proved that
ifan is a conditionally convergent series and r is any real number what soever, then there is a rearrangement ofan that has a sum equal to r.
A proof of this fact is outlined in Exercise 40.
11.6 EXERCISES
1. What can you say about the seriesan in each of the following
cases? 5.
1n1 4
1n 6. 4
a lim an18 nl an
c liman1 1 nl an
b lim an10.8
n1 sn
n1 nn!
nl
an
7.
9.
k2k 3
k1
8.
n
10. 1n n
n1 sn32
sin 4n 12. n
n1 100
n
228 Determine whether the series is absolutely convergent, conditionally convergent, or divergent.
1n 1.1 n1 n4
n2 2.
1n e1n 11. 3
n1 2n
n1 n
10n n22n
10n 3.
n0 n!
1n1
n1
2n n4
n1 n142n1 n1 n!
13.
n1 4
14.1n1
4.
720
CHAPTER 11 INFINITE SEQUENCES AND SERIES
15. 17.
23. 25.
26. 27.
28.
30.
n1
n2
n1
n1
n1
1n arctan n n2
1n ln n
cosn 3 n!
n21 n 2
16.
18.
20.
22.
3cosn n1 n232
Ratio Test. As usual, we let Rn be the remainder after n terms,
thatis, Rn an1 an2 an3
a If rnis a decreasing sequence and rn11, show, by summing a geometric series, that
Rnan1
1rn1
n! n
n1 n
2n
19.
n1 nn
2n5n
b If rnis an increasing sequence, show that a n1
Rn1L
a Find the partial sum s5 of the seriesn
21.
2n 1
n2 n1 n
n
1 n2
n2 ln n
35.
36.
37.
38.
n1 1n2 . Use Exer cise 34 to estimate the error in using s5 as an approximation
1
13 135 1357
n
24.
13! 5!7!n1 1352n1
to the sum of the series.
b Find a value of n so that sn is within 0.00005 of the sum.
Use this value of n to approximate the sum of the series. Use the sum of the first 10 terms to approximate the sum of
1 2n1! 2262610261014
the series
Use Exercise 34 to estimate the error.
Prove the Root Test. Hint for part i: Take any number r such that Lr1 and use the fact that there is an integer N such thatsn anrwhenevernN.
Around 1910, the Indian mathematician Srinivasa Ramanujan discovered the formula
12s24n!110326390n 9801 n0 n!4 3964n
William Gosper used this series in 1985 to compute the first 17 million digits of .
a Verify that the series is convergent.
b How many correct decimal places of do you get if you
use just the first term of the series? What if you use two terms?
Given any seriesan, we define a seriesan whose terms are all the positive terms ofan and a seriesan whose terms are all the negative terms ofan. To be specific, we let
anan an anan an 22
Noticethatifan 0,thenan an andan 0,whereasif an 0,thenan an andan 0.
a Ifan is absolutely convergent, show that both of the series
an andan are convergent.
b Ifan is conditionally convergent, show that both of the
seriesan andan are divergent.
Prove that ifan is a conditionally convergent series and
r is any real number, then there is a rearrangement ofan whose sum is r. Hints: Use the notation of Exercise 39.
Take just enough positive terms an so that their sum is greater than r. Then add just enough negative terms an so that the cumulative sum is less than r. Continue in this manner and use Theorem 11.2.6.
5 58 58112462n
n1
n1
Determine whetheran converges or diverges. A seriesan is defined by the equations
a1 1 an12cosn an sn
Determine whetheran converges or diverges.
For which of the following series is the Ratio Test inconclusive that is, it fails to give a definite answer?
n n1 2n
581114
5n1
a1 2 an14n3 an
n!
58113n2
2 n n !
1 n
The terms of a series are defined recursively by the equations
29.
31.
32.
34.
40.
1
n
a
n1 n
3n1 c
n1 sn
For which positive integers k is the following series
convergent?
n!2 n1 k n!
a Show that n0 xnn! converges for all x. b Deduce that limnl xnn!0 for all x.
Letan be a series with positive terms and let rnan1 an. Suppose that limnl rnL1, so an converges by the
n1
sn 1n
39.
3
b
d
2
n
n1
2
33.
SECTION 11.7 STRATEGY FOR TESTING SERIES721
11.7 STRATEGY FOR TESTING SERIES
We now have several ways of testing a series for convergence or divergence; the problem is to decide which test to use on which series. In this respect, testing series is similar to integrating functions. Again there are no hard and fast rules about which test to apply to a given series, but you may find the following advice of some use.
It is not wise to apply a list of the tests in a specific order until one finally works. That would be a waste of time and effort. Instead, as with integration, the main strategy is to classify the series according to its form.
1. If the series is of the form1n p, it is a pseries, which we know to be convergent if p1anddivergentif p1.
2. If the series has the formar n1 orar n, it is a geometric series, which converges if r 1 and diverges if r 1. Some preliminary algebraic manipulation may be required to bring the series into this form.
3. If the series has a form that is similar to a pseries or a geometric series, then
one of the comparison tests should be considered. In particular, if an is a rational function or an algebraic function of n involving roots of polynomials, then the series should be compared with a pseries. Notice that most of the series in Exer cises 11.4 have this form. The value of p should be chosen as in Section 11.4 by keeping only the highest powers of n in the numerator and denominator. The com parison tests apply only to series with positive terms, but ifan has some negative terms, then we can apply the Comparison Test toanand test for absolute convergence.
4. If you can see at a glance that limnl an0, then the Test for Divergence should be used.
5. If the series is of the form1n1bn or1nbn , then the Alternating Series Test is an obvious possibility.
6. Series that involve factorials or other products including a constant raised to the nth power are often conveniently tested using the Ratio Test. Bear in mind that an1an l 1 as n lfor all pseries and therefore all rational or algebraic functions of n. Thus the Ratio Test should not be used for such series.
7. If an is of the form bn n, then the Root Test may be useful.
8. If anf n, where x f x dx is easily evaluated, then the Integral Test is effective
1
assuming the hypotheses of this test are satisfied.
In the following examples we dont work out all the details but simply indicate which tests should be used.
VEXAMPLE1 n1 n1 2n1
Sincean l1 0asnl,weshouldusetheTestforDivergence. M 2
E X A M P L E 2s n 31 n1 3n3 4n2 2
Since an is an algebraic function of n, we compare the given series with a pseries. The
722
CHAPTER 11 INFINITE SEQUENCES AND SERIES
7.
k1 k2! 10. n2en3
k1
k ln k k1
1n cosh n
comparison series for the Limit Comparison Test isbn, where sn3 n32 1
bn3n33n33n32 M Since the integral x xex2 dx is easily evaluated, we use the Integral Test. The Ratio Test
V EXAMPLE 3nen2
also works.
M
n1
1
n1
Since the series is alternating, we use the Alternating Series Test. M
V EXAMPLE 52k k1 k!
Since the series involves k!, we use the Ratio Test. M E X A M P L E 61
EXAMPLE 4
1n
n 3 n41
Test.
11.7 EXERCISES
138 Test the series for convergence or divergence.
M
n1 23n
Since the series is closely related to the geometric series13n, we use the Comparison
1.
3.
5.
n1 n31n
n
n1 n 4.1n
n1 n
tan1n
1
2n1n 2. 2n
22n 21. n
sn2 1 22. 32
n1n2n5 nn
n1
n2 n1
n1
n!
n2 2n1 n
n21
6. 5 n1
25.n1 e
2
n
26.
2 n2
1 2n1
2kk! 8.
23.
24.n sin1n n1
n
30. 1j
j1 j5
n1
1
n1 5e1n
2
nsln n
27.
29.
28.
n2
9. k2ek k1
11.n2
3
n1 n
sj
1 n1 n ln n
12.
n1
sin n n1
n1
5k
n2
31.32. kk
n1
n! n
4n
13.3 n n1 n!
15. 17. 19.
14. sin2n
k1 3 4 33.sin1n
n
n!
n1 12n21
34.1
n1 nncos2n
n
n0 2583n2
16.3
n1 n 1
n1 sn
nn 1
1n 21n n1
1 n1 18.
2
35.n1 36. lnnlnn
n1 lnnk5n
n2 sn1
n2
1n 20. 37. sn 21 38. sn 21 k
n1 sn k1 5 n1 n1
N TRIGONOMETRIC SERIES
A power series is a series in which each term is a power function. A trigonometric series
an cosnxbn sinnx
n0
is a series whose terms are trigonometric func tions. This type of series is discussed on the website
www.stewartcalculus.com
Click on Additional Topics and then on Fourier Series.
11.8
POWER SERIES
A power series is a series of the form
cnxn c0 c1xc2x2 c3x3n0
where x is a variable and the cns are constants called the coefficients of the series. For each fixed x, the series 1 is a series of constants that we can test for convergence or divergence. A power series may converge for some values of x and diverge for other values of x. The sum of the series is a function
fxc0 c1xc2x2 cnxn
whose domain is the set of all x for which the series converges. Notice that f resembles a polynomial. The only difference is that f has infinitely many terms.
For instance, if we take cn1 for all n, the power series becomes the geometric series
xn 1xx2 xnn0
which converges when 1×1 and diverges when x 1 see Equation 11.2.5. More generally, a series of the form
cnxan c0 c1xac2xa2
n0
is called a power series in xa or a power series centered at a or a power series about a. Notice that in writing out the term corresponding to n0 in Equations 1 and 2 we have adopted the convention that xa01 even when xa. Notice also that when xa, all of the terms are 0 for n1 and so the power series 2 always converges when xa.
V EXAMPLE 1 For what values of x is the seriesn! x n convergent? n0
SOLUTION We use the Ratio Test. If we let an, as usual, denote the nth term of the series, thenan n!xn.Ifx0,wehave
lim an1lim n1!xn1lim n1xnl an nl n!xn nl
By the Ratio Test, the series diverges when x0. Thus the given series converges only
1
SECTION 11.8 POWER SERIES
723
2
N Notice that
n1!n1nn1. . .321
n1n!
when x0 .
V EXAMPLE 2 For what values of x does the series
SOLUTION Let anx3nn. Then
an1x3n1 n n
M
x3n n1 n
converge?
1 x3l x3
an n1 x3
11 n
as nl
724
CHAPTER 11 INFINITE SEQUENCES AND SERIES
N Notice how closely the computergenerated model which involves Bessel functions and cosine functions matches the photograph of a vibrating rubber membrane.
an1an
1n1x2n1 22nn!222n1n1!21nx2n
x2n2 22nn!222n2n12n!2x2n
x2
4n12 l 01 forallx
By the Ratio Test, the given series is absolutely convergent, and therefore convergent, when x3 1 and divergent when x3 1. Now
x31 ? 1×31 ? 2×4
so the series converges when 2×4 and diverges when x2 or x4.
The Ratio Test gives no information when x3 1 so we must consider x2
and x4 separately. If we put x4 in the series, it becomes1n, the harmonic series, which is divergent. If x2, the series is1nn, which converges by the Alternating Series Test. Thus the given power series converges for 2×4. M
We will see that the main use of a power series is that it provides a way to represent some of the most important functions that arise in mathematics, physics, and chemistry. In particular, the sum of the power series in the next example is called a Bessel function, after the German astronomer Friedrich Bessel 17841846, and the function given in Exer cise 35 is another example of a Bessel function. In fact, these functions first arose when Bessel solved Keplers equation for describing planetary motion. Since that time, these functions have been applied in many different physical situations, including the tempera ture distribution in a circular plate and the shape of a vibrating drumhead.
EXAMPLE 3 Find the domain of the Bessel function of order 0 defined by J0x 1nx2n
n0 22nn!2 SOLUTION Let an1nx2n22nn!2. Then
National Film Board of Canada
Thus, by the Ratio Test, the given series converges for all values of x. In other words, the domain of the Bessel function J0 is , . M
Recall that the sum of a series is equal to the limit of the sequence of partial sums. So when we define the Bessel function in Example 3 as the sum of a series we mean that, for every real number x,
snxn 1i x2i i0 22ii!2
x2 x4 s2x1 464
x2 x4 x6
s3x1 4642304 s4x1 4642304147,456
J0xlim snx where nl
The first few partial sums are
x2 s0x1 s1x1 4
x2 x4 x6 x8
FIGURE 1
y
1
s 01x
Js s
Figure 1 shows the graphs of these partial sums, which are polynomials. They are all approximations to the function J0, but notice that the approximations become better when more terms are included. Figure 2 shows a more complete graph of the Bessel function.
For the power series that we have looked at so far, the set of values of x for which the series is convergent has always turned out to be an interval a finite interval for the geometric series and the series in Example 2, the infinite interval ,in Example 3, and a collapsed interval 0, 00 in Example 1. The following theorem, proved in Appendix F, says that this is true in general.
Partial sums of the Bessel function Jy
1
yJ x 10 10
The number R in case iii is called the radius of convergence of the power series. By convention, the radius of convergence is R0 in case i and R in case ii. The interval of convergence of a power series is the interval that consists of all values of x for which the series converges. In case i the interval consists of just a single point a. In case ii the interval is , . In case iii note that the inequality xa R can be rewrit ten as aRxaR. When x is an endpoint of the interval, that is, xaR, anything can happenthe series might converge at one or both endpoints or it might diverge at both endpoints. Thus in case iii there are four possibilities for the interval of convergence:
aR, aR aR, aR aR, aR aR, aR The situation is illustrated in Figure 3.
convergence for xaR
aR a aR
divergence for xaR
We summarize here the radius and interval of convergence for each of the examples already considered in this section.
SECTION 11.8 POWER SERIES725
sTM s
THEOREM For a given power seriescnxan, there are only three
n0
iii There is a positive number R such that the series converges if xa R and diverges if xa R.
3
possibilities:
i The series converges only when xa.
ii The series converges for all x.
FIGURE 2
0x
FIGURE 3
Series
Radius of convergence
Interval of convergence
Geometric series Example 1 Example 2 Example 3
xn n0
n! x n n0
x3n
n1 n
1n x 2 n
2n 2 n0 2 n!
R1 R0 R1 R
1, 1 0 2, 4 ,
726
CHAPTER 11 INFINITE SEQUENCES AND SERIES
In general, the Ratio Test or sometimes the Root Test should be used to determine the radius of convergence R. The Ratio and Root Tests always fail when x is an endpoint of the interval of convergence, so the endpoints must be checked with some other test.
EXAMPLE 4 Find the radius of convergence and interval of convergence of the series3n x n
n0 sn1 SOLUTION Let an3nxnsn1 . Then
an13n1xn1 sn1 n1 nn3x
an sn2 3 x n2 311n xl 3x as nl
12n
By the Ratio Test, the given series converges if 3 x 1 and diverges if 3 x 1.
Thus it converges if x 1 and diverges if x 1 . This means that the radius of con 33
vergence is R1 . 3
We know the series converges in the interval1 , 1 , but we must now test for con 33
vergence at the endpoints of this interval. If x1 , the series becomes 3
3n1n1 1 1 1 1
3
n0 sn1 n0 sn1 s1 s2 s3 s4
which diverges. Use the Integral Test or simply observe that it is a pseries with
p1 1.Ifx1,theseriesis 23
3n 1 n1n
3 n0 sn1 n0
sn1
which converges by the Alternating Series Test. Therefore the given power series con
verges when 1×1, so the interval of convergence is 1, 1 . M 3333
V EXAMPLE 5 Find the radius of convergence and interval of convergence of the seriesnx2n
n0 3n1 SOLUTION If annx2n3n1, then
an1n1x2n1 3n1 n2n
an 3 nx2 11x2l x2 asnl
n33
Using the Ratio Test, we see that the series converges if x2 31 and it diverges if x231. So it converges if x2 3 and diverges if x2 3. Thus the radius of convergence is R3.
The inequality x2 3 can be written as 5×1, so we test the series at the endpoints 5 and 1. When x5, the series is
n3n
1 1nn
3
SECTION 11.8 POWER SERIES727
n1
which also diverges by the Test for Divergence. Thus the series converges only when
5×1, so the interval of convergence is 5, 1. M
n0 3
which diverges by the Test for Divergence 1nn doesnt converge to 0. When x1,
the series is
n3n1n
n0
n1 3
n0 3 n0
11.8 EXERCISES
1. What is a power series?
2. a What is the radius of convergence of a power series? How do you find it?
b What is the interval of convergence of a power series? How do you find it?
n!2x1n
n1
n2xn
23.
24.
4x1n
n1 2462nx2n
25.
n1 n
2
26.
n2 nlnn
2
27.
4.n!xn
328 Find the radius of convergence and interval of convergence of the series.
xn
n1 1352n1
xn1nxn
3.
n1 sn
1n1 x n
n0 n1
6. snxn n1
8. nnxn n1
28.
30.
5.
9.
11. 13.
17. n1
3
n1 1352n1c4n
7.
n1 nxn
n0 n!
If n0 n is convergent, does it follow that the following series are convergent?
ac 2n b n
29.
n2 xn n1 2n
n1
10nxn n3
n0 Suppose that
n n0 cn x
n0
converges when x4 and diverges
1n
2n x n
10.
c 4n n
12.n1
when x6. What can be said about the convergence or diver gence of the following series?
xn n5n
4 n1 sn
5n acn bcn8 n0 n0
xn x2n
1 n 14.1n nnn
n
n2 4lnn n0 2n!
nnx2 16.1n x3
2
n0n1 n0 2n1
ccn3 d1 cn 9 n0 n0
31. If k is a positive integer, find the radius of convergence of the series
15.
3nx4n sn
n
18. x1
kn! xn
n
n0
n !k
n n 19. x2 20. 3×2
32. Let p and q be real numbers with pq. Find a power series whose interval of convergence is
a p, q b p, q
c p, q d p, q
33. Isitpossibletofindapowerserieswhoseintervalof convergence is 0, ? Explain.
n
n1 n n1 n3
21.n xan, b0 22.nx4n n1 bn n1 n31
n1 4
n
n
728CHAPTER 11 INFINITE SEQUENCES AND SERIES
; 34.
35.
Graph the first several partial sums snx of the series n0 xn, CAS together with the sum function f x11x, on a com
mon screen. On what interval do these partial sums appear to
be converging to f x?
c If your CAS has builtin Airy functions, graph A on the same screen as the partial sums in part b and observe how the partial sums approximate A.
A function f is defined by
fx12xx2 2×3 x4
that is, its coefficients are c2n1 and c2n12 for all n0. Find the interval of convergence of the series and find an explicit formula for f x.
38. If fxn0 cnxn,wherecn4 cn foralln0,findthe interval of convergence of the series and a formula for f x.
39. Show that if limnl sn cn c, where c0, then the radius of convergence of the power seriescn x n is R1c.
40. Suppose that the power seriescnxan satisfies cn0 for all n. Show that if limnl cncn1 exists, then it is equal to the radius of convergence of the power series.
41. Suppose the seriesc x n has radius of convergence 2 and the n
seriesdn x n has radius of convergence 3. What is the radius of convergence of the series cndnxn?
42. Suppose that the radius of convergence of the power seriescn x n is R. What is the radius of convergence of the power seriescn x2n?
The function J1 defined by J1x
n0
1nx2n1 n!n1!22n1
;
is called the Bessel function of order 1.
a Find its domain.
b Graph the first several partial sums on a common
screen.
CAS c If your CAS has builtin Bessel functions, graph J1 on the
;
is called the Airy function after the English mathematician and astronomer Sir George Airy 18011892.
a Find the domain of the Airy function.
b Graph the first several partial sums on a common screen.
36.
same screen as the partial sums in part b and observe how the partial sums approximate J1.
The function A defined by
x3 x6 x9
Ax123 2356 235689
11.9
REPRESENTATIONS OF FUNCTIONS AS POWER SERIES
In this section we learn how to represent certain types of functions as sums of power series by manipulating geometric series or by differentiating or integrating such a series. You might wonder why we would ever want to express a known function as a sum of infinitely many terms. We will see later that this strategy is useful for integrating functions that dont have elementary antiderivatives, for solving differential equations, and for approximating func tions by polynomials. Scientists do this to simplify the expressions they deal with; com puter scientists do this to represent functions on calculators and computers.
We start with an equation that we have seen before:
N A geometric illustration of Equation 1 is shown in Figure 1. Because the sum of a series is the limit of the sequence of partial sums, we have
1
1xx2 x3xn x1
1
lim snx 1x nl
We first encountered this equation in Example 5 in Section 11.2, where we obtained it by observing that it is a geometric series with a1 and rx. But here our point of view is different. We now regard Equation 1 as expressing the function f x11x as a sum of a power series.
snx1xx2 xn
37.
1
1x n0
where
is the nth partial sum. Notice that as n increases, snx becomes a better approxima tionto fxfor1x1.
FIGURE 1
1 and some partial sums 1x
s
y s
s f
sTM
1 0 1x
SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES729
V EXAMPLE 1 Express 11x 2as the sum of a power series and find the interval of convergence.
SOLUTION Replacing x by x2 in Equation 1, we have 1 1
2 2x2n 1x 1xn0
1nx2n 1×2 x4 x6 x8
n0
Because this is a geometric series, it converges when x21, that is, x21, or
x 1. Therefore the interval of convergence is 1, 1. Of course, we could have determined the radius of convergence by applying the Ratio Test, but that much work is unnecessary here. M
EXAMPLE 2 Find a power series representation for 1×2.
SOLUTION In order to put this function in the form of the left side of Equation 1 we first
factor a 2 from the denominator:
111
2x 21 x 21x
22
1 xn1n
n1xn
N Its legitimate to move x 3 across the sigma sign because it doesnt depend on n. Use Theorem 11.2.8i with cx 3.
x3 11n1n x3 x3n1 xn n1 xn3
2n0 2 n02
This series converges when x2 1, that is, x 2. So the interval of convergence
is 2, 2. M EXAMPLE 3 Find a power series representation of x3x2.
SOLUTION Since this function is just x3 times the function in Example 2, all we have to do is to multiply that series by x3:
x2x222
n0
1×3 1×4 1×51 x6
n0
2 4 8 16
Another way of writing this series is as follows:
3n1
x 1 xn
x2 n3 2n2
As in Example 2, the interval of convergence is 2, 2.
DIFFERENTIATION AND INTEGRATION OF POWER SERIES
M
The sum of a power series is a function f xn0 cnxan whose domain is the inter val of convergence of the series. We would like to be able to differentiate and integrate such functions, and the following theorem which we wont prove says that we can do so by differentiating or integrating each individual term in the series, just as we would for a polynomial. This is called termbyterm differentiation and integration.
730CHAPTER 11 INFINITE SEQUENCES AND SERIES
THEOREM If the power seriescnxan has radius of convergence R0, then the function f defined by
fxc0 c1xac2xa2cnxan
n0
is differentiable and therefore continuous on the interval aR, aR and
i fxc1 2c2xa3c3xa2ncnxan1
n1
xa3
c2 3
The radii of convergence of the power series in Equations i and ii are both R.
2
xa2 ii yfxdxCc0xac1 2
xan1 C cn
n0 n1
N Inpartii,xc0 dxc0xC1 iswrittenas c0xaC,whereCC1 ac0,soall the terms of the series have the same form.
NOTE 1 Equations i and ii in Theorem 2 can be rewritten in the form
iii
d dx n0
cnxancnxan d
n0 dx
iv y cnxan dx ycnxandx
n0 n0
We know that, for finite sums, the derivative of a sum is the sum of the derivatives and the integral of a sum is the sum of the integrals. Equations iii and iv assert that the same is true for infinite sums, provided we are dealing with power series. For other types of series of functions the situation is not as simple; see Exercise 36.
NOTE 2 Although Theorem 2 says that the radius of convergence remains the same when a power series is differentiated or integrated, this does not mean that the interval of convergence remains the same. It may happen that the original series converges at an end point, whereas the differentiated series diverges there. See Exercise 37.
NOTE 3 The idea of differentiating a power series term by term is the basis for a power ful method for solving differential equations. We will discuss this method in Chapter 17.
EXAMPLE 4 In Example 3 in Section 11.8 we saw that the Bessel function J0x 1nx2n
n0 22nn!2
is defined for all x. Thus, by Theorem 2, J0 is differentiable for all x and its derivative is
found by termbyterm differentiation as follows:
d 1nx2n1n 2nx2n1
J0x 2n 22n 2 M n0 dx 2 n! n1 2 n!
SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES731
V EXAMPLE 5 Express 11×2 as a power series by differentiating Equation 1. What is the radius of convergence?
SOLUTION Differentiating each side of the equation
1xx2 x3xn
1
1x n0
1
2 12x3x2nxn1
weget
If we wish, we can replace n by n1 and write the answer as
1x n1
1
2 n1xn
1x n0
According to Theorem 2, the radius of convergence of the differentiated series is the
same as the radius of convergence of the original series, namely, R1. M EXAMPLE 6 Find a power series representation for ln1x and its radius of
convergence.
SOLUTION We notice that, except for a factor of 1, the derivative of this function is
11x. So we integrate both sides of Equation 1: ln1xy 1 dxy1xx2 dx
1x
x2 x3xn1xn
xCCC x1 23 n0n1n1n
To determine the value of C we put x0 in this equation and obtain ln10C. Thus C0 and
x2 x3 xn
ln1xx x 1
23 n1n
The radius of convergence is the same as for the original series: R1. M
Notice what happens if we put x1 in the result of Example 6. Since ln 1ln 2, 22
we see that
ln211 111 2 8 24 64 n1 n2n
V EXAMPLE 7 Find a power series representation for f xtan1x.
SOLUTION Weobservethatfx11x2andfindtherequiredseriesbyintegrating
the power series for 11x 2found in Example 1.
tan1xy 1 dxy1x2 x4 x6 dx
1×2 Cx357
x3 x5 x7
732CHAPTER 11 INFINITE SEQUENCES AND SERIES
N The power series for tan1x obtained in Exam ple 7 is called Gregorys series after the Scottish mathematician James Gregory 16381675, who had anticipated some of Newtons discoveries. We have shown that Gregorys series is valid when 1×1, but it turns out although it isnt easy to prove that it is also valid when
x1. Notice that when x1 the series becomes
1111 4 357
This beautiful result is known as the Leibniz formula for .
To find C we put x0 and obtain Ctan1 00. Therefore
x3x5x7x2n11n
tan1xx
Since the radius of convergence of the series for 11x 2is 1, the radius of conver
gence of this series for tan1x is also 1. M EXAMPLE 8
a Evaluate x11x7dx as a power series.
b Use part a to approximate x0.5 11x7dx correct to within 107.
0
SOLUTION
a The first step is to express the integrand, 11×7, as the sum of a power series. As in Example 1, we start with Equation 1 and replace x by x7:
357 n0 2n1
N This example demonstrates one way in
which power series representations are useful. Integrating 11x 7by hand is incredibly dif ficult. Different computer algebra systems return different forms of the answer, but they are all extremely complicated. If you have a CAS, try
it yourself. The infinite series answer that we obtain in Example 8a is actually much easier to deal with than the finite answer provided by a CAS.
1nx7n 1×7 x14n0
Now we integrate term by term:
y
1x
7 dxy1nx7ndxC 1n n0 n0
x8 x15 x22 Cx
7n1
1 1
7 7x7n
1x 1xn0
1 x7n1
8 15 22 This series converges for x71, that is, for x 1.
b In applying the Fundamental Theorem of Calculus, it doesnt matter which anti derivative we use, so lets use the antiderivative from part a with C0:
0.5 1×8 x15 x22 12 y 7dxx
01x 815220 1111 1n
282815215222227n127n1
This infinite series is the exact value of the definite integral, but since it is an alternating series, we can approximate the sum using the Alternating Series Estimation Theorem.
If we stop adding after the term with n3, the error is smaller than the term with
n4:
16.41011 29229
So we have
y0.5 1dx11110.49951374 M 0 1×7 2 828 15215 22222
11.9 EXERCISES
1. If the radius of convergence of the power series n0 cn xn is 10, what is the radius of convergence of the series
n1 ncnxn1?Why?
2. Suppose you know that the series n0 bnxn converges for
x 2. What can you say about the following series? Why?
b n x n1 n0 n1
310 Find a power series representation for the function and determine the interval of convergence.
1518 Find a power series representation for the function and determine the radius of convergence.
SECTION 11.9 REPRESENTATIONS OF FUNCTIONS AS POWER SERIES733
5.
11. fx
3
x2 x2
12. fx
x2 2×2 x1
8.
17.
fxln5x x 3
f xx22
x2
16. fx12×2
18. f xarctanx3
3. fx 1 1x
2 fx 3x
7. fx x
9×2
9.fx1x 1x
3
19. fx x
x2 16
fxln1x 1x
20. fxlnx2 4 22. fxtan12x
4. fx
6. fx x10
fx x 2×21
10. fx x2
a3 x3
; 1922 Find a power series representation for f, and graph f and several partial sums snx on the same screen. What happens as n increases?
1×4 1
21.
2326 Evaluate the indefinite integral as a power series. What is the radius of convergence?
ln1t 24. y t dt
26. ytan1x2dx
2730 Use a power series to approximate the definite integral to
t y1t8 dt
six decimal places.
27. 29.
31. 32.
33.
23.
1112 Express the function as the sum of a power series by first using partial fractions. Find the interval of convergence.
25. yxtan1xdx x3
y0.2 1
0 1×5
28. y0.4 ln1x4dx 0
13.
a Use differentiation to find a power series representation for
fx1
1×2
What is the radius of convergence?
b Use part a to find a power series for
fx1
1×3
c Use part b to find a power series for fxx2
dx
x arctan3x dx
0.3 x2
0 01x
0.1
y
decimal places.
Show that the function
30. y
4 dx
Use the result of Example 6 to compute ln 1.1 correct to five
1nx2n fx
What is the radius of convergence?
b Use part a to find a power series for fxx ln1x.
c Use part a to find a power series for fxlnx21.
15.
n0 2n! is a solution of the differential equation
14. a
1×3
Find a power series representation for f xln1x.
f xf x0
a Show that J0 the Bessel function of order 0 given in
Example 4 satisfies the differential equation x2J0xxJ0xx2J0x0
b Evaluate x01 J0x d x correct to three decimal places.
734CHAPTER 11 INFINITE SEQUENCES AND SERIES
34.
The Bessel function of order 1 is defined by
1n x 2 n1
38. a
Starting with the geometric series n0 xn, find the sum of the series
n0
J1x
a Show that J1 satisfies the differential equation
nx n1
n1
2 n1
x 1
Find the sum of each of the following series.
n!n1!2 22n
xJ1xxJ1xx 1J1x0 b Show that J0xJ1x.
a Showthatthefunction
fx xn n0 n!
is a solution of the differential equation
fxfx b Show that fxex.
Let fnxsin nxn2. Show that the seriesfnx converges for all values of x but the series of derivativesfnx diverges when x2n , n an integer. For what values of x does the seriesfnx converge?
Let
xn fx 2
n1n
Find the intervals of convergence for f, f , and f .
i nxn, x1 iin n1 n1 2
b
c Find the sum of each of the following series.
i nn1xn, x1
35.
n2
n2n
n2
39.
40.
n1 2
Use the power series for tan1x to prove the following expres
ii
n2 2
iii
sion for
a
b
n
as the sum of an infinite series:
1n 2s3 2n13n
n
36.
n0
By completing the square, show that
dx
y12
0 xx1 3s3
By factoring x31 as a sum of cubes, rewrite the integral in part a. Then express 1×31 as the sum of a power series and use it to prove the following formula for :
2
37.
3s31 21
n
4 n0 8n 3n1 3n2
11.10 TAYLOR AND MACLAURIN SERIES
In the preceding section we were able to find power series representations for a certain restricted class of functions. Here we investigate more general problems: Which functions have power series representations? How can we find such representations?
We start by supposing that f is any function that can be represented by a power series
fxc0 c1xac2xa2 c3xa3 c4xa4xaR
Lets try to determine what the coefficients cn must be in terms of f. To begin, notice that if we put xa in Equation 1, then all terms after the first one are 0 and we get
fac0
By Theorem 11.9.2, we can differentiate the series in Equation 1 term by term:
fxc1 2c2xa3c3xa2 4c4xa3xaR
and substitution of xa in Equation 2 gives
f ac1
1
2
SECTION 11.10 TAYLOR AND MACLAURIN SERIES735
Now we differentiate both sides of Equation 2 and obtain
fx2c2 23c3xa34c4xa2xaR
Again we put xa in Equation 3. The result is
f a2c2
Lets apply the procedure one more time. Differentiation of the series in Equation 3 gives
fx23c3 234c4xa345c5xa2xaR
and substitution of xa in Equation 4 gives fa23c3 3!c3
By now you can see the pattern. If we continue to differentiate and substitute xa, we obtain
fna234ncn n!cn Solving this equation for the nth coefficient cn, we get
cnfna n!
This formula remains valid even for n0 if we adopt the conventions that 0!1 and f 0f. Thus we have proved the following theorem.
3
4
5
THEOREM If f has a power series representation expansion at a, that is, if
fx cnxan n0
then its coefficients are given by the formula
cnfna n!
xaR
Substituting this formula for cn back into the series, we see that if f has a power series expansion at a, then it must be of the following form.
fxfn a xan n0 n!
fa fa xa fa xa2fa xa31! 2! 3!
6
736CHAPTER 11 INFINITE SEQUENCES AND SERIES
TAYLOR AND MACLAURIN
The Taylor series is named after the English mathematician Brook Taylor 16851731 and the Maclaurin series is named in honor of the Scot tish mathematician Colin Maclaurin 16981746 despite the fact that the Maclaurin series is really just a special case of the Taylor series. But the idea of representing particular functions as sums of power series goes back to Newton, and the general Taylor series was known to the Scot tish mathematician James Gregory in 1668 and to the Swiss mathematician John Bernoulli in the 1690s. Taylor was apparently unaware of the work of Gregory and Bernoulli when he published his discoveries on series in 1715 in his book Methodus incrementorum directa et inversa. Maclaurin series are named after Colin Maclau rin because he popularized them in his calculus textbook Treatise of Fluxions published in 1742.
The series in Equation 6 is called the Taylor series of the function f at a or about a or centered at a. For the special case a0 the Taylor series becomes
This case arises frequently enough that it is given the special name Maclaurin series.
NOTE
V EXAMPLE 1 Find the Maclaurin series of the function fxex and its radius of convergence.
SOLUTION If fxex, then fnxex, so fn0e01 for all n. Therefore the Taylor series for f at 0 that is, the Maclaurin series is
fn0 nxn x x2 x3 x1
fx fn0xnf0 f0x f0x2 n0n! 1!2!
7
We have shown that if f can be represented as a power series about a, then f is equal to the sum of its Taylor series. But there exist functions that are not equal to the sum of their Taylor series. An example of such a function is given in Exercise 70.
2! 3! To find the radius of convergence we let anxnn!. Then
n0 n! n0 n! 1!
an1xn1 n! x nl01
an n1! x n1
so, by the Ratio Test, the series converges for all x and the radius of convergence
is R. M The conclusion we can draw from Theorem 5 and Example 1 is that if ex has a power
series expansion at 0, then
ex xn n0 n!
So how can we determine whether ex does have a power series representation?
Lets investigate the more general question: Under what circumstances is a function equal to the sum of its Taylor series? In other words, if f has derivatives of all orders, when
is it true that
fxfn a xan n0 n!
As with any convergent series, this means that f x is the limit of the sequence of partial sums. In the case of the Taylor series, the partial sums are
T nx n fi a xai i0 i!
fa fa xa fa xa2fna xan 1! 2! n!
y
0,1
0x
yTx
FIGURE 1
N As n increases, Tnx appears to approach ex in Figure 1. This suggests that e x is equal to the sum of its Taylor series.
Notice that Tn is a polynomial of degree n called the nthdegree Taylor polynomial of f at a. For instance, for the exponential function fxex, the result of Example 1 shows that the Taylor polynomials at 0 or Maclaurin polynomials with n1, 2, and 3 are
SECTION 11.10 TAYLOR AND MACLAURIN SERIES737
yTTMx
yyTx
yTTMx yTx
T1x1x
x2 T2x1x 2!
x2 x3 T3x1x 2!3!
The graphs of the exponential function and these three Taylor polynomials are drawn in Figure 1.
In general, f x is the sum of its Taylor series if fxlim Tnx
nl
If we let
RnxfxTnx
then Rnx is called the remainder of the Taylor series. If we can somehow show that
limnl Rnx0, then it follows that
lim Tnxlim fxRnxfxlim Rnxfx
nl nl nl We have therefore proved the following.
so that
fxTnxRnx
THEOREM If f xTnxRnx, where Tn is the nthdegree Taylor polyno mialof f ataand
lim Rnx0 nl
for xa R, then f is equal to the sum of its Taylor series on the interval xa R.
8
In trying to show that limnl Rnx0 for a specific function f, we usually use the following fact.
To see why this is true for n1, we assume that fx M. In particular, we have fxM,sofor axadwehave
xx
y f t dty M dt
aa
An antiderivative of fis f , so by Part 2 of the Fundamental Theorem of Calculus, we have
TAYLORS INEQUALITY Iff n1x M for xa d, then the remainder Rnx of the Taylor series satisfies the inequality
Rnx M xan1 for xa d n1!
9
fxfaMxa or fxfaMxa
738CHAPTER 11 INFINITE SEQUENCES AND SERIES
N As alternatives to Taylors Inequality, we have the following formulas for the remainder term. If f n1 is continuous on an interval I and xI,
then
This is called the integral form of the remainder term. Another formula, called Lagranges form of the remainder term, states that there is a number z between x and a such that
Rnxf n1z xan1 n1!
This version is an extension of the Mean Value Theorem which is the case n0.
Proofs of these formulas, together with dis cussions of how to use them to solve the exam ples of Sections 11.10 and 11.11, are given on the website
www.stewartcalculus.com
Click on Additional Topics and then on Formulas for the Remainder Term in Taylor series.
Thus
xx
y ftdty faMtadt
aa
fxfafaxaM xa2 2
fxfafaxaM xa2 2
Rnx 1 yxxtnfn1tdt n! a
But R1xfxT1xfxfafaxa. So M
R1x 2 xa2 A similar argument, using f xM, shows that
R1xM xa2 2
So R1xMxa2 2
Although we have assumed that xa, similar calculations show that this inequality is also true for xa.
This proves Taylors Inequality for the case where n1. The result for any n is proved in a similar way by integrating n1 times. See Exercise 69 for the case n2.
NOTE In Section 11.11 we will explore the use of Taylors Inequality in approxi mating functions. Our immediate use of it is in conjunction with Theorem 8.
In applying Theorems 8 and 9 it is often helpful to make use of the following fact.
This is true because we know from Example 1 that the seriesx nn! converges for all x and so its nth term approaches 0.
V EXAMPLE 2 Prove that ex is equal to the sum of its Maclaurin series.
SOLUTION If fxex, then fn1xex for all n. If d is any positive number and x d, then fn1x exed. So Taylors Inequality, with a0 and Med, says that
ed
Rnx n1! xn1 for x d
Notice that the same constant Med works for every value of n. But, from Equa tion 10, we have
ed xn1
lim xn1 ed lim 0 nl n1! nl n1!
xn
lim0 for every real number x nl n!
10
N In 1748 Leonard Euler used Equation 12 to find the value of e correct to 23 digits. In 2003 Shigeru Kondo, again using the series in 12, computed e to more than 50 billion decimal places. The special techniques employed to speed up the computation are explained on the web page
numbers.computation.free.fr
fn 2n0 n!
e 2 n0 n!
SECTION 11.10 TAYLOR AND MACLAURIN SERIES739
It follows from the Squeeze Theorem that limnl Rnx 0 and therefore limnl Rnx0 for all values of x. By Theorem 8, ex is equal to the sum of its Maclaurin series, that is,
In particular, if we put x1 in Equation 11, we obtain the following expression for the number e as a sum of an infinite series:
EXAMPLE 3 Find the Taylor series for fxex at a2.
SOLUTION We have f n2e 2 and so, putting a2 in the definition of a Taylor series
6, we get
M
x xn
e forallx
n0 n!
11
e 11111 n0 n! 1! 2! 3!
12
x2n
Again it can be verified, as in Example 1, that the radius of convergence is R. As in
Example 2 we can verify that limnl Rnx0, so
e2 n0 n!
We have two power series expansions for ex, the Maclaurin series in Equation 11 and the Taylor series in Equation 13. The first is better if we are interested in values of x near 0 and the second is better if x is near 2.
EXAMPLE 4 Find the Maclaurin series for sin x and prove that it represents sin x for all x. SOLUTION We arrange our computation in two columns as follows:
x2n
ex
x2n forallx M
13
fxsin x fxcos x
f xsin x
fxcos x f4xsin x
f00 f01 f 00
f01 f400
Since the derivatives repeat in a cycle of four, we can write the Maclaurin series as follows:
f0 f0 x f0 x2f0 x31! 2! 3!
x3x5x7x2n1
x3!
5!
1n
7! n0 2n1!
740CHAPTER 11 INFINITE SEQUENCES AND SERIES
N Figure2showsthegraphofsinxtogether with its Taylor or Maclaurin polynomials
Since fn1xissinxorcosx,weknowthatfn1x1forallx.Sowecan take M1 in Taylors Inequality:
M n1 xn1 Rnx n1!xn1!
By Equation 10 the right side of this inequality approaches 0 as n l , so
Rnx l 0 by the Squeeze Theorem. It follows that Rnx l 0 as n l , so sin x
is equal to the sum of its Maclaurin series by Theorem 8. M
T1xx T3xx 3!
T5xx3!5!
Notice that, as n increases, Tnx becomes a
x3
x3 x5
14
better approximation to sin x. y
1
ysin x
0
FIGURE 2
T
We state the result of Example 4 for future reference.
x3 x5 x7 sinxx 3!5!7!
15
1
T
x
T
1n
for all x
n0
x 2 n1 2n1!
EXAMPLE 5 Find the Maclaurin series for cos x.
SOLUTION We could proceed directly as in Example 4 but its easier to differentiate the
Maclaurin series for sin x given by Equation 15:
d dx3 x5 x7
cosx dx sinx dx x 3!5!7!
3×2 5×4 7×6 x2 x4 x6
N The Maclaurin series for e x, sin x, and cos x that we found in Examples 2, 4, and 5 were dis covered, using different methods, by Newton. These equations are remarkable because they say we know everything about each of these functions if we know all its derivatives at the single number 0.
1 3!5!7! 12!4!6!
Since the Maclaurin series for sin x converges for all x, Theorem 2 in Section 11.9 tells
us that the differentiated series for cos x also converges for all x. Thus
x2 x4 x6 cosx1 2!4!6!
16
1n
for all x
n0
x 2 n 2n!
EXAMPLE 6 Find the Maclaurin series for the function f xx cos x.
SOLUTION Instead of computing derivatives and substituting in Equation 7, its easier to
M
M
multiply the series for cos x Equation 16 by x:x2n
xcos xx 1n
n0 2n!
1n
n0
x2n1 2n!
EXAMPLE 7 Represent f xsin x as the sum of its Taylor series centered at 3.
SOLUTION Arranging our work in columns, we have fxsin x
fs3 32
SECTION 11.10 TAYLOR AND MACLAURIN SERIES741
N We have obtained two different series repre sentations for sin x, the Maclaurin series in Example 4 and the Taylor series in Example 7. It is best to use the Maclaurin series for values of x near 0 and the Taylor series for x near 3. Notice that the third Taylor polynomial T3 in Fig ure 3 is a good approximation to sin x near 3 but not as good near 0. Compare it with the third Maclaurin polynomial T3 in Figure 2, where the opposite is true.
fxcos x
f xsin x
f1 32
fs3 32
1 f 3 2
y
f 32 f 3 x32! x33!
fxcosx
and this pattern repeats indefinitely. Therefore the Taylor series at
3 is
x3 3
f 3 f 31!
s3
2 21! x3 22! x3 23! x3
The proof that this series represents sin x for all x is very similar to that in Example 4. Just replace x by x3 in 14. We can write the series in sigma notation if we separate the terms that contain s3 :
1ns3 2n1n2n1
sinx xx M
ysin x
0x 3
T
1 s32 13
FIGURE 3
n0 22n! 3
n0 22n1! 3
The power series that we obtained by indirect methods in Examples 5 and 6 and in Section 11.9 are indeed the Taylor or Maclaurin series of the given functions because Theorem 5 asserts that, no matter how a power series representation f x cnxan is obtained, it is always true that cnf nan!. In other words, the coefficients are uniquely determined.
EXAMPLE 8 Find the Maclaurin series for fx1xk, where k is any real number. SOLUTION Arranging our work in columns, we have
fx1xk fxk1xk1
f xkk11xk2
f xkk1k21xk3
f01 f0k
f 0kk1
f 0kk1k2
.. .. ..
f nxkk1kn11xkn f n0kk1kn1 Therefore the Maclaurin series of f x1xk is
f n0kk1kn1 n!xn n! xn
n0 n0
742
CHAPTER 11 INFINITE SEQUENCES AND SERIES
This series is called the binomial series. If its nth term is an, then
an1kk1kn1knxn1 n! n
M
an n1! kk1kn1x 1k
knx n xlx asnl n1 11
n
Thus, by the Ratio Test, the binomial series converges if x 1 and diverges
if x 1.
The traditional notation for the coefficients in the binomial series is
k kk1k2kn1 n n!
and these numbers are called the binomial coefficients.
The following theorem states that 1xk is equal to the sum of its Maclaurin series.
It is possible to prove this by showing that the remainder term Rnx approaches 0, but that turns out to be quite difficult. The proof outlined in Exercise 71 is much easier.
Although the binomial series always converges when x 1, the question of whether or not it converges at the endpoints, 1, depends on the value of k. It turns out that the series converges at 1 if 1k0 and at both endpoints if k0. Notice that if k is a positive integer and nk, then the expression for nkcontains a factor kk, so nk 0 for nk. This means that the series terminates and reduces to the ordinary Binomial Theorem when k is a positive integer. See Reference Page 1.
V EXAMPLE 9 Find the Maclaurin series for the function f x1 and its radius of convergence. s4x
SOLUTION We write f x in a form where we can use the binomial series:
11 1x 12
s4x 41x21x 2 4 44
17
THE BINOMIAL SERIES If k is any real number and x 1, then
1xk kxn 1kx kk1x2kk1k2x3n0n 2! 3!
11
SECTION 11.10 TAYLOR AND MACLAURIN SERIES
743
Using the binomial series with k1 and with x replaced by x4, we have 2
1 1 x12 11xn 12
s4x 2 4 2n0 n 4
1 1x 13×2 135×3 122222
2 2 4 2! 4 3! 4
135 1n1xn222 2
n! 4
1 11 x13 x 2135 x 31352n1 x n
TABLE 1
Important Maclaurin Series and Their Radii of Convergence
2 8 2!82 3!83 n!8n
We know from 17 that this series converges when x4 1, that is, x 4, so the
radius of convergence is R4. M We collect in the following table, for future reference, some important Maclaurin series
that we have derived in this section and the preceding one.
1
1x n0
xn 1xx2 x3
R1
R R
Rx2n1 x3 x5 x7
xxn xx2x3 e1
sin x
cos x
1n
x2n x2 x4 x6
n0 n!
n0
1! 2! 3! x2n1 x3
x5
x7 7!
x2n1! 3!
1n1
5! n0 2n! 2! 4! 6!
tan1x 1n xR1
n0 2n1 357
1xk kxn 1kx kk1x2kk1k2x3R1 n0n 2! 3!
TEC Module 11.1011.11 enables you to see how successive Taylor polynomials approach the original function.
One reason that Taylor series are important is that they enable us to integrate functions that we couldnt previously handle. In fact, in the introduction to this chapter we men tioned that Newton often integrated functions by first expressing them as power series and then integrating the series term by term. The function fxex2 cant be integrated by techniques discussed so far because its antiderivative is not an elementary function see Section 7.5. In the following example we use Newtons idea to integrate this function.
744
CHAPTER 11 INFINITE SEQUENCES AND SERIES
V EXAMPLE 10
a Evaluate x ex2 dx as an infinite series.
b Evaluate x1 ex2 dx correct to within an error of 0.001. 0
SOLUTION
a First we find the Maclaurin series for fxex2. Although its possible to use the direct method, lets find it simply by replacing x with x2 in the series for ex given in Table 1. Thus, for all values of x,
x2n ex2
x2n 1n 1
x2 x4 x6
1! 2! 3!
n0 n! Now we integrate term by term:
n0 n!
2x2x4x6 x2n yex dxy 1 1ndx
1!
2!
3! x5
52!
n!
1n
x3
31!
x7
73!
x2n1 2n1n!
Cx
This series converges for all x because the original series for ex2 converges for all x.
b The Fundamental Theorem of Calculus gives
1×2 x3 x5 x7 x9 1 ye dx x
N We can take C0 in the antiderivative in part a.
3 10 42
216
0 31! 1111
52! 73! 94! 1
1 0.7475 216
0
111 1 3 10 42
The Alternating Series Estimation Theorem shows that the error involved in this approxi mation is less than
11 0.001 M 115! 1320
Another use of Taylor series is illustrated in the next example. The limit could be found with lHospitals Rule, but instead we use a series.
EXAMPLE 11 Evaluate lim ex1x . xl0 x2
SOLUTION Using the Maclaurin series for ex, we have
ex1x
lim 2lim xl0 x xl0
xx2x311!2!3! 1x
2
N Some computer algebra systems compute limits in this way.
x2 x3 x4
lim 2!3!4!
xl0 x2
1xx2x3 1 lim xl0 2 3! 4! 5! 2
x
because power series are continuous functions.
M
SECTION 11.10 TAYLOR AND MACLAURIN POLYNOMIALS745 MULTIPLICATION AND DIVISION OF POWER SERIES
If power series are added or subtracted, they behave like polynomials Theorem 11.2.8 shows this. In fact, as the following example illustrates, they can also be multiplied and divided like polynomials. We find only the first few terms because the calculations for the later terms become tedious and the initial terms are the most important ones.
EXAMPLE 12 Find the first three nonzero terms in the Maclaurin series for a ex sin x and b tan x.
SOLUTION
a Using the Maclaurin series for ex and sin x in Table 1, we have
x xx2x3 x3
esinx 11!2!3! x3! We multiply these expressions, collecting like terms just as for polynomials:
1x1x2 1×326
x 1×36
x x2 1×3 1×426
1×3 1×466
x x21x3 3
Thus
b Using the Maclaurin series in Table 1, we have
exsinxxx2 1×33
x3 x5
x 3!5!
x2 x4
12!4!
x1x32 x53 15
11×21 x4 x1x31 x5
sinx tan xcos x
We use a procedure like long division:
2 24
6 120
Thus
x1x31 x52 24
1×31 x53 30
1×3 1×5 36
2 x515
tanxx1x32 x5M 3 15
Although we have not attempted to justify the formal manipulations used in Exam ple 12, they are legitimate. There is a theorem which states that if both f x cn x n and txbnxn converge for x R and the series are multiplied as if they were polyno mials, then the resulting series also converges for x R and represents fxtx. For
division we require b00; the resulting series converges for sufficiently small x .
746CHAPTER 11 INFINITE SEQUENCES AND SERIES
11.10 EXERCISES
1. If f xn0 bnx5n for all x, write a formula for b8.
2. The graph of f is shown. y
f
1 01x
a Explain why the series
1.60.8×10.4×12 0.1×13
is not the Taylor series of f centered at 1. b Explain why the series
2.80.5×21.5×22 0.1×23
is not the Taylor series of f centered at 2.
3. If f n0n1! for n0, 1, 2, . . . , find the Maclaurin
series for f and its radius of convergence.
4. Find the Taylor series for f centered at 4 if
f n41n n! 3nn1
What is the radius of convergence of the Taylor series?
512 Find the Maclaurin series for f x using the definition
of a Maclaurin series. Assume that f has a power series expan sion. Do not show that Rnx l 0. Also find the associated radius of convergence.
17. 19.
21. 22. 23. 24.
fxcosx, a fx1sx, a9
18. fxsinx, a 2 20. fxx2, a1
Prove that the series obtained in Exercise 7 represents sin x for all x.
Prove that the series obtained in Exercise 18 represents sin x for all x.
Prove that the series obtained in Exercise 11 represents sinh x for all x.
Prove that the series obtained in Exercise 12 represents cosh x for all x.
2528 Use the binomial series to expand the function as a power series. State the radius of convergence.
25. s1x 1
2×3
26. 1
1×4
28. 1×23
27.
2938 Use a Maclaurin series in Table 1 to obtain the Maclaurin series for the given function.
fxsin x
fxex e2x
fxxcos1x2 2
xsin x
fx x3 1
6
30. fxcos x2 32. fxex 2ex 34. fxx2tan1x3
36. fx x2 s2x
2
if x0 if x0
; 3942 Find the Maclaurin series of f by any method and its radius of convergence. Graph f and its first few Taylor polynomials on the same screen. What do you notice about the relationship between these polynomials and f ?
29. 31.
37.
38.
x s4x2
33.
35.
fx
f xsin2x Hint: Use sin2x1 1cos 2x.
5.
7.
9. 11.
fx1x2 fxsin x
fxe5x fxsinh x
6. fxln1x
8. fxcos3x 10. fxxex 12. fxcosh x
1320 Find the Taylor series for of a. Assume that f has a power that Rnx l 0.
f x centered at the given value series expansion. Do not show
fxcosx2 41. fxxex
40. fxex2cos x 42. fxln1x2
13. fxx4 3×2 1, a1 14. fxxx3, a2
fxex, a3 16. fx1x, a3
43. Use the Maclaurin series for ex to calculate e0.2 correct to five decimal places.
39.
15.
44. Use the Maclaurin series for sin x to compute sin 3 correct to five decimal places.
a Use the binomial series to expand 1s1x 2 .
b Use part a to find the Maclaurin series for sin1x.
61.yx 62. sin x
yexln1x
45.
46. a
b Use part a to estimate 1s4 1.1 correct to three decimal
1n
n0 n!
64.
66.
n0 6 2n!3n
SECTION 11.10 TAYLOR AND MACLAURIN SERIES747
Expand 1s4 1x as a power series.
6368 Find the sum of the series.x4n
1n 2n2n
places.
4750 Evaluate the indefinite integral as an infinite series.
47. y x cosx3 dx 48. y ex1 dx x
49. ycosx1dx 50. yarctanx2dx x
5154 Use series to approximate the definite integral to within the indicated accuracy.
51. y1 x cosx3dx three decimal places 0
0.213 3
52. y tan x sinxdx five decimal places
0
2n1 n0 4 2n1!
4 6
55. lim xtan1x xl0 x3
sin xx1 x3 lim 6 xl0 x5
fx e1x2
is not equal to its Maclaurin series.
Graph the function in part a and comment on its behavior near the origin.
0
if x0 if x0
54. y0.5 x2ex2 dx 0
0
error 0.001 5557 Use series to evaluate the limit.
;
b
Use the following steps to prove 17.
65.
67.
68.
69.
70.
71.
72.
1n2n1
n n05n!
3 9 2781 2! 3! 4!
ln 22 ln 23 1ln2 2!3!
Prove Taylors Inequality for n2, that is, prove that iff x M for xa d, then
0.4
53. y s1x dx error510
a
R2x M xa3 forxad 6
Show that the function defined by
58. Use the series in Example 12b to evaluate
tan xx lim 3
xl0 x
We found this limit in Example 4 in Section 4.4 using lHospi
tals Rule three times. Which method do you prefer?
5962 Use multiplication or division of power series to find the first three nonzero terms in the Maclaurin series for the function.
yex2 cosx 60. ysecx
c Deduce that tx1xk.
In Exercise 53 in Section 10.2 it was shown that the length of
theellipsexasin ,ybcos ,whereab0,is
s1e2 sin2 d
56. lim 1cosx xl0 1xex
aLettx nk n ktx
63.
n0x . Differentiate this series to show that
57.
tx 1x 1×1
b Let hx1xktx and show that hx0.
2 0
L4ay
where esa2b2 a is the eccentricity of the ellipse. Expand the integrand as a binomial series and use the result of Exercise 46 in Section 7.1 to express L as a series in powers of the eccentricity up to the term in e6.
59.
748CHAPTER 11 INFINITE SEQUENCES AND SERIES
CAS AN ELUSIVE LIMIT
L A B O R AT O R Y PROJECT
This project deals with the function
f xsintan xtansin x
arcsinarctan xarctanarcsin x
1. Use your computer algebra system to evaluate f x for x1, 0.1, 0.01, 0.001, and 0.0001.
Does it appear that f has a limit as xl0?
2. UsetheCAStograph f nearx0.Doesitappearthat f hasalimitasxl0?
3. Try to evaluate limx l 0 fx with lHospitals Rule, using the CAS to find derivatives of the numerator and denominator. What do you discover? How many applications of lHospitals Rule are required?
4. Evaluate limx l 0 fx by using the CAS to find sufficiently many terms in the Taylor series of the numerator and denominator. Use the command taylor in Maple or Series in Mathematica.
5. Use the limit command on your CAS to find limx l 0 fx directly. Most computer algebra systems use the method of Problem 4 to compute limits.
6. In view of the answers to Problems 4 and 5, how do you explain the results of Problems 1 and 2?
HOW NEWTON DISCOVERED THE BINOMIAL SERIES
WRITING PROJECT
The Binomial Theorem, which gives the expansion of abk, was known to Chinese mathe maticians many centuries before the time of Newton for the case where the exponent k is a positive integer. In 1665, when he was 22, Newton was the first to discover the infinite series expansion of abk when k is a fractional exponent positive or negative. He didnt publish his discovery, but he stated it and gave examples of how to use it in a letter now called the epistola prior dated June 13, 1676, that he sent to Henry Oldenburg, secretary of the Royal Society of London, to transmit to Leibniz. When Leibniz replied, he asked how Newton had discovered the binomial series. Newton wrote a second letter, the epistola posterior of Octo ber 24, 1676, in which he explained in great detail how he arrived at his discovery by a very indirect route. He was investigating the areas under the curves y1x 2 n2 from 0 to x for n0, 1, 2, 3, 4, . . . . These are easy to calculate if n is even. By observing patterns and inter polating, Newton was able to guess the answers for odd values of n. Then he realized he could get the same answers by expressing 1x 2 n2 as an infinite series.
Write a report on Newtons discovery of the binomial series. Start by giving the statement of the binomial series in Newtons notation see the epistola prior on page 285 of 4 or page 402 of 2. Explain why Newtons version is equivalent to Theorem 17 on page 742. Then read Newtons epistola posterior page 287 in 4 or page 404 in 2 and explain the patterns that Newton discovered in the areas under the curves y1x 2 n2. Show how he was able to guess the areas under the remaining curves and how he verified his answers. Finally, explain how these discoveries led to the binomial series. The books by Edwards 1 and Katz 3 contain commentaries on Newtons letters.
1. C. H. Edwards, The Historical Development of the Calculus New York: SpringerVerlag, 1979, pp. 178187.
2. John Fauvel and Jeremy Gray, eds., The History of Mathematics: A Reader London: MacMillan Press, 1987.
3. Victor Katz, A History of Mathematics: An Introduction New York: HarperCollins, 1993, pp. 463466.
4. D. J. Struik, ed., A Sourcebook in Mathematics, 12001800 Princeton, NJ: Princeton University Press, 1969.
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS749
11.11
APPLICATIONS OF TAYLOR POLYNOMIALS
In this section we explore two types of applications of Taylor polynomials. First we look at how they are used to approximate functionscomputer scientists like them because polynomials are the simplest of functions. Then we investigate how physicists and engi neers use them in such fields as relativity, optics, blackbody radiation, electric dipoles, the velocity of water waves, and building highways across a desert.
APPROXIMATING FUNCTIONS BY POLYNOMIALS
Suppose that f x is equal to the sum of its Taylor series at a: fxfn a xan
n0 n!
In Section 11.10 we introduced the notation Tnx for the nth partial sum of this series
and called it the nthdegree Taylor polynomial of f at a. Thus
T nx n fi a xai i0 i!
fa fa xa fa xa2fna xan 1! 2! n!
Since f isthesumofitsTaylorseries,weknowthatTnxlfxasnlandsoTn can be used as an approximation to f : f xTnx.
Notice that the firstdegree Taylor polynomial
T1xfafaxa
is the same as the linearization of f at a that we discussed in Section 3.10. Notice also that T1 and its derivative have the same values at a that f and fhave. In general, it can be shown that the derivatives of Tn at a agree with those of f up to and including derivatives of order n see Exercise 38.
To illustrate these ideas lets take another look at the graphs of yex and its first few Taylor polynomials, as shown in Figure 1. The graph of T1 is the tangent line to yex at 0, 1; this tangent line is the best linear approximation to ex near 0, 1. The graph of T2 is the parabola y1xx22, and the graph of T3 is the cubic curve y1xx22x36,whichisacloserfittotheexponentialcurveyex thanT2. The next Taylor polynomial T4 would be an even better approximation, and so on.
The values in the table give a numerical demonstration of the convergence of the Taylor polynomials Tnx to the function yex. We see that when x0.2 the convergence is very rapid, but when x3 it is somewhat slower. In fact, the farther x is from 0, the more slowly Tnx converges to ex.
When using a Taylor polynomial Tn to approximate a function f , we have to ask the questions: How good an approximation is it? How large should we take n to be in order to achieve a desired accuracy? To answer these questions we need to look at the absolute value of the remainder:
FIGURE 1
y
yyTx yTTMx
yTx 0,1
0x
x0.2
x3.0
T2x T4x T6x T8x T10x
1.220000 1.221400 1.221403 1.221403 1.221403
8.500000 16.375000 19.412500 20.009152 20.079665
ex
1.221403
20.085537
Rnx fxTnx
750
CHAPTER 11 INFINITE SEQUENCES AND SERIES
There are three possible methods for estimating the size of the error:
1. If a graphing device is available, we can use it to graph Rnxand thereby esti mate the error.
2. If the series happens to be an alternating series, we can use the Alternating Series Estimation Theorem.
3. In all cases we can use Taylors Inequality Theorem 11.10.9, which says that iff n1x M, then
Rnx M xan1 n1!
V EXAMPLE 1
a Approximate the function f xs3 x by a Taylor polynomial of degree 2 at a8. b How accurate is this approximation when 7×9?
SOLUTION
a
fx s3 xx 13 f8 2 fx1x23 f8 1
3 12
fx2x53 f8 1 9 144
f x10 x83 27
Thus the seconddegree Taylor polynomial is
T2xf8 f8 x8 f8 x82 1! 2!
2 1 x8 1 x82 12 288
The desired approximation is
s3 x T2x2 1 x8 1 x82 12 288
b The Taylor series is not alternating when x8, so we cant use the Alternating Series Estimation Theorem in this example. But we can use Taylors Inequality with n2 and a8 :
R2x Mx83 3!
where fx M. Because x7, we have x83783 and so fx10 1 10 1 0.0021
27 x83 27 783
Therefore we can take M0.0021. Also 7×9, so 1×81 and
x8 1. Then Taylors Inequality gives
R2x 0.0021130.00210.0004
3! 6
Thus, if 7×9, the approximation in part a is accurate to within 0.0004. M
2.5
0
FIGURE 2
0.0003
yRTMx
Lets use a graphing device to check the calculation in Example 1. Figure 2 shows that thegraphsofys3 x andyT2xareveryclosetoeachotherwhenxisnear8.Fig ure 3 shows the graph of R2xcomputed from the expression
TTM yx
R2xs3 xT2x R2x 0.0003
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS751
15
We see from the graph that
79 0
FIGURE 3
when 7×9. Thus the error estimate from graphical methods is slightly better than the error estimate from Taylors Inequality in this case.
V EXAMPLE 2
a What is the maximum error possible in using the approximation
x3 x5 sinxx 3!5!
when 0.3×0.3? Use this approximation to find sin 12 correct to six decimal places.
b For what values of x is this approximation accurate to within 0.00005?
SOLUTION
a Notice that the Maclaurin series
x3 x5 x7 sinxx 3!5!7!
is alternating for all nonzero values of x, and the successive terms decrease in size becausex 1, so we can use the Alternating Series Estimation Theorem. The error in approximating sin x by the first three terms of its Maclaurin series is at most
x7x7 7! 5040
If 0.3×0.3, then x 0.3, so the error is smaller than 0.37
50404.3108 To find sin 12 we first convert to radian measure.
sin 12sin12sin180 15
3 15 1 0.20791169 15 15 3! 15 5!
Thus, correct to six decimal places, sin 120.207912. b The error will be smaller than 0.00005 if
x7
50400.00005
752CHAPTER 11 INFINITE SEQUENCES AND SERIES
TEC Module 11.1011.11 graphically shows the remainders in Taylor polynomial approximations.
Solving this inequality for x, we get
x7 0.252 or x0.25217 0.821
So the given approximation is accurate to within 0.00005 when x 0.82. M What if we use Taylors Inequality to solve Example 2? Since f 7xcos x, we
4.310 yRx
0
R6x 1 x7 7!
havef 7x 1 and so
So we get the same estimates as with the Alternating Series Estimation Theorem.
What about graphical methods? Figure 4 shows the graph of
R6xsinxx1x31 x5 6 120
and we see from it that R6x 4.3108 when x 0.3. This is the same estimate that we obtained in Example 2. For part b we want R6x 0.00005, so we graph both yR6xand y0.00005 in Figure 5. By placing the cursor on the right intersection point we find that the inequality is satisfied when x 0.82. Again this is the same esti mate that we obtained in the solution to Example 2.
If we had been asked to approximate sin 72 instead of sin 12 in Example 2, it would have been wise to use the Taylor polynomials at a3 instead of a0 because they are better approximations to sin x for values of x close to 3. Notice that 72 is close to 60 or 3 radians and the derivatives of sin x are easy to compute at 3.
Figure 6 shows the graphs of the Maclaurin polynomial approximations
T1xx T3xx x3 3!
0.3
FIGURE 4
0.3
0.00006 y0.00005
yRx
1 1 0
FIGURE 5
x3 x5 T5xx 3!5!
x3x5x7 T7xx 3!5!7!
to the sine curve. You can see that as n increases, Tnx is a good approximation to sin x on a larger and larger interval.
y
One use of the type of calculation done in Examples 1 and 2 occurs in calculators and computers. For instance, when you press the sin or ex key on your calculator, or when a computer programmer uses a subroutine for a trigonometric or exponential or Bessel func tion, in many machines a polynomial approximation is calculated. The polynomial is often a Taylor polynomial that has been modified so that the error is spread more evenly through out an interval.
APPLICATIONS TO PHYSICS
Taylor polynomials are also used frequently in physics. In order to gain insight into an equation, a physicist often simplifies a function by considering only the first two or three terms in its Taylor series. In other words, the physicist uses a Taylor polynomial as an
FIGURE 6
T
0x
ysin x T T
T
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS753
approximation to the function. Taylors Inequality can then be used to gauge the accuracy of the approximation. The following example shows one way in which this idea is used in special relativity.
V EXAMPLE 3 In Einsteins theory of special relativity the mass of an object moving with velocity v is
m m0
s1v2c2
where m0 is the mass of the object when at rest and c is the speed of light. The kinetic energy of the object is the difference between its total energy and its energy at rest:
Kmc2 m0c2
a Show that when v is very small compared with c, this expression for K agrees with
classical Newtonian physics: K1 m0v2. 2
b Use Taylors Inequality to estimate the difference in these expressions for K when v 100 ms.
SOLUTION
a Using the expressions given for K and m, we get m0c2
Kmc2 m0c2s1v2c2 m0c2 2 v212
N The upper curve in Figure 7 is the graph of the expression for the kinetic energy K of an object with velocity v in special relativity. The lower curve shows the function used for K in classical Newtonian physics. When v is much smallerthanthespeedoflight,thecurvesare practically identical.
K
Kmcm c
K21 m0c
FIGURE 7
m0c 1 c2 1
With xv2c2, the Maclaurin series for 1×12 is most easily computed as a
binomial series with k1 . Notice thatx 1 because vc. Therefore we have 2
1×1211 x2
1 32 2
1 3 5
x22 2 3!
2
x3
2! 11x3x25 x3
and
2 8 16
2 1 v2 3 v4 5 v6
Km0c 12 c2 8 c4 16 c61
21 v2 3 v4 5 v6m0c 2c28c416c6
If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get
Km0c21 v21m0v2 2c2 2
b If xv2c2, fxm0c21x121, and M is a number such thatf x M, then we can use Taylors Inequality to write
R1x M x2 2!
Wehavefx3m0c21x52 andwearegiventhatv100ms,so 4
3m0 c2 3m0 c2
fx 41v2c252411002c252 M
754
CHAPTER 11 INFINITE SEQUENCES AND SERIES
Thus, with c3108 ms,
1 3m0c2 1004
R1x 2411002c252c44.171010m0
So when v 100 ms, the magnitude of the error in using the Newtonian expression
for kinetic energy is at most 4.21010 m0. M
Another application to physics occurs in optics. Figure 8 is adapted from Optics, 4th ed., by Eugene Hecht San Francisco: AddisonWesley, 2002, page 153. It depicts a wave from the point source S meeting a spherical interface of radius R centered at C. The ray SA is refracted toward P.
r
i Lo
V
A
h
Rt Li
1
S
so
FIGURE 8 n
sinTM
CP
Refraction at a spherical interface
N Here we use the identity
Using Fermats principle that light travels so as to minimize the time taken, Hecht derives the equation
n1 n2 1n2si n1so o i R i o
where n1 and n2 are indexes of refraction and o, i, so, and si are the distances indicated in Figure 8. By the Law of Cosines, applied to triangles ACS and ACP, we have
o sR2 so R2 2Rso Rcos i sR2 si R2 2Rsi Rcos
Because Equation 1 is cumbersome to work with, Gauss, in 1841, simplified it by using the linear approximation cos1 for small values of . This amounts to using the Taylor polynomial of degree 1. Then Equation 1 becomes the following simpler equation as you are asked to show in Exercise 34a:
n1 n2 n2n1 so si R
The resulting optical theory is known as Gaussian optics, or firstorder optics, and has become the basic theoretical tool used to design lenses.
A more accurate theory is obtained by approximating cos by its Taylor polynomial of degree 3 which is the same as the Taylor polynomial of degree 2. This takes into account rays for which is not so small, that is, rays that strike the surface at greater distances h above the axis. In Exercise 34b you are asked to use this approximation to derive the
2
cos
cos
3
Courtesy of Eugene Hecht
SECTION 11.11 APPLICATIONS OF TAYLOR POLYNOMIALS
755
more accurate equation
n1 n2 n2 n1 2n1 1 12 n2 1 12 h
so si R 2so so R 2si R si
The resulting optical theory is known as thirdorder optics.
Other applications of Taylor polynomials to physics and engineering are explored in
Exercises 32, 33, 35, 36, and 37 and in the Applied Project on page 757.
11.11 EXERCISES
; 1.
; 2.
a Find the Taylor polynomials up to degree 6 for fxcos x centered at a0. Graph f and these
polynomials on a common screen.
b Evaluate f and these polynomials at x4, 2,
and .
c Comment on how the Taylor polynomials converge
to fx.
a Find the Taylor polynomials up to degree 3 for fx1x centered at a1. Graph f and these
polynomials on a common screen.
b Evaluate f and these polynomials at x0.9 and 1.3. c Comment on how the Taylor polynomials converge
; c 13. 14. 15. 16. 17.
20. 21. 22.
23. 24. 25.
26.
Check your result in part b by graphing Rnx. fxsx, a4, n2, 4×4.2 fxx2, a1, n2, 0.9×1.1 fxx23, a1, n3, 0.8×1.2
fxsinx, a 6, n4, 0x 3 fxsecx, a0, n2, 0.2×0.2
fxln12x, a1, n3, 0.5×1.5 fxex2, a0, n3, 0x0.1
fxxlnx, a1, n3, 0.5×1.5 fxxsinx, a0, n4, 1×1 f xsinh 2x, a0, n5, 1×1
Use the information from Exercise 5 to estimate cos 80 cor rect to five decimal places.
Use the information from Exercise 16 to estimate sin 38 correct to five decimal places.
Use Taylors Inequality to determine the number of terms of the Maclaurin series for ex that should be used to estimate e0.1 to within 0.00001.
How many terms of the Maclaurin series for ln1x do you need to use to estimate ln 1.4 to within 0.001?
to fx.
; 310 Find the Taylor polynomial Tnx for the function f at the
number a. Graph f and T3 on the same screen. 3.fx1x, a2
4. fxxex, a0 fxcosx, a 2
6. fxex sinx, a0
7. fxarcsinx, a0
8. fxlnx, a1 x
fxxe2x, a0
10. fxtan1x, a1
CAS 1112 Use a computer algebra system to find the Taylor poly nomials Tn centered at a for n2, 3, 4, 5. Then graph these polynomials and f on the same screen.
11.fxcotx, a 4 12.fxs31x2, a0
1322
a Approximate f by a Taylor polynomial with degree n at the number a.
b Use Taylors Inequality to estimate the accuracy of the approximation f xTnx when x lies in the given interval.
; 2729 Use the Alternating Series Estimation Theorem or Taylors Inequality to estimate the range of values of x for which the given approximation is accurate to within the stated error. Check your answer graphically.
x3
27. sin xx6 error 0.01
x2 x4
28. cosx1 224 error0.005
x3 x5
29. arctan xx35 error 0.05
5.
4
18.
19.
9.
756CHAPTER 11 INFINITE SEQUENCES AND SERIES
30. Suppose you know that
f n4
1n n! 3nn1
Equation 4 for thirdorder optics. Hint: Use the first two terms in the binomial series for o1 and i1. Also, use
sin .
If a water wave with length L moves with velocity v across a
body of water with depth d, as in the figure, then v2tL tanh 2 d
and the Taylor series of f centered at 4 converges to f x for all x in the interval of convergence. Show that the fifth degree Taylor polynomial approximates f 5 with error less than 0.0002.
A car is moving with speed 20 ms and acceleration 2 ms2 at a given instant. Using a seconddegree Taylor polynomial, estimate how far the car moves in the next second. Would it be reasonable to use this polynomial to estimate the distance traveled during the next minute?
32. The resistivity of a conducting wire is the reciprocal of the conductivity and is measured in units of ohmmeters m. The resistivity of a given metal depends on the temperature according to the equation
t20 e t20
where t is the temperature in C. There are tables that list the values of called the temperature coefficient and 20 the resistivity at 20C for various metals. Except at very low temperatures, the resistivity varies almost linearly with tem perature and so it is common to approximate the expression for t by its first or seconddegree Taylor polynomial
at t20.
a Find expressions for these linear and quadratic
approximations.
b For copper, the tables give0.0039C and
201.7108 m. Graph the resistivity of copper and the linear and quadratic approximations for 250Ct1000C.
a b
c
2L
If the water is deep, show that vstL2 .
If the water is shallow, use the Maclaurin series for tanh to show that vstd . Thus in shallow water the veloc ity of a wave tends to be independent of the length of the wave.
Use the Alternating Series Estimation Theorem to show that if L10d, then the estimate v2td is accurate to within 0.014tL.
35.
31.
d
L
36.
The period of a pendulum with length L that makes a maxi
mum angle
0 with the vertical is
L2 dx T4 y
t 0 s1k2 sin2x
1 where ksin2
;
;
c For what values of t does the linear approximation agree with the exponential expression to within one percent?
0and t is the acceleration due to gravity. In Exercise 40 in Section 7.7 we approximated this integral
using Simpsons Rule.
a Expand the integrand as a binomial series and use the
result of Exercise 46 in Section 7.1 to show that
L12 1232 123252T2 t 122 k22242 k4224262 k6
If 0 is not too large, the approximation T2 sLt, obtained by using only the first term in the series, is often used. A better approximation is obtained by using two terms:
L12 T2 t 14 k
b Notice that all the terms in the series after the first one have coefficients that are at most 1. Use this fact to com
4
pare this series with a geometric series and show that
33. An electric dipole consists of two electric charges of equal magnitude and opposite sign. If the charges are q and q and are located at a distance d from each other, then the electric field E at the point P in the figure is
Eqq
D2 Dd2
By expanding this expression for E as a series in powers of dD, show that E is approximately proportional to 1D3 when P is far away from the dipole.
q q
P 34. a Derive Equation 3 for Gaussian optics from Equation 1
D d L12 L43k2
2 t 1kT2 4
by approximating cos in Equation 2 by its firstdegree
Taylor polynomial.
b Show that if cos is replaced by its thirddegree Taylor
polynomial in Equation 2, then Equation 1 becomes
t 44k
c Use the inequalities in part b to estimate the period of a
pendulum with L1 meter and compare with the estimate T2
2
010. How does it sLt ? What if
0 42?
37. If a surveyor measures differences in elevation when making plans for a highway across a desert, corrections must be made for the curvature of the earth.
a If R is the radius of the earth and L is the length of the
highway, show that the correction is CR secLRR
38. Show that Tn and f have the same derivatives at a up to order n.
39. In Section 4.8 we considered Newtons method for approxi mating a root r of the equation f x0, and from an initial approximation x1 we obtained successive approxi mations x2, x3, . . . , where
xn1 xnfxn fxn
Use Taylors Inequality with n1, axn, and xr to show that if f x exists on an interval I containing r, xn , andxn1,andfxM,fxKforallxI,then
b Use a Taylor polynomial to show that L2 5L4
APPLIED PROJECT RADIATION FROM THE STARS757
C2R24R3
c Compare the corrections given by the formulas in parts a and b for a highway that is 100 km long. Take the radius of the earth to be 6370 km.
R
L
R
xn1 r M xn r2 C 2K
This means that if xn is accurate to d decimal places, then xn1 is accurate to about 2d decimal places. More precisely, if the error at stage n is at most 10m, then the error at stage n1 is at most M2K102m.
APPLIED PROJECT
RADIATION FROM THE STARS
Any object emits radiation when heated. A blackbody is a system that absorbs all the radiation that falls on it. For instance, a matte black surface or a large cavity with a small hole in its wall like a blastfurnace is a blackbody and emits blackbody radiation. Even the radiation from the sun is close to being blackbody radiation.
Proposed in the late 19th century, the RayleighJeans Law expresses the energy density of blackbody radiation of wavelength as
f8 kT 4
where is measured in meters, T is the temperature in kelvins K, and k is Boltzmanns con stant. The RayleighJeans Law agrees with experimental measurements for long wavelengths but disagrees drastically for short wavelengths. The law predicts that f las l 0 but experiments have shown that f l 0. This fact is known as the ultraviolet catastrophe.
In 1900 Max Planck found a better model known now as Plancks Law for blackbody radiation:
where
f 8hc5 ehc kT 1
is measured in meters, T is the temperature in kelvins, and hPlancks constant6.62621034 Js
cspeed of light2.997925108 ms
kBoltzmanns constant1.38071023 JK 1. Use lHospitals Rule to show that
lim f 0 and lim f 0 l0 l
for Plancks Law. So this law models blackbody radiation better than the RayleighJeans Law for short wavelengths.
Luke Dodd, Photo Researchers, Inc.
758CHAPTER 11 INFINITE SEQUENCES AND SERIES
2. ; 3.
4. ; 5.
Use a Taylor polynomial to show that, for large wavelengths, Plancks Law gives approxi mately the same values as the RayleighJeans Law.
Graph f as given by both laws on the same screen and comment on the similarities and differences. Use T5700 K the temperature of the sun. You may want to change from meters to the more convenient unit of micrometers: 1 m106 m.
Use your graph in Problem 3 to estimate the value of for which f is a maximum under Plancks Law.
Investigate how the graph of f changes as T varies. Use Plancks Law. In particular, graph f for the stars Betelgeuse T3400 K, Procyon T6400 K, and Sirius T9200 K
as well as the sun. How does the total radiation emitted the area under the curve vary with T ? Use the graph to comment on why Sirius is known as a blue star and Betelgeuse as a red star.
11 REVIEW
CONCEPT CHECK
1. a
b What is a convergent series?
c What does limnl an3 mean?
d What does n1 an3 mean?
What is a convergent sequence?
b If a series is convergent by the Comparison Test, how do you estimate its sum?
c If a series is convergent by the Alternating Series Test, how do you estimate its sum?
8. a Write the general form of a power series.
b What is the radius of convergence of a power series? c What is the interval of convergence of a power series?
9. Suppose f x is the sum of a power series with radius of con vergence R.
a How do you differentiate f ? What is the radius of conver gence of the series for f ?
b How do you integrate f ? What is the radius of convergence of the series for x f x dx?
2. a
b What is a monotonic sequence?
c What can you say about a bounded monotonic sequence?
What is a bounded sequence?
3. a
b What is a pseries? Under what circumstances is it
What is a geometric series? Under what circumstances is it convergent? What is its sum?
convergent?
4. Supposean3 and sn is the nth partial sum of the series.
What is limnl an? What is limnl sn?
5. State the following.
a The Test for Divergence
b The Integral Test
c The Comparison Test
d The Limit Comparison Test
e The Alternating Series Test
f The Ratio Test
g The Root Test
10. a
Write an expression for the nthdegree Taylor polynomial
6. a
b What can you say about such a series?
c What is a conditionally convergent series?
of f centered at a.
b Write an expression for the Taylor series of f centered at a.
c Write an expression for the Maclaurin series of f.
d How do you show that f x is equal to the sum of its
Taylor series?
e State Taylors Inequality.
11. Write the Maclaurin series and the interval of convergence for each of the following functions.
7. a
If a series is convergent by the Integral Test, how do you estimate its sum?
12. Write the binomial series expansion of 1xk. What is the radius of convergence of this series?
What is an absolutely convergent series?
a11x b ex
d cos x e tan1x
c sinx
TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. If limnl an0, then an is convergent.
11. If 1 1, then limn ln0.
12. Ifan is divergent, thenanis divergent.
13. Iffx2xx21x3convergesforallx, 3
then f 02.
14. If anand bnare divergent, then anbnis divergent.
15. If anand bnare divergent, then an bnis divergent.
16. If an is decreasing and an0 for all n, then an is convergent.
17. If an0 andan converges, then1nan converges.
18. If an0 and limnl an1an1, then limnl an0.
19. 0.99999…1
20. If an Aand bn B,then anbn AB.
2. The seriessin 1 n1 n
is convergent.
3. If limnl anL, then limnl a2n1L.
CHAPTER 11 REVIEW
759
4. Ifcn6n is convergent, thencn2n is convergent.
5. Ifcn6n is convergent, thencn6n is convergent.
6. If cnxn diverges when x6, then it diverges when x10.
7. The Ratio Test can be used to determine whether1n 3 converges.
8. The Ratio Test can be used to determine whether1n! converges.
9. If 0anbn and bn diverges, then an diverges.
1n 1
10.
n0 n! e
EXERCISES
18 Determine whether the sequence is convergent or divergent. If it is convergent, find its limit.
n1
17.cos3n n1 11.2n
n1 n1
2n3 1. an12n3
9n1 2. an10n
18.n1
n2n
12n2n
3.
5. 7.
9.
; 10.
52n
n3
an1n2 4.ancosn2
annsinn 6.anlnn n21 sn
13n4n 8. 10nn!
A sequence is defined recursively by the equations a11,
an1 1an 4.Showthatanisincreasingandan 2 3
for all n. Deduce that anis convergent and find its limit. Show that limnl n4en0 and use a graph to find the
smallest value of N that corresponds to 0.1 in the pre cise definition of a limit.
19.n1
n
5 n!
1352n1
20.
n1 n9
2n
21.1n1
sn1sn1 n
sn
n1 n1 n1
22.
2326 Determine whether the series is conditionally conver
gent, absolutely convergent, or divergent.
23.1n1n 13 n1
1nn13n 25.2n1
24.
1n1n 3 n1
1nsn ln n
1122 Determine whether the series is convergent or divergent. 11. n 12. n21
n1 2
26.n2
n1 n31 n1 n31
3 n n1
13.n n1 5
2731 Find the sum of the series. 14. 1 27. 3
3n
1
n1 nn3
n
28.
30.
n1 sn1 n1 1 n
2
15.16. ln 29. tan1n1tan1n
1n n
n2 nslnn n1 3n1 n1
2n
n0 3 2n!
760CHAPTER 11 INFINITE SEQUENCES AND SERIES
e2 e3 e4
31. 1e2!3!4!
32. Express the repeating decimal 4.17326326326 . . . as a fraction.
33. Showthatcoshx11x2 forallx. 2
n
34. For what values of x does the series n1 ln x converge?
50. fxxe2x
52. fx10x
54. fx13x5
mal places.
5758
49. fxln1x 51. fxsinx4
53. fx1s4 16x
y ex x
dx as an infinite series. 0
1n1 35. Find the sum of the series 5
decimal places. n1 n
correct to four
55. Evaluate
56. Use series to approximate x1 s1x4 dx correct to two deci
36. a Find the partial sum s5 of the series n1 1n6 and estimate the error in using it as an approximation to the sum of the series.
b Find the sum of this series correct to five decimal places.
37. Use the sum of the first eight terms to approximate the sum of the series n1 25n1. Estimate the error involved in this approximation.
;
;
a Approximate f by a Taylor polynomial with degree n at the number a.
b Graph f and Tn on a common screen.
c Use Taylors Inequality to estimate the accuracy of the approxi
mation f xTnx when x lies in the given interval. d Check your result in part c by graphing Rnx.
57. fxsx, a1, n3, 0.9×1.1 58. fxsecx, a0, n2, 0x 6
59. Use series to evaluate the following limit. lim sinxx
xl0 x3
60. The force due to gravity on an object with mass m at a
height h above the surface of the earth is F mtR2
Rh2
where R is the radius of the earth and t is the acceleration due
to gravity.
a Express F as a series in powers of hR.
b Observe that if we approximate F by the first term in the
series, we get the expression Fmt that is usually used when h is much smaller than R. Use the Alternating Series Estimation Theorem to estimate the range of values of h for which the approximation Fmt is accurate to within one percent. Use R6400 km.
38. a Show that the series
nn n1 2n!
is convergent.
nn
b Deduce that lim0. nl 2n!
39. Prove that if the series n1 an is absolutely convergent, then theseries
n1
is also absolutely convergent.
n1an
4043 Find the radius of convergence and interval of convergence of the series.
40. 42.
44.
45. 46.
xn 1n
x2n 41.
n
n1 n25n2nx2n
n1 n4n
2nx3n
43. ;
n1 n2! n0 sn3 Find the radius of convergence of the series
2n!xn n1 n!2
Find the Taylor series of fxsin x at a6. Find the Taylor series of fxcos x at a3.
61. Suppose that f x n
4754 Find the Maclaurin series for f and its radius of conver gence. You may use either the direct method definition of a Maclaurin series or known series such as geometric series, binomial series, or the Maclaurin series for ex, sin x, and tan1x.
x2
47. fx 1x 48. fxtan1x2
c0 c2 c4 0 b If f is an even function, show that
c1 c3 c5 0
2n! 62. If fxex2, show that f 2n0n! .
n0cnx forallx. a If f is an odd function, show that
PROBLEMS PLUS
P
4
2
1. If fxsinx3,find f150.
2. A function f is defined by
PTM
81 A 1 P
Where is f continuous?
3. a Showthattan1xcot1x2cotx.
1tanx n1 2n 2n
P
fxlim x2n1 nl x2n 1
b Find the sum of the series
22
P
FIGURE FOR PROBLEM 4
4. Let Pnbe a sequence of points determined as in the figure. ThusAP1 1, Pn Pn1 2n1, and angle APn Pn1 is a right angle. Find limn lPn APn1.
5. To construct the snowflake curve, start with an equilateral triangle with sides of length 1. Step 1 in the construction is to divide each side into three equal parts, construct an equilateral triangle on the middle part, and then delete the middle part see the figure. Step 2 is to repeat step 1 for each side of the resulting polygon. This process is repeated at each succeeding step. The snowflake curve is the curve that results from repeating this process indefinitely.
a Let sn, ln, and pn represent the number of sides, the length of a side, and the total length of the nth approximating curve the curve obtained after step n of the construction, respec tively. Find formulas for sn, ln, and pn.
b Showthatpn lasnl.
c Sum an infinite series to find the area enclosed by the snowflake curve.
Note: Parts b and c show that the snowflake curve is infinitely long but encloses only a finite area.
6. Find the sum of the series 11111111
2 3 4 6 8 9 12
where the terms are the reciprocals of the positive integers whose only prime factors are 2s and 3s.
1
2
3
FIGURE FOR PROBLEM 5
7. a
Show that for xy1,
arctanxarctanyarctan xy
if the left side lies between2 and 2. b Show that
arctan 120arctan 1 119 239
1xy
4
c Deduce the following formula of John Machin 16801751:
4 arctan 1arctan 1
5 2394
d Use the Maclaurin series for arctan to show that 0.197395560arctan 10.197395562
e Show that
5
0.004184075arctan 10.004184077 239
761
PROBLEMS PLUS
f Deduce that, correct to seven decimal places,
3.1415927
Machin used this method in 1706 to find correct to 100 decimal places. Recently, with the aid of computers, the value of has been computed to increasingly greater accuracy. Yasumada Kanada of the University of Tokyo recently computed the value of to a trillion decimal places!
8. a Prove a formula similar to the one in Problem 7a but involving arccot instead of arctan. b Find the sum of the series
arccotn2 n1
n0
9. Find the interval of convergence of n1 n3xn and find its sum.
10. Ifa0 a1 a2 ak 0,showthat
lima0sn a1sn1 a2sn2 aksnk0
nl
If you dont see how to prove this, try the problemsolving strategy of using analogy see page 76. Try the special cases k1 and k2 first. If you can see how to prove the asser tion for these cases, then you will probably see how to prove it in general.
11. Findthesumoftheseries ln 1 2 .1
12. Suppose you have a large supply of books, all the same size, and you stack them at the edge of a table, with each book extending farther beyond the edge of the table than the one beneath it. Show that it is possible to do this so that the top book extends entirely beyond the table. In fact, show that the top book can extend any distance at all beyond the edge of the table if the stack is high enough. Use the following method of stacking: The top book extends half its length beyond the second book. The second book extends a quarter of its length beyond the third. The third extends onesixth of its length beyond the fourth, and so on. Try it yourself with a deck of cards. Consider centers of mass.
13. If the curve ye x10 sin x, x0, is rotated about the xaxis, the resulting solid looks like an infinite decreasing string of beads.
a Find the exact volume of the nth bead. Use either a table of integrals or a computer
algebra system.
b Find the total volume of the beads.
n2 n
1 1114 2
86
FIGURE FOR PROBLEM 12
14. If p1, evaluate the expression 1111
2p 3p 4p 1111
FIGURE FOR PROBLEM 15
15. Suppose that circles of equal diameter are packed tightly in n rows inside an equilateral tri angle. The figure illustrates the case n4. If A is the area of the triangle and An is the total area occupied by the n rows of circles, show that
lim An
nl A 2s3
2p 3p 4p
762
PROBLEMS PLUS
P P PTM
P
P P P
FIGURE FOR PROBLEM 18
16.
17.
18.
19.
20.
21.
22.
A sequence anis defined recursively by the equations
a0 a1 1 nn1an n1n2an1 n3an2
Find the sum of the series n0 an.
Taking the value of xx at 0 to be 1 and integrating a series term by term, show that
y1xxdx1n1 0 n1nn
Starting with the vertices P10, 1, P21, 1, P31, 0, P40, 0 of a square, we construct further points as shown in the figure: P5 is the midpoint of P1P2, P6 is the midpoint of P2P3, P7 is the midpoint of P3 P4, and so on. The polygonal spiral path P1 P2 P3 P4 P5 P6 P7 . . . approaches a point P inside the square.
a IfthecoordinatesofPn arexn,yn,showthat1xn xn1 xn2 xn3 2andfinda 2
similar equation for the ycoordinates. b Find the coordinates of P.
If fxm0 cmxm has positive radius of convergence and efxn0 dnxn, show that n
ndn icidni n1 i1
Rightangled triangles are constructed as in the figure. Each triangle has height 1 and its base is the hypotenuse of the preceding triangle. Show that this sequence of triangles makes indefi nitely many turns around P by showing thatn is a divergent series.
Considertheserieswhosetermsarethereciprocalsofthepositiveintegersthatcanbewritten in base 10 notation without using the digit 0. Show that this series is convergent and the sum is less than 90.
a Show that the Maclaurin series of the function
P
P
P
1 1
P1 FIGURE FOR PROBLEM 20
1
1
TM x
1 fx2 isfnxn
1xx n1
b
23. Let
where fn is the nth Fibonacci number, that is, f11, f21, and fnfn1fn2
for n3. Hint: Write x1xx2c0c1xc2 x2. . . and multiply both sides of this equation by 1xx2.
By writing f x as a sum of partial fractions and thereby obtaining the Maclaurin series in a different way, find an explicit formula for the nth Fibonacci number.
x3 x6 x9 u13!6!9!
x4 x7 x10
vx4!7!10!
x2 x5 x8
w2!5!8!
Showthatu3 v3 w3 3uvw1.
24. Prove that if n1, the nth partial sum of the harmonic series is not an integer.
Hint: Let 2k be the largest power of 2 that is less than or equal to n and let M be the product of all odd integers that are less than or equal to n. Suppose that snm, an integer. Then M2ksnM2km. The right side of this equation is even. Prove that the left side is odd by showing that each of its terms is an even integer, except for the last one.
763
764
12
VECTORS AND THE GEOMETRY OF SPACE
LISBON
Wind velocity is a vector because it has both magnitude and direction. Pictured are velocity vectors showing
the wind pattern over the North Atlantic and Western Europe on February 28, 2007. Larger arrows indicate stronger winds.
In this chapter we introduce vectors and coordinate systems for threedimensional space. This will be the setting for our study of the calculus of functions of two variables in Chapter 14 because the graph of such a function is a surface in space. In this chapter we will see that vectors provide particularly simple descriptions of lines and planes
in space.
LONDON
PARIS
MADRID
ROME
OSLO
BERLIN
x
FIGURE 1
Coordinate axes
z
Righthand rule
y
x
FIGURE 2
z
z
z O
12.1 THREEDIMENSIONAL COORDINATE SYSTEMS
To locate a point in a plane, two numbers are necessary. We know that any point in the plane can be represented as an ordered pair a, b of real numbers, where a is the xcoordinate and b is the ycoordinate. For this reason, a plane is called twodimensional. To locate a point in space, three numbers are required. We represent any point in space by an ordered triple a, b, c of real numbers.
In order to represent points in space, we first choose a fixed point O the origin and three directed lines through O that are perpendicular to each other, called the coordinate axes and labeled the xaxis, yaxis, and zaxis. Usually we think of the x and yaxes as being horizontal and the zaxis as being vertical, and we draw the orientation of the axes as in Figure 1. The direction of the zaxis is determined by the righthand rule as illus trated in Figure 2: If you curl the fingers of your right hand around the zaxis in the direc tion of a 90 counterclockwise rotation from the positive xaxis to the positive yaxis, then your thumb points in the positive direction of the zaxis.
The three coordinate axes determine the three coordinate planes illustrated in Fig ure 3a. The xyplane is the plane that contains the x and yaxes; the yzplane contains the y and zaxes; the xzplane contains the x and zaxes. These three coordinate planes divide space into eight parts, called octants. The first octant, in the foreground, is deter mined by the positive axes.
y
z
Pa, b, c
c
y b
Because many people have some difficulty visualizing diagrams of threedimensional figures, you may find it helpful to do the following see Figure 3b. Look at any bottom corner of a room and call the corner the origin. The wall on your left is in the xzplane, the wall on your right is in the yzplane, and the floor is in the xyplane. The xaxis runs along the intersection of the floor and the left wall. The yaxis runs along the intersection of the floor and the right wall. The zaxis runs up from the floor toward the ceiling along the inter section of the two walls. You are situated in the first octant, and you can now imagine seven other rooms situated in the other seven octants three on the same floor and four on the floor below, all connected by the common corner point O.
Now if P is any point in space, let a be the directed distance from the yzplane to P, let b be the distance from the xzplane to P, and let c be the distance from the xyplane to P. We represent the point P by the ordered triple a, b, c of real numbers and we call a, b, and c the coordinates of P; a is the xcoordinate, b is the ycoordinate, and c is the zcoordinate. Thus, to locate the point a, b, c, we can start at the origin O and move a units along the xaxis, then b units parallel to the yaxis, and then c units parallel to the zaxis as in Figure 4.
FIGURE 3
b
x
OO yx
a Coordinate planes
y
x
FIGURE 4
aO
765
yzplane
right wall
xzplane
left wall
xyplane
floor
766
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
zz
z 0
y
The point Pa, b, c determines a rectangular box as in Figure 5. If we drop a perpen dicular from P to the xyplane, we get a point Q with coordinates a, b, 0 called the pro jection of P on the xyplane. Similarly, R0, b, c and Sa, 0, c are the projections of P on the yzplane and xzplane, respectively.
As numerical illustrations, the points 4, 3, 5 and 3, 2, 6 are plotted in Fig ure 6.
0, 0, c
0
a, 0, 0
x
FIGURE 5
3
Pa,b,c 0 5 2
Sa, 0, c
R0, b, c 4 xy
3
0, b, 0
4, 3, 5
6
x
3, 2, 6
Qa, b, 0
y
FIGURE 6
The Cartesian product x,y,zx,y,z is the set of all ordered triples of real numbers and is denoted by 3. We have given a onetoone correspon dence between points P in space and ordered triples a, b, c in 3. It is called a three dimensional rectangular coordinate system. Notice that, in terms of coordinates, the first octant can be described as the set of points whose coordinates are all positive.
In twodimensional analytic geometry, the graph of an equation involving x and y is a curve in 2. In threedimensional analytic geometry, an equation in x, y, and z represents a surface in 3.
V EXAMPLE 1 What surfaces in 3 are represented by the following equations? a z3 b y5
SOLUTION
a The equation z3 represents the set x, y, zz3, which is the set of all points in 3 whose zcoordinate is 3. This is the horizontal plane that is parallel to the xyplane and three units above it as in Figure 7a.
zzy
3 x0yx5y
5 0x
0
FIGURE 7
a z3, a plane in R b y5, a plane in R c y5, a line in R
b The equation y5 represents the set of all points in 3 whose ycoordinate is 5. This is the vertical plane that is parallel to the xzplane and five units to the right of it as in Figure 7b. M
z
y 0
x
NOTE When an equation is given, we must understand from the context whether it rep resents a curve in 2 or a surface in 3. In Example 1, y5 represents a plane in 3, but of course y5 can also represent a line in2 if we are dealing with twodimensional ana lytic geometry. See Figure 7b and c.
In general, if k is a constant, then xk represents a plane parallel to the yzplane, yk is a plane parallel to the xzplane, and zk is a plane parallel to the xyplane. In Figure 5, the faces of the rectangular box are formed by the three coordinate planes x0 the yzplane, y0 the xzplane, and z0 the xyplane, and the planes xa, yb, and zc.
V EXAMPLE 2 Describe and sketch the surface in 3 represented by the equation yx.
SOLUTION The equation represents the set of all points in 3 whose x and ycoordinates are equal, that is, x, x, zx, z. This is a vertical plane that intersects the xyplane in the line yx, z0. The portion of this plane that lies in the first octant is sketched in Figure 8. M
The familiar formula for the distance between two points in a plane is easily extended to the following threedimensional formula.
To see why this formula is true, we construct a rectangular box as in Figure 9, where P1 and P2 are opposite vertices and the faces of the box are parallel to the coordinate planes. If Ax2, y1, z1 and Bx2, y2, z1 are the vertices of the box indicated in the figure, then
P1Ax2 x1 ABy2 y1 BP2z2 z1 Because triangles P1BP2 and P1AB are both rightangled, two applications of the Pythago
rean Theorem give
P1P22 P1B2 BP22
and P1B2P1A2AB2 Combining these equations, we get
P1P2 2P1A2AB2BP2 2
x2 x12 y2 y12 z2 z12 x2 x12 y2 y12 z2 z12
Therefore P1P2 sx2x12y2y12z2z12
SECTION 12.1 THREEDIMENSIONAL COORDINATE SYSTEMS767
FIGURE 8
The plane yx
DISTANCE FORMULA IN THREE DIMENSIONS The distance P1P2between the points P1x1, y1, z1 and P2x2, y2, z2is
P1P2sx2 x12 y2 y12 z2 z12
z
0
FIGURE 9
P,,z PTM,fi,zTM
x
A, , z
B, fi, z
y
768
z
0 x
FIGURE 10
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 3 The distance from the point P2, 1, 7 to the point Q1, 3, 5 is
Px, y, z r
Ch,k,l
PQs122 312 572 s144 3 M V EXAMPLE 4 Find an equation of a sphere with radius r and center Ch, k, l.
SOLUTION By definition, a sphere is the set of all points Px, y, z whose distance from
C is r. See Figure 10. Thus P is on the sphere if and only if PC r. Squaring both sides, we have PC2r2 or
xh2 yk2 zl2 r2 M The result of Example 4 is worth remembering.
y
EQUATION OF A SPHERE An equation of a sphere with center Ch, k, l and radius r is
xh2 yk2 zl2 r2
In particular, if the center is the origin O, then an equation of the sphere is
x2 y2 z2 r2
z
SOLUTION The inequalities can be rewritten as
1×2 y2 z2 4 1sx2y2z2 2
EXAMPLE 5 Showthatx2 y2 z2 4x6y2z60istheequationofa sphere, and find its center and radius.
SOLUTION We can rewrite the given equation in the form of an equation of a sphere if we complete squares:
x2 4x4y2 6y9z2 2z16491 x22 y32 z12 8
Comparing this equation with the standard form, we see that it is the equation of a sphere with center 2, 3, 1 and radius s82 s2 . M
EXAMPLE 6 What region in 3 is represented by the following inequalities? 1×2 y2 z2 4 z0
0 xy
FIGURE 11
1 2
so they represent the points x, y, z whose distance from the origin is at least 1 and at most 2. But we are also given that z0, so the points lie on or below the xyplane.
Thus the given inequalities represent the region that lies between or on the spheres
x2 y2 z2 1andx2 y2 z2 4andbeneathoronthexyplane.Itissketched in Figure 11. M
SECTION 12.1 THREEDIMENSIONAL COORDINATE SYSTEMS769
12.1 EXERCISES
1. Suppose you start at the origin, move along the xaxis a distance of 4 units in the positive direction, and then move downward a distance of 3 units. What are the coordinates of your position?
2. Sketch the points 0, 5, 2, 4, 0, 1, 2, 4, 6, and 1, 1, 2 on a single set of coordinate axes.
3. Which of the points P6, 2, 3, Q5, 1, 4, and R0, 3, 8 is closest to the xzplane? Which point lies in the yzplane?
4. What are the projections of the point 2, 3, 5 on the xy, yz, and xzplanes? Draw a rectangular box with the origin and
2, 3, 5 as opposite vertices and with its faces parallel to the coordinate planes. Label all vertices of the box. Find the length of the diagonal of the box.
5. Describe and sketch the surface in 3 represented by the equa tionxy2.
6. a What does the equation x4 represent in 2? What does it represent in 3? Illustrate with sketches.
b What does the equation y3 represent in 3? What does z5 represent? What does the pair of equations y3,
z5 represent? In other words, describe the set of points x, y, z such that y3 and z5. Illustrate with a sketch.
78 Find the lengths of the sides of the triangle PQR. Is it a right triangle? Is it an isosceles triangle?
1518 Show that the equation represents a sphere, and find its center and radius.
15. x2 y2 z2 6x4y2z11
16. x2 y2 z2 8x6y2z170 17. 2×2 2y2 2z2 8x24z1
18. 4×2 4y2 4z2 8x16y1
19. a Prove that the midpoint of the line segment from P1x1, y1, z1 to P2x2, y2, z2is
x1x2, y1y2,z1z2 222
b Find the lengths of the medians of the triangle with vertices A1, 2, 3, B2, 0, 5, and C4, 1, 5.
20. Find an equation of a sphere if one of its diameters has end points 2, 1, 4 and 4, 3, 10.
Find equations of the spheres with center 2, 3, 6 that touch a the xyplane, b the yzplane, c the xzplane.
22. Find an equation of the largest sphere with center 5, 4, 9 that is contained in the first octant.
2332 Describe in words the region of 3 represented by the equa tion or inequality.
7. P3, 2, 3,
8. P2, 1, 0,
Q7, 0, 1, Q4, 1, 1,
R1, 2, 1 R4, 5, 4
23. y4 25. x3
0z6
29. x2 y2 z2 3
x2 z2 9
24. x10
26. y0
28. z2 1
30. xz
32. x2 y2 z2 2z
9. Determine whether the points lie on straight line. a A2, 4, 2, B3, 7, 2, C1, 3, 3
b D0, 5, 5, E1, 2, 4, F3, 4, 2
10. Find the distance from 3, 7, 5 to each of the following.
a The xyplane c The xzplane e The yaxis
b The yzplane d The xaxis f The zaxis
3336 Write inequalities to describe the region.
33. The region between the yzplane and the vertical plane x5
34. The solid cylinder that lies on or below the plane z8 and on or above the disk in the xyplane with center the origin and radius 2
The region consisting of all points between but not on the spheres of radius r and R centered at the origin, whererR
36. The solid upper hemisphere of the sphere of radius 2 centered at the origin
21.
27.
11. Find an equation of the sphere with center 1, 4, 3 and radius 5. What is the intersection of this sphere with the x zplane?
12. Find an equation of the sphere with center 2, 6, 4 and radius 5. Describe its intersection with each of the coordinate planes.
Find an equation of the sphere that passes through the point 4, 3, 1 and has center 3, 8, 1.
14. Find an equation of the sphere that passes through the origin and whose center is 1, 2, 3.
13.
31.
35.
770CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 37. The figure shows a line L1 in space and a second line L2,
which is the projection of L1 on the xyplane. In other z
words, the points on L2 are directly beneath, or above, the points on L 1.
a Find the coordinates of the point P on the line L1.
b Locate on the diagram the points A, B, and C, where
the line L1 intersects the xyplane, the yzplane, and the xzplane, respectively.
38. Consider the points P such that the distance from P to
A1, 5, 3 is twice the distance from P to B6, 2, 2. Show that the set of all such points is a sphere, and find its center and radius.
Find an equation of the set of all points equidistant from the points A1, 5, 3 and B6, 2, 2. Describe the set.
40. Find the volume of the solid that lies inside both of the spheres x2 y2 z2 4x2y4z50
and x2 y2 z2 4
L P
1 101
LTM
x
y
12.2 VECTORS
Bu
D
The term vector is used by scientists to indicate a quantity such as displacement or veloc ity or force that has both magnitude and direction. A vector is often represented by an arrow or a directed line segment. The length of the arrow represents the magnitude of the vector and the arrow points in the direction of the vector. We denote a vector by printing a letter in boldface v or by putting an arrow above the letter vl.
For instance, suppose a particle moves along a line segment from point A to point B.
v
The corresponding displacement vector v, shown in Figure 1, has initial point A the tail Cl
and terminal point B the tip and we indicate this by writing vAB. Notice that the vec Al
FIGURE 1
Equivalent vectors
tor uCD has the same length and the same direction as v even though it is in a differ ent position. We say that u and v are equivalent or equal and we write uv. The zero vector, denoted by 0, has length 0. It is the only vector with no specific direction.
COMBINING VECTORS
l
A
FIGURE 2
resulting displacement vector AC is called the sum of AB and BC and we write lll
C
Suppose a particle moves from A to B, so its displacement vector is AB. Then the particle l
39.
B
combined effect of these displacements is that the particle has moved from A to C. The lll
changes direction and moves from B to C, with displacement vector BC as in Figure 2. The
ACABBC
In general, if we start with vectors u and v, we first move v so that its tail coincides with the tip of u and define the sum of u and v as follows.
DEFINITION OF VECTOR ADDITION If u and v are vectors positioned so the initial point of v is at the terminal point of u, then the sum uv is the vector from the initial point of u to the terminal point of v.
SECTION 12.2 VECTORS771 The definition of vector addition is illustrated in Figure 3. You can see why this defini
tion is sometimes called the Triangle Law.
u
uv uu
vv v
ab
FIGURE 5
V EXAMPLE 1 Draw the sum of the vectors a and b shown in Figure 5.
SOLUTION First we translate b and place its tail at the tip of a, being careful to draw a copy of b that has the same length and direction. Then we draw the vector ab see Figure 6a starting at the initial point of a and ending at the terminal point of the copy of b.
Alternatively, we could place b so it starts where a starts and construct ab by the Parallelogram Law as in Figure 6b.
FIGURE 3 The Triangle Law FIGURE 4 The Parallelogram Law
In Figure 4 we start with the same vectors u and v as in Figure 3 and draw another copy of v with the same initial point as u. Completing the parallelogram, we see that uvvu. This also gives another way to construct the sum: If we place u and v so they start at the same point, then uv lies along the diagonal of the parallelogram with u and v as sides. This is called the Parallelogram Law.
TEC Visual 12.2 shows how the Triangle b and Parallelogram Laws work for various
vectors a and b.
aa
ab ab
b
DEFINITION OF SCALAR MULTIPLICATION If c is a scalar and v is a vector, then the scalar multiple cv is the vector whose length is ctimes the length of v and whose direction is the same as v if c0 and is opposite to v if c0. If c0 or v0, then cv0.
v 2v 21v v 1.5v
FIGURE 7
Scalar multiples of v
It is possible to multiply a vector by a real number c. In this context we call the real number c a scalar to distinguish it from a vector. For instance, we want 2v to be the same vector as vv, which has the same direction as v but is twice as long. In general, we mul tiply a vector by a scalar as follows.
This definition is illustrated in Figure 7. We see that real numbers work like scaling fac tors here; thats why we call them scalars. Notice that two nonzero vectors are parallel if they are scalar multiples of one another. In particular, the vector v1v has the same length as v but points in the opposite direction. We call it the negative of v.
By the difference uv of two vectors we mean
uvuv
FIGURE 6 a
b
M
vu
uv
772
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
y Ox
aka, aTMl
a, aTM, a
a
FIGURE 8
So we can construct uv by first drawing the negative of v, v, and then adding it to u by the Parallelogram Law as in Figure 8a. Alternatively, since vuvu, the vec tor uv, when added to v, gives u. So we could construct uv as in Figure 8b by means of the Triangle Law.
v
u
uv Drawing uv a
uv u
b
v
v
EXAMPLE 2 If a and b are the vectors shown in Figure 9, draw a2b.
SOLUTION We first draw the vector 2b pointing in the direction opposite to b and twice as long. We place it with its tail at the tip of a and then use the Triangle Law to draw
a2b as in Figure 10.
a
FIGURE 9
COMPONENTS
b
a2b
2b FIGURE 10
a
M
For some purposes its best to introduce a coordinate system and treat vectors algebra ically. If we place the initial point of a vector a at the origin of a rectangular coordinate system, then the terminal point of a has coordinates of the form a1, a2 or a1, a2, a3, depending on whether our coordinate system is two or threedimensional see Figure 11. These coordinates are called the components of a and we write
aa1, a2 or aa1, a2, a3
We use the notationa1, a2for the ordered pair that refers to a vector so as not to confuse it with the ordered pair a1, a2 that refers to a point in the plane.
For instance, the vectors shown in Figure 12 are all equivalent to the vector
minal point is reached from the initial point by a displacement of three units to the right and two upward. We can think of all these geometric vectors as representations of the
FIGURE 12
z
a, aTM
a
Ol
OP3, 2 whose terminal point is P3, 2. What they have in common is that the ter
x
aka, aTM, al FIGURE 11
y
y
4,
5
1,
3
P
3, 2
0
x
Representations of the vector ak3, 2l
z
SECTION 12.2 VECTORS773 l
Pa, aTM, a
point Pa1, a2, a3. See Figure 13. Lets consider any other representation AB of a, where the initial point is Ax1, y1, z1 and the terminal point is Bx2, y2, z2 . Then we must have x1 a1 x2, y1 a2 y2, and z1 a3 z2 and so a1 x2 x1, a2 y2 y1, and a3z2z1. Thus we have the following result.
algebraic vector a3, 2. The particular representation OP from the origin to the point
position vector of P
P3, 2 is called the position vector of the point P. l
In three dimensions, the vector aOP a1, a2, a3is the position vector of the l
O x Ax, y, z
Bxa,yaTM,za y FIGURE 13
1
Given the points Ax1, y1, z1 and Bx2, y2, z2 , the vector a with represen l
ax2 x1,y2 y1,z2 z1
tation AB is
Representations of aka, aTM, al
y
ab, aTMbTM
V EXAMPLE 3 Find the vector represented by the directed line segment with initial point
A2, 3, 4 and terminal point B2, 1, 1. SOLUTION By 1, the vector corresponding to AB is
a22, 13, 144, 4, 3 M
The magnitude or length of the vector v is the length of any of its representations and is denoted by the symbol vorv . By using the distance formula to compute the length of a segment OP, we obtain the following formulas.
l
The length of the twodimensional vector aa1, a2 isa s a 12a 2 2
The length of the threedimensional vector aa1, a2, a3 isa s a 12a 2 2a 32
ab b
How do we add vectors algebraically? Figure 14 shows that if aa1, a2and bTM b b1,b2,thenthesumisab a1 b1,a2 b2,atleastforthecasewherethe components are positive. In other words, to add algebraic vectors we add their compo nents. Similarly, to subtract vectors we subtract components. From the similar triangles in aTM Figure 15 we see that the components of ca are ca1 and ca2. So to multiply a vector by a
b
0 a b x
a
aTM
scalar we multiply each component by that scalar.
Ifa a1,a2 andb b1,b2,then
ab a1 b1,a2 b2 ab a1 b1,a2 b2
caca1, ca2 Similarly, for threedimensional vectors,
a1,a2,a3b1,b2,b3a1 b1,a2 b2,a3 b3 a1,a2,a3b1,b2,b3a1 b1,a2 b2,a3 b3 ca1, a2, a3 ca1, ca2, ca3
FIGURE 14
a
ca
caTM
a
ca
FIGURE 15
aTM
774
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
N Vectors in n dimensions are used to list vari ous quantities in an organized way. For instance, the components of a sixdimensional vector
p p1,p2,p3,p4,p5,p6
might represent the prices of six different ingre dients required to make a particular product. Fourdimensional vectorsx, y, z, tare used in relativity theory, where the first three compo nents specify a position in space and the fourth represents time.
We denote by V2 the set of all twodimensional vectors and by V3 the set of all three dimensional vectors. More generally, we will later need to consider the set Vn of all ndimensional vectors. An ndimensional vector is an ordered ntuple:
a a1,a2,…,an
where a1, a2, . . . , an are real numbers that are called the components of a. Addition and scalar multiplication are defined in terms of components just as for the cases n2 and n3.
abc
bc
l
V EXAMPLE 4 If a4, 0, 3 and b2, 1, 5, find aand the vectors ab, ab, 3b, and 2a5b.
SOLUTION
as42 02 32 s25 5
4, 0, 32, 1, 5
ab
42, 01, 352, 1, 8
4, 0, 32, 1, 5
32, 1, 532, 31, 356, 3, 15
ab
42, 01, 356, 1, 2
3b2a5b
24, 0, 352, 1, 5
8, 0, 610, 5, 252, 5, 31 M
PROPERTIES OF VECTORS
then
1. abba
3. a0a
5. cabcacb 7. cdacda
If a, b, and c are vectors in Vn and c and d are scalars,
2. abcabc 4. aa0
6. cdacada
8. 1aa
Q abc
c
These eight properties of vectors can be readily verified either geometrically or alge braically. For instance, Property 1 can be seen from Figure 4 its equivalent to the Paral lelogram Law or as follows for the case n2:
ab a1,a2b1,b2a1 b1,a2 b2b1 a1,b2 a2b1,b2a1,a2 ba
We can see why Property 2 the associative law is true by looking at Figure 16 and
applying the Triangle Law several times: The vector PQ is obtained either by first con structing ab and then adding c or by adding a to the vector bc.
Three vectors in V3 play a special role. Let
FIGURE 16
P
ab
a
b
i1, 0, 0 j0, 1, 0 k0, 0, 1
These vectors i, j, and k are called the standard basis vectors. They have length 1 and point in the directions of the positive x, y, and zaxes. Similarly, in two dimensions we define i1, 0 and j0, 1. See Figure 17.
yz
SECTION 12.2 VECTORS775
0, 1
j 0ix
k i
j
FIGURE 17
Standard basis vectors in VTM and V
1,0 x
a b
y
y
a
0ai x
If aa1, a2, a3, then we can write
aa1, a2, a3a1, 0, 00, a2, 00, 0, a3
a11, 0, 0a20, 1, 0a30, 0, 1 aa1ia2 ja3k
Thus any vector in V3 can be expressed in terms of i , j, and k. For instance, 1, 2, 6i2j6k
Similarly, in two dimensions, we can write
a, aTM aTMj
a aaiaTMj z
a a1, a2 a1 ia2 j
See Figure 18 for the geometric interpretation of Equations 3 and 2 and compare with
Figure 17.
EXAMPLE 5 Ifai2j3kandb4i7k,expressthevector2a3binterms
of i, j, and k.
SOLUTION Using Properties 1, 2, 5, 6, and 7 of vectors, we have
2a3b2i2j3k34i7k
2i4j6k12i21k14i4j15k M
A unit vector is a vector whose length is 1. For instance, i, j, and k are all unit vec tors. In general, if a0, then the unit vector that has the same direction as a is
u1aa a a
In order to verify this, we let c1a. Then uca and c is a positive scalar, so u has the same direction as a. Also
u ca ca 1 a 1 a
a, aTM, a
a
ai
x aTMj
ak
y
2
3
b aaiaTMjak FIGURE 18
4
776
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
EXAMPLE 6 Find the unit vector in the direction of the vector 2ij2k.
SOLUTION The given vector has length
2ij2ks22 12 22 s9 3
so, by Equation 4, the unit vector with the same direction is
12ij2k2i1 j2k M 3 333
APPLICATIONS
Vectors are useful in many aspects of physics and engineering. In Chapter 13 we will see how they describe the velocity and acceleration of objects moving in space. Here we look at forces.
A force is represented by a vector because it has both a magnitude measured in pounds or newtons and a direction. If several forces are acting on an object, the resultant force experienced by the object is the vector sum of these forces.
EXAMPLE 7 A 100lb weight hangs from two wires as shown in Figure 19. Find the tensions forces T1 and T2 in both wires and their magnitudes.
50 T
FIGURE 19
50 T
50
FIGURE 20
32
TTM SOLUTION We first express T1 and T2 in terms of their horizontal and vertical components.
100
From Figure 20 we see that
T1T1 cos 50 iT1 sin 50 j
T2T2 cos 32 iT2 sin 32 j .
The resultant T1T2 of the tensions counterbalances the weight w and so we must have T1 T2 w100j
Thus
T1 cos 50T2 cos 32iT1 sin 50T2 sin 32j100j Equating components, we get
T1 cos 50T2 cos 320 T1 sin 50T2 sin 32100
32 TTM
32
5
6
w
Solving the first of these equations for T2and substituting into the second, we get T1 sin 50T1cos 50 sin 32100
cos 32 So the magnitudes of the tensions are
T110085.64 lb sin 50tan 32 cos 50
and T2T1 cos 5064.91 lb cos 32
Substituting these values in 5 and 6, we obtain the tension vectors
T155.05i65.60j T255.05i34.40j M
12.2 EXERCISES
1. Are the following quantities vectors or scalars? Explain. a The cost of a theater ticket
b The current in a river
c The initial flight path from Houston to Dallas
d The population of the world
2. What is the relationship between the point 4, 7 and the
vector 4, 7? Illustrate with a sketch.
Name all the equal vectors in the parallelogram shown.
AB E
DC
4. Write each combination of vectors as a single vector. ll ll
9. A1, 3, A0, 3, 1,
B2, 2
B2, 3, 1
10. A2, 1, B0, 6
3.
SECTION 12.2 VECTORS777
12. A4, 0, 2, 1316 Find the sum of the given vectors and illustrate
B4, 2, 1
5, 7 0, 4, 0
geometrically. 13. 1, 4, 15. 0, 1, 2,
6, 2 0, 0, 3
14. 2, 1, 16. 1, 0, 2,
a ab c 2a
e 2ab
b ab d 1 b
f b3a
1720 Findab,2a3b,a,andab. 17. a 5,12, b 3,6
18. a4ij, bi2j
19. ai2j3k, b2ij5k 20. a2i4j4k, b2jk
a PQQR b RPPS ll lll
2123 Find a unit vector that has the same direction as the given vector.
c QS PS
d RS SP PQ P
S
a uv c vw
b uv
d wvu
uv
Q
R
21. 3i7j
8ij4k
22. 4, 2, 4
23.
5. Copy the vectors in the figure and use them to draw the following vectors.
24. Find a vector that has the same direction as 2, 4, 2 but has length 6.
If v lies in the first quadrant and makes an angle 3 with the positive xaxis andv 4, find v in component form.
26. If a child pulls a sled through the snow on a level path with a force of 50 N exerted at an angle of 38 above the horizontal, find the horizontal and vertical components of the force.
27. A quarterback throws a football with angle of elevation 40 and speed 60 fts. Find the horizontal and vertical components of the velocity vector.
2829 Find the magnitude of the resultant force and the angle it makes with the positive xaxis.
25.
6. Copy the vectors in the figure and use them to draw the following vectors.
w
2
28.
y 20 lb 29. y
45 300 N 60
200 N
b
a 030x0x
712 Find a vector a with representation given by the directed line ll
16 lb
30. The magnitude of a velocity vector is called speed. Suppose that a wind is blowing from the direction N45W at a speed of 50 kmh. This means that the direction from which the wind blows is 45 west of the northerly direction. A pilot is steering
segment AB. Draw AB and the equivalent representation starting at the origin.
7. A2, 3, B2, 1 8. A2, 2, B5, 3
11.
778CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
a plane in the direction N60E at an airspeed speed in still air of 250 kmh. The true course, or track, of the plane is the direction of the resultant of the velocity vectors of the plane and the wind. The ground speed of the plane is the magnitude of the resultant. Find the true course and the ground speed of the plane.
31. A woman walks due west on the deck of a ship at 3 mih. The ship is moving north at a speed of 22 mih. Find the speed and direction of the woman relative to the surface of the water.
32. Ropes 3 m and 5 m in length are fastened to a holiday decora tion that is suspended over a town square. The decoration has a mass of 5 kg. The ropes, fastened at different heights, make angles of 52 and 40 with the horizontal. Find the tension in each wire and the magnitude of each tension.
52 40 3m 5m
33. A clothesline is tied between two poles, 8 m apart. The line is quite taut and has negligible sag. When a wet shirt with a mass of 0.8 kg is hung at the middle of the line, the midpoint is pulled down 8 cm. Find the tension in each half of the clothesline.
34. The tension T at each end of the chain has magnitude 25 N. What is the weight of the chain?
37 37
35. Find the unit vectors that are parallel to the tangent line to the parabola yx 2 at the point 2, 4.
36. a Find the unit vectors that are parallel to the tangent line to the curve y2 sin x at the point6, 1.
b Find the unit vectors that are perpendicular to the tangent line.
c Sketch the curve y2 sin x and the vectors in parts a and b, all starting at6, 1.
37. If A, B, and C are the vertices of a triangle, find lll
a Draw the vectors a3, 2, b2, 1, and c7, 1.
b Show, by means of a sketch, that there are scalars s and t suchthatcsatb.
c Use the sketch to estimate the values of s and t. d Find the exact values of s and t.
40. Suppose that a and b are nonzero vectors that are not parallel and c is any vector in the plane determined by a and b. Give a geometric argument to show that c can be written as
csatb for suitable scalars s and t. Then give an argu ment using components.
Ifr x,y,z andr0x0,y0,z0,describethesetofall pointsx,y,zsuchthatrr01.
42. If rx, y, r1x1, y1, and r2x2, y2, describe the setofallpointsx,ysuchthatrr1rr2k, wherekr1 r2.
43. Figure 16 gives a geometric demonstration of Property 2 of vectors. Use components to give an algebraic proof of this fact for the case n2.
44. Prove Property 5 of vectors algebraically for the case n3. Then use similar triangles to give a geometric proof.
Use vectors to prove that the line joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
46. Suppose the three coordinate planes are all mirrored and a
light ray given by the vector aa1, a2, a3 first strikes the xzplane, as shown in the figure. Use the fact that the angle of incidence equals the angle of reflection to show that the direc tion of the reflected ray is given by ba1, a2, a3 . Deduce that, after being reflected by all three mutually perpendicular mirrors, the resulting ray is parallel to the initial ray. American space scientists used this principle, together with laser beams and an array of corner mirrors on the moon, to calculate very precisely the distance from the earth to the moon.
ABBCCA.
z
b a
38. Let C be the point on the line segment AB that is twice ll
y
as far from B as it is from A. If aOA, bOB, and
l21 cOC,showthatc3 a3 b.
x
39.
41.
45.
12.3 THE DOT PRODUCT
So far we have added two vectors and multiplied a vector by a scalar. The question arises: Is it possible to multiply two vectors so that their product is a useful quantity? One such product is the dot product, whose definition follows. Another is the cross product, which is discussed in the next section.
Thus, to find the dot product of a and b, we multiply corresponding components and add. The result is not a vector. It is a real number, that is, a scalar. For this reason, the dot product is sometimes called the scalar product or inner product. Although Definition 1 is given for threedimensional vectors, the dot product of twodimensional vectors is defined in a similar fashion:
SECTION 12.3 THE DOT PRODUCT779
DEFINITION If aa1, a2, a3 and bb1, b2, b3, then the dot product of a and b is the number ab given by
aba1b1 a2b2 a3b3
1
V EXAMPLE 1
a1, a2b1, b2a1b1a2b2 2, 43, 123412
1, 7, 4 6, 2,1 16724 1 6 22
i2j3k2jk1022317 M The dot product obeys many of the laws that hold for ordinary products of real num
bers. These are stated in the following theorem.
These properties are easily proved using Definition 1. For instance, here are the proofs of Properties 1 and 3:
1 . aaa 12a 2 2a 32 a2
3. abc a1,a2,a3b1 c1,b2 c2,b3 c3
a1b1 c1a2b2 c2a3b3 c3 a1b1 a1c1 a2b2 a2c2 a3b3 a3c3
a1b1a2b2a3b3a1c1a2c2a3c3 abac
The proofs of the remaining properties are left as exercises. M
The dot product ab can be given a geometric interpretation in terms of the angle between a and b, which is defined to be the angle between the representations of a and
PROPERTIES OF THE DOT PRODUCT
scalar, then
1. aaa2
3. abcabac 5. 0a0
If a, b, and c are vectors in V3 and c is a
2. abba
4. cabcabacb
2
780
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
z
ab 0 aA
b that start at the origin, where 0 . In other words, is the angle between the ll
B
line segments OA and OB in Figure 1. Note that if a and b are parallel vectors, then0 or.
b
The formula in the following theorem is used by physicists as the definition of the dot product.
3
THEOREM If is the angle between the vectors a and b, then ababcos
x
FIGURE 1
y
PROOF
If we apply the Law of Cosines to triangle OAB in Figure 1, we get AB2OA2OB22OAOBcos
4
Observe that the Law of Cosines still applies in the limiting cases when
a0 or b0. But OA a, OB b, and AB ab, so Equation 4 becomes
ab2a2b22abcos
Using Properties 1, 2, and 3 of the dot product, we can rewrite the left side of this equa
5
0 or
, or
tion as follows:
Therefore Equation 5 gives
ab2 abab aaabbabb
a2 2abb2
a2 2abb2 a2 b2 2abcos Thus 2ab2abcos
or ababcos M EXAMPLE 2 If the vectors a and b have lengths 4 and 6, and the angle between them is
3, find ab.
SOLUTION Using Theorem 3, we have
ababcos 3461 12 M 2
The formula in Theorem 3 also enables us to find the angle between two vectors.
V EXAMPLE 3 Find the angle between the vectors a2, 2, 1 and b5, 3, 2. SOLUTION Since
6
COROLLARY If is the angle between the nonzero vectors a and b, then
cosab ab
as22 22 12 3 and bs52 32 22 s38
and since
we have, from Corollary 6,
a b3s38 Sotheanglebetweenaandbis2
cos11.46 or 84 3s38
M
SECTION 12.3 THE DOT PRODUCT
781
ab2523122 cosab2
a b b
aFIGURE 2
ab0 ab0 ab0
Two nonzero vectors a and b are called perpendicular or orthogonal if the angle between them is2. Then Theorem 3 gives
ababcos 20
and conversely if ab0, then cos0, so2. The zero vector 0 is considered to be perpendicular to all vectors. Therefore we have the following method for determin ing whether two vectors are orthogonal.
EXAMPLE 4 Show that 2i2jk is perpendicular to 5i4j2k. SOLUTION Since
2i2jk5i4j2k2524120
these vectors are perpendicular by 7. M
Becausecos 0if02andcos 0if 2,weseethatab is positive for2 and negative for2. We can think of ab as measuring the extent to which a and b point in the same direction. The dot product ab is positive if a and b point in the same general direction, 0 if they are perpendicular, and negative if they point in generally opposite directions see Figure 2. In the extreme case where a and b point in exactly the same direction, we have0, so cos1 and
abab
If a and b point in exactly opposite directions, thenand so cos1 and
abab.
DIRECTION ANGLES AND DIRECTION COSINES
The direction angles of a nonzero vector a are the angles , , and in the interval 0,that a makes with the positive x, y, and zaxes. See Figure 3.
The cosines of these direction angles, cos , cos , and cos , are called the direction cosines of the vector a. Using Corollary 6 with b replaced by i , we obtain
cosaia1 ai a
a
b
TEC Visual 12.3A shows an animation of Figure 2.
7
Two vectors a and b are orthogonal if and only if ab0.
z
a
x
ca
ay
8
FIGURE 3
This can also be seen directly from Figure 3.
782
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE Similarly, we also have
9
cos a2 cos a3 a a
By squaring the expressions in Equations 8 and 9 and adding, we see that
cos2cos2cos21 We can also use Equations 8 and 9 to write
10
Therefore
aa1, a2, a3acos , acosacos , cos , cos
1 acos , cos , cos a
, acos
11
which says that the direction cosines of a are the components of the unit vector in the direc tion of a.
EXAMPLE 5 Find the direction angles of the vector a1, 2, 3. SOLUTION Since a s122232s14, Equations 8 and 9 give
cos1 s14
cos2 s14
cos12 58 s14
cos3 s14
cos13 37
M
and so
cos11 74 s14
PROJECTIONS
s14
ll
Figure 4 shows representations PQ and PR of two vectors a and b with the same initial l
point P. If S is the foot of the perpendicular from R to the line containing PQ, then the l
vector with representation PS is called the vector projection of b onto a and is denoted by proja b. You can think of it as a shadow of b.
R
R
TEC Visual 12.3B shows how Figure 4
changes when we vary a and b. bb
aa
Q
The scalar projection of b onto a also called the component of b along a is defined to be the signed magnitude of the vector projection, which is the numberbcos , where
FIGURE4 PSQS
Vector projections proja b proja b
P
R
b
SECTION 12.3 THE DOT PRODUCT783 is the angle between a and b. See Figure 5. This is denoted by compa b. Observe that
it is negative if 2 . The equation
ababcosabcos
shows that the dot product of a and b can be interpreted as the length of a times the scalar projection of b onto a. Since
bcos ab a b a a
the component of b along a can be computed by taking the dot product of b with the unit vector in the direction of a. We summarize these ideas as follows.
a
b coscompa b FIGURE 5
Scalar projection
PSQ
Scalar projection of b onto a: compa bab a
Vectorprojectionofbontoa: proja babaab a a a a2
Notice that the vector projection is the scalar projection times the unit vector in the direc tion of a.
V EXAMPLE 6 Find the scalar projection and vector projection of b1, 1, 2 onto a2, 3, 1.
SOLUTION Since a s223212s14, the scalar projection of b onto a is compa b ab2131123
a s14 s14
The vector projection is this scalar projection times the unit vector in the direction of a:
projab3 a3a3,9,3 M s14 a 14 7 14 14
F
S
vector is DPQ. The work done by this force is defined to be the product of the com ponent of the force along D and the distance moved:
PQ
WFcos D But then, from Theorem 3, we have
WFDcos FD
R
tion, as in Figure 6. If the force moves the object from P to Q, then the displacement l
One use of projections occurs in physics in calculating work. In Section 6.4 we defined
the work done by a constant force F in moving an object through a distance d as WFd,
but this applies only when the force is directed along the line of motion of the object.
Suppose, however, that the constant force is a vector FPR pointing in some other direc
l
FIGURE 6
D
12
784
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
Thus the work done by a constant force F is the dot product FD, where D is the displace
35
ment vector.
EXAMPLE 7 A wagon is pulled a distance of 100 m along a horizontal path by a constant force of 70 N. The handle of the wagon is held at an angle of 35 above the horizontal. Find the work done by the force.
SOLUTION If F and D are the force and displacement vectors, as pictured in Figure 7, then the work done is
WFDF Dcos 35
70100 cos 355734 Nm5734 J M
FIGURE 7
F
35
D
EXAMPLE 8 AforceisgivenbyavectorF3i4j5kandmovesaparticlefrom the point P2, 1, 0 to the point Q4, 6, 2. Find the work done.
l
WFD3, 4, 52, 5, 26201036
If the unit of length is meters and the magnitude of the force is measured in newtons, then the work done is 36 joules.
SOLUTION The displacement vector is DPQ2, 5, 2, so by Equation 12, the work done is
M
12.3 EXERCISES
1. Which of the following expressions are meaningful? Which are meaningless? Explain.
1112 If u is a unit vector, find uv and uw. 12.
u
a abc c abc e abc
b abc
d abc f a bc
uv
w
2. Find the dot product of two vectors if their lengths are 6 and 1 and the angle between them is 4.
w
3
310 Findab.
3. a2, 1 ,
4. a 2, 3,
7. a i2j 3k, b5i9k
8. a4j3k, b2i4j6k
9.a6, b5, theanglebetweenaandbis23
10. a 3, b s6, the angle between a and b is 45
b5, 12 b0.7, 1.2
b6, 3, 8 6. a s, 2s, 3s, bt, t, 5t
5. a 4, 1, 1 , 4
14. A street vendor sells a hamburgers, b hot dogs, and c soft drinks on a given day. He charges 2 for a hamburger, 1.50 for a hot dog, and 1 for a soft drink. If Aa, b, c and
P2, 1.5, 1, what is the meaning of the dot product AP?
1520 Find the angle between the vectors. First find an exact expression and then approximate to the nearest degree.
15. a8, 6, bs7, 3 16. as3 , 1, b0, 5
3
11.
13. a Showthatijjkki0. b Show that iijjkk1.
v
17. a3, 1, 5,
18. a4, 0, 2,
b2, 4, 3 b2, 1, 0
2
19.
SECTION 12.3 THE DOT PRODUCT 39. a2ij4k, bj1k
785
ajk, bi2j3k 20. ai2j2k, b4i3k
2122 Find, correct to the nearest degree, the three angles of the triangle with the given vertices.
21. A1, 0, B3, 6, C1, 4
22. D0, 1, 1, E2, 4, 3, F1, 2, 1
2324 Determine whether the given vectors are orthogonal, parallel, or neither.
23. a a5, 3, 7, b6, 8, 2
b a4, 6, b3, 2
c ai2j5k, b3i4jk da2i6j4k, b3i9j6k
24. a u3, 9, 6, v4, 12, 8 buij2k, v2ijk
c ua, b, c, vb, a, 0
25. Use vectors to decide whether the triangle with vertices
P1, 3, 2, Q2, 0, 4, and R6, 2, 5 is rightangled.
26. For what values of b are the vectors 6, b, 2 and b, b2, b orthogonal?
Find a unit vector that is orthogonal to both ij and ik.
28. Find two unit vectors that make an angle of 60 with
v3, 4.
2933 Find the direction cosines and direction angles of the vector. Give the direction angles correct to the nearest degree.
40. aijk, bijk
Show that the vector ortha bbproja b is orthogonal to a.
It is called an orthogonal projection of b.
42. For the vectors in Exercise 36, find ortha b and illustrate by
drawing the vectors a, b, proja b, and ortha b.
If a3, 0, 1, find a vector b such that compa b2.
44. Suppose that a and b are nonzero vectors.
a Under what circumstances is compa bcompb a? b Under what circumstances is proja bprojb a?
45. FindtheworkdonebyaforceF8i6j9kthatmoves an object from the point 0, 10, 8 to the point 6, 12, 20 along a straight line. The distance is measured in meters and the force in newtons.
46. A tow truck drags a stalled car along a road. The chain makes an angle of 30 with the road and the tension in the chain is 1500 N. How much work is done by the truck in pulling the car 1 km?
47. A sled is pulled along a level path through snow by a rope. A 30lb force acting at an angle of 40 above the horizontal moves the sled 80 ft. Find the work done by the force.
48. A boat sails south with the help of a wind blowing in the direc tion S36E with magnitude 400 lb. Find the work done by the wind as the boat moves 120 ft.
Use a scalar projection to show that the distance from a point P1x1, y1 to the line axbyc0 is
ax1 by1 c sa2 b2
Use this formula to find the distance from the point 2, 3 to the line 3x4y50.
50. Ifr x,y,z,a a1,a2,a3,andb b1,b2,b3,show that the vector equation rarb0 represents a sphere, and find its center and radius.
51. Find the angle between a diagonal of a cube and one of its edges.
52. Find the angle between a diagonal of a cube and a diagonal of one of its faces.
53. A molecule of methane, CH4, is structured with the four hydro gen atoms at the vertices of a regular tetrahedron and the car bon atom at the centroid. The bond angle is the angle formed by the HCH combination; it is the angle between the lines that join the carbon atom to two of the hydrogen atoms. Show that the bond angle is about 109.5. Hint: Take the vertices of the tetrahedron to be the points 1, 0, 0, 0, 1, 0 ,
27.
29. 3, 4, 5
31. 2i3j6k
33. c,c,c, wherec0
30. 1, 2, 1 32. 2ij2k
4 and3540 Find the scalar and vector projections of b onto a.
34. If a vector has direction anglesthird direction angle .
35. a3, 4,
36. a1, 2,
37. a3, 6, 2, b1, 2, 3
38. a2, 3, 6, b5, 1, 4
b5, 0 b4, 1
3, find the
41.
43.
49.
786CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 0, 0, 1, and 1, 1, 1 as shown in the figure. Then the centroid
Use Theorem 3 to prove the CauchySchwarz Inequality:
ab ab The Triangle Inequality for vectors is
ab a b
a Give a geometric interpretation of the Triangle Inequality. b Use the CauchySchwarz Inequality from Exercise 57 to
prove the Triangle Inequality. Hint: Use the fact that ab2 ababanduseProperty3ofthe dot product.
The Parallelogram Law states that
ab2 ab2 2a2 2b2
a Give a geometric interpretation of the Parallelogram Law. b Prove the Parallelogram Law. See the hint in Exercise 58.
Show that if uv and uv are orthogonal, then the vectors u and v must have the same length.
is 1 , 1 , 1 . 222
z
H 58.
x
CHH
y
H
54. If cabba, where a, b, and c are all nonzero vectors, show that c bisects the angle between a and b.
55. Prove Properties 2, 4, and 5 of the dot product Theorem 2.
59.
56. Suppose that all sides of a quadrilateral are equal in length and
opposite sides are parallel. Use vector methods to show that the 60. diagonals are perpendicular.
12.4 THE CROSS PRODUCT
The cross product ab of two vectors a and b, unlike the dot product, is a vector. For this reason it is also called the vector product. Note that ab is defined only when a and b are threedimensional vectors.
This may seem like a strange way of defining a product. The reason for the particular form of Definition 1 is that the cross product defined in this way has many useful proper ties, as we will soon see. In particular, we will show that the vector ab is perpendicu lar to both a and b.
In order to make Definition 1 easier to remember, we use the notation of determinants. A determinant of order 2 is defined by
a badbc cd
Forexample,2 1241614 6 4
A determinant of order 3 can be defined in terms of secondorder determinants as follows:
DEFINITION If aa1, a2, a3 and bb1, b2, b3, then the cross product of a and b is the vector
ab a2b3 a3b2,a3b1 a1b3,a1b2 a2b1
1
2
a1 a2 a3
b2 b3c2 c3
b1 b3c1 c3
b1 b2c1 c2
c1 c2 c3
a2
a3
b1 b2 b3a1
57.
3
SECTION 12.4 THE CROSS PRODUCT787
Observe that each term on the right side of Equation 2 involves a number ai in the first row of the determinant, and ai is multiplied by the secondorder determinant obtained from the left side by deleting the row and column in which ai appears. Notice also the minus sign in the second term. For example,
1 2 10 1 3 1 3 0 3 0 1 1 4 2 2 5 2 1 5 4
5 4 2
104265112038
If we now rewrite Definition 1 using secondorder determinants and the standard basis vectors i, j, and k, we see that the cross product of the vectors aa1 ia2 ja3 k and bb1 ib2 jb3 k is
aba2 a3 ia1 a3 ja1 a2 k b2 b3 b1 b3 b1 b2
In view of the similarity between Equations 2 and 3, we often write
ijk
4
b1 b2 b3
ab a1 a2 a3
Although the first row of the symbolic determinant in Equation 4 consists of vectors, if we expand it as if it were an ordinary determinant using the rule in Equation 2, we obtain Equation 3. The symbolic formula in Equation 4 is probably the easiest way of remember ing and computing cross products.
V EXAMPLE 1 If a1, 3, 4 and b2, 7, 5, then
ijk
2 7 5
ab13 4
3 4 i1 4 j1 3 k 7 5 2 5 2 7
1528 i58 j76 k43 i13 jk M V EXAMPLE 2 Showthataa0foranyvectorainV3.
SOLUTION If aa1, a2, a3 , then
ijk
a1 a2 a3
aa a1 a2 a3
a2a3a3a2ia1a3a3a1ja1a2a2a1k 0i0j0k0 M
788
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
One of the most important properties of the cross product is given by the following
theorem.
PROOF In order to show that ab is orthogonal to a, we compute their dot product as follows:
abaa2 a3 a1 a1 a3 a2 a1 a2 a3 b2 b3 b1 b3 b1 b2
a1a2b3a3b2a2a1b3a3b1a3a1b2a2b1a1a2b3a1b2a3a1a2b3b1a2a3a1b2a3b1a2a3 0
A similar computation shows that abb0. Therefore ab is orthogonal to both a and b. M
If a and b are represented by directed line segments with the same initial point as in Figure 1, then Theorem 5 says that the cross product ab points in a direction perpen dicular to the plane through a and b. It turns out that the direction of ab is given by the righthand rule: If the fingers of your right hand curl in the direction of a rotation through an angle less than 180 from a to b, then your thumb points in the direction of ab.
Now that we know the direction of the vector ab, the remaining thing we need to complete its geometric description is its length ab . This is given by the following theorem.
axb ab
FIGURE 1
TEC Visual 12.4 shows how ab changes as b changes.
6
THEOREM If is the angle between a and b so 0 , then ab absin
Geometric characterization of ab
Since a vector is completely determined by its magnitude and direction, we can now say that ab is the vector that is perpendicular to both a and b, whose orientation is deter
PROOF
From the definitions of the cross product and length of a vector, we have
ab2a2b3a3b22a3b1a1b32a1b2a2b12
a2b322a2a3b2b3a32b2a32b122a1a3b1b3a12b23
5
THEOREM The vector ab is orthogonal to both a and b.
a12b22a1a2b1b2a2b12
a12 a2 a32b12 b2 b32a1b1 a2b2 a3b32 a2b2 ab2
a2b2 a2b2cos2
a2b21cos2
a2b2 sin2
0 when
M
Taking square roots and observing that ssin2 0,wehave
byTheorem12.3.3
sin because sin ab absin
7
COROLLARY Two nonzero vectors a and b are parallel if and only if ab0
b b sina
FIGURE 2
The geometric interpretation of Theorem 6 can be seen by looking at Figure 2. If a and b are represented by directed line segments with the same initial point, then they determine a parallelogram with base a , altitude b sin , and area
Aabsin ab
Thus we have the following way of interpreting the magnitude of a cross product.
EXAMPLE 3 Find a vector perpendicular to the plane that passes through the points P1, 4, 6, Q2, 5, 1, and R1, 1, 1.
mined by the righthand rule, and whose length is a b sin physicists define ab.
PROOF Two nonzero vectors a and b are parallel if and only if sin 0,soab0andthereforeab0.
. In fact, that is exactly how
0 or . In either case M
The length of the cross product ab is equal to the area of the parallelogram determined by a and b.
ll
l
ijk
SECTION 12.4 THE CROSS PRODUCT789
ll ll
SOLUTION The vector PQPR is perpendicular to both PQ and PR and is therefore per pendicular to the plane through P, Q, and R. We know from 12.2.1 that
PQ21 i54 j16 k3 ij7 k l
PR11 i14 j16 k5 j5 k We compute the cross product of these vectors:
535 i150 j150 k40 i15 j15 k
So the vector 40, 15, 15 is perpendicular to the given plane. Any nonzero scalar multiple of this vector, such as 8, 3, 3, is also perpendicular to the plane. M
EXAMPLE 4 Find the area of the triangle with vertices P1, 4, 6, Q2, 5, 1, and R1, 1, 1.
PQPRs402 152 152 5s82 l l
The area A of the triangle PQR is half the area of this parallelogram, that is, 5 s82 . M 2
PQPR 3 1 7 0 5 5
ll
SOLUTION In Example 3 we computed that PQPR40, 15, 15. The area of the parallelogram with adjacent sides PQ and PR is the length of this cross product:
790
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
If we apply Theorems 5 and 6 to the standard basis vectors i, j, and k using2,
we obtain
ijk jki kij
jik kji ikj
whereas
Observe that
Thus the cross product is not commutative. Also
9
ijji
iijikj
iij0j0
So the associative law for multiplication does not usually hold; that is, in general,
abcabc
However, some of the usual laws of algebra do hold for cross products. The following the
orem summarizes the properties of vector products.
THEOREM If a, b, and c are vectors and c is a scalar, then
1. abba
2. cabcabacb 3. abcabac
4. abcacbc
5. abcabc
6. abcacbabc
8
These properties can be proved by writing the vectors in terms of their components and using the definition of a cross product. We give the proof of Property 5 and leave the remaining proofs as exercises.
PROOF OF PROPERTY 5 Ifa a1,a2,a3,b b1,b2,b3,andc c1,c2,c3,then
abca1b2c3b3c2a2b3c1b1c3a3b1c2b2c1
a1b2c3a1b3c2a2b3c1a2b1c3a3b1c2a3b2c1
a2b3a3b2 c1a3b1a1b3 c2a1b2a2b1c3
abc M
TRIPLE PRODUCTS
The product abc that occurs in Property 5 is called the scalar triple product of the vectors a, b, and c. Notice from Equation 9 that we can write the scalar triple product as a determinant:
10
a1 a2 a3 abcb1 b2 b3
c1 c2 c3
bxc h a
c
FIGURE 3
b
The geometric significance of the scalar triple product can be seen by considering the parallelepiped determined by the vectors a, b, and c. See Figure 3. The area of the base parallelogram is Abc. If is the angle between a and bc, then the height h of the parallelepiped is ha cos . We must use cosinstead of cos in case
2. Therefore the volume of the parallelepiped is VAhbcacos abc
Thus we have proved the following formula.
If we use the formula in 11 and discover that the volume of the parallelepiped determined by a, b, and c is 0, then the vectors must lie in the same plane; that is, they are coplanar.
V EXAMPLE 5 Use the scalar triple product to show that the vectors a1, 4, 7, b2, 1, 4, and c0, 9, 18 are coplanar.
SOLUTION We use Equation 10 to compute their scalar triple product:
1 4 7
SECTION 12.4 THE CROSS PRODUCT791
The volume of the parallelepiped determined by the vectors a, b, and c is the magnitude of their scalar triple product:
Vabc
11
abc 2 1 4
0 9 18
FIGURE 4
11 442 472 1 9 18 0 18 0 9
1184367180
Therefore, by 11, the volume of the parallelepiped determined by a, b, and c is 0. This
means that a, b, and c are coplanar. M
The product abc that occurs in Property 6 is called the vector triple product of a, b, and c. Property 6 will be used to derive Keplers First Law of planetary motion in Chapter 13. Its proof is left as Exercise 46.
TORQUE
The idea of a cross product occurs often in physics. In particular, we consider a force F acting on a rigid body at a point given by a position vector r. For instance, if we tighten a bolt by applying a force to a wrench as in Figure 4, we produce a turning effect. The torque relative to the origin is defined to be the cross product of the position and force vectors
rF
and measures the tendency of the body to rotate about the origin. The direction of the torque vector indicates the axis of rotation. According to Theorem 6, the magnitude of the
r
F
792
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE torque vector is
0.25 m
FIGURE 5
75 40 N
rF rFsin
where is the angle between the position and force vectors. Observe that the only com ponent of F that can cause a rotation is the one perpendicular to r, that is, F sin . The magnitude of the torque is equal to the area of the parallelogram determined by r and F.
EXAMPLE 6 A bolt is tightened by applying a 40N force to a 0.25m wrench as shown in Figure 5. Find the magnitude of the torque about the center of the bolt.
SOLUTION The magnitude of the torque vector is
rF rFsin 750.2540 sin 75
10 sin 759.66 Nm
If the bolt is rightthreaded, then the torque vector itself is
n9.66n
where n is a unit vector directed down into the page. M
12.4 EXERCISES
17 Find the cross product ab and verify that it is orthogonal
14 15 Finduvand determine whether uv is directed into the page or out of the page.
to both a and b.
1. a6, 0, 2,
2. a1, 1, 1,
b0, 8, 0 b2, 4, 6
14.
u5 60
15.
3. ai3j2k, bi5k
4. aj7k, b2ij4k
5. aijk, b1ij1k
u6 150
v8
v10
22
The figure shows a vector a in the xyplane and a vector b in the direction of k. Their lengths are a 3 and b 2.
a Findab.
b Use the righthand rule to decide whether the components
of ab are positive, negative, or 0. z
6. aietjetk, b2ietjetk at, t2, t3, b1, 2t, 3t2
8. Ifai2kandbjk,findab.Sketcha,b,and ab as vectors starting at the origin.
912 Find the vector, not with determinants, but by using proper ties of cross products.
9. ijk 10. ki2j
11. jkki 12. ijij
State whether each expression is meaningful. If not, explain why. If so, state whether it is a vector or a scalar.
b
a
x
y
7.
13.
a abc
c abc
e abcd
b abc
d abc
f abcd
that abcabc.
Find two unit vectors orthogonal to both 1, 1, 1 and
0, 4, 4.
16.
17. 18.
19.
If aIf a
1, 2, 1 and b0, 1, 3, find ab and ba. 3, 1, 2, b1, 1, 0, and c0, 0, 4, show
20. Find two unit vectors orthogonal to both ijk and2ik.
21. Showthat0a0a0foranyvectorainV3.
22. Showthatabb0forallvectorsaandbinV3.
23. Prove Property 1 of Theorem 8.
24. Prove Property 2 of Theorem 8.
25. Prove Property 3 of Theorem 8.
26. Prove Property 4 of Theorem 8.
27. Find the area of the parallelogram with vertices A2, 1, B0, 4, C4, 2, and D2, 1.
28. Find the area of the parallelogram with vertices K1, 2, 3, L1, 3, 6, M3, 8, 6, and N3, 7, 3.
2932 a Find a nonzero vector orthogonal to the plane through the points P, Q, and R, and b find the area of triangle PQR.
SECTION 12.4 THE CROSS PRODUCT793 40. Find the magnitude of the torque about P if a 36lb force is
29.
31.
d ab 3334 Find the volume of the parallelepiped determined by the a
37. Use the scalar triple product to verify that the vectors u i5j2k,v3ij,andw5i9j4k are coplanar.
38. Use the scalar triple product to determine whether the points A1, 3, 2, B3, 1, 6, C5, 2, 0, and D3, 6, 4 lie in the same plane.
39. A bicycle pedal is pushed by a foot with a 60N force as shown. The shaft of the pedal is 18 cm long. Find the magnitude of the torque about P.
where aQR, bQS, and cQP.
b Use the formula in part a to find the distance from the
43.
applied as shown.
P
4 ft
4 ft
30
36 lb
P1, 0, 0, 30. P2, 1, 5,
P0, 2, 0, 32. P1, 3, 1,
Q0, 2, 0, Q1, 3, 4,
R0, 0, 3 R3, 0, 6
41. A wrench 30 cm long lies along the positive yaxis and grips a bolt at the origin. A force is applied in the direction 0, 3, 4 at the end of the wrench. Find the magnitude of the force needed to supply 100 Nm of torque to the bolt.
42. Let v5j and let u be a vector with length 3 that starts at the origin and rotates in the xyplane. Find the maximum and minimum values of the length of the vector uv. In what direction does uv point?
a Let P be a point not on the line L that passes through the points Q and R. Show that the distance d from the point P to the line L is
Q4, 1, 2, Q0, 5, 2,
R5, 3, 1 R4, 3, 1
vectors a, b, and c. 33. a6, 3, 1, 34. a i j k,
ll
b0, 1, 2, bijk,
c4, 2, 5 cij k
where aQR and bQP.
b Use the formula in part a to find the distance from
the point P1, 1, 1 to the line through Q0, 6, 8 and R1, 4, 7.
44. a Let P be a point not on the plane that passes through the points Q, R, and S. Show that the distance d from P to the
planeis abc dab
point P2, 1, 4 to the plane through the points Q1, 0, 0, R0, 2, 0, and S0, 0, 3.
45. Provethatabab2ab. 46. Prove Property 6 of Theorem 8, that is,
abcacbabc 47. Use Exercise 46 to prove that
abcbcacab0 48. Prove that
abcdac bcad bd
49. Suppose that a0.
a Ifabac,doesitfollowthatbc?
3536 Find the volume of the parallelepiped with adjacent edges PQ, PR, and PS.
35. P2, 0, 1,
36. P3, 0, 1,
Q4, 1, 0, Q1, 2, 5,
R3, 1, 1, R5, 1, 1,
S2, 2, 2 S0, 4, 2
lll
60 N
10
70
P
794CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
b If abac, does it follow that bc?
c Ifabacandabac,doesitfollow
that bc?
50. If v1, v2, and v3 are noncoplanar vectors, let
These vectors occur in the study of crystallography. Vectors
of the form n1v1n2v2n3v3, where each ni is an integer, form a lattice for a crystal. Vectors written similarly in terms of k1, k2, and k3 form the reciprocal lattice.
a Show that ki is perpendicular to vj if ij.
b Show that kivi1 for i1, 2, 3.
cShowthatk1k2k3 1 . v1 v2 v3
k1
v2v3
v1 v2 v3
k2v1v2
v3v1
v1 v2 v3
k3
v1 v2 v3
THE GEOMETRY OF A TETRAHEDRON
DISCOVERY PROJECT
Q
P
S
R
A tetrahedron is a solid with four vertices, P, Q, R, and S, and four triangular faces as shown in the figure.
1. Let v1, v2, v3, and v4 be vectors with lengths equal to the areas of the faces opposite the vertices P, Q, R, and S, respectively, and directions perpendicular to the respective faces and pointing outward. Show that
v1 v2 v3 v4 0
2. The volume V of a tetrahedron is onethird the distance from a vertex to the opposite face, times the area of that face.
a Find a formula for the volume of a tetrahedron in terms of the coordinates of its vertices
P, Q, R, and S.
b Find the volume of the tetrahedron whose vertices are P1, 1, 1, Q1, 2, 3, R1, 1, 2,
and S3, 1, 2.
3. Suppose the tetrahedron in the figure has a trirectangular vertex S. This means that the three angles at S are all right angles. Let A, B, and C be the areas of the three faces that meet at S, and let D be the area of the opposite face PQR. Using the result of Problem 1, or otherwise, show that
D2 A2 B2 C2
This is a threedimensional version of the Pythagorean Theorem.
12.5
L
Px, y, z
Ov FIGURE 1
x
z
P x , y , z
a rr
EQUATIONS OF LINES AND PLANES
A line in the xyplane is determined when a point on the line and the direction of the line its slope or angle of inclination are given. The equation of the line can then be written using the pointslope form.
Likewise, a line L in threedimensional space is determined when we know a point P0x0, y0, z0 on L and the direction of L. In three dimensions the direction of a line is con veniently described by a vector, so we let v be a vector parallel to L. Let Px, y, z be an arbitrary point on L and let r0 and r be the position vectors of P0 and P that is, they have representations OAP0 and OAP . If a is the vector with representation PA0 P , as in Figure 1, then the Triangle Law for vector addition gives rr0a. But, since a and v are parallel vectors, there is a scalar t such that atv. Thus
y
1
rr0tv
z
t0
t0
t0
SECTION 12.5 EQUATIONS OF LINES AND PLANES795 which is a vector equation of L. Each value of the parameter t gives the position vector
r of a point on L. In other words, as t varies, the line is traced out by the tip of the vec tor r. As Figure 2 indicates, positive values of t correspond to points on L that lie on one side of P0, whereas negative values of t correspond to points that lie on the other side of P0.
If the vector v that gives the direction of the line L is written in component form as
v a,b,c, then we have tv ta,tb,tc. We can also write r x,y,z and
r0x0, y0, z0 , so the vector equation 1 becomes
y
L
rx
FIGURE 2
x,y,zx0 ta,y0 tb,z0 tc
Two vectors are equal if and only if corresponding components are equal. Therefore we
have the three scalar equations:
where t. These equations are called parametric equations of the line L through the point P0x0, y0, z0 and parallel to the vector va, b, c. Each value of the parameter t gives a point x, y, z on L.
EXAMPLE 1
a Find a vector equation and parametric equations for the line that passes through the point 5, 1, 3 and is parallel to the vector i4j2k.
b Find two other points on the line.
SOLUTION
a Here r05, 1, 35ij3k and vi4j2k, so the vector equa tion 1 becomes
r5ij3kti4j2k
or r5t i14t j32t k Parametric equations are
x5t y14t z32t
b Choosing the parameter value t1 gives x6, y5, and z1, so 6, 5, 1 is a
point on the line. Similarly, t1 gives the point 4, 3, 5. M
The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. For instance, if, instead of 5, 1, 3, we choose the point 6, 5, 1 in Example 1, then the para metric equations of the line become
x6t y54t z12t
Or, if we stay with the point 5, 1, 3 but choose the parallel vector 2i8j4k, we
arrive at the equations
x52t y18t z34t
In general, if a vector va, b, c is used to describe the direction of a line L, then the numbers a, b, and c are called direction numbers of L. Since any vector parallel to v
N Figure 3 shows the line L in Example 1 and its relation to the given point and to the vector that gives its direction.
L
5, 1, 3
x
r
z
vi4j2k
FIGURE 3
y
2
xx0 at yy0 bt zz0 ct
796
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
3
xx0yy0zz0 abc
N Figure 4 shows the line L in Example 2 and the point P where it intersects the xyplane.
could also be used, we see that any three numbers proportional to a, b, and c could also be used as a set of direction numbers for L.
Another way of describing a line L is to eliminate the parameter t from Equations 2. If none of a, b, or c is 0, we can solve each of these equations for t, equate the results, and obtain
These equations are called symmetric equations of L. Notice that the numbers a, b, and c that appear in the denominators of Equations 3 are direction numbers of L, that is, com ponents of a vector parallel to L. If one of a, b, or c is 0, we can still eliminate t. For instance, if a0, we could write the equations of L as
yy0 zz0 xx0 bc
This means that L lies in the vertical plane xx0.
EXAMPLE 2
a Find parametric equations and symmetric equations of the line that passes through the points A2, 4, 3 and B3, 1, 1.
b At what point does this line intersect the xyplane?
SOLUTION
z 1
B124 x P1
a We are not explicitly given a vector parallel to the line, but observe that the vector v l
y
with representation AB is parallel to the line and
v32, 14, 131, 5, 4
FIGURE 4
L
A
Thus direction numbers are a1, b5, and c4. Taking the point 2, 4, 3 as P0, we see that parametric equations 2 are
x2t y45t z34t and symmetric equations 3 are
x2y4z3 1 5 4
b The line intersects the xyplane when z0, so we put z0 in the symmetric equa tions and obtain
x2y43 1 5 4
This gives x11 and y1, so the line intersects the xyplane at the point 11, 1 , 0. M 4444
In general, the procedure of Example 2 shows that direction numbers of the line L through the points P0x0, y0, z0and P1x1, y1, z1 are x1x0, y1y0, and z1z0 and so symmetric equations of L are
xx0yy0zz0 x1 x0 y1 y0 z1 z0
N The lines L1 and L 2 in Example 3, shown in Figure 5, are skew lines.
z L5
5
5 10
xy 5
Often, we need a description, not of an entire line, but of just a line segment. How, for instance, could we describe the line segment AB in Example 2? If we put t0 in the para metric equations in Example 2a, we get the point 2, 4, 3 and if we put t1 we get 3, 1, 1. So the line segment AB is described by the parametric equations
x2t y45t z34t 0t1 or by the corresponding vector equation
rt2t, 45t, 34t 0t1
In general, we know from Equation 1 that the vector equation of a line through the tip of the vector r0 in the direction of a vector v is rr0tv. If the line also passes through the tip of r1, then we can take vr1r0 and so its vector equation is
rr0 tr1 r01tr0 tr1
The line segment from r0 to r1 is given by the parameter interval 0t1.
V EXAMPLE 3 Show that the lines L1 and L2 with parametric equations x1t y23t z4t
x2s y3s z34s
are skew lines; that is, they do not intersect and are not parallel and therefore do not lie
in the same plane.
SOLUTION The lines are not parallel because the corresponding vectors 1, 3, 1 and 2, 1, 4 are not parallel. Their components are not proportional. If L1 and L2 had a
point of intersection, there would be values of t and s such that 1 t2s
23t3s 4 t34s
But if we solve the first two equations, we get t11 and s8, and these values dont 55
satisfy the third equation. Therefore there are no values of t and s that satisfy the three equations, so L 1 and L 2 do not intersect. Thus L 1 and L 2 are skew lines. M
PLANES
Although a line in space is determined by a point and a direction, a plane in space is more difficult to describe. A single vector parallel to a plane is not enough to convey the direction of the plane, but a vector perpendicular to the plane does completely specify its direction. Thus a plane in space is determined by a point P0x0, y0, z0 in the plane and a vector n that is orthogonal to the plane. This orthogonal vector n is called a normal vector. Let Px, y, z be an arbitrary point in the plane, and let r0 and r be the position vectors of P0 and P. Then the vector rr0 is represented by PA0 P. See Figure 6. The nor mal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal
Px, y, z r rr
SECTION 12.5 EQUATIONS OF LINES AND PLANES797
LTM
FIGURE 5
z
n
0 r
4
The line segment from r0 to r1 is given by the vector equation rt1tr0 tr1 0t1
P x , y , zxy
FIGURE 6
798
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE torr0 andsowehave
5
6
7
nrnr0
axx0byy0czz00
z
0,0,3
0, 4, 0
which can be rewritten as
Either Equation 5 or Equation 6 is called a vector equation of the plane.
To obtain a scalar equation for the plane, we write na, b, c, rx, y, z, and
r0x0, y0, z0 . Then the vector equation 5 becomes
a,b,cxx0,yy0,zz0 0
or
Equation 7 is the scalar equation of the plane through P0x0, y0, z0with normal vector na, b, c.
V EXAMPLE 4 Find an equation of the plane through the point 2, 4, 1 with normal vector n2, 3, 4. Find the intercepts and sketch the plane.
SOLUTION Puttinga2,b3,c4,x0 2,y0 4,andz0 1inEquation7,we see that an equation of the plane is
2x23y44z10 or 2x3y4z12
To find the xintercept we set yz0 in this equation and obtain x6. Similarly,
the yintercept is 4 and the zintercept is 3. This enables us to sketch the portion of the plane that lies in the first octant see Figure 7. M
By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation of a plane as
where dax0by0cz0 . Equation 8 is called a linear equation in x, y, and z. Conversely, it can be shown that if a, b, and c are not all 0, then the linear equation 8 rep resents a plane with normal vector a, b, c. See Exercise 77.
EXAMPLE 5 Find an equation of the plane that passes through the points P1, 3, 2,
6,0,0 y x
FIGURE 7
8
Q3, 1, 6, and R5, 2, 0.
SOLUTION The vectors a and b corresponding to PQ and PR are
a2, 4, 4 b4, 1, 2
nrr0 0
axbyczd0
ll
N Figure 8 shows the portion of the plane in Example 5 that is enclosed by triangle PQR.
z
Q3, 1, 6
P1,3,2
SECTION 12.5 EQUATIONS OF LINES AND PLANES799 Since both a and b lie in the plane, their cross product ab is orthogonal to the plane
and can be taken as the normal vector. Thus
y R5, 2, 0
FIGURE 8
With the point P1, 3, 2 and the normal vector n, an equation of the plane is 12x120y314z20
x
FIGURE 9
nTM n
SOLUTION We substitute the expressions for x, y, and z from the parametric equations into the equation of the plane:
423t54t25t18
This simplifies to 10t20, so t2. Therefore the point of intersection occurs when the parameter value is t2. Then x2324, y428,
z523 and so the point of intersection is 4, 8, 3. M
Two planes are parallel if their normal vectors are parallel. For instance, the planes x2y3z4 and 2x4y6z3 are parallel because their normal vectors are n11, 2, 3 and n22, 4, 6 and n22n1. If two planes are not parallel, then they intersect in a straight line and the angle between the two planes is defined as the acute angle between their normal vectors see angle in Figure 9.
V EXAMPLE 7
a Findtheanglebetweentheplanesxyz1andx2y3z1. b Find symmetric equations for the line of intersection L of these two planes.
SOLUTION
a The normal vectors of these planes are
N Figure 10 shows the planes in Example 7 and their line of intersection L.
x2y3z1 L
ijk
nab 2 4 4 12i20j14k
4 1 2
or
EXAMPLE 6 Find the point at which the line with parametric equations x23t,
6x10y7z50 M y4t, z5t intersects the plane 4x5y2z18.
xyz1
6 4
z 20 2 4
2 0
FIGURE 10cos1 s4272
b We first need to find a point on L. For instance, we can find the point where the line intersects the xyplane by setting z0 in the equations of both planes. This gives the equations xy1 and x2y1, whose solution is x1, y0. So the point
1, 0, 0 lies on L.
and so, if
n11, 1, 1 n21, 2, 3
is the angle between the planes, Corollary 12.3.6 gives
n1 n2 111213 2 cosn1n2 s111s149s42
0 2
y22x 2
800CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
Now we observe that, since L lies in both planes, it is perpendicular to both of the
N Another way to find the line of intersection is to solve the equations of the planes for two of the variables in terms of the third, which can be taken as the parameter.
ijk
vn1n2 1 1 1 5i2j3k
2 1 z0 1 2
FIGURE 11
x1 y 5 2
and so the symmetric equations of L can be written as
x1yz M
5 2 3
NOTE Since a linear equation in x, y, and z represents a plane and two nonparallel planes intersect in a line, it follows that two linear equations can represent a line. The pointsx,y,zthatsatisfybotha1xb1yc1zd1 0anda2xb2yc2zd2 0 lie on both of these planes, and so the pair of linear equations represents the line of inter section of the planes if they are not parallel. For instance, in Example 7 the line L was givenasthelineofintersectionoftheplanesxyz1andx2y3z1.The symmetric equations that we found for L could be written as
x1 y and yz 52 23
which is again a pair of linear equations. They exhibit L as the line of intersection of the planes x15y2 and y2z3. See Figure 11.
In general, when we write the equations of a line in the symmetric form
xx0yy0zz0 abc
we can regard the line as the line of intersection of the two planes
xx0yy0 and yy0zz0 ab bc
EXAMPLE 8 Find a formula for the distance D from a point P1x1, y1, z1 to the plane axbyczd0.
SOLUTION Let P0x0, y0, z0be any point in the given plane and let b be the vector corresponding to PA0 P1. Then
bx1 x0,y1 y0,z1 z0
From Figure 12 you can see that the distance D from P1 to the plane is equal to the absolute value of the scalar projection of b onto the normal vector na, b, c. See Section 12.3. Thus
2y3z 1 0
1012 y12x
N Figure 11 shows how the line L in Example 7 can also be regarded as the line of intersection of planes derived from its symmetric equations.
P
FIGURE 12
Dcompnb n
ax1 x0by1 y0cz1 z0
sa2 b2 c2
ax1 by1 cz1ax0 by0 cz0 sa2 b2 c2
P
L
normal vectors. Thus a vector v parallel to L is given by the cross product
bnD nb
1 2 3
SECTION 12.5 EQUATIONS OF LINES AND PLANES801 Since P0 lies in the plane, its coordinates satisfy the equation of the plane and so we
haveax0 by0 cz0 d0.ThustheformulaforDcanbewrittenas
EXAMPLE 9 Find the distance between the parallel planes 10x2y2z5 and 5xyz1.
SOLUTION First we note that the planes are parallel because their normal vectors
M
9
D ax1 by1 cz1 d sa2 b2 c2
10, 2, 2 and 5, 1, 1 are parallel. To find the distance D between the planes,
we choose any point on one plane and calculate its distance to the other plane. In par
ticular, if we put yz0 in the equation of the first plane, we get 10×5 and so
1 , 0, 0 is a point in this plane. By Formula 9, the distance between 1 , 0, 0 and the 22
plane 5xyz10 is
51 10101
3 s3 2
D s521212 So the distance between the planes is s36.
3s3 6 z4t
z34s
M
2
EXAMPLE 10 In Example 3 we showed that the lines L1: x1t y23t
L2: x2s y3s are skew. Find the distance between them.
SOLUTION Since the two lines L1 and L2 are skew, they can be viewed as lying on two parallel planes P1 and P2. The distance between L1 and L2 is the same as the distance between P1 and P2, which can be computed as in Example 9. The common normal vector to both planes must be orthogonal to both v1 1, 3, 1the direction of L 1and v22, 1, 4 the direction of L2 . So a normal vector is
ijk
nv1 v2 1 3 113i6j5k
214
If we put s0 in the equations of L2, we get the point 0, 3, 3 on L2 and so an equa
tion for P2 is
13x06y35z30 or 13x6y5z30
If we now set t0 in the equations for L1, we get the point 1, 2, 4 on P1. So the distance between L1 and L2 is the same as the distance from 1, 2, 4 to
13x6y5z30. By Formula 9, this distance is
D 13162543 8 0.53 M s1326252 s230
802CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
12.5 EXERCISES
1. Determine whether each statement is true or false.
a Two lines parallel to a third line are parallel.
b Two lines perpendicular to a third line are parallel. c Two planes parallel to a third plane are parallel.
d Two planes perpendicular to a third plane are parallel. e Two lines parallel to a plane are parallel.
f Two lines perpendicular to a plane are parallel.
g Two planes parallel to a line are parallel.
h Two planes perpendicular to a line are parallel.
iTwo planes either intersect or are parallel.
j Two lines either intersect or are parallel.
k A plane and a line either intersect or are parallel.
25 Find a vector equation and parametric equations for the line. 2. The line through the point 6, 5, 2 and parallel to the
16. a Find parametric equations for the line through 2, 4, 6 that is perpendicular to the plane xy3z7.
b In what points does this line intersect the coordinate planes?
17. Find a vector equation for the line segment from 2, 1, 4 to 4, 6, 1.
18. Find parametric equations for the line segment from 10, 3, 1 to 5, 6, 3.
1922 Determine whether the lines L1 and L2 are parallel, skew, or intersecting. If they intersect, find the point of intersection.
vector 1, 3, 2L : 32
L1:
L2: 20. L1:
x6t, y19t, z3t x12s, y43s, zs
x12t, y3t, z2t x1s, y4s, z13s
3. The line through the point 2, 2.4, 3.5 and parallel to the 21. L1: xy1z2 vector3i2jk 123
4. The line through the point 0, 14, 10 and parallel to the line L2: x3y2z1 x12t,y63t,z39t 4 3 2
5. The line through the point 1, 0, 6 and perpendicular to the 22. L1: x1y3z2 planex3yz5 2 2 1
L2: x2y6z2
612 Find parametric equations and symmetric equations for the 1 1 3
line.
6. The line through the origin and the point 1, 2, 3
24. The plane through the point 4, 0, 3 and with normal vector j2k
25. The plane through the point 1, 1, 1 and with normal vector ijk
26. The plane through the point 2, 8, 10 and perpendicular to the line x1t, y2t, z43t
27. The plane through the origin and parallel to the plane 2xy3z1
28. The plane through the point 1, 6, 5 and parallel to the plane xyz20
29. The plane through the point 4, 2, 3 and parallel to the plane 3x7z12
30. Theplanethatcontainsthelinex32t,yt,z8t and is parallel to the plane 2x4y8z17
The plane through the points 0, 1, 1, 1, 0, 1, and 1, 1, 0
7. The line through the points 1, 3, 2 and 4, 3, 0 23. The plane through the point 6, 3, 2 and perpendicular to the
8. The line through the points 6, 1, 3 and 2, 4, 5 The line through the points 0, 1 , 1 and 2, 1, 3
9.
2338 Find an equation of the plane. vector 2, 1, 5
2
10. The line through 2, 1, 0 and perpendicular to both ij
and jk
11. The line through 1, 1, 1 and parallel to the line
x21yz3 2
12. Thelineofintersectionoftheplanesxyz1 and xz0
Is the line through 4, 6, 1 and 2, 0 3 parallel to the line through 10, 18, 4 and 5, 3, 14?
14. Is the line through 4, 1, 1 and 2, 5, 3 perpendicular to the line through 3, 2, 0 and 5, 1, 4?
15. a Find symmetric equations for the line that passes through the point 1, 5, 6 and is parallel to the vector
1, 2, 3.
13.
b Find the points in which the required line in part a inter 32. The plane through the origin and the points 2, 4, 6 sects the coordinate planes. and 5, 1, 3
19.
31.
33. The plane through the points 3, 1, 2, 8, 2, 4, and 1, 2, 3
34. The plane that passes through the point 1, 2, 3 and contains the line x3t, y1t, z2t
35. The plane that passes through the point 6, 0, 2 and contains the line x42t, y35t, z74t
36. The plane that passes through the point 1, 1, 1 and contains the line with symmetric equations x2y3z
37. The plane that passes through the point 1, 2, 1 and contains the line of intersection of the planes xyz2 and
2xy3z1
38. The plane that passes through the line of intersection of the planes xz1 and y2z3 and is perpendicular to the planexy2z1
3942 Use intercepts to help sketch the plane.
SECTION 12.5 EQUATIONS OF LINES AND PLANES803 5758 Find symmetric equations for the line of intersection of the
planes.
57. 5x2y2z1, 4xyz6 58. z2xy5, z4x3y5
59. Find an equation for the plane consisting of all points that are equidistant from the points 1, 0, 2 and 3, 4, 0.
60. Find an equation for the plane consisting of all points that are equidistant from the points 2, 5, 5 and 6, 3, 1.
Find an equation of the plane with xintercept a, yintercept b, and zintercept c.
62. a Find the point at which the given lines intersect: r 1, 1, 0 t1, 1, 2
r2, 0, 2s1, 1, 0
b Find an equation of the plane that contains these lines.
63. Find parametric equations for the line through the point 0, 1, 2 that is parallel to the plane xyz2 and perpendicular to the line x1t, y1t, z2t.
64. Find parametric equations for the line through the point 0, 1, 2 that is perpendicular to the line x1t,
y1t, z2t and intersects this line.
65. Which of the following four planes are parallel? Are any of them identical?
61.
39. 2x5yz10 41. 6x3y4z6
40. 3xy2z6 42. 6x5y3z15
4345 Find the point at which the line intersects the given plane.
43. x3t, y2t, z5t; xy2z9
44. x12t, y4t, z23t; x2yz10
45. xy12z; 4xy3z8
46. Where does the line through 1, 0, 1 and 4, 2, 2 intersect the plane xyz6?
47. Find direction numbers for the line of intersection of the planes xyz1 and xz0.
48. Find the cosine of the angle between the planes xyz0 and x2y3z1.
4954 Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them.
x4y3z1, 3x6y7z0
50. 2z4yx, 3x12y6z1
51. xyz1, xyz1
52. 2x3y4z5, x6y4z3
53. x4y2z, 8y12x4z
54. x2y2z1, 2xy2z1
5556 a Find parametric equations for the line of intersection of the planes and b find the angle between the planes.
55. xyz1, x2y2z1 56. 3x2yz1, 2xy3z3
P1: 4x2y6z3
P3: 6x3y9z5
66. Which of the following four lines are parallel? Are any of them
identical?
L1: x1t, yt, z25t
L2: x1y21z
L3: x1t, y4t, z1t L4: r2, 1, 3t2, 2, 10
6768 Use the formula in Exercise 43 in Section 12.4 to find the distance from the point to the given line.
49.
P2: 4x2y2z6
P4: z2xy3
67. 4, 1, 2;
68. 0, 1, 3;
x1t, y32t, z43t x2t, y62t, z3t
6970 Find the distance from the point to the given plane.
69. 1, 2, 4, 3x2y6z5
70. 6, 3, 5, x2y4z8
7172 Find the distance between the given parallel planes. 71. 2x3yz4, 4x6y2z3
804CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 72. 6z4y2x, 9z13x6y
73. Show that the distance between the parallel planes axbyczd1 0andaxbyczd2 0is
Dd1d2sa2 b2 c2
74. Find equations of the planes that are parallel to the plane x2y2z1 and two units away from it.
75. Show that the lines with symmetric equations xyz and x1y2z3 are skew, and find the distance between these lines.
76. Find the distance between the skew lines with para metric equations x1t, y16t, z2t, and x12s, y515s, z26s.
77. Ifa,b,andcarenotall0,showthattheequation axbyczd0representsaplaneand a,b,c is a normal vector to the plane.
Hint: Suppose a0 and rewrite the equation in the form ax dby0cz00
a
78. Give a geometric description of each family of planes. a xyzc b xycz1
c ycos zsin 1
L A B O R AT O R Y PROJECT
PUTTING 3D IN PERSPECTIVE
Computer graphics programmers face the same challenge as the great painters of the past: how to represent a threedimensional scene as a flat image on a twodimensional plane a screen or a canvas. To create the illusion of perspective, in which closer objects appear larger than those farther away, threedimensional objects in the computers memory are projected onto a rect angular screen window from a viewpoint where the eye, or camera, is located. The viewing volumethe portion of space that will be visibleis the region contained by the four planes that pass through the viewpoint and an edge of the screen window. If objects in the scene extend beyond these four planes, they must be truncated before pixel data are sent to the screen. These planes are therefore called clipping planes.
1. Suppose the screen is represented by a rectangle in the yzplane with vertices 0, 400, 0 and 0, 400, 600, and the camera is placed at 1000, 0, 0. A line L in the scene passes through the points 230, 285, 102 and 860, 105, 264. At what points should L be clipped by the clipping planes?
2. If the clipped line segment is projected on the screen window, identify the resulting line segment.
3. Use parametric equations to plot the edges of the screen window, the clipped line segment, and its projection on the screen window. Then add sight lines connecting the viewpoint to each end of the clipped segments to verify that the projection is correct.
4. A rectangle with vertices 621, 147, 206, 563, 31, 242, 657, 111, 86, and
599, 67, 122 is added to the scene. The line L intersects this rectangle. To make the rect angle appear opaque, a programmer can use hidden line rendering, which removes portions of objects that are behind other objects. Identify the portion of L that should be removed.
CYLINDERS AND QUADRIC SURFACES
We have already looked at two special types of surfaces: planes in Section 12.5 and spheres in Section 12.1. Here we investigate two other types of surfaces: cylinders and quadric surfaces.
In order to sketch the graph of a surface, it is useful to determine the curves of intersec tion of the surface with planes parallel to the coordinate planes. These curves are called traces or crosssections of the surface.
12.6
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES805
z
xy
FIGURE 1
The surface z is a parabolic cylinder.
CYLINDERS
A cylinder is a surface that consists of all lines called rulings that are parallel to a given line and pass through a given plane curve.
V EXAMPLE 1 Sketch the graph of the surface zx2.
SOLUTION Notice that the equation of the graph, zx2, doesnt involve y. This means that any vertical plane with equation yk parallel to the xzplane intersects the graph in a curve with equation zx2. So these vertical traces are parabolas. Figure 1 shows how the graph is formed by taking the parabola zx2 in the xzplane and moving it in the direction of the yaxis. The graph is a surface, called a parabolic cylinder, made up of infinitely many shifted copies of the same parabola. Here the rulings of the cylinder are parallel to the yaxis. M
We noticed that the variable y is missing from the equation of the cylinder in Exam ple 1. This is typical of a surface whose rulings are parallel to one of the coordinate axes. If one of the variables x, y, or z is missing from the equation of a surface, then the surface is a cylinder.
EXAMPLE 2 Identify and sketch the surfaces.
a x2 y2 1 b y2 z2 1
SOLUTION
a Since z is missing and the equations x2y21, zk represent a circle with radius 1 in the plane zk, the surface x2y21 is a circular cylinder whose axis is the zaxis. See Figure 2. Here the rulings are vertical lines.
b In this case x is missing and the surface is a circular cylinder whose axis is the xaxis. See Figure 3. It is obtained by taking the circle y2z21, x0 in the yzplane and moving it parallel to the xaxis.
0
zz
y
FIGURE 3 z1
NOTE When you are dealing with surfaces, it is important to recognize that an equa tion like x2y21 represents a cylinder and not a circle. The trace of the cylinder x2 y2 1inthexyplaneisthecirclewithequationsx2 y2 1,z0.
QUADRIC SURFACES
A quadric surface is the graph of a seconddegree equation in three variables x, y, and z. The most general such equation is
Ax2 By2 Cz2 DxyEyzFxzGxHyIzJ0
0
x
y
x
FIGURE 2 1
M
806
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
where A, B, C, . . . , J are constants, but by translation and rotation it can be brought into
z
0,0,2
z2
2z2 k2
one of the two standard forms
Ax2 By2 Cz2 J0 or Ax2 By2 Iz0
Quadric surfaces are the counterparts in three dimensions of the conic sections in the plane. See Section 10.5 for a review of conic sections.
EXAMPLE 3 Use traces to sketch the quadric surface with equation 2 y2 z2
SOLUTION By substituting z0, we find that the trace in the xyplane is x2y291, which we recognize as an equation of an ellipse. In general, the horizontal trace in the plane zk is
2y2 k2 x914 zk
which is an ellipse, provided that k24, that is, 2k2. Similarly, the vertical traces are also ellipses:
x941
y2 94
1k2
xk yk
if 1k1 if3k3
1,0,0
x
FIGURE 4
0 0, 3, 0
y
x4 1 9
y z
The ellipsoid94 1
Figure 4 shows how drawing some traces indicates the shape of the surface. Its called an ellipsoid because all of its traces are ellipses. Notice that it is symmetric with respect to each coordinate plane; this is a reflection of the fact that its equation involves only even powers of x, y, and z. M
EXAMPLE 4 Use traces to sketch the surface z4×2y2.
SOLUTION If we put x0, we get zy2, so the yzplane intersects the surface in a parabola. If we put xk a constant, we get zy24k2. This means that if we slice the graph with any plane parallel to the yzplane, we obtain a parabola that opens upward. Similarly, if yk, the trace is z4×2k2, which is again a parabola that opens upward. If we put zk, we get the horizontal traces 4×2y2k, which we recognize as a family of ellipses. Knowing the shapes of the traces, we can sketch the graph in Figure 5. Because of the elliptical and parabolic traces, the quadric surface
z4×2y2 is called an elliptic paraboloid.
z
0
FIGURE 5
The surface z4 is an elliptic paraboloid. Horizontal traces are ellipses;
vertical traces are parabolas. x
y M
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES807 V EXAMPLE 5 Sketch the surface zy2x2.
SOLUTION The traces in the vertical planes xk are the parabolas zy2k2, which open upward. The traces in yk are the parabolas zx2k2, which open down ward. The horizontal traces are y2x2k, a family of hyperbolas. We draw the fami lies of traces in Figure 6, and we show how the traces appear when placed in their correct planes in Figure 7.
0
1
1 y 0 x 0 x
2
Traces in xk are zk
z 2z y
1 1 1
FIGURE6
Vertical traces are parabolas; horizontal traces are hyperbolas. All traces are labeled with the value of k.
1 Traces in zk are k
Traces in yk are zk zzz
1 y y 0y
FIGURE 7
Traces moved to their correct planes
TEC In Module 12.6A you can investi gate how traces determine the shape of a surface.
FIGURE 8
The surface z is a hyperbolic paraboloid.
0
1 0 1 1
x 1x x
1
Traces in yk Traces in zk
In Figure 8 we fit together the traces from Figure 7 to form the surface zy2x2,
a hyperbolic paraboloid. Notice that the shape of the surface near the origin resembles that of a saddle. This surface will be investigated further in Section 14.7 when we dis cuss saddle points.
z
0
y
x
Traces in xk
z2
SOLUTION The trace in any horizontal plane zk is the ellipse
x2 k2
4 y2 1 4 zk
x2
EXAMPLE 6 Sketch the surface 4y241.
M
808
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
but the traces in the xz and yzplanes are the hyperbolas
x
z
2, 0, 0 xy
FIGURE 9
The idea of using traces to draw a surface is employed in threedimensional graphing software for computers. In most such software, traces in the vertical planes xk and yk are drawn for equally spaced values of k, and parts of the graph are eliminated using hidden line removal. Table 1 shows computerdrawn graphs of the six basic types of quadric surfaces in standard form. All surfaces are symmetric with respect to the zaxis. If a quadric surface is symmetric about a different axis, its equation changes accordingly.
Surface
Equation
Surface
Cone
z
xy
Equation
z2 x2 y2 c2 a2 b2
Horizontal traces are ellipses.
Vertical traces in the planes xk and yk are hyperbolas if k0 but are pairs of lines if k0.
x2 y2 z2
a2 b2 c2 1
Horizontal traces are ellipses. Vertical traces are hyperbolas.
The axis of symmetry corresponds to the variable whose coefficient is negative.
x2 y2 z2 a2 b2 c2 1
Horizontal traces in zk are ellipses if kc or kc.
Vertical traces are hyperbolas.
The two minus signs indicate two sheets.
y
z
x
z
0, 1, 0
x0
This surface is called a hyperboloid of one sheet and is sketched in Figure 9. M
Graphs of quadric surfaces
TABLE 1
Ellipsoid
z xy
Elliptic Paraboloid
z
Hyperbolic Paraboloid
y
y
x2
a2 b2 c2 1
y2 z2
All traces are ellipses.
x2 z2 z2
441 y0 and y241
If abc, the ellipsoid is a sphere.
z x 2 y 2 ca2b2
Horizontal traces are ellipses. Vertical traces are parabolas.
The variable raised to the first power indicates the axis of the paraboloid.
z x 2 y 2 ca2b2
Horizontal traces are hyperbolas.
Vertical traces are parabolas.
The case where c0 is x illustrated.
Hyperboloid of One Sheet
z
x
Hyperboloid of Two Sheets
y
TEC In Module 12.6B you can see how changing a, b, and c in Table 1 affects the shape of the quadric surface.
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES809 V EXAMPLE 7 Identify and sketch the surface 4×2y22z240.
SOLUTION Dividing by 4, we first put the equation in standard form: 2 y2 z2
Comparing this equation with Table 1, we see that it represents a hyperboloid of two sheets, the only difference being that in this case the axis of the hyperboloid is the yaxis. The traces in the xy and yzplanes are the hyperbolas
2 y2 y2 z2
x41 z0 and 421 x0
The surface has no trace in the xzplane, but traces in the vertical planes yk for k 2 are the ellipses
x421
z
0,2,0
0
which can be written as
2 z2 k2 x241 yk
x2z2
k2 k2 1yk
0,2,0 y 42z40
x
FIGURE 10
41241
These traces are used to make the sketch in Figure 10. M
EXAMPLE 8 Classify the quadric surface x22z26xy100. SOLUTION By completing the square we rewrite the equation as
y1x32 2z2
Comparing this equation with Table 1, we see that it represents an elliptic paraboloid. Here, however, the axis of the paraboloid is parallel to the yaxis, and it has been shifted so that its vertex is the point 3, 1, 0. The traces in the plane yk k1 are the ellipses
x32 2z2 k1 yk
The trace in the xyplane is the parabola with equation y1x32, z0. The
paraboloid is sketched in Figure 11.
z 0
3, 1, 0
y
FIGURE 11
2z6xy100
x
M
810
CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
APPLICATIONS OF QUADRIC SURFACES
A satellite dish reflects signals to the focus of a paraboloid.
Nuclear reactors have cooling towers in the shape of hyperboloids.
Hyperboloids produce gear transmission.
6. yz4
8. x2 y2 1
Examples of quadric surfaces can be found in the world around us. In fact, the world itself is a good example. Although the earth is commonly modeled as a sphere, a more accurate model is an ellipsoid because the earths rotation has caused a flattening at the poles. See Exercise 47.
Circular paraboloids, obtained by rotating a parabola about its axis, are used to collect and reflect light, sound, and radio and television signals. In a radio telescope, for instance, signals from distant stars that strike the bowl are reflected to the receiver at the focus and are therefore amplified. The idea is explained in Problem 18 on page 268. The same prin ciple applies to microphones and satellite dishes in the shape of paraboloids.
Cooling towers for nuclear reactors are usually designed in the shape of hyperboloids of one sheet for reasons of structural stability. Pairs of hyperboloids are used to transmit rotational motion between skew axes. The cogs of gears are the generating lines of the hyperboloids. See Exercise 49.
12.6 EXERCISES
1. aWhatdoestheequationyx2 representasacurvein2?
bWhatdoesitrepresentasasurfacein3? c What does the equation zy2 represent?
2. a Sketch the graph of yex as a curve in 2. bSketchthegraphofyex asasurfacein3. c Describe and sketch the surface zey.
38 Describe and sketch the surface.
3. y2 4z2 4 4. z4x2
5. xy2 0 7. zcosx
9. a Find and identify the traces of the quadric surface
x2 y2 z2 1andexplainwhythegraphlookslikethe graph of the hyperboloid of one sheet in Table 1.
bIfwechangetheequationinpartatox2 y2 z2 1, how is the graph affected?
c What if we change the equation in part a to x2 y2 2yz2 0?
Corbis
David BurnettPhoto Researchers, Inc
10. a Find and identify the traces of the quadric surface
x2y2z21 and explain why the graph looks like the graph of the hyperboloid of two sheets in Table 1.
bIftheequationinpartaischangedtox2 y2 z2 1, what happens to the graph? Sketch the new graph.
1120 Use traces to sketch and identify the surface.
SECTION 12.6 CYLINDERS AND QUADRIC SURFACES811 2936 Reduce the equation to one of the standard forms, classify
the surface, and sketch it.
29. z2 4×2 9y2 36
31. x2y2 3z2
33. 4×2 y2 4z2 4y24z360 34. 4y2 z2 x16y4z200 35. x2 y2 z2 4x2y2z40 36. x2 y2 z2 2x2y4z20
; 3740 Use a computer with threedimensional graphing software to graph the surface. Experiment with viewpoints and with domains for the variables until you get a good view of the surface.
11. xy2 4z2
13. x2 y2 4z2
15. x2 4y2 z2 4 17. 36×2 y2 36z2 36 19. yz2 x2
12. 9×2 y2 z2 0
14. 25×2 4y2 z2 100 16. 4×2 9y2 z0
18. 4×2 16y2 z2 16 20. xy2z2
2128 Match the equation with its graph labeled IVIII. Give reasons for your choices.
37. 4×2 y2 z2 1 39. 4×2 y2 z2 0
38. x2 y2 z0
40. x2 6x4y2 z0
21. x2 4y2 9z2 1 23. x2 y2 z2 1 25. y2x2 z2
27. x2 2z2 1
22. 9×2 4y2 z2 1 24. x2 y2 z2 1 26. y2 x2 2z2
28. yx2 z2
41. Sketch the region bounded by the surfaces zsx2y2 andx2 y2 1for1z2.
42. Sketch the region bounded by the paraboloids zx 2y 2 andz2x2 y2.
43. Find an equation for the surface obtained by rotating the parabola yx2 about the yaxis.
44. Find an equation for the surface obtained by rotating the line x3y about the xaxis.
45. Find an equation for the surface consisting of all points that are equidistant from the point 1, 0, 0 and the plane x1. Identify the surface.
46. Find an equation for the surface consisting of all points P for which the distance from P to the xaxis is twice the distance from P to the yzplane. Identify the surface.
47. Traditionally, the earths surface has been modeled as a sphere, but the World Geodetic System of 1984 WGS84 uses an ellipsoid as a more accurate model. It places the center of the earth at the origin and the north pole on the positive zaxis. The distance from the center to the poles is 6356.523 km and
the distance to a point on the equator is 6378.137 km.
I z II z xyxy
III z
IV z
VII
zz
y xy
x
V z VI z
xyxy
a Find an equation of the earths surface as used by WGS84.
b Curves of equal latitude are traces in the planes zk. What is the shape of these curves?
c Meridians curves of equal longitude are traces in planes of the form ymx. What is the shape of these meridians?
y x
x
y
48. A cooling tower for a nuclear reactor is to be constructed in the shape of a hyperboloid of one sheet see the photo on
page 810. The diameter at the base is 280 m and the minimum
VIII
30. x2 2y2 3z2
32. 4xy2 4z2 0
812CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE diameter, 500 m above the base, is 200 m. Find an equation
for the tower.
49. Show that if the point a, b, c lies on the hyperbolic paraboloid zy2x2, then the lines with parametric equations
xat, ybt, zc2bat and xat,
ybt, zc2bat both lie entirely on this parabo loid. This shows that the hyperbolic paraboloid is what is called a ruled surface; that is, it can be generated by the motion of a straight line. In fact, this exercise shows that through each point on the hyperbolic paraboloid there are two
generating lines. The only other quadric surfaces that are ruled surfaces are cylinders, cones, and hyperboloids of one sheet.
50. Show that the curve of intersection of the surfaces
x2 2y2 z2 3x1and2x2 4y2 2z2 5y0 lies in a plane.
12 REVIEW
1. What is the difference between a vector and a scalar?
2. How do you add two vectors geometrically? How do you add them algebraically?
3. Ifaisavectorandcisascalar,howiscarelatedtoa geometrically? How do you find ca algebraically?
4. How do you find the vector from one point to another?
5. How do you find the dot product ab of two vectors if you
know their lengths and the angle between them? What if you know their components?
6. How are dot products useful?
7. Write expressions for the scalar and vector projections of b
onto a. Illustrate with diagrams.
8. How do you find the cross product ab of two vectors if you know their lengths and the angle between them? What if you know their components?
9. How are cross products useful?
10. a How do you find the area of the parallelogram determined
by a and b?
b How do you find the volume of the parallelepiped
determined by a, b, and c?
TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. ForanyvectorsuandvinV3,uvvu.
2. ForanyvectorsuandvinV3,uvvu.
3. ForanyvectorsuandvinV3,uvvu.
4. For any vectors u and v in V3 and any scalar k, kuvkuv.
5. For any vectors u and v in V3 and any scalar k, kuvkuv.
; 51.
11. 12. 13.
14. 15.
16.
17.
18. 19.
6. 7. 8. 9.
10.
Graph the surfaces zx2y2 and z1y2 on a common screen using the domain x 1.2, y 1.2 and observe the curve of intersection of these surfaces. Show that the projection of this curve onto the xyplane is an ellipse.
How do you find a vector perpendicular to a plane?
How do you find the angle between two intersecting planes?
Write a vector equation, parametric equations, and symmetric equations for a line.
Write a vector equation and a scalar equation for a plane.
a How do you tell if two vectors are parallel?
b How do you tell if two vectors are perpendicular? c How do you tell if two planes are parallel?
a Describe a method for determining whether three points P, Q, and R lie on the same line.
b Describe a method for determining whether four points P, Q, R, and S lie in the same plane.
a How do you find the distance from a point to a line? b How do you find the distance from a point to a plane? c How do you find the distance between two lines?
What are the traces of a surface? How do you find them?
Write equations in standard form of the six types of quadric surfaces.
For any vectors u, v, and w in V3, uvwuwvw.
For any vectors u, v, and w in V3, uvwuvw.
For any vectors u, v, and w in V3, uvwuvw.
ForanyvectorsuandvinV3,uvu0. ForanyvectorsuandvinV3,uvvuv.
CONCEPT CHECK
11. Thecrossproductoftwounitvectorsisaunitvector.
12. AlinearequationAxByCzD0representsaline
in space.
13. Thesetofpointsx,y,zx2 y2 1isacircle.
14. Ifu u1,u2 andv v1,v2,thenuv u1v1,u2v2.
EXERCISES
1. a Find an equation of the sphere that passes through the point 6, 2, 3 and has center 1, 2, 1.
b Find the curve in which this sphere intersects the yzplane.
c Find the center and radius of the sphere
x2 y2 z2 8x2y6z10
2. Copy the vectors in the figure and use them to draw each of the following vectors.
15. Ifuv0,thenu0orv0.
16. Ifuv0,thenu0orv0.
17. Ifuv0,anduv0,thenu0orv0.
18. IfuandvareinV ,then uvu v .
CHAPTER 12 REVIEW813
a ab
b ab
a
c 1a 2
d 2ab
3
7. Suppose that uvw2. Find
a uvw b uwv c vuw d uvv
8. Show that if a, b, and c are in V3, then
abbccaabc2
9. Find the acute angle between two diagonals of a cube. 10. Given the points A1, 0, 1, B2, 3, 0, C1, 1, 4, and
D0, 3, 2, find the volume of the parallelepiped with adjacent edges AB, AC, and AD.
11. a Find a vector perpendicular to the plane through the points A1, 0, 0, B2, 0, 1, and C1, 4, 3.
b Find the area of triangle ABC.
12. A constant force F3i5j10k moves an object along the line segment from 1, 0, 2 to 5, 3, 8. Find the work done if the distance is measured in meters and the force in newtons.
13. A boat is pulled onto shore using two ropes, as shown in the diagram. If a force of 255 N is needed, find the magnitude of the force in each rope.
20 255 N 30
14. Find the magnitude of the torque about P if a 50N force is applied as shown.
50 N30 40 cm
P
3. Ifuandvarethevectorsshowninthefigure,finduvand uv. Is uv directed into the page or out of it?
b
v3 45
u2 4. Calculate the given quantity if
aij2k a2a3b
b3i2jk b b
cj5k
cab
ebc
gcc
i compa b
k The angle between a and b correct to the nearest degree
5. Find the values of x such that the vectors 3, 2, x and 2x, 4, x are orthogonal.
6. Find two unit vectors that are orthogonal to both j2k and i2j3k.
d ab
f abc h abcj proja b
814CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE
1517 Find parametric equations for the line.
15. The line through 4, 1, 2 and 1, 1, 5
16. The line through 1, 0, 1 and parallel to the line 1x41yz2
b Find, correct to the nearest degree, the angle between these planes.
25. Find an equation of the plane through the line of intersection of theplanesxz1andy2z3andperpendiculartothe planexy2z1.
26. a Find an equation of the plane that passes through the points A2, 1, 1, B1, 1, 10, and C1, 3, 4.
b Find symmetric equations for the line through B that is perpendicular to the plane in part a.
c A second plane passes through 2, 0, 4 and has normal vector 2, 4, 3. Show that the acute angle between the planes is approximately 43.
d Find parametric equations for the line of intersection of the two planes.
27. Find the distance between the planes 3xy4z2 and 3xy4z24.
2836 Identify and sketch the graph of each surface.
32
17. The line through 2, 2, 4 and perpendicular to the plane 2xy5z12
1820 Find an equation of the plane.
18. The plane through 2, 1, 0 and parallel to x4y3z1
19. The plane through 3, 1, 1, 4, 0, 2, and 6, 3, 1
20. The plane through 1, 2, 2 that contains the line x2t, y3t, z13t
21. Find the point in which the line with parametric equations x2t,y13t,z4tintersectstheplane
2xyz2.
22. Find the distance from the origin to the line x1t, y2t, z12t.
23. Determine whether the lines given by the symmetric equations
x1y2z3 234
and x1y3z5 6 1 2
are parallel, skew, or intersecting.
24. a Showthattheplanesxyz1and
2x3y4z5 are neither parallel nor perpendicular.
28. x3
30. yz2
32. 4xy2z4
34. y2 z2 1×2
35. 4×2 4y2 8yz2 0 36. xy2 z2 2y4z5
29. xz
31. x2 y2 4z2
33. 4×2 y2 4z2 4
37. An ellipsoid is created by rotating the ellipse 4×2y216 about the xaxis. Find an equation of the ellipsoid.
38. A surface consists of all points P such that the distance from P to the plane y1 is twice the distance from P to the point
0, 1, 0. Find an equation for this surface and identify it.
PROBLEMS PLUS
1m
1 m
FIGURE FOR PROBLEM 1
1.
2. 3.
4.
5.
Each edge of a cubical box has length 1 m. The box contains nine spherical balls with the same radius r. The center of one ball is at the center of the cube and it touches the other eight balls. Each of the other eight balls touches three sides of the box. Thus the balls are tightly packed in the box. See the figure. Find r. If you have trouble with this problem, read about the problemsolving strategy entitled Use Analogy on page 76.
Let B be a solid box with length L, width W, and height H. Let S be the set of all points that are a distance at most 1 from some point of B. Express the volume of S in terms of L, W, and H.
LetLbethelineofintersectionoftheplanescxyzcandxcycz1, where c is a real number.
a Find symmetric equations for L.
b As the number c varies, the line L sweeps out a surface S. Find an equation for the curve
of intersection of S with the horizontal plane zt the trace of S in the plane zt.
c Find the volume of the solid bounded by S and the planes z0 and z1.
A plane is capable of flying at a speed of 180 kmh in still air. The pilot takes off from an airfield and heads due north according to the planes compass. After 30 minutes of flight time, the pilot notices that, due to the wind, the plane has actually traveled 80 km at an angle 5 east of north.
a What is the wind velocity?
b In what direction should the pilot have headed to reach the intended destination?
Suppose a block of mass m is placed on an inclined plane, as shown in the figure. The blocks descent down the plane is slowed by friction; if is not too large, friction will prevent the block from moving at all. The forces acting on the block are the weight W, where W mt t is the acceleration due to gravity; the normal force N the normal component of the reac tionary force of the plane on the block, whereN n; and the force F due to friction, which acts parallel to the inclined plane, opposing the direction of motion. If the block is at rest and is increased,Fmust also increase until ultimatelyFreaches its maximum, beyond which the block begins to slide. At this angle s , it has been observed thatFis proportional to n. Thus, when Fis maximal, we can say that F s n, where s is
called the coefficient of static friction and depends on the materials that are in contact.
a ObservethatNFW0anddeducethat s tan s.
b Suppose that, fors , an additional outside force H is applied to the block, horizontally
from the left, and letH h. If h is small, the block may still slide down the plane; if h is large enough, the block will move up the plane. Let hmin be the smallest value of h that allows the block to remain motionless so thatFis maximal.
By choosing the coordinate axes so that F lies along the xaxis, resolve each force into components parallel and perpendicular to the inclined plane and show that
hminsin mtcos n and hmin cossnmtsin
c Show that hminmt tans
Does this equation seem reasonable? Does it make sense fors ? As l 90 ?
Explain.
d Let hmax be the largest value of h that allows the block to remain motionless. In which
direction is F heading? Show that
hmax mttans
Does this equation seem reasonable? Explain.
N
F W
1m
FIGURE FOR PROBLEM 5
815
13
VECTOR FUNCTIONS
816
Tangent vectors show the direction in which a space curve proceeds at any point.
The functions that we have been using so far have been realvalued functions. We now study functions whose values are vectors because such functions are needed to describe curves and surfaces in space. We will also use vectorvalued functions to describe the motion of objects through space. In particular, we will use them to derive Keplers laws of planetary motion.
13.1 VECTOR FUNCTIONS AND SPACE CURVES
In general, a function is a rule that assigns to each element in the domain an element in the range. A vectorvalued function, or vector function, is simply a function whose domain is a set of real numbers and whose range is a set of vectors. We are most interested in vec tor functions r whose values are threedimensional vectors. This means that for every num ber t in the domain of r there is a unique vector in V3 denoted by rt. If f t, tt, and ht are the components of the vector rt, then f , t, and h are realvalued functions called the component functions of r and we can write
rt ft, tt, htftittjhtk
We usually use the letter t to denote the independent variable because it represents time in
most applications of vector functions. EXAMPLE 1 If
rtt3, ln3t, stthen the component functions are
ftt3 ttln3t htst
By our usual convention, the domain of r consists of all values of t for which the expres sion for rt is defined. The expressions t3, ln3t, and st are all defined when
3t0 and t0. Therefore the domain of r is the interval 0, 3. M
The limit of a vector function r is defined by taking the limits of its component func tions as follows.
If rt ft, tt, ht, then
lim rtlim f t, lim tt, lim ht
tla tla tla tla provided the limits of the component functions exist.
1
N If lim t l a rtL, this definition is equiva lent to saying that the length and direction of the vector rt approach the length and direction of the vector L.
Equivalently, we could have used an
functions obey the same rules as limits of realvalued functions see Exercise 43.
definition see Exercise 45. Limits of vector
EXAMPLE2 Findlimrt,wherert1t3itetj sintk. tl0 t
SOLUTION According to Definition 1, the limit of r is the vector whose components are the limits of the component functions of r:
lim rtlim1t3ilim tetjlim sin tk tl0 tl0 tl0 tl0t
ik by Equation 3.3.2 M 817
818
CHAPTER 13 VECTOR FUNCTIONS
A vector function r is continuous at a if
z
C
0 x
FIGURE 1
Pft, gt, ht rtkft, gt, htl
y
lim rtra tla
In view of Definition 1, we see that r is continuous at a if and only if its component func tions f , t, and h are continuous at a.
There is a close connection between continuous vector functions and space curves. Suppose that f , t, and h are continuous realvalued functions on an interval I. Then the set C of all points x, y, z in space, where
xf t ytt zht
and t varies throughout the interval I, is called a space curve. The equations in 2 are called parametric equations of C and t is called a parameter. We can think of C as being traced out by a moving particle whose position at time t isf t, tt, ht. If we now con sider the vector function rt f t, tt, ht, then rt is the position vector of the point P f t, tt, ht on C. Thus any continuous vector function r defines a space curve C that is traced out by the tip of the moving vector rt, as shown in Figure 1.
V EXAMPLE 3 Describe the curve defined by the vector function rt1t, 25t, 16t
SOLUTION The corresponding parametric equations are
x1t y25t z16t
which we recognize from Equations 12.5.2 as parametric equations of a line passing through the point 1, 2, 1 and parallel to the vector 1, 5, 6. Alternatively, we could observe that the function can be written as rr0tv, where r01, 2, 1 and
v1, 5, 6, and this is the vector equation of a line as given by Equation 12.5.1. M
Plane curves can also be represented in vector notation. For instance, the curve given by the parametric equations xt22t and yt1 see Example 1 in Section 10.1 could also be described by the vector equation
rt t2 2t,t1 t2 2tit1j where i1, 0 and j0, 1.
V EXAMPLE 4 Sketch the curve whose vector equation is rtcos t isin t jt k
SOLUTION The parametric equations for this curve are
xcos t ysin t zt
Since x2y2cos2tsin2t1, the curve must lie on the circular cylinder
x2y21. The point x, y, z lies directly above the point x, y, 0, which moves counterclockwise around the circle x2y21 in the xyplane. See Example 2 in Section 10.1. Since zt, the curve spirals upward around the cylinder as t increases. The curve, shown in Figure 2, is called a helix. M
C is traced out by the tip of a moving position vector rt.
TEC Visual 13.1A shows several curves being traced out by position vectors, including those in Figures 1 and 2.
z
0, 1, 2
2
x
FIGURE 2
1, 0, 0
y
FIGURE 3
N Figure 4 shows the line segment PQ in Example 5.
z
Q2, 1, 3
x
FIGURE 4
The corkscrew shape of the helix in Example 4 is familiar from its occurrence in coiled springs. It also occurs in the model of DNA deoxyribonucleic acid, the genetic material of living cells. In 1953 James Watson and Francis Crick showed that the structure of the DNA molecule is that of two linked, parallel helixes that are intertwined as in Figure 3.
In Examples 3 and 4 we were given vector equations of curves and asked for a geo metric description or sketch. In the next two examples we are given a geometric descrip tion of a curve and are asked to find parametric equations for the curve.
EXAMPLE 5 Find a vector equation and parametric equations for the line segment that joins the point P1, 3, 2 to the point Q2, 1, 3.
SOLUTION In Section 12.5 we found a vector equation for the line segment that joins the tip of the vector r0 to the tip of the vector r1:
rt1tr0 tr1 0t1
See Equation 12.5.4. Here we take r01, 3, 2 and r12, 1, 3 to obtain a
P1,3,2
y
0t1 0t1
SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES819
vector equation of the line segment from P to Q:
rt1t1, 3, 2t2, 1, 3
or rt1t, 34t, 25t The corresponding parametric equations are
x1t y34t z25t
V EXAMPLE 6 Find a vector function that represents the curve of intersection of the
cylinderx2 y2 1andtheplaneyz2.
SOLUTION Figure 5 shows how the plane and the cylinder intersect, and Figure 6 shows the
curve of intersection C, which is an ellipse. zz
yz2
1
0, 1, 3
C
1,0,2
1,0,2
0, 1, 1
0 xyxy
FIGURE 5 FIGURE 6
0t1 M
820
CHAPTER 13 VECTOR FUNCTIONS
The projection of C onto the xyplane is the circle x2y21, z0. So we know
from Example 2 in Section 10.1 that we can write
xcos t ysin t 0t2 From the equation of the plane, we have
x
FIGURE 7 A toroidal spiral z
y
z2y2sin t So we can write parametric equations for C as
xcos t ysin t z2sin t The corresponding vector equation is
0t2
rtcos t isin t j2sin t k
This equation is called a parametrization of the curve C. The arrows in Figure 6 indicate
0t2
the direction in which C is traced as the parameter t increases. M
USING COMPUTERS TO DRAW SPACE CURVES
z Space curves are inherently more difficult to draw by hand than plane curves; for an accu rate representation we need to use technology. For instance, Figure 7 shows a computer
generated graph of the curve with parametric equations
x4sin 20t cos t y4sin 20t sin t zcos 20t
Its called a toroidal spiral because it lies on a torus. Another interesting curve, the tre foil knot, with equations
x2cos 1.5t cos t y2cos 1.5t sin t zsin 1.5t
is graphed in Figure 8. It wouldnt be easy to plot either of these curves by hand.
Even when a computer is used to draw a space curve, optical illusions make it difficult to get a good impression of what the curve really looks like. This is especially true in
Figure 8. See Exercise 44. The next example shows how to cope with this problem. EXAMPLE 7 Use a computer to draw the curve with vector equation rtt, t2, t3 .
xy This curve is called a twisted cubic.
FIGURE 8 A trefoil knot
SOLUTION We start by using the computer to plot the curve with parametric equations xt,yt2,zt3 for2t2.TheresultisshowninFigure9a,butitshardto see the true nature of the curve from that graph alone. Most threedimensional computer graphing programs allow the user to enclose a curve or surface in a box instead of dis playing the coordinate axes. When we look at the same curve in a box in Figure 9b, we have a much clearer picture of the curve. We can see that it climbs from a lower corner of the box to the upper corner nearest us, and it twists as it climbs.
z 2 66
2z0
x 6 2
a
6 z0
0 1 2 3 42 82 1 0 1 2 80 1 2 3 4 yxy
d
FIGURE 9 Views of the twisted cubic
TEC InVisual 13.1B you can rotate the box in Figure 9 to see the curve from any viewpoint.
e f
We get an even better idea of the curve when we view it from different vantage points. Part c shows the result of rotating the box to give another viewpoint. Parts d, e, and f show the views we get when we look directly at a face of the box. In par ticular, part d shows the view from directly above the box. It is the projection of the curve on the xyplane, namely, the parabola yx2. Part e shows the projection on
the xzplane, the cubic curve zx3. Its now obvious why the given curve is called a twisted cubic. M
Another method of visualizing a space curve is to draw it on a surface. For instance, the twisted cubic in Example 7 lies on the parabolic cylinder yx2. Eliminate the parame ter from the first two parametric equations, xt and yt2. Figure 10 shows both the cylinder and the twisted cubic, and we see that the curve moves upward from the origin along the surface of the cylinder. We also used this method in Example 4 to visualize the helix lying on the circular cylinder see Figure 2.
A third method for visualizing the twisted cubic is to realize that it also lies on the cylin der zx3. So it can be viewed as the curve of intersection of the cylinders yx2 and zx3. See Figure 11.
z
x
y
FIGURE 10
TEC Visual 13.1C shows how curves arise as intersections of surfaces.
FIGURE 11
8
4 z0 4
SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES821
6
4y02026 2
y42x 0y2420x
b c
28 8 14 4 0xz0 z0 14 4
8101 02 x y
4
822CHAPTER 13 VECTOR FUNCTIONS
N Some computer algebra systems provide us with a clearer picture of a space curve by enclos ing it in a tube. Such a plot enables us to see whether one part of a curve passes in front of or behind another part of the curve. For example, Figure 13 shows the curve of Figure 12b as ren dered by the tubeplot command in Maple.
We have seen that an interesting space curve, the helix, occurs in the model of DNA. Another notable example of a space curve in science is the trajectory of a positively charged particle in orthogonally oriented electric and magnetic fields E and B. Depending on the initial velocity given the particle at the origin, the path of the particle is either a space curve whose projection on the horizontal plane is the cycloid we studied in Section 10.1 Figure 12a or a curve whose projection is the trochoid investigated in Exercise 40 in Section 10.1 Figure 12b.
BB EE
tt
a rtktsint, 1cost, tl
FIGURE 12
Motion of a charged particle in orthogonally oriented electric and magnetic fields
b rtkt32 sint, 132 cost, tl
For further details concerning the physics involved and animations of the trajectories of
the particles, see the following websites:
N www.phy.ntnu.edu.twjavaemFieldemField.html
2. rt
36 Find the limit.
t2 t2
rtt2it4jt6k
14. rtcosticostjsintk
1518 Find a vector equation and parametric equations for the line segment that joins P to Q.
isintjln9t2k
N
www.physics.ucla.edu plasmaexpBeam
FIGURE 13
13.1 EXERCISES
12 Find the domain of the vector function.
1. rts4t2 , e3t, lnt1
9. rtt, cos 2t, sin 2t 11. rt1, cos t, 2 sin t
10. rt1t, 3t, t 12. rtt2 itj2k
3. lim cos t, sin t, t ln t tl0
5. lim e ijcos 2t k
6. lim arctan t, e2t, ln t tl t
15. P0, 0, 0, 16. P1, 0, 1, 17. P1, 1, 2, 18. P2, 4, 0,
Q1, 2, 3 Q2, 3, 1
Q4, 1, 7 Q6, 1, 2
et1 s1t1 4. lim ,
3 tl0tt1t
,
3t t2
13.
2
tl0 sint
1924 Match the parametric equations with the graphs labeled IVI. Give reasons for your choices.
19. xcos 4t, yt, zsin 4t 20. xt, yt2, zet
714 Sketch the curve with the given vector equation. Indicate with an arrow the direction in which t increases.
7. rtsint,t 8. rtt3,t2
xt, y11t2, zt2
22. xet cos10t, yet sin10t, zet
; 33. ; 34.
Graph the curve with parametric equations
x1cos 16t cos t, y1cos 16t sin t,
z1cos 16t. Explain the appearance of the graph by showing that it lies on a cone.
Graph the curve with parametric equations
xs10.25 cos2 10t cos t ys10.25 cos2 10t sin t z0.5 cos10t
Explain the appearance of the graph by showing that it lies on a sphere.
21.
23. xcos t,
24. xcos t,
ysin t, ysin t,
zsin 5t zln t
CHAPTER 13 VECTOR FUNCTIONS AND SPACE CURVES823
Iz IIz
xy IIIz IVz
xy
x
Vz VIz
y
y
x
x
y
xy
35. Show that the curve with parametric equations xt2, y13t,z1t3 passesthroughthepoints1,4,0 and 9, 8, 28 but not through the point 4, 7, 6.
3638 Find a vector function that represents the curve of intersection of the two surfaces.
36. Thecylinderx2 y2 4andthesurfacezxy Theconezsx2 y2 andtheplanez1y
38. The paraboloid z4×2y2 and the parabolic cylinder yx2
37.
25. Show that the curve with parametric equations xt cos t, ytsint,ztliesontheconez2 x2 y2,andusethis fact to help sketch the curve.
26. Show that the curve with parametric equations xsin t,
ycos t, zsin2t is the curve of intersection of the surfaces zx2 and x2y21. Use this fact to help sketch the curve.
27. At what points does the curve rtt i2tt2 k inter sect the paraboloid zx2y2?
28. At what points does the helix rtsin t, cos t, t intersect thespherex2 y2 z2 5?
; 2932 Use a computer to graph the curve with the given vector equation. Make sure you choose a parameter domain and view points that reveal the true nature of the curve.
29. rtcos t sin 2t, sin t sin 2t, cos 2t
30. rtt2, ln t, t
31. rtt,tsint,tcost
32. rtt, et, cos t
; ; 40.
Try to sketch by hand the curve of intersection of the circular cylinder x2y24 and the parabolic cylinder zx2. Then find parametric equations for this curve and use these equations and a computer to graph the curve.
Try to sketch by hand the curve of intersection of the parabolic cylinder yx2 and the top half of the ellipsoid x24y24z216. Then find parametric equations for this curve and use these equations and a computer to graph the curve.
39.
41. If two objects travel through space along two different curves, its often important to know whether they will collide. Will a missile hit its moving target? Will two aircraft collide? The curves might intersect, but we need to know whether the objects are in the same position at the same time. Suppose the trajectories of two particles are given by the vector functions
r1tt2, 7t12, t2 r2t4t3, t2, 5t6
for t0. Do the particles collide?
42. Two particles travel along the space curves
r1tt, t2, t3 r2t12t, 16t, 114t
Do the particles collide? Do their paths intersect?
43. Suppose u and v are vector functions that possess limits as
t l a and let c be a constant. Prove the following properties of limits.
a lim utvtlim utlim vt tla tla tla
824CHAPTER 13 VECTOR FUNCTIONS
b lim cutc lim ut that the projection of the curve onto the xyplane has polar
tla tla
c lim utvtlim utlim vt
tla tla tla
d lim utvtlim utlim vt
tlatlatla; 44. The view of the trefoil knot shown in Figure 8 is accurate, but
it doesnt reveal the whole story. Use the parametric equations x2cos 1.5t cos t
y2cos 1.5t sin t
zsin 1.5t
to sketch the curve by hand as viewed from above, with gaps
indicating where the curve passes over itself. Start by showing
coordinates r2cos 1.5t andt, so r varies between 1 and 3. Then show that z has maximum and minimum values when the projection is halfway between r1 and r3.
When you have finished your sketch, use a computer to draw the curve with viewpoint directly above and compare with your sketch. Then use the computer to draw the curve from several other viewpoints. You can get a better impression of the curve if you plot a tube with radius 0.2 around the curve. Use the tubeplot command in Maple.
45. Showthatlimtla rtbifandonlyifforevery0 there is a number0 such that
if 0ta then rtbDERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS
Later in this chapter we are going to use vector functions to describe the motion of plan ets and other objects through space. Here we prepare the way by developing the calculus of vector functions.
DERIVATIVES
The derivative r of a vector function r is defined in much the same way as for real valued functions:
if this limit exists. The geometric significance of this definition is shown in Figure 1. If the
points P and Q have position vectors rt and rth, then PQ represents the vector rthrt, which can therefore be regarded as a secant vector. If h0, the scalar multiple 1hrthrt has the same direction as rthrt. As h l 0, it appears that this vector approaches a vector that lies on the tangent line. For this reason, the vector rt is called the tangent vector to the curve defined by r at the point P, pro vided that rt exists and rt0. The tangent line to C at P is defined to be the line through P parallel to the tangent vector rt. We will also have occasion to consider the unit tangent vector, which is
13.2
TEC Visual 13.2 shows an animation of Figure 1.
z
rth 0
C
x
y
z
C
0
rthrt h
rth
rt rt
rthrt
P rt
Q
a The secant vector
1
dr rtlim rthrt dt hl0 h
l
rat PQ
Tt
rt
The following theorem gives us a convenient method for computing the derivative of a vector function r: just differentiate each component of r.
THEOREM Ifrtft,tt,htftittjhtk,wheref,t,and h are differentiable functions, then
rt ft, tt, htft itt jht k
2
x
y
FIGURE 1
b The tangent vector
SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS825
PROOF
rtlim 1 rttrt tl0 t
lim 1 ftt,ttt,httft,tt,ht tl0 t
lim fttft, ttttt, httht tl0 t t t
lim fttft, lim ttttt, lim httht tl0 t tl0 t tl0 t
f t, tt, htM
V EXAMPLE 1
a Find the derivative of rt1t3itet jsin 2t k. b Find the unit tangent vector at the point where t0.
SOLUTION
a According to Theorem 2, we differentiate each component of r: rt3t2 i1tet j2 cos 2t k
y 2
01x FIGURE 2
r0 s14 s5 s5
EXAMPLE 2 For the curve rtst i2t j, find rt and sketch the position
b Since r0i and r0j2k, the unit tangent vector at the point 1, 0, 0 is T0 r0 j2k 1 j 2 k M
r1 1, 1 ra1
vector r1 and the tangent vector r1. SOLUTION We have
11 rt 2st ij and r1 2 ij
The curve is a plane curve and elimination of the parameter from the equations xst , y2t gives y2x2, x0. In Figure 2 we draw the position vector r1ij starting at the origin and the tangent vector r1 starting at the corresponding point 1, 1.
V EXAMPLE 3 Find parametric equations for the tangent line to the helix with para metric equations
x2 cos t ysin t zt
at the point 0, 1, 2.
SOLUTION The vector equation of the helix is rt2 cos t, sin t, t, so
M
rt2 sin t, cos t, 1
826
CHAPTER 13 VECTOR FUNCTIONS
The parameter value corresponding to the point 0, 1, 2 is t2, so the tangent vector there is r 22, 0, 1. The tangent line is the line through 0, 1, 2 parallel to the vector 2, 0, 1, so by Equations 12.5.2 its parametric equations are
x2t
12
z8 4
y1
z 2 t
M
N The helix and the tangent line in Example 3 are shown in Figure 3.
FIGURE 3
N In Section 13.4 we will see how rt and rt can be interpreted as the velocity and acceleration vectors of a particle moving through space with position vector rt at time t.
0 2 10.5y00.512 x
0
Just as for realvalued functions, the second derivative of a vector function r is the derivative of r, that is, rr. For instance, the second derivative of the function in Example 3 is
rt2 cos t, sin t, 0
DIFFERENTIATION RULES
The next theorem shows that the differentiation formulas for realvalued functions have their counterparts for vectorvalued functions.
3
and f
is a realvalued function. Then
utvtutvt
cutcut
ftutftutftut
utvtutvtutvt utvtutvtutvt uftftuft ChainRule
1. 2. 3. 4. 5. 6.
d dt d dt d dt d dt d dt d dt
THEOREM Suppose u and v are differentiable vector functions, c is a scalar,
This theorem can be proved either directly from Definition 1 or by using Theorem 2 and the corresponding differentiation formulas for realvalued functions. The proof of Formula 4 follows; the remaining proofs are left as exercises.
SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS827 PROOF OF FORMULA 4 Let
and so
ut f1t, f2t, f3tvt t1t, t2t, t3t3
Then utvtf1t t1tf2t t2tf3t t3t fi t ti t i1
so the ordinary Product Rule gives
d d3 3d
utvtfit tit
dt i1 i1 dt
3
fit titf i tt it f itt it
dt
utvtutvt
V EXAMPLE 4 Show that if rt c a constant, then rt is orthogonal to rt for
all t.
SOLUTION Since
rtrtrt2c2 and c2 is a constant, Formula 4 of Theorem 3 gives
0 d rtrtrtrtrtrt2rtrt dt
Thus rtrt0, which says that rt is orthogonal to rt.
Geometrically, this result says that if a curve lies on a sphere with center the origin,
i1 33
fit tit fit tit i1 i1
n n n
lim f ti t itti t jhti t k
nl
i1 i1 i1
M
then the tangent vector rt is always perpendicular to the position vector rt. M INTEGRALS
The definite integral of a continuous vector function rt can be defined in much the same way as for realvalued functions except that the integral is a vector. But then we can express the integral of r in terms of the integrals of its component functions f, t, and h as follows. We use the notation of Chapter 5.
yb n rt dtlim rtit
a nl i1
yb rt dtyb ft dtiyb tt dtjyb ht dtk aaaa
828
CHAPTER 13 VECTOR FUNCTIONS
1.
This means that we can evaluate an integral of a vector function by integrating each com ponent function.
We can extend the Fundamental Theorem of Calculus to continuous vector functions as follows:
yb rt dtRtbaRbRa a
where R is an antiderivative of r, that is, Rtrt. We use the notation x rt dt for indefinite integrals antiderivatives.
EXAMPLE 5 If rt2 cos t isin t j2t k, then
y rt dty 2 cos t dtiy sin t dtjy 2t dtk
2 sin t icos t jt2 kC where C is a vector constant of integration, and
y2 22
0 rt dt2 sin t icos t jt2 k02ij4 k M
13.2 EXERCISES
The figure shows a curve C given by a vector function rt.
a Draw the vectors r4.5r4 and r4.2r4.
b Draw the vectors
b Draw the vector r1 starting at 1, 1 and compare it with the vector
r1.1r1 0.1
Explain why these vectors are so close to each other in length and direction.
38
a Sketch the plane curve with the given vector equation.
b Find rt.
c Sketch the position vector rt and the tangent vector rt for
the given value of t.
rtt2,t21, t1
4. rt 1t,st, t1
r4.5r4 0.5
r4.2r4 0.2
and
c Write expressions for r4 and the unit tangent vector T4.
d Draw the vector T4. y
C r4.5
R
1
P
Q
3.
r4.2 r4
01x
Make a large sketch of the curve described by the vector function rtt2, t, 0t2, and draw the vectors r1, r1.1, and r1.1r1.
5. rtsin t i2 cos t j, 6.rtetietj, t0 7.rtetie3tj, t0
8. rt1costi2sintj, t 6
916 Find the derivative of the vector function. 9. rtt sin t, t2, t cos 2t
t4
2. a
10. rttan t, sec t, 1t2
11. rtije4t k
12. rtsin1tis1t2 jk
13. rtet2ijln13tk
14. rtat cos 3t ib sin3t jc cos3t k
rtatbt2c
16. rttabtc
1720 Find the unit tangent vector Tt at the point with the given value of the parameter t.
17. rttet, 2 arctan t, 2et, t0 18.rt4stit2jtk, t1
dt
35. 2 3 sin2t cos t i3 sin t cos2t j2 sin t cos t k dt
15.
SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS 3338 Evaluate the integral.
y1324 33. 0 16t i9t j25t kdt
34. y142t jk
829
rtcosti3tj2 sin2tk, t0 20. rt2 sin t i2 cos t jtan t k, t
36. y2t2itst1jtsin tkdt 1
37. yeti2tjlntkdt
38. ycos tisin tjtkdt
39. Findrtifrt2ti3t2jst kandr1ij.
40. Find rt if rtt iet jtet k and r0ijk.
41. Prove Formula 1 of Theorem 3.
42. Prove Formula 3 of Theorem 3.
43. Prove Formula 5 of Theorem 3.
44. Prove Formula 6 of Theorem 3.
45. If utsin t, cos t, t and vtt, cos t, sin t, use Formula 4 of Theorem 3 to find
d
dt utvt
46. If u and v are the vector functions in Exercise 45, use Formula 5 of Theorem 3 to find
d utvt dt
47. Show that if r is a vector function such that r exists, then d rtrtrtrt
dt
48. Find an expression for d utvtwt. dt
d1
If rt0, show that dt rt rtrtrt.
Hint: rtrtrt2
50. If a curve has the property that the position vector rt is always perpendicular to the tangent vector rt, show that the curve lies on a sphere with center the origin.
51. If utrtrtrt, show that utrtrtrt
19.
0 1t2 1t2 y
0
4
21. If rtt, t2, t3 , find rt, T1, rt, and rtrt.
22. If rte2t, e2t, te2t, find T0, r0, and rtrt.
2326 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point.
23. x12st, yt3t,
24. xet, ytet, ztet2;
xet cos t, yet sin t,
26. xln t, y2st, zt2;
zt3t; 1, 0, 0
zet; 0, 2, 1
3, 0, 2 1, 0, 1
25.
; 2729 Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. Illustrate by graphing both the curve and the tangent line on a common screen.
27. 28. 29.
30.
;
31. 32.
xt, yet, z2tt2; 0, 1, 0 x2 cos t, y2 sin t, z4 cos 2t; xt cos t, yt, zt sin t;,
s3, 1, 2 , 0
a Find the point of intersection of the tangent lines to the curve rtsin t, 2 sin t, cos t at the points where t0 and t0.5.
b Illustrate by graphing the curve and both tangent lines.
The curves r1tt, t2, t3and r2tsin t, sin 2t, t intersect at the origin. Find their angle of intersection correct to the nearest degree.
At what point do the curves r1tt, 1t, 3t2and r2s3s, s2, s2intersect? Find their angle of intersection correct to the nearest degree.
49.
830
CHAPTER 13 VECTOR FUNCTIONS
13.3
ARC LENGTH AND CURVATURE
In Section 10.2 we defined the length of a plane curve with parametric equations xf t, ytt, atb, as the limit of lengths of inscribed polygons and, for the case where
fand t are continuous, we arrived at the formula
yb yb dx2 dy2 L sft2 tt2 dtdt a a dt dt
The length of a space curve is defined in exactly the same way see Figure 1. Suppose that the curve has the vector equation rt f t, tt, ht, atb, or, equivalently, the parametric equations xf t, ytt, zht, where f , t, and h are continuous. If the curve is traversed exactly once as t increases from a to b, then it can be shown that its length is
1
z
0 x
FIGURE 1
y
The length of a space curve is the limit of lengths of inscribed polygons.
2
Lyb sft2tt2ht2 dt a
yb dx2 dy2 dz2
a dtdtdt dt
3
Lyb rtdt a
N Figure 2 shows the arc of the helix whose length is computed in Example 1.
z
1, 0, 2
Notice that both of the arc length formulas 1 and 2 can be put into the more com pact form
because, for plane curves rtf t itt j,
rt ftittj sft2tt2
and for space curves rtftittjhtk,
rt ftittjhtk sft2tt2ht2
V EXAMPLE 1 Find the length of the arc of the circular helix with vector equation rtcos t isin t jt k from the point 1, 0, 0 to the point 1, 0, 2 .
SOLUTION Sincertsinticostjk,wehave
rt ssin t2cos2t1s2
The arc from 1, 0, 0 to 1, 0, 2is described by the parameter interval 0t2 and so, from Formula 3, we have
1,0,0 xy00
4
Ly2 rtdty2 s2dt2s2 M
FIGURE 2
A single curve C can be represented by more than one vector function. For instance, the twisted cubic
r1tt, t2, t31t2
5
6
z
st
rt
ra
0
x
FIGURE 3
C
y
could also be represented by the function
r2ueu, e2u, e3u 0uln 2
where the connection between the parameters t and u is given by teu. We say that Equations 4 and 5 are parametrizations of the curve C. If we were to use Equation 3 to compute the length of C using Equations 4 and 5, we would get the same answer. In gen eral, it can be shown that when Equation 3 is used to compute arc length, the answer is independent of the parametrization that is used.
Now we suppose that C is a curve given by a vector function rtftittjhtk atb
where r is continuous and C is traversed exactly once as t increases from a to b. We define its arc length function s by
sta rudua dududu du yt yt dx2 dy2 dz2
Thus st is the length of the part of C between ra and rt. See Figure 3. If we differ entiate both sides of Equation 6 using Part 1 of the Fundamental Theorem of Calculus, we obtain
ds rt dt
7
SECTION 13.3 ARC LENGTH AND CURVATURE831
It is often useful to parametrize a curve with respect to arc length because arc length arises naturally from the shape of the curve and does not depend on a particular coordinate system. If a curve rt is already given in terms of a parameter t and st is the arc length function given by Equation 6, then we may be able to solve for t as a function of s: tts. Then the curve can be reparametrized in terms of s by substituting for t : rrts. Thus, if s3 for instance, rt3 is the position vector of the point 3 units of length along the curve from its starting point.
EXAMPLE 2 Reparametrize the helix rtcos t isin t jt k with respect to arc length measured from 1, 0, 0 in the direction of increasing t.
SOLUTION The initial point 1, 0, 0 corresponds to the parameter value t0. From Example 1 we have
ds rts2 dt
andso sstyt ruduyt s2 dus2t 00
Therefore tss2 and the required reparametrization is obtained by substituting for t: rtscosss2isinss2jss2k M
CURVATURE
A parametrization rt is called smooth on an interval I if r is continuous and rt0 on I. A curve is called smooth if it has a smooth parametrization. A smooth curve has no sharp corners or cusps; when the tangent vector turns, it does so continuously.
832z
CHAPTER 13 VECTOR FUNCTIONS
If C is a smooth curve defined by the vector function r, recall that the unit tangent vec
tor Tt is given by
0C x
FIGURE 4
Ttrt rt
Unit tangent vectors at equally spaced points on C
TEC Visual 13.3A shows animated unit tangent vectors, like those in Figure 4, for a variety of plane curves and space curves.
y
and indicates the direction of the curve. From Figure 4 you can see that Tt changes direc tion very slowly when C is fairly straight, but it changes direction more quickly when C bends or twists more sharply.
The curvature of C at a given point is a measure of how quickly the curve changes direc tion at that point. Specifically, we define it to be the magnitude of the rate of change of the unit tangent vector with respect to arc length. We use arc length so that the curvature will be independent of the parametrization.
The curvature is easier to compute if it is expressed in terms of the parameter t instead of s, so we use the Chain Rule Theorem 13.2.3, Formula 6 to write
dTdT ds and dTdTdt dt dsdt ds dsdt
But dsdtrtfrom Equation 7, so
V EXAMPLE 3 Show that the curvature of a circle of radius a is 1a.
SOLUTION We can take the circle to have center the origin, and then a parametrization is
8
DEFINITION The curvature of a curve is
dTds
where T is the unit tangent vector.
9
t Tt rt
Therefore so
rta cos t ia sin t j
rta sin t ia cos t j and rt a
Tt rt sinticostj rt
Ttcos t isin t j
tTt1 M
rt a
The result of Example 3 shows that small circles have large curvature and large circles have small curvature, in accordance with our intuition. We can see directly from the defi
and
This givesTt 1, so using Equation 9, we have
SECTION 13.3 ARC LENGTH AND CURVATURE833
nition of curvature that the curvature of a straight line is always 0 because the tangent vec tor is constant.
Although Formula 9 can be used in all cases to compute the curvature, the formula given by the following theorem is often more convenient to apply.
PROOF SinceTrrandrdsdt,wehave rrT ds T
dt
so the Product Rule Theorem 13.2.3, Formula 3 gives
rd2s Tds T dt2 dt
10
THEOREM The curvature of the curve given by the vector function r is
t rtrt rt3
Using the fact that TT0 see Example 2 in Section 12.4, we have r rds2TT
dt
Now Tt 1 for all t, so T and T are orthogonal by Example 4 in Section 13.2.
Therefore, by Theorem 12.4.6,
Thus and
r r ds2 TT ds2 TT ds2 T dt dt dt
T r rr r dsdt2r 2
T rr M r r3
EXAMPLE 4 Find the curvature of the twisted cubic rtt, t2, t3at a general point and at 0, 0, 0.
SOLUTION We first compute the required ingredients:
rt1, 2t, 3t2rt0, 2, 6t
rts14t2 9t4
ijk
rtrt 1 2t 3t2 6t2i6tj2k
0 2 6t
rtrts36t4 36t2 4 2s9t4 9t2 1
834
CHAPTER 13 VECTOR FUNCTIONS
Theorem 10 then gives
rtrt t rt3
At the origin, where t0, the curvature is
2s19t29t414t2 9t432
02. M
FIGURE 5
The curvature at 0, 0 is
2217320.03. Observe from the expression for x or the graph of in Fig
The parabola y and its curvature function
to become flatter as x l . M THE NORMAL AND BINORMAL VECTORS
At a given point on a smooth space curve rt, there are many vectors that are orthogonal to the unit tangent vector Tt. We single out one by observing that, because Tt 1 for all t, we have TtTt0 by Example 4 in Section 13.2, so Tt is orthogonal to Tt. Note that Tt is itself not a unit vector. But if r is also smooth, we can define the principal unit normal vector Nt or simply unit normal as
NtTt Tt
The vector BtTtNt is called the binormal vector. It is perpendicular to both T and N and is also a unit vector. See Figure 6.
EXAMPLE 6 Find the unit normal and binormal vectors for the circular helix rtcos t isin t jt k
y 2
0
y
ykx
For the special case of a plane curve with equation yf x, we choose x as the parameter and write rxxifxj. Then rxifxj and rxfxj. Since ijk and jj0, we have rxrxfxk. We also have rx s1 f x2 and so, by Theorem 10,
EXAMPLE 5 Find the curvature of the parabola yx2 at the points 0, 0, 1, 1, and 2, 4.
SOLUTION Since y2x and y2, Formula 11 gives x y2
x
11
x fx2 32 1 f x
1
02. At 1, 1 it is 125320.18. At 2, 4 it is ure 5 that x l 0 as x l . This corresponds to the fact that the parabola appears
1y2 32 14×2 32
N We can think of the normal vector as indi cating the direction in which the curve is turning at each point.
Tt Nt
Bt
FIGURE 6
N Figure 7 illustrates Example 6 by showing
the vectors T, N, and B at two locations on the helix. In general, the vectors T, N, and B, start ing at the various points on a curve, form a set of orthogonal vectors, called the TNB frame, that moves along the curve as t varies. This TNB frame plays an important role in the branch of mathematics known as differential geometry and in its applications to the motion of spacecraft.
z
rt s2
SECTION 13.3 ARC LENGTH AND CURVATURE SOLUTION We first compute the ingredients needed for the unit normal vector:
835
rtsin t icos t jk
Tt rt1 sinticostjk
rt s2
Tt 1 costisintj
Tt 1 s2
T
B
s2 Tt
cos t isin t jcos t, sin t, 0 the zaxis. The binormal vector is
1i j k1
BtTtNts2 sin t cos t 1s2 sin t, cos t, 1 M
cost sint 0
The plane determined by the normal and binormal vectors N and B at a point P on a curve C is called the normal plane of C at P. It consists of all lines that are orthogonal to the tangent vector T. The plane determined by the vectors T and N is called the oscu lating plane of C at P. The name comes from the Latin osculum, meaning kiss. It is the plane that comes closest to containing the part of the curve near P. For a plane curve, the osculating plane is simply the plane that contains the curve.
The circle that lies in the osculating plane of C at P, has the same tangent as C at P, lies on the concave side of C toward which N points, and has radius1 the reciprocal of the curvature is called the osculating circle or the circle of curvature of C at P. It is the circle that best describes how C behaves near P; it shares the same tangent, normal, and curvature at P.
V EXAMPLE 7 Find the equations of the normal plane and osculating plane of the helix in Example 6 at the point P0, 1, 2.
SOLUTION The normal plane at P has normal vector r 21, 0, 1, so an equation 1x00y11 z 2 0 or zx 2
Nt
This shows that the normal vector at a point on the helix is horizontal and points toward
N
T
B N
Tt
y x
FIGURE 7
TEC Visual 13.3B shows how the TNB frame moves along several curves.
N Figure 8 shows the helix and the osculating plane in Example 7.
z zx2
P x
FIGURE 8
is
The osculating plane at P contains the vectors T and N, so its normal vector is
TNB.FromExample6wehave Bt1 sin t, cos t, 1
1 , 0, 1
y
2 s2 s2 1x00y11z 20 or zx 2 M
A simpler normal vector is 1, 0, 1, so an equation of the osculating plane is
s2
B
836CHAPTER 13 VECTOR FUNCTIONS
EXAMPLE 8 Find and graph the osculating circle of the parabola yx2 at the origin.
y
osculating circle
y
SOLUTION From Example 5 the curvature of the parabola at the origin is 02. So the radius of the osculating circle at the origin is 11 and its center is 0, 1 . Its equation
is therefore
For the graph in Figure 9 we use parametric equations of this circle:
x1 cost y1 1 sint M 222
We summarize here the formulas for unit tangent, unit normal and binormal vectors, and curvature.
22
x2 y12 1 24
21 01x
TEC Visual 13.3C shows how the oscu lating circle changes as a point moves along a curve.
13.3 EXERCISES
16 Find the length of the curve.
1. rt2 sin t, 5t, 2 cos t, 2.rt2t,t2,1t3, 0t1
FIGURE 9
Ttrt NtTt BtTtNt rtTt
dTTtrtrt ds rt rt3
10t10 rts2tiet jet k, 0t1
12. Find, correct to four decimal places, the length of the curve of intersection of the cylinder 4×2y24 and the plane xyz2.
1314 Reparametrize the curve with respect to arc length mea suredfromthepointwheret0inthedirectionofincreasingt.
13. rt2ti13tj 54tk
14. rte2t cos 2t i2 je2t sin 2t k
15. Suppose you start at the point 0, 0, 3 and move 5 units along the curve x3 sin t, y4t, z3 cos t in the positive direction. Where are you now?
3
4. rtcos t isin t jln cos t k, 0t4 rt it2 jt3k, 0t1
6. rt12ti8t32 j3t2k, 0t1
79 Find the length of the curve correct to four decimal places.
Use your calculator to approximate the integral. 7.rtst,t,t2, 1t4
16. Reparametrize the curve rt 2
1i
2t j t2 1
8. 9.
; 10. 11.
rtt, ln t, t ln t, 1t2
rtsin t, cos t, tan t, 0t4
Graph the curve with parametric equations xsin t,
ysin 2t, zsin 3t. Find the total length of this curve correct to four decimal places.
Let C be the curve of intersection of the parabolic cylinder x22y and the surface 3zxy. Find the exact length of C from the origin to the point 6, 18, 36.
t2 1
3.
5.
with respect to arc length measured from the point 1, 0 in the direction of increasing t. Express the reparametrization in its simplest form. What can you conclude about the curve?
1720
a Find the unit tangent and unit normal vectors Tt and Nt. b Use Formula 9 to find the curvature.
rt2 sin t, 5t, 2 cos t
17.
18. rtt2, sin tt cos t, cos tt sin t,
19. rts2t,et,et
20. rtt,1t2,t2 2
2123 Use Theorem 10 to find the curvature.
21. rtt2 it k
22. rttitj1t2k
23. rt3t i4 sin t j4 cos t k
t0
3637 Two graphs, a and b, are shown. One is a curve yfx and the other is the graph of its curvature function yx. Identify each curve and explain your choices.
SECTION 13.3 ARC LENGTH AND CURVATURE837
37.
24. Find the curvature of rtet cos t, et sin t, t at the point 1, 0, 0.
25. Findthecurvatureofrt t,t2,t3 atthepoint1,1,1.
36.
CAS 38.
CAS 39.
40.
yy
aa bb
xx
a Graph the curve rtsin 3t, sin 2t, sin 3t. At how many points on the curve does it appear that the curvature has a local or absolute maximum?
b Use a CAS to find and graph the curvature function. Does this graph confirm your conclusion from part a?
The graph of rtt3 sin t, 13 cos t, t is shown in 22
Figure 12b in Section 13.1. Where do you think the curva ture is largest? Use a CAS to find and graph the curvature function. For which values of t is the curvature largest?
Use Theorem 10 to show that the curvature of a plane para metric curve xft, ytt is
xyyxx 2y 232
; 26.
2729 Use Formula 11 to find the curvature.
27. y2xx2 28. ycosx 29. y4x52 3031 At what point does the curve have maximum curvature?
What happens to the curvature as x l ?
30. ylnx 31. yex
32. Find an equation of a parabola that has curvature 4 at the origin.
33. a Is the curvature of the curve C shown in the figure greater at P or at Q? Explain.
b Estimate the curvature at P and at Q by sketching the osculating circles at those points.
Graph the curve with parametric equations
xt y4t32 zt2 and find the curvature at the point 1, 4, 1.
where the dots indicate derivatives with respect to t. 41 42 Use the formula in Exercise 40 to find the curvature.
41. xet cost, yet sint 42. x1t3, ytt2
y
1
P
C
rtt , 3 t , t, 1, 3, 1
44. rtcos t, sin t, ln cos t, 1, 0, 0
4344 Find the vectors T, N, and B at the given point. 223 2
43.
Q 01x
; 3435 Use a graphing calculator or computer to graph both the curve and its curvature function x on the same screen. Is the graph of what you would expect?
34. yx4 2×2 35. yx2
4546 Find equations of the normal plane and osculating plane of the curve at the given point.
45. x2 sin 3t , yt, z2 cos 3t; 0, , 2 46. xt, yt2, zt3; 1, 1, 1
; 47. Find equations of the osculating circles of the ellipse
9×24y236 at the points 2, 0 and 0, 3. Use a graphing calculator or computer to graph the ellipse and both oscu lating circles on the same screen.
838CHAPTER 13 VECTOR FUNCTIONS
; 48.
CAS 50.
52. Show that the curvature of a plane curve is d ds , where is the angle between T and i; that is, is the angle of inclination of the tangent line. This shows that the defini tion of curvature is consistent with the definition for plane curves given in Exercise 69 in Section 10.2.
53. a Show that dBds is perpendicular to B.
b Show that dBds is perpendicular to T.
c Deduce from parts a and b that dBds sN for
some number s called the torsion of the curve. The
torsion measures the degree of twisting of a curve. d Show that for a plane curve the torsion is s0.
54. The following formulas, called the FrenetSerret formulas, are of fundamental importance in differential geometry:
55.
56. 57.
Use the FrenetSerret formulas to prove each of the following. Primes denote derivatives with respect to t. Start as in the proof of Theorem 10.
a rsT s2N b rr s3B
Find equations of the osculating circles of the parabola
y1 x 2 at the points 0, 0 and 1, 1 . Graph both osculating
22
circles and the parabola on the same screen. Atwhatpointonthecurvext3,y3t,zt4 isthe
normal plane parallel to the plane 6x6y8z1?
Is there a point on the curve in Exercise 49 where the osculating plane is parallel to the plane xyz1? Note: You will need a CAS for differentiating, for simplify ing, and for computing a cross product.
Show that the curvature is related to the tangent and normal vectors by the equation
crs 2s3T3 ss s2N s3B rrr
drr2
curve rtt, 1 t2, 1 t3. dT 23
Show that the circular helix rta cos t, a sin t, bt, where a and b are positive constants, has constant curvature and constant torsion. Use the result of Exercise 55d.
Use the formula in Exercise 55d to find the torsion of the
dsN
58. Find the curvature and torsion of the curve xsinh t, ycosh t, zt at the point 0, 1, 0.
59. The DNA molecule has the shape of a double helix see Figure 3 on page 819. The radius of each helix is about
10 angstroms 1 A108 cm. Each helix rises about 34 A during each complete turn, and there are about 2.9108 complete turns. Estimate the length of each helix.
60. Lets consider the problem of designing a railroad track to make a smooth transition between sections of straight track. Existing track along the negative xaxis is to be joined smoothly to a track along the line y1 for x1.
1.dTdsN 2.dNdsT B 3.dBdsN
Formula 1 comes from Exercise 51 and Formula 3 comes from Exercise 53. Use the fact that NBT to deduce Formula 2 from Formulas 1 and 3.
13.4
rthrt
a
; b
Find a polynomial PPx of degree 5 such that the function F defined by
0 if x0 Fx Px if 0x1 1 if x1
is continuous and has continuous slope and continuous curvature.
Use a graphing calculator or computer to draw the graph of F.
z
O
x
rat Q
rt
h
MOTION IN SPACE: VELOCITY AND ACCELERATION
In this section we show how the ideas of tangent and normal vectors and curvature can be used in physics to study the motion of an object, including its velocity and acceleration, along a space curve. In particular, we follow in the footsteps of Newton by using these methods to derive Keplers First Law of planetary motion.
Suppose a particle moves through space so that its position vector at time t is rt. Notice from Figure 1 that, for small values of h, the vector
rthrt h
49.
51.
C
P
rth
1
FIGURE 1
y
approximates the direction of the particle moving along the curve rt. Its magnitude mea sures the size of the displacement vector per unit time. The vector 1 gives the average
y
a1 0x
FIGURE 2
TEC Visual 13.4 shows animated velocity and acceleration vectors for objects moving along various curves.
N Figure 3 shows the path of the particle in Example 2 with the velocity and acceleration vectors when t1.
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION839 velocity over a time interval of length h and its limit is the velocity vector vt at time t:
Thus the velocity vector is also the tangent vector and points in the direction of the tangent line.
The speed of the particle at time t is the magnitude of the velocity vector, that is, vt . This is appropriate because, from 2 and from Equation 13.3.7, we have
vt rt dsrate of change of distance with respect to time dt
As in the case of onedimensional motion, the acceleration of the particle is defined as the derivative of the velocity:
atvtrt
EXAMPLE 1 The position vector of an object moving in a plane is given by
rtt3 it2 j. Find its velocity, speed, and acceleration when t1 and illustrate geometrically.
SOLUTION The velocity and acceleration at time t are
vtrt3t2 i2t j
v1 1, 1
atrt6t i2 j vt s3t222t2s9t44t2
and the speed is
z
v 1
When t1, we have
v13i2 j a16i2 j v1 s13
These velocity and acceleration vectors are shown in Figure 2. M EXAMPLE 2 Find the velocity, acceleration, and speed of a particle with position vector
a 1
rtt2, et, tet . SOLUTION
y
x
FIGURE 3
The vector integrals that were introduced in Section 13.2 can be used to find position vectors when velocity or acceleration vectors are known, as in the next example.
1
vtrt2t, et, 1tet
atvt2, et, 2tet
vt s4t2e2t1t2e2t M
2
vtlim rthrt rt hl0 h
840
CHAPTER 13 VECTOR FUNCTIONS
N The expression for rt that we obtained in Example 3 was used to plot the path of the particleinFigure4for0 t3.
6
z4 2
V EXAMPLE 3 A moving particle starts at an initial position r01, 0, 0 with initial velocity v0ijk. Its acceleration is at4t i6t jk. Find its velocity and position at time t.
SOLUTION Since atvt, we have
vty at dty 4t i6t jk dt
2t2 i3t2 jt kC
To determine the value of the constant vector C, we use the fact that v0ijk.
The preceding equation gives v0C, so Cijk and
vt2t2 i3t2 jt kijk 2t2 1i3t2 1jt1k
Since vtrt, we have rtyvtdt
y2t2 1i3t2 1jt1kdt 2t3 tit3 tj1t2 tkD
1, 0, 0
0 0x
32
Putting t0, we find that Dr0i, so the position at time t is given by
0
5y101520 20
rt2t3 t1it3 tj1t2 tk M 3 2
In general, vector integrals allow us to recover velocity when acceleration is known and position when velocity is known:
vtvt0yt audu rtrt0yt vudu t0 t0
If the force that acts on a particle is known, then the acceleration can be found from Newtons Second Law of Motion. The vector version of this law states that if, at any time t, a force Ft acts on an object of mass m producing an acceleration at, then
Ftmat
EXAMPLE 4 An object with mass m that moves in a circular path with constant angular
speed has position vector rta cos t ia sin t j. Find the force acting on the object and show that it is directed toward the origin.
SOLUTION To find the force, we first need to know the acceleration: vtrta sin tia cos tj
atvta 2cos tia 2sin tj Therefore Newtons Second Law gives the force as
Ftmatm 2acos tiasin tj
FIGURE 4
N The angular speed of the object moving with
position P isd dt, where shown in Figure 5.
y
P 0 x
is the angle
FIGURE 5
y
va
0dx FIGURE 6
Notice that Ftm 2rt. This shows that the force acts in the direction opposite to the radius vector rt and therefore points toward the origin see Figure 5. Such a force is called a centripetal centerseeking force. M
V EXAMPLE 5 A projectile is fired with angle of elevation and initial velocity v0. See Figure 6. Assuming that air resistance is negligible and the only external force is due to gravity, find the position function rt of the projectile. What value of maximizes the range the horizontal distance traveled?
3
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION841
SOLUTION We set up the axes so that the projectile starts at the origin. Since the force due to gravity acts downward, we have
Fmamt j
where ta 9.8 ms2. Thus
at j
Since vta, we have
where Cv0v0. Therefore
vttt jC rtvttt jv0
Integrating again, we obtain
But Dr00, so the position vector of the projectile is given by
rt1tt2 jtv0 2
If we write v0 v0 the initial speed of the projectile, then v0 v0cos iv0sin j
and Equation 3 becomes
rtv0cos tiv0sin t1tt2j 2
The parametric equations of the trajectory are therefore
The horizontal distance d is the value of x when y0. Setting y0, we obtain t0 or t2v0 sin t. This second value of t then gives
rt1tt2jtv0 D 2
N If you eliminate t from Equations 4, you will see that y is a quadratic function of x. So the path of the projectile is part of a parabola.
4
xv0cos t yv0 sin t1tt2 2
2v0 sin v022 sin cos dxv0costt
Clearly, d has its maximum value when sin 21, that is,
v02 sin 2t
4.
M
842
CHAPTER 13 VECTOR FUNCTIONS
V EXAMPLE 6 A projectile is fired with muzzle speed 150 ms and angle of elevation 45 from a position 10 m above ground level. Where does the projectile hit the ground, and with what speed?
SOLUTION If we place the origin at ground level, then the initial position of the projectile is 0, 10 and so we need to adjust Equations 4 by adding 10 to the expression for y. With v0150 ms,45, and t9.8 ms2, we have
x150 cos 4t75s2 t
y10150 sin 4t1 9.8t21075s2 t4.9t2
Impact occurs when y0, that is, 4.9t275s2t100. Solving this quadratic equation and using only the positive value of t, we get
t75s2s11,25019621.74 9.8
Then x75s2 21.742306, so the projectile hits the ground about 2306 m away. The velocity of the projectile is
vtrt75s2 i75s29.8tj So its speed at impact is
v21.74 s75s2 275s29.821.742151 ms M TANGENTIAL AND NORMAL COMPONENTS OF ACCELERATION
When we study the motion of a particle, it is often useful to resolve the acceleration into two components, one in the direction of the tangent and the other in the direction of the normal. If we write vvfor the speed of the particle, then
Ttrtvtv rt vt v
2
5
vvT
avvTvT
and so
If we differentiate both sides of this equation with respect to t, we get
If we use the expression for the curvature given by Equation 13.3.9, then we have
T T so T v r v
The unit normal vector was defined in the preceding section as NTT, so 6 gives TTNvN
and Equation 5 becomes
6
7
avT v2N
FIGURE 7
This resolution is illustrated in Figure 7.
Lets look at what Formula 7 says. The first thing to notice is that the binormal vector
B is absent. No matter how an object moves through space, its acceleration always lies in the plane of T and N the osculating plane. Recall that T gives the direction of motion and N points in the direction the curve is turning. Next we notice that the tangential com ponent of acceleration is v, the rate of change of speed, and the normal component of acceleration is v2, the curvature times the square of the speed. This makes sense if we think of a passenger in a cara sharp turn in a road means a large value of the curvature
, so the component of the acceleration perpendicular to the motion is large and the pas senger is thrown against a car door. High speed around the turn has the same effect; in fact, if you double your speed, aN is increased by a factor of 4.
Although we have expressions for the tangential and normal components of accelera tion in Equations 8, its desirable to have expressions that depend only on r, r, and r. To this end we take the dot product of vvT with a as given by Equation 7:
aT
T
a N
aN
Writing aT and aN for the tangential and normal components of acceleration, we have aaT TaN N
8
Therefore
vavTvT v2N
vvTT v3TN
vv since TT1 and TN0
aT v vartrt v rt
9
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION843
where
aTv and
aN v2
Using the formula for curvature given by Theorem 13.3.10, we have
aNv2rtrtrt2rtrt rt3 rt
EXAMPLE 7 A particle moves with position function rtt2, t2, t3 . Find the tangen tial and normal components of acceleration.
SOLUTION rtt2 it2 jt3 k rt2t i2t j3t2 k
rt2i2j6t k rts8t2 9t4
Therefore Equation 9 gives the tangential component as
rtrt 8t18t3 aT rt s8t29t4
10
844
CHAPTER 13 VECTOR FUNCTIONS
ijk
Since rtrt 2t 2t 3t2 6t2 i6t2 j
2 2 6t
Equation 10 gives the normal component as
rtrt 6s2 t2
aNrts8t29t4 M
KEPLERS LAWS OF PLANETARY MOTION
We now describe one of the great accomplishments of calculus by showing how the mate rial of this chapter can be used to prove Keplers laws of planetary motion. After 20 years of studying the astronomical observations of the Danish astronomer Tycho Brahe, the German mathematician and astronomer Johannes Kepler 15711630 formulated the fol lowing three laws.
KEPLERS LAWS
1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
2. The line joining the sun to a planet sweeps out equal areas in equal times.
3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
In his book Principia Mathematica of 1687, Sir Isaac Newton was able to show that these three laws are consequences of two of his own laws, the Second Law of Motion and the Law of Universal Gravitation. In what follows we prove Keplers First Law. The remaining laws are proved as exercises with hints.
Since the gravitational force of the sun on a planet is so much larger than the forces exerted by other celestial bodies, we can safely ignore all bodies in the universe except the sun and one planet revolving about it. We use a coordinate system with the sun at the ori gin and we let rrt be the position vector of the planet. Equally well, r could be the position vector of the moon or a satellite moving around the earth or a comet moving around a star. The velocity vector is vr and the acceleration vector is ar. We use the following laws of Newton:
Second Law of Motion: Fma
Law of Gravitation: F GMm r GMm u
where F is the gravitational force on the planet, m and M are the masses of the planet and the sun, G is the gravitational constant, rr , and u1rr is the unit vector in the direction of r.
We first show that the planet moves in one plane. By equating the expressions for F in Newtons two laws, we find that
a GM r r3
r3 r2
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION845 and so a is parallel to r. It follows that ra0. We use Formula 5 in Theorem 13.2.3
to write
Therefore
d rvrvrv dt
vvra000 rvh
where h is a constant vector. We may assume that h0; that is, r and v are not parallel. This means that the vector rrt is perpendicular to h for all values of t, so the planet always lies in the plane through the origin perpendicular to h. Thus the orbit of the planet is a plane curve.
To prove Keplers First Law we rewrite the vector h as follows:
hrvrrr ur u
rururur2uurruur2uu
Then
GM uuuuuu by Theorem 12.4.8, Property 6
Butuuu21and,sinceut 1,itfollowsfromExample4inSection13.2that
ah GM ur2uuGMuuu r2
z
h
c r y v
u
ahGM u Integrating both sides of this equation, we get
uu0. Therefore
and so vhvhahGM u
11
x
FIGURE 8
vhGM uc
where c is a constant vector.
At this point it is convenient to choose the coordinate axes so that the standard basis
vector k points in the direction of the vector h. Then the planet moves in the xyplane. Since both vh and u are perpendicular to h, Equation 11 shows that c lies in the xyplane. This means that we can choose the x and yaxes so that the vector i lies in the direction of c, as shown in Figure 8.
If is the angle between c and r, then r,are polar coordinates of the planet. From Equation 11 we have
rvhrGM ucGM rurc GMruurccos GMrrccos
846
CHAPTER 13 VECTOR FUNCTIONS
where cc. Then
r rvh1 rvh GMccos GM 1ecos
where ecGM. But
rvhrvhhhh2 h2
where hh. So
Writing dh2c, we obtain the equation
r ed
1e cos
Comparing with Theorem 10.6.6, we see that Equation 12 is the polar equation of a conic section with focus at the origin and eccentricity e. We know that the orbit of a planet is a closed curve and so the conic must be an ellipse.
This completes the derivation of Keplers First Law. We will guide you through the der ivation of the Second and Third Laws in the Applied Project on page 848. The proofs of these three laws show that the methods of this chapter provide a powerful tool for describ ing some of the laws of nature.
12
h 2GMr 1ecos
eh 2c1ecos
13.4 EXERCISES
1. The table gives coordinates of a particle moving through space along a smooth curve.
a Find the average velocities over the time intervals 0, 1,
0.5, 1, 1, 2, and 1, 1.5.
b Estimate the velocity and speed of the particle at t1.
d Draw an approximation to the vector v2 and estimate the speed of the particle at t2.
y
r2.4
2 1
r2 r1.5
t
x
y
z
0 0.5 1.0 1.5 2.0
2.7 3.5 4.5 5.9 7.3
9.8 7.2 6.0 6.4 7.8
3.7 3.3 3.0 2.8 2.7
2. The figure shows the path of a particle that moves with position vector rt at time t.
a Draw a vector that represents the average velocity of the
particle over the time interval 2t2.4.
b Draw a vector that represents the average velocity over the
time interval 1.5t2.
c Write an expression for the velocity vector v2.
012x
38 Find the velocity, acceleration, and speed of a particle with the given position function. Sketch the path of the particle and draw the velocity and acceleration vectors for the specified value of t.
3. rt1 t2, t, t2 2
4. rt2t, 4st , t1
5. rt3 cos t i2 sin t j,
6. rtet ie2t j, t0
7. rttit2j2k, t1
8. rtt i2 cos tjsin tk,
3
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION847
26. A gun is fired with angle of elevation 30. What is the
muzzle speed if the maximum height of the shell is 500 m?
27. A gun has muzzle speed 150 ms. Find two angles of eleva tion that can be used to hit a target 800 m away.
28. A batter hits a baseball 3 ft above the ground toward the center field fence, which is 10 ft high and 400 ft from home plate. The ball leaves the bat with speed 115 fts at an angle 50 above the horizontal. Is it a home run? In other words, does the ball clear the fence?
29. A medieval city has the shape of a square and is protected
by walls with length 500 m and height 15 m. You are the commander of an attacking army and the closest you can get to the wall is 100 m. Your plan is to set fire to the city by cat apulting heated rocks over the wall with an initial speed of 80 ms. At what range of angles should you tell your men to set the catapult? Assume the path of the rocks is perpendicu lar to the wall.
30. A ball with mass 0.8 kg is thrown southward into the air with a speed of 30 ms at an angle of 30 to the ground. A west wind applies a steady force of 4 N to the ball in an easterly direction. Where does the ball land and with what speed?
t0
914 Find the velocity, acceleration, and speed of a particle with
11.
t
the given position function.
9. rtt21, t3, t21
10. rt2 cos t, 3t, 2 sin t
rts2tiet jet k
12. rtt2 ilntjtk
13. rtetcostisintjtk
14. rttsintitcostjt2 k
1516 Find the velocity and position vectors of a particle that has the given acceleration and the given initial velocity and position.
15. ati2j, v0k, r0i
16. at2i6tj12t2k, v0i, r0jk
1718
a Find the position vector of a particle that has the given accel eration and the specified initial velocity and position.
; b Use a computer to graph the path of the particle.
17. at2t isin t jcos 2t k, v0i, r0j
18. attiet jet k, v0k, r0jk
The position function of a particle is given by
rtt2, 5t, t216t. When is the speed a minimum?
20. What force is required so that a particle of mass m has the positionfunctionrtt3it2jt3k?
21. A force with magnitude 20 N acts directly upward from the xyplane on an object with mass 4 kg. The object starts at the origin with initial velocity v0ij. Find its position function and its speed at time t.
Show that if a particle moves with constant speed, then the velocity and acceleration vectors are orthogonal.
23. A projectile is fired with an initial speed of 500 ms and angle of elevation 30. Find a the range of the projectile, b the maximum height reached, and c the speed at impact.
24. Rework Exercise 23 if the projectile is fired from a position 200 m above the ground.
A ball is thrown at an angle of 45 to the ground. If the ball lands 90 m away, what was the initial speed of the ball?
; 31.
Water traveling along a straight portion of a river normally flows fastest in the middle, and the speed slows to almost zero at the banks. Consider a long straight stretch of river flowing north, with parallel banks 40 m apart. If the maxi mum water speed is 3 ms, we can use a quadratic function as a basic model for the rate of water flow x units from the west bank: fx3 x40x.
a A boat proceeds at a constant speed of 5 ms from a point A on the west bank while maintaining a heading perpen dicular to the bank. How far down the river on the oppo site bank will the boat touch shore? Graph the path of the boat.
b Suppose we would like to pilot the boat to land at the point B on the east bank directly opposite A. If we main tain a constant speed of 5 ms and a constant heading, find the angle at which the boat should head. Then graph the actual path the boat follows. Does the path seem realistic?
19.
400
22.
32. Another reasonable model for the water speed of the river in Exercise 31 is a sine function: f x3 sin x40. If a boater would like to cross the river from A to B with con stant heading and a constant speed of 5 ms, determine the angle at which the boat should head.
3338 Find the tangential and normal components of the acceler ation vector.
33. rt3tt3i3t2 j
34. rt1tit2 2tj
rtcos t isin t jt k
36. rttit2 j3tk
35.
25.
848CHAPTER 13 VECTOR FUNCTIONS t t
37.rteis2tje k 38. rtticos2tjsin2tk
41.
42.
The position function of a spaceship is
39. The magnitude of the acceleration vector a is 10 cms2. Use the figure to estimate the tangential and normal components of a.
y
a
0x
40. If a particle with mass m moves with position vector rt, then its angular momentum is defined as Ltmrtvt and its torque as tmrtat. Show that Ltt. Deduce that if t0 for all t, then Lt is constant. This is the law of conservation of angular momentum.
KEPLERS LAWS
and the coordinates of a space station are 6, 4, 9. The captain wants the spaceship to coast into the space station. When should the engines be turned off?
A rocket burning its onboard fuel while moving through space has velocity vt and mass mt at time t. If the exhaust gases escape with velocity ve relative to the rocket, it can be deduced from Newtons Second Law of Motion that
m dvdm ve dt dt
a Show that vtv0ln m0 ve . mt
b For the rocket to accelerate in a straight line from rest to twice the speed of its own exhaust gases, what fraction of its initial mass would the rocket have to burn as fuel?
rt3t i2ln t j
4 k
7
2 t1
APPLIED PROJECT
y
rt
At
0
rt
Johannes Kepler stated the following three laws of planetary motion on the basis of masses of data on the positions of the planets at various times.
Kepler formulated these laws because they fitted the astronomical data. He wasnt able to see why they were true or how they related to each other. But Sir Isaac Newton, in his Principia Mathematica of 1687, showed how to deduce Keplers three laws from two of Newtons own laws, the Second Law of Motion and the Law of Universal Gravitation. In Section 13.4 we proved Keplers First Law using the calculus of vector functions. In this project we guide you through the proofs of Keplers Second and Third Laws and explore some of their consequences.
KEPLERS LAWS
1. A planet revolves around the sun in an elliptical orbit with the sun at one focus.
2. The line joining the sun to a planet sweeps out equal areas in equal times.
3. The square of the period of revolution of a planet is proportional to the cube of the length of the major axis of its orbit.
1.
Use the following steps to prove Keplers Second Law. The notation is the same as in the proof of the First Law in Section 13.4. In particular, use polar coordinates so that rrcos irsin j.
a Showthathr2 d k. dt
b Deduce that r2 dh. dt
c IfAAtistheareasweptoutbytheradiusvectorrrtinthetimeintervalt0,t as in the figure, show that
dA1r2 d dt 2 dt
x
CHAPTER 13 REVIEW849
d Deduce that
dA1 hconstant dt 2
This says that the rate at which A is swept out is constant and proves Keplers Second Law.
2. Let T be the period of a planet about the sun; that is, T is the time required for it to travel once around its elliptical orbit. Suppose that the lengths of the major and minor axes of the ellipse are 2a and 2b.
2GM is
a Use part d of Problem 1 to show that T2 abh. h2 b2
b ShowthatGMed a.
c Use parts a and b to show that T2GM a3.
42
This proves Keplers Third Law. Notice that the proportionality constant 4 independent of the planet.
3. The period of the earths orbit is approximately 365.25 days. Use this fact and Keplers Third Law to find the length of the major axis of the earths orbit. You will need the mass of the sun, M1.991030 kg, and the gravitational constant, G6.671011 Nm2kg2 .
4. Its possible to place a satellite into orbit about the earth so that it remains fixed above a given location on the equator. Compute the altitude that is needed for such a satellite. The earths mass is 5.981024 kg; its radius is 6.37106 m. This orbit is called the Clarke Geosynchronous Orbit after Arthur C. Clarke, who first proposed the idea in 1945. The first such satellite, Syncom 2, was launched in July 1963.
13 REVIEW
1. What is a vector function? How do you find its derivative and
its integral?
2. What is the connection between vector functions and space curves?
3. How do you find the tangent vector to a smooth curve at a point? How do you find the tangent line? The unit tangent vector?
4. If u and v are differentiable vector functions, c is a scalar, and f is a realvalued function, write the rules for differentiating
CONCEPT CHECK
the following vector functions.
a utvt b cut
d utvt e utvt
c f t ut f u f t
6. a What is the definition of curvature?
b Write a formula for curvature in terms of rt and Tt. c Write a formula for curvature in terms of rt and rt. d Write a formula for the curvature of a plane curve with
equation yf x.
7. a Write formulas for the unit normal and binormal vectors of a smooth space curve rt.
b What is the normal plane of a curve at a point? What is the osculating plane? What is the osculating circle?
8. a How do you find the velocity, speed, and acceleration of a particle that moves along a space curve?
b Write the acceleration in terms of its tangential and normal components.
9. State Keplers Laws.
5. How do you find the length of a space curve given by a vector function rt?
850CHAPTER 13 VECTOR FUNCTIONS TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. The curve with vector equation rtt3 i2t3 j3t3 k is a line.
2. The derivative of a vector function is obtained by differen tiating each component function.
3. If ut and vt are differentiable vector functions, then d utvtutvt
5. If Tt is the unit tangent vector of a smooth curve, then the curvature is d Tdt .
6. The binormal vector is BtNtTt.
7. Suppose f is twice continuously differentiable. At an inflection
point of the curve yf x, the curvature is 0. 8. If t0 for all t, the curve is a straight line. 9. If rt 1 for all t, then rtis a constant.
10. If rt 1 for all t, then rt is orthogonal to rt for all t.
11. The osculating circle of a curve C at a point has the same tangent vector, normal vector, and curvature as C at that point.
12. Different parametrizations of the same curve result in identical tangent vectors at a given point on the curve.
dt
4. If rt is a differentiable vector function, then
d rt rt dt
EXERCISES
1. a Sketch the curve with vector function
11. For the curve given by rt1 t3, 1 t2, t, find
a the unit tangent vector b the unit normal vector c the curvature
12. Findthecurvatureoftheellipsex3cost,y4sintatthe points 3, 0 and 0, 4.
13. Find the curvature of the curve yx 4 at the point 1, 1.
; 14. Find an equation of the osculating circle of the curve
yx4x2 at the origin. Graph both the curve and its oscu lating circle.
15. Find an equation of the osculating plane of the curve xsin 2t, yt, zcos 2t at the point 0, , 1.
16. ThefigureshowsthecurveCtracedbyaparticlewithposi tion vector rt at time t.
a Draw a vector that represents the average velocity of the
particle over the time interval 3t3.2.
b Write an expression for the velocity v3.
c Write an expression for the unit tangent vector T3 and
rtt icos t jsin b Find rt and rt.
t k
t0
2. Let rts2t, et1t, lnt1. a Find the domain of r.
b Find limt l 0 rt.
c Find rt.
32
3. Find a vector function that represents the curve of intersection ofthecylinderx2 y2 16andtheplanexz5.
; 4.
5. Ifrtt2itcos tjsin tk,evaluatex1rtdt.
Find parametric equations for the tangent line to the curve
x2 sin t, y2 sin 2t , z2 sin 3t at the point 1, s3, 2. Graph the curve and the tangent line on a common screen.
6. LetCbethecurvewithequationsx2t3,y2t1, zln t. Find a the point where C intersects the xzplane, b parametric equations of the tangent line at 1, 1, 0, and c an equation of the normal plane to C at 1, 1, 0.
7. Use Simpsons Rule with n6 to estimate the length of the arc of the curve with equations xt2, yt3, zt4, 0t3.
8. Find the length of the curve rt2t32, cos 2t, sin 2t, 0t1.
9. Thehelixr1tcostisintjtkintersectsthecurve r2t1tit2 jt3 k at the point 1, 0, 0. Find the angle of intersection of these curves.
10. Reparametrizethecurvertet iet sintjet costk with respect to arc length measured from the point 1, 0, 1 in the direction of increasing t.
draw it.
y 1
r3
C
0
0
r3.2 1
x
17. A particle moves with position function
rtt ln t it jet k. Find the velocity, speed, and acceleration of the particle.
18. A particle starts at the origin with initial velocity ij3k. Its acceleration is at6t i12t2 j6t k. Find its position function.
19. An athlete throws a shot at an angle of 45 to the horizontal at an initial speed of 43 fts. It leaves his hand 7 ft above the ground.
a Where is the shot 2 seconds later?
b How high does the shot go? c Where does the shot land?
20. Find the tangential and normal components of the acceleration vector of a particle with position function
rtt i2t jt2 k
21. A disk of radius 1 is rotating in the counterclockwise direction
at a constant angular speed . A particle starts at the center of the disk and moves toward the edge along a fixed radius so that its position at time t, t0, is given by rttRt, where
Rtcos tisin tj
a Show that the velocity v of the particle is vcos tisin tjtvd
22.
CHAPTER 13 REVIEW851 c Determine the Coriolis acceleration of a particle that moves
on a rotating disk according to the equation
rtetcos tietsin tj
In designing transfer curves to connect sections of straight rail road tracks, its important to realize that the acceleration of the train should be continuous so that the reactive force exerted by the train on the track is also continuous. Because of the formu las for the components of acceleration in Section 13.4, this will be the case if the curvature varies continuously.
a
A logical candidate for a transfer curve to join existing tracks given by y1 for x0 and ys2x for
x1s2 might be the function fxs1x2,
0x1s2, whose graph is the arc of the circle shown in the figure. It looks reasonable at first glance. Show that the function
if x0
if 0x1s2 if x1s2
1 Fx s1x2
where vdRt is the velocity of a point on the edge of
the disk.
b Show that the acceleration a of the particle is
a2vd tad
where adRt is the acceleration of a point on the rim of the disk. The extra term 2vd is called the Coriolis accel eration; it is the result of the interaction of the rotation of the disk and the motion of the particle. One can obtain a physical demonstration of this acceleration by walking toward the edge of a moving merrygoround.
y 1
0
yFx
1 2
y
y0 x 0
yx
transfer curve
1 x
s2x
;
b
is continuous and has continuous slope, but does not have continuous curvature. Therefore f is not an appropriate transfer curve.
Find a fifthdegree polynomial to serve as a transfer curve between the following straight line segments: y0 for
x0 and yx for x1. Could this be done with a fourthdegree polynomial? Use a graphing calculator or computer to sketch the graph of the connected function and check to see that it looks like the one in the figure.
PROBLEMS PLUS
y 1. vt
x
A particle P moves with constant angular speed around a circle whose center is at the origin and whose radius is R. The particle is said to be in uniform circular motion. Assume that the motion is counterclockwise and that the particle is at the point R, 0 when t0. The position vectorattimet0isrtRcos tiRsin tj.
a Find the velocity vector v and show that vr0. Conclude that v is tangent to the circle and points in the direction of the motion.
b Show that the speedvof the particle is the constant R. The period T of the particle is the time required for one complete revolution. Conclude that
T2R2 v
c Find the acceleration vector a. Show that it is proportional to r and that it points toward the origin. An acceleration with this property is called a centripetal acceleration. Show that the magnitude of the acceleration vector isa R 2.
d Suppose that the particle has mass m. Show that the magnitude of the force F that is required to produce this motion, called a centripetal force, is
F mv2 R
A circular curve of radius R on a highway is banked at an angle so that a car can safely traverse the curve without skidding when there is no friction between the road and the tires. The loss of friction could occur, for example, if the road is covered with a film of water or ice. The rated speed vR of the curve is the maximum speed that a car can attain without skidding. Suppose a car of mass m is traversing the curve at the rated speed vR. Two forces are acting on the car: the vertical force, mt, due to the weight of the car, and a force F exerted by, and normal to, the road. See the figure.
The vertical component of F balances the weight of the car, so that F cosmt. The horizontal component of F produces a centripetal force on the car so that, by Newtons Sec ond Law and part d of Problem 1,
Fsin mvR2 R
a Show that vR2Rt tan .
b Find the rated speed of a circular curve with radius 400 ft that is banked at an angle of 12. c Suppose the design engineers want to keep the banking at 12, but wish to increase the
rated speed by 50. What should the radius of the curve be?
A projectile is fired from the origin with angle of elevation and initial speed v0. Assuming that air resistance is negligible and that the only force acting on the projectile is gravity, t, we showed in Example 5 in Section 13.4 that the position vector of the projectile is
rtv0cos tiv0sin t1tt2j 2
We also showed that the maximum horizontal distance of the projectile is achieved when45 and in this case the range is Rv02t.
a At what angle should the projectile be fired to achieve maximum height and what is the maximum height?
b Fix the initial speed v0 and consider the parabola x22RyR20, whose graph is shown in the figure. Show that the projectile can hit any target inside or on the boundary of the region bounded by the parabola and the xaxis, and that it cant hit any target out side this region.
c Suppose that the gun is elevated to an angle of inclination in order to aim at a target that is suspended at a height h directly over a point D units downrange. The target is released at the instant the gun is fired. Show that the projectile always hits the target, regardless of the value v0, provided the projectile does not hit the ground before D.
v
r
FIGURE FOR PROBLEM 1
2.
F
mg
FIGURE FOR PROBLEM 2
y
3.
R 0 Rx y
0x D
FIGURE FOR PROBLEM 3
852
PROBLEMS PLUS
y
va
4. a
A projectile is fired from the origin down an inclined plane that makes an angle with the horizontal. The angle of elevation of the gun and the initial speed of the projectile are
and v0, respectively. Find the position vector of the projectile and the parametric equations of the path of the projectile as functions of the time t. Ignore air resistance.
x
b Show that the angle of elevation that will maximize the downhill range is the angle halfway between the plane and the vertical.
c Suppose the projectile is fired up an inclined plane whose angle of inclination is . Show that, in order to maximize the uphill range, the projectile should be fired in the direction halfway between the plane and the vertical.
d In a paper presented in 1686, Edmond Halley summarized the laws of gravity and projectile motion and applied them to gunnery. One problem he posed involved firing a projectile to hit a target a distance R up an inclined plane. Show that the angle at which the projectile should be fired to hit the target but use the least amount of energy is the same as the angle in part c. Use the fact that the energy needed to fire the projectile is proportional to the square of the initial speed, so minimizing the energy is equivalent to minimizing the initial speed.
5. A ball rolls off a table with a speed of 2 fts. The table is 3.5 ft high.
a Determine the point at which the ball hits the floor and find its speed at the instant of
impact.
b Find the angle between the path of the ball and the vertical line drawn through the point
of impact. See the figure.
c Suppose the ball rebounds from the floor at the same angle with which it hits the floor, but
loses 20 of its speed due to energy absorbed by the ball on impact. Where does the ball strike the floor on the second bounce?
6. Find the curvature of the curve with parametric equations
xyt sin1 2d yyt cos1 2d
FIGURE FOR PROBLEM 4
3.5 ft
FIGURE FOR PROBLEM 5
; 7.
2
If a projectile is fired with angle of elevation and initial speed v, then parametric equations for its trajectory are
xvcos t yvsin t1tt2 2
See Example 5 in Section 13.4. We know that the range horizontal distance traveled
is maximized when45. What value of maximizes the total distance traveled by the projectile? State your answer correct to the nearest degree.
2 00
8. A cable has radius r and length L and is wound around a spool with radius R without over lapping. What is the shortest length along the spool that is covered by the cable?
853
14
PARTIAL DERIVATIVES
Functions of two variables can be visualized by means of level curves, which con nect points where the function takes on a given value. Atmospheric pressure at a given time is a function of longitude and latitude and is measured in millibars. Here the level curves are called isobars and those pictured join locations that had the same pressure on March 7, 2007. The curves labeled 1028, for instance, connect points with pressure 1028 mb. Surface winds tend to flow from areas of high pressure across the isobars toward areas of low pressure, and are strongest where the isobars are tightly packed.
854
So far we have dealt with the calculus of functions of a single variable. But, in the real world, physical quantities often depend on two or more variables, so in this chapter we turn our attention to functions of several variables and extend the basic ideas of differential calculus to such functions.
14.1 FUNCTIONS OF SEVERAL VARIABLES
In this section we study functions of two or more variables from four points of view:
N verbally
N numerically N algebraically N visually
by a description in words by a table of values
by an explicit formula by a graph or level curves
FUNCTIONS OF TWO VARIABLES
The temperature T at a point on the surface of the earth at any given time depends on the longitude x and latitude y of the point. We can think of T as being a function of the two variables x and y, or as a function of the pair x, y. We indicate this functional dependence by writing Tfx, y.
The volume V of a circular cylinder depends on its radius r and its height h. In fact, we knowthatV r2h.WesaythatVisafunctionofrandh,andwewriteVr,h r2h.
We often write zf x, y to make explicit the value taken on by f at the general point x, y. The variables x and y are independent variables and z is the dependent variable. Compare this with the notation yf x for functions of a single variable.
A function of two variables is just a function whose domain is a subset of 2 and whose range is a subset of . One way of visualizing such a function is by means of an arrow dia gram see Figure 1, where the domain D is represented as a subset of the xyplane.
DEFINITION A function f of two variables is a rule that assigns to each ordered pair of real numbers x, y in a set D a unique real number denoted by f x, y. The set D is the domain of f and its range is the set of values that f takes on, that is,fx, y x, yD.
y
x,y
fa,b
0 x 0fx,yz
D
a,b
f
FIGURE 1
If a function f is given by a formula and no domain is specified, then the domain of f is understood to be the set of all pairs x, y for which the given expression is a well defined real number.
EXAMPLE 1 For each of the following functions, evaluate f 3, 2 and find the domain.
a fx,y sxy1 x1
b fx,yxlny2 x
f 3, 2s321s6 31 2
SOLUTION
a
855
856CHAPTER 14 PARTIAL DERIVATIVES
xy10
y
x1
10 x 1
Domain of fx,yxy1 x1
The expression for f makes sense if the denominator is not 0 and the quantity under the square root sign is nonnegative. So the domain of f is
Dx,yxy10, x1
The inequality xy10, or yx1, describes the points that lie on or above the line yx1, while x1 means that the points on the line x1 must be excluded from the domain. See Figure 2.
b f3, 23 ln2233 ln 10
Sincelny2 xisdefinedonlywheny2 x0,thatis,xy2,thedomainof f is
Dx, y xy2. This is the set of points to the left of the parabola xy2. See Figure 3. M
Not all functions are given by explicit formulas. The function in the next example is described verbally and by numerical estimates of its values.
EXAMPLE 2 In regions with severe winter weather, the windchill index is often used to describe the apparent severity of the cold. This index W is a subjective temperature that depends on the actual temperature T and the wind speed v. So W is a function of T and v, and we can write Wf T, v. Table 1 records values of W compiled by the NOAA National Weather Service of the US and the Meteorological Service of Canada.
Wind speed kmh
FIGURE 2
y
0x
FIGURE 3
Domain of fx, yx lnx
TABLE 1
Windchill index as a function of air temperature and wind speed
N THE NEW WINDCHILL INDEX
A new windchill index was introduced in
November of 2001 and is more accurate than the old index at measuring how cold it feels when its windy. The new index is based on a model of how fast a human face loses heat. It was devel oped through clinical trials in which volunteers were exposed to a variety of temperatures and wind speeds in a refrigerated wind tunnel.
x
Tv
5
10
15
20
25
30
40
50
60
70
80
5
4
3
2
1
1
0
1
1
2
2
3
0
2
3
4
5
6
6
7
8
9
9
10
5
7
9
11
12
12
13
14
15
16
16
17
10
13
15
17
18
19
20
21
22
23
23
24
15
19
21
23
24
25
26
27
29
30
30
31
20
24
27
29
30
32
33
34
35
36
37
38
25
30
33
35
37
38
39
41
42
43
44
45
30
36
39
41
43
44
46
48
49
50
51
52
35
41
45
48
49
51
52
54
56
57
58
60
40
47
51
54
56
57
59
61
63
64
65
67
For instance, the table shows that if the temperature is 5C and the wind speed is 50 kmh, then subjectively it would feel as cold as a temperature of about 15C with no wind. So
f5, 5015 M
EXAMPLE 3 In 1928 Charles Cobb and Paul Douglas published a study in which they modeled the growth of the American economy during the period 18991922. They con
Actual temperature C
TABLE 2
sidered a simplified view of the economy in which production output is determined by the amount of labor involved and the amount of capital invested. While there are many other factors affecting economic performance, their model proved to be remarkably accurate. The function they used to model production was of the form
PL, KbL K1
where P is the total production the monetary value of all goods produced in a year,
L is the amount of labor the total number of personhours worked in a year, and K is the amount of capital invested the monetary worth of all machinery, equipment, and buildings. In Section 14.3 we will show how the form of Equation 1 follows from cer tain economic assumptions.
Cobb and Douglas used economic data published by the government to obtain Table 2. They took the year 1899 as a baseline, and P, L, and K for 1899 were each assigned the value 100. The values for other years were expressed as percentages of the 1899 figures.
Cobb and Douglas used the method of least squares to fit the data of Table 2 to the function
PL, K1.01L0.75K0.25
See Exercise 75 for the details.
If we use the model given by the function in Equation 2 to compute the production in
the years 1910 and 1920, we get the values
P147, 2081.011470.752080.25161.9
P194, 4071.011940.754070.25235.8
which are quite close to the actual values, 159 and 231.
The production function 1 has subsequently been used in many settings, ranging
from individual firms to global economic questions. It has become known as the CobbDouglas production function. Its domain is L, KL0, K0 because
L and K represent labor and capital and are therefore never negative. M
EXAMPLE 4 Findthedomainandrangeoftx,ys9x2 y2. SOLUTION The domain of t is
Dx,y9x2 y2 0x,yx2 y2 9
which is the disk with center 0, 0 and radius 3. See Figure 4. The range of t is
zzs9x2 y2,x,yD Since z is a positive square root, z0. Also
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES857
. Year
P
L
K
1899
1900
1901
1902
1903
1904
1905
1906
1907
1908
1909
1910
1911
1912
1913
1914
1915
1916
1917
1918
1919
1920
1921
1922
100
101
112
122
124
122
143
152
151
126
155
159
153
177
184
169
189
225
227
223
218
231
179
240
100
105
110
117
122
121
125
134
140
123
143
147
148
155
156
152
156
183
198
201
196
194
146
161
100
107
114
122
131
138
149
163
176
185
198
208
216
226
236
244
266
298
335
366
387
407
417
431
1
2
y
9
3 3x
FIGURE 4
Domain of gx, y9
So the range is
9×2 y2 9 ? s9x2 y2 3
z0z30, 3 M
858
CHAPTER 14 PARTIAL DERIVATIVES GRAPHS
z
S
Another way of visualizing the behavior of a function of two variables is to consider its graph.
x, y, fx, y
fx,y
D x,y,0 y
0
FIGURE 5
Just as the graph of a function f of one variable is a curve C with equation yf x, so the graph of a function f of two variables is a surface S with equation zf x, y. We can visualize the graph S of f as lying directly above or below its domain D in the xyplane. See Figure 5.
EXAMPLE 5 Sketch the graph of the function f x, y63x2y.
SOLUTION Thegraphof f hastheequationz63x2y,or3x2yz6,which represents a plane. To graph the plane we first find the intercepts. Putting yz0 in the equation, we get x2 as the xintercept. Similarly, the yintercept is 3 and the zintercept is 6. This helps us sketch the portion of the graph that lies in the first octant. See Figure 6.
x
z
0, 0, 6
z 0,0,3
so it is a plane. In much the same way that linear functions of one variable are important in singlevariable calculus, we will see that linear functions of two variables play a central role in multivariable calculus.
V EXAMPLE 6 Sketchthegraphoftx,ys9x2 y2.
SOLUTION The graph has equation zs9x2y2 . We square both sides of this equa tiontoobtainz2 9×2 y2,orx2 y2 z2 9,whichwerecognizeasanequa tion of the sphere with center the origin and radius 3. But, since z0, the graph of t is just the top half of this sphere see Figure 7. M
FIGURE 6
M
2, 0, 0
x
0, 3, 0
y
The function in Example 5 is a special case of the function
fx, yaxbyc
which is called a linear function. The graph of such a function has the equation
zaxbyc or axbyzc0
0
x
FIGURE 7
0, 3, 0 3, 0, 0
y
Graph of gx, y9
DEFINITION If f is a function of two variables with domain D, then the graph of f isthesetofallpointsx,y,zin3 suchthatzfx,yandx,yisinD.
FIGURE 9
Graph of hx, y4
x
yM
FIGURE 8
M
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES859
NOTE An entire sphere cant be represented by a single function of x and y. As we saw in Example 6, the upper hemisphere of the sphere x2y2z29 is represented by the function tx, ys9x2y2 . The lower hemisphere is represented by the function hx,ys9x2 y2.
EXAMPLE 7 Use a computer to draw the graph of the CobbDouglas production function PL, K1.01L0.75K0.25.
SOLUTION Figure 8 shows the graph of P for values of the labor L and capital K that lie between 0 and 300. The computer has drawn the surface by plotting vertical traces. We see from these traces that the value of the production P increases as either L or K increases, as is to be expected.
300
200
P 100 0
300 200 100 200 K 00100L
300
V EXAMPLE 8 Find the domain and range and sketch the graph of hx, y4×2y2.
SOLUTION Notice that hx, y is defined for all possible ordered pairs of real numbers x, y, so the domain is 2, the entire xyplane. The range of h is the set 0,of all nonnega tiverealnumbers.Noticethatx2 0andy2 0,sohx,y0forallxandy.
The graph of h has the equation z4×2y2, which is the elliptic paraboloid that we sketched in Example 4 in Section 12.6. Horizontal traces are ellipses and vertical traces are parabolas see Figure 9.
z
Computer programs are readily available for graphing functions of two variables. In most such programs, traces in the vertical planes xk and yk are drawn for equally spaced values of k and parts of the graph are eliminated using hidden line removal.
860
CHAPTER 14 PARTIAL DERIVATIVES
x
FIGURE 10
y
x
y
a fx, y3e z
b fx, y3e z
Figure 10 shows computergenerated graphs of several functions. Notice that we get an especially good picture of a function when rotation is used to give views from different vantage points. In parts a and b the graph of f is very flat and close to the xyplane except near the origin; this is because ex 2 y 2 is very small when x or y is large.
zz
c fx, ysin xsin y
LEVEL CURVES
d fx, ysin x sin y
x
y xy
So far we have two methods for visualizing functions: arrow diagrams and graphs. A third method, borrowed from mapmakers, is a contour map on which points of constant eleva tion are joined to form contour curves, or level curves.
DEFINITION The level curves of a function f of two variables are the curves with equations f x, yk, where k is a constant in the range of f .
Alevelcurve fx,ykisthesetofallpointsinthedomainof f atwhich f takeson a given value k. In other words, it shows where the graph of f has height k.
You can see from Figure 11 the relation between level curves and horizontal traces. The level curves f x, yk are just the traces of the graph of f in the horizontal plane zk projected down to the xyplane. So if you draw the level curves of a function and visual ize them being lifted up to the surface at the indicated height, then you can mentally piece
x
z 45
0 x
fx, y20
y
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES
861
LONESOME MTN.
A B
k45
k40 k35
k30 k25
FIGURE 11
FIGURE 12
TEC Visual 14.1A animates Figure 11 by showing level curves being lifted up to graphs of functions.
together a picture of the graph. The surface is steep where the level curves are close together. It is somewhat flatter where they are farther apart.
One common example of level curves occurs in topographic maps of mountainous regions, such as the map in Figure 12. The level curves are curves of constant elevation above sea level. If you walk along one of these contour lines, you neither ascend nor descend. Another common example is the temperature function introduced in the opening paragraph of this section. Here the level curves are called isothermals and join locations with the same temperature. Figure 13 shows a weather map of the world indicating the average January temperatures. The isothermals are the curves that separate the colored bands. The isobars in the atmospheric pressure map on page 854 provide another example of level curves.
k20
FIGURE 13
World mean sealevel temperatures
in January in degrees Celsius
Tarbuck, Atmosphere: Introduction to Meteorology, 4th Edition,1989. Reprinted by permission of Pearson Education, Inc., Upper Saddle River, NJ.
4000
4500
5000
5500
5000
4500
k
e
e
r
esom
ne
oC
L
862y
5 4 3 2 1
CHAPTER 14 PARTIAL DERIVATIVES
50 70
80
012345x
FIGURE 14
y 0x
Similarly, we estimate that
EXAMPLE 10 Sketch the level curves of the function f x, y63x2y for the
50
EXAMPLE 9 A contour map for a function f is shown in Figure 14. Use it to estimate the values of f 1, 3 and f 4, 5.
SOLUTION The point 1, 3 lies partway between the level curves with zvalues 70 and 80. We estimate that
Contour map of gx, y9
M
7800
60
60
FIGURE 16
values k6, 0, 6, 12. SOLUTION The level curves are
EXAMPLE 12 Sketch some level curves of the function hx, y4×2y2. SOLUTION The level curves are
22 x2y2
4x y k or k4 k 1
f 1, 373
f 4, 556 M
63x2yk or 3x2yk60 This is a family of lines with slope 3. The four particular level curves with
k6, 0, 6, and 12 are 3x2y120, 3x2y60, 3x2y0, and
3x2y60. They are sketched in Figure 15. The level curves are equally spaced parallel lines because the graph of f is a plane see Figure 6. M
2
FIGURE 15
Contour map of
fx, y63x2y
V EXAMPLE 11 Sketch the level curves of the function
tx, ys9x2y2 for k0, 1, 2, 3
SOLUTION The level curves are
s9x2 y2 k or x2 y2 9k2
This is a family of concentric circles with center 0, 0 and radius s9k2 . The cases k0, 1, 2, 3 are shown in Figure 16. Try to visualize these level curves lifted up to form a surface and compare with the graph of t a hemisphere in Figure 7. See TEC Visual 14.1A.
y
0
k3k2 k1
k0 3,0 x
k6 k12
k6 k0
TEC Visual 14.1B demonstrates the connection between surfaces and their contour maps.
FIGURE 17
The graph of hx, y4 is formed by lifting the level curves.
x
y
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES863
which, for k0, describes a family of ellipses with semiaxes sk2 and sk . Figure 17a shows a contour map of h drawn by a computer with level curves corresponding to
k0.25, 0.5, 0.75, . . . , 4. Figure 17b shows these level curves lifted up to the graph of h an elliptic paraboloid where they become horizontal traces. We see from Figure 17 how the graph of h is put together from the level curves.
a Contour map
EXAMPLE 13 Plot level curves for the CobbDouglas production function of Example 3.
SOLUTION In Figure 18 we use a computer to draw a contour plot for the CobbDouglas production function
PL, K1.01L0.75K0.25 K
FIGURE 18 100
200
300 L
300
200
100
220 180
140 100
x
z
b Horizontal traces are raised level curves M
Level curves are labeled with the value of the production P. For instance, the level curve labeled 140 shows all values of the labor L and capital investment K that result in a pro duction of P140. We see that, for a fixed value of P, as L increases K decreases, and M vice versa.
For some purposes, a contour map is more useful than a graph. That is certainly true in Example 13. Compare Figure 18 with Figure 8. It is also true in estimating function val ues, as in Example 9.
y
864
CHAPTER 14 PARTIAL DERIVATIVES
FIGURE 19
d fx, y 3y 1
Figure 19 shows some computergenerated level curves together with the corresponding computergenerated graphs. Notice that the level curves in part c crowd together near the origin. That corresponds to the fact that the graph in part d is very steep near the origin.
yzz
x
x
c Level curves of fx, y 3y 1
y
x
y
a Level curves of fx, yxye yz
FUNCTIONS OF THREE OR MORE VARIABLES
b Two views of fx, yxye
x
A function of three variables, f , is a rule that assigns to each ordered triple x, y, z in a domain D3 a unique real number denoted by fx, y, z. For instance, the temperature T at a point on the surface of the earth depends on the longitude x and latitude y of the point and on the time t, so we could write Tfx, y, t.
EXAMPLE 14 Find the domain of f if
fx, y, zlnzyxy sin z
SOLUTION The expression for f x, y, z is defined as long as zy0, so the domain of f is
Dx, y, z3 zy
This is a halfspace consisting of all points that lie above the plane zy. M
x
FIGURE 20
Functions of any number of variables can be considered. A function of n variables is a rule that assigns a number zf x1, x2, . . . , xnto an ntuple x1, x2, . . . , xnof real numbers. We denote by n the set of all such ntuples. For example, if a company uses n different ingredients in making a food product, ci is the cost per unit of the ith ingredient, and xi units of the ith ingredient are used, then the total cost C of the ingredients is a func tion of the n variables x1, x2, . . . , xn:
3 Cfx1,x2,…,xnc1x1 c2x2 cnxn
The function f is a realvalued function whose domain is a subset of n. Sometimes we will use vector notation to write such functions more compactly: If xx1, x2, . . . , xn , we often write fx in place of fx1,x2,…,xn. With this notation we can rewrite the function defined in Equation 3 as
fxcx
wherec c1,c2,…,cn andcxdenotesthedotproductofthevectorscandxinVn. In view of the onetoone correspondence between points x1, x2, . . . , xn in n and their position vectors xx1, x2, . . . , xnin Vn, we have three ways of looking at a func
tion f defined on a subset of n:
1. As a function of n real variables x1, x2, …, xn
2. Asafunctionofasinglepointvariablex1,x2,…,xn
3. Asafunctionofasinglevectorvariablexx1,x2,…,xn
We will see that all three points of view are useful.
z
z9
z4
14.1 EXERCISES
1. In Example 2 we considered the function Wf T, v, where W is the windchill index, T is the actual temperature, and v is the wind speed. A numerical representation is given in Table 1. a What is the value of f 15, 40? What is its meaning?
b Describe in words the meaning of the question For what value of v is f 20, v30 ? Then answer the question.
c Describe in words the meaning of the question For what value of T is f T, 2049 ? Then answer the question.
d What is the meaning of the function Wf 5, v? Describe the behavior of this function.
e What is the meaning of the function Wf T, 50? Describe the behavior of this function.
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES865
Its very difficult to visualize a function f of three variables by its graph, since that would lie in a fourdimensional space. However, we do gain some insight into f by exam ining its level surfaces, which are the surfaces with equations f x, y, zk, where k is a constant. If the point x, y, z moves along a level surface, the value of f x, y, z remains fixed.
EXAMPLE 15 Find the level surfaces of the function fx,y,zx2 y2 z2
SOLUTION The level surfaces are x2y2z2k, where k0. These form a family of concentric spheres with radius sk . See Figure 20. Thus, as x, y, z varies over any sphere with center O, the value of f x, y, z remains fixed. M
y
z1
866CHAPTER 14 PARTIAL DERIVATIVES
2. The temperaturehumidity index I or humidex, for short is the perceived air temperature when the actual temperature is T and the relative humidity is h, so we can write If T, h. The fol lowing table of values of I is an excerpt from a table compiled by the National OceanicAtmospheric Administration.
TABLE 4
Duration hours
t
5
10
15
20
30
40
50
10
2
2
2
2
2
2
2
15
4
4
5
5
5
5
5
20
5
7
8
8
9
9
9
30
9
13
16
17
18
19
19
40
14
21
25
28
31
33
33
50
19
29
36
40
45
48
50
60
24
37
47
54
62
67
69
TABLE 3
Apparent temperature as a function of temperature and humidity
Relative humidity
Th
20
30
40
50
60
70
80
77
78
79
81
82
83
85
82
84
86
88
90
93
90
87
90
93
96
100
106
95
93
96
101
107
114
124
100
99
104
110
120
132
144
a What is the value of f95, 70? What is its meaning?
b For what value of h is f90, h100?
c For what value of T is fT, 5088?
d What are the meanings of the functions If 80, h
and If 100, h? Compare the behavior of these two functions of h.
3. Verify for the CobbDouglas production function PL, K 1.01L 0.75K 0.25
discussed in Example 3 that the production will be doubled if both the amount of labor and the amount of capital are doubled. Determine whether this is also true for the general production function
PL, KbL K1
4. The windchill index W discussed in Example 2 has been
modeled by the following function:
WT, v13.120.6215T11.37v 0.160.3965Tv 0.16 Check to see how closely this model agrees with the values in
Table 1 for a few values of T and v.
the behavior of this function.
c What is the meaning of the function hf v, 30? Describe
7. Let fx, yx2e3xy. a Evaluate f 2, 0. c Find the range of f.
6. Let fx,ylnxy1.
a Evaluate f 1, 1.
c Find and sketch the domain of f. d Find the range of f.
b Find the domain of f. 8. Find and sketch the domain of the function
fx,ys1xy2.Whatistherangeof f?
9. Let f x, y, zeszx2y2 .
a Evaluate f 2, 1, 6. b Find the domain of f. c Find the range of f.
10. Lettx,y,zln25x2 y2 z2.
a Evaluate t2, 2, 4. b Find the domain of t. c Find the range of t.
1120 Find and sketch the domain of the function. 11. fx,ysxy
12. f x, ysxy fx,yln9x2 9y2
14. fx,ysyxlnyx
15. fx,ys1x2 s1y2 16. fx,ysy s25x2 y2
13.
b Evaluate f e, 1.
5.
The wave heights h in the open sea depend on the speed v
of the wind and the length of time t that the wind has been
blowing at that speed. Values of the function hf v, t are 1x recorded in feet in Table 4.
a What is the value of f 40, 15? What is its meaning?
b What is the meaning of the function hf 30, t? Describe
syx2 17. f x, y2
18. fx,yarcsinx2 y2 2
19. fx,y,zs1x2 y2 z2
20. fx,y,zln164x2 4y2 z2
the behavior of this function.
Actual temperature F
Wind speed knots
2129 Sketch the graph of the function.
32. Two contour maps are shown. One is for a function f whose graph is a cone. The other is for a function t whose graph is a paraboloid. Which is which, and why?
I y II y xx
33. Locate the points A and B in the map of Lonesome Mountain Figure 12. How would you describe the terrain near A? NearB?
34. Make a rough sketch of a contour map for the function whose graph is shown.
21.
23.
25. 27. 28. 29.
30.
fx,y3 fx,y104x5y
fx,yy2 1 fx,y4x2y21 fx,ys16x2 16y2 fx,ysx2y2
22. fx,yy
24. fx,ycosx
26. fx,y3x2 y2
Match the function with its graph labeled IVI.Give reasons for your choices.
b fx, yxy
d fx, yx2y22
xyxy
a fx, yx y c fx, y1
1×2 y2 e fx, yxy2
f fx, ysinx y I z II z
III z
x
V z
IV z yx
y
xy VI z
3538 A contour map of a function is shown. Use it to make a rough sketch of the graph of f.
xyxyx
31. A contour map for a function f is shown. Use it to estimate the values of f 3, 3 and f 3, 2. What can you say about the shape of the graph?
y
170605040 01 30x
37.y5 38.y 4
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES867
35. y 36. y 14
20 0 x 10
13 12 11
z
8
6 4
x
8
3 2
1 210
3
2 1
00 2345 3 x 1
868CHAPTER 14 PARTIAL DERIVATIVES
3946 Draw a contour map of the function showing several level
6164 Describe the level surfaces of the function. f x, y, zx3y5z
62. fx,y,zx2 3y2 5z2 63. fx,y,zx2 y2 z2 64. fx,y,zx2 y2
6566 Describe how the graph of t is obtained from the graph of f.
curves.
39. fx,yy2x2
41. fx,yylnx fx,yyex
40. fx,yx3 y
42. fx,yeyx
44. fx,yy secx
46. fx,yyx2 y2
49. A thin metal plate, located in the xyplane, has temperature Tx, y at the point x, y. The level curves of T are called isothermals because at all points on an isothermal the temper ature is the same. Sketch some isothermals if the temperature function is given by
Tx,y1001x2 2y2
50. If Vx, y is the electric potential at a point x, y in the xyplane, then the level curves of V are called equipotential curves because at all points on such a curve the electric potential is the same. Sketch some equipotential curves if Vx, ycsr2x2y2 , where c is a positive constant.
; 5154 Use a computer to graph the function using various domains and viewpoints. Get a printout of one that, in your opin ion, gives a good view. If your software also produces level curves, then plot some contour lines of the same function and compare with the graph.
51. fx,yex2 e2y2
52. fx, y13×2y2e1x2y2
43.
45. fx,ysy2 x2
4748 Sketch both a contour map and a graph of the function
and compare them.
47. fx,yx2 9y2
48. fx,ys369x2 4y2
a tx,yfx,y2 c tx,yfx,y
66. a tx,yfx2,y
c tx,yfx3,y4
b tx,y2fx,y d tx,y2fx,y
b tx,yfx,y2
23 53. f x, yxyx
54. fx, yxy3yx3
monkey saddle dog saddle
; 6768 Use a computer to graph the function using various domains and viewpoints. Get a printout that gives a good view of the peaks and valleys. Would you say the function has a maxi mum value? Can you identify any points on the graph that you might consider to be local maximum points? What about local minimum points?
67. fx,y3xx4 4y2 10xy 68. fx,yxyex2y2
; 6970 Use a computer to graph the function using various domains and viewpoints. Comment on the limiting behavior of the function. What happens as both x and y become large? What happens as x, y approaches the origin?
69. fx,y xy 70. fx,y xy x2 y2 x2 y2
; 71. Use a computer to investigate the family of functions
fx, yecx2y2. How does the shape of the graph depend
on c?
; 72. Use a computer to investigate the family of surfaces
zax2by2ex2y2
How does the shape of the graph depend on the numbers a
and b?
; 73. Use a computer to investigate the family of surfaces
zx2y2cxy. In particular, you should determine the transitional values of c for which the surface changes from one type of quadric surface to another.
5560 Match the function a with its graph labeled AF on page 869 and b with its contour map labeled IVI. Give reasons for your choices.
zsinxy 57. zsinxy
59. z1x21y2
56. zex cosy
58. zsinxsiny
55.
60. z
xy 1×2 y2
61.
65.
SECTION 14.1 FUNCTIONS OF SEVERAL VARIABLES Graphs and Contour Maps for Exercises 5560
869
AzBzCz
x
y
y
x
y
x
DzEzFz x
I y
y
xyx
II y III y
y
xxx
IVy VyVIy
xxx
870CHAPTER 14 PARTIAL DERIVATIVES
; 74. Graph the functions fx, ysx2y2
; 75. a fx,ysinsx2 y2
c Deduce that the CobbDouglas production function is P1.01L0.75K0.25.
and
fx,ylnsx2 y2 fx, y
fx, yesx2y2
1 b
Show that, by taking logarithms, the general Cobb Douglas function PbL K1 can be expressed as
lnPlnb lnL K K
If we let xlnLK and ylnPK, the equation in part a becomes the linear equation yxln b. Use Table 2 in Example 3 to make a table of values of lnLK and lnPK for the years 18991922. Then use a graphing calculator or computer to find the least squares regression line through the points lnLK, lnPK.
sx2y2
In general, if t is a function of one variable, how is the graph
of
obtained from the graph of t?
14.2
f x, ytsx 2y 2
LIMITS AND CONTINUITY
Lets compare the behavior of the functions
sinx2 y2 x2 y2 fx,y x2 y2 and tx,y x2 y2
TABLE 1
as x and y both approach 0 and therefore the point x, y approaches the origin. Values of fx, y TABLE 2 Values of tx, y
xy
1.0
0.5
0.2
0
0.2
0.5
1.0
1.0
0.000
0.600
0.923
1.000
0.923
0.600
0.000
0.5
0.600
0.000
0.724
1.000
0.724
0.000
0.600
0.2
0.923
0.724
0.000
1.000
0.000
0.724
0.923
0
1.000
1.000
1.000
1.000
1.000
1.000
0.2
0.923
0.724
0.000
1.000
0.000
0.724
0.923
0.5
0.600
0.000
0.724
1.000
0.724
0.000
0.600
1.0
0.000
0.600
0.923
1.000
0.923
0.600
0.000
xy
1.0
0.5
0.2
0
0.2
0.5
1.0
1.0
0.455
0.759
0.829
0.841
0.829
0.759
0.455
0.5
0.759
0.959
0.986
0.990
0.986
0.959
0.759
0.2
0.829
0.986
0.999
1.000
0.999
0.986
0.829
0
0.841
0.990
1.000
1.000
0.990
0.841
0.2
0.829
0.986
0.999
1.000
0.999
0.986
0.829
0.5
0.759
0.959
0.986
0.990
0.986
0.959
0.759
1.0
0.455
0.759
0.829
0.841
0.829
0.759
0.455
Tables 1 and 2 show values of f x, y and tx, y, correct to three decimal places, for points x, y near the origin. Notice that neither function is defined at the origin. It appears that as x, y approaches 0, 0, the values of f x, y are approaching 1 whereas the values of tx, y arent approaching any number. It turns out that these guesses based on numerical evidence are correct, and we write
sinx2 y2 lim 2 2
and lim
x, y l 0, 0
x2 y2
x
does not exist
1 In general, we use the notation
2 2
x, y l 0, 0 xy
y
lim
x, yla, b
fx, yL
Other notations for the limit in Definition 1 are
lim fx,yL and fx,ylL as x,yla,b xla
ylb
Notice thatf x, yLis the distance between the numbers f x, y and L, and sxa2yb2 is the distance between the point x, y and the point a, b. Thus Definition 1 says that the distance between f x, y and L can be made arbitrarily small by making the distance from x, y to a, b sufficiently small but not 0. Figure 1 illustrates Definition 1 by means of an arrow diagram. If any small interval L, L is given around L, then we can find a disk D with center a, b and radius0 such that f maps all the points in D except possibly a, b into the interval L, L.
yz
x a,b D FIGURE 2
Another illustration of Definition 1 is given in Figure 2 where the surface S is the graph of f.If0isgiven,wecanfind 0suchthatifx,yisrestrictedtolieinthedisk D and x, ya, b, then the corresponding part of S lies between the horizontal planes zL and zL.
For functions of a single variable, when we let x approach a, there are only two pos sible directions of approach, from the left or from the right. We recall from Chapter 2 that if limxla fxlimxla fx, then limxl a fx does not exist.
For functions of two variables the situation is not as simple because we can let x, y approach a, b from an infinite number of directions in any manner whatsoever see Figure 3 as long as x, y stays within the domain of f.
SECTION 14.2 LIMITS AND CONTINUITY871
to indicate that the values of f x, y approach the number L as the point x, y approaches the point a, b along any path that stays within the domain of f. In other words, we can make the values of f x, y as close to L as we like by taking the point x, y sufficiently close to the point a, b, but not equal to a, b. A more precise definition follows.
DEFINITION Let f be a function of two variables whose domain D includes points arbitrarily close to a, b. Then we say that the limit of f x, y as x, y approaches a, b is L and we write
lim fx, yL x, yla, b
if for every number 0 there is a corresponding number0 such that
if x,yD and 0sxa2 yb2then fx,yL
1
x,y
D LS
LL 0
0 a,b x f z 0 LLL
FIGURE 1
y
b 0ax
FIGURE 3
y
872
CHAPTER 14 PARTIAL DERIVATIVES
FIGURE 4
at the beginning of this section. EXAMPLE 2 If fx, yxy
M
y f1
Then y0 gives fx, 0x2x21 for all x0, so
f x, y l 1 as x, y l 0, 0 along the xaxis
y2
We now approach along the yaxis by putting x0. Then f0, yy21 for
y f0
f0
yx x
f x, y l 0 as x, y l 0, 0 along the xaxis Ifx0,then f0,y0y2 0,so
f x, y l 0 as x, y l 0, 0 along the yaxis
Although we have obtained identical limits along the axes, that does not show that the
Definition 1 says that the distance between f x, y and L can be made arbitrarily small by making the distance from x, y to a, b sufficiently small but not 0. The definition refers only to the distance between x, y and a, b. It does not refer to the direction of approach. Therefore, if the limit exists, then f x, y must approach the same limit no mat ter how x, y approaches a, b. Thus if we can find two different paths of approach along which the function f x, y has different limits, then it follows that limx, y l a, b f x, y does not exist.
If fx,ylL1 asx,yla,balongapathC1 and fx,ylL2 as
x, y l a, b along a path C2, where L1L2, then limx, yla, b fx, y does not exist.
x2 y2
V EXAMPLE 1 Show that lim 2 2 does not exist.
x, yl0, 0 xy
SOLUTION Let f x, yx2y2 x2y2 . First lets approach 0, 0 along the xaxis.
f1
all y0, so x
fx, y l 1 as x, y l 0, 0 along the yaxis
See Figure 4. Since f has two different limits along two different lines, the given limit does not exist. This confirms the conjecture we made on the basis of numerical evidence
, does lim fx, y exist? x, yl0, 0
x2y2
SOLUTION If y0, then fx, 00×20. Therefore
f 21
given limit is 0. Lets now approach 0, 0 along another line, say yx. For all x0, x2 1
Therefore
f x, y l 1 2
fx,xx2x2 2
as x, y l 0, 0 along yx
FIGURE 5
See Figure 5. Since we have obtained different limits along different paths, the given limit does not exist. M
TEC InVisual 14.2 a rotating line on the surface in Figure 6 shows different limits at the origin from different directions.
FIGURE 6
fx,y xy
zy
x
N Figure 7 shows the graph of the function in Example 3. Notice the ridge above the parabola xy2.
0.5 z0
0.5 2
FIGURE 7
SOLUTION With the solution of Example 2 in mind, lets try to save time by letting
x, y l 0, 0 along any nonvertical line through the origin. Then ymx, where m is the slope, and
xmx2 m2x3 m2x fx,yfx,mx x2 mx4x2 m4x41m4x2
So fx, y l 0 as x, y l 0, 0 along ymx
Thus f has the same limiting value along every nonvertical line through the origin. But that does not show that the given limit is 0, for if we now let x, y l 0, 0 along the parabola xy2, we have
SECTION 14.2 LIMITS AND CONTINUITY873 Figure 6 sheds some light on Example 2. The ridge that occurs above the line yx cor
responds to the fact that f x, y1 for all points x, y on that line except the origin. 2
V EXAMPLE 3 If fx,y 2 x
xy2y
4 ,does lim fx,yexist?x, y l 0, 0
2 y2y2 y4 1 fx,yfy ,y y22 y42y42
02
2 y so
Now lets look at limits that do exist. Just as for functions of one variable, the calcula tion of limits for functions of two variables can be greatly simplified by the use of proper ties of limits. The Limit Laws listed in Section 2.3 can be extended to functions of two variables: The limit of a sum is the sum of the limits, the limit of a product is the product of the limits, and so on. In particular, the following equations are true.
0 2 x2
fx, y l 1 as x, y l 0, 0 along xy2
Since different paths lead to different limiting values, the given limit does not exist. M
2 lim xa x, yla, b
The Squeeze Theorem also holds. EXAMPLE 4 Find lim 3x2y
SOLUTION As in Example 3, we could show that the limit along any line through the
origin is 0. This doesnt prove that the given limit is 0, but the limits along the parabolas
x, yl0, 0 x2y2
lim yb lim cc x, yla, b x, yla, b
if it exists.
874
CHAPTER 14 PARTIAL DERIVATIVES
yx2 andxy2 alsoturnouttobe0,sowebegintosuspectthatthelimitdoesexist and is equal to 0.
Let0.Wewanttofind 0suchthat
if 0sx2y2 then3x2y0
x2 y2
thatis, if 0sx2 y2then 3x2y
x2 y2 Butx2 x2 y2 sincey2 0,sox2x2 y21andtherefore
N Another way to do Example 4 is to use the Squeeze Theorem instead of Definition 1. From 2 it follows that
lim 3 y 0x, y l 0, 0
and so the first inequality in 3 shows that the given limit is 0.
3x2y3y 3sy23sx2y2 x2 y2
Thusifwechoose 3andlet0sx2 y2,then
3x2y 03sx2y2 3 3
3
x2 y2 3 Hence, by Definition 1,
3x2y
lim2 20 M
Recall that evaluating limits of continuous functions of a single variable is easy. It can be accomplished by direct substitution because the defining property of a continuous function islimxla fxfa.Continuousfunctionsoftwovariablesarealsodefinedbythedirect substitution property.
CONTINUITY
x, yl0, 0 xy
DEFINITION A function f of two variables is called continuous at a, b if lim fx, yfa, b
x, yla, b
We say f is continuous on D if f is continuous at every point a, b in D.
4
The intuitive meaning of continuity is that if the point x, y changes by a small amount, then the value of f x, y changes by a small amount. This means that a surface that is the graph of a continuous function has no hole or break.
Using the properties of limits, you can see that sums, differences, products, and quo tients of continuous functions are continuous on their domains. Lets use this fact to give examples of continuous functions.
A polynomial function of two variables or polynomial, for short is a sum of terms of the form cxmyn, where c is a constant and m and n are nonnegative integers. A rational function is a ratio of polynomials. For instance,
fx,yx4 5x3y2 6xy4 7y6
N Figure 8 shows the graph of the continuous function in Example 8.
EXAMPLE 8 Let
FIGURE 8
x, yl0, 0
Therefore f is continuous at 0, 0, and so it is continuous on 2. M
z
if x, y0, 0
lim 3x2y0f0, 0 x, yl0, 0 x2y2
y x
0
We know f is continuous for x, y0, 0 since it is equal to a rational function there.
is a polynomial, whereas
Also, from Example 4, we have
lim fx, y
3x2y fx,y x2y2
SECTION 14.2 LIMITS AND CONTINUITY875
tx,y 2xy1 x2 y2
is a rational function.
The limits in 2 show that the functions fx, yx, tx, yy, and hx, yc are
continuous. Since any polynomial can be built up out of the simple functions f , t, and h by multiplication and addition, it follows that all polynomials are continuous on 2. Likewise, any rational function is continuous on its domain because it is a quotient of continuous functions.
V EXAMPLE 5 Evaluate lim x2y3x3y23x2y. x, yl1, 2
SOLUTION Since fx, yx2y3x3y23x2y is a polynomial, it is continuous every where, so we can find the limit by direct substitution:
lim x2y3 x3y2 3x2y12 23 13 22 312211 M x, yl1, 2
x2 y2
EXAMPLE 6 Where is the function fx, yx2y2 continuous?
SOLUTION The function f is discontinuous at 0, 0 because it is not defined there.
Since f is a rational function, it is continuous on its domain, which is the set
Dx, y x, y0, 0. M
M
EXAMPLE 7 Let x2y2 tx,y x2y2
if x, y0, 0 0 if x, y0, 0
Here t is defined at 0, 0 but t is still discontinuous there because limx, y l 0, 0 tx, y does not exist see Example 1.
Just as for functions of one variable, composition is another way of combining two con tinuous functions to get a third. In fact, it can be shown that if f is a continuous function of two variables and t is a continuous function of a single variable that is defined on the range of f, then the composite function htf defined by hx, ytfx, y is also a continuous function.
if x, y0, 0
876
CHAPTER 14 PARTIAL DERIVATIVES
EXAMPLE 9 Where is the function hx, yarctanyx continuous?
2 z0 2
1 2
SOLUTION The function f x, yyx is a rational function and therefore continuous except on the line x0. The function ttarctan t is continuous everywhere. So the composite function
tfx, yarctanyxhx, y
is continuous except where x0. The graph in Figure 9 shows the break in the graph of
h above the yaxis. M FUNCTIONS OF THREE OR MORE VARIABLES
Everything that we have done in this section can be extended to functions of three or more variables. The notation
lim fx, y, zL x, y, zla, b, c
means that the values of f x, y, z approach the number L as the point x, y, z approaches the point a, b, c along any path in the domain of f. Because the distance between two pointsx,y,zanda,b,cin3 isgivenbysxa2 yb2 zc2,wecan write the precise definition as follows: For every number 0 there is a corresponding number0 such that
if x, y, z is in the domain of f and 0sxa2yb2zc2then fx,y,zL
The function f is continuous at a, b, c if
lim fx, y, zfa, b, c
2 1
0x
y01221
FIGURE 9
The function hx, yarctanyx is discontinuous where x0.
For instance, the function
x, y, zla, b, c
fx, y, z1
x2 y2 z2 1
is a rational function of three variables and so is continuous at every point in 3 except where x2y2z21. In other words, it is discontinuous on the sphere with center the origin and radius 1.
If we use the vector notation introduced at the end of Section 14.1, then we can write the definitions of a limit for functions of two or three variables in a single compact form as follows.
5 If f is defined on a subset D of n, then limxla fxL means that for every number 0 there is a corresponding number0 such that
if xD and 0xa then fxL
Notice that if n1, then xx and aa, and 5 is just the definition of a limit for functions of a single variable. For the case n2, we have xx, y, aa, b, and xasxa2 yb2, so 5 becomes Definition 1. If n3, then xx, y, z, aa, b, c, and 5 becomes the definition of a limit of a function of three variables. In each case the definition of continuity can be written as
lim fxfa xla
14.2 EXERCISES
1. Suppose that limx, y l 3, 1 f x, y6. What can you say
about the value of f 3, 1? What if f is continuous?
2. Explain why each function is continuous or discontinuous.
a The outdoor temperature as a function of longitude,
latitude, and time
b Elevation height above sea level as a function of longi
tude, latitude, and time
c The cost of a taxi ride as a function of distance traveled
and time
34 Use a table of numerical values of fx, y for x, y near the origin to make a conjecture about the value of the limit of f x, y as x, y l 0, 0. Then explain why your guess is correct.
3. fx,y x2y3 x3y2 5 4. fx,y 2xy 2xy x2 2y2
522 Find the limit, if it exists, or show that the limit does not exist.
24.
lim
x, yl0, 0
xy3 x2y6
SECTION 14.2 LIMITS AND CONTINUITY
877
2526 Find hx, yt fx, y and the set on which h is continuous.
ttt2 st, fx,y2x3y6 26. tttlnt, fx,y 1xy
1x2y2
; 2728 Graph the function and observe where it is discontinuous.
Then use the formula to explain what you have observed.
27. fx,ye
1 xy 1 fx,y1x2y2
5. 7.
11.
13.
15.
17. 19. 20. 21. 22.
6. lim
exy cosxy
29.
31. 33.
35. 36.
lim 5x3x2y2
x, yl1, 2 x,yl1,1
sinxy xy Fx,y ex y2 30. Fx,y 1×2 y2
4xy lim228.
1y2 2
2 Fx,yarctanx y 32. Fx,yex yxy2
2938 Determine the set of points at which the function is continuous.
x, yl2, 1 x3y y4
x3y xy cos y
lim ln
x, yl1, 0 x xy
ss
lim4 4
10. 12. 14. 16. 18.
x2sin2y lim 2 2
Gx,ylnx2 y2 4 34. Gx,ytan1xy2 sy
x, yl0, 0 lim
x, yl0, 0 lim
2x y 6x3y
fx,y,z x2 y2 z2 fx,y,zsxyz
3×2y2 x, yl0, 0 sx2y2
x, yl0, 0 2x4y4 44
x, yl0, 0
lim xy
lim
x, yl0, 0
xy x2y2
25.
28.
9.
x2yey lim4 2
x2 sin2y lim 22
x2y3 fx, y2×2y2
if x, y0, 0 if x, y0, 0
37.
x, yl0, 0 x4y
x2y xy4
1
fx,y x2xyy2
lim 2 2 x, yl0, 0 sxy
xy
x2y2
x, yl0, 0
lim 2 8
3941 Use polar coordinates to find the limit. If r,are
polar coordinates of the point x, y with r0, note that r l 0 as x, y l 0, 0.
11 exy sin z2
x2 2y2 3z2 x2y2z2
x, yl0, 0
xy
38.
if x, y0, 0 0 if x, y0, 0
lim
x, y, zl3, 0, 1
lim
x, y, zl0, 0, 0
lim
x, y, zl0, 0, 0
lim
x, y, zl0, 0, 0
22
xyyz xz x2y2z4 yz
x24y29z2
39.
40. 41.
lim
x, yl0, 0
lim
x, yl0, 0
lim
x, yl0, 0
x3y3 x2y2
x2 y2lnx2 y2 ex2y21
x2y2
; 2324 Use a computer graph of the function to explain why the limit does not exist.
23. lim 2×23xy4y2 x, yl0, 0 3×25y2
878CHAPTER 14 PARTIAL DERIVATIVES
; 42. At the beginning of this section we considered the function
fx,y sinx2 y2 x2y2
and guessed that fx, y l 1 as x, y l 0, 0 on the basis of numerical evidence. Use polar coordinates to confirm the value of the limit. Then graph the function.
44.
45. 46.
Let
fx,y 0 if y0 or yx4
4
a Show that f x, y l 0 as x, y l 0, 0 along any path through 0, 0 of the form ymxa with a4.
b Despite part a, show that f is discontinuous at 0, 0. c Show that f is discontinuous on two entire curves.
Show that the function f given by f xxis continuous onn. Hint:Considerxa2 xaxa.
IfcVn,showthatthefunction f givenby fxcxis continuous on n.
43. Graph and discuss the continuity of the function
sin xy fx,y xy
if xy0 if xy0
PARTIAL DERIVATIVES
1 if0yx
; 1
14.3
TABLE 1
Heat index I as a function of temperature and humidity
On a hot day, extreme humidity makes us think the temperature is higher than it really is, whereas in very dry air we perceive the temperature to be lower than the thermom eter indicates. The National Weather Service has devised the heat index also called the temperaturehumidity index, or humidex, in some countries to describe the combined effects of temperature and humidity. The heat index I is the perceived air temperature when the actual temperature is T and the relative humidity is H. So I is a function of T and H and we can write If T, H . The following table of values of I is an excerpt from a table compiled by the National Weather Service.
Relative humidity
Actual temperature F
If we concentrate on the highlighted column of the table, which corresponds to a rela tive humidity of H70, we are considering the heat index as a function of the single variable T for a fixed value of H. Lets write tT f T, 70. Then tTdescribes how the heat index I increases as the actual temperature T increases when the relative humidity is 70. The derivative of t when T96F is the rate of change of I with respect to T when T96F:
t96lim t96ht96lim f 96h, 70f 96, 70 hl0hhl0h
TH
50
55
60
65
70
75
80
85
90
90
96
98
100
103
106
109
112
115
119
92
100
103
105
108
112
115
119
123
128
94
104
107
111
114
118
122
127
132
137
96
109
113
116
121
125
130
135
141
146
98
114
118
123
127
133
138
144
150
157
100
119
124
129
135
141
147
154
161
168
SECTION 14.3 PARTIAL DERIVATIVES879 We can approximate t96 using the values in Table 1 by taking h2 and 2:
t96t98t96f 98, 70f 96, 701331254 222
t96t94t96f 94, 70f 96, 701181253.5 2 2 2
Averaging these values, we can say that the derivative t96 is approximately 3.75. This means that, when the actual temperature is 96F and the relative humidity is 70, the apparent temperature heat index rises by about 3.75F for every degree that the actual temperature rises!
Now lets look at the highlighted row in Table 1, which corresponds to a fixed temper ature of T96F. The numbers in this row are values of the function GHf96, H, which describes how the heat index increases as the relative humidity H increases when the actual temperature is T96F. The derivative of this function when H70 is the rate of change of I with respect to H when H70:
G70lim G70hG70lim f 96, 70hf 96, 70 hl0hhl0h
By taking h5 and 5, we approximate G70 using the tabular values: G70G75G70f 96, 75f 96, 701301251
555
G70G65G70f 96, 65f 96, 701211250.8
5 5 5
By averaging these values we get the estimate G700.9. This says that, when the tem perature is 96F and the relative humidity is 70, the heat index rises about 0.9F for every percent that the relative humidity rises.
In general, if f is a function of two variables x and y, suppose we let only x vary while keeping y fixed, say yb, where b is a constant. Then we are really considering a func tion of a single variable x, namely, txf x, b. If t has a derivative at a, then we call it the partial derivative of f with respect to x at a, b and denote it by fxa, b. Thus
By the definition of a derivative, we have
1
fxa, bta where txfx, b
talim tahta hl0 h
and so Equation 1 becomes
2
fxa,blim fah,bfa,b hl0 h
880
CHAPTER 14 PARTIAL DERIVATIVES
Similarly, the partial derivative of f with respect to y at a, b, denoted by fya, b, is obtained by keeping x fixed xa and finding the ordinary derivative at b of the func tion Gyfa, y:
With this notation for partial derivatives, we can write the rates of change of the heat index I with respect to the actual temperature T and relative humidity H when T96F and H70 as follows:
fT 96, 703.75 fH96, 700.9
If we now let the point a, b vary in Equations 2 and 3, fx and fy become functions of
two variables.
3
fya,blim fa,bhfa,b hl0 h
If f is a function of two variables, its partial derivatives are the functions fx and fy defined by
fxx,ylim fxh,yfx,y hl0 h
fyx,ylim fx,yhfx,y hl0 h
4
There are many alternative notations for partial derivatives. For instance, instead of fx we can write f1 or D1 f to indicate differentiation with respect to the first variable or
fx. But here fx cant be interpreted as a ratio of differentials.
NOTATIONS FOR PARTIAL DERIVATIVES If zf x, y, we write fxx,yfxf fx,y z f1 D1fDxf
fyx,yfyf fx,y z f2 D2fDyf yy y
xx x
To compute partial derivatives, all we have to do is remember from Equation 1 that the partial derivative with respect to x is just the ordinary derivative of the function t of a single variable that we get by keeping y fixed. Thus we have the following rule.
RULE FOR FINDING PARTIAL DERIVATIVES OF zf x, y
1. To find fx , regard y as a constant and differentiate f x, y with respect to x. 2. To find fy , regard x as a constant and differentiate f x, y with respect to y.
SECTION 14.3 PARTIAL DERIVATIVES EXAMPLE 1 If fx, yx3x2y32y2, find fx2, 1 and fy2, 1.
SOLUTION Holding y constant and differentiating with respect to x, we get fxx, y3×22xy3
and so fx2, 1322221316 Holding x constant and differentiating with respect to y, we get
fyx,y3x2y2 4y
fy2, 132212418
INTERPRETATIONS OF PARTIAL DERIVATIVES
881
M
To give a geometric interpretation of partial derivatives, we recall that the equation zfx, y represents a surface S the graph of f . If fa, bc, then the point Pa, b, c lies on S. By fixing yb, we are restricting our attention to the curve C1 in which the ver tical plane yb intersects S. In other words, C1 is the trace of S in the plane yb. Likewise, the vertical plane xa intersects S in a curve C2. Both of the curves C1 and C2 pass through the point P. See Figure 1.
z
S
0
T
C TTM Pa,b,c CTM
FIGURE 1
The partial derivatives of f at a, b are the slopes of the tangents to C and CTM.
xy
a, b, 0
Notice that the curve C1 is the graph of the function txf x, b, so the slope of its tangent T1 at P is tafxa, b. The curve C2 is the graph of the function Gyfa, y, so the slope of its tangent T2 at P is Gbfya, b.
Thus the partial derivatives fx a, b and fy a, b can be interpreted geometrically as the slopes of the tangent lines at Pa, b, c to the traces C1 and C2 of S in the planes yb and xa.
As we have seen in the case of the heat index function, partial derivatives can also be interpreted as rates of change. If zf x, y, then zx represents the rate of change of z with respect to x when y is fixed. Similarly, zy represents the rate of change of z with respect to y when x is fixed.
EXAMPLE 2 If fx, y4x22y2, find fx1, 1 and fy1, 1 and interpret these num bers as slopes.
SOLUTION We have
fxx, y2x fyx, y4y fx1, 12 fy1, 14
882
CHAPTER 14 PARTIAL DERIVATIVES
FIGURE 4
Thegraphof f istheparaboloidz4x2 2y2 andtheverticalplaney1inter sects it in the parabola z2x2, y1. As in the preceding discussion, we label it C1 in Figure 2. The slope of the tangent line to this parabola at the point 1, 1, 1 is
fx1, 12. Similarly, the curve C2 in which the plane x1 intersects the parabo loid is the parabola z32y2, x1, and the slope of the tangent line at 1, 1, 1 is
fy1, 14. See Figure 3. zz
z42
C CTM
z42
y1 x1 21,1y2 y
x x 1,1
FIGURE 2 FIGURE 3 M
Figure 4 is a computerdrawn counterpart to Figure 2. Part a shows the plane y1 intersecting the surface to form the curve C1 and part b shows C1 and T1. We have used the vector equations rtt, 1, 2t2 for C1 and rt1t, 1, 12t for T1. Similarly, Figure 5 corresponds to Figure 3.
44
33 z2 z2 11
00 1000 10 y12x y12x
a b
44
33 z2 z2 11
1, 1, 1
1, 1, 1
00 1000 10 FIGURE5 y12x y12x
V EXAMPLE3 Iffx,ysinx ,calculate f and f. 1y x y
SOLUTION Using the Chain Rule for functions of one variable, we have
SECTION 14.3 PARTIAL DERIVATIVES
883
f x
f y
cosx x cosx1 1y x 1y 1y 1y
cosx x cosxx
1y y 1y 1y 1y2
M
N Some computer algebra systems can plot surfaces defined by implicit equations in three variables. Figure 6 shows such a plot of the surface defined by the equation in Example 4.
V EXAMPLE 4 Find zx and zy if z is defined implicitly as a function of x and y by the equation
FIGURE 6
x3 y3 z3 6xyz1
SOLUTION To find zx, we differentiate implicitly with respect to x, being careful to treat
y as a constant:
3×2 3z2 z 6yz6xy z 0 x x
Solving this equation for zx, we obtain
z x22yz
x z2 2xy Similarly, implicit differentiation with respect to y gives
z y22xz M y z2 2xy
FUNCTIONS OF MORE THAN TWO VARIABLES
Partial derivatives can also be defined for functions of three or more variables. For example, if f is a function of three variables x, y, and z, then its partial derivative with respect to x is defined as
fxx,y,zlim fxh,y,zfx,y,z hl0 h
and it is found by regarding y and z as constants and differentiating f x, y, z with respect to x. If wfx, y, z, then fxwx can be interpreted as the rate of change of w with respect to x when y and z are held fixed. But we cant interpret it geometrically because the graph of f lies in fourdimensional space.
In general, if u is a function of n variables, uf x1, x2, . . . , xn , its partial derivative with respect to the ith variable xi is
u lim fx1,…,xi1,xi h,xi1,…,xnfx1,…,xi,…,xn xi hl0 h
884
CHAPTER 14 PARTIAL DERIVATIVES
and we also write
uffxifiDi f xi xi
EXAMPLE5 Findfx,fy,andfziffx,y,zexylnz.
SOLUTION Holding y and z constant and differentiating with respect to x, we have
Similarly, fyxe lnz and fz z M HIGHER DERIVATIVES
If f is a function of two variables, then its partial derivatives fx and fy are also functions of two variables, so we can consider their partial derivatives fxx, fxy, fyx, and fyy, which are called the second partial derivatives of f. If zf x, y, we use the following notation:
fxx fxx f11 f 2f2z x x x2 x2
fxy fxy f12 f 2f2z y x y x y x
fyx fyx f21 f 2f2z x y x y x y
fyy fyy f22 f 2f2z y y y2 y2
Thus the notation fx y or 2 fy x means that we first differentiate with respect to x and then with respect to y, whereas in computing fyx the order is reversed.
EXAMPLE 6 Find the second partial derivatives of fx,yx3 x2y3 2y2
fx yexylnz
xy exy
SOLUTION In Example 1 we found that fxx,y3x2 2xy3
Therefore
fxx 3×2 2xy36x2y3 x
fyx 3x2y2 4y6xy2 x
fyx,y3x2y2 4y
fxy 3×2 2xy36xy2
y
fyy 3x2y2 4y6x2y4 M y
N Figure 7 shows the graph of the function f inExample6andthegraphsofitsfirstand secondorder partial derivatives for 2×2, 2y2. Notice that these graphs are con sistent with our interpretations of fx and fy as slopes of tangent lines to traces of the graph of f. For instance, the graph of f decreases if we start at 0, 2 and move in the positive xdirection. This is reflected in the negative values of fx . You should compare the graphs of fy x and fyy with the graph of fy to see the relationships.
20
z0
20
40
2
1 0 y
1
z 20
2 2 1
0 12 x
40 20 0 20
2 1 y0
40
z
20 40 40 20 20 z0z0z0 20 20 40 0 12 40
20
2 1 y0
1
0 12 221 x
2 1 y0 1 221 x 2 1 y0 1
fxyfyx fyy
FIGURE 7
fxx
N Alexis Clairaut was a child prodigy in mathematics: he read lHospitals textbook
on calculus when he was ten and presented a paper on geometry to the French Academy of Sciences when he was 13. At the age of 18, Clairaut published Recherches sur les courbes a double courbure, which was the first systematic treatise on threedimensional analytic geometry and included the calculus of space curves.
Notice that fxyfyx in Example 6. This is not just a coincidence. It turns out that the mixed partial derivatives fx y and fy x are equal for most functions that one meets in practice. The following theorem, which was discovered by the French mathematician Alexis Clairaut 17131765, gives conditions under which we can assert that fxyfyx.The proof is given in Appendix F.
CLAIRAUTS THEOREM Suppose f is defined on a disk D that contains the point a, b. If the functions fx y and fy x are both continuous on D, then
fx ya, bfyxa, b
Partial derivatives of order 3 or higher can also be defined. For instance,
fxyy fxyy 2f3f y y x y2 x
fx
1
0 12 221 x
0
2 1 y0 1
fy
0 12 221 x
0 12 221 x
SECTION 14.3 PARTIAL DERIVATIVES
885
f
886
CHAPTER 14 PARTIAL DERIVATIVES
and using Clairauts Theorem it can be shown that fxyyfyxyfyyx if these functions are
continuous.
V EXAMPLE 7 Calculate fxx yz if f x, y, zsin3xyz. SOLUTION fx3 cos3xyz
fxx9 sin3xyz fxxy 9zcos3xyz
fxxyz9cos3xyz9yzsin3xyz M PARTIAL DIFFERENTIAL EQUATIONS
Partial derivatives occur in partial differential equations that express certain physical laws. For instance, the partial differential equation
2u2u0 x2 y2
is called Laplaces equation after Pierre Laplace 17491827. Solutions of this equation are called harmonic functions; they play a role in problems of heat conduction, fluid flow, and electric potential.
EXAMPLE 8 Show that the function ux, yex sin y is a solution of Laplaces equation.
SOLUTION ux exsiny uy excosy uxx exsiny uyy exsiny
uxx uyy ex sinyex siny0
Therefore u satisfies Laplaces equation. M
The wave equation
2ua2 2u t2 x2
describes the motion of a waveform, which could be an ocean wave, a sound wave, a light ux,t wave, or a wave traveling along a vibrating string. For instance, if ux, t represents the dis placement of a vibrating violin string at time t and at a distance x from one end of the string as in Figure 8, then ux, t satisfies the wave equation. Here the constant a depends
x
FIGURE 8
on the density of the string and on the tension in the string.
EXAMPLE 9 Verify that the function ux, tsinxat satisfies the wave equation.
SOLUTION uxcosxat
utacosxat
So u satisfies the wave equation.
uxxsinxat
utta2 sinxata2uxx
M
SECTION 14.3 PARTIAL DERIVATIVES887 THE COBBDOUGLAS PRODUCTION FUNCTION
In Example 3 in Section 14.1 we described the work of Cobb and Douglas in modeling the total production P of an economic system as a function of the amount of labor L and the capital investment K. Here we use partial derivatives to show how the particular form of their model follows from certain assumptions they made about the economy.
If the production function is denoted by PPL, K, then the partial derivative PL is the rate at which production changes with respect to the amount of labor. Economists call it the marginal production with respect to labor or the marginal productivity of labor. Likewise, the partial derivative PK is the rate of change of production with respect to capital and is called the marginal productivity of capital. In these terms, the assumptions made by Cobb and Douglas can be stated as follows.
i If either labor or capital vanishes, then so will production.
ii The marginal productivity of labor is proportional to the amount of production per unit of labor.
iii The marginal productivity of capital is proportional to the amount of production per unit of capital.
Because the production per unit of labor is PL, assumption ii says that P P
L L
for some constant . If we keep K constant KK0 , then this partial differential equa
tion becomes an ordinary differential equation:
dP P dL L
If we solve this separable differential equation by the methods of Section 9.3 see also Exercise 79, we get
PL, K0 C1K0 L
Notice that we have written the constant C1 as a function of K0 because it could depend on the value of K0.
Similarly, assumption iii says that
P P K K
and we can solve this differential equation to get
PL0, KC2L0K
Comparing Equations 6 and 7, we have
PL, KbL K
5
6
7
8
888
CHAPTER 14 PARTIAL DERIVATIVES
14.3 EXERCISES
The temperature T at a location in the Northern Hemisphere depends on the longitude x, latitude y, and time t, so we can write Tf x, y, t. Lets measure time in hours from the beginning of January.
a What are the meanings of the partial derivatives Tx, Ty, and Tt?
b Honolulu has longitude 158 W and latitude 21 N . Suppose that at 9:00 AM on January 1 the wind is blowing hot air to the northeast, so the air to the west and south is warm and the air to the north and east is cooler. Would you expect fx158, 21, 9, fy158, 21, 9, and ft158, 21, 9 to be positive or negative? Explain.
2. At the beginning of this section we discussed the function
If T, H , where I is the heat index, T is the temperature, and H is the relative humidity. Use Table 1 to estimate
fT 92, 60 and fH 92, 60. What are the practical interpretations of these values?
3. The windchill index W is the perceived temperature when the actual temperature is T and the wind speed is v, so we can write Wf T, v. The following table of values is an excerpt from Table 1 in Section 14.1.
Wind speed kmh
4.
b In general, what can you say about the signs of WT and Wv?
c What appears to be the value of the following limit? lim W
vl v
The wave heights h in the open sea depend on the speed v of the wind and the length of time t that the wind has been blowing at that speed. Values of the function hf v, t are recorded in feet in the following table.
Duration hours
1.
where b is a constant that is independent of both L and K. Assumption i shows that0 and 0.
Notice from Equation 8 that if labor and capital are both increased by a factor m, then PmL,mKbmLmK m bLK m PL,K
If 1, then PmL, mK mPL, K , which means that production is also increased by a factor of m. That is why Cobb and Douglas assumed that 1 and therefore
PL, KbL K1
This is the CobbDouglas production function that we discussed in Section 14.1.
vt
5
10
15
20
30
40
50
10
2
2
2
2
2
2
2
15
4
4
5
5
5
5
5
20
5
7
8
8
9
9
9
30
9
13
16
17
18
19
19
40
14
21
25
28
31
33
33
50
19
29
36
40
45
48
50
60
24
37
47
54
62
67
69
Tv
20
30
40
50
60
70
10
18
20
21
22
23
23
15
24
26
27
29
30
30
20
30
33
34
35
36
37
25
37
39
41
42
43
44
a Estimate the values of fT 15, 30 and fv15, 30. What are the practical interpretations of these values?
a What are the meanings of the partial derivatives hv and ht?
b Estimate the values of fv40, 15 and ft40, 15. What are the practical interpretations of these values?
c What appears to be the value of the following limit?
lim h tl t
Actual temperature C
Wind speed knots
58 Determine the signs of the partial derivatives for the function f whose graph is shown.
SECTION 14.3 PARTIAL DERIVATIVES889 10. A contour map is given for a function f. Use it to estimate
x
5. a fx1, 2
6. a fx1, 2
7. a fxx1, 2
8. a fxy1, 2
2
y
14 16
z
1
4
2
4 2
y 30
6
8
9. The following surfaces, labeled a, b, and c, are graphs of a function f and its partial derivatives fx and fy . Identify each surface and give reasons for your choices.
8
4 z0 4 8
15. fx,yy5 3xy
16. fx,yx4y3 8x2y 18. fx,tsx lnt
20. ztanxy
22. f x, yx y
24. wevuv2
26. fx,tarctanxst 28. fx,yyx cost2dt
y
30. fx,y,zxsinyz
32. wzexyz 34. uxyz
36. fx,y,z,t xy2
t2z
8
4 z0
38. usinx1 2×2 nxn
3942 Find the indicated partial derivatives. 39. fx, ylnxsx2y2 ; fx3, 4
fx 2, 3
; fy 2, 1, 1
4 c
8 321
40. f x, yarctan yx;
y
a
0 2
3 2 1 0 1 2 3 2 x y
4 z0
4 b 2 3 2 1 0 1 2 3 2 0x
23. wsin cos
25. fr,srlnr2 s2
27. utewt
29. fx,y,zxz5x2y3z4
31. wlnx2y3z 33. uxy sin1yz
35. fx,y,z,txyz2 tanyt 3 7 . us x 12x 2 2x n2
b b b b
fy1, 2 fy1, 2 fyy1, 2
fxy1, 2
11. If fx, y164×2y2, find fx1, 2 and fy1, 2 and inter pret these numbers as slopes. Illustrate with either handdrawn sketches or computer plots.
12. If fx, ys4x24y2 , find fx1, 0 and fy1, 0 and interpret these numbers as slopes. Illustrate with either hand drawn sketches or computer plots.
; 1314 Find fx and fy and graph f, fx, and fy with domains and viewpoints that enable you to see the relationships between them.
13. fx,yx2 y2 x2y 14. fx,yxex2y2 1538 Find the first partial derivatives of the function.
2 0 x2
y xyz
0123
41. f x, y, z
y
fx2, 1 and fy2, 1.
17. fx, tet cos 19. z2x3y10
21. fx,y xy xy
x
10 12
1 318 x
890CHAPTER 14 PARTIAL DERIVATIVES
42. fx, y, zssin2xsin2ysin2z; fz0, 0, 4
4344 Use the definition of partial derivatives as limits 4 to find fxx, y and fyx, y.
x
46. yzlnxz
48. sinxyzx2y3z
b zfxy b zfxy
Use the table of values of f x, y to estimate the values of fx 3, 2, fx 3, 2.2, and fx y 3, 2.
x
y
1.8
2.0
2.2
2.5
12. 5
10. 2
9.3
3.0
18. 1
17. 5
15. 9
3.5
20. 0
22. 4
26. 1
43. fx,yxy2 x3y 44. fx,y
4548 Use implicit differentiation to find zx and zy.
45. x2 y2 z2 3xyz 47. xzarctanyz
4950 Find zx and zy.
49. a zfxty
50. azfxty czfxy
70. Level curves are shown for a function f. Determine whether the following partial derivatives are positive or negative at the point P.
a fx
xy2
b fy e fyy
c fxx
d fxy
y
10 8 6
4
2
x
5156 Find all the second partial derivatives.
51. fx,yx3y5 2x4y 53. wsu2 v2
55. zarctan xy 1xy
5760 Verify that the conclusion of Clairauts Theorem holds, that is,uxy uyx.
52. fx,ysin2mxny xy
71. Verify that the function ue heat conduction equation u t
2k2t sin kx is a solution of the 2u x x .
P
54. vxy 56. vexey
72. Determine whether each of the following functions is a solution of Laplaces equation uxxuyy0.
a ux2 y2 b ux2 y2
c ux3 3xy2 d ulnsx2 y2
e usinx coshycosx sinhy fuex cosyey cosx
73. Verify that the function u1sx2y2z2 is a solution of the threedimensional Laplace equation uxxuyyuzz0.
74. Show that each of the following functions is a solution of the wave equation ut ta2uxx.
a usinkx sinakt b uta2t2x2
c uxat6 xat6
d usinxatlnxat
75. If f and t are twice differentiable functions of a single vari
able, show that the function
ux, tf xattxat
is a solution of the wave equation given in Exercise 74.
76. If uea1x1a2 x2an xn, where a12a2an21, show that
2u2u2u ux 12x 2 2x n2
77. Verify that the function zlnexey is a solution of the differential equations
zz1 x y
57. uxsinx2y 59. ulnsx2 y2
58. ux4y2 2xy5 60. uxyey
6168 Find the indicated partial derivative. 432
61.fx,y3xyxy; fxxy, fyyy 62. fx, tx2ect; fttt, ftxx
63. fx, y, zcos4x3y2z;
fxyz,
fyzz
64. fr, s, tr lnrs2t3;
frss,
frst
65.uer sin; 66.zusvw; uvw 67.w x ; 3w , 3w
3u r2
3z y2z zyx
6u
x y 2 z 3
x2 y
69.
68. uxaybzc;
and 2z 2z 2z 2 220
You are told that there is a function f whose partial derivatives are fxx,yx4yand fyx,y3xy.Shouldyou believe it?
88. The paraboloid z6xx 22 y 2 intersects the plane
x1 in a parabola. Find parametric equations for the tangent line to this parabola at the point 1, 2, 4. Use a computer to graph the paraboloid, the parabola, and the tangent line on the same screen.
89. The ellipsoid 4×22y2z216 intersects the plane y2 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point 1, 2, 2.
90. In a study of frost penetration it was found that the temperature T at time t measured in days at a depth x measured in feet can be modeled by the function
Tx,tT0T1exsint x
where2 365 and is a positive constant.
a Find Tx. What is its physical significance?
b Find Tt. What is its physical significance?
c Show that T satisfies the heat equation TtkTxx for a cer
tain constant k.
dIf 0.2,T0 0,andT1 10,useacomputerto
graph Tx, t.
e What is the physical significance of the termx in the
expression sin tx?
91. Use Clairauts Theorem to show that if the thirdorder partial
derivatives of f are continuous, then fxyy fyxy fyyx
92. a How many nthorder partial derivatives does a function of two variables have?
b If these partial derivatives are all continuous, how many of them can be distinct?
c Answer the question in part a for a function of three variables.
93. If fx, yxx2y232esinx2y, find fx1, 0.
Hint: Instead of finding fxx, y first, note that its easier to use Equation 1 or Equation 2.
94. If fx, ys3 x3y3 , find fx0, 0.
x y xy
78. Show that the CobbDouglas production function PbL K ;
satisfies the equation
LPKPP L K
79. Show that the CobbDouglas production function satisfies
SECTION 14.3 PARTIAL DERIVATIVES891
PL, K0 C1K0 L See Equation 5.
80. The temperature at a point x, y on a flat metal plate is given byTx,y601x2 y2,whereT ismeasuredinC and x, y in meters. Find the rate of change of temperature with respect to distance at the point 2, 1 in a the xdirection and b the ydirection.
The total resistance R produced by three conductors with resis tances R1, R2, R3 connected in a parallel electrical circuit is given by the formula
1111 R R1 R2 R3
Find RR1.
82. The gas law for a fixed mass m of an ideal gas at absolute tem perature T, pressure P, and volume V is PVmRT, where R is the gas constant. Show that
P V T1 V T P
by solving the differential equation
dP P dL L
;
83. For the ideal gas of Exercise 82, show that T P VmR
T T
84. The windchill index is modeled by the function
W13.120.6215T11.37v0.160.3965Tv0.16
where T is the temperature C and v is the wind speed kmh. When T15C and v30 kmh, by how much would you expect the apparent temperature W to drop if the actual temperature decreases by 1C? What if the wind speed increases by 1 kmh?
85. The kinetic energy of a body with mass m and velocity v is K1 mv2. Show that
95. Let
fx,y
0
a Use a computer to graph f.
b Find fxx, y and fyx, y when x, y0, 0.
c Find fx0, 0 and fy0, 0 using Equations 2 and 3.
d Show that fxy0, 01 and fyx0, 01.
e Does the result of part d contradict Clairauts Theorem?
Use graphs of fx y and fy x to illustrate your answer.
x3yxy3 x2 y2
if x, y0, 0 if x, y0, 0
2
K 2K ; m v2 K
87.
81.
86.
If a, b, c are the sides of a triangle and A, B, C are the opposite angles, find Aa, Ab, Ac by implicit differentiation of CAS the Law of Cosines.
892
CHAPTER 14 PARTIAL DERIVATIVES
z
TTM x
FIGURE 1
T C
0
The tangent plane contains the tangent lines T and TTM.
N Note the similarity between the equation of a tangent plane and the equation of a tangent line:
yy0 fx0xx0
P
C TM
14.4
TANGENT PLANES AND LINEAR APPROXIMATIONS
One of the most important ideas in singlevariable calculus is that as we zoom in toward a point on the graph of a differentiable function, the graph becomes indistinguishable from its tangent line and we can approximate the function by a linear function. See Sec tion 3.10. Here we develop similar ideas in three dimensions. As we zoom in toward a point on a surface that is the graph of a differentiable function of two variables, the surface looks more and more like a plane its tangent plane and we can approximate the function by a linear function of two variables. We also extend the idea of a differential to functions of two or more variables.
TANGENT PLANES
Suppose a surface S has equation zf x, y, where f has continuous first partial deriva tives, and let Px0, y0, z0be a point on S. As in the preceding section, let C1 and C2 be the curves obtained by intersecting the vertical planes yy0 and xx0 with the surface S. Then the point P lies on both C1 and C2. Let T1 and T2 be the tangent lines to the curves C1 and C2 at the point P. Then the tangent plane to the surface S at the point P is defined to be the plane that contains both tangent lines T1 and T2. See Figure 1.
We will see in Section 14.6 that if C is any other curve that lies on the surface S and passes through P, then its tangent line at P also lies in the tangent plane. Therefore you can think of the tangent plane to S at P as consisting of all possible tangent lines at P to curves that lie on S and pass through P. The tangent plane at P is the plane that most closely approximates the surface S near the point P.
We know from Equation 12.5.7 that any plane passing through the point Px0, y0, z0has an equation of the form
Axx0Byy0Czz00
By dividing this equation by C and letting aAC and bBC, we can write it in
the form
1 zz0 axx0byy0
If Equation 1 represents the tangent plane at P, then its intersection with the plane yy0
must be the tangent line T1. Setting yy0 in Equation 1 gives zz0 axx0 yy0
and we recognize these as the equations in pointslope form of a line with slope a. But from Section 14.3 we know that the slope of the tangent T1 is fxx0, y0 . Therefore afxx0, y0.
Similarly, putting xx0 in Equation 1, we get zz0byy0, which must repre sent the tangent line T2, so bfyx0, y0 .
2 Suppose f has continuous partial derivatives. An equation of the tangent plane to the surface zfx, y at the point Px0, y0, z0 is
zz0 fxx0,y0xx0fyx0,y0yy0
y
TEC Visual 14.4 shows an animation of Figures 2 and 3.
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS893 V EXAMPLE 1 Find the tangent plane to the elliptic paraboloid z2×2y2 at the
point 1, 1, 3.
SOLUTION Let fx, y2×2y2. Then
fxx, y4x fyx, y2y fx1, 14 fy1, 12
Then 2 gives the equation of the tangent plane at 1, 1, 3 as z34x12y1
or z4x2y3 M
Figure 2a shows the elliptic paraboloid and its tangent plane at 1, 1, 3 that we found in Example 1. In parts b and c we zoom in toward the point 1, 1, 3 by restricting the domain of the function fx, y2×2y2. Notice that the more we zoom in, the flatter the graph appears and the more it resembles its tangent plane.
40 40 20 20 20
z0 z0 20 20 20
40 z0
420 024 2 0 2 0 1 y2442x y220x y221x
FIGURE 3
Zooming in toward 1, 1
on a contour map of
fx, y2 0.5
1.5
1.2
1.05
1.5
1.2
0.8
1.05
0.95
0
In Figure 3 we corroborate this impression by zooming in toward the point 1, 1 on a contour map of the function fx, y2×2y2. Notice that the more we zoom in, the more the level curves look like equally spaced parallel lines, which is characteristic of a plane.
a b c FIGURE 2 The elliptic paraboloid z2 appears to coincide with its tangent plane as we zoom in toward 1, 1, 3.
894
CHAPTER 14 PARTIAL DERIVATIVES
LINEAR APPROXIMATIONS
FIGURE 4
You can verify see Exercise 46 that its partial derivatives exist at the origin and, in fact,
fx0, 00 and fy0, 00, but fx and fy are not continuous. The linear approximation
wouldbe fx,y0,butfx,y1 atallpointsonthelineyx.Soafunctionoftwo 2
variables can behave badly even though both of its partial derivatives exist. To rule out such behavior, we formulate the idea of a differentiable function of two variables.
Recall that for a function of one variable, yf x, if x changes from a to ax, we defined the increment of y as
yfaxfa
fx, y
xy
if x, y0, 0,
f0,00
zy
In Example 1 we found that an equation of the tangent plane to the graph of the function fx, y2×2y2 at the point 1, 1, 3 is z4x2y3. Therefore, in view of the
visual evidence in Figures 2 and 3, the linear function of two variables
Lx, y4x2y3
is a good approximation to f x, y when x, y is near 1, 1. The function L is called the
linearization of f at 1, 1 and the approximation
fx, y4x2y3
is called the linear approximation or tangent plane approximation of f at 1, 1. For instance, at the point 1.1, 0.95 the linear approximation gives
f 1.1, 0.9541.120.9533.3
which is quite close to the true value of f 1.1, 0.9521.120.9523.3225. But if we take a point farther away from 1, 1, such as 2, 3, we no longer get a good approxi mation. In fact, L2, 311 whereas f 2, 317.
In general, we know from 2 that an equation of the tangent plane to the graph of a function f of two variables at the point a, b, f a, b is
zfa, bfxa, bxafya, byb The linear function whose graph is this tangent plane, namely
3 Lx, yfa, bfxa, bxafya, byb is called the linearization of f at a, b and the approximation
4 fx, yfa, bfxa, bxafya, byb
is called the linear approximation or the tangent plane approximation of f at a, b. We have defined tangent planes for surfaces zf x, y, where f has continuous first partial derivatives. What happens if fx and fy are not continuous? Figure 4 pictures such a
function; its equation is
x
xy fx,yx2 y2
if x, y0, 0 0 if x, y0, 0
N This is Equation 3.4.7.
In Chapter 3 we showed that if f is differentiable at a, then
5 yfaxx where l0 as xl0
Now consider a function of two variables, zf x, y, and suppose x changes from a to ax and y changes from b to by. Then the corresponding increment of z is
6 zfax, byfa, b
Thus the increment z represents the change in the value of f when x, y changes from a, b to ax, by. By analogy with 5 we define the differentiability of a func tion of two variables as follows.
7 DEFINITION If zf x, y, then f is differentiable at a, b if z can be expressed in the form
zfxa, b xfya, b y1 x2 y where 1 and 2 l 0 as x, y l 0, 0.
Definition 7 says that a differentiable function is one for which the linear approxima tion 4 is a good approximation when x, y is near a, b. In other words, the tangent plane approximates the graph of f well near the point of tangency.
Its sometimes hard to use Definition 7 directly to check the differentiability of a func tion, but the next theorem provides a convenient sufficient condition for differentiability.
8 THEOREM If the partial derivatives fx and fy exist near a, b and are continu ous at a, b, then f is differentiable at a, b.
V EXAMPLE 2 Show that fx, yxexy is differentiable at 1, 0 and find its lineariza tion there. Then use it to approximate f 1.1, 0.1.
SOLUTION The partial derivatives are
fxx, yexyxyexy
fx1, 01
Both fx and fy are continuous functions, so f is differentiable by Theorem 8. The
linearization is
Lx, yf1, 0fx1, 0x1fy1, 0y011x11yxy
The corresponding linear approximation is
N Theorem 8 is proved in Appendix F.
N Figure 5 shows the graphs of the function f and its linearization L in Example 2.
6
z4 2
0
fyx, yx2exy fy1, 01
FIGURE 5
Compare this with the actual value of f 1.1, 0.11.1e 0.110.98542. M
xexy xy
1x 01 0y 1 so f 1.1, 0.11.10.11
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS895
896
CHAPTER 14 PARTIAL DERIVATIVES
y
0 aaIx x
tangent line
yfafaaxa
EXAMPLE 3 At the beginning of Section 14.3 we discussed the heat index perceived temperature I as a function of the actual temperature T and the relative humidity H and gave the following table of values from the National Weather Service.
Relative humidity
Actual temperature F
Find a linear approximation for the heat index IfT, H when T is near 96F and H is near 70. Use it to estimate the heat index when the temperature is 97F and the relative humidity is 72.
SOLUTION We read from the table that f 96, 70125. In Section 14.3 we used the tabu lar values to estimate that fT 96, 703.75 and fH 96, 700.9. See pages 878 79. So the linear approximation is
fT, Hf96, 70fT96, 70T96fH96, 70H701253.75T960.9H70
DIFFERENTIALS
For a differentiable function of one variable, yfx, we define the differential dx to be an independent variable; that is, dx can be given the value of any real number. The differ ential of y is then defined as
dyfx dx
See Section 3.10. Figure 6 shows the relationship between the increment y and the dif ferential dy: y represents the change in height of the curve yfx and dy represents the change in height of the tangent line when x changes by an amount dxx.
For a differentiable function of two variables, zf x, y, we define the differentials dx and dy to be independent variables; that is, they can be given any values. Then the dif ferential dz, also called the total differential, is defined by
TH
50
55
60
65
70
75
80
85
90
90
96
98
100
103
106
109
112
115
119
92
100
103
105
108
112
115
119
123
128
94
104
107
111
114
118
122
127
132
137
96
109
113
116
121
125
130
135
141
146
98
114
118
123
127
133
138
144
150
157
100
119
124
129
135
141
147
154
161
168
In particular,
Therefore, when T97F and H72, the heat index is
9
f 97, 721253.7510.92130.55
I131F M
y dxIx dy
Iy
FIGURE 6
Compare with Equation 9. Sometimes the notation d f is used in place of dz.
10
dzfxx,ydxfyx,ydy z dx z dy x y
N In Example 4, dz is close to z because the tangent plane is a good approximation to the surface zx23xyy2 near 2, 3, 13. See Figure 8.
60
40 z 20 0
V EXAMPLE 4
a Ifzfx,yx2 3xyy2,findthedifferentialdz.
b If x changes from 2 to 2.05 and y changes from 3 to 2.96, compare the values of z and dz.
SOLUTION
a Definition 10 gives
dz z dx z dy2x3ydx3x2ydy x y
b Puttingx2,dxx0.05,y3,anddyy0.04,weget dz22330.0532230.040.65
The increment of z is
zf2.05, 2.96f2, 3
2.05232.052.962.962 2232332 0.6449
Notice that zdz but dz is easier to compute. M
EXAMPLE 5 The base radius and height of a right circular cone are measured as 10 cm and 25 cm, respectively, with a possible error in measurement of as much as 0.1 cm in
205 4 3 2 1 0 4 2 0 xy
FIGURE 8
FIGURE 7
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS897 If we take dxxxa and dyyyb in Equation 10, then the differen
tial of z is
dzfxa, bxafya, byb
So, in the notation of differentials, the linear approximation 4 can be written as
fx, yfa, bdz
Figure 7 is the threedimensional counterpart of Figure 6 and shows the geometric inter pretation of the differential dz and the increment z: dz represents the change in height of the tangent plane, whereas z represents the change in height of the surface zf x, y when x, y changes from a, b to ax, by.
a, b, fa, b
0 fa,b
x
a, b, 0
aIx, bIy, faIx, bIy
Iz dz
fa,b
y aIx, bIy, 0
Iydy
zfa, bfxa, bxafya, byb
z
surface zfx, y
Ixdx
tangent plane
898
CHAPTER 14 PARTIAL DERIVATIVES
each. Use differentials to estimate the maximum error in the calculated volume of the
cone.
SOLUTION The volume V of a cone with base radius r and height h is Vr2h3. So the
differential of V is
V V 2rh r2 dVrdrhdh 3 dr 3 dh
Since each error is at most 0.1 cm, we have r 0.1, h 0.1. To find the largest error in the volume we take the largest error in the measurement of r and of h. Therefore we take dr0.1 and dh0.1 along with r10, h25. This gives
dV 500 0.1 100 0.120 33
33
Thus the maximum error in the calculated volume is about 20 cm63 cm . M
FUNCTIONS OF THREE OR MORE VARIABLES
Linear approximations, differentiability, and differentials can be defined in a similar man ner for functions of more than two variables. A differentiable function is defined by an expression similar to the one in Definition 7. For such functions the linear approximation is
fx, y, zfa, b, cfxa, b, cxafya, b, cybfza, b, czc and the linearization Lx, y, z is the right side of this expression.
If wfx, y, z, then the increment of w is
wfxx, yy, zzfx, y, z
The differential dw is defined in terms of the differentials dx, dy, and dz of the independ ent variables by
dww dxw dyw dz x y z
EXAMPLE 6 The dimensions of a rectangular box are measured to be 75 cm, 60 cm, and 40 cm, and each measurement is correct to within 0.2 cm. Use differentials to esti mate the largest possible error when the volume of the box is calculated from these measurements.
SOLUTION If the dimensions of the box are x, y, and z, its volume is Vxyz and so dV V dx V dy V dzyzdxxzdyxydz
We are given that x 0.2, y 0.2, and z 0.2. To find the largest error in the volume, we therefore use dx0.2, dy0.2, and dz0.2 together with x75, y60, and z40:
VdV60400.275400.275600.21980
Thus an error of only 0.2 cm in measuring each dimension could lead to an error of as muchas1980cm3 inthecalculatedvolume!Thismayseemlikealargeerror,butits only about 1 of the volume of the box. M
x y z
14.4 EXERCISES
16 Find an equation of the tangent plane to the given surface at
19. Find the linear approximation of the function fx, ys20x27y2 at 2, 1 and use it to
approximate f 1.95, 1.08.
; 20. Find the linear approximation of the function
f x, ylnx3y at 7, 2 and use it to approximate
f 6.9, 2.06. Illustrate by graphing f and the tangent plane.
Find the linear approximation of the function
fx, y, zsx2y2z2 at 3, 2, 6 and use it to
approximate the number s3.0221.9725.992 .
22. The wave heights h in the open sea depend on the speed v of the wind and the length of time t that the wind has been blowing at that speed. Values of the function hf v, t are recorded in feet in the following table.
Duration hours
the specified point.
1. z4×2y22y, 1, 2, 4
2. z3x122 y327,
3. zsxy, 1, 1, 1
4. zy ln x, 1, 4, 0
5. zy cosxy, 2, 2, 2
6. zex2y2, 1, 1, 1
2, 2, 12
SECTION 14.4 TANGENT PLANES AND LINEAR APPROXIMATIONS
899
; 78 Graph the surface and the tangent plane at the given point. Choose the domain and viewpoint so that you get a good view of both the surface and the tangent plane. Then zoom in until the surface and the tangent plane become indistinguishable.
7. zx2xy3y2, 1, 1, 5 8. zarctanxy2, 1, 1, 4
CAS 910 Draw the graph of f and its tangent plane at the given point. Use your computer algebra system both to compute the partial derivatives and to graph the surface and its tangent plane. Then zoom in until the surface and the tangent plane become indistinguishable.
xy sinxy
9. fx, y1x2y2 , 1, 1, 0
10. fx, yexy10sxsysxy , 1, 1, 3e0.1
1116 Explain why the function is differentiable at the given point. Then find the linearization Lx, y of the function at that point.
Use the table to find a linear approximation to the wave height function when v is near 40 knots and t is near
20 hours. Then estimate the wave heights when the wind has been blowing for 24 hours at 43 knots.
23. Use the table in Example 3 to find a linear approximation to the heat index function when the temperature is near 94F and the relative humidity is near 80. Then estimate the heat index when the temperature is 95F and the relative humidity is 78.
24. The windchill index W is the perceived temperature when the actual temperature is T and the wind speed is v, so we can write Wf T, v. The following table of values is an excerpt from Table 1 in Section 14.1.
Wind speed km h
vt
5
10
15
20
30
40
50
20
5
7
8
8
9
9
9
30
9
13
16
17
18
19
19
40
14
21
25
28
31
33
33
50
19
29
36
40
45
48
50
60
24
37
47
54
62
67
69
fx, yxsy, fx, yx3y4,
1, 4 1, 1
12. 13.
14. 15. 16.
17. 2×3 32x12y 18. sycos2x 11y 4y1 2
f x, yx , 2, 1 xy
fx, ysxe4y , 3, 0 f x, yexy cos y,, 0
Tv
20
30
40
50
60
70
10
18
20
21
22
23
23
15
24
26
27
29
30
30
20
30
33
34
35
36
37
25
37
39
41
42
43
44
fx, ysin2x3y, 3, 2
1718 Verify the linear approximation at 0, 0.
21.
11.
Use the table to find a linear approximation to the windchill
Actual temperature C Wind speed knots
900CHAPTER 14 PARTIAL DERIVATIVES
index function when T is near 15C and v is near 50 kmh. Then estimate the windchill index when the temperature is 17C and the wind speed is 55 kmh.
2530 Find the differential of the function.
25. zx3 lny2 27. mp5q3
29. R 2cos
If z5×2y2 and x, y changes from 1, 2 to 1.05, 2.1,
compare the values of z and dz.
32. Ifzx2 xy3y2 andx,ychangesfrom3,1to
2.96, 0.95, compare the values of z and dz.
33. The length and width of a rectangle are measured as 30 cm and 24 cm, respectively, with an error in measurement of at most 0.1 cm in each. Use differentials to estimate the maximum error in the calculated area of the rectangle.
34. The dimensions of a closed rectangular box are measured as 80 cm, 60 cm, and 50 cm, respectively, with a possible error of 0.2 cm in each dimension. Use differentials to estimate the maximum error in calculating the surface area of the box.
Use differentials to estimate the amount of tin in a closed tin can with diameter 8 cm and height 12 cm if the tin is 0.04 cm thick.
36. Use differentials to estimate the amount of metal in a closed cylindrical can that is 10 cm high and 4 cm in diameter if the metal in the top and bottom is 0.1 cm thick and the metal in the sides is 0.05 cm thick.
A boundary stripe 3 in. wide is painted around a rectangle whose dimensions are 100 ft by 200 ft. Use differentials to approximate the number of square feet of paint in the stripe.
38. The pressure, volume, and temperature of a mole of an ideal gas are related by the equation PV8.31T, where P is mea sured in kilopascals, V in liters, and T in kelvins. Use differen tials to find the approximate change in the pressure if the volume increases from 12 L to 12.3 L and the temperature decreases from 310 K to 305 K.
39. If R is the total resistance of three resistors, connected in par allel, with resistances R1 , R2 , R3 , then
1111 R R1 R2 R3
If the resistances are measured in ohms as R125 ,
R2 40,andR3 50,withapossibleerrorof0.5in each case, estimate the maximum error in the calculated value of R.
40. Four positive numbers, each less than 50, are rounded to the first decimal place and then multiplied together. Use differen tials to estimate the maximum possible error in the computed product that might result from the rounding.
41. A model for the surface area of a human body is given by
S0.1091w0.425h0.725, where w is the weight in pounds, h is the height in inches, and S is measured in square feet. If the errors in measurement of w and h are at most 2, use differ entials to estimate the maximum percentage error in the calculated surface area.
42. Suppose you need to know an equation of the tangent plane to a surface S at the point P2, 1, 3. You dont have an equation for S but you know that the curves
r 1t23 t , 1t 2, 34 tt 2
r2u1u2, 2u31, 2u1
both lie on S. Find an equation of the tangent plane at P.
4344 Show that the function is differentiable by finding values of 1 and 2 that satisfy Definition 7.
fx,yx2 y2 44. fx,yxy5y2
Prove that if f is a function of two variables that is differen tiable at a, b, then f is continuous at a, b.
Hint: Show that
lim f ax, byf a, b
31.
26. vy cos xy 28. Tv
1uvw 30.wxyexz
35.
37.
43.
45.
46. a
x, yl0, 0
The functionxy
was graphed in Figure 4. Show that fx0, 0 and fy0, 0 both exist but f is not differentiable at 0, 0. Hint: Use the result of Exercise 45.
if x, y0, 0 0 if x, y0, 0
fx,y x2y2
b Explain why fx and fy are not continuous at 0, 0.
1
SECTION 14.5 THE CHAIN RULE901
14.5 THE CHAIN RULE
Recall that the Chain Rule for functions of a single variable gives the rule for differenti ating a composite function: If yf x and xtt, where f and t are differentiable func tions, then y is indirectly a differentiable function of t and
dydy dx dt dxdt
For functions of more than one variable, the Chain Rule has several versions, each of them giving a rule for differentiating a composite function. The first version Theorem 2 deals with the case where zf x, y and each of the variables x and y is, in turn, a func tion of a variable t. This means that z is indirectly a function of t, zftt, ht, and the Chain Rule gives a formula for differentiating z as a function of t. We assume that f is dif ferentiable Definition 14.4.7. Recall that this is the case when fx and fy are continuous Theorem 14.4.8.
THE CHAIN RULE CASE 1 Suppose that zf x, y is a differentiable func tion of x and y, where xtt and yht are both differentiable functions of t. Then z is a differentiable function of t and
dzf dxf dy dt x dt y dt
2
PROOF A change of t in t produces changes of x in x and y in y. These, in turn, pro duce a change of z in z, and from Definition 14.4.7 we have
z f x f y1x2y x y
where 1 l 0 and 2 l 0 as x, y l 0, 0. If the functions 1 and 2 are not defined at 0, 0, we can define them to be 0 there. Dividing both sides of this equation by t, we have
z f x f y x y tx ty t1 t2 t
If we now let t l 0, then xttttt l 0 because t is differentiable and therefore continuous. Similarly, y l 0. This, in turn, means that 1 l 0 and 2 l 0, so
dzlim z dt tl0 t
f lim xf lim y lim1lim x lim 2lim y x tl0 t y tl0 t tl0 tl0 t tl0 tl0 t
f dxf dy0dx0dy x dt y dt dt dt
f dxf dy x dt y dt
M
902CHAPTER 14 PARTIAL DERIVATIVES
Since we often write zx in place of fx, we can rewrite the Chain Rule in the form
N Notice the similarity to the definition of the differential:
dzz dxz dy x y
dzz dxz dy dt x dt y dt
y
0,1
C
EXAMPLE 1 If zx2y3xy4, where xsin 2t and ycos t, find dzdt when t0. SOLUTION The Chain Rule gives
dzz dxz dy dt x dt y dt
2xy3y42 cos 2tx212xy3sin t
Its not necessary to substitute the expressions for x and y in terms of t. We simply
observe that when t0, we have xsin 00 and ycos 01. Therefore
dz 032 cos 000sin 06 M
dt t0
The derivative in Example 1 can be interpreted as the rate of change of z with respect to t as the point x, y moves along the curve C with parametric equations xsin 2t, ycos t. See Figure 1. In particular, when t0, the point x, y is 0, 1 and dzdt6 is the rate of increase as we move along the curve C through 0, 1. If, for instance, zTx, yx2 y3xy4 represents the temperature at the point x, y, then the compos ite function zTsin 2t, cos t represents the temperature at points on C and the deriva tive dzdt represents the rate at which the temperature changes along C.
V EXAMPLE 2 The pressure P in kilopascals, volume V in liters, and temperature T in kelvins of a mole of an ideal gas are related by the equation PV8.31T. Find the rate at which the pressure is changing when the temperature is 300 K and increasing at a rate of 0.1 Ks and the volume is 100 L and increasing at a rate of 0.2 Ls.
SOLUTION If t represents the time elapsed in seconds, then at the given instant we have T300, dTdt0.1, V100, dVdt0.2. Since
P8.31 T V
the Chain Rule gives
dPP dTP dV8.31 dT8.31T dV dt T dt V dt V dt V2 dt
8.31 0.18.31300 0.20.04155 100 1002
The pressure is decreasing at a rate of about 0.042 kPas. M
We now consider the situation where zf x, y but each of x and y is a function of two variables s and t: xts, t, yhs, t. Then z is indirectly a function of s and t and we
FIGURE 1
The curve xsin 2t, ycos t
x
SECTION 14.5 THE CHAIN RULE903 wish to find zs and zt. Recall that in computing zt we hold s fixed and compute
the ordinary derivative of z with respect to t. Therefore we can apply Theorem 2 to obtain zz xz y
t x t y t
A similar argument holds for zs and so we have proved the following version of the
Chain Rule.
THE CHAIN RULE CASE 2 Suppose that zf x, y is a differentiable func tion of x and y, where xts, t and yhs, t are differentiable functions of s and t. Then
zz xz y zz xz y s x s y s t x t y t
3
EXAMPLE3 Ifzexsiny,wherexst2 andys2t,findzsandzt. SOLUTION Applying Case 2 of the Chain Rule, we get
zz xz y ex sinyt2ex cosy2st s x s y s
t2est 2 sins2t2stest 2 coss2t
zz xz y ex siny2stex cosys2
z
x x x y y y
s t s t stst
FIGURE 2
Case 2 of the Chain Rule contains three types of variables: s and t are independent variables, x and y are called intermediate variables, and z is the dependent variable. Notice that Theorem 3 has one term for each intermediate variable and each of these terms resembles the onedimensional Chain Rule in Equation 1.
To remember the Chain Rule, its helpful to draw the tree diagram in Figure 2. We draw branches from the dependent variable z to the intermediate variables x and y to indicate that z is a function of x and y. Then we draw branches from x and y to the independent variables s and t. On each branch we write the corresponding partial derivative. To find zs, we find the product of the partial derivatives along each path from z to s and then add these products:
zz xz y s x s y s
Similarly, we find zt by using the paths from z to t.
Now we consider the general situation in which a dependent variable u is a function of
n intermediate variables x1, . . . , xn, each of which is, in turn, a function of m independent variables t1, . . . , tm. Notice that there are n terms, one for each intermediate variable. The proof is similar to that of Case 1.
z x
z y
t x t y t
2stest 2 sins2ts2est 2 coss2t M
904
CHAPTER 14 PARTIAL DERIVATIVES
THE CHAIN RULE GENERAL VERSION Suppose that u is a differentiable func tion of the n variables x1, x2, . . . , xn and each xj is a differentiable function of the m variables t1, t2, …, tm. Then u is a function of t1, t2, …, tm and
uu x1u x2u xn ti x1 ti x2 ti xn ti
for each i1, 2, . . . , m.
4
w
V EXAMPLE 4 Write out the Chain Rule for the case where wf x, y, z, t and xxu, v, yyu, v, zzu, v, and ttu, v.
SOLUTION We apply Theorem 4 with n4 and m2. Figure 3 shows the tree diagram. Although we havent written the derivatives on the branches, its understood that if a branch leads from y to u, then the partial derivative for that branch is yu. With the aid of the tree diagram, we can now write the required expressions:
ww xw yw zw t u x u y u z u t u
ww xw yw zw t M v x v y v z v t v
V EXAMPLE 5 Ifux4yy2z3,wherexrset,yrs2et,andzr2ssint,findthe value of us when r2, s1, t0.
SOLUTION With the help of the tree diagram in Figure 4, we have uu xu yu z
s x s y s z s
4x3yretx42yz32rset3y2z2r2 sin t
When r2, s1, and t0, we have x2, y2, and z0, so
u64216400192 M
z t uvuvuvuv
FIGURE 3
u
x y z
rstrstrst
FIGURE 4
x
y
s
EXAMPLE 6 If ts, tfs2t2, t2s2 and f is differentiable, show that t satisfies
the equation
t ts t0 s t
SOLUTION Letxs2 t2 andyt2 s2.Thents,tfx,yandtheChainRule gives
tf xf yf2sf2s s x s y s x y
tf xf yf2tf2t t x t y t x y
Therefore
ttst 2stf2stf2stf2stf 0 M
s t x y x y
SECTION 14.5 THE CHAIN RULE905 EXAMPLE 7 If zf x, y has continuous secondorder partial derivatives and
xr2 s2 andy2rs,findazrandb2zr2. SOLUTION
a The Chain Rule gives
zz xz yz2rz2s r x r y r x y
b Applying the Product Rule to the expression in part a, we get
5
2z 2rz2sz
r2 r x y2z2r z 2s z
z x
xy
rsrs
FIGURE 5
x r x r y But, using the Chain Rule again see Figure 5, we have
zzx zy2z 2r2z 2s 2
r x x x r y x r x yx
zzx zy2z 2r2z 2s
r y x y r y y r xy y
Putting these expressions into Equation 5 and using the equality of the mixed second
order derivatives, we obtain2z2z2r 2r2z2s 2z 2s 2r 2z 2s2z
2
r2 x x2 yx xy y2 2 z 4r2 2z 8rs 2z 4s2 2z
M
x x2 xy y2 IMPLICIT DIFFERENTIATION
The Chain Rule can be used to give a more complete description of the process of implicit differentiation that was introduced in Sections 3.5 and 14.3. We suppose that an equa tion of the form Fx, y0 defines y implicitly as a differentiable function of x, that is, yfx, where Fx, fx0 for all x in the domain of f. If F is differentiable, we can apply Case 1 of the Chain Rule to differentiate both sides of the equation Fx, y0 with respect to x. Since both x and y are functions of x, we obtain
F dxF dy0 x dx y dx
But dxdx1, so if Fy0 we solve for dydx and obtain
6
F
dy x Fx dx F Fy y
906
CHAPTER 14 PARTIAL DERIVATIVES
N The solution to Example 8 should be compared to the one in Example 2 in Section 3.5.
so Equation 6 gives
Now we suppose that z is given implicitly as a function zf x, y by an equation of the form Fx, y, z0. This means that Fx, y, fx, y0 for all x, y in the domain of f. If F and f are differentiable, then we can use the Chain Rule to differentiate the equa tion Fx, y, z0 as follows:
To derive this equation we assumed that Fx, y0 defines y implicitly as a function of x. The Implicit Function Theorem, proved in advanced calculus, gives conditions under which this assumption is valid: It states that if F is defined on a disk containing a, b, where Fa, b0, Fya, b0, and Fx and Fy are continuous on the disk, then the equation Fx, y0 defines y as a function of x near the point a, b and the derivative of this function is given by Equation 6.
EXAMPLE 8 Findyifx3 y3 6xy. SOLUTION The given equation can be written as
Fx,yx3 y3 6xy0
dy Fx 3×2 6y x2 2y
2 2 M
dx Fy 3y 6x y 2x
F xF yF z0
But
so this equation becomes
x x
x1
x
y x and
z x
y0
x FF z0
x zx
If Fz0, we solve for zx and obtain the first formula in Equations 7. The formula
for zy is obtained in a similar manner.
Again, a version of the Implicit Function Theorem gives conditions under which our assumption is valid: If F is defined within a sphere containing a, b, c, where Fa, b, c0, Fza, b, c0, and Fx, Fy, and Fz are continuous inside the sphere, then the equation Fx, y, z0 defines z as a function of x and y near the point a, b, c and this function is differentiable, with partial derivatives given by 7.
F F z x z y x F y F z z
7
SECTION 14.5 THE CHAIN RULE907 EXAMPLE 9 Find z and z ifx3 y3 z3 6xyz1.
z Fx 3×2 6yz x2 2yz x Fz 3z26xy z22xy
z Fy 3y2 6xz y2 2xz M
x y
SOLUTION Let Fx, y, zx3y3z36xyz1. Then, from Equations 7, we have
N The solution to Example 9 should be compared to the one in Example 4 in Section14.3.
14.5 EXERCISES
16 Use the Chain Rule to find dzdt or dwdt.
1. zx2 y2 xy, xsint, yet
2. zcosx4y, x5t4, y1t
3. zs1x2 y2, xlnt, ycost
4. ztan1yx, xet, y1et
wxeyz, xt2, y1t, z12t
6. wlnsx2 y2 z2, xsint, ycost, ztant
712 Use the Chain Rule to find zs and zt. 7. zx2y3, xscost, yssint
8. zarcsinxy, xs2 t2, y12st
9.zsin cos , st2, s2t
10. zex2y, xst, yts
11. zercos, rst, ss2t2
12. ztanuv, u2s3t, v3s2t
13. If zf x, y, where f is differentiable, and
y
Fz 3z2 6xy z2 2xy
14. Let Ws, tFus, t, vs, t, where F, u, and v are differentiable, and
5.
u1, 02 us1, 02 ut1, 06
Fu2, 31 Find Ws1, 0 and Wt1, 0.
v1, 03 vs1, 05 vt1, 04
Fv2, 310
f
t
fx
fy
0, 0
3
6
4
8
1, 2
6
3
2
5
xtt
t32
t35
fx2, 76 find dzdt when t3.
yht h37
h34 fy2, 78
15. Suppose f is a differentiable function of x and y, and tu,vfeu sinv,eu cosv.Usethetableofvalues to calculate tu0, 0 and tv0, 0.
16. Suppose f is a differentiable function of x and y, and tr,sf2rs,s2 4r.Usethetableofvaluesin Exercise 15 to calculate tr1, 2 and ts1, 2.
1720 Use a tree diagram to write out the Chain Rule for the given case. Assume all functions are differentiable.
17. ufx,y, wherexxr,s,t, yyr,s,t
18. Rfx,y,z,t, wherexxu,v,w, yyu,v,w,
zzu,v,w, ttu,v,w
19. wfr,s,t, whererrx,y, ssx,y, ttx,y
20. tfu,v,w, whereuup,q,r,s, vvp,q,r,s, wwp, q, r, s
908CHAPTER 14 PARTIAL DERIVATIVES
2126 Use the Chain Rule to find the indicated partial derivatives.
37.
The speed of sound traveling through ocean water with salinity 35 parts per thousand has been modeled by the equation
C1449.24.6T0.055T 20.00029T 30.016D
where C is the speed of sound in meters per second, T is the temperature in degrees Celsius, and D is the depth below the ocean surface in meters. A scuba diver began a leisurely dive into the ocean water; the divers depth and the surrounding water temperature over time are recorded in the following graphs. Estimate the rate of change with respect to time of the speed of sound through the ocean water experienced by the diver 20 minutes into the dive. What are the units?
21. zx2 xy3, xuv2 w3, yuvew; z, z, z whenu2,v1,w0
u v w
22. usr2 s2, ryxcost, sxysint;
u, u, u whenx1,y2,t0 x y t
23. Rlnu2 v2 w2,
ux2y, v2xy, w2xy;
R, R whenxy1 xy
24. Mxeyz2, x2uv, yuv, zuv;
, when u3, v1 v
D T 16
20 14 15 12 10 10
58 10 20 30 40t
min
M M
u
25. ux2 yz, xprcos , yprsin , zpr;
10 20 30 40t min
u u u
p, r,when p2,r3, 0
26. Ywtan1uv, urs, vst, wtr; Y Y Y
38.
The radius of a right circular cone is increasing at a rate of
1.8 ins while its height is decreasing at a rate of 2.5 ins. At what rate is the volume of the cone changing when the radius is 120 in. and the height is 140 in.?
The length , width w, and height h of a box change with time. At a certain instant the dimensions are 1 m and
wh2 m, andand w are increasing at a rate of 2 ms while h is decreasing at a rate of 3 ms. At that instant find the rates at which the following quantities are changing.
a The volume
b The surface area
c The length of a diagonal
The voltage V in a simple electrical circuit is slowly decreasing as the battery wears out. The resistance R is slowly increasing as the resistor heats up. Use Ohms Law, VIR, to find how the current I is changing at the moment when R400 ,
I0.08 A, dVdt0.01 Vs, and dRdt0.03 s.
The pressure of 1 mole of an ideal gas is increasing at a rate of 0.05 kPas and the temperature is increasing at a rate of 0.15 Ks. Use the equation in Example 2 to find the rate of change of the volume when the pressure is 20 kPa and the temperature is 320 K.
Car A is traveling north on Highway 16 and car B is traveling west on Highway 83. Each car is approaching the intersection of these highways. At a certain moment, car A is 0.3 km from the intersection and traveling at 90 kmh while car B is 0.4 km from the intersection and traveling at 80 kmh. How fast is the distance between the cars changing at that moment?
One side of a triangle is increasing at a rate of 3 cms and a second side is decreasing at a rate of 2 cms. If the area of the
r, s, t whenr1,s0,t1 2730 Use Equation 6 to find dydx.
27. sxy 1x2y 29. cosxyxey
28. y5 x2y3 1yex2
30. sinxcosysinx cosy
39.
3134 Use Equations 7 to find zx and zy.
31. x2 y2 z2 3xyz 33. xzarctanyz
xyzcosxyz 34. yzlnxz
40.
41.
42.
43.
32.
The temperature at a point x, y is Tx, y, measured in degrees
Celsius. A bug crawls so that its position after t seconds is
given by xs1t , y21 t, where x and y are measured 3
in centimeters. The temperature function satisfies Tx2, 34 and Ty2, 33. How fast is the temperature rising on the bugs path after 3 seconds?
36. Wheat production W in a given year depends on the average temperature T and the annual rainfall R. Scientists estimate that the average temperature is rising at a rate of 0.15Cyear and rainfall is decreasing at a rate of 0.1 cmyear. They also estimate that, at current production levels, WT2
and WR8.
a What is the significance of the signs of these partial
derivatives?
b Estimate the current rate of change of wheat production,
dWdt.
35.
triangle remains constant, at what rate does the angle between the sides change when the first side is 20 cm long, the second side is 30 cm, and the angle is 6?
44. If a sound with frequency fs is produced by a source traveling along a line with speed vs and an observer is traveling with speed vo along the same line from the opposite direction toward the source, then the frequency of the sound heard by the observer is
50.
51. 52. 53.
54.
SECTION 14.5 THE CHAIN RULE909 Ifufx,y,wherexes costandyes sint,showthat
45.
focvo fs
Compare with Example 7. Ifzfx,y,wherexrcos andyrsin ,find
a zr, b z , and c 2zr. Ifzfx,y,wherexrcos andyrsin ,showthat
2z 2z 2z 1 2z 1 z 22222
2 2e2s
2u 2u 2u 2u
xy st Ifzfx,y,wherexr2 s2 andy2rs,find2zrs.
22
cvs
where c is the speed of sound, about 332 ms. This is the Doppler effect. Suppose that, at a particular moment, you
are in a train traveling at 34 ms and accelerating at 1.2 ms2. A train is approaching you from the opposite direction on the other track at 40 ms, accelerating at 1.4 ms2, and sounds its whistle, which has a frequency of 460 Hz. At that instant, what is the perceived frequency that you hear and how fast is it changing?
4548 Assume that all the given functions are differentiable. Ifzfx,y,wherexrcos andyrsin ,afindzr
x y r r rr Supposezfx,y,wherexts,tandyhs,t.
a Show that
2z 2z t 2x 2
b Find a similar formula for 2zs t.
y 2
x 2
2z x y 2z
2 x y t ty x t2 y t2
2
t
z 2xz 2y
t
and z and b show that
z2 z2 z2 1 z2
xyr r2
46. Ifufx,y,wherexes costandyes sint,showthat
55.
A function f is called homogeneous of degree n if it satisfies the equation ftx, tytnfx, y for all t, where n is a positive integer and f has continuous secondorder partial derivatives.
a Verify that fx, yx2y2xy25y3 is homogeneous
of degree 3.
b Show that if f is homogeneous of degree n, then
x f y f nfx,y
x y
Hint: Use the Chain Rule to differentiate ftx, ty with respect to t.
47.
x y s t
u 2 u 2 u 2 u 2 e2s
Ifzfxy,showthat zz 0. x y
48. Ifzfx,y,wherexstandyst,showthat z2 z2 z z
56.
57. 58.
If f is homogeneous of degree n, show that
x2 2f 2xy 2f y2 2f nn1fx,y
xy st
4954 Assume that all the given functions have continuous
x2 xy y2
If f is homogeneous of degree n, show that
secondorder partial derivatives.
49. Show that any function of the form
zfxattxat is a solution of the wave equation
fxtx, tytn1fxx, y
Suppose that the equation Fx, y, z0 implicitly defines each of the three variables x, y, and z as functions of the other two: zfx, y, ytx, z, xhy, z. If F is differentiable and Fx , Fy , and Fz are all nonzero, show that
z x y1 x y z
a2 Hint:Letuxat,vxat.
2z t2
2z x2
910
CHAPTER 14 PARTIAL DERIVATIVES
60
50
14.6
0 50 100 150 200 Distance in miles
60
Las Vegas
DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR
The weather map in Figure 1 shows a contour map of the temperature function Tx, y for the states of California and Nevada at 3:00 PM on a day in October. The level curves, or isothermals, join locations with the same temperature. The partial derivative Tx at a loca tion such as Reno is the rate of change of temperature with respect to distance if we travel east from Reno; Ty is the rate of change of temperature if we travel north. But what if we want to know the rate of change of temperature when we travel southeast toward Las Vegas, or in some other direction? In this section we introduce a type of derivative, called a directional derivative, that enables us to find the rate of change of a function of two or more variables in any direction.
DIRECTIONAL DERIVATIVES
Recall that if zf x, y, then the partial derivatives fx and fy are defined as fxx0, y0lim fx0h, y0fx0, y0
Reno
San Francisco 70
70
Los Angeles
80
FIGURE 1
y
1
hl0 h
fyx0, y0lim fx0, y0hfx0, y0
hl0 h
u sin
cos 0x
and represent the rates of change of z in the x and ydirections, that is, in the directions of the unit vectors i and j.
Suppose that we now wish to find the rate of change of z at x0, y0in the direction of an arbitrary unit vector ua, b. See Figure 2. To do this we consider the surface S with equation zfx, y the graph of fand we let z0fx0, y0. Then the point Px0, y0, z0lies on S. The vertical plane that passes through P in the direction of u inter sects S in a curve C. See Figure 3. The slope of the tangent line T to C at the point P is the rate of change of z in the direction of u.
x , y
FIGURE 2
A unit vector uka, blkcos, sin l
TEC Visual 14.6A animates Figure 3 by rotating u and therefore T .
FIGURE 3
z
T Px , y , z
y
x
1997 USA Today
SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR911 If Qx, y, z is another point on C and P, Q are the projections of P, Q on the xyplane,
then the vector PBQ is parallel to u and so
PBQhuha, hb
forsomescalarh.Thereforexx0 ha,yy0 hb,soxx0 ha,yy0 hb, and
zzz0fx0 ha,y0 hbfx0,y0 hhh
If we take the limit as h l 0, we obtain the rate of change of z with respect to distance in the direction of u, which is called the directional derivative of f in the direction of u.
DEFINITION The directional derivative of f at x0, y0in the direction of a unit vector ua, b is
Du fx0,y0lim fx0 ha,y0 hbfx0,y0 hl0 h
if this limit exists.
2
By comparing Definition 2 with Equations 1, we see that if ui1, 0, then Di ffx and if uj0, 1, then Dj ffy. In other words, the partial derivatives of f with respect to x and y are just special cases of the directional derivative.
EXAMPLE 1 Use the weather map in Figure 1 to estimate the value of the directional derivative of the temperature function at Reno in the southeasterly direction.
SOLUTION The unit vector directed toward the southeast is uijs2 , but we wont need to use this expression. We start by drawing a line through Reno toward the south east. See Figure 4.
60
50
Reno
San Francisco
70
60
Las Vegas
70
Los Angeles
80
FIGURE 4
0 50 100 150 200 Distance in miles
We approximate the directional derivative Du T by the average rate of change of the temperature between the points where this line intersects the isothermals T50 and
1997 USA Today
912
CHAPTER 14 PARTIAL DERIVATIVES
T60. The temperature at the point southeast of Reno is T60F and the temperature at the point northwest of Reno is T50F. The distance between these points looks to be about 75 miles. So the rate of change of the temperature in the southeasterly direction is
DuT 605010 0.13Fmi M 75 75
When we compute the directional derivative of a function defined by a formula, we gen erally use the following theorem.
PROOF If we define a function t of the single variable h by thfx0 ha,y0 hb
then, by the definition of a derivative, we have
t0lim tht0 lim fx0 ha,y0 hbfx0,y0 hl0 h hl0 h
Du fx0,y0 Ontheotherhand,wecanwritethfx,y,wherexx0 ha,yy0 hb,sothe
Chain Rule Theorem 14.5.2 gives
th f dxf dy fxx,yafyx,yb x dh y dh
If we now put h0, then xx0, yy0, and
t0fxx0, y0afyx0, y0b
Comparing Equations 4 and 5, we see that
Du fx0, y0fxx0, y0afyx0, y0b M If the unit vector u makes an angle with the positive xaxis as in Figure 2, then we
can write ucos , sinand the formula in Theorem 3 becomes Du fx, yfxx, y cosfyx, y sin
EXAMPLE 2 Find the directional derivative Du f x, y if fx,yx3 3xy4y2
and u is the unit vector given by angle6. What is Du f 1, 2?
THEOREM If f is a differentiable function of x and y, then f has a directional derivative in the direction of any unit vector ua, b and
Du fx, yfxx, yafyx, yb
3
4
5
6
N The directional derivative Du f 1, 2 in Example 2 represents the rate of change of z in the direction of u. This is the slope of the tan gent line to the curve of intersection of the surface zx33xy4y2 and the vertical plane through 1, 2, 0 in the direction of u shown in Figure 5.
z
Du fx,yfxx,ycos 6 fyx,ysin 6
2 s3 1
0 1,2,0 y x 6 u
FIGURE 5
THE GRADIENT VECTOR
SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR SOLUTION Formula 6 gives
Therefore
913
2
2
3x 3y 2 3x8y2 13s3x2 3x83s3y
2
Du f 1, 21 3 s3 123183 s3 2133 s3
M
Notice from Theorem 3 that the directional derivative can be written as the dot product of two vectors:
7 Du fx, yfxx, yafyx, yb
fxx, y, fyx, ya, b
fxx, y, fyx, yu
The first vector in this dot product occurs not only in computing directional derivatives but in many other contexts as well. So we give it a special name the gradient of fand a spe cial notation grad f or f, which is read del f .
8 DEFINITION If f is a function of two variables x and y, then the gradient of f is the vector function f defined by
fx,y fxx,y, fyx,yf i f j x y
EXAMPLE3 Iffx,ysinxexy,then
fx,y fx, fycosxyexy,xexy
and f0, 12, 0 M With this notation for the gradient vector, we can rewrite the expression 7 for the
directional derivative as
9 Du fx,yfx,yu
This expresses the directional derivative in the direction of u as the scalar projection of the gradient vector onto u.
914CHAPTER 14 PARTIAL DERIVATIVES
N Thegradientvectorf2,1inExample4 is shown in Figure 6 with initial point 2, 1. Also shown is the vector v that gives the direc tion of the directional derivative. Both of these vectors are superimposed on a contour plot of thegraphof f.
y
f2, 1
FIGURE 6
V EXAMPLE 4 Findthedirectionalderivativeofthefunction fx,yx2y3 4yatthe point 2, 1 in the direction of the vector v2i5j.
SOLUTION We first compute the gradient vector at 2, 1: fx,y2xy3i3x2y2 4j
f 2, 14 i8 j
Note that v is not a unit vector, but since v s29 , the unit vector in the direction
v
of v is
x uv2i5j
2, 1
v s29 s29
Du f2,1f2,1u4i8j2
4285 32 s29 s29
FUNCTIONS OF THREE VARIABLES
Therefore, by Equation 9, we have
s29
i 5 s29
j
For functions of three variables we can define directional derivatives in a similar manner. Again Du f x, y, z can be interpreted as the rate of change of the function in the direction of a unit vector u.
M
DEFINITION The directional derivative of f at x0, y0, z0in the direction of a unit vector ua, b, c is
Du fx0,y0,z0lim fx0 ha,y0 hb,z0 hcfx0,y0,z0 hl0 h
if this limit exists.
10
If we use vector notation, then we can write both definitions 2 and 10 of the direc tional derivative in the compact form
where x0x0, y0 if n2 and x0x0, y0, z0 if n3. This is reasonable because the vector equation of the line through x0 in the direction of the vector u is given by xx0tu Equation 12.5.1 and so fx0hu represents the value of f at a point on this line.
11
Du fx0lim fx0hufx0 hl0 h
SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR915 If fx, y, z is differentiable and ua, b, c, then the same method that was used to
prove Theorem 3 can be used to show that
Du fx,y,zfxx,y,zafyx,y,zbfzx,y,zc
For a function f of three variables, the gradient vector, denoted by f or grad f, is
fx, y, z fxx, y, z, fyx, y, z, fzx, y, z
or, for short,
Then, just as with functions of two variables, Formula 12 for the directional derivative can be rewritten as
V EXAMPLE 5 If fx, y, zx sin yz, a find the gradient of f and b find the direc tional derivative of f at 1, 3, 0 in the direction of vi2 jk.
SOLUTION
a The gradient of f is
fx, y, z fxx, y, z, fyx, y, z, fzx, y, z
sinyz,xzcosyz,xycosyz
b At 1, 3, 0 we have f1, 3, 00, 0, 3. The unit vector in the direction of
12
13
ffx,fy,fz f i f j f k x y z
14
vi2 jk is Therefore Equation 14 gives
u1i2j1k s6 s6 s6
Du f1, 3, 0f1, 3, 0u
3k1 i 2 j 1 k
s6 2 MAXIMIZING THE DIRECTIONAL DERIVATIVE
Du fx,y,zfx,y,zu
s6 s6 s6
3 1 3 M
Suppose we have a function f of two or three variables and we consider all possible direc tional derivatives of f at a given point. These give the rates of change of f in all possible directions. We can then ask the questions: In which of these directions does f change fastest and what is the maximum rate of change? The answers are provided by the follow ing theorem.
916CHAPTER 14 PARTIAL DERIVATIVES
TEC Visual 14.6B provides visual confirmation of Theorem 15.
15 THEOREM Suppose f is a differentiable function of two or three variables. The maximum value of the directional derivative Du fx is fxand it occurs when u has the same direction as the gradient vector fx.
PROOF From Equation 9 or 14 we have
Du ffufucos fcos
where is the angle between f and u. The maximum value of cos is 1 and this occurs when0. Therefore the maximum value of Du f is f and it occurs when
0, that is, when u has the same direction as f. M EXAMPLE 6
a If fx, yxey, find the rate of change of f at the point P2, 0 in the direction from P to Q1, 2.
b In what direction does f have the maximum rate of change? What is this maximum rate of change?
SOLUTION
a We first compute the gradient vector:
fx,y fx, fyey,xey f2, 01, 2
l
of f in the direction from P to Q is
Du f 2, 0 f 2, 0u 1, 23 , 4
13 24 1 55
b According to Theorem 15, f increases fastest in the direction of the gradient vector f2, 01, 2. The maximum rate of change is
f2, 0 1, 2s5 M EXAMPLE 7 Suppose that the temperature at a point x, y, z in space is given by
Tx, y, z801x22y23z2 , where T is measured in degrees Celsius and x, y, z in meters. In which direction does the temperature increase fastest at the point 1, 1, 2? What is the maximum rate of increase?
y 2
Q
1 01P3x
FIGURE 7
N At 2, 0 the function in Example 6 increases fastest in the direction of the gradient vector f 2, 01, 2. Notice from Figure 7 that this vector appears to be perpendicular to the level curve through 2, 0. Figure 8 shows the graph of f and the gradient vector.
20
15 z 10 5
001 12 x230y
FIGURE 8
f2, 0
2
The unit vector in the direction of PQ1.5, 2 is u3, 4 , so the rate of change
SOLUTION The gradient of T is TT iT jT k
x y z
160x i 320y j 480z k
1×2 2y2 3z22 1×2 2y2 3z22 1×2 2y2 3z22
160 xi2yj3zk 1×2 2y2 3z22
55
55
SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR917 At the point 1, 1, 2 the gradient vector is
T1, 1, 2160 i2 j6 k5 i2 j6 k 256 8
By Theorem 15 the temperature increases fastest in the direction of the gradient vector T 1, 1, 25 i2 j6 k or, equivalently, in the direction of i2 j6 k or
the unit vector i2 j6 ks41. The maximum rate of increase is the length of the gradient vector:
T 1, 1, 2 5i2 j6 k 5 s41 88
Therefore the maximum rate of increase of temperature is 5 s414Cm. M 8
TANGENT PLANES TO LEVEL SURFACES
Suppose S is a surface with equation Fx, y, zk, that is, it is a level surface of a func tion F of three variables, and let Px0, y0, z0be a point on S. Let C be any curve that lies on the surface S and passes through the point P. Recall from Section 13.1 that the curve C is described by a continuous vector function rtxt, yt, zt. Let t0 be the param eter value corresponding to P; that is, rt0x0, y0, z0. Since C lies on S, any point xt, yt, zt must satisfy the equation of S, that is,
Fxt, yt, ztk
If x, y, and z are differentiable functions of t and F is also differentiable, then we can use
the Chain Rule to differentiate both sides of Equation 16 as follows:
F dxF dyF dz0 x dt y dt z dt
But, since FFx , Fy , Fzand rtxt, yt, zt, Equation 17 can be written in terms of a dot product as
Frt0
In particular, when tt0 we have rt0x0, y0, z0 , so
Fx0, y0, z0rt00
Equation 18 says that the gradient vector at P, Fx0, y0, z0 , is perpendicular to the tangent vector rt0 to any curve C on S that passes through P. See Figure 9. If Fx0, y0, z0 0, it is therefore natural to define the tangent plane to the level surface Fx, y, zk at Px0, y0, z0as the plane that passes through P and has normal vector Fx0, y0, z0 . Using the standard equation of a plane Equation 12.5.7, we can write the equation of this tangent plane as
16
8
17
z
FIGURE 9
Fx , y , z
tangent plane
y
18
x
19
Fxx0, y0, z0xx0Fyx0, y0, z0yy0Fzx0, y0, z0zz00
918
CHAPTER 14 PARTIAL DERIVATIVES
N Figure 10 shows the ellipsoid, tangent plane, and normal line in Example 8.
4 2 0
z 2 4 6
y 0 FIGURE 10
Therefore we have
2 2
0 2
x x2y1z3 M
The normal line to S at P is the line passing through P and perpendicular to the tan gent plane. The direction of the normal line is therefore given by the gradient vector Fx0, y0, z0and so, by Equation 12.5.3, its symmetric equations are
20 xx0yy0zz0 Fxx0, y0, z0Fyx0, y0, z0Fzx0, y0, z0
In the special case in which the equation of a surface S is of the form zf x, y that is, S is the graph of a function f of two variables, we can rewrite the equation as
Fx, y, zf x, yz0 and regard S as a level surface with k0 of F. Then
Fxx0, y0, z0 fxx0, y0Fyx0, y0, z0 fyx0, y0Fzx0, y0, z01
so Equation 19 becomes
fxx0, y0xx0fyx0, y0yy0zz00
which is equivalent to Equation 14.4.2. Thus our new, more general, definition of a tangent
plane is consistent with the definition that was given for the special case of Section 14.4. V EXAMPLE 8 Find the equations of the tangent plane and normal line at the point
2, 1, 3 to the ellipsoid
x2 z2
4 y29 3
SOLUTION The ellipsoid is the level surface with k3 of the function
x2 z2 y249
Fzx,y,z 2z xyz3
Then Equation 19 gives the equation of the tangent plane at 2, 1, 3 as 1x22y12 z30
which simplifies to 3x6y2z180.
By Equation 20, symmetric equations of the normal line are
Fx,y,z
Fxx,y,z x 29
Fyx,y,z2y
F 2, 1, 31 F 2, 1, 32 F 2, 1, 32
1 2 2 3
3
SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR919 SIGNIFICANCE OF THE GRADIENT VECTOR
We now summarize the ways in which the gradient vector is significant. We first consider a function f of three variables and a point Px0, y0, z0in its domain. On the one hand, we know from Theorem 15 that the gradient vector fx0, y0, z0 gives the direction of fastest increase of f. On the other hand, we know that fx0, y0, z0 is orthogonal to the level sur face S of f through P. Refer to Figure 9. These two properties are quite compatible intu itively because as we move away from P on the level surface S, the value of f does not change at all. So it seems reasonable that if we move in the perpendicular direction, we get the maximum increase.
In like manner we consider a function f of two variables and a point Px0, y0in its domain. Again the gradient vector f x0, y0gives the direction of fastest increase of f. Also, by considerations similar to our discussion of tangent planes, it can be shown that fx0, y0 is perpendicular to the level curve fx, yk that passes through P. Again this is intuitively plausible because the values of f remain constant as we move along the curve. See Figure 11.
y
fx, yk
0x
FIGURE 11
fx , yPx ,y
level curve
curve of 300 steepest 200 ascent 100
FIGURE 12
If we consider a topographical map of a hill and let f x, y represent the height above sea level at a point with coordinates x, y, then a curve of steepest ascent can be drawn as in Figure 12 by making it perpendicular to all of the contour lines. This phenomenon can also be noticed in Figure 12 in Section 14.1, where Lonesome Creek follows a curve of steepest descent.
Computer algebra systems have commands that plot sample gradient vectors. Each gra dient vectorf a, b is plotted starting at the point a, b. Figure 13 shows such a plot called a gradient vector field for the function fx, yx2y2 superimposed on a con tour map of f. As expected, the gradient vectors point uphill and are perpendicular to the level curves.
y
0369
x
9
6
3
FIGURE 13
920CHAPTER 14 PARTIAL DERIVATIVES
1.
14.6 EXERCISES
Level curves for barometric pressure in millibars are shown for 6:00 AM on November 10, 1998. A deep low with pressure 972 mb is moving over northeast Iowa. The distance along the red line from K Kearney, Nebraska to S Sioux City, Iowa is 300 km. Estimate the value of the directional derivative of the pressure function at Kearney in the direction of Sioux City. What are the units of the directional derivative?
710
a Find the gradient of f.
b Evaluate the gradient at the point P.
c Find the rate of change of f at P in the direction of the
10081012 1016
vector u.
7. f x, ysin2x3y,
8. fx, yy2x, P1, 2,
P6, 4, u1 s3 ij 2
u1 2is5 j 3
u 2 ,2 , 1333
236 u7 , 7 , 7
10201024 9. 10.
fx, y, zxe 2 y z,
f x, y, zsxyz ,
1004 1000
P 3, 0, 2, P1, 3, 1,
1012
K
1008
From Meteorology Today, 8E by C. Donald Ahrens 2007 Thomson BrooksCole.
2. The contour map shows the average maximum temperature for November 2004 in C. Estimate the value of the directional derivative of this temperature function at Dubbo, New South Wales, in the direction of Sydney. What are the units?
996 992
988 984
980
1117 Find the directional derivative of the function at the given point in the direction of the vector v.
976 972
fx, y12xsy,
12. fx, ylnx2y2,
13. tp, qp4p2q3,
14. tr, stan1rs, 1, 2, 15. fx, y, zxeyyezzex, 16. fx, y, zsxyz, 3, 2, 6, 17. tx, y, zx2y3z32,
v4, 3 v1, 2
vi3j v5i10j
0, 0, 0, v5, 1, 2 v1, 2, 2
11.
2, 1,
1, 1, 2, 18. Use the figure to estimate Du f 2, 2.
y
u
v2 jk
3, 4, 2, 1,
0 100 200 300 Distance in kilometres
30
24
Sydney
2, 2
Dubbo
f2, 2 0x
19. Find the directional derivative of f x, ysxy at P2, 8 in the direction of Q5, 4.
20. Find the directional derivative of f x, y, zxyyzzx at P1, 1, 3 in the direction of Q2, 4, 5.
2126 Find the maximum rate of change of f at the given point and the direction in which it occurs.
21. fx, yy2x, 2, 4
22. fp, qqeppeq, 0, 0
f x, ysinxy, 1, 0
24. f x, y, zxyz, 1, 1, 1
25. fx, y, zsx2y2z2 , 3, 6, 2 26. f x, y, ztanx2y3z, 5, 1, 1
27
18
Copyright Commonwealth of Australia. Reproduced by permission.
24
3. A table of values for the windchill index Wf T, v is given in Exercise 3 on page 888. Use the table to estimate the value of Du f20, 30, where uijs2.
46 Find the directional derivative of f at the given point in the direction indicated by the angle .
21
4. fx, yx2y3y4, 2, 1,
5. f x, yyex, 0, 4,2
6. fx, yx sinxy, 2, 0,
4
3
3
23.
S
27.
a Show that a differentiable function f decreases most rapidly at x in the direction opposite to the gradient vector, that is, in the direction off x.
b Use the result of part a to find the direction in which the function fx, yx4yx2y3 decreases fastest at the point 2, 3.
28. Find the directions in which the directional derivative of f x, yyexy at the point 0, 2 has the value 1.
Find all points at which the direction of fastest change of the function fx,yx2 y2 2x4yisij.
30. Near a buoy, the depth of a lake at the point with coordinates x,yisz2000.02×2 0.001y3,wherex,y,andzare measured in meters. A fisherman in a small boat starts at the point 80, 60 and moves toward the buoy, which is located at 0, 0. Is the water under the boat getting deeper or shallower when he departs? Explain.
31. The temperature T in a metal ball is inversely proportional to the distance from the center of the ball, which we take to be the origin. The temperature at the point 1, 2, 2 is 120.
a Find the rate of change of T at 1, 2, 2 in the direction
toward the point 2, 1, 3.
b Show that at any point in the ball the direction of greatest
increase in temperature is given by a vector that points toward the origin.
32. The temperature at a point x, y, z is given by Tx, y, z200ex 23y 29z 2
where T is measured in C and x, y, z in meters.
a Find the rate of change of temperature at the point
P2, 1, 2 in the direction toward the point 3, 3, 3.
b In which direction does the temperature increase fastest
atP?
c Find the maximum rate of increase at P.
33. Suppose that over a certain region of space the electrical poten tialV isgivenbyVx,y,z5x2 3xyxyz.
a Find the rate of change of the potential at P3, 4, 5 in the
direction of the vector vijk.
b In which direction does V change most rapidly at P?
c What is the maximum rate of change at P?
34. Suppose you are climbing a hill whose shape is given by the equation z10000.005x 20.01y 2, where x, y, and z are measured in meters, and you are standing at a point with coor dinates 60, 40, 966. The positive xaxis points east and the positive yaxis points north.
a If you walk due south, will you start to ascend or descend? At what rate?
b If you walk northwest, will you start to ascend or descend? At what rate?
c In which direction is the slope largest? What is the rate of ascent in that direction? At what angle above the horizontal does the path in that direction begin?
35.
36.
Let f be a function of two variables that has continuous
partial derivatives and consider the points A1, 3, B3, 3,
SECTION 14.6 DIRECTIONAL DERIVATIVES AND THE GRADIENT VECTOR921
29.
37.
38.
Show that the operation of taking the gradient of a function has the given property. Assume that u and v are differentiable func tions of x and y and that a, b are constants.
C1, 7, and D6, 15. The directional derivative of f at A in the l
direction of the vector AB is 3 and the directional derivative at l
A in the direction of AC is 26. Find the directional derivative of l
f at A in the direction of the vector AD.
For the given contour map draw the curves of steepest ascent
starting at P and at Q.
P
Q
a aubvaubv cuvuuv
v v2
b uvuvvu
d un nun1u
60
50
20 40 30
Sketch the gradient vectorf 4, 6 for the function f whose level curves are shown. Explain how you chose the direction and length of this vector.
y
6 3 4,6
4 1
0
5
1
35 0246x
2
3944 Find equations of a the tangent plane and b the normal line to the given surface at the specified point.
39. 2x22y12z3210, 3, 3, 5 40. yx2z2, 4, 7, 3
41. x22y2z2yz2, 2, 1, 1
42. xz4 arctanyz, 1, 1, 1
z1xey cos z, 1, 0, 0 44. yzlnxz, 0, 0, 1
43.
922CHAPTER 14 PARTIAL DERIVATIVES
; 4546 Use a computer to graph the surface, the tangent plane, and the normal line on the same screen. Choose the domain care fully so that you avoid extraneous vertical planes. Choose the viewpoint so that you get a good view of all three objects.
56.
58. 59. 60.
61.
45. xyyzzx3,
46. xyz6, 1, 2, 3
1, 1, 1
Show that every normal line to the sphere x2y2z2r2 passes through the center of the sphere.
Show that the sum of the x, y, and zintercepts of any tangentplanetothesurfacesx sy sz sc isa constant.
Show that the pyramids cut off from the first octant by any tangent planes to the surface xyz1 at points in the first octant must all have the same volume.
Find parametric equations for the tangent line to the curve of intersection of the paraboloid zx 2y 2 and the ellipsoid 4×2y2z29 at the point 1, 1, 2.
a Theplaneyz3intersectsthecylinderx2 y2 5 in an ellipse. Find parametric equations for the tangent line to this ellipse at the point 1, 2, 1.
b Graph the cylinder, the plane, and the tangent line on the same screen.
a Two surfaces are called orthogonal at a point of inter section if their normal lines are perpendicular at that point. Show that surfaces with equations Fx, y, z0 and Gx, y, z0 are orthogonal at a point P where F0andG0ifandonlyif
FxGx FyGy FzGz 0 atP
b Use part a to show that the surfaces z2x2y2 and x2y2z2r2 are orthogonal at every point of intersection. Can you see why this is true without using calculus?
a Show that the function fx, ys3 xy is continuous and the partial derivatives fx and fy exist at the origin but the directional derivatives in all other directions do not exist.
;
47. If fx, yxy, find the gradient vector f3, 2 and use it to find the tangent line to the level curve f x, y6 at the point 3, 2. Sketch the level curve, the tangent line, and the gradient vector.
48. If tx, yx2y24x, find the gradient vector t1, 2 and use it to find the tangent line to the level curve
tx, y1 at the point 1, 2. Sketch the level curve, the tangent line, and the gradient vector.
49. Show that the equation of the tangent plane to the ellipsoid x2a2y2b2z2c21 at the point x0, y0, z0 can be written as
xx0yy0 zz0 1 a2 b2 c2
50. Find the equation of the tangent plane to the hyperboloid x2a2y2b2z2c21 at x0, y0, z0 and express it in a form similar to the one in Exercise 49.
51. Show that the equation of the tangent plane to the elliptic paraboloid zcx2a2y2b2 at the point x0, y0, z0can be written as
;
2xx02yy0zz0 a2 b2 c
62.
63.
64.
b Graph f near the origin and comment on how the graph confirms part a.
Suppose that the directional derivatives of f x, y are known at a given point in two nonparallel directions given by unit vectors u and v. Is it possible to find f at this point? If so, how would you do it?
Show that if zfx, y is differentiable at x0x0, y0, then
lim fxfx0fx0xx0 0 xlx0 xx0
Hint: Use Definition 14.4.7 directly.
57.
52. At what point on the paraboloid yx2z2 is the tangent planeparalleltotheplanex2y3z1?
53. Are there any points on the hyperboloid x2y2z21 where the tangent plane is parallel to the plane zxy?
54. Show that the ellipsoid 3×22y2z29 and the sphere x2 y2 z2 8x6y8z240aretangenttoeach other at the point 1, 1, 2. This means that they have a com mon tangent plane at the point.
55. Show that every plane that is tangent to the cone x 2y 2z2 passes through the origin.
14.7 MAXIMUM AND MINIMUM VALUES
As we saw in Chapter 4, one of the main uses of ordinary derivatives is in finding maxi mum and minimum values. In this section we see how to use partial derivatives to locate maxima and minima of functions of two variables. In particular, in Example 6 we will see how to maximize the volume of a box without a lid if we have a fixed amount of cardboard to work with.
z
local maximum
x
SECTION 14.7 MAXIMUM AND MINIMUM VALUES923 Look at the hills and valleys in the graph of f shown in Figure 1. There are two points
a, b where f has a local maximum, that is, where f a, b is larger than nearby values of f x, y. The larger of these two values is the absolute maximum. Likewise, f has two local minima, where f a, b is smaller than nearby values. The smaller of these two values is the
absolute minimum. y
1 DEFINITION A function of two variables has a local maximum at a, b if fx, yfa, b when x, y is near a, b. This means that fx, yfa, b for
all points x, y in some disk with center a, b. The number f a, b is called a local maximum value. If fx, yfa, b when x, y is near a, b, then f has a local minimum at a, b and f a, b is a local minimum value.
If the inequalities in Definition 1 hold for all points x, y in the domain of f, then f has an absolute maximum or absolute minimum at a, b.
2 THEOREM If f has a local maximum or minimum at a, b and the firstorder partial derivatives of f exist there, then fxa, b0 and fya, b0.
PROOF Let txf x, b. If f has a local maximum or minimum at a, b, then t has a local maximum or minimum at a, so ta0 by Fermats Theorem see Theorem 4.1.4. But tafxa, b see Equation 14.3.1 and so fxa, b0. Similarly, by applying Fermats Theorem to the function Gyfa, y, we obtain fya, b0. M
If we put fxa, b0 and fya, b0 in the equation of a tangent plane Equation 14.4.2, we get zz0. Thus the geometric interpretation of Theorem 2 is that if the graph of f has a tangent plane at a local maximum or minimum, then the tangent plane must be horizontal.
A point a, b is called a critical point or stationary point of f if fxa, b0 and fya, b0, or if one of these partial derivatives does not exist. Theorem 2 says that if f has a local maximum or minimum at a, b, then a, b is a critical point of f. However, as
in singlevariable calculus, not all critical points give rise to maxima or minima. At a crit ical point, a function could have a local maximum or a local minimum or neither.
EXAMPLE 1 Let fx,yx2 y2 2x6y14.Then
fxx, y2x2 fyx, y2y6
These partial derivatives are equal to 0 when x1 and y3, so the only critical point is 1, 3. By completing the square, we find that
fx,y4x12 y32
Sincex12 0andy32 0,wehave fx,y4forallvaluesofxandy. Therefore f 1, 34 is a local minimum, and in fact it is the absolute minimum
of f. This can be confirmed geometrically from the graph of f, which is the elliptic paraboloid with vertex 1, 3, 4 shown in Figure 2. M
absolute minimum
FIGURE 1
local minimum
N Notice that the conclusion of Theorem 2 can be stated in the notation of gradient vectors as f a, b0.
z
1, 3, 4 0
x
y
FIGURE 2
z2x6y14
absolute maximum
924
CHAPTER 14 PARTIAL DERIVATIVES
EXAMPLE 2 Find the extreme values of fx, yy2x2.
x
FIGURE 3
z
z
SOLUTION Since fx2x and fy2y, the only critical point is 0, 0. Notice that forpointsonthexaxiswehavey0,so fx,yx2 0ifx0.However,for pointsontheyaxiswehavex0,so fx,yy2 0ify0.Thuseverydisk with center 0, 0 contains points where f takes positive values as well as points where
f takes negative values. Therefore f 0, 00 cant be an extreme value for f, so f has no extreme value. M
Example 2 illustrates the fact that a function need not have a maximum or minimum value at a critical point. Figure 3 shows how this is possible. The graph of f is the hyper bolic paraboloid zy2x2, which has a horizontal tangent plane z0 at the origin. You can see that f 0, 00 is a maximum in the direction of the xaxis but a minimum
y in the direction of the yaxis. Near the origin the graph has the shape of a saddle and so 0, 0 is called a saddle point of f.
We need to be able to determine whether or not a function has an extreme value at a critical point. The following test, which is proved at the end of this section, is analogous to the Second Derivative Test for functions of one variable.
3 SECOND DERIVATIVES TEST Suppose the second partial derivatives of f are continuous on a disk with center a, b, and suppose that fxa, b0 and
fya, b0 that is, a, b is a critical point of f . Let
DDa, bfxxa, bfyya, bfxya, b2
a IfD0and fxxa,b0,then fa,bisalocalminimum. b IfD0and fxxa,b0,then fa,bisalocalmaximum. c If D0, then f a, b is not a local maximum or minimum.
NOTE 1 In case c the point a, b is called a saddle point of f and the graph of f crosses its tangent plane at a, b.
NOTE 2 If D0, the test gives no information: f could have a local maximum or local minimum at a, b, or a, b could be a saddle point of f.
NOTE 3 To remember the formula for D, its helpful to write it as a determinant: Dfxx fxy fxx fyy fxy2
fyx fyy
V EXAMPLE 3 Find the local maximum and minimum values and saddle points of fx,yx4 y4 4xy1.
SOLUTION We first locate the critical points:
fx 4×3 4y fy 4y3 4x
Setting these partial derivatives equal to 0, we obtain the equations x3 y0 and y3 x0
To solve these equations we substitute yx3 from the first equation into the second one. This gives
0x9 xxx8 1xx4 1×4 1xx2 1×2 1×4 1
z
x
FIGURE 4
zxy4xy1
y
so there are three real roots: x0, 1, 1. The three critical points are 0, 0, 1, 1, and 1, 1.
Next we calculate the second partial derivatives and Dx, y:
fxx 12×2 fxy 4 fyy 12y2
Dx, yfxx fyyfxy2144x2y216
Since D0, 0160, it follows from case c of the Second Derivatives Test that the origin is a saddle point; that is, f has no local maximum or minimum at 0, 0. Since D1, 11280 and fxx1, 1120, we see from case a of the test that
f 1, 11 is a local minimum. Similarly, we have D1, 11280 and fxx1, 1120, so f1, 11 is also a local minimum.
N A contour map of the function f in Example 3 is shown in Figure 5. The level curves near 1, 1 and 1, 1 are oval in shape and indicate that as we move away from 1, 1 or 1, 1 in any direction the values of f are increasing. The level curves near 0, 0, on the other hand, resemble hyperbolas. They reveal that as we move away from the origin where the value of f is 1, the values of f decrease in some directions but increase in other directions. Thus the contour map suggests the presence of the minima and saddle point that we found in Example 3.
FIGURE 5
TEC In Module 14.7 you can use contour maps to estimate the locations of critical points.
0.5 0
0.5 10.9
1.1
1.5 2
3
x
The graph of f is shown in Figure 4.
M
SECTION 14.7 MAXIMUM AND MINIMUM VALUES925
EXAMPLE 4 Find and classify the critical points of the function fx,y10x2y5x2 4y2 x4 2y4
Also find the highest point on the graph of f. SOLUTION The firstorder partial derivatives are
fx 20xy10x4x3 fy 10×2 8y8y3 So to find the critical points we need to solve the equations
4 2x10y52×20
5×2 4y4y3 0
x0 or 10y52x2 0
In the first case x0, Equation 5 becomes 4y1y20, so y0 and we have the critical point 0, 0.
5
From Equation 4 we see that either
y
926
CHAPTER 14 PARTIAL DERIVATIVES
3
FIGURE 6
2.7
Inthesecondcase10y52x2 0,weget
6 x2 5y2.5
and,puttingthisinEquation5,wehave25y12.54y4y3 0.Sowehaveto solve the cubic equation
7 4y3 21y12.50 Using a graphing calculator or computer to graph the function
ty4y3 21y12.5
as in Figure 6, we see that Equation 7 has three real roots. By zooming in, we can find
the roots to four decimal places:
y2.5452 y0.6468 y1.8984 Alternatively, we could have used Newtons method or a rootfinder to locate these
roots. From Equation 6, the corresponding xvalues are given by xs5y2.5
If y2.5452, then x has no corresponding real values. If y0.6468, then
x0.8567. If y1.8984, then x2.6442. So we have a total of five critical points, which are analyzed in the following chart. All quantities are rounded to two decimal places.
TEC Visual 14.7 shows several families of surfaces.The surface in Figures 7 and 8 is a member of one of these families.
x
FIGURE 7
x
y
y
Critical point 0, 0
2.64, 1.90 0.86, 0.65
Value of f 0.00
8.50 1.48
fxx
10.00 55.93 5.87
D
80.00 2488.72 187.64
Conclusion
local maximum local maximum saddle point
Figures 7 and 8 give two views of the graph of f and we see that the surface opens downward. This can also be seen from the expression for f x, y: The dominant terms are x42y4 when xand yare large. Comparing the values of f at its local maxi mum points, we see that the absolute maximum value of f is f 2.64, 1.908.50. In other words, the highest points on the graph of f are 2.64, 1.90, 8.50.
zz
FIGURE 8
M
SECTION 14.7 MAXIMUM AND MINIMUM VALUES y
2
3 3 1
927
N The five critical points of the function f in Example 4 are shown in red in the contour map of f in Figure 9.
FIGURE 9
37 1
0.8 1.48
V EXAMPLE 5 Find the shortest distance from the point 1, 0, 2 to the plane x2yz4.
SOLUTION The distance from any point x, y, z to the point 1, 0, 2 is dsx12 y2 z22
but if x, y, z lies on the plane x2yz4, then z4x2y and so we have dsx12 y2 6x2y2.Wecanminimizedbyminimizingthesimpler expression
d2 fx,yx12 y2 6x2y2 By solving the equations
fx 2x126x2y4x4y140 fy 2y46x2y4x10y240
wefindthattheonlycriticalpointis11,5.Since fxx 4, fxy 4,and fyy 10,we 63
haveDx,yfxx fyy fxy2 240and fxx 0,sobytheSecondDerivativesTest f has a local minimum at 11, 5 . Intuitively, we can see that this local minimum is actually
an absolute minimum because there must be a point on the given plane that is closest to 1, 0, 2. If x11 and y5, then
x
63
63
dsx12 y2 6x2y2 s52 52 52 5s6 6366
The shortest distance from 1, 0, 2 to the plane x2yz4 is 5 s6. M 6
V EXAMPLE 6 A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box.
SOLUTION Let the length, width, and height of the box in meters be x, y, and z, as shown in Figure 10. Then the volume of the box is
Vxyz
x We can express V as a function of just two variables x and y by using the fact that the
area of the four sides and the bottom of the box is 2xz2yzxy12
N Example 5 could also be solved using vectors. Compare with the methods of Section 12.5.
z
FIGURE 10
y
3
102030
928
CHAPTER 14 PARTIAL DERIVATIVES
Solving this equation for z, we get z12xy2xy, so the expression for V
becomes
We compute the partial derivatives:
Vxy 12xy12xyx2y2
2xy V y2122xyx2
2xy
V x2122xyy2
x2xy2
If V is a maximum, then VxVy0, but x0 or y0 gives V0, so we
y2xy2 122xyx2 0 122xyy2 0
a Closed sets
These imply that x2y2 and so xy. Note that x and y must both be positive in this problem.Ifweputxyineitherequationweget123x2 0,whichgivesx2, y2, and z12222221.
We could use the Second Derivatives Test to show that this gives a local maximum
of V, or we could simply argue from the physical nature of this problem that there must be an absolute maximum volume, which has to occur at a critical point of V, so it must occur when x2, y2, z1. Then V2214, so the maximum volume of the box is 4 m3 . M
ABSOLUTE MAXIMUM AND MINIMUM VALUES
For a function f of one variable the Extreme Value Theorem says that if f is continuous on a closed interval a, b, then f has an absolute minimum value and an absolute maxi mum value. According to the Closed Interval Method in Section 4.1, we found these by evaluating f not only at the critical numbers but also at the endpoints a and b.
There is a similar situation for functions of two variables. Just as a closed interval con tains its endpoints, a closed set in 2 is one that contains all its boundary points. A bound ary point of D is a point a, b such that every disk with center a, b contains points in D and also points not in D. For instance, the disk
Dx,yx2 y2 1
which consists of all points on and inside the circle x2y21, is a closed set because it contains all of its boundary points which are the points on the circle x2y21. But if even one point on the boundary curve were omitted, the set would not be closed. See Figure 11.
A bounded set in 2 is one that is contained within some disk. In other words, it is finite in extent. Then, in terms of closed and bounded sets, we can state the following coun terpart of the Extreme Value Theorem in two dimensions.
must solve the equations
b Sets that are not closed
FIGURE 11
EXTREME VALUE THEOREM FOR FUNCTIONS OF TWO VARIABLES If f is continu ous on a closed, bounded set D in 2, then f attains an absolute maximum value
f x1, y1 and an absolute minimum value f x2, y2at some points x1, y1 and x2, y2 in D.
8
y L 2,2 0, 2
L
0,0 L 3,0 x
FIGURE 12
9
0
D
To find the extreme values guaranteed by Theorem 8, we note that, by Theorem 2, if f has an extreme value at x1, y1, then x1, y1 is either a critical point of f or a boundary point of D. Thus we have the following extension of the Closed Interval Method.
9 To find the absolute maximum and minimum values of a continuous function f on a closed, bounded set D:
1. Find the values of f at the critical points of f in D.
2. Find the extreme values of f on the boundary of D.
3. The largest of the values from steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.
EXAMPLE 7 Find the absolute maximum and minimum values of the function fx, yx22xy2y on the rectangle Dx, y 0x3, 0y2.
SOLUTION Since f is a polynomial, it is continuous on the closed, bounded rectangle D, so Theorem 8 tells us there is both an absolute maximum and an absolute minimum. According to step 1 in 9, we first find the critical points. These occur when
fx 2x2y0 fy 2×20
so the only critical point is 1, 1, and the value of f there is f 1, 11.
In step 2 we look at the values of f on the boundary of D, which consists of the four
line segments L1, L2, L3, L4 shown in Figure 12. On L1 we have y0 and fx, 0x2 0x3
This is an increasing function of x, so its minimum value is f 0, 00 and its maxi mum value is f 3, 09. On L 2 we have x3 and
f3, y94y 0y2
This is a decreasing function of y, so its maximum value is f 3, 09 and its minimum
value is f3, 21. On L3 we have y2 and
fx, 2x24x4 0x3
By the methods of Chapter 4, or simply by observing that f x, 2×22, we see that the minimum value of this function is f 2, 20 and the maximum value is
f0, 24. Finally, on L4 we have x0 and
f 0, y2y 0y2
with maximum value f 0, 24 and minimum value f 0, 00. Thus, on the bound ary, the minimum value of f is 0 and the maximum is 9.
In step 3 we compare these values with the value f 1, 11 at the critical point and conclude that the absolute maximum value of f on D is f3, 09 and the absolute minimum value is f 0, 0f 2, 20. Figure 13 shows the graph of f. M
L
30 LTM
2
FIGURE 13
fx, y2xy2y
3, 2 LTM
SECTION 14.7 MAXIMUM AND MINIMUM VALUES929
930
CHAPTER 14 PARTIAL DERIVATIVES
We close this section by giving a proof of the first part of the Second Derivatives Test.
Part b has a similar proof.
PROOF OF THEOREM 3, PART A We compute the secondorder directional derivative of f in
the direction of uh, k. The firstorder derivative is given by Theorem 14.6.3: Du ffx hfy k
1.
Applying this theorem a second time, we have
Du2 fDuDu fDu fhDu fk x y
fx x hfy x kh fx y hfy y kkfxx h22 fxy hkfyy k2
If we complete the square in this expression, we obtain
Wearegiventhat fxxa,b0andDa,b0.But fxx andDfxx fyy fx2y arecon tinuous functions, so there is a disk B with center a, b and radius0 such that
fxxx, y0 and Dx, y0 whenever x, y is in B. Therefore, by looking at Equation 10, we see that Du2 fx, y0 whenever x, y is in B. This means that if C is the curve obtained by intersecting the graph of f with the vertical plane through Pa, b, f a, b in the direction of u, then C is concave upward on an interval of length 2 . This is true in the direction of every vector u, so if we restrict x, y to lie in B, the graph of f lies above its horizontal tangent plane at P. Thus f x, yf a, b whenever x, y is in B. This shows that f a, b is a local minimum. M
10
fxy2 k2 22
Du ffxx h
kfxx fyy fxyfxx fxx
by Clairauts Theorem
14.7 EXERCISES
Suppose 1, 1 is a critical point of a function f with contin uous second derivatives. In each case, what can you say about f ?
a fxx1, 14, fx y1, 11, fyy1, 12
b fxx1, 14, fx y1, 13, fyy1, 12
2. Suppose 0, 2 is a critical point of a function t with contin uous second derivatives. In each case, what can you say
reasoning. Then use the Second Derivatives Test to confirm your predictions.
3. fx,y4x3 y3 3xy y
1
3.2
3.7
1 44.2 1 x
about t?
a txx0, 21,
b txx0, 21, c txx0, 24,
txy0, 26, txy0, 22, tx y0, 26,
tyy0, 21 tyy0, 28 tyy0, 29
3.7
5
012 3.2
6
34 Use the level curves in the figure to predict the location of the critical points of f and whether f has a saddle point or a local maximum or minimum at each critical point. Explain your
1
4. fx,y3xx3 2y2 y4 y
22. 23.
24.
fx,yxyex2y2 fx,ysinxsinysinxy,
0x2 ,0y2 fx,ysinxsinycosxy,
0x 4, 0y 4
1.5 1
SECTION 14.7 MAXIMUM AND MINIMUM VALUES
931
2.9
2.7
2.5 1.5
1.7 1.9
1 1 x 1
518 Find the local maximum and minimum values and saddle points of the function. If you have threedimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.
; 2528 Use a graphing device as in Example 4 or Newtons method or a rootfinder to find the critical points of f correct to three decimal places. Then classify the critical points and find the highest or lowest points on the graph.
5. fx,y92x4yx2 4y2
6. fx,yx3y12x2 8y
7. fx,yx4 y4 4xy2
8. fx,ye4yx2y2
9. fx,y1xyxy
10. fx,y2x3 xy2 5×2 y2
11. fx,yx3 12xy8y3
12. fx,yxy 11 xy
13. fx,yexcosy
14. fx,yycosx
15. fx,yx2 y2ey2x2
16. fx,yeyy2 x2
17. fx,yy2 2ycosx, 1×7
18. fx,ysinx siny,x ,
y
25. fx,yx4 5×2 y2 3×2
26. fx,y510xy4x2 3yy4
27. fx,y2x4x2 y2 2xy2 x4 y4 28. fx,yex y4 x3 4cosy
2936 Find the absolute maximum and minimum values of f on the set D.
29. fx, y14x5y, D is the closed triangular region with vertices 0, 0, 2, 0, and 0, 3
30. fx,y3xyx2y, Distheclosedtriangular region with vertices 1, 0, 5, 0, and 1, 4
fx,yx2 y2 x2y4, Dx,yx1,y1
32. fx,y4x6yx2 y2,
Dx, y0x4, 0y5
33. fx,yx4 y4 4xy2,
Dx, y0x3, 0y2
34. fx,yxy2, Dx,yx0, y0,x2 y2 3
35. fx,y2x3 y4, Dx,yx2 y2 1
36. fx, yx33xy312y, D is the quadrilateral whose vertices are 2, 3, 2, 3, 2, 2, and 2, 2.
37. For functions of one variable it is impossible for a continuous function to have two local maxima and no local minimum. But for functions of two variables such functions exist. Show that the function
fx,yx2 12 x2yx12
has only two critical points, but has local maxima at both of them. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible.
38. If a function of one variable is continuous on an interval and has only one critical number, then a local maximum has to be
31.
19. Showthatfx,yx2 4y2 4xy2hasaninfinite number of critical points and that D0 at each one. Then show that f has a local and absolute minimum at each critical point.
20. Show that fx, yx2yex2y2 has maximum values at
1, 1s2and minimum values at 1, 1s2 . Show also that f has infinitely many other critical points and D0 at each of them. Which of them give rise to maximum values? Minimum values? Saddle points?
; 2124 Use a graph andor level curves to estimate the local maximum and minimum values and saddle points of the function. Then use calculus to find these values precisely.
21. fx,yx2 y2 x2y2
;
;
1
0.5
0
2
1
1.5
932CHAPTER 14 PARTIAL DERIVATIVES
an absolute maximum. But this is not true for functions of two
variables. Show that the function
fx,y3xey x3 e3y
has exactly one critical point, and that f has a local maximum there that is not an absolute maximum. Then use a computer to produce a graph with a carefully chosen domain and viewpoint to see how this is possible.
39. Find the shortest distance from the point 2, 1, 1 to the plane xyz1.
40. Findthepointontheplane xyz4thatisclosesttothe point 1, 2, 3.
Find the points on the cone z2x2y2 that are closest to the point 4, 2, 0.
42. Find the points on the surface y29xz that are closest to the origin.
Find three positive numbers whose sum is 100 and whose product is a maximum.
44. Find three positive numbers whose sum is 12 and the sum of whose squares is as small as possible.
45. Find the maximum volume of a rectangular box that is inscribed in a sphere of radius r.
46. Find the dimensions of the box with volume 1000 cm3 that has minimal surface area.
47. Find the volume of the largest rectangular box in the first octant with three faces in the coordinate planes and one vertex in the plane x2y3z6.
48. Find the dimensions of the rectangular box with largest volume if the total surface area is given as 64 cm2 .
49. Find the dimensions of a rectangular box of maximum volume such that the sum of the lengths of its 12 edges is a constant c.
50. The base of an aquarium with given volume V is made of slate and the sides are made of glass. If slate costs five times as much per unit area as glass, find the dimensions of the aquar ium that minimize the cost of the materials.
A cardboard box without a lid is to have a volume of 32,000 cm3. Find the dimensions that minimize the amount of cardboard used.
52. A rectangular building is being designed to minimize
heat loss. The east and west walls lose heat at a rate of
10 unitsm2 per day, the north and south walls at a rate of
8 unitsm2 per day, the floor at a rate of 1 unitm2 per day, and the roof at a rate of 5 unitsm2 per day. Each wall must be at least 30 m long, the height must be at least 4 m, and the volume must be exactly 4000 m3.
a Find and sketch the domain of the heat loss as a function of
the lengths of the sides.
53. 54.
b Find the dimensions that minimize heat loss. Check both the critical points and the points on the boundary of the domain.
c Could you design a building with even less heat loss
if the restrictions on the lengths of the walls were removed?
If the length of the diagonal of a rectangular box must be L, what is the largest possible volume?
Three alleles alternative versions of a gene A, B, and O determine the four blood types A AA or AO, B BB or BO, O OO, and AB. The HardyWeinberg Law states that the pro portion of individuals in a population who carry two different alleles is
P2pq2pr2rq
where p, q, and r are the proportions of A, B, and O in the
population.Usethefactthat pqr1toshowthatPis
at most 2.
41.
43.
55.
Suppose that a scientist has reason to believe that two quan tities x and y are related linearly, that is, ymxb, at least approximately, for some values of m and b. The scientist performs an experiment and collects data in the form of points x1, y1, x2, y2 , . . . , xn, yn , and then plots these points. The points dont lie exactly on a straight line, so the scientist wants to find constants m and b so that the line ymxb fits the points as well as possible. See the figure.
3
y
0x
Let diyimxib be the vertical deviation of the point xi, yi from the line. The method of least squares determines m and b so as to minimize ni1 di2, the sum of the squares of these deviations. Show that, according to this method, the line of best fit is obtained when
nn mxi bnyi
di ,
mxib
xi, yi
51.
56.
Thus the line is found by solving these two equations in the two unknowns m and b. See Section 1.2 for a further discus sion and applications of the method of least squares.
Find an equation of the plane that passes through the point 1, 2, 3 and cuts off the smallest volume in the first octant.
i1 i1 nnn
mx i2bx i x i y i i1 i1 i1
DISCOVERY PROJECT QUADRATIC APPROXIMATIONS AND CRITICAL POINTS933
DESIGNING A DUMPSTER
APPLIED PROJECT
For this project we locate a trash dumpster in order to study its shape and construction. We then attempt to determine the dimensions of a container of similar design that minimize construction cost.
1. First locate a trash dumpster in your area. Carefully study and describe all details of its con struction, and determine its volume. Include a sketch of the container.
2. While maintaining the general shape and method of construction, determine the dimensions such a container of the same volume should have in order to minimize the cost of construc tion. Use the following assumptions in your analysis:
N The sides, back, and front are to be made from 12gauge 0.1046 inch thick steel sheets, which cost 0.70 per square foot including any required cuts or bends.
N The base is to be made from a 10gauge 0.1345 inch thick steel sheet, which costs 0.90 per square foot.
N Lids cost approximately 50.00 each, regardless of dimensions.
N Welding costs approximately 0.18 per foot for material and labor combined. Give justification of any further assumptions or simplifications made of the details of
construction.
3. Describe how any of your assumptions or simplifications may affect the final result.
4. If you were hired as a consultant on this investigation, what would your conclusions be? Would you recommend altering the design of the dumpster? If so, describe the savings that would result.
QUADRATIC APPROXIMATIONS AND CRITICAL POINTS
DISCOVERY PROJECT
The Taylor polynomial approximation to functions of one variable that we discussed in Chap ter 11 can be extended to functions of two or more variables. Here we investigate quadratic approximations to functions of two variables and use them to give insight into the Second Derivatives Test for classifying critical points.
In Section 14.4 we discussed the linearization of a function f of two variables at a point a, b:
Lx, yfa, bfxa, bxafya, byb
Recall that the graph of L is the tangent plane to the surface zf x, y at a, b, f a, b and the corresponding linear approximation is f x, yLx, y. The linearization L is also called the firstdegree Taylor polynomial of f at a, b.
1. If f has continuous secondorder partial derivatives at a, b, then the seconddegree Taylor polynomial of f at a, b is
Qx, yfa, bfxa, bxafya, byb
1 fxxa,bxa2 fxya,bxayb1 fyya,byb2
22
and the approximation f x, yQx, y is called the quadratic approximation to f at a, b. Verify that Q has the same first and secondorder partial derivatives as f at a, b.
934
CHAPTER 14 PARTIAL DERIVATIVES
;
;
at 0, 0.
5. a
Suppose f is any function with continuous secondorder partial derivatives such that
f 0, 00 and 0, 0 is a critical point of f. Write an expression for the seconddegree
Find the first and seconddegree Taylor polynomials L and Q of fx, yex2y2
2. a
b Graph f, L, and Q. Comment on how well L and Q approximate f.
Find the first and seconddegree Taylor polynomials L and Q for fx, yxey at 1, 0.
3. a
b Compare the values of L, Q, and f at 0.9, 0.1.
c Graph f, L, and Q. Comment on how well L and Q approximate f.
4. In this problem we analyze the behavior of the polynomial fx, yax2bxycy2 without using the Second Derivatives Test by identifying the graph as a paraboloid. a Bycompletingthesquare,showthatifa0,then
fx,yax2 bxycy2 a x yb 2
4acb2
2
b LetD4acb2.ShowthatifD0anda0,then f hasalocalminimum at 0, 0.
c ShowthatifD0anda0,then f hasalocalmaximumat0,0.
d Show that if D0, then 0, 0 is a saddle point.
Taylor polynomial, Q, of f at 0, 0.
b What can you conclude about Q from Problem 4?
c In view of the quadratic approximation f x, yQx, y, what does part b suggest
about f ?
2a
4a
y2
14.8
fx, y11 fx, y10 fx, y9 fx, y8 fx, y7
T E C Visual 14.8 animates Figure 1 for both level curves and level surfaces.
LAGRANGE MULTIPLIERS
In Example 6 in Section 14.7 we maximized a volume function Vxyz subject to the constraint 2xz2yzxy12, which expressed the side condition that the surface area was 12 m2. In this section we present Lagranges method for maximizing or minimizing a general function f x, y, z subject to a constraint or side condition of the form tx, y, zk.
Its easier to explain the geometric basis of Lagranges method for functions of two variables. So we start by trying to find the extreme values of f x, y subject to a constraint of the form tx, yk. In other words, we seek the extreme values of f x, y when the point x, y is restricted to lie on the level curve tx, yk. Figure 1 shows this curve together with several level curves of f. These have the equations f x, yc, where c7, 8, 9, 10, 11. To maximize f x, y subject to tx, yk is to find the largest value of c such that the level curve f x, yc intersects tx, yk. It appears from Figure 1 that this happens when these curves just touch each other, that is, when they have a common tan gent line. Otherwise, the value of c could be increased further. This means that the nor mal lines at the point x0, y0where they touch are identical. So the gradient vectors are parallel; that is, fx0, y0tx0, y0 for some scalar .
This kind of argument also applies to the problem of finding the extreme values of f x, y, z subject to the constraint tx, y, zk. Thus the point x, y, z is restricted to lie on the level surface S with equation tx, y, zk. Instead of the level curves in Figure 1,
we consider the level surfaces f x, y, zc and argue that if the maximum value of f is f x0, y0, z0 c, then the level surface f x, y, zc is tangent to the level surface tx, y, zk and so the corresponding gradient vectors are parallel.
y
gx, yk 0x
FIGURE 1
N Lagrange multipliers are named after the FrenchItalian mathematician JosephLouis Lagrange 17361813. See page 283 for a biographical sketch of Lagrange.
N In deriving Lagranges method we assumed that t0. In each of our examples you can check that t0 at all points where
tx, y, zk. See Exercise 21 for what can go wrong if t0.
This intuitive argument can be made precise as follows. Suppose that a function f has an extreme value at a point Px0, y0, z0on the surface S and let C be a curve with vector equation rtxt, yt, zt that lies on S and passes through P. If t0 is the parameter value corresponding to the point P, then rt0x0, y0, z0 . The composite function htf xt, yt, zt represents the values that f takes on the curve C. Since f has an extreme value at x0, y0, z0, it follows that h has an extreme value at t0, so ht00. But if f is differentiable, we can use the Chain Rule to write
0ht0
fxx0, y0, z0 xt0 fyx0, y0, z0 yt0 fzx0, y0, z0 zt0fx 0 , y 0 , z 0 rt 0
This shows that the gradient vector f x0, y0, z0is orthogonal to the tangent vector rt0to every such curve C. But we already know from Section 14.6 that the gradient vector of t, tx0, y0, z0 , is also orthogonal to rt0for every such curve. See Equation 14.6.18. This means that the gradient vectors f x0, y0, z0and tx0, y0, z0must be parallel. Therefore, if tx0, y0, z0 0, there is a number such that
The number in Equation 1 is called a Lagrange multiplier. The procedure based on Equation 1 is as follows.
SECTION 14.8 LAGRANGE MULTIPLIERS935
1
fx0, y0, z0tx0, y0, z0
METHOD OF LAGRANGE MULTIPLIERS To find the maximum and minimum values of f x, y, z subject to the constraint tx, y, zk assuming that these extreme values exist and t0 on the surface tx, y, zk:
a Find all values of x, y, z, and such that
fx, y, ztx, y, z and tx, y, zk
b Evaluate f at all the points x, y, z that result from step a. The largest of these values is the maximum value of f ; the smallest is the minimum value of f.
If we write the vector equation ftions in step a become
fx tx fy ty
t in terms of its components, then the equa fz tz tx,y,zk
This is a system of four equations in the four unknowns x, y, z, and , but it is not neces sary to find explicit values for .
For functions of two variables the method of Lagrange multipliers is similar to the method just described. To find the extreme values of fx,y subject to the constraint tx, yk, we look for values of x, y, and such that
fx, ytx, y and tx, yk
936
CHAPTER 14 PARTIAL DERIVATIVES
N Another method for solving the system of equations 25 is to solve each of Equations 2,
There are no general rules for solving systems of equations. Sometimes some ingenuity is required. In the present example you might notice that if we multiply 2 by x, 3 by y, and 4 by z, then the left sides of these equations will be identical. Doing this, we have
xyz 2xzxy xyz 2yzxy xyz 2xz2yz
Weobservethat 0because 0wouldimplyyzxzxy0from2,3,and 4 and this would contradict 5. Therefore, from 6 and 7, we have
2xzxy2yzxy
which gives xzyz. But z0 since z0 would give V0, so xy. From 7 and
8 we have
2yzxy2xz2yz
which gives 2xzxy and so since x0 y2z. If we now put xy2z in 5,
we get
4z2 4z2 4z2 12
Since x, y, and z are all positive, we therefore have z1 and so x2 and y2. This agrees with our answer in Section 14.7. M
3, and 4 for expressions.
and then to equate the resulting
This amounts to solving three equations in three unknowns:
fx tx fy ty tx,yk
Our first illustration of Lagranges method is to reconsider the problem given in
Example 6 in Section 14.7.
V EXAMPLE 1 A rectangular box without a lid is to be made from 12 m2 of cardboard. Find the maximum volume of such a box.
SOLUTION As in Example 6 in Section 14.7, we let x, y, and z be the length, width, and height, respectively, of the box in meters. Then we wish to maximize
subject to the constraint
Vxyz
tx, y, z2xz2yzxy12
Using the method of Lagrange multipliers, we look for values of x, y, z, and Vt and tx, y, z12. This gives the equations
such that
Vxtx which become
Vyty Vztz 2xz2yzxy12
yz 2zy xz 2zx
xy 2x2y 2xz2yzxy12
2
3
4
5
6
7
8
N In geometric terms, Example 2 asks for the highest and lowest points on the curve C in Fig ure 2 that lies on the paraboloid zx22y2 and directly above the constraint circle
SECTION 14.8 LAGRANGE MULTIPLIERS937 V EXAMPLE 2 Find the extreme values of the function fx, yx22y2 on the
circle x2y21.
SOLUTION We are asked for the extreme values of f subject to the constraint
tx, yx2y21. Using Lagrange multipliers, we solve the equations ft and tx, y1, which can be written as
x2 y2 1.
z
z2
fx tx
fy ty
2x2x
4y2y
x2 y2 1
tx,y1
or as
x
FIGURE 2
1
C
y
From9wehavex0or 1.Ifx0,then11givesy1.If 1,then y0 from 10, so then 11 gives x1. Therefore f has possible extreme values at the points 0, 1, 0, 1, 1, 0, and 1, 0. Evaluating f at these four points, we find that
f 0, 12 f 0, 12 f 1, 01 f 1, 01 Therefore the maximum value of f on the circle x2y21 is f0, 12 and the
minimum value is f 1, 01. Checking with Figure 2, we see that these values look reasonable. M
EXAMPLE 3 Findtheextremevaluesof fx,yx2 2y2 onthediskx2 y2 1.
SOLUTION According to the procedure in 14.7.9, we compare the values of f at the criti cal points with values at the points on the boundary. Since fx2x and fy4y, the only critical point is 0, 0. We compare the value of f at that point with the extreme values on the boundary from Example 2:
f 0, 00 f 1, 01 f 0, 12 Thereforethemaximumvalueof f onthediskx2 y2 1is f0,12andthe
minimum value is f 0, 00. M EXAMPLE 4 Find the points on the sphere x2y2z24 that are closest to and
farthest from the point 3, 1, 1.
SOLUTION The distance from a point x, y, z to the point 3, 1, 1 is
dsx32 y12 z12
but the algebra is simpler if we instead maximize and minimize the square of the
distance:
d2 fx,y,zx32 y12 z12 The constraint is that the point x, y, z lies on the sphere, that is,
9
10
11
N The geometry behind the use of Lagrange multipliers in Example 2 is shown in Figure 3. The extreme values of f x, yx22y2 correspond to the level curves that touch the circlex2 y2 1.
y
0x
21
22
FIGURE 3
tx,y,zx2 y2 z2 4
938
CHAPTER 14 PARTIAL DERIVATIVES
N Figure 4 shows the sphere and the nearest point P in Example 4. Can you see how to find the coordinates of P without using calculus?
z
3, 1, 1
FIGURE 4
hc
C
gk
FIGURE 5
s11 s11 s11 s11 s11 s11
Its easy to see that f has a smaller value at the first of these points, so the closest point
is 6s11, 2s11, 2s11and the farthest is 6s11, 2s11, 2s11 . M TWO CONSTRAINTS
g
Suppose now that we want to find the maximum and minimum values of a function f x, y, z subject to two constraints side conditions of the form tx, y, zk and hx, y, zc. Geometrically, this means that we are looking for the extreme values of f f when x, y, z is restricted to lie on the curve of intersection C of the level surfaces tx, y, zk and hx, y, zc. See Figure 5. Suppose f has such an extreme value at a point Px0, y0, z0. We know from the beginning of this section that f is orthogonal to C at P. But we also know that t is orthogonal to tx, y, zk and h is orthogonal to hx, y, zc, so t and h are both orthogonal to C. This means that the gradient vector f x0, y0, z0is in the plane determined by tx0, y0, z0and hx0, y0, z0 . We assume
that these gradient vectors are not zero and not parallel. So there are numbers and
P h
According to the method of Lagrange multipliers, we solve f
t, t4. This gives
2x32x 2y12y 2z12z
12 13 14 15
12, 13, and 14, and then substitute these values into 15. From 12 we have
x3x or x13 or x3 1
Note that 1 0 because1 is impossible from 12. Similarly, 13 and 14 give
y1 z1 1 1
Therefore, from 15, we have
32 12 12
1 2 1 2 1 2 4
whichgives1 2 11,1 s112,so 4
1s11 2
x2 y2 z2 4
The simplest way to solve these equations is to solve for x, y, and z in terms of from
These values of then give the corresponding points x, y, z: xPy 6 , 2 , 2and6 , 2 , 2
N The cylinder x2y21 intersects the plane xyz1 in an ellipse Figure 6. Example 5 asks for the maximum value of f when x, y, z is restricted to lie on the ellipse.
4 3 2
z1 0 1
2 1 y0 1
FIGURE 6
16 fx0, y0, z0tx0, y0, z0hx0, y0, z0
In this case Lagranges method is to look for extreme values by solving five equations in the five unknowns x, y, z, , and . These equations are obtained by writing Equation 16 in terms of its components and using the constraint equations:
fx tx hx fy ty hy fz tz hz tx, y, zk hx, y, zc
V EXAMPLE 5 Find the maximum value of the function fx, y, zx2y3z on the curveofintersectionoftheplanexyz1andthecylinderx2 y2 1.
SOLUTION We maximize the function f x, y, zx2y3z subject to the constraints tx,y,zxyz1andhx,y,zx2 y2 1.TheLagrangeconditionis
called Lagrange multipliers such that
Putting
gives y52
and so 229, 4
t
h, so we solve the equations 12x
22y 3 xyz1 x2 y2 1
SECTION 14.8 LAGRANGE MULTIPLIERS939
f
17 18 19 20 21
s292. Then x2s29, y5s29, and, from 20, z1xy17s29.Thecorrespondingvaluesof f are
2 2 5 31 7 3s29 s29 s29 s29
Therefore the maximum value of f on the given curve is 3s29 . M
3 from 19 in 17, we get 2×2, so x1 . Similarly, 18 . Substitution in 21 then gives
125 1 242
940CHAPTER 14 PARTIAL DERIVATIVES
14.8 EXERCISES
Pictured are a contour map of f and a curve with equation tx, y8. Estimate the maximum and minimum values of f subject to the constraint that tx, y8. Explain your reasoning.
y
gx, y8
15. fx,y,zx2y; xyz1, y2 z2 4 16. fx,y,z3xy3z;
xyz0, x2 2z2 1
17. fx,y,zyzxy; xy1, y2 z2 1
1819 Find the extreme values of f on the region described by the inequality.
18. fx,y2x2 3y2 4×5, x2 y2 16 fx,yexy, x2 4y2 1
1.
50 40
0 30x
70 60
10
19.
20 20.
;2. a Use a graphing calculator or computer to graph the circle ; x2y21. On the same screen, graph several curves of
the form x2yc until you find two that just touch the circle. What is the significance of the values of c for these
two curves?
b Use Lagrange multipliers to find the extreme values of
fx,yx2 ysubjecttotheconstraintx2 y2 1. Compare your answers with those in part a.
317 Use Lagrange multipliers to find the maximum and minimum values of the function subject to the given constraints.
3. fx,yx2 y2; xy1
4. fx,y4x6y; x2 y2 13
5. fx,yx2y; x2 2y2 6
6. fx,yexy; x3 y3 16
7. fx,y,z2x6y10z; x2 y2 z2 35
8. fx,y,z8x4z; x2 10y2 z2 5
9. fx,y,zxyz; x2 2y2 3z2 6
10. fx,y,zx2y2z2; x2 y2 z2 1
11. fx,y,zx2 y2 z2; x4 y4 z4 1
12. fx,y,zx4 y4 z4; x2 y2 z2 1
13. fx,y,z,txyzt; x2 y2 z2 t2 1
14. fx1,x2,…,xnx1 x2 xn; x12 x2 xn2 1
Consider the problem of maximizing the function fx,y2x3ysubjecttotheconstraintsx sy 5. a Try using Lagrange multipliers to solve the problem.
b Does f 25, 0 give a larger value than the one in part a?
c Solve the problem by graphing the constraint equation
and several level curves of f.
d Explain why the method of Lagrange multipliers fails to
solve the problem.
e What is the significance of f 9, 4?
21. Consider the problem of minimizing the function f x, yx onthecurvey2 x4 x3 0apiriform.
a Try using Lagrange multipliers to solve the problem.
b Show that the minimum value is f 0, 00 but the
Lagrange conditionf 0, 0t0, 0 is not satisfied
for any value of .
c Explain why Lagrange multipliers fail to find the mini
mum value in this case.
a If your computer algebra system plots implicitly defined curves, use it to estimate the minimum and maximum val ues of fx, yx3y33xy subject to the constraint x32y329 by graphical methods.
b Solve the problem in part a with the aid of Lagrange multipliers. Use your CAS to solve the equations numeri cally. Compare your answers with those in part a.
CAS 22.
23. The total production P of a certain product depends on the amount L of labor used and the amount K of capital invest ment. In Sections 14.1 and 14.3 we discussed how the Cobb Douglas model PbL K1 follows from certain economic assumptions, where b and are positive constants and1. If the cost of a unit of labor is m and the cost of a unit of capital is n, and the company can spend only p dollars as its total budget, then maximizing the production P is subject to the constraint mLnKp. Show that the maximum production occurs when
Lp and K1p mn
24. Referring to Exercise 23, we now suppose that the pro duction is fixed at bL K1Q, where Q is a constant. What values of L and K minimize the cost function CL, KmLnK?
Use Lagrange multipliers to prove that the rectangle with maximum area that has a given perimeter p is a square.
26. Use Lagrange multipliers to prove that the triangle with maximum area that has a given perimeter p is equilateral.
Hint: Use Herons formula for the area: Asssxsysz
where sp2 and x, y, z are the lengths of the sides.
2739 Use Lagrange multipliers to give an alternate solution to the indicated exercise in Section 14.7.
CAS 4344 Find the maximum and minimum values of f subject to the given constraints. Use a computer algebra system to solve the system of equations that arises in using Lagrange multipliers. If your CAS finds only one solution, you may need to use additional commands.
43. fx,y,zyexz; 9×2 4y2 36z2 36, xyyz1 44. fx,y,zxyz; x2 y2 z, x2 z2 4
25.
APPLIED PROJECT ROCKET SCIENCE941
27. Exercise 39 29. Exercise 41 31. Exercise 43 33. Exercise 45
Exercise 47 37. Exercise 49 39. Exercise 53
28. Exercise 40 30. Exercise 42 32. Exercise 44 34. Exercise 46 36. Exercise 48 38. Exercise 50
a
b
Find the maximum value of
fx1,x2,…,xnsn x1x2xn
given that x1, x2, . . . , xn are positive numbers and
x1 x2 xn c,wherecisaconstant. Deduce from part a that if x1, x2, . . . , xn are positive numbers, then
sn x 1 x 2x nx 1x 2x n n
This inequality says that the geometric mean of n numbers is no larger than the arithmetic mean of the numbers. Under what circumstances are these two means equal?
Maximize ni1 xi yi subject to the constraints ni1 xi21 and ni1 yi21.
Put
35.
40. Find the maximum and minimum volumes of a rectangular box whose surface area is 1500 cm2 and whose total edge length is 200 cm.
41. The plane xy2z2 intersects the paraboloid
zx2y2 in an ellipse. Find the points on this ellipse that are nearest to and farthest from the origin.
42. The plane 4x3y8z5 intersects the cone z2x2y2inanellipse.
a Graph the cone, the plane, and the ellipse.
b Use Lagrange multipliers to find the highest and lowest
46. a b
xi ai s aj2
and yi bi s bj2
;
aibisa sbj2 j2
points on the ellipse.
ROCKET SCIENCE
45.
to show that
for any numbers a1, . . . , an, b1, . . . , bn. This inequality is known as the CauchySchwarz Inequality.
APPLIED PROJECT
Many rockets, such as the Pegasus XL currently used to launch satellites and the Saturn V that first put men on the moon, are designed to use three stages in their ascent into space. A large first stage initially propels the rocket until its fuel is consumed, at which point the stage is jettisoned to reduce the mass of the rocket. The smaller second and third stages function similarly in order to place the rockets payload into orbit about the earth. With this design, at least two stages are required in order to reach the necessary velocities, and using three stages has proven to be a good compromise between cost and performance. Our goal here is to determine the individual masses of the three stages, which are to be designed in such a way as to minimize the total mass of the rocket while enabling it to reach a desired velocity.
942CHAPTER 14 PARTIAL DERIVATIVES
For a singlestage rocket consuming fuel at a constant rate, the change in velocity resulting from the acceleration of the rocket vehicle has been modeled by
Vcln1 1SMr PMr
where Mr is the mass of the rocket engine including initial fuel, P is the mass of the payload,
S is a structural factor determined by the design of the rocket specifically, it is the ratio of the mass of the rocket vehicle without fuel to the total mass of the rocket with payload, and c is the constant speed of exhaust relative to the rocket.
Now consider a rocket with three stages and a payload of mass A. Assume that outside forces are negligible and that c and S remain constant for each stage. If Mi is the mass of the ith stage, we can initially consider the rocket engine to have mass M1 and its payload to have mass
M2M3A; the second and third stages can be handled similarly.
1. Show that the velocity attained after all three stages have been jettisoned is given by
vf clnM1 M2 M3 A lnM2 M3 A lnM3 A
SM1 M2 M3 A SM2 M3 A SM3 A
2. We wish to minimize the total mass MM1M2M3 of the rocket engine subject totheconstraintthatthedesiredvelocityvf fromProblem1isattained.Themethodof Lagrange multipliers is appropriate here, but difficult to implement using the current expres sions. To simplify, we define variables Ni so that the constraint equation may be expressed as vfcln N1ln N2ln N3. Since M is now difficult to express in terms of the Nis, we wish to use a simpler function that will be minimized at the same place as M. Show that
M1 M2 M3 A1SN1 M2 M3 A 1SN1
and conclude that
M2 M3 A1SN2 M3A 1SN2
M3 A1SN3 A 1SN3
1S3N1N2N3
1SN11SN21SN3
MA A
3. Verify that lnMAA is minimized at the same location as M; use Lagrange multipliers and the results of Problem 2 to find expressions for the values of Ni where the minimum occurs subject to the constraint vfcln N1ln N2ln N3 . Hint: Use properties of logarithms to help simplify the expressions.
4. Find an expression for the minimum value of M as a function of vf .
5. If we want to put a threestage rocket into orbit 100 miles above the earths surface, a final velocity of approximately 17,500 mih is required. Suppose that each stage is built with a structural factor S0.2 and an exhaust speed of c6000 mih.
a Find the minimum total mass M of the rocket engines as a function of A.
b Find the mass of each individual stage as a function of A. They are not equally sized!
6. The same rocket would require a final velocity of approximately 24,700 mih in order to escape earths gravity. Find the mass of each individual stage that would minimize the total mass of the rocket engines and allow the rocket to propel a 500pound probe into deep space.
Courtesy of Orbital Sciences Corporation
HYDROTURBINE OPTIMIZATION
APPLIED PROJECT HYDROTURBINE OPTIMIZATION943
APPLIED PROJECT
The Katahdin Paper Company in Millinocket, Maine, operates a hydroelectric generating station on the Penobscot River. Water is piped from a dam to the power station. The rate at which the water flows through the pipe varies, depending on external conditions.
The power station has three different hydroelectric turbines, each with a known and unique power function that gives the amount of electric power generated as a function of the water flow arriving at the turbine. The incoming water can be apportioned in different volumes to each turbine, so the goal is to determine how to distribute water among the turbines to give the maxi mum total energy production for any rate of flow.
Using experimental evidence and Bernoullis equation, the following quadratic models were determined for the power output of each turbine, along with the allowable flows of operation:
where
KW118.890.1277Q14.08105Q12 1701.6106QT2KW224.510.1358Q24.69105Q2 1701.6106QT2KW327.020.1380Q33.84105Q32 1701.6106QT2250Q1 1110, 250Q2 1110, 250Q3 1225
Qiflow through turbine i in cubic feet per second KWipower generated by turbine i in kilowatts
QTtotal flow through the station in cubic feet per second
1. If all three turbines are being used, we wish to determine the flow Qi to each turbine that will give the maximum total energy production. Our limitations are that the flows must sum to the total incoming flow and the given domain restrictions must be observed. Consequently, use Lagrange multipliers to find the values for the individual flows as functions of QT
that maximize the total energy production KW1KW2KW3 subject to the constraints Q1Q2Q3QT and the domain restrictions on each Qi .
2. For which values of QT is your result valid?
3. For an incoming flow of 2500 ft3s, determine the distribution to the turbines and verify
by trying some nearby distributions that your result is indeed a maximum.
4. Until now we have assumed that all three turbines are operating; is it possible in some situa tions that more power could be produced by using only one turbine? Make a graph of the three power functions and use it to help decide if an incoming flow of 1000 ft3s should be distributed to all three turbines or routed to just one. If you determine that only one turbine should be used, which one would it be? What if the flow is only 600 ft3s?
5. Perhaps for some flow levels it would be advantageous to use two turbines. If the incoming flow is 1500 ft3s, which two turbines would you recommend using? Use Lagrange multi pliers to determine how the flow should be distributed between the two turbines to maxi mize the energy produced. For this flow, is using two turbines more efficient than using all three?
6. If the incoming flow is 3400 ft3s, what would you recommend to the company?
944CHAPTER 14 PARTIAL DERIVATIVES
14 REVIEW
CONCEPT CHECK
1. a What is a function of two variables?
b Describe three methods for visualizing a function of two
variables.
2. What is a function of three variables? How can you visualize such a function?
12. If z is defined implicitly as a function of x and y by an equation of the form Fx, y, z0, how do you find zx and zy?
13. a Write an expression as a limit for the directional derivative of f at x0, y0 in the direction of a unit vector ua, b. How do you interpret it as a rate? How do you interpret it geometrically?
b If f is differentiable, write an expression for Du fx 0 , y0in termsoffxandfy.
14. a Define the gradient vector f for a function f of two or three variables.
b Express Du f in terms of f.
c Explain the geometric significance of the gradient.
15. What do the following statements mean?
3. What does
mean? How can you show that such a limit does not exist?
f x, yL
4. a What does it mean to say that f is continuous at a, b?
lim
x, yla, b
b If f is continuous on 2, what can you say about its graph?
5. a Write expressions for the partial derivatives fxa, b and fya, b as limits.
b How do you interpret fxa, b and fya, b geometrically? How do you interpret them as rates of change?
c If f x, y is given by a formula, how do you calculate fx and fy ?
6. What does Clairauts Theorem say?
7. How do you find a tangent plane to each of the following types of surfaces?
a A graph of a function of two variables, zf x, y
b A level surface of a function of three variables,
Fx, y, zk
8. Define the linearization of f at a, b. What is the corre sponding linear approximation? What is the geometric interpretation of the linear approximation?
9. a What does it mean to say that f is differentiable at a, b?
b How do you usually verify that f is differentiable?
10. If zfx, y, what are the differentials dx, dy, and dz?
11. State the Chain Rule for the case where zfx, y and x and y are functions of one variable. What if x and y are functions of two variables?
TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. fya,blim fa,yfa,b ylb yb
2. There exists a function f with continuous secondorder partial derivatives such that fxx, yxy2 and
fyx, yxy2.
a b c d e
16. a b
f has a local maximum at a, b.
f has an absolute maximum at a, b. f has a local minimum at a, b.
f has an absolute minimum at a, b. f has a saddle point at a, b.
If f has a local maximum at a, b, what can you say about its partial derivatives at a, b?
What is a critical point of f ?
17. State the Second Derivatives Test.
18. a What is a closed set in 2? What is a bounded set? b State the Extreme Value Theorem for functions of two
variables.
c How do you find the values that the Extreme Value
Theorem guarantees?
19. Explain how the method of Lagrange multipliers works
in finding the extreme values of f x, y, z subject to the constraint tx, y, zk. What if there is a second constraint hx,y,zc?
3. fxy2f x y
4. Dk fx,y,zfzx,y,z
5. If fx, y l L as x, y l a, b along every straight line
througha,b,thenlimx,yla,b fx,yL.
6. If fxa, b and fya, b both exist, then f is differentiable at a, b.
7. If f has a local minimum at a, b and f is differentiable at a, b, then fa, b0.
EXERCISES
12 Find and sketch the domain of the function. 1. fx,ylnxy1
2. fx,ys4x2 y2 s1x2
34 Sketch the graph of the function. 3. fx,y1y2
4. fx,yx2 y22
56 Sketch several level curves of the function.
5. fx,ys4x2y2
6. fx,yex y
7. Make a rough sketch of a contour map for the function whose graph is shown.
z
x2 2y
8. A contour map of a function f is shown. Use it to make a rough sketch of the graph of f.
y
1
1.5 2
4
x
10. If 2, 1 is a critical point of f and
fxx2, 1 fyy2, 1 fx y2, 12
then f has a saddle point at 2, 1.
11. If fx,ysinxsiny,thens2 Du fx,ys2.
12. If f x, y has two local maxima, then f must have a local minimum.
8. If f is a function, then lim
fx, yf2, 5 9. If fx,ylny,thenfx,y1y.
x, y l 2, 5
910 Evaluate the limit or show that it does not exist. 9. lim 2xy 10. lim 2xy
x, yl1, 1 x22y2 x, yl0, 0 x22y2
CHAPTER 14 REVIEW945
11. A metal plate is situated in the xyplane and occupies the rectangle 0x10, 0y8, where x and y are measured in meters. The temperature at the point x, y in the plate is
Tx, y, where T is measured in degrees Celsius. Temperatures at equally spaced points were measured and recorded in the table.
a Estimate the values of the partial derivatives Tx6, 4
and Ty6, 4. What are the units?
b Estimate the value of Du T6, 4, where uijs2.
Interpret your result.
c Estimate the value of Txy6, 4.
xy
0 30 38 45 51 55 2 52 56 60 62 61 4 78 74 72 68 66 6 98 87 80 75 71 8 96 90 86 80 75
10 92 92 91 87 78
12. Find a linear approximation to the temperature function Tx, y in Exercise 11 near the point 6, 4. Then use it to estimate the temperature at the point 5, 3.8.
1317 Find the first partial derivatives.
13. fx, ys2xy2
15. tu,vutan1v
17. Tp,q,rplnqer
14. uer sin 2
x
yz
02468
16. w
946CHAPTER 14 PARTIAL DERIVATIVES
18. The speed of sound traveling through ocean water is a func tion of temperature, salinity, and pressure. It has been modeled by the function
C1449.24.6T0.055T 20.00029T 31.340.01T S350.016D
where C is the speed of sound in meters per second, T is the temperature in degrees Celsius, S is the salinity the concen tration of salts in parts per thousand, which means the num ber of grams of dissolved solids per 1000 g of water, and D is the depth below the ocean surface in meters. Compute CT, CS, and CD when T10C, S35 parts per thousand, and D100 m. Explain the physical signifi cance of these partial derivatives.
1922 Find all second partial derivatives of f.
35. Ifux2y3 z4,wherexp3p2,ypep,and zp sin p, use the Chain Rule to find dudp.
36. Ifvx2sinyyexy,wherexs2tandyst,usethe Chain Rule to find vs and vt when s0 and t1.
37. Supposezfx,y,wherexts,t,yhs,t,
t1, 23, ts1, 21, tt1, 24, h1, 26,
hs1, 25, ht1, 210, fx3, 67, and fy3, 68. Find zs and zt when s1 and t2.
38. Use a tree diagram to write out the Chain Rule for the case where wf t, u, v, tt p, q, r, s, uu p, q, r, s, and vv p, q, r, s are all differentiable functions.
39. If zyfx2y2, where f is differentiable, show that y zx zx
x y
40. The length x of a side of a triangle is increasing at a rate of 3 ins, the length y of another side is decreasing at a rate of 2 ins, and the contained angle is increasing at a rate of 0.05 radians. How fast is the area of the triangle changing whenx40in,y50in,and6?
41. Ifzfu,v,whereuxy,vyx,and f hascontinuous second partial derivatives, show that
2z z2v
44. a When is the directional derivative of f a maximum? b When is it a minimum?
c When is it 0?
d When is it half of its maximum value?
4546 Find the directional derivative of f at the given point in the indicated direction.
45. fx, y2sxy2, 1, 5,
in the direction toward the point 4, 1
46. f x, y, zx 2 yx s1z , 1, 2, 3, inthedirectionofv2ij2k
47. Findthemaximumrateofchangeof fx,yx2ysy at the point 2, 1. In which direction does it occur?
48. Findthedirectioninwhich fx,y,zzexy increasesmost rapidly at the point 0, 1, 2. What is the maximum rate of increase?
49. The contour map shows wind speed in knots during Hurri cane Andrew on August 24, 1992. Use it to estimate the
19. fx,y4x3 xy2 21. fx,y,zxkylzm
20. zxe2y
22. vrcoss2t
23. Ifzxyxeyx,showthatxz yz xyz.
x 24. Ifzsinxsint,showthat
z 2zz 2z x xt t x2
y
2529 Find equations of a the tangent plane and b the normal line to the given surface at the specified point.
25. z3×2y22x, 1, 2, 1
26. zex cos y, 0, 0, 1
27. x22y23z23, 2, 1, 1
28. xyyzzx3, 1, 1, 1
29. sinxyzx2y3z, 2, 1, 0
4uv
x2 y2 uv v
2z 2zy2
x2
42. Ifyz4 x2z3 exyz,find z and z.
x y
43. Findthegradientofthefunction fx,y,zz2exsy.
; 30.
31. Find the points on the hyperboloid x24y2z24 where
Use a computer to graph the surface zx2y4 and its tangent plane and normal line at 1, 1, 2 on the same screen. Choose the domain and viewpoint so that you get a good view of all three objects.
the tangent plane is parallel to the plane 2x2yz5.
32. Findduifuln1se2t.
33. Find the linear approximation of the function fx,y,zx3sy2 z2 atthepoint2,3,4anduseit
to estimate the number 1.983s3.0123.972 .
34. The two legs of a right triangle are measured as 5 m and
12 m with a possible error in measurement of at most
0.2 cm in each. Use differentials to estimate the maximum error in the calculated value of a the area of the triangle and b the length of the hypotenuse.
value of the directional derivative of the wind speed at Homestead, Florida, in the direction of the eye of the hurricane.
; 58. Use a graphing calculator or computer or Newtons method or a computer algebra system to find the critical points of
fx,y1210y2x2 8xyy4 correcttothree decimal places. Then classify the critical points and find the highest point on the graph.
5962 Use Lagrange multipliers to find the maximum and mini mum values of f subject to the given constraints.
CHAPTER 14 REVIEW947
60 70 80 70 5565 7565
60
55 50
45 40
35 30
Homestead
Key West
59. fx,yx2y; x2 y2 1
60. fx,y11; 11 1
0
10 20 30 40 Distance in miles
62. fx,y,zx2 2y2 3z2; xyz1, xy2z2
63. Find the points on the surface xy2z32 that are closest to the origin.
64. A package in the shape of a rectangular box can be mailed by the US Postal Service if the sum of its length and girth the perimeter of a crosssection perpendicular to the length is at most 108 in. Find the dimensions of the package with largest volume that can be mailed.
65. A pentagon is formed by placing an isosceles triangle on a rectangle, as shown in the figure. If the pentagon has fixed perimeter P, find the lengths of the sides of the pentagon that maximize the area of the pentagon.
66. A particle of mass m moves on the surface zf x, y. Let xxt and yyt be the x and ycoordinates of the particle at time t.
a Find the velocity vector v and the kinetic energy
K1 mv2 of the particle. 2
b Determine the acceleration vector a.
c Letzx2 y2 andxttcost,yttsint.Find
the velocity vector, the kinetic energy, and the acceleration vector.
50. Find parametric equations of the tangent line at the point 2, 2, 4 to the curve of intersection of the surface z2x2 y2 andtheplanez4.
5154 Find the local maximum and minimum values and saddle points of the function. If you have threedimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function.
51. fx,yx2 xyy2 9x6y10 52. fx,yx3 6xy8y3
53. fx,y3xyx2yxy2
54. fx,yx2 yey2
5556 Find the absolute maximum and minimum values of f on the set D.
55. fx,y4xy2 x2y2 xy3; Distheclosedtriangular region in the xyplane with vertices 0, 0, 0, 6, and 6, 0
56. fx,yex2y2x2 2y2; Disthediskx2 y2 4
; 57. Use a graph andor level curves to estimate the local maximum and minimum values and saddle points of
fx,yx3 3xy4 2y2.Thenusecalculustofind these values precisely.
x y x2 y2
61. fx,y,zxyz; x2 y2 z2 3
NOAAAOMLHurricane Research Division
PROBLEMS PLUS
1. A rectangle with length L and width W is cut into four smaller rectangles by two lines parallel to the sides. Find the maximum and minimum values of the sum of the squares of the areas of the smaller rectangles.
2. Marinebiologistshavedeterminedthatwhenasharkdetectsthepresenceofbloodinthewater, it will swim in the direction in which the concentration of the blood increases most rapidly. Based on certain tests, the concentration of blood in parts per million at a point Px, y on the surface of seawater is approximated by
Cx, yex22y2 104
where x and y are measured in meters in a rectangular coordinate system with the blood source at the origin.
a Identify the level curves of the concentration function and sketch several members of this
family together with a path that a shark will follow to the source.
b Suppose a shark is at the point x0, y0when it first detects the presence of blood in the
water. Find an equation of the sharks path by setting up and solving a differential equation.
3. A long piece of galvanized sheet metal with width w is to be bent into a symmetric form with three straight sides to make a rain gutter. A crosssection is shown in the figure.
a Determine the dimensions that allow the maximum possible flow; that is, find the dimen
sions that give the maximum possible crosssectional area.
b Would it be better to bend the metal into a gutter with a semicircular crosssection?
x x w2x
4. For what values of the number r is the function
xyzr fx,y,z x2 y2 z2
0 if x,y,z0
continuous on3 ?
5. Suppose f is a differentiable function of one variable. Show that all tangent planes to the
surface zx fyx intersect in a common point.
6. a Newtons method for approximating a root of an equation f x0 see Section 4.8
can be adapted to approximating a solution of a system of equations f x, y0 and
tx, y0. The surfaces zf x, y and ztx, y intersect in a curve that intersects the
if x,y,z0
948
PROBLEMS PLUS
xyplane at the point r, s, which is the solution of the system. If an initial approximation x1, y1 is close to this point, then the tangent planes to the surfaces at x1, y1 intersect in a straight line that intersects the xyplane in a point x2, y2 , which should be closer to r, s. Compare with Figure 2 in Section 4.8. Show that
x2 x1fty fyt and y2 y1fxtftx fxtyfytx fxtyfytx
where f , t, and their partial derivatives are evaluated at x1, y1. If we continue this proce
dure, we obtain successive approximations xn, yn .
b It was Thomas Simpson 17101761 who formulated Newtons method as we know it
today and who extended it to functions of two variables as in part a. See the biography of Simpson on page 502. The example that he gave to illustrate the method was to solve the system of equations
xx yy 1000 xy yx 100
In other words, he found the points of intersection of the curves in the figure. Use the method of part a to find the coordinates of the points of intersection correct to six deci mal places.
y 4
x xy y1000
024x
7. Iftheellipsex2a2 y2b2 1istoenclosethecirclex2 y2 2y,whatvaluesofaandb minimize the area of the ellipse?
8. Among all planes that are tangent to the surface xy2z21, find the ones that are farthest from the origin.
xyyx100 2
949
950
15
MULTIPLE INTEGRALS
A double integral of a positive function is a volume, which is the limit of sums of volumes of rectangular columns.
In this chapter we extend the idea of a definite integral to double and triple integrals of functions of two or three variables. These ideas are then used to compute volumes, masses, and centroids of more general regions than we were able to consider in Chapters 6 and 8. We also use double integrals to calculate probabilities when two random variables are involved.
We will see that polar coordinates are useful in computing double integrals over some types of regions. In a similar way, we will introduce two new coordinate systems in threedimensional spacecylindrical coordinates and spherical coordinatesthat greatly simplify the computation of triple integrals over certain commonly occurring solid regions.
15.1
DOUBLE INTEGRALS OVER RECTANGLES
In much the same way that our attempt to solve the area problem led to the definition of a definite integral, we now seek to find the volume of a solid and in the process we arrive at the definition of a double integral.
REVIEW OF THE DEFINITE INTEGRAL
First lets recall the basic facts concerning definite integrals of functions of a single vari able. If f x is defined for axb, we start by dividing the interval a, b into n sub intervals xi1, xiof equal width xban and we choose sample points xi in these subintervals. Then we form the Riemann sum
n
fx i x
i1
and take the limit of such sums as n lto obtain the definite integral of f from a to b:
fxdxlim fxi x yb n
In the special case where f x0, the Riemann sum can be interpreted as the sum of the areas of the approximating rectangles in Figure 1, and xb f x dx represents the area under the curve yfx from a to b. a
1
2
a nl i1
y
0 axx TM x
VOLUMES AND DOUBLE INTEGRALS
In a similar manner we consider a function f of two variables defined on a closed rectangle Ra,bc,dx,y2axb, cyd
and we first suppose that f x, y0. The graph of f is a surface with equation zf x, y. Let S be the solid that lies above R and under the graph of f, that is,
Sx,y,z30zfx,y, x,yR
See Figure 2. Our goal is to find the volume of S.
The first step is to divide the rectangle R into subrectangles. We accomplish this by
dividing the interval a, b into m subintervals xi1, xiof equal width xbam and dividing c, dinto n subintervalsyj1, yjof equal width ydcn. By draw
fx i
FIGURE 1
a0cd bRy
FIGURE 2
Ix
xi1 xi xn1 b x xx
in
z
zfx, y
x
951
952
CHAPTER 15 MULTIPLE INTEGRALS
ing lines parallel to the coordinate axes through the endpoints of these subintervals, as in
Figure 3, we form the subrectangles
Rij xi1,xiyj1,yjx,yxi1 xxi, yj1 yyj
each with area Ax y. y
d
yj yj1
c
Rij
xi, yj
xTM, yTM
Iy
xij, yij
FIGURE 3
Dividing R into subrectangles
0 axi1xi bx Ix
If we choose a sample point xij , yijin each Rij, then we can approximate the part of S that lies above each Rij by a thin rectangular box or column with base Rij and height f xij , yijas shown in Figure 4. Compare with Figure 1. The volume of this box is the
height of the box times the area of the base rectangle:
fx i j , y i j A
If we follow this procedure for all the rectangles and add the volumes of the corresponding
a
See Figure 5. This double sum means that for each subrectangle we evaluate f at the cho sen point and multiply by the area of the subrectangle, and then we add the results.
zz
0 c fxij, yij 0 dyy
boxes, we get an approximation to the total volume of S: mn
Vfx i j , y i j A i1 j1
xbx Rij
FIGURE 4
FIGURE 5
3
N The meaning of the double limit in Equation 4 is that we can make the double sum as close as we like to the number V for any choice of
xij , yijin Rijby taking m and n sufficiently large.
SECTION 15.1 DOUBLE INTEGRALS OVER RECTANGLES953 Our intuition tells us that the approximation given in 3 becomes better as m and n
become larger and so we would expect that
mn
V lim fxij,yijA
m,nl i1 j1
We use the expression in Equation 4 to define the volume of the solid S that lies under the graph of f and above the rectangle R. It can be shown that this definition is consistent with our formula for volume in Section 6.2.
Limits of the type that appear in Equation 4 occur frequently, not just in finding vol umes but in a variety of other situations as wellas we will see in Section 15.5even when f is not a positive function. So we make the following definition.
5
DEFINITION The double integral of f over the rectangle R is fx,ydA lim fxij,yijA
if this limit exists.
yy m n
R
m,nl i1 j1
N Notice the similarity between Definition 5 and the definition of a single integral in Equation 2.
N Although we have defined the double integral by dividing R into equalsized subrectangles, we couldhaveusedsubrectanglesRij ofunequal size. But then we would have to ensure that all of their dimensions approach 0 in the limiting process.
The precise meaning of the limit in Definition 5 is that for every number 0 there is an integer N such that
fx,ydA fxij,yijAyymn
R i1 j1
for all integers m and n greater than N and for any choice of sample points xij , yijin Rij. A function f is called integrable if the limit in Definition 5 exists. It is shown in courses on advanced calculus that all continuous functions are integrable. In fact, the double inte gral of f exists provided that f is not too discontinuous. In particular, if f is bounded thatis,thereisaconstantM suchthatfx,yM forallx,yinR,and f iscon tinuous there, except on a finite number of smooth curves, then f is integrable over R. The sample point xij , yijcan be chosen to be any point in the subrectangle Rij, but if we choose it to be the upper righthand corner of Rij namely xi, yj , see Figure 3, then
the expression for the double integral looks simpler:
By comparing Definitions 4 and 5, we see that a volume can be written as a double integral:
4
6
fx,ydA lim fxi,yjA yy m n
R
m,nl i1 j1
If f x, y0, then the volume V of the solid that lies above the rectangle R and below the surface zf x, y is
Vyy f x, y dA R
954
CHAPTER 15 MULTIPLE INTEGRALS
The sum in Definition 5,
y 1,2
2 2,2
RTM RTMTM
1 1,1 2,1
R RTM
012x
FIGURE 6
z 16
2 x
FIGURE 7
mn
fx i j , y i j A
i1 j1
is called a double Riemann sum and is used as an approximation to the value of the double integral. Notice how similar it is to the Riemann sum in 1 for a function of a single variable. If f happens to be a positive function, then the double Riemann sum represents the sum of volumes of columns, as in Figure 5, and is an approximation to the volume under the graph of f and above the rectangle R.
V EXAMPLE 1 Estimate the volume of the solid that lies above the square
R0, 20, 2 and below the elliptic paraboloid z16x22y2. Divide R into four equal squares and choose the sample point to be the upper right corner of each square Rij. Sketch the solid and the approximating rectangular boxes.
SOLUTION The squares are shown in Figure 6. The paraboloid is the graph of
fx, y16x22y2 and the area of each square is 1. Approximating the volume
z162
2
y
f1, 1Af1, 2Af2, 1Af2, 2A
131711014134
This is the volume of the approximating rectangular boxes shown in Figure 7. M
We get better approximations to the volume in Example 1 if we increase the number of squares. Figure 8 shows how the columns start to look more like the actual solid and the corresponding approximations become more accurate when we use 16, 64, and 256 squares. In the next section we will be able to show that the exact volume is 48.
The Riemann sum approximations to the volume under z162 become more accurate as m and n increase.
a mn4, VA41.5
V EXAMPLE2 IfRx,y1x1,2y2,evaluatetheintegral
yys1x2 dA R
FIGURE 8
by the Riemann sum with mn2, we have 22
Vf xi, yj A i1 j1
b mn8, VA44.875
c mn16, VA46.46875
z
0, 0, 1
0, 2, 0
SOLUTION It would be very difficult to evaluate this integral directly from Definition 5 but, because s1x20, we can compute the integral by interpreting it as a volume. If zs1x2,thenx2 z2 1andz0,sothegivendoubleintegralrepresentsthe volume of the solid S that lies below the circular cylinder x2z21 and above the rectangle R. See Figure 9. The volume of S is the area of a semicircle with radius 1 times the length of the cylinder. Thus
SECTION 15.1 DOUBLE INTEGRALS OVER RECTANGLES955
S
x
1, 0, 0
y
yys1x2 dA1 2
R
FIGURE 9
1242 M
THE MIDPOINT RULE
The methods that we used for approximating single integrals the Midpoint Rule, the Trapezoidal Rule, Simpsons Rule all have counterparts for double integrals. Here we consider only the Midpoint Rule for double integrals. This means that we use a double Rie mann sum to approximate the double integral, where the sample point xij , yijin Rij is chosen to be the center xi, yj of Rij. In other words, xi is the midpoint of xi1, xi and yj is the midpoint ofyj1, yj .
MIDPOINT RULE FOR DOUBLE INTEGRALS
fx, y dAfxi, yjA yy mn
i1 j1
where xi is the midpoint of xi1, xi and yj is the midpoint of yj1, yj.
R
V EXAMPLE 3 Use the Midpoint Rule with mn2 to estimate the value of the integralxxx3y2dA,whereRx,y 0x2,1y2.
y
the centers of the four subrectangles shown in Figure 10. So x11, x23, y15, and
71
y24. The area of each subrectangle is A2. Thus
x3y dA fxi,yjA yy2 22
SOLUTION In using the Midpoint Rule with mn2, we evaluate fx, yx3y2 at
R
2 2,2 32
1 012x
FIGURE 10
R
Thus we have
i1 j1
fx1, y1Afx1, y2Afx2, y1Afx2, y2A
f 1 , 5Af 1 , 7Af 3 , 5Af 3 , 7A 24242424
224
RTM
RTMTM
R
RTM
671 1391 511 1231
16 2 16 2 16 2
16 2
9511.875 8
yy x3y2dA11.875 R
M
NOTE In the next section we will develop an efficient method for com puting double integrals and then we will see that the exact value of the double integral in Example 3 is 12. Remember that the interpretation of a double integral as a volume is valid only when the integrand f is a positive function. The integrand in Example 3 is not a positive function, so its integral is not a volume. In Examples 2 and 3 in Section 15.2 we will discuss how to interpret integrals of functions that are not always positive in terms of volumes. If we keep dividing each subrectangle in Figure 10 into four smaller ones with
956
CHAPTER 15 MULTIPLE INTEGRALS
similar shape, we get the Midpoint Rule approximations displayed in the chart in the mar
Number of
subrectangles
Midpoint Rule
approximations
1
4
16
64
256
1024
11.5000 11.8750 11.9687 11.9922 11.9980 11.9995
gin. Notice how these approximations approach the exact value of the double integral, 12. AVERAGE VALUE
Recall from Section 6.5 that the average value of a function f of one variable defined on an interval a, b is
fave1 yb fxdx ba a
In a similar fashion we define the average value of a function f of two variables defined on a rectangle R to be
fave 1 yyfx,ydA AR R
ARfave yyfx,ydA R
FIGURE 11
says that the box with base R and height fave has the same volume as the solid that lies under the graph of f. If zf x, y describes a mountainous region and you chop off the tops of the mountains at height fave , then you can use them to fill in the valleys so that the region becomes completely flat. See Figure 11.
EXAMPLE 4 The contour map in Figure 12 shows the snowfall, in inches, that fell on the state of Colorado on December 20 and 21, 2006. The state is in the shape of a rectangle that measures 388 mi west to east and 276 mi south to north. Use the contour map to estimate the average snowfall for the entire state of Colorado on those days.
where AR is the area of R.
If f x, y0, the equation
20 16
12 40 3632
12 16
12 8
44
28
24
403632
28 24
0 4812162024
32 28
FIGURE 12
y
276
SECTION 15.1 DOUBLE INTEGRALS OVER RECTANGLES957
SOLUTION Lets place the origin at the southwest corner of the state. Then 0x388,
0y276, and f x, y is the snowfall, in inches, at a location x miles to the east and y miles to the north of the origin. If R is the rectangle that represents Colorado, then the average snowfall for the state on December 2021 was
fave 1 yyfx,ydA AR R
where AR388276. To estimate the value of this double integral, lets use the Mid point Rule with mn4. In other words, we divide R into 16 subrectangles of equal size, as in Figure 13. The area of each subrectangle is
A1 3882766693 mi2 16
40 3632
44
28
12 16
20
16
12
24 40
36 32
28 24
8 121620 24
32 28
12
0
4
8
FIGURE 13
388 x Using the contour map to estimate the value of f at the center of each subrectangle,
0
we get
Therefore
fx, y dAfxi, yjA yy 44
R
i1 j1
A0158722518.511
4.5281713.5121517.5136693207
fave669320712.9 388276
On December 2021, 2006, Colorado received an average of approximately 13 inches of snow. M
958
CHAPTER 15 MULTIPLE INTEGRALS
PROPERTIES OF DOUBLE INTEGRALS
N Double integrals behave this way because the double sums that define them behave this way.
We list here three properties of double integrals that can be proved in the same manner as in Section 5.2. We assume that all of the integrals exist. Properties 7 and 8 are referred to as the linearity of the integral.
yyfx, ytx, y dAyyfx, y dAyytx, y dA RRR
1.
sample points to be lower left corners.
b Use the Midpoint Rule to estimate the integral in part a.
8
yy c f x, y dAc yy f x, y dA RR
where c is a constant
If fx,ytx,yforallx,yinR,then
yy f x, y dAyy tx, y dA
RR
b Estimate the double integral with mn4 by choosing the sample points to be the points farthest from the origin.
9
15.1 EXERCISES
a Estimate the volume of the solid that lies below
the surface zxy and above the rectangle Rx, y 0x6, 0y4
Use a Riemann sum with m3, n2, and take the sample point to be the upper right corner of each square.
b Use the Midpoint Rule to estimate the volume of the solid
xy
0
1
2
3
4
1.0
2
0
3
6
5
1.5
3
1
4
8
6
2.0
4
3
0
5
8
2.5
5
5
3
1
4
3.0
7
8
6
3
0
in part a.
2. If R1, 30, 2, use a Riemann sum with m4,
n2 to estimate the value of xx y22×2 dA. Take the R
sample points to be the upper left corners of the squares.
3. a Use a Riemann sum with mn2 to estimate the value
6. A 20ftby30ft swimming pool is filled with water. The depth is measured at 5ft intervals, starting at one corner of the pool, and the values are recorded in the table. Estimate the volume of water in the pool.
7. Let V be the volume of the solid that lies under the graph of fx, ys52x2y2 and above the rectangle given by 2×4, 2y6. We use the lines x3 and y4 to
of xx sinxy dA, where R0, 0, . Take the R
0
5
10
15
20
25
30
0
5 10 15 20
2 2 2 2 2
3 3 4 3 2
4 4 6 4 2
6 7 8 5 2
7
8 10 6 3
8 10 12 8 4
8
8 10 7 4
4. a Estimate the volume of the solid that lies below the surface zx2y2 and above the rectangle R0, 20, 4. Use a Riemann sum with mn2 and choose the sample points to be lower right corners.
b Use the Midpoint Rule to estimate the volume in part a.
5. A table of values is given for a function f x, y defined on R1, 30, 4.
a Estimate xxR f x, y dA using the Midpoint Rule with
mn2.
7
divide R into subrectangles. Let L and U be the Riemann sums computed using lower left corners and upper right corners, respectively. Without calculating the numbers V, L, and U, arrange them in increasing order and explain your reasoning.
8. The figure shows level curves of a function f in the square R0, 20, 2. Use the Midpoint Rule with mn2 to estimate xxR f x, y dA. How could you improve your estimate?
24
20 16
16
32 44 44
24
13.
SECTION 15.2 ITERATED INTEGRALS959
28
32
24
20 24 28
40 44 3236 48
5652
y 2
1
0
32 36
4 3
567
2 1
440 48
9. A contour map is shown for a function f on the square
R0, 40, 4.
a Use the Midpoint Rule with mn2 to estimate the
0x5, 0y3
5256
1
2
x
1113 Evaluate the double integral by first identifying it as the volume of a solid.
11. xxR 3 dA, Rx, y2x2, 1y6
12. xxR 5x dA, xx 42y dA,
R
14. The integral xxR s9y2 dA, where R0, 40, 2, represents the volume of a solid. Sketch the solid.
15. Use a programmable calculator or computer or the sum command on a CAS to estimate
yy s1xey dA R
where R0, 10, 1. Use the Midpoint Rule with the following numbers of squares of equal size: 1, 4, 16, 64, 256, and 1024.
valueofxxR fx,ydA.
b Estimate the average value of f.
y 4
2
024x
10. The contour map shows the temperature, in degrees Fahrenheit, at 4:00 PM on February 26, 2007, in Colorado. The state measures 388 mi east to west and 276 mi north to south. Use the Midpoint Rule with mn4 to estimate the average temperature in Colorado at that time.
15.2
Rx, y
R0, 10, 1
10
0
0
10
20
30
10
20
30
16. Repeat Exercise 15 for the integral xxR sinxsydA. If f is a constant function, f x, yk, and
17.
Ra,bc,d,showthatxxR kdAkbadc. 18. Use the result of Exercise 17 to show that
0yysin xcos ydA 1 R 32
where R0, 1 1, 1. 442
ITERATED INTEGRALS
Recall that it is usually difficult to evaluate single integrals directly from the definition of an integral, but the Fundamental Theorem of Calculus provides a much easier method. The evaluation of double integrals from first principles is even more difficult, but in this sec
960
CHAPTER 15 MULTIPLE INTEGRALS
tion we see how to express a double integral as an iterated integral, which can then be eval uated by calculating two single integrals.
Suppose that f is a function of two variables that is integrable on the rectangle Ra,bc,d. We use the notation xd fx,ydy to mean that x is held fixed and
c
f x, y is integrated with respect to y from yc to yd. This procedure is called par
tial integration with respect to y. Notice its similarity to partial differentiation. Now xd fx, y dy is a number that depends on the value of x, so it defines a function of x:
c
Axyd fx,ydy c
If we now integrate the function A with respect to x from xa to xb, we get yb Ax dxyb yd fx, y dydx
aac
The integral on the right side of Equation 1 is called an iterated integral. Usually the brackets are omitted. Thus
yb yd fx,ydydxyb yd fx,ydydx ac ac
means that we first integrate with respect to y from c to d and then with respect to x from a to b.
Similarly, the iterated integral
yd yb fx,ydxdyyd yb fx,ydxdy ca ca
means that we first integrate with respect to x holding y fixed from xa to xb and then we integrate the resulting function of y with respect to y from yc to yd. Notice that in both Equations 2 and 3 we work from the inside out.
EXAMPLE 1 Evaluate the iterated integrals.
a y3 y2 x2ydydx b y2 y3 x2ydxdy
01 10
SOLUTION
a Regarding x as a constant, we obtain
y2 y2y2 22 12 x2ydy x2 x2 x2 3×2 1 2y1222
Thus the function A in the preceding discussion is given by Ax3 x2 in this example.
1
2
3
We now integrate this function of x from 0 to 3:
y3 y2 2 y3 y2 2
2
0 1xydydx0
y332 x3327
1xydydx
0 xdx2 2 2
0
SECTION 15.2 ITERATED INTEGRALS b Here we first integrate with respect to x:
y2y3 2 y2y3 2y2x3 x3 10xydxdy1 0xydxdy1 3y dy
961
M
Notice that in Example 1 we obtained the same answer whether we integrated with respect to y or x first. In general, it turns out see Theorem 4 that the two iterated integrals in Equations 2 and 3 are always equal; that is, the order of integration does not matter. This is similar to Clairauts Theorem on the equality of the mixed partial derivatives.
The following theorem gives a practical method for evaluating a double integral by expressing it as an iterated integral in either order.
x0 y2 y2 2 27
1 9ydy9 22 1
FUBINIS THEOREM If f is continuous on the rectangle Rx, yaxb, cyd , then
yyfx,ydAyb yd fx,ydydxyd yb fx,ydxdy
4
ac ca
More generally, this is true if we assume that f is bounded on R, f is discontin uous only on a finite number of smooth curves, and the iterated integrals exist.
R
N Theorem 4 is named after the Italian mathe matician Guido Fubini 18791943, who proved a very general version of this theorem in 1907. But the version for continuous functions was known to the French mathematician Augustin Louis Cauchy almost a century earlier.
z
0 Ax xay
The proof of Fubinis Theorem is too difficult to include in this book, but we can at least give an intuitive indication of why it is true for the case where f x, y0. Recall that if f is positive, then we can interpret the double integral xxR f x, y dA as the volume V of the solid S that lies above R and under the surface zf x, y. But we have another for
mula that we used for volume in Chapter 6, namely,
Vyb Ax dx a
where Ax is the area of a crosssection of S in the plane through x perpendicular to the xaxis. From Figure 1 you can see that Ax is the area under the curve C whose equation is zfx, y, where x is held constant and cyd. Therefore
xb
FIGURE 1
TEC Visual 15.2 illustrates Fubinis Theorem by showing an animation of Figures 1 and 2.
z
0
and we have
Axyd fx,ydy c
yyfx,ydAVyb Axdxyb yd fx,ydydx
aac
c
y
d y
R
A similar argument, using crosssections perpendicular to the yaxis as in Figure 2, shows
that yyfx,ydAyd yb fx,ydxdy R ca
x
FIGURE2
962CHAPTER 15 MULTIPLE INTEGRALS
V EXAMPLE 2 Evaluate the double integral xx x3y2dA, where
N Notice the negative answer in Example 2; nothing is wrong with that. The function f in that example is not a positive function, so its integral doesnt represent a volume. From Figure 3 we see that f is always negative on R, so the value of the integral is the negative of the volume that lies above the graph of f and below R.
R
Rx,y 0x2,1y2.ComparewithExample3inSection15.1.
SOLUTION 1 Fubinis Theorem gives
0 z4
8 zx3
R
first, we have
120 0.5
FIGURE 3
0 y 1.5 22 x
x2
2 3xy2 dy
1
1
1 x0
y2 26y2dy2y2y32112 M
N For a function f that takes on both positive andnegativevalues,xxR fx,ydAisadiffer ence of volumes: V1V2, where V1 is the vol ume above R and below the graph of f and V2 is the volume below R and above the graph. The fact that the integral in Example 3 is 0 means that these two volumes V1 and V2 are equal. See Figure 4.
SOLUTION 2 If we reverse the order of integration, we get yy y2 y
1 z0 1
andso
ycosxyy 1 y y0
zy sinxy 1
01y232
FIGURE 4
x
yy yy y R
x3y2dA 2 2x3y2dydx 2xyy3y2 01 0 y1dx
R
2 x2 y
y2 x2 2×7 dx 7x
12
SOLUTION 2 Again applying Fubinis Theorem, but this time integrating with respect to x
02
0
yyx3y2dAy2 y2 x3y2dxdy
1
V EXAMPLE 3 Evaluate xxR y sinxy dA, where R1, 20, .
10
SOLUTION 1 If we first integrate with respect to x, we get
yy y sinxy dAy y2 y sinxy dx dyy cosxyx2 dy
R
01 0 x1y cos 2ycos y dy
0
1 sin 2ysin y0 20
ysinxydA 1 0 ysinxydydx R
To evaluate the inner integral, we use integration by parts with
uy dudy
0 ysinxydy y
x
xx 0 cosxydy
dvsinxy dy v cosxy
cos x1sinxyy xx2 y0
cos xsin x x x2
N In Example 2, Solutions 1 and 2 are equally straightforward, but in Example 3 the first solu tion is much easier than the second one. There fore, when we evaluate double integrals, it is wise to choose the order of integration that gives simpler integrals.
16
12 z8 4
y2y sin x2 ysinxydydx
If we now integrate the first term by parts with u1x and dvcos dudxx2,vsin x,and
x dx, we get
Therefore andso
y cos xdxsin xysin xdx x x x2
y cos xsin xdx sin x xx2 x
10 x1 sin 2
SECTION 15.2 ITERATED INTEGRALS963
2 sin 0
V EXAMPLE 4 Find the volume of the solid S that is bounded by the elliptic paraboloid
M
00 0 R
FIGURE 5
3330
y22x
3 x0
y288 88 4 2
x2 2y2 z16,theplanesx2andy2,andthethreecoordinateplanes.
SOLUTION We first observe that S is the solid that lies under the surface z16x22y2 and above the square R0, 20, 2. See Figure 5. This solid was considered in Example 1 in Section 15.1, but we are now in a position to evaluate the double integral using Fubinis Theorem. Therefore
Vyy16x2 2y2dAy2 y2 16×2 2y2dxdy
00 y16x1x32y2x dy
0
4y2dy y y348 M
0
In the special case where f x, y can be factored as the product of a function of x only and a function of y only, the double integral of f can be written in a particularly simple form. To be specific, suppose that fx, ytxhy and Ra, bc, d. Then Fubinis Theorem gives
yyfx,ydAydybtxhydxdyyd ybtxhydx dy
ca ca
In the inner integral, y is a constant, so hy is a constant and we can write
yd yb txhy dxdyyd hyyb tx dxdyyb tx dx yd hy dy cacaac
since xb tx dx is a constant. Therefore, in this case, the double integral of f can be writ a
5 yytxhydAyb txdxyd hydy whereRa,bc,d
11 2 x2
R
ten as the product of two single integrals:
R
ac
964
CHAPTER 15 MULTIPLE INTEGRALS
EXAMPLE 5 If R0, 20, 2, then, by Equation 5,
N Thefunctionfx,ysinxcosyin Example 5 is positive on R, so the integral repre sents the volume of the solid that lies above R and below the graph of f shown in Figure 6.
FIGURE 6
15.2 EXERCISES
12 Find x5 fx, y dx and x1 fx, y dy. 00
1. fx,y12x2y3 2. fx,yyxey 314 Calculate the iterated integral.
z
0 x
y
3. y3 y1 14xydxdy 4. y1y2 4×3 9x2y2dydx 10 01
R
R
yy x dA, R 1xy
yy xyex2y dA, R
R0, 10, 1 R0, 10, 2
5.y2y2xsinydydx 0 0
6.y2y5 cosydxdy 6 1
y2 y1 y1 y2 xex
y4 y2 x y 13
9.dydx 10. yyex3ydxdy 11yx 00
11. y1 y1 uv5 du dv 12. y1 y1 xysx2y2 dy dx 00 00
13.y2yrsin2 d dr 14.y1y1sstdsdt 00 00
1522 Calculate the double integral.
15. yy6x2y3 5y4dA, Rx,y0x3, 0y1
R
16. yycosx2ydA, Rx,y0x , 0y 2 R
7. 2xy8 dxdy
00 01y
yyxy2
17. x2 1 dA,
R
23.
24. 25.
26.
y1 y1 4x2y dx dy 00
y1y12x2y2dydx 00
Find the volume of the solid that lies under the plane 3x2yz12 and above the rectangle Rx,y0x1, 2y3.
Find the volume of the solid that lies under the hyperbolic paraboloid z4x2y2 and above the square
R1, 10, 2.
8. dy dx
Rx,y0x1, 3y3
yy sin x cos y dAy 2 sin x dx y 2 cos y dy
R
00
cos x 2 sin y 2111 M 00
18.
19.
20. 21. 22.
integral.
yy1x2
1y2 dA, Rx,y 0x1, 0y1 yyxsinxydA, R0, 60, 3
yy x dA, R1, 20, 1
x2 y2
2324 Sketch the solid whose volume is given by the iterated
R
27.
Find the volume of the solid lying under the elliptic paraboloidx24y29z1andabovetherectangle R1, 12, 2.
28. Find the volume of the solid enclosed by the surface z1ex sinyandtheplanesx1,y0,y , and z0.
CAS 34. Graph the solid that lies between the surfaces
zex2 cosx2 y2andz2x2 y2 forx1,
y 1. Use a computer algebra system to approximate the volume of this solid correct to four decimal places.
3536 Find the average value of f over the given rectangle.
fx, yx2y, R has vertices 1, 0, 1, 5, 1, 5, 1, 0
29. Find the volume of the solid enclosed by the surface zxsec2yandtheplanesz0,x0,x2,y0, 36. and y4.
fx, yeysxey , R0, 40, 1 Use your CAS to compute the iterated integrals
SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS965
30. Find the volume of the solid in the first octant bounded by thecylinderz16x2 andtheplaney5.
31. Find the volume of the solid enclosed by the paraboloid z2x2 y22 andtheplanesz1,x1,x1, y0, and y4.
CAS 37.
38.
y1 y1 xy dy dx and 0 0 xy3
y1 y1 xy dx dy 0 0 xy3
; 32. CAS 33.
Graph the solid that lies between the surface
z2xyx2 1andtheplanezx2yandisbounded by the planes x0, x2, y0, and y4. Then find its volume.
Use a computer algebra system to find the exact value of the integral xxR x5y3exy dA, where R0, 10, 1. Then use the CAS to draw the solid whose volume is given by the integral.
15.3
Do the answers contradict Fubinis Theorem? Explain what is happening.
a In what way are the theorems of Fubini and Clairaut similar?
b If fx, y is continuous on a, bc, d and tx, ya c fs, t dt ds
yx yy foraxb,cyd,showthattxy tyx fx,y.
DOUBLE INTEGRALS OVER GENERAL REGIONS
For single integrals, the region over which we integrate is always an interval. But for double integrals, we want to be able to integrate a function f not just over rectangles but also over regions D of more general shape, such as the one illustrated in Figure 1. We sup pose that D is a bounded region, which means that D can be enclosed in a rectangular region R as in Figure 2. Then we define a new function F with domain R by
Fx, yf x, y if x, y is in D
0 if x, y is in R but not in D
yy
D
0x0x
FIGURE 1 FIGURE 2
1
35.
R
D
966
CHAPTER 15 MULTIPLE INTEGRALS
z
graph of f 0
y
graph of F 0
If F is integrable over R, then we define the double integral of f over D by
x
FIGURE 3
D
z
Definition 2 makes sense because R is a rectangle and so xxR Fx, y dA has been previ ously defined in Section 15.1. The procedure that we have used is reasonable because the values of Fx, y are 0 when x, y lies outside D and so they contribute nothing to the inte gral. This means that it doesnt matter what rectangle R we use as long as it contains D.
FIGURE 5
y d
c
Some type I regions
ygTMx
In order to evaluate xxD f x, y dA when D is a region of type I, we choose a rectangle Ra, bc, d that contains D, as in Figure 6, and we let F be the function given by Equation 1; that is, F agrees with f on D and F is 0 outside D. Then, by Fubinis Theorem,
yyfx,ydAyyFx,ydAyb yd Fx,ydydx
In the case where f x, y0, we can still interpret xxD f x, y dA as the volume of the solid that lies above D and under the surface zf x, y the graph of f . You can see that this is reasonable by comparing the graphs of f and F in Figures 3 and 4 and remember ing that xxR Fx, y dA is the volume under the graph of F.
x
FIGURE 4
y
Figure 5.
D
y
Figure 4 also shows that F is likely to have discontinuities at the boundary points of D. Nonetheless, if f is continuous on D and the boundary curve of D is well behaved in a sense outside the scope of this book, then it can be shown that xxR Fx, y dA exists and therefore xxD f x, y dA exists. In particular, this is the case for the following types of regions.
A plane region D is said to be of type I if it lies between the graphs of two continuous functions of x, that is,
Dx,yaxb, t1xyt2x
where t1 and t2 are continuous on a, b. Some examples of type I regions are shown in
y ygTMx DD
ygx ygx
y ygTMx D
ygx
2
yyfx, y dAyyFx, y dA where F is given by Equation 1 DR
ygTMx
0abx0abx0abx
D
ygx
0axbx
FIGURE 6
ac
Observe that Fx, y0 if yt1x or yt2x because x, y then lies outside D. Therefore
yd Fx, y dyyt2x Fx, y dyyt2x fx, y dy c t1x t1x
DR
y d
SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS967 because Fx, yf x, y when t1xyt2x. Thus we have the following formula
that enables us to evaluate the double integral as an iterated integral.
If f is continuous on a type I region D such that Dx,yaxb, t1xyt2x
then yy fx, y dAyb yt2x fx, y dy dx D a t1x
3
xhTMy 0x
y d
0x c
c
D
xhy D xhTMy
xhy
The integral on the right side of 3 is an iterated integral that is similar to the ones we considered in the preceding section, except that in the inner integral we regard x as being constant not only in f x, y but also in the limits of integration, t1x and t2x.
We also consider plane regions of type II, which can be expressed as Dx,ycyd, h1yxh2y
where h1 and h2 are continuous. Two such regions are illustrated in Figure 7. Using the same methods that were used in establishing 3, we can show that
V EXAMPLE 1 Evaluate xxD x2y dA, where D is the region bounded by the parabolasy2x2 andy1x2.
SOLUTION The parabolas intersect when 2×21×2, that is, x21, so x1. We note that the region D, sketched in Figure 8, is a type I region but not a type II region and we can write
Dx,y1x1, 2×2 y1x2
Since the lower boundary is y2×2 and the upper boundary is y1x2, Equation 3 gives
yyfx,ydAyd yh2y fx,ydxdy
c h1y where D is a type II region given by Equation 4.
5
D
FIGURE 7
Some type II regions
y
1, 2
D
y1 1, 2 y2
4
1 1
FIGURE 8
x
y1 y1x2
yy
D
y1 y1x2 x2ydA 1 2×2
x2ydydx
xyy2
dx
y1 x1x21x22x2x22x22dx
1
y1 3×4 x3 2×2 x1dx
1
y2 x
2
x5 x4 x3 x2 32 1 1
3 542 32×15
1 M
968y
CHAPTER 15 MULTIPLE INTEGRALS
y
x
2
Therefore another expression for V is
Vyyx2 y2dAy4 ysyx2 y2dxdy
FIGURE 11
y 15796035
V EXAMPLE 3 Evaluate xx xy dA, where D is the region bounded by the line yx1
2, 4 y2x
y D
NOTE When we set up a double integral as in Example 1, it is essential to draw a diagram. Often it is helpful to draw a vertical arrow as in Figure 8. Then the limits of integration for the inner integral can be read from the diagram as follows: The arrow starts at the lower boundary yt1x, which gives the lower limit in the integral, and the arrow ends at the upper boundary yt2x, which gives the upper limit of integration. For a type II region the arrow is drawn horizontally from the left boundary to the right boundary.
EXAMPLE 2 Find the volume of the solid that lies under the paraboloid zx2y2 and above the region D in the xyplane bounded by the line y2x and the parabola yx2.
012x
FIGURE 9
D as a type I region y
4 2,4 x 12 y
xy
D
0x
FIGURE 10
D as a type II region
N Figure 11 shows the solid whose volume
is calculated in Example 2. It lies above the xyplane,belowtheparaboloidzx2 y2, and between the plane y2x and the parabolic cylinder yx 2.
z
y2x
SOLUTION 1 From Figure 9 we see that D is a type I region and Dx,y0x2, x2 y2x
Therefore the volume under zx2y2 and above D is Vyyx2 y2dAy2 y2x x2 y2dydx
z
1 D 02y
D 0 x2
y2y3y2x y22×3 x23
x2y dx x22x x2x2dx 0 3yx2 0 3 3
y2x6 14×3 x7 x5 7×42 216 3×4 3 dx2156 35
00
SOLUTION 2 From Figure 10 we see that D can also be written as a type II region: Dx,y0y4, 1yxsy
D and the parabola y22x6.
y4 x3 xsy
y4 y32 y3 y3 y52
3 y2x dy
dy
2 y 522 y 7213 y 4 216
3 242
0 x2y 0 4
1
SOLUTION The region D is shown in Figure 12. Again D is both type I and type II, but the description of D as a type I region is more complicated because the lower boundary con sists of two parts. Therefore we prefer to express D as a type II region:
Dx,y2y4, 1y2 3xy1 2
M
FIGURE 12
a D as a type I region Then 5 gives
y 2×6
1, 2
1, 2 2
b D as a type II region
y4 x2 xy1
2 y dy
SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS969 yy
y2x6 5, 4 x 2 3 yx1
5, 4 xy1
30 x 0 x
yy
D
y4 yy1
xydA 2 1y23xydxdy 2
2
2 x2y23 1y4 yy121y232dy
1
2 2
14y32y28y dy y4 y5
1y6 y3 4
2 24y4234y2 36
2
z
0, 0, 2
24 2
x2yz2
T
0
FIGURE 13
y
1
If we had expressed D as a type I region using Figure 12a, then we would have obtained
yyxydAy1 ys2x6 xydydxy5 ys2x6 xydydx D 3 s2x6 1 x1
but this would have involved more work than the other method. M EXAMPLE 4 Find the volume of the tetrahedron bounded by the planes x2yz2,
x2y, x0, and z0.
SOLUTION In a question such as this, its wise to draw two diagrams: one of the three dimensional solid and another of the plane region D over which it lies. Figure 13 shows the tetrahedron T bounded by the coordinate planes x0, z0, the vertical plane
x2y, and the plane x2yz2. Since the plane x2yz2 intersects the xyplane whose equation is z0 in the line x2y2, we see that T lies above the triangular region D in the xyplane bounded by the lines x2y, x2y2, and x0. See Figure 14.
The plane x2yz2 can be written as z2x2y, so the required volume lies under the graph of the function z2x2y and above
Dx,y0x1, x2y1x2
x2y
y 0, 1, 0
1, 21 , 0
x
x2y2
or y1x2
D
yx2
1, 21
01x
FIGURE 14
970
CHAPTER 15 MULTIPLE INTEGRALS Therefore
Vyy2x2ydAy1 y1x2 2x2ydydx
D
y1 0
0
1 x311
x2 2x1dx x2 x0303
y1x2 2yxyy2
0 x2 yx2 dx
2xx1212 x24 dx y1 xx2 x2 x2
M
y
D
yx
01x
FIGURE 15
D as a type I region y
1
D
x0 xy
0x
FIGURE 16
D as a type II region
V EXAMPLE 5 Evaluate the iterated integral x1 x1 siny2dy dx. 0x
SOLUTION If we try to evaluate the integral as it stands, we are faced with the task of first evaluating x sin y 2dy. But its impossible to do so in finite terms since x sin y 2dy is not an elementary function. See the end of Section 7.5. So we must change the order of integration. This is accomplished by first expressing the given iterated integral as a double integral. Using 3 backward, we have
y1 y1 siny2dydxyysiny2dA
0x
y1
D Dx,y0x1, xy1
where
We sketch this region D in Figure 15. Then from Figure 16 we see that an alternative
description of D is
This enables us to use 5 to express the double integral as an iterated integral in the
Dx,y0y1, 0xy y1 y1 siny2dydxyysiny2dA
reverse order:
0x D
y1 yy siny2dxdyy1 xsiny2xy dy
0
00 0 x0 y1 ysiny2dy1 cosy21
1 1cos 1 M 2
PROPERTIES OF DOUBLE INTEGRALS
We assume that all of the following integrals exist. The first three properties of double inte grals over a region D follow immediately from Definition 2 and Properties 7, 8, and 9 in Section 15.1.
yyfx, ytx, y dAyyfx, y dAyytx, y dA DDD
yy c f x, y dAc yy f x, y dA DD
20
6
7
y
D
aac
If DD1D2, where D1 and D2 dont overlap except perhaps on their boundaries
FIGURE 17
SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS971 If fx,ytx,yforallx,yinD,then
yy f x, y dAyy tx, y dA DD
8
The next property of double integrals is similar to the property of single integrals given by the equation xb fx dxxc fx dxxb fx dx.
D
DTM
see Figure 17, then
Property 9 can be used to evaluate double integrals over regions D that are neither type I nor type II but can be expressed as a union of regions of type I or type II. Figure 18 illus trates this procedure. See Exercises 51 and 52.
yy
D 0x0x
a D is neither type I nor type II. b DDDTM, D is type I, DTM is type II.
The next property of integrals says that if we integrate the constant function f x, y1
over a region D, we get the area of D:
y
0x
9
yyfx, y dAyyfx, y dAyyfx, y dA D D1 D2
z
z1
FIGURE 18
DTM D
0 x
FIGURE 19
Cylinder with base D and height 1
Figure 19 illustrates why Equation 10 is true: A solid cylinder whose base is D and whose height is 1 has volume AD1AD, but we know that we can also write its volume as xxD 1 dA.
Finally, we can combine Properties 7, 8, and 10 to prove the following property. See Exercise 57.
10
yy 1 dAAD D
11
If mfx, yM for all x, y in D, then
mADyy f x, y dAMAD D
972
CHAPTER 15 MULTIPLE INTEGRALS
EXAMPLE 6 Use Property 11 to estimate the integral xx esin x cos y dA, where D is the disk
5.
D
SOLUTION Since1sinx1and1cosy1,wehave1sinxcosy1and therefore
e1 esinxcosy e1 e
Thus, using me11e, Me, and AD22 in Property 11, we obtain
with center the origin and radius 2.
15.3 EXERCISES
16 Evaluate the iterated integral.
1. y4 ysy xy2 dxdy 0 0
3. y1 yx 12ydydx 0 x2
2. y1 y2 xydydx 0 2x
4. y2 y2y xydxdy 0 y
4 yyesinxcosydA4 e M eD
yy 2xy dA, D
D is bounded by the circle with center the origin and radius 2 18. yy 2xy dA, D is the triangular region with vertices 0, 0,
D
1, 2, and 0, 3
1928 Find the volume of the given solid.
17.
y2 ycos esin drd 6. y1 yv s1v2 dudv 00 00
718 Evaluate the double integral.
7. yyy2 dA, Dx,y1y1, y2xy
D
19.
20.
21.
22.
23.
24. 25.
26. 27. 28.
29.
Undertheplanex2yz0andabovetheregion bounded by yx and yx4
Under the surface z2xy2 and above the region bounded byxy2 andxy3
Under the surface zxy and above the triangle with vertices 1, 1, 4, 1, and 1, 2
Enclosed by the paraboloid zx23y2 and the planes x0, y1, yx, z0
Bounded by the coordinate planes and the plane 3x2yz6
Boundedbytheplaneszx,yx,xy2,andz0 Enclosed by the cylinders zx 2, yx 2 and the planes
z0,y4
Bounded by the cylinder y2z24 and the planes x2y,
x0, z0 in the first octant
Bounded by the cylinder x2y21 and the planes yz,
x0,z0inthefirstoctant Boundedbythecylindersx2 y2 r2 andy2 z2 r2
Use a graphing calculator or computer to estimate the xcoordinates of the points of intersection of the curves yx4 and y3xx2. If D is the region bounded by these curves, estimate xxD x dA.
8. yy y dA, Dx,y0x1,0yx2 5
Dx1
9. yyxdA, Dx,y0x , 0ysinx
D
10. yyx3 dA, Dx,y1xe, 0ylnx D
11. yyy2exy dA, Dx,y0y4, 0xy D
12. xsy2 x2 dA, Dx,y 0y1, 0xy yy
D
yyxcosydA, Disboundedbyy0, yx2, x1 D
13.
14. yyxydA, D
Disboundedbyysx andyx2
D is the triangular region with vertices 0, 2, 1, 1, 3, 2
16. yy xy2 dA, D is enclosed by x0 and xs1y2 D
15. yyy3 dA, D
;
;30. Find the approximate volume of the solid in the first octant that is bounded by the planes yx, z0, and zx and the cylinder ycos x. Use a graphing device to estimate the points of intersection.
3132 Find the volume of the solid by subtracting two volumes.
31. The solid enclosed by the parabolic cylinders 1 y1x2,yx2 1andtheplanesxyz2,
2x2yz100
SECTION 15.3 DOUBLE INTEGRALS OVER GENERAL REGIONS973
5152 Express D as a union of regions of type I or type II and evaluate the integral.
yyx2dA 52. yyydA DD
yy
1,1 1 xyA yx1
D
0
32. The solid enclosed by the parabolic cylinder yx2 and the planes z3y, z2y
3334 Sketch the solid whose volume is given by the iterated integral.
1
1 1x 0x
1
1
y1 y1x 33. 0 0
y1 y1x2 34. 0 0
5354 Use Property 11 to estimate the value of the integral.
1xy dy dx
CAS 3538 Use a computer algebra system to find the exact volume
of the solid.
35. Under the surface zx3y4xy2 and above the region
boundedbythecurvesyx3 xandyx2 xforx0
36. Between the paraboloids z2x 2y 2 and
z8x2 2y2 andinsidethecylinderx2 y2 1
37. Enclosedbyz1x2 y2 andz0
38. Enclosedbyzx2 y2 andz2y
3944 Sketch the region of integration and change the order of integration.
1x dy dx
53. yy ex 2 y 2 2 dA, Q is the quartercircle with center the origin Q
and radius 1 in the first quadrant 2
54. yy sin4xy dA, T is the triangle enclosed by the lines T
y0,y2x,andx1
5556 Find the average value of f over region D.
39. y4ysx fx, y dy dx 0 0
1 0
4550 Evaluate the integral by reversing the order of integration.
55. f x, yxy, D is the triangle with vertices 0, 0, 1, 0, and 1, 3
56. fx,yxsiny, Disenclosedbythecurvesy0, yx2,andx1
57. Prove Property 11.
In evaluating a double integral over a region D, a sum of
iterated integrals was obtained as follows:
yyfx,ydAy1 y2y fx,ydxdyy3 y3y fx,ydxdy
00 10
D
Sketch the region D and express the double integral as an
41. y3ys9y2 0 s9y2
40. y1y4 fx, y dy dx 0 4x
42. y3 ys9y fx, y dx dy 0 0
44. y1 y4 fx,ydydx 0 arctan x
43.
fx, y dx dy y2 ylnx fx,ydydx
iterated integral with reversed order of integration. 59. Evaluate xx x2 tan xy34 dA, where
51.
58.
45.
46. ys ys cosx2dxdy 0 y
D
Dx, y x2y22. Hint: Exploit the fact that D is symmetric with respect to both axes.
y1 y3 ex2dxdy 0 3y
y4y2111
47. 0 sx y 1dydx 48. 0 x exydydx
49. y1 y 2 cos x s1cos2x dx dy 0 arcsin y
60. Use symmetry to evaluate xxD 23x4y dA, where D is the region bounded by the square with vertices 5, 0 and 0, 5.
61. ComputexxDs1x2y2dA,whereDisthedisk
x2y21, by first identifying the integral as the volume of a solid.
62. Graphthesolidboundedbytheplanexyz1and the paraboloid z4x2y2 and find its exact volume. Use your CAS to do the graphing, to find the equations of the boundary curves of the region of integration, and to eval uate the double integral.
3
yy
50.y8y2 ex4dxdy
3
0 sy
CAS
974
CHAPTER 15 MULTIPLE INTEGRALS
y
FIGURE 1
Pr, Px,y
a Rsr, 0 r 1, 0 2d
Recall from Figure 2 that the polar coordinates r,
15.4
DOUBLE INTEGRALS IN POLAR COORDINATES
Suppose that we want to evaluate a double integral xxR f x, y dA, where R is one of the regions shown in Figure 1. In either case the description of R in terms of rectangular coor dinates is rather complicated but R is easily described using polar coordinates.
yy 1
4
R 0x
R
0
of a point are related to the rect
1 x b Rsr, 1 r 2, 0 d
angular coordinates x, y by the equations
r
y Oxx
FIGURE 2
See Section 10.3.
The regions in Figure 1 are special cases of a polar rectangle
Rr, arb,
which is shown in Figure 3. In order to compute the double integral xxR f x, y dA, where R is a polar rectangle, we divide the interval a, b into m subintervals ri1, riof equal width rbam and we divide the interval,into n subintervalsj1, jof equal width n. Then the circles rri and the raysj divide the polar rectangle R into the small polar rectangles shown in Figure 4.
rb
R
Rij I
j
j1
ri,j
ra
a
rri
rri1
Dividing R into polar subrectangles
r2 x2 y2 xrcos yrsin
OaO
FIGURE 3 Polar rectangle FIGURE 4
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES The center of the polar subrectangle
975
has polar coordinates
Rij r, ri1 rri, j1 j ri 1ri1 ri j1 j1j
22
We compute the area of Rij using the fact that the area of a sector of a circle with radius r and central angle is 1 r 2 . Subtracting the areas of two such sectors, each of which has
central angle
2
jj1, we find that the area of Rij is
1212122
Ai 2ri2ri12riri1
1riri1riri1 rir 2
Although we have defined the double integral xxR f x, y dA in terms of ordinary rect angles, it can be shown that, for continuous functions f, we always obtain the same answer using polar rectangles. The rectangular coordinates of the center of Rij are ri cos j, ri sin j, so a typical Riemann sum is
mn mn
fricos j,risin jAifricos j,risin jrir
i1 j1 i1 j1
Ifwewritetr, rfrcos ,rsin ,thentheRiemannsuminEquation1canbewrit
ten as
mn
tri, j r
i1 j1
which is a Riemann sum for the double integral
1
Therefore we have
yy
R
yybtr, drd a
lim fri cos j, ri sin jAi m n
fx, y dA
lim m n tri, jr yybtr, drd
m,nl i1 j1
m,nl i1 j1 a
y yb frcos ,rsin rdrd a
CHANGE TO POLAR COORDINATES IN A DOUBLE INTEGRAL If f is con tinuousonapolarrectangleRgivenby0arb, ,where 02 ,then
yyfx,ydAy yb frcos ,rsin rdrd
2
R
a
976
CHAPTER 15 MULTIPLE INTEGRALS
The formula in 2 says that we convert from rectangular to polar coordinates in a double integral by writing xr cos and yr sin , using the appropriate limits ofintegration for r and , and replacing dA by r dr d . Be careful not to forget the additional factor r on the right side of Formula 2. A classical method for remembering this is shown in Figure 5, where the infinitesimal polar rectangle can be thought of as an ordinary rect
angle with dimensions r d
and dr and therefore has area dAr dr d .
FIGURE 5
O
EXAMPLE 1 Evaluate xx 3x4y 2dA, where R is the region in the upper halfplane R
r
dA
d
dr
rd
boundedbythecirclesx2 y2 1andx2 y2 4. SOLUTION The region R can be described as
Rx,yy0, 1×2 y2 4
It is the halfring shown in Figure 1b, and in polar coordinates it is given by 1r2,
0
. Therefore, by Formula 2, yy3x4y2dAyy23rcos 4r2sin2 rdrd
R
01
yy23r2cos 4r3sin2 drd 01
y 7 cos 0
yr3cos r4sin2d7cos 15sin2 d r1
00
r2 y
N Here we use the trigonometric identity
15 1cos 2d 2
sin2
1 1cos 22
1515 15
See Section 7.2 for advice on integrating trigonometric functions.
z 0,0,1
M7 sin24 sin 2 2
0
V EXAMPLE 2 Find the volume of the solid bounded by the plane z0 and the parabo
loidz1x2 y2.
SOLUTION If we put z0 in the equation of the paraboloid, we get x2y21. This means that the plane intersects the paraboloid in the circle x2y21, so the solid lies under the paraboloid and above the circular disk D given by x2y21 see Figures6and1a.InpolarcoordinatesDisgivenby0r1,0 2 .Since 1×2 y2 1r2,thevolumeis
Vyy1x2 y2dAy2 y1 1r2rdrd
y
00
y2 y1 3 r2 r4 1
x
FIGURE6
D
0 d 0rrdr2 24 2
0
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES977 If we had used rectangular coordinates instead of polar coordinates, then we would have
obtained
V 1×2 y2dA yy
y1 ys1x2 1 s1x 2
1×2 y2dydx
which is not easy to evaluate because it involves finding x1x232 dx. M
What we have done so far can be extended to the more complicated type of region shown in Figure 7. Its similar to the type II rectangular regions considered in Section 15.3. In fact, by combining Formula 2 in this section with Formula 15.3.5, we obtain the fol lowing formula.
D
D
rhTM a
If f is continuous on a polar region of the form
Dr,, h1 rh2
then yyfx,ydAyyh2 frcos ,rsin rdrd
3
D
h1
a
O rh
FIGURE 7
Dsr, a, h r hTMd
In particular, taking fx, y1, h1 0, and h2 hin this formula, we see thattheareaoftheregionDboundedby,,andrh is
ADyy 1 dAy yhr dr d
D
0
r2hy 1 2 y d2hd
and this agrees with Formula 10.4.3.
V EXAMPLE 3 Use a double integral to find the area enclosed by one loop of the four
leaved rose rcos 2 .
SOLUTION From the sketch of the curve in Figure 8, we see that a loop is given by the
1r2 d 1 202
cos22 d
20
4
4
region
So the area is
Dr, 4 4, 0rcos 2ADyydAy4 ycos2 rdrd
D
y 44
4 0 cos 2
y 44
FIGURE 8
1 y 4 44
1cos 4d
1 1 sin 444 4 4
8
M
978y
CHAPTER 15 MULTIPLE INTEGRALS
x11
or r2cos
V EXAMPLE 4 Find the volume of the solid that lies under the paraboloid zx2y2, above the xyplane, and inside the cylinder x2y22x.
SOLUTION The solid lies above the disk D whose boundary circle has equation x2y22x or, after completing the square,
x12 y2 1
See Figures 9 and 10. In polar coordinates we have x2y2r2 and xrcos , so
the boundary circle becomes r22rcos , or r2cos . Thus the disk D is given by Dr, 2 2, 0r2 cos
D 012x
FIGURE 9
z
and, by Formula 3, we have4 2 cos
Vyyx2 y2dAy2 y2cos r2rdrd y2 r d
x0
2
FIGURE 10
15.4 EXERCISES
14 A region R is shown. Decide whether to use polar coordinates orrectangularcoordinatesandwritexxR fx,ydAasaniterated integral, where f is an arbitrary continuous function on R.
56 Sketch the region whose area is given by the integral and eval uatetheintegral.
y
23sin 21 sin 422 3
M
5. y2 y7 r dr d 6. y 2 y4 cos r dr d 1.y2.y4 00
4 04x
1 y1
714 Evaluate the given integral by changing to polar coordinates.
7. xxD xy dA,
where D is the disk with center the origin and radius 3
8. xxR xy dA, where R is the region that lies to the left of the
yaxisbetweenthecirclesx2 y2 1andx2 y2 4
9. xx cosx2y2 dA, where R is the region that lies above the
1 0 1x 3.y4.y R 22
16 3
xaxis within the circle xy9 10.xxRs4x2y2 dA,
1 0 1x
0 x
whereRx,yx2 y2 4, x0
11. xx ex2y2 dA, where D is the region bounded by the
y 2 d8 0
4 y 2 cos d8 0
d
20 2 4
0
1cos 2 2
3 280222
D
y 2 442 cos
2 y 2 12 cos 21 1cos 4d
2
D
semicircle xs4y2 and the yaxis
12. xx yex dA, where R is the region in the first quadrant enclosed
R
bythecirclex2 y2 25
xxR arctan yx dA,
whereRx,y1x2 y2 4, 0yx
1518 Use a double integral to find the area of the region. One loop of the rose rcos 3
16. The region enclosed by the curve r43 cos
33.
34.
A swimming pool is circular with a 40ft diameter. The depth is constant along eastwest lines and increases linearly from
2 ft at the south end to 7 ft at the north end. Find the volume of water in the pool.
An agricultural sprinkler distributes water in a circular pattern of radius 100 ft. It supplies water to a depth of er feet per hour at a distance of r feet from the sprinkler.
a If 0R100, what is the total amount of water supplied
per hour to the region inside the circle of radius R centered
at the sprinkler?
b Determine an expression for the average amount of water
per hour per square foot supplied to the region inside the circle of radius R.
13.
14. xx x dA, where D is the region in the first quadrant that lies
15.
D 2 2 2
between the circles xy4 and xy2x
2
SECTION 15.4 DOUBLE INTEGRALS IN POLAR COORDINATES979
17. The region within both of the circles rcos
18. The region inside the cardioid r1cos
circle r3 cos
and rsin and outside the
35.
1927 Use polar coordinates to find the volume of the given solid.
19. Undertheconezsx2y2 andabovethediskx2 y2 4
20. Below the paraboloid z182×22y2 and above the x yplane
21. Enclosed by the hyperboloid x2y2z21 and the planez2
16 and outside the
Use polar coordinates to combine the sum
y1 yx ys2 yx y2 ys4x2 1s2 s1x2 xydydx 1 0 xydydx s2 0
xydydx
22. Inside the sphere xyz cylinder x2y24
222
lim yyex2y2dA al Da
where Da is the disk with radius a and center the origin. Show that
y y ex2y2dA
An equivalent definition of the improper integral in part a is
yy ex 2y 2dAlim yy ex 2y 2dA al
2 Sa
where Sa is the square with vertices a, a. Use this to
show that
Deduce that
23. A sphere of radius a
24. Bounded by the paraboloid z12×22y2 and the
plane z7 in the first octant
Above the cone zsx2y2 and below the sphere
x2 y2 z2 1
26. Bounded by the paraboloids z3x 23y 2 and
z4x2 y2
27. Inside both the cylinder x2y24 and the ellipsoid
4×2 4y2 z2 64
28. a A cylindrical drill with radius r1 is used to bore a hole through the center of a sphere of radius r2. Find the volume of the ringshaped solid that remains.
b Express the volume in part a in terms of the height h of the ring. Notice that the volume depends only on h, not on r1 or r2.
2932 Evaluate the iterated integral by converting to polar coordinates.
b
c
d
y ex2 dx y ey2 dy
y ex2dxs
36.
We define the improper integral over the entire plane 2 yy x2y2y y x2y2
into one double integral. Then evaluate the double integral.
a
I e dAe dydx 2
25.
29. y3 ys9x2 sinx2 y2dydx 3 0
31. y1 ys2y2 xydxdy 0 y
30. ya y0 x2ydxdy 0 sa2y2
32. y2 ys2xx2 sx2 y2 dydx 0 0
By making the change of variable ts2 x, show that y ex22 dxs2
This is a fundamental result for probability and statistics.
37.
Use the result of Exercise 36 part c to evaluate the following integrals.
a y x2ex2 dx b y sx ex dx 00
980
CHAPTER 15 MULTIPLE INTEGRALS
15.5
APPLICATIONS OF DOUBLE INTEGRALS
We have already seen one application of double integrals: computing volumes. Another geometric application is finding areas of surfaces and this will be done in Section 16.6. In this section we explore physical applications such as computing mass, electric charge, cen ter of mass, and moment of inertia. We will see that these physical ideas are also impor tant when applied to probability density functions of two random variables.
DENSITY AND MASS
In Section 8.3 we were able to use single integrals to compute moments and the center of mass of a thin plate or lamina with constant density. But now, equipped with the double integral, we can consider a lamina with variable density. Suppose the lamina occupies a region D of the xyplane and its density in units of mass per unit area at a point x, y in D is given by x, y, where is a continuous function on D. This means that
x,ylim m A
where m and A are the mass and area of a small rectangle that contains x, y and the limit is taken as the dimensions of the rectangle approach 0. See Figure 1.
To find the total mass m of the lamina, we divide a rectangle R containing D into sub rectangles Rij of equal size as in Figure 2 and consider x, y to be 0 outside D. If we choose a point xij , yijin Rij, then the mass of the part of the lamina that occupies Rij is approximately xij , yijA, where A is the area of Rij. If we add all such masses, we get an approximation to the total mass:
kl
m x i j , y i j A
i1 j1
If we now increase the number of subrectangles, we obtain the total mass m of the lamina
as the limiting value of the approximations:
Physicists also consider other types of density that can be treated in the same manner. For example, if an electric charge is distributed over a region D and the charge density in units of charge per unit area is given by x, y at a point x, y in D, then the total charge Q is given by
Qyy x,ydA D
EXAMPLE 1 Charge is distributed over the triangular region D in Figure 3 so that the charge density at x, y is x, yxy, measured in coulombs per square meter Cm2 . Find the total charge.
y
x,y D
0x
FIGURE 1
y
0x
FIGURE 2
x,y R ijij ij
1
m lim k l xij,yijAyy x,ydA
2
k,ll i1 j1
D
y 1
0x
FIGURE 3
Qyy x,ydAy1y1
y1 D
1, 1
xydydx
dx 1 x121x2dx
SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS SOLUTION From Equation 2 and Figure 3 we have
981
Thus the total charge is 5 C. 24
MOMENTS AND CENTERS OF MASS
M
y
0 2 y1x
D
1 xy
y
0 1x 2 y1
y1x
11 2 3 12×3 x41 5 202xxdx2 34 24
y
0 2
0
3
Mxlim yij xij,yijA y x,ydA m n yy
D
m,nl i1j1
x,y
D
4
Mylim xij xij,yijA x x,ydA m n yy
D
m,nl i1j1
FIGURE 4
In Section 8.3 we found the center of mass of a lamina with constant density; here we con sider a lamina with variable density. Suppose the lamina occupies a region D and has den sity function x, y. Recall from Chapter 8 that we defined the moment of a particle about an axis as the product of its mass and its directed distance from the axis. We divide D into small rectangles as in Figure 2. Then the mass of Rij is approximately xij , yijA, so we can approximate the moment of Rij with respect to the xaxis by
xij,yijAyij
If we now add these quantities and take the limit as the number of subrectangles becomes
large, we obtain the moment of the entire lamina about the xaxis:
Similarly, the moment about the yaxis is
As before, we define the center of mass x, y so that mxMy and myMx. The physi cal significance is that the lamina behaves as if its entire mass is concentrated at its center of mass. Thus the lamina balances horizontally when supported at its center of mass see Figure 4.
The coordinates x, y of the center of mass of a lamina occupying the region D and having density function x, y are
xMy 1yyx x,ydA yMx 1yyy x,ydA mmD mmD
5
where the mass m is given by
myy x,ydA D
982
CHAPTER 15 MULTIPLE INTEGRALS
V EXAMPLE 2 Find the mass and center of mass of a triangular lamina with vertices
0, 0, 1, 0, and 0, 2 if the density function is x, y13xy.
SOLUTION The triangle is shown in Figure 5. Note that the equation of the upper boundary
y 0,2
is y22x. The mass of the lamina is
myy x,ydAy1y22x13xydydx
y22x
311 D 00
8 , 16
2 y22x y
1 y3xy y
01,0x 0 2y0 0 303
D
FIGURE 5
y
3 1 dx4 11x2dx4 xx 8
Then the formulas in 5 give
x1yyx x,ydA3y1y22xx3x2xydydx m 800
2 y22 x y
D
3 1 xy3x2yx y
8 0 2 y0
y
1 xx3dx 0
y1yyy x,ydA3y1y22xy3xyy2dydx m 800
dx3 2
3 x2 x4 1 3 2248
0
D
3 y1 y2 y2 y3 y22x
3xdx1
y1 0
79x3x2 5x3dx
80 2 2 3y0 1 x2 x41
4
4 7×9 2×35 4 The center of mass is at the point 3, 11 .
0
1116
8 16
M
y a
a
V EXAMPLE 3 The density at any point on a semicircular lamina is proportional to the distance from the center of the circle. Find the center of mass of the lamina.
SOLUTION Lets place the lamina as the upper half of the circle x2y2a2. See Fig ure 6. Then the distance from a point x, y to the center of the circle the origin is sx2y2 . Therefore the density function is
x,yK x2y2 s
D
a 0 ax
FIGURE 6
0, 3a2
where K is some constant. Both the density function and the shape of the lamina suggest that we convert to polar coordinates. Then sx2y2r and the region D is given by 0ra,0.Thusthemassofthelaminais
m x,ydA Ksx2 y2 dA Krrdrd yy yyyya
00 yya2 r3a Ka3
DD
K 0 d 0 r drK 33
0
Both the lamina and the density function are symmetric with respect to the yaxis, so the
SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS983 center of mass must lie on the yaxis, that is, x0. The ycoordinate is given by
y1yyyx,ydA 3 yyarsinrrdrd mD Ka3 0 0
3 ysin d yar3dr 3 cos 0r4a a30 0 a3 40
N Compare the location of the center of mass in Example 3 with Example 4 in Section 8.3, where we found that the center of mass of a lamina with the same shape but uniform density is
32a4 3a a3 42
Therefore the center of mass is located at the point 0, 3a2 MOMENT OF INERTIA
. M
located at the point 0, 4a3
.
The moment of inertia also called the second moment of a particle of mass m about an axis is defined to be mr2, where r is the distance from the particle to the axis. We extend this concept to a lamina with density function x, y and occupying a region D by pro ceeding as we did for ordinary moments. We divide D into small rectangles, approximate the moment of inertia of each subrectangle about the xaxis, and take the limit of the sum as the number of subrectangles becomes large. The result is the moment of inertia of the lamina about the xaxis:
Similarly, the moment of inertia about the yaxis is
It is also of interest to consider the moment of inertia about the origin, also called the polar moment of inertia:
Note that I0IxIy.
V EXAMPLE 4 Find the moments of inertia Ix, Iy, and I0 of a homogeneous disk D with
density x, y, center the origin, and radius a.
SOLUTION The boundary of D is the circle x2y2a2 and in polar coordinates D is
6
7
Ix lim m n yij2 xij,yijAyyy2 x,ydA
m,nl i1j1
D
Iy lim m n xij2 xij,yijAyyx2 x,ydA
8
m,nl i1j1
D
I0 lim m n xij2yij2xij,yijAyyx2y2x,ydA
m,nl i1j1
D
984
CHAPTER 15 MULTIPLE INTEGRALS
describedby0 2 ,0ra.LetscomputeI0 first:
I0yyx2y2 dA y2 yar2rdrd
9
00 y2ya3 r4aa4
D
0d0rdr2 42
0
Instead of computing Ix and Iy directly, we use the facts that IxIyI0 and IxIy
from the symmetry of the problem. Thus
I0 a4
IxIy2 4 M
In Example 4 notice that the mass of the disk is
mdensityareaa2
so the moment of inertia of the disk about the origin like a wheel about its axle can be
written as
a4 1 2 2 1 2 I0 2 2 aa2ma
Thus if we increase the mass or the radius of the disk, we thereby increase the moment of inertia. In general, the moment of inertia plays much the same role in rotational motion that mass plays in linear motion. The moment of inertia of a wheel is what makes it diffi cult to start or stop the rotation of the wheel, just as the mass of a car is what makes it dif ficult to start or stop the motion of the car.
The radius of gyration of a lamina about an axis is the number R such that mR2I
where m is the mass of the lamina and I is the moment of inertia about the given axis. Equation 9 says that if the mass of the lamina were concentrated at a distance R from the axis, then the moment of inertia of this point mass would be the same as the moment of inertia of the lamina.
In particular, the radius of gyration y with respect to the xaxis and the radius of gyra tion x with respect to the yaxis are given by the equations
my2 Ix mx2 Iy
Thusx, yis the point at which the mass of the lamina can be concentrated without chang ing the moments of inertia with respect to the coordinate axes. Note the analogy with the center of mass.
10
V EXAMPLE 5 Find the radius of gyration about the xaxis of the disk in Example 4. SOLUTION As noted, the mass of the disk is ma2, so from Equations 10 we have
Ix 1 a4 a2 y24
m a2 4
Therefore the radius of gyration about the xaxis is y1 a , which is half the radius of
the disk.
2
M
PROBABILITY
SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS985
In Section 8.5 we considered the probability density function f of a continuous random variable X. This means that f x0 for all x, x f x dx1, and the probability that X
lies between a and b is found by integrating f from a to b: PaXbyb fxdx
a
Now we consider a pair of continuous random variables X and Y, such as the lifetimes of two components of a machine or the height and weight of an adult female chosen at ran dom. The joint density function of X and Y is a function f of two variables such that the probability that X, Y lies in a region D is
PX, Y Dyy f x, y dA D
In particular, if the region is a rectangle, the probability that X lies between a and b and Y lies between c and d is
See Figure 7.
FIGURE 7
PaXb, cYdyb yd fx,ydydx ac
z
a
zfx, y c
The probability that X lies between bd a and b and Y lies between c and d xD
is the volume that lies above the rectangle Da, bxc, d and below the graph of the joint density function.
y
Because probabilities arent negative and are measured on a scale from 0 to 1, the joint density function has the following properties:
f x, y0 yy f x, y dA1 2
As in Exercise 36 in Section 15.4, the double integral over 2 is an improper integral defined as the limit of double integrals over expanding circles or squares and we can write
yyfx,ydAy y fx,ydxdy1
2
986
CHAPTER 15 MULTIPLE INTEGRALS
EXAMPLE 6 If the joint density function for X and Y is given by
fx,yCx2y if 0x10, 0y10 0 otherwise
find the value of the constant C. Then find PX7, Y2.
SOLUTION We find the value of C by ensuring that the double integral of f is equal to 1. Because f x, y0 outside the rectangle 0, 100, 10, we have
y y y10y10 fx, y dy dx
0 0
0
y10 0
y10 xyy2
Therefore1500C1andsoC 1 . 1500
Cx2y dy dxC Cy10 10x100dx1500C
y0
dx
Now we can compute the probability that X is at most 7 and Y is at least 2: PX7,Y2y7 y fx,ydydxy7 y10 1 x2ydydx
2 0 2 1500
1 y7 xyy2y10dx 1 y7 8x96dx
8680.5787 1500
1500 0 0
ft0
where is the mean waiting time. In the next example we consider a situation with two
independent waiting times.
EXAMPLE 7 The manager of a movie theater determines that the average time movie goers wait in line to buy a ticket for this weeks film is 10 minutes and the average time they wait to buy popcorn is 5 minutes. Assuming that the waiting times are independent, find the probability that a moviegoer waits a total of less than 20 minutes before taking his or her seat.
SOLUTION Assuming that both the waiting time X for the ticket purchase and the waiting time Y in the refreshment line are modeled by exponential probability density functions, we can write the individual density functions as
y2 1500
if t0 1et if t0
f1x0 if x0 f2y0 if y0
1 ex10 if x0 1ey5 if y0 10 5
M
Suppose X is a random variable with probability density function f1x and Y is a ran dom variable with density function f2y. Then X and Y are called independent random variables if their joint density function is the product of their individual density functions:
fx, yf1x f2y
In Section 8.5 we modeled waiting times by using exponential density functions
y 20
0
We are asked for the probability that XY20:
SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS Since X and Y are independent, the joint density function is the product:
if x0, y0 otherwise
PXY20PX, YD where D is the triangular region shown in Figure 8. Thus
987
1 ex10ey5 fx, yf1xf2y50
xy20 D
PXY20yyfx,ydAy20y20x 1ex10ey5dydx 020x D0050
1 y20 ex105ey5y20x dx
10 0
1 y20 ex10e4ex10dx 10 0
FIGURE 8
y0
1 y20 ex101ex205dx
50 0
1e4 2e2 0.7476
This means that about 75 of the moviegoers wait less than 20 minutes before taking
their seats. M EXPECTED VALUES
Recall from Section 8.5 that if X is a random variable with probability density function f, then its mean is
y xfxdx
Now if X and Y are random variables with joint density function f, we define the Xmean and Ymean, also called the expected values of X and Y, to be
1 yyxfx,ydA 2
Notice how closely the expressions for 1 and
of a lamina with density function in Equations 3 and 4. In fact, we can think of proba bility as being like continuously distributed mass. We calculate probability the way we cal culate massby integrating a density function. And because the total probability mass is 1, the expressions for x and y in 5 show that we can think of the expected values of X and Y, 1 and 2, as the coordinates of the center of mass of the probability distribution.
In the next example we deal with normal distributions. As in Section 8.5, a single ran dom variable is normally distributed if its probability density function is of the form
fx 1 ex 22 2 s2
where is the mean and is the standard deviation.
11
2 yyyfx,ydA 2
2 in 11 resemble the moments Mx and My
988
CHAPTER 15 MULTIPLE INTEGRALS
1500 1000 500 0
5.95 6 4x y 6.05 4.05
FIGURE 9
Graph of the bivariate normal joint density function in Example 8
3.95
f1x1 ex420.0002 f2y1 e y620.0002 0.01 s2 0.01 s2
Since X and Y are independent, the joint density function is the product:
1 fx, yf1xf2y0.0002
15.5 EXERCISES
1. Electric charge is distributed over the rectangle 1×3, 0y2 so that the charge density at x, y is
x, y2xyy2 measured in coulombs per square meter. Find the total charge on the rectangle.
4. Dx, y 0xa, 0yb;
5. D is the triangular region with vertices 0, 0, 2, 1, 0, 3;
22 2. Electric charge is distributed over the disk xy
4 so thatthechargedensityatx,yis x,yxyx2 y2
and2xy6; x,yx2
7. Disboundedbyyex,y0,x0,andx1; x,yy 8. Disboundedbyysx,y0,andx1; x,yx
9. Dx, y 0ysin xL, 0xL; x, yy
10. D is bounded by the parabolas yx2 and xy2; x, ysx
measured in coulombs per square meter. Find the total charge on the disk.
310 Find the mass and center of mass of the lamina that occupies the region D and has the given density function .
3. Dx,y0x2,1y1; x,yxy2
EXAMPLE 8 A factory produces cylindrically shaped roller bearings that are sold as having diameter 4.0 cm and length 6.0 cm. In fact, the diameters X are normally distrib uted with mean 4.0 cm and standard deviation 0.01 cm while the lengths Y are normally distributed with mean 6.0 cm and standard deviation 0.01 cm. Assuming that X and Y are independent, write the joint density function and graph it. Find the probability that a bearing randomly chosen from the production line has either length or diameter that differs from the mean by more than 0.02 cm.
SOLUTION We are given that X and Y are normally distributed with 14.0, 26.0, and 120.01. So the individual density functions for X and Y are
e e 5000 5000×4 y6
e
22
A graph of this function is shown in Figure 9.
Lets first calculate the probability that both X and Y differ from their means by less
than 0.02 cm. Using a calculator or computer to estimate the integral, we have
P3.98X4.02, 5.98Y6.02y4.02 y6.02 fx, y dy dx 3.98 5.98
5000 y4.02 y6.02 e5000x42 y62 dy dx 3.98 5.98
0.91
Then the probability that either X or Y differs from its mean by more than 0.02 cm is
approximately
10.910.09
M
22 x4 0.0002 y6 0.0002
x, ycxy
6. D is the triangular region enclosed by the lines x0, yx,
x, yxy
11. A lamina occupies the part of the disk x2y21 in the first quadrant. Find its center of mass if the density at any point is proportional to its distance from the xaxis.
12. Find the center of mass of the lamina in Exercise 11 if the density at any point is proportional to the square of its distance from the origin.
13. The boundary of a lamina consists of the semicircles ys1x2 andys4x2 togetherwiththeportions of the xaxis that join them. Find the center of mass of the lamina if the density at any point is proportional to its dis tance from the origin.
14. Find the center of mass of the lamina in Exercise 13 if the density at any point is inversely proportional to its distance from the origin.
Find the center of mass of a lamina in the shape of an isos celes right triangle with equal sides of length a if the density at any point is proportional to the square of the distance from the vertex opposite the hypotenuse.
16. A lamina occupies the region inside the circle x2y22y but outside the circle x2y21. Find the center of mass if the density at any point is inversely proportional to its dis tance from the origin.
17. Find the moments of inertia Ix , Iy , I0 for the lamina of Exercise 7.
18. Find the moments of inertia Ix , Iy , I0 for the lamina of Exercise 12.
19. Find the moments of inertia Ix , Iy , I0 for the lamina of Exercise 15.
20. Consider a square fan blade with sides of length 2 and the lower left corner placed at the origin. If the density of the blade is x, y10.1x, is it more difficult to rotate the blade about the xaxis or the yaxis?
CAS 2122 Use a computer algebra system to find the mass, center of mass, and moments of inertia of the lamina that occupies the region D and has the given density function.
SECTION 15.5 APPLICATIONS OF DOUBLE INTEGRALS989 The joint density function for a pair of random variables X
28.
29.
30.
CAS 31.
32.
a Verifythat
4xy if 0x1, 0y1
is a joint density function.
b If X and Y are random variables whose joint density func
tion is the function f in part a, find
i PX1 ii PX1,Y1
27.
15.
andYis
fx,y Cx1y if 0x1, 0y2
0 otherwise
a Find the value of the constant C. b Find PX1, Y1.
c FindPXY1.
f x, y0 otherwise
function
222
c Find the expected values of X and Y.
Suppose X and Y are random variables with joint density
0.1e0.5×0.2y if x0, y0 0 otherwise
a Verify that f is indeed a joint density function. b Find the following probabilities.
i PY1 ii PX2,Y4 c Find the expected values of X and Y.
a A lamp has two bulbs of a type with an average lifetime of 1000 hours. Assuming that we can model the proba bility of failure of these bulbs by an exponential density function with mean1000, find the probability that both of the lamps bulbs fail within 1000 hours.
b Another lamp has just one bulb of the same type as in part a. If one bulb burns out and is replaced by a bulb of the same type, find the probability that the two bulbs fail within a total of 1000 hours.
Suppose that X and Y are independent random variables, where X is normally distributed with mean 45 and standard deviation 0.5 and Y is normally distributed with mean 20 and standard deviation 0.1.
a Find P40X50, 20Y25.
b Find P4X452100Y2022.
Xavier and Yolanda both have classes that end at noon and they agree to meet every day after class. They arrive at the coffee shop independently. Xaviers arrival time is X and Yolandas arrival time is Y, where X and Y are measured in minutes after noon. The individual density functions are
fx, y
21. Dx,y0ysinx, 0x ; 22. Disenclosedbythecardioidr1cos ;
x,ysx2 y2
x,yxy
CAS 2326 A lamina with constant density x, yoccupies the given region. Find the moments of inertia Ix and Iy and the radii of gyration x and y.
23. The rectangle 0xb, 0yh
24. The triangle with vertices 0, 0, b, 0, and 0, h
25. The part of the disk x2y2a2 in the first quadrant
26. Theregionunderthecurveysinxfromx0tox
f1x
ex 0
if x0 if x0
f2y
1 y 50
0
if 0y10 otherwise
Xavier arrives sometime after noon and is more likely to arrive promptly than late. Yolanda always arrives by 12:10 P M and is more likely to arrive late than promptly. After Yolanda arrives, shell wait for up to half an hour for Xavier, but he wont wait for her. Find the probability that they meet.
990CHAPTER 15 MULTIPLE INTEGRALS
33. When studying the spread of an epidemic, we assume that the probability that an infected individual will spread the disease to an uninfected individual is a function of the distance between them. Consider a circular city of radius 10 mi in which the population is uniformly distributed. For an uninfected indi vidual at a fixed point Ax0, y0 , assume that the probability function is given by
a Suppose the exposure of a person to the disease is the sum of the probabilities of catching the disease from all members of the population. Assume that the infected people are uniformly distributed throughout the city, with k infected individuals per square mile. Find a double integral that represents the exposure of a person residing at A.
b Evaluate the integral for the case in which A is the center of the city and for the case in which A is located on the edge of the city. Where would you prefer to live?
f P1 20dP, A 20
where dP, A denotes the distance between P and A.
15.6
TRIPLE INTEGRALS
z
B
x
z
x
y
Just as we defined single integrals for functions of one variable and double integrals for functions of two variables, so we can define triple integrals for functions of three variables. Lets first deal with the simplest case where f is defined on a rectangular box:
Bx,y,zaxb, cyd, rzs
The first step is to divide B into subboxes. We do this by dividing the interval a, b into l subintervals xi1, xi of equal width x, dividing c, d into m subintervals of width y, and dividing r, s into n subintervals of width z. The planes through the endpoints of these subintervals parallel to the coordinate planes divide the box B into lmn subboxes
Bijkxi1, xiyj1, yjzk1, zk
which are shown in Figure 1. Each subbox has volume Vx y z.
Then we form the triple Riemann sum
integral 15.1.5, we define the triple integral as the limit of the triple Riemann sums in 2.
Again, the triple integral always exists if f is continuous. We can choose the sample point to be any point in the subbox, but if we choose it to be the point xi, yj, zkwe get a simplerlooking expression for the triple integral:
Just as for double integrals, the practical method for evaluating triple integrals is to express them as iterated integrals as follows.
Bijk
Iz Iy Ix
l m n i1 j1 k1
fx ij k , y i j k , z ij k V
where the sample point xijk, yijk, zijkis in Bi jk. By analogy with the definition of a double
3
DEFINITION The triple integral of f over the box B is
if this limit exists.
yyy
f x, y, z dV
lim f xijk, yijk, zijkV l m n
B
l, m, nl i1 j1 k1
FIGURE 1
y
1
2
fx,y,zdV lim fxi,yj,zkV yyy l m n
B
l, m, nl i1 j1 k1
yyyfx,y,zdVyb ys yd fx,y,zdydzdx
arc
V EXAMPLE 1 Evaluate the triple integral xxx xyz2 dV, where B is the rectangular box
B
integrate with respect to x, then y, and then z, we obtain
x1 x0
yyy 2 B
y3y2 y1
3 2 yz2
x2yz2 2
0 y1
dydz
xyzdV0 1 0xyzdxdydz
y3 3z z 27 0 4dz4 4
M
SECTION 15.6 TRIPLE INTEGRALS991
FUBINIS THEOREM FOR TRIPLE INTEGRALS If f is continuous on the rectan gular box Ba, bc, dr, s, then
yyyfx,y,zdVys yd yb fx,y,zdxdydz
4
B
rca
The iterated integral on the right side of Fubinis Theorem means that we integrate first with respect to x keeping y and z fixed, then we integrate with respect to y keeping z fixed, and finally we integrate with respect to z. There are five other possible orders in which we can integrate, all of which give the same value. For instance, if we integrate with respect to y, then z, and then x, we have
B Bx,y,z0x1, 1y2, 0z3
given by
SOLUTION We could use any of the six possible orders of integration. If we choose to
2 y3y2
012dydz 4 dz y
yy 3 233
0 1 y2z2 y2
0
x
z
E
0
D
zuTMx,y
zux,y
y
Now we define the triple integral over a general bounded region E in three dimensional space a solid by much the same procedure that we used for double integrals 15.3.2. We enclose E in a box B of the type given by Equation 1. Then we define a function F so that it agrees with f on E but is 0 for points in B that are outside E. By definition,
yyyfx, y, z dVyyyFx, y, z dV EB
This integral exists if f is continuous and the boundary of E is reasonably smooth. The triple integral has essentially the same properties as the double integral Properties 69 in Section 15.3.
We restrict our attention to continuous functions f and to certain simple types of regions. A solid region E is said to be of type 1 if it lies between the graphs of two continuous func tions of x and y, that is,
Ex,y,zx,yD, u1x,yzu2x,y
where D is the projection of E onto the xyplane as shown in Figure 2. Notice that the upper boundary of the solid E is the surface with equation zu2x, y, while the lower boundary is the surface zu1x, y.
FIGURE 2
A type 1 solid region
5
992
CHAPTER 15 MULTIPLE INTEGRALS
By the same sort of argument that led to Formula 15.3.3, it can be shown that if E is a
z
zu TMx, y
type 1 region given by Equation 5, then
The meaning of the inner integral on the right side of Equation 6 is that x and y are held fixed, and therefore u1x, y and u2x, y are regarded as constants, while f x, y, z is inte grated with respect to z.
In particular, if the projection D of E onto the xyplane is a type I plane region as in Figure 3, then
6
E
xb
Ex,y,z axb, t1xyt2x, u1x,yzu2x,y and Equation 6 becomes
If, on the other hand, D is a type II plane region as in Figure 4, then Ex,y,zcyd, h1yxh2y, u1x,yzu2x,y
and Equation 6 becomes
d
y
xhTMy
zux,y a0
ygx D ygTMx y
FIGURE 3
A type 1 solid region where the projection D is a type I plane region
z
E
0c
x
FIGURE 4
7
yyyfx, y, z dVyyyu2x, y fx, y, z dzdA
u1x, y
ED
yyyfx, y, z dVyb yt2x yu2x, y fx, y, z dz dy dx
E
zuTMx,y zux,y
xhy
D
A type 1 solid region with a type II projection
EXAMPLE 2 Evaluate xxx z dV, where E is the solid tetrahedron bounded by the four E
8
planes x0, y0, z0, and xyz1.
a t1x u1x, y
yyyfx, y, z dVyd yh2y yu2x, y fx, y, z dz dx dy
c h1y u1x, y
E
SOLUTION When we set up a triple integral its wise to draw two diagrams: one of the solid region E see Figure 5 and one of its projection D on the xyplane see Figure 6. The lower boundary of the tetrahedron is the plane z0 and the upper
z
z1xy
y
0, 0, 1
1
y1x
ED
0
0, 1, 0
z0 y 0 y0 1 x
1,0,0 x
FIGURE 5
FIGURE 6
boundary is the plane xyz1 or z1xy, so we use u1x, y0 and u2x, y1xy in Formula 7. Notice that the planes xyz1 and z0 intersect in the line xy1 or y1x in the xyplane. So the projection of E is the triangular region shown in Figure 6, and we have
Ex,y,z0x1, 0y1x, 0z1xy
This description of E as a type 1 region enables us to evaluate the integral as follows:
9
10
yyy
E
y1 y1x y1xy zdV0 0 0
y1 y1x z2 z1xy
2 dydx 00 z0
SECTION 15.6 TRIPLE INTEGRALS993
zdzdydx
2 1 y1 1xy3y1x
1 y1 y1x2 0 0
1×3 dx
1xy dy dx23 dx 0 y0
1
6 4 0 24
y1 60
11×4 1 1
M
A solid region E is of type 2 if it is of the form Ex,y,zy,zD, u1y,zxu2y,z
where, this time, D is the projection of E onto the yzplane see Figure 7. The back sur face is xu1y, z, the front surface is xu2y, z, and we have
yyyfx, y, z dVyyyu2y, z fx, y, z dxdA
z 0
xEy xuy,z
xuTMy,z
FIGURE 7 A type 2 region
Finally, a type 3 region is of the form
yuTMx,z
E
yux, z
ED
u1 y, z
z
D
0
x
D
y
11
FIGURE 8 A type 3 region
Ex,y,zx,zD, u1x,zyu2x,z
where D is the projection of E onto the xzplane, yu1x, z is the left surface, and
yu2x, z is the right surface see Figure 8. For this type of region we have
yyyfx, y, z dVyyyu2x, z fx, y, z dydA
ED
u1x, z
994
CHAPTER 15 MULTIPLE INTEGRALS
In each of Equations 10 and 11 there may be two possible expressions for the integral depending on whether D is a type I or type II plane region and corresponding to Equa tions 7 and 8.
V EXAMPLE 3 Evaluate xxxE sx2z2 dV, where E is the region bounded by the parabo loidyx2 z2 andtheplaney4.
SOLUTION The solid E is shown in Figure 9. If we regard it as a type 1 region, then we need to consider its projection D1 onto the xyplane, which is the parabolic region in Figure10.Thetraceofyx2 z2 intheplanez0istheparabolayx2.
TEC Visual 15.6 illustrates how solid regions including the one in Figure 9 project onto coordinate planes.
FIGURE 9
Region of integration
y
D
y
0x FIGURE 10
Projection on xyplane z
z4
Projection on xzplane
Themostdifficultstepinevaluatingatriple integral is setting up an expression for the region of integration such as Equation 9 in Example 2. Remember that the limits of integration in the inner integral contain at most two variables, the limits of integration in the middle integral con tain at most one variable, and the limits of inte gration in the outer integral must be constants.
z
0
x
yz E
4y
y4
Fromyx2 z2 weobtainzsyx2,sothelowerboundarysurfaceofEis zsyx2 andtheuppersurfaceiszsyx2.ThereforethedescriptionofEas a type 1 region is
Ex,y,z2x2,x2y4,syx2 zsyx2 and so we obtain
yyysx2z2dVy2 y4ysyx2
E
sx2z2dzdydx
2 x2 syx2
D
x2z2 Although this integral could be written as
yy
Although this expression is correct, it is extremely difficult to evaluate. So lets instead consider E as a type 3 region. As such, its projection D3 onto the xzplane is the diskx2 z2 4showninFigure11.
Then the left boundary of E is the paraboloid yx2z2 and the right boundary is theplaney4,sotakingu1x,zx2 z2 andu2x,z4inEquation11,wehave
yyysx2 z2 dVyy y4 sx2 z2 dy dA 4×2 z2sx2 z2 dA
0x
D3
2 2 E D3
FIGURE 11
y2 ys4x2 4×2 z2sx2z2 dzdx 2 s4x 2
its easier to convert to polar coordinates in the xzplane: xr cos gives
yyysx2 z2 dVyy4x2 z2sx2 z2 dA
, zr sin
. This
E
D3
4r2r r dr d y2 y2
y2 0
d
y2 0
4r2r4 dr
0 0
128
15
2 4r 3
3
5 2
r
5 0
M
APPLICATIONS OF TRIPLE INTEGRALS
Recall that if f x0, then the single integral xb f x dx represents the area under the a
curve yfx from a to b, and if fx, y0, then the double integral xxD fx, y dA rep resents the volume under the surface zf x, y and above D. The corresponding inter pretation of a triple integral xxxE f x, y, z dV, where f x, y, z0, is not very useful because it would be the hypervolume of a fourdimensional object and, of course, that is very difficult to visualize. Remember that E is just the domain of the function f ; the graph of f lies in fourdimensional space. Nonetheless, the triple integral xxxE f x, y, z dV can be interpreted in different ways in different physical situations, depending on the phys ical interpretations of x, y, z and f x, y, z.
Lets begin with the special case where f x, y, z1 for all points in E. Then the triple integral does represent the volume of E:
12
For example, you can see this in the case of a type 1 region by putting f x, y, z1 in Formula6:
u2x,y dz dA
yyy1dVyy EDD
y
yy
u2x,yu1x,ydA
and from Section 15.3 we know this represents the volume that lies between the surfaces
zu1x, y and zu2x, y.
EXAMPLE 4 Use a triple integral to find the volume of the tetrahedron T bounded by the
planes x2yz2, x2y, x0, and z0.
SOLUTION The tetrahedron T and its projection D on the xyplane are shown in Figures 12 and 13. The lower boundary of T is the plane z0 and the upper boundary is the plane x2yz2, that is, z2x2y.
z
0, 0, 2
T
0
y 1
x2y2
or y1 x 2 D 1, 21
yx 2
u1x, y
SECTION 15.6 TRIPLE INTEGRALS995
x2yz2 0, 1, 0
1, 21 , 0
x
Therefore we have
x2y
FIGURE 12
01x
FIGURE 13
y
VTyyydVy1 y1x2 y2x2y dzdydx
0 x2 0
y1 y1x2 2x2ydydx1
T
VEyyy dV E
0 x2 3 by the same calculation as in Example 4 in Section 15.3.
996
CHAPTER 15 MULTIPLE INTEGRALS
Notice that it is not necessary to use triple integrals to compute volumes. They
simply give an alternative method for setting up the calculation. M
All the applications of double integrals in Section 15.5 can be immediately extended to triple integrals. For example, if the density function of a solid object that occupies the region E is x, y, z, in units of mass per unit volume, at any given point x, y, z, then its mass is
myyy x,y,zdV E
and its moments about the three coordinate planes are
Myzyyyx x,y,zdV Mxzyyyy x,y,zdV
EE
Mxy yyyz x,y,zdV E
The center of mass is located at the point x, y, z, where
xMyz yMxz zMxy
mmm
If the density is constant, the center of mass of the solid is called the centroid of E. The moments of inertia about the three coordinate axes are
Ix yyyy2 z2 x,y,zdV Iy yyyx2 z2 x,y,zdV EE
Iz yyyx2 y2 x,y,zdV E
As in Section 15.5, the total electric charge on a solid object occupying a region E and having charge density x, y, z is
Qyyy x,y,zdV E
If we have three continuous random variables X, Y, and Z, their joint density function is a function of three variables such that the probability that X, Y, Z lies in E is
PX, Y, Z E yyy f x, y, z dV E
In particular,
PaXb, cYd, rZsyb yd ys fx,y,zdzdydx
acr
The joint density function satisfies
fx,y,z0 y y y fx,y,zdzdydx1
13
14
15
16
zx
z
SECTION 15.6 TRIPLE INTEGRALS997 V EXAMPLE 5 Find the center of mass of a solid of constant density that is bounded by
theparaboliccylinderxy2 andtheplanesxz,z0,andx1.
SOLUTION The solid E and its projection onto the xyplane are shown in Figure 14. The lower and upper surfaces of E are the planes z0 and zx, so we describe E as a type 1 region:
Ex,y,z1y1, y2 x1, 0zx Then,ifthedensityis x,y,z ,themassis
myyy dVy1 y1yx dzdxdy 1 y2 0
y1 1y4dy 0
E
x
0 1
y
y
E 1 xy2
x D
x1
y1y1 y1 x2x1
1 y2xdxdy 2 dy
0x
y1 1y4dy 2 1
Therefore the center of mass is
y55
FIGURE 14
y51 4 0
Because of the symmetry of E and about the xzplane, we can immediately say that Mxz0 and therefore y0. The other moments are
Myzyyyx dVy1 y1yxx dzdxdy
1 y2 0
y1 y1 2
1 y2xdxdy
2 y1
1y6dy
E
y1 x3x1
3 dy
1 xy2
2y71
4 30 3707
y Mxyyyyz dVy1 y1yxz dzdxdy
E
1 y2 0
1 1 z dxdy y1 y1x2dxdy y y2 zx
1 y2 2 z0 2 1 y2y11y6dy2
307
x, y, zMyz , Mxz , Mxy5, 0, 5M
mmm
7 14
998CHAPTER 15 MULTIPLE INTEGRALS
15.6 EXERCISES
1. Evaluate the integral in Example 1, integrating first with
20. 21. 22.
The solid bounded by the cylinder yx2 and the planes z0,z4,andy9
The solid enclosed by the cylinder x2y29 and the planes yz5 and z1
a Express the volume of the wedge in the first octant that is cutfromthecylindery2 z2 1bytheplanesyxand x1 as a triple integral.
b Use either the Table of Integrals on Reference Pages 610 or a computer algebra system to find the exact value of the triple integral in part a.
respect to y, then z, and then x.
2. Evaluate the integral xxxx zy 3 dV, where
E
Ex,y,z1x1, 0y2, 0z1
22
using three different orders of integration. 38 Evaluate the iterated integral.
3. y1 yz yxz 6xzdydxdz 4. y1 y2x yy 2xyzdzdydx 000 0x0
The solid enclosed by the paraboloid xyz and the plane x16
5. 3 1 s1z2 zey dxdzdy 6. 1 z y zey2 dxdydz
yyy yyy CAS
CAS
000 000
7. y2 yy yx cosxyzdzdxdy 000
8. ys yx yxz x2 sinydydzdx 000
918 Evaluate the triple integral. 9. xxxE 2x dV, where
24. a In the Midpoint Rule for triple integrals we use a triple Riemann sum to approximate a triple integral over a box B, where f x, y, z is evaluated at the center xi, yj, zk
of the box Bi j k . Use the Midpoint Rule to estimate xxxBsx2y2z2 dV,whereBisthecubedefinedby 0x4, 0y4, 0z4.DivideBintoeight cubes of equal size.
b Use a computer algebra system to approximate the integral in part a correct to the nearest integer. Compare with the answer to part a.
2526 Use the Midpoint Rule for triple integrals Exercise 24 to estimate the value of the integral. Divide B into eight subboxes of equal size.
1
26. xxxB sinxy2z3 dV, where Bx,y,z0x4, 0y2, 0z1
2728 Sketch the solid whose volume is given by the iterated integral.
27. y1 y1x y22zdydzdx 28. y2 y2y y4y2 dxdzdy 000 000
2932 Express the integral xxxE f x, y, z dV as an iterated integral in six different ways, where E is the solid bounded by the given surfaces.
29. y4x2 4z2, y0
30. y2 z2 9, x2, x2
31. yx2, z0, y2z4
32. x2, y2, z0, xy2z2
Ex,y,z0y2, 0xs4y2, 0zy 10. xxx yzcosx5dV,where
E
Ex,y,z0x1, 0yx, xz2x
xxxE 6xydV, whereEliesundertheplanez1xy and above the region in the xyplane bounded by the curves ysx,y0,andx1
12. xxxE ydV, whereEisboundedbytheplanesx0,y0, z0, and 2x2yz4
25. xxxB ln1xyz dV,where Bx,y,z0x4, 0y8, 0z4
13. xxx x 2e y dV, where E is bounded by the parabolic cylinder E
z1y2 andtheplanesz0,x1,andx1
14. xxxE xy dV, where E is bounded by the parabolic cylinders
yx2 andxy2 andtheplanesz0andzxy 15. xxx x2 dV, where T is the solid tetrahedron with vertices
T
0, 0, 0, 1, 0, 0, 0, 1, 0, and 0, 0, 1
16. xxxT xyz dV, where T is the solid tetrahedron with vertices
0, 0, 0, 1, 0, 0, 1, 1, 0, and 1, 0, 1
17. xxxE x dV, where E is bounded by the paraboloid
x4y24z2 and the plane x4
18. xxxzdV, whereEisboundedbythecylindery2 z2 9
E andtheplanesx0,y3x,andz0inthefirstoctant
23.
11.
1922 Use a triple integral to find the volume of the given solid.
The tetrahedron enclosed by the coordinate planes and the plane 2xyz4
19.
33. The figure shows the region of integration for the integral y1 y1 y1y fx,y,zdzdydx
SECTION 15.6 TRIPLE INTEGRALS999 40. E is the tetrahedron bounded by the planes x0, y0,
z0, xyz1; x, y, zy
4144 Assume that the solid has constant density k.
41. Find the moments of inertia for a cube with side length L if one vertex is located at the origin and three edges lie along the coordinate axes.
42. Find the moments of inertia for a rectangular brick with dimen sions a, b, and c and mass M if the center of the brick is situ ated at the origin and the edges are parallel to the coordinate axes.
43. Find the moment of inertia about the zaxis of the solid cylin derx2 y2 a2, 0zh.
44. Find the moment of inertia about the zaxis of the solid cone sx2y2zh.
4546 Set up, but do not evaluate, integral expressions for
a the mass, b the center of mass, and c the moment of inertia about the zaxis.
45. The solid of Exercise 21; x, y, zsx2y2 46. Thehemispherex2 y2 z2 1, z0;
x,y,zsx2 y2 z2
CAS 47. Let E be the solid in the first octant bounded by the cylinder x2 y2 1andtheplanesyz,x0,andz0withthe densityfunction x,y,z1xyz.Useacomputer algebra system to find the exact values of the following quan tities for E.
a The mass
b The center of mass
c The moment of inertia about the zaxis
CAS 48. If E is the solid of Exercise 18 with density function
x, y, zx2y2, find the following quantities, correct
to three decimal places.
a The mass
b The center of mass
c The moment of inertia about the zaxis
49. The joint density function for random variables X, Y, and Z is fx,y,zCxyzif0x2, 0y2, 0z2,and f x, y, z0 otherwise.
a Find the value of the constant C. bFindPX1,Y1,Z1. c FindPXYZ1.
50. Suppose X, Y, and Z are random variables with joint density function f x, y, zCe0.5×0.2y0.1z if x0, y0, z0, and f x, y, z0 otherwise.
a Find the value of the constant C. bFindPX1,Y1.
c FindPX1,Y1,Z1.
0 sx 0
Rewrite this integral as an equivalent iterated integral in the
five other orders.
0
yx
five other orders.
z 1
z1y
1y
x
34. The figure shows the region of integration for the integral y1 y1x2 y1x
0 0 0 f x, y, z dy dz dx
Rewrite this integral as an equivalent iterated integral in the
z
1
z1
0
1y
3536 Write five other iterated integrals that are equal to the given iterated integral.
35. y1 y1 yy fx,y,zdzdxdy 0y0
36. y1 yx2 yy fx,y,zdzdydx 000
3740 Find the mass and center of mass of the solid E with the given density function .
37. E is the solid of Exercise 11; x, y, z2
38. E is bounded by the parabolic cylinder z1y2 and the
planes xz1, x0, and z0; x, y, z4
39. Eisthecubegivenby0xa, 0ya, 0za; x,y,zx2 y2 z2
x
1
y1x
1000CHAPTER 15 MULTIPLE INTEGRALS
5152 The average value of a function fx, y, z over a solid
is a density
52. Find the average value of the function fx, y, zx2zy2z over the region enclosed by the paraboloid z1x2y2 and the plane z0.
53. Find the region E for which the triple integral yyy1x2 2y2 3z2dV
E
region E is defined to be
fave1 yyyfx,y,zdV
VE E
where VE is the volume of E. For instance, if
function, then ave is the average density of E.
51.
Find the average value of the function f x, y, zxyz over the cube with side length L that lies in the first octant with one vertex at the origin and edges parallel to the coordinate axes.
is a maximum.
FIGURE 1
VOLUMES OF HYPERSPHERES
DISCOVERY PROJECT
In this project we find formulas for the volume enclosed by a hypersphere in ndimensional space.
1. Use a double integral and trigonometric substitution, together with Formula 64 in the Table of Integrals, to find the area of a circle with radius r.
2. Use a triple integral and trigonometric substitution to find the volume of a sphere with radius r.
3. Use a quadruple integral to find the hypervolume enclosed by the hypersphere
x2y2z2w2r2 in 4. Use only trigonometric substitution and the reduction formulas for x sinnx dx or x cosnx dx.
4. Use an ntuple integral to find the volume enclosed by a hypersphere of radius r in ndimensional space n. Hint: The formulas are different for n even and n odd.
15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES
y
O
r
Pr, Px,y y
In plane geometry the polar coordinate system is used to give a convenient description of certain curves and regions. See Section 10.3. Figure 1 enables us to recall the connection between polar and Cartesian coordinates. If the point P has Cartesian coordinates x, y and polar coordinates r, , then, from the figure,
xr cos yr sin
y
x x
r2x2y2 tan x
In three dimensions there is a coordinate system, called cylindrical coordinates, that is similar to polar coordinates and gives convenient descriptions of some commonly occur ring surfaces and solids. As we will see, some triple integrals are much easier to evaluate in cylindrical coordinates.
z
SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES1001 CYLINDRICAL COORDINATES
In the cylindrical coordinate system, a point P in threedimensional space is represented by the ordered triple r, , z, where r and are polar coordinates of the projection of P onto the xyplane and z is the directed distance from the xyplane to P. See Figure 2.
To convert from cylindrical to rectangular coordinates, we use the equations
whereas to convert from rectangular to cylindrical coordinates, we use
EXAMPLE 1
a Plot the point with cylindrical coordinates 2, 2 3, 1 and find its rectangular coordinates.
b Find cylindrical coordinates of the point with rectangular coordinates 3, 3, 7.
SOLUTION
a The point with cylindrical coordinates 2, 2 3, 1 is plotted in Figure 3. From
Pr, ,z
zr
FIGURE 2
The cylindrical coordinates of a point
z
0
2 3
FIGURE 3
z
O
y x r, ,0
1
2
r2x2y2 tany zz x
2, 2, 1 3
1 2
y
Equations 1, its rectangular coordinates are x2 cos 221
1 y2 sin 22 s3s3
3 2 32
x
z1
Thus the point is 1, s3 , 1 in rectangular coordinates.
b From Equations 2 we have
rs32 32 3s2
tan31 so72n
c, 0, 0 x
FIGURE 4
rc, a cylinder
0
y
Therefore one set of cylindrical coordinates is 3 s2 , 7 4, 7. Another is
3s2, 4,7.Aswithpolarcoordinates,thereareinfinitelymanychoices. M
Cylindrical coordinates are useful in problems that involve symmetry about an axis, and the zaxis is chosen to coincide with this axis of symmetry. For instance, the axis of the circular cylinder with Cartesian equation x2y2c2 is the zaxis. In cylindrical coordi nates this cylinder has the very simple equation rc. See Figure 4. This is the reason for the name cylindrical coordinates.
0, c, 0
xr cos yr sin zz
34 z7
1002
x
FIGURE 5
zr, a cone z
CHAPTER 15 MULTIPLE INTEGRALS
V EXAMPLE 2 Describe the surface whose equation in cylindrical coordinates is zr.
z
0
SOLUTION The equation says that the zvalue, or height, of each point on the surface is the same as r, the distance from the point to the zaxis. Because doesnt appear, it can vary. So any horizontal trace in the plane zk k0 is a circle of radius k. These traces suggest that the surface is a cone. This prediction can be confirmed by converting the equation into rectangular coordinates. From the first equation in 2 we have
z2 r2 x2 y2
We recognize the equation z2x2y2 by comparison with Table 1 in Section 12.6 as
being a circular cone whose axis is the zaxis. See Figure 5. M EVALUATING TRIPLE INTEGRALS WITH CYLINDRICAL COORDINATES
Suppose that E is a type 1 region whose projection D on the xyplane is conveniently described in polar coordinates see Figure 6. In particular, suppose that f is continuous and
Ex,y,zx,yD, u1x,yzu2x,y where D is given in polar coordinates by
Dr,,h1rh2 We know from Equation 15.6.6 that
rh
x
FIGURE 6
z
FIGURE 7
0
y
zuTMx,y
zux,y
Db y rhTM
a
3 fx,y,zdV yyy
yy yu2x, y
u1x,y fx,y,zdz dA
But we also know how to evaluate double integrals in polar coordinates. In fact, combin
ing Equation 3 with Equation 15.4.3, we obtain
4 yyyfx,y,zdVyyh2 yu2rcos ,rsin frcos ,rsin ,zrdzdrd
ED
h1u1rcos ,rsin
Formula 4 is the formula for triple integration in cylindrical coordinates. It says that we convert a triple integral from rectangular to cylindrical coordinates by writing xr cos , yr sin , leaving z as it is, using the appropriate limits of integration for z, r, and , and replacing dV by r dz dr d . Figure 7 shows how to remember this. It is worthwhile to use this formula when E is a solid region easily described in cylindrical coordinates, and especially when the function fx, y, z involves the expression x2y2.
V EXAMPLE 3 A solid E lies within the cylinder x2y21, below the plane z4, and above the paraboloid z1x2y2. See Figure 8. The density at any point is proportional to its distance from the axis of the cylinder. Find the mass of E.
d
dz
Volume element in cylindrical coordinates: dVr dz dr d
E
r dr rd
z
z4
0, 0, 4
0, 0, 1
SECTION 15.7 TRIPLE INTEGRALS IN CYLINDRICAL COORDINATES1003 SOLUTION In cylindrical coordinates the cylinder is r1 and the paraboloid is z1r2,
so we can write
Er, ,z0 2 , 0r1, 1r2 z4
Since the density at x, y, z is proportional to the distance from the zaxis, the density
z1r
y
of E is
fx,y,zKsx2 y2 Kr myyyKsx2y2 dV
E
y2 y1 y4 Krrdzdrd
0 0 1r2
y2 y1Kr241r2drd
00
Ky2 dy13r2r4dr 00
function is
where K is the proportionality constant. Therefore, from Formula 15.6.13, the mass
0
x 1,0,0
FIGURE 8
0
3 r51 12K
2Kr55 M
z
z2
EXAMPLE 4 Evaluate y2 ys4x2 y2 x2y2dz dy dx. 2 s4x2 sx2y2
SOLUTION This iterated integral is a triple integral over the solid region Ex,y,z2x2, s4x2 ys4x2, sx2 y2 z2
and the projection of E onto the xyplane is the disk x2y24. The lower surface of Eistheconezsx2 y2 anditsuppersurfaceistheplanez2.SeeFigure9. This region has a much simpler description in cylindrical coordinates:
Er, ,z0 2 , 0r2, rz2 Therefore, we have
2
2
FIGURE 9
z 2 y
y2 ys4x2 y2 x2 y2dzdydxyyyx2 y2dV
E
y2 y2 y2 r2rdzdrd 00r
y2 dy2r32rdr 00
14 152 16
2 2r5r05 M
x
2 s4x2 sx2y2
1004CHAPTER 15 MULTIPLE INTEGRALS 15.7 EXERCISES
12 Plot the point whose cylindrical coordinates are given. Then find the rectangular coordinates of the point.
20. Evaluate xxxE x dV, where E is enclosed by the planes z0 andzxy5andbythecylindersx2 y2 4and x2 y2 9.
1. a 2,
2. a 1,
4, 1 , e
b 4,b 1, 3
3, 5 2, 2
Evaluate xxx x 2 dV, where E is the solid that lies within the 2E2
34 Change from rectangular to cylindrical coordinates. a 1, 1, 4 b 1, s3 , 2
4. a 2s3, 2, 1 b 4, 3, 2
56 Describe in words the surface whose equation is given.
5.4 6. r5
78 Identify the surface whose equation is given.
7. z4r2 8. 2r2 z2 1 910 Write the equations in cylindrical coordinates.
a zx2y2 b x2y22y
10. a 3x2yz6 bx2 y2 z2 1
1112 Sketch the solid described by the given inequalities.
cylinderx y 1,abovetheplanez0,andbelowthe conez2 4×2 4y2.
3.
22. Find the volume of the solid that lies within both the cylinder x2 y2 1andthespherex2 y2 z2 4.
9.
23. a
b Find the centroid of E the center of mass in the case
Find the volume of the region E bounded by the parabo loidszx2 y2 andz363x2 3y2.
where the density is constant.
24. a
b Illustrate the solid of part a by graphing the sphere and
;
Find the volume of the solid that the cylinder ra cos cuts out of the sphere of radius a centered at the origin.
the cylinder on the same screen.
25. Find the mass and center of mass of the solid S bounded by theparaboloidz4x2 4y2 andtheplanezaa0if S has constant density K.
26. FindthemassofaballBgivenbyx2 y2 z2 a2 ifthe density at any point is proportional to its distance from the z axis.
11. 12.
13.
0r2,22, 0z1 02, rz2
A cylindrical shell is 20 cm long, with inner radius 6 cm and outer radius 7 cm. Write inequalities that describe the shell in an appropriate coordinate system. Explain how you have positioned the coordinate system with respect to the shell.
2728 Evaluate the integral by changing to cylindrical coordinates. 27. y2 ys4y2 y2 xz dz dx dy
17.
Use a graphing device to draw the solid enclosed by the paraboloidszx2 y2 andz5x2 y2.
and evaluate the integral.
15. y4 y2 y4 rdzd dr 16. y2 y2 y9r2 rdzdrd 00r 000
1726 Use cylindrical coordinates.
Evaluate xxxE sx 2y 2 dV, where E is the region that lies inside the cylinder x2y216 and between the planes z5 and z4.
2 s4y 2 sx 2y 2
28. y3 ys9x2 y9x2y2 sx2y2 dz dy dx
30 0
; 14.
1516 Sketch the solid whose volume is given by the integral
29. When studying the formation of mountain ranges, geologists estimate the amount of work required to lift a mountain from sea level. Consider a mountain that is essentially in the shape of a right circular cone. Suppose that the weight density of the material in the vicinity of a point P is tP and the height is hP.
a Find a definite integral that represents the total work done in forming the mountain.
b Assume that Mount Fuji in Japan is in the shape of a right circular cone with radius 62,000 ft, height 12,400 ft, and density a constant 200 lbft3 . How much work was done in forming Mount Fuji if the land was initially at sea level?
P
18. Evaluate xxx x3xy2dV, where E is the solid in the first E
octant that lies beneath the paraboloid z1x 2y 2. 19. Evaluate xxx ez dV, where E is enclosed by the paraboloid
21.
E
z1x2 y2,thecylinderx2 y2 5,andthexyplane.
SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES1005 THE INTERSECTION OF THREE CYLINDERS
DISCOVERY PROJECT
The figure shows the solid enclosed by three circular cylinders with the same diameter that inter sect at right angles. In this project we compute its volume and determine how its shape changes if the cylinders have different diameters.
1. Sketch carefully the solid enclosed by the three cylinders x2y21, x2z21, and
y2z21. Indicate the positions of the coordinate axes and label the faces with the equa tions of the corresponding cylinders.
2. Find the volume of the solid in Problem 1.
Use a computer algebra system to draw the edges of the solid. Illustrate with a handdrawn sketch or a computer graph.
CAS 3.
4. What happens to the solid in Problem 1 if the radius of the first cylinder is different from 1?
5. If the first cylinder is x2y2a2, where a1, set up, but do not evaluate, a double inte gral for the volume of the solid. What if a1?
15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES
Another useful coordinate system in three dimensions is the spherical coordinate system.
It simplifies the evaluation of triple integrals over regions bounded by spheres or cones.
SPHERICAL COORDINATES
The spherical coordinates, ,of a point P in space are shown in Figure 1, whereOPis the distance from the origin to P, is the same angle as in cylindrical coor
dinates, and
is the angle between the positive zaxis and the line segment OP. Note that 00
z
P, ,
O
x
The spherical coordinates of a point
FIGURE 1
y
1006
CHAPTER 15 MULTIPLE INTEGRALS
x
The spherical coordinate system is especially useful in problems where there is symmetry about a point, and the origin is placed at this point. For example, the sphere with center the origin and radius c has the simple equationc see Figure 2; this is the reason for the name spherical coordinates. The graph of the equationc is a vertical halfplane see Figure 3, and the equationc represents a halfcone with the zaxis as its axis see Figure 4.
zzzz c
0000c ycyyy
FIGURE 2 c, a sphere z
Q
Px,y,z
zP, ,
The relationship between rectangular and spherical coordinates can be seen from Fig ure 5. From triangles OPQ and OPP we have
z cos r sin
But xr cos and yr sin , so to convert from spherical to rectangular coordinates,
we use the equations
Also, the distance formula shows that
We use this equation in converting from rectangular to spherical coordinates.
V EXAMPLE 1 The point 2, 4, 3 is given in spherical coordinates. Plot the point and find its rectangular coordinates.
xx
x FIGURE 3c, a halfplane FIGURE 4c, a halfcone
2c
0c 2
O
x
x
FIGURE 5
r y
y
Pax,y,0
1
x sin cos y sin sin z cos
2
2 x2 y2 z2
z
3
O
SOLUTION We plot the point in Figure 6. From Equations 1 we have
xsin cos2 sin cos2s31 3
2, 4, 3
2
3 42s22 ysin sin2 sin sin2 s3 13
4 y FIGURE 6
zcos Thus the point 2, 4,
3 4 2 s2 22 cos21 1
x
32
3 is s32 , s32 , 1 in rectangular coordinates. M
WARNING There is not universal agree ment on the notation for spherical coordinates. Most books on physics reverse the meanings
of and anduserinplaceof .
SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES1007 V EXAMPLE 2 The point 0, 2 s3 , 2 is given in rectangular coordinates. Find spheri
cal coordinates for this point. SOLUTION From Equation 2 we have
and so Equations 1 give
sx2 y2 z2 s0124 4 cos z21 2
cosx 0sin
423
TEC In Module 15.8 you can investigate families of surfaces in cylindrical and spheri cal coordinates.
2
3 2 because y2s30. Therefore spherical coordinates of the
z
isin kII
k Irii sink
Note that
given point are 4, 2, 2 3. M
EVALUATING TRIPLE INTEGRALS WITH SPHERICAL COORDINATES
In the spherical coordinate system the counterpart of a rectangular box is a spherical wedge
E, , a b, ,c d
wherea0, 2 ,anddc.Althoughwedefinedtripleintegralsbydivid ing solids into small boxes, it can be shown that dividing a solid into small spherical wedges always gives the same result. So we divide E into smaller spherical wedges Eijk by means of equally spaced spheresi , halfplanesj , and halfconesk . Figure 7 shows that Eijk is approximately a rectangular box with dimensions, iarc ofacirclewithradius i,angle ,and i sin k arcofacirclewithradius i sin k, angle. So an approximation to the volume of Eijk is given by
Vijk iisin k 2isin k
In fact, it can be shown, with the aid of the Mean Value Theorem Exercise 45, that the
x
y
0
i I
ri I i sink I
FIGURE 7
volume of Eijk is given exactly by
Vijki sin k
2
wherei, j, kis some point in Eijk. Let xijk, yijk, zijk be the rectangular coordinates of
this point. Then
f x, y, z dVlim f xijk, yijk, zijk Vijk l m n
yyy
E
But this sum is a Riemann sum for the function
F,,fsin cos, sin sin, cos2sin
Consequently, we have arrived at the following formula for triple integration in spherical coordinates.
lim
l,m,nl i1 j1 k1
l,m,nl i1 j1 k1
lmn
fi sin kcos j,isin ksin j,i cos ki2sin k
I
1008
CHAPTER 15 MULTIPLE INTEGRALS
3
yyy f x, y, z dV
E
ydyybf sin cos , sin sin , cos2sin d d d ca
where E is a spherical wedge given by
E, , a b, ,c d
z
d
y
sin d
d
0
d
Formula 3 says that we convert a triple integral from rectangular coordinates to spher ical coordinates by writing
x
FIGURE 8
Volume element in spherical coordinates: dV sin d dd
x sin cos
using the appropriate limits of integration, and replacing dV by
z cos
2 sin d d d
y sin sin
illustrated in Figure 8.
This formula can be extended to include more general spherical regions such as
E, ,,c d,t1,t2,
In this case the formula is the same as in 3 except that the limits of integration for are t1 , andt2 , .
Usually, spherical coordinates are used in triple integrals when surfaces such as cones and spheres form the boundary of the region of integration.
V EXAMPLE 3 Evaluate xxx ex2y2z232 dV, where B is the unit ball: B
Bx,y,zx2 y2 z2 1
SOLUTION Since the boundary of B is a sphere, we use spherical coordinates:
B , , 0 1, 0 2 , 0 In addition, spherical coordinates are appropriate because
. This is
Thus 3 gives
B
000
ysin d y2 d y1 2e3d 000
31
cos 2 1e 4 e1
x2y2z2 2 yyyex2y2z232dVyy2 y1e232 2sin ddd
NOTE It would have been extremely awkward to evaluate the integral in Example 3 without spherical coordinates. In rectangular coordinates the iterated integral would have been
y1 ys1x 2 ys1x2y2 ex2y2z232 dz dy dx 1 s1x 2 s1x2y2
0303
M
N Figure 10 gives another look this time drawn by Maple at the solid of Example 4.
SOLUTION Notice that the sphere passes through the origin and has center 0, 0, 1 . We
write the equation of the sphere in spherical coordinates as
2 cos or cos The equation of the cone can be written as
FIGURE 10
TEC Visual 15.8 shows an animation of Figure 11.
Figure 11 shows how E is swept out if we integrate first with respect to
, then
, and
FIGURE 9
x
z 0, 0, 1
zz
4 z
y
SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES1009 V EXAMPLE 4 Use spherical coordinates to find the volume of the solid that lies above
theconezsx2 y2 andbelowthespherex2 y2 z2 z.SeeFigure9.
cos s 2sin2 cos22sin2 sin2sin
This gives sincos , or4. Therefore the description of the solid E in
spherical coordinates is
E , , 0 2 , 04, 0 cos
2
F I G U R E 1 1
xyxyxy
varies from 0 to cos varies from 0 to 4 varies from 0 to 2. while and are constant. while is constant.
M
then
. The volume of E is
VEyyydVy2 y4ycos 2sin d d d
2 4 y
sin
2 cos4
000 y2y4 3cos
E
0d0sin3 d
0 cos3 d
4 30 3408
zzz
1010CHAPTER 15 MULTIPLE INTEGRALS
15.8 EXERCISES
12 Plot the point whose spherical coordinates are given. Then
1920 Set up the triple integral of an arbitrary continuous function f x, y, z in cylindrical or spherical coordinates over the solid
find the rectangular coordinates of the point.
a 1, 0, 0 b 2, 3, 4
2. a 5, , 2 b 4, 3 4, 3 34 Change from rectangular to spherical coordinates.
3. a 1, s3, 2s3b 0, 1, 1 4. a 0, s3 , 1 b 1, 1, s6
56 Describe in words the surface whose equation is given. 3 6. 3
shown.
19.z 20.z 3
y1 xx2
2134 Use spherical coordinates.
21. Evaluatexxxx2 y2 z22dV,whereBistheballwith
B
y
1.
5.
2
78 Identify the surface whose equation is given. 7.sin sin 8. 2 sin2 sin2
910 Write the equation in spherical coordinates.
center the origin and radius 5.
22. Evaluate xxx 9x2y2dV, where H is the solid
cos2
9
H
hemispherex2 y2 z2 9,z0.
23. Evaluate xxxE z dV, where E lies between the spheres
x2 y2 z2 1andx2 y2 z2 4inthefirstoctant.
9.a z2x2y2
10. a x2 2xy2 z2 0
bx2z29
b x2y3z1
24. Evaluate xxx esx2y2z2 dV, where E is enclosed by the sphere 2 2 E2
13. 14.
15.
1, 3 4 2,csc
x y z 9inthefirstoctant.
25. Evaluate xxx x 2 dV, where E is bounded by the x zplane
30.
E andthehemispheresys9x2z2 and
1114 Sketch the solid described by the given inequalities. 11. 2, 02, 02
12.23, 2
ys16x2 z2.
26. Evaluate xxxE xyz dV, where E lies between the spheres
2and 4andabovethecone3.
27. Find the volume of the part of the balla that lies between
the cones6 and3.
28. Find the average distance from a point in a ball of radius a to
its center.
29. a Find the volume of the solid that lies above the cone
3andbelowthesphere 4cos . b Find the centroid of the solid in part a.
Find the volume of the solid that lies within the sphere
x2y2z24, above the xyplane, and below the cone zsx2 y2.
31. Find the centroid of the solid in Exercise 25.
32. Let H be a solid hemisphere of radius a whose density at any point is proportional to its distance from the center of the base.
a Find the mass of H.
b Find the center of mass of H.
c Find the moment of inertia of H about its axis.
A solid lies above the cone zsx2y2 and below the sphere x2y2z2z. Write a description of the solid in terms of inequalities involving spherical coordinates.
16. a Find inequalities that describe a hollow ball with diameter 30 cm and thickness 0.5 cm. Explain how you have positioned the coordinate system that you have chosen.
b Suppose the ball is cut in half. Write inequalities that describe one of the halves.
1718 Sketch the solid whose volume is given by the integral and evaluate the integral.
y6 y2 y3 000
y2 y y2 2 18. 0 2 1
2sin ddd sin d d d
17.
33. a
b Find the moment of inertia of the solid in part a about a
Find the centroid of a solid homogeneous hemisphere of radius a.
diameter of its base.
34. Find the mass and center of mass of a solid hemisphere of radius a if the density at any point is proportional to its distance from the base.
3538 Use cylindrical or spherical coordinates, whichever seems more appropriate.
CAS 43.
The surfaces11 sin m sin n have been used as 5
models for tumors. The bumpy sphere with m6 and
n5 is shown. Use a computer algebra system to find the volume it encloses.
35.
36. CAS 37.
Find the volume and centroid of the solid E that lies above the cone zsx2y2 and below the sphere x2 y2 z2 1.
Find the volume of the smaller wedge cut from a sphere of radius a by two planes that intersect along a diameter at an angle of 6.
Evaluate xxxE z dV, where E lies above the paraboloid zx2 y2 andbelowtheplanez2y.Useeitherthe Table of Integrals on Reference Pages 610 or a computer algebra system to evaluate the integral.
sin
3940 Evaluate the integral by changing to spherical coordinates.
44.
45.
Show that
y y y sx2 y2 z2 ex2y2z2dxdydz2
The improper triple integral is defined as the limit of a triple integral over a solid sphere as the radius of the sphere increases indefinitely.
a Use cylindrical coordinates to show that the volume of the solid bounded above by the sphere r2z2a2 and belowbytheconezrcot 0 or0,where 0 0 2,is
2 a3
V 3 1cos 0
b Deduce that the volume of the spherical wedge given by 12, 12, 12is
V 23 13cos 1cos 22 1 3
c Use the Mean Value Theorem to show that the volume in part b can be written as
V 2 sin
wherelies between 1 and 2,lies between 1 and 2,21,21,and21.
a Find the volume enclosed by the torus b Use a computer to draw the torus.
.
;38.
39. y1 ys1x2 ys2x2y2 xy dz dy dx
; 41.
0 0 sx2y2
40. ya ysa2y2 ysa2x2y2 x2zy2zz3 dz dx dy
a sa y sa x y
Use a graphing device to draw a silo consisting of a cylinder
with radius 3 and height 10 surmounted by a hemisphere.
22
222
42. The latitude and longitude of a point P in the Northern Hemi sphere are related to spherical coordinates , , as follows. We take the origin to be the center of the earth and the posi tive zaxis to pass through the North Pole. The positive xaxis passes through the point where the prime meridian the meridian through Greenwich, England intersects the equator. Then the latitude of P is90 and the longitude is
360. Find the greatcircle distance from Los Angeles lat. 34.06 N, long. 118.25 W to Montreal lat. 45.50 N, long. 73.60 W. Take the radius of the earth to be 3960 mi. A great circle is the circle of intersection of a sphereandaplanethroughthecenterofthesphere.
SECTION 15.8 TRIPLE INTEGRALS IN SPHERICAL COORDINATES1011
1012
CHAPTER 15 MULTIPLE INTEGRALS
ROLLER DERBY
APPLIED PROJECT
h
Suppose that a solid ball a marble, a hollow ball a squash ball, a solid cylinder a steel bar, and a hollow cylinder a lead pipe roll down a slope. Which of these objects reaches the bottom first? Make a guess before proceeding.
To answer this question, we consider a ball or cylinder with mass m, radius r, and moment of
inertia I about the axis of rotation. If the vertical drop is h, then the potential energy at the top
is mth. Suppose the object reaches the bottom with velocity v and angular velocity , so vr.
The kinetic energy at the bottom consists of two parts: 1 mv2 from translation moving down the 2
a
slope and 1 I 2 from rotation. If we assume that energy loss from rolling friction is negligible, 2
then conservation of energy gives 1. Show that
mth1mv2 1I 2 22
v2 2th whereI I 1I mr2
2. If yt is the vertical distance traveled at time t, then the same reasoning as used in Problem 1 shows that v22ty1I at any time t. Use this result to show that y satisfies the differential equation
dy dt
2t sin sy 1I
where is the angle of inclination of the plane.
3. By solving the differential equation in Problem 2, show that the total travel time is
2h1I
T
This shows that the object with the smallest value of I wins the race.
t sin
4. Show that I1 for a solid cylinder and I1 for a hollow cylinder.
2
2
5. Calculate I for a partly hollow ball with inner radius a and outer radius r. Express your
answer in terms of bar. What happens as a l 0 and as a l r?
6. Show that I2 for a solid ball and I2 for a hollow ball. Thus the objects finish in the
53
following order: solid ball, solid cylinder, hollow ball, hollow cylinder.
15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS
In onedimensional calculus we often use a change of variable a substitution to simplify
an integral. By reversing the roles of x and u, we can write the Substitution Rule 5.5.6 as yb fxdxyd ftutudu
ac
where xtu and atc, btd. Another way of writing Formula 1 is as follows:
yb fx dxyd fxu dx du a c du
1
2
SECTION 15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS1013 A change of variables can also be useful in double integrals. We have already seen one
example of this: conversion to polar coordinates. The new variables r and the old variables x and y by the equations
xr cos yr sin
and the change of variables formula 15.4.2 can be written as
yyfx,ydAyyfrcos ,rsin rdrd RS
are related to
where S is the region in the r plane that corresponds to the region R in the xyplane. More generally, we consider a change of variables that is given by a transformation T
or, as we sometimes write,
xxu, v yyu, v
We usually assume that T is a C1 transformation, which means that t and h have contin uous firstorder partial derivatives.
A transformation T is really just a function whose domain and range are both subsets of 2. If Tu1, v1x1, y1, then the point x1, y1 is called the image of the point u1, v1. If no two points have the same image, T is called onetoone. Figure 1 shows the effect of a transformation T on a region S in the uvplane. T transforms S into a region R in the xyplane called the image of S, consisting of the images of all points in S.
y
T T!
0u0x
from the uvplane to the xyplane:
where x and y are related to u and v by the equations
Tu, vx, y
xtu, v yhu, v
FIGURE 1
3
S u,
R
x, y
If T is a onetoone transformation, then it has an inverse transformation T 1 from the xyplane to the uvplane and it may be possible to solve Equations 3 for u and v in terms of x and y:
uGx, y vHx, y
V EXAMPLE 1 A transformation is defined by the equations xu2 v2 y2uv
FindtheimageofthesquareSu,v0u1, 0v1.
SOLUTION The transformation maps the boundary of S into the boundary of the image. So we begin by finding the images of the sides of S. The first side, S1, is given by v0
1014
CHAPTER 15 MULTIPLE INTEGRALS
0, 1
1, 1
S
S S STM
0 S1,0 u T
0u1. See Figure 2. From the given equations we have xu2, y0, and so
0x1. Thus S1 is mapped into the line segment from 0, 0 to 1, 0 in the xyplane. The second side, S2, is u1 0v1 and, putting u1 in the given equations, we get
Eliminating v, we obtain
x1v2 y2
x1 4
y2v 0x1
4
which is part of a parabola. Similarly, S3 is given by v1 0u1, whose image is
y
0,2 y2
the parabolic arc
x 4 1
x1 4
1,0 x
x41 1×0
Finally,S4 isgivenbyu00v1whoseimageisxv2,y0,thatis,
1×0. Notice that as we move around the square in the counterclockwise direc tion, we also move around the parabolic region in the counterclockwise direction. The image of S is the region R shown in Figure 2 bounded by the xaxis and the parabolas given by Equations 4 and 5. M
Now lets see how a change of variables affects a double integral. We start with a small rectangle S in the uvplane whose lower left corner is the point u0 , v0and whose dimen sions are u and v. See Figure 3.
5
1,0 0
FIGURE 2
R
y
uu
I S u , Iu
0u0x
T
ru , x , y
R ru,
FIGURE 3
The image of S is a region R in the xyplane, one of whose boundary points is x0, y0 Tu0, v0 . The vector
ru, vtu, vihu, vj
is the position vector of the image of the point u, v. The equation of the lower side of S is vv0, whose image curve is given by the vector function ru, v0. The tangent vector at x0, y0to this image curve is
ru tuu0,v0ihuu0,v0j x i y j u u
Similarly, the tangent vector at x0, y0to the image curve of the left side of S namely, uu0 is
rv tvu0,v0ihvu0,v0j x i y j v v
r u ,I b
ru ,R a
r u Iu, FIGURE 4
SECTION 15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS1015 We can approximate the image region RTS by a parallelogram determined by the
secant vectors
aru0 u,v0ru0,v0 shown in Figure 4. But
6
bru0,v0 vru0,v0 rulim ru0 u,v0ru0,v0
ul0 u
ru0 u,v0ru0,v0uru
ru0, v0vru0, v0 v rv
Ir r u ,
andso Similarly
Iuru FIGURE 5
This means that we can approximate R by a parallelogram determined by the vectors u ru and v rv. See Figure 5. Therefore we can approximate the area of R by the area of this parallelogram, which, from Section 12.4, is
uruvrvru rvuv Computing the cross product, we obtain
i j k x y x x
x y 0 u u u v rurvu uk k
x y y y x y 0 v v u v
v v
The determinant that arises in this calculation is called the Jacobian of the transformation
and is given a special notation.
7
DEFINITION
The Jacobian of the transformation T given by xtu, v and x x
x,y u v xy xy u,v y y uv vu
u v
yhu, v is
N The Jacobian is named after the German mathematician Carl Gustav Jacob Jacobi 18041851. Although the French mathematician Cauchy first used these special determinants involving partial derivatives, Jacobi developed them into a method for evaluating multiple integrals.
With this notation we can use Equation 6 to give an approximation to the area A of R:
Ax, y u v u, v
where the Jacobian is evaluated at u0 , v0 .
8
1016
CHAPTER 15 MULTIPLE INTEGRALS
Next we divide a region S in the uvplane into rectangles Sij and call their images in the
FIGURE 6
xyplane Rij. See Figure 6. y
Sij
T
0u0x Applying the approximation 8 to each Rij, we approximate the double integral of f
I
I
u
S
R
Rij
ui, j
xi, yj
over R as follows:
yy mn
for the integral
R
f x, y dA
i1 j1
f xi, yjA
f tui, vj , hui, vj
m n i1 j1
x,y u, v
u v
where the Jacobian is evaluated at ui , vj . Notice that this double sum is a Riemann sum
yy f tu, v, hu, vx, ydu dv S u, v
The foregoing argument suggests that the following theorem is true. A full proof is given in books on advanced calculus.
CHANGE OF VARIABLES IN A DOUBLE INTEGRAL Suppose that T is a C1 trans formation whose Jacobian is nonzero and that maps a region S in the uvplane onto a region R in the xyplane. Suppose that f is continuous on R and that R and S are type I or type II plane regions. Suppose also that T is onetoone, except perhaps on the boundary of S. Then
yy f x, y dAyy f xu, v, yu, vx, ydu dv R S u, v
9
Theorem 9 says that we change from an integral in x and y to an integral in u and v by expressing x and y in terms of u and v and writing
dAx, y du dv u, v
Notice the similarity between Theorem 9 and the onedimensional formula in Equation 2. Instead of the derivative dxdu, we have the absolute value of the Jacobian, that is, x, yu, v.
a
rb
As a first illustration of Theorem 9, we show that the formula for integration in polar coordinates is just a special case. Here the transformation T from the r plane to the xyplane is given by
xtr, rcos yhr, rsin
and the geometry of the transformation is shown in Figure 7. T maps an ordinary rectangle
in the r plane to a polar rectangle in the xyplane. The Jacobian of T is
SECTION 15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS1017
ra
S
a
0r ab
T
rb R
a
x x
x,y rcos rsin
y
ra
a
rcos2 rsin2 r0
x, yfx,ydxdy frcos ,rsinr,drd
y y sin
r,
Thus Theorem 9 gives
r cos
yy
yy
r
0xRS
FIGURE 7
The polar coordinate transformation
y yb frcos ,rsin rdrd a
which is the same as Formula 15.4.2.
V EXAMPLE 2 Use the change of variables xu2v2, y2uv to evaluate the integral
xx y dA, where R is the region bounded by the xaxis and the parabolas y244x R
andy2 44x,y0.
SOLUTION The region R is pictured in Figure 2 on page 1014. In Example 1 we discov ered that TSR, where S is the square 0, 10, 1. Indeed, the reason for making the change of variables to evaluate the integral is that S is a much simpler region than R. First we need to compute the Jacobian:
4u24v20
yyydAyy2uvx,ydAy1y1 2uv4u2 v2dudv
x x
x, y u v 2u 2v
u, v y y2v2u
u v Therefore, by Theorem 9,
RS
u, v 0 0
8y1y1 u3vuv3dudv8y11u4v1u2v3u1 dv
000
42
u0
y 12 v4 v 3d v v 2v 4102 M 0
1018
CHAPTER 15 MULTIPLE INTEGRALS
NOTE Example 2 was not a very difficult problem to solve because we were given a suitable change of variables. If we are not supplied with a transformation, then the first step is to think of an appropriate change of variables. If f x, y is difficult to integrate, then the form of f x, y may suggest a transformation. If the region of integration R is awkward, then the transformation should be chosen so that the corresponding region S in the uvplane has a convenient description.
EXAMPLE 3 Evaluate the integral xx exyxy dA, where R is the trapezoidal region with R
vertices 1, 0, 2, 0, 0, 2, and 0, 1.
SOLUTION Since it isnt easy to integrate exyxy, we make a change of variables sug
gested by the form of this function:
uxy vxy
These equations define a transformation T1 from the xyplane to the uvplane. Theorem 9 talks about a transformation T from the uvplane to the xyplane. It is obtained by solving Equations 10 for x and y:
x1 uv y1 uv 22
x x
The Jacobian of T is
u v
To find the region S in the uvplane corresponding to R, we note that the sides of R lie on
x,y u v 1 1
2 21
u,v y y 1 1 2 22
y0 xy2 x0 xy1
the lines
and, from either Equations 10 or Equations 11, the image lines in the uvplane are
uv v2 uv v1
Thus the region S is the trapezoidal region with vertices 1, 1, 2, 2, 2, 2, and
1, 1 shown in Figure 8. Since
Su,v1v2, vuv
2, 2
u 1, 1
2 1
2, 2
10
11
S
u 1, 1
T
0u
T!
y
0 x R S u,v
Theorem 9 gives
xy1
yy
yy
x, y
euv du dv
1
2
exyxy dA
y2yv euv1dudv1y2veuvuv dv
1 y2 ee1vdv3ee1 M 21 4
1 2
R xy2
22
1 v 1
uv
FIGURE8
SECTION 15.9 CHANGE OF VARIABLES IN MULTIPLE INTEGRALS1019 TRIPLE INTEGRALS
There is a similar change of variables formula for triple integrals. Let T be a transfor mation that maps a region S in uvwspace onto a region R in xyzspace by means of the equations
xtu, v, w yhu, v, w zku, v, w The Jacobian of T is the following 33 determinant:
x x x
12
u v w
x,y,zy y y
u, v, w u
z z z
v w u v w
Under hypotheses similar to those in Theorem 9, we have the following formula for triple integrals:
V EXAMPLE 4 Use Formula 13 to derive the formula for triple integration in spherical coordinates.
13
yyyfx,y,zdVyyyfxu,v,w,yu,v,w,zu,v,wx,y,z dudvdw R S u, v, w
SOLUTION Here the change of variables is given by
x sin cos
We compute the Jacobian as follows:
cossin sin cos cossinsin cos cos sin
sin cossin sin cos cos, , sin sinsin cos cos sin
x, y, z
cos 0sin
y sin sin
z cos
sin cos sin sin
cos2sin cos sin22sin cos cos2 sinsin2 cos2sin2 sin2
2sin cos22sin sin22sin Since 0 , we have sin0. Therefore
x,y,z2sin2sin, ,
and Formula 13 gives
yyyfx,y,zdVyyyf sin cos , sin sin , cos2sin d d d
RS
sin sinsin cos
which is equivalent to Formula 15.8.3.
M
1020CHAPTER 15 MULTIPLE INTEGRALS
15.9 EXERCISES
16 Find the Jacobian of the transformation.
1. x5uv, yu3v 2. xuv, yuv
3. xersin , yercos 4. xest, yest
5. xuv, yvw, zwu
6. xvw2, ywu2, zuv2
710 Find the image of the set S under the given transformation.
7. Su,v0u3, 0v2;
x2u3v, yuv
8. Sisthesquareboundedbythelinesu0,u1,v0,
v1; xv, yu1v2
9. S is the triangular region with vertices 0, 0, 1, 1, 0, 1;
xu2, yv
10. Sisthediskgivenbyu2 v2 1; xau, ybv
1116 Use the given transformation to evaluate the integral.
11. xxR x3y dA, where R is the triangular region with
vertices 0, 0, 2, 1, and 1, 2; x2uv, yu2v
12. xxR 4 x8y dA, where R is the parallelogram with
vertices 1, 3, 1, 3, 3, 1, and 1, 5;
x1uv, y1v3u 44
xx y 2 dA, where R is the region bounded by the curves R
; 16.
17. a Evaluate xxx dV, where E is the solid enclosed by the
xy1,xy2,xy2 1,xy2 2; uxy, vxy2. Illustrate by using a graphing calculator or computer to draw R.
E
ellipsoid x2a2y2b2z2c21. Use the transfor
18.
mationxau, ybv, zcw.
b The earth is not a perfect sphere; rotation has resulted in
flattening at the poles. So the shape can be approximated by an ellipsoid with ab6378 km and c6356 km. Use part a to estimate the volume of the earth.
If the solid of Exercise 17a has constant density k, find its moment of inertia about the zaxis.
1923 Evaluate the integral by making an appropriate change of variables.
19. yy x2y dA, where R is the parallelogram enclosed by R 3xy
thelines x2y0, x2y4,3xy1,and 3xy8
20. xx xyex2y2 dA, where R is the rectangle enclosed by the R
lines xy0, xy2, xy0, and xy3 21. yy cos yx dA, where R is the trapezoidal region
R yx
13.
xx x 2 dA, where R is the region bounded by the ellipse R2 2
with vertices 1, 0, 2, 0, 0, 2, and 0, 1
22. xx sin9x24y2dA, where R is the region in the first
9x 4y 36; x2u, y3v
14. xx x2xyy2dA, where R is the region bounded
R22 quadrant bounded by the ellipse 9x4y1
R
bytheellipsex2 xyy2 2;
23. xx exydA, whereRisgivenbytheinequality xy 1
24. Let f be continuous on 0, 1 and let R be the triangular
R
xs2us23v, ys2us23v
15. xxR xy dA, where R is the region in the first quadrant bounded bythelinesyxandy3xandthehyperbolasxy1, xy3; xuv,yv
region with vertices 0, 0, 1, 0, and 0, 1. Show that
fxydA 0 ufudu yy y1
R
f x, y dA
3. How do you change from rectangular coordinates to polar coor dinates in a double integral? Why would you want to make the change?
4. If a lamina occupies a plane region D and has density function x, y, write expressions for each of the following in terms of
double integrals.
a The mass
b The moments about the axes
c The center of mass
d The moments of inertia about the axes and the origin
5. Let f be a joint density function of a pair of continuous random variables X and Y.
a Write a double integral for the probability that X lies
between a and b and Y lies between c and d.
TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. y2 y6 x2 sinxydxdyy6 y2 x2 sinxydydx 1 0 0 1
2. y1yx sxy2 dydxyxy1 sxy2 dxdy 00 00
3. y2y4 x2ey dydxy2 x2 dxy4 ey dy 1313
4. y1 y1 ex2y2 sinydxdy0 1 0
if D is a type I region?
c What is a type II region? How do you evaluate
xxD fx, y dA if D is a type II region?
d What properties do double integrals have?
15 REVIEW
CHAPTER 15 REVIEW
1021
CONCEPT CHECK
1. Suppose f is a continuous function defined on a rectangle Ra,bc,d.
a Write an expression for a double Riemann sum of f.
If f x, y0, what does the sum represent?
b Write the definition of xxR f x, y dA as a limit.
c What is the geometric interpretation of xxR f x, y dA if
b What properties does f possess?
c What are the expected values of X and Y ?
b How do you evaluate xxxB f x, y, z dV?
c How do you define xxxE fx, y, z dV if E is a bounded solid region that is not a box?
d What is a type 1 solid region? How do you evaluate xxxE fx, y, z dV if E is such a region?
e What is a type 2 solid region? How do you evaluate xxxE fx, y, z dV if E is such a region?
f What is a type 3 solid region? How do you evaluate xxxE fx, y, z dV if E is such a region?
7. Suppose a solid object occupies the region E and has density function x, y, z. Write expressions for each of the following. a The mass
b The moments about the coordinate planes
c The coordinates of the center of mass d The moments of inertia about the axes
8. a How do you change from rectangular coordinates to cylin drical coordinates in a triple integral?
b How do you change from rectangular coordinates to spherical coordinates in a triple integral?
c In what situations would you change to cylindrical or spherical coordinates?
9. a If a transformation T is given by xtu, v, yhu,v,whatistheJacobianofT?
b How do you change variables in a double integral? c How do you change variables in a triple integral?
5. IfDisthediskgivenbyx2 y2 4,then yys4x2y2 dA16
D
6. y4 y1 x2 sysinx2y2dxdy9
f x, y0? What if f takes on both positive and negative values?
d How do you evaluate xxR f x, y dA?
e What does the Midpoint Rule for double integrals say?
f Write an expression for the average value of f.
How do you define xxD f x, y dA if D is a bounded region that is not a rectangle?
2. a
b What is a type I region? How do you evaluate xxD
6. a Write the definition of the triple integral of f over a rectangular box B.
3
10
7. The integral
y2 y2y2dzdrd 00r
represents the volume enclosed by the cone zs x 2y 2 and the plane z2.
8. The integral xxxE kr 3 dz dr d represents the moment of inertia about the zaxis of a solid E with constant density k.
1022CHAPTER 15 MULTIPLE INTEGRALS EXERCISES
1. A contour map is shown for a function f on the square
R0, 30, 3. Use a Riemann sum with nine terms to estimate the value of xxR f x, y dA. Take the sample points to be the upper right corners of the squares.
y 3
2 1
0 123x
2. Use the Midpoint Rule to estimate the integral in Exercise 1. 38 Calculate the iterated integral.
3. y2 y2 y2xeydxdy 4. y1 y1 yexy dxdy 10 00
5. y1 yx cosx2dydx 6. y1 yex 3xy2 dydx 00 0x
y y1 ys1y2 y1 yy y1
7. 0 0 0 ysinxdzdydx 8. 0 0 x 6xyzdzdxdy
910 Write xx f x, y dA as an iterated integral, where R is the R
1314 Calculate the iterated integral by first reversing the order of integration.
13.
14.
0 sy x
3
dx dy
y1 y1 0 x
y1 y1 yex2
cosy2dy dx
1528 Calculate the value of the multiple integral.
15. xx yexy dA, where Rx, y 0x2, 0y3 R
7 8 9 10
3 2
45
6
1
16. xxD xydA, whereDx,y0y1, y2 xy2 17.yy y dA,
D 1×2
where D is bounded by ysx , y0, x1
18. yy 1
D 1x
dA, where D is the triangular region with vertices 0, 0, 1, 1, and 0, 1
19. xxD y dA, where D is the region in the first quadrant bounded by theparabolasxy2 andx8y2
20. xxD y dA, where D is the region in the first quadrant that lies above the hyperbola xy1 and the line yx and below the liney2
21. xxD x2 y232dA, whereDistheregioninthefirst quadrant bounded by the lines y0 and ys3x and the circlex2 y2 9
22. xxD x dA, where D is the region in the first quadrant that lies betweenthecirclesx2 y2 1andx2 y2 2
23. xxxE xydV, where
Ex,y,z0x3, 0yx, 0zxy
24. xxxT xy dV, where T is the solid tetrahedron with vertices 0, 0, 0, 1 , 0, 0, 0, 1, 0, and 0, 0, 1
2
region shown and f is an arbitrary continuous function on R.
yy 25. xxx y2z2 dV, where E is bounded by the paraboloid
9. 10. E
3
44
x1y2 z2 andtheplanex0
26. xxxE zdV, whereEisboundedbytheplanesy0,z0,
xy2andthecylindery2 z2 1inthefirstoctant
27. xxxE yz dV, where E lies above the plane z0, below the plane
R 42024x404x
11. Describe the region whose area is given by the integral y 2 ysin 2 r dr d
00
12. Describe the solid whose volume is given by the integral
y2y2y2 2sin d d d 001
and evaluate the integral.
2
R
zy,andinsidethecylinderx2 y2 4
28. xxx z3sx2y2z2 dV, where H is the solid hemisphere that
H
lies above the xyplane and has center the origin and radius 1
2934 Find the volume of the given solid.
29. Under the paraboloid zx 24y 2 and above the rectangle
R0, 21, 4
30. Under the surface zx 2 y and above the triangle in the xyplane with vertices 1, 0, 2, 1, and 4, 0
31. 32. 33. 34.
35.
36.
37.
38. 39.
40.
;41. CAS 42.
43.
The solid tetrahedron with vertices 0, 0, 0, 0, 0, 1, 0, 2, 0, and 2, 2, 0
Boundedbythecylinderx2 y2 4andtheplanesz0 and yz3
One of the wedges cut from the cylinder x29y2a2 by the planes z0 and zmx
Above the paraboloid zx2y2 and below the halfcone zsx2 y2
Consider a lamina that occupies the region D bounded by the parabola x1y2 and the coordinate axes in the first quadrant with density function x, yy.
a Find the mass of the lamina.
b Find the center of mass.
c Find the moments of inertia and radii of gyration about
the x and yaxes.
A lamina occupies the part of the disk x2y2a2 that lies in the first quadrant.
a Find the centroid of the lamina.
b Find the center of mass of the lamina if the density func
tion is x, yxy2.
Find the centroid of a right circular cone with height h
and base radius a. Place the cone so that its base is in the xyplane with center the origin and its axis along the positive z axis.
Find the moment of inertia of the cone in Exercise 37 about its axis the zaxis.
Use polar coordinates to evaluate
y3 ys9x2 x3xy2 dy dx 0 s9x 2
Use spherical coordinates to evaluate
y2 ys4y2 ys4x2y2 y2sx2y2z2 dz dx dy 2 0 s4x 2y 2
IfDistheregionboundedbythecurvesy1x2 and
ye x, find the approximate value of the integral xx y 2 dA.
D
Use a graphing device to estimate the points of intersection of the curves.
Find the center of mass of the solid tetrahedron with vertices 0, 0, 0, 1, 0, 0, 0, 2, 0, 0, 0, 3 and density function
x,y,zx2 y2 z2.
The joint density function for random variables X and Y is
fx,y Cxy if 0x3, 0y2 0 otherwise
a Find the value of the constant C. b Find PX2, Y1.
c FindPXY1.
44. A lamp has three bulbs, each of a type with average lifetime 800 hours. If we model the probability of failure of the bulbs by an exponential density function with mean 800, find the probability that all three bulbs fail within a total of 1000 hours.
45. Rewrite the integral
y1 y1 y1y fx,y,zdzdydx
1 x2 0
as an iterated integral in the order dx dy dz.
46. Give five other iterated integrals that are equal to y2 yy3 yy2 fx,y,zdzdxdy
000
47. Usethetransformationuxy, vxytoevaluate xxR xyxy dA, where R is the square with vertices 0, 2, 1, 1, 2, 2, and 1, 3.
48. Usethetransformationxu2,yv2,zw2to find the volume of the region bounded by the surface sxsysz1 and the coordinate planes.
49. Use the change of variables formula and an appropriate trans formation to evaluate xxR xy dA, where R is the square with vertices 0, 0, 1, 1, 2, 0, and 1, 1.
50. The Mean Value Theorem for double integrals says that if f is a continuous function on a plane region D that is of type I or II, then there exists a point x0, y0in D such that
yy f x, y dAf x0, y0AD D
Use the Extreme Value Theorem 14.7.8 and Property 15.3.11 of integrals to prove this theorem. Use the proof of the singlevariable version in Section 6.5 as a guide.
51. Suppose that f is continuous on a disk that contains the point a, b. Let Dr be the closed disk with center a, b and radius r. Use the Mean Value Theorem for double integrals see Exercise 50 to show that
CHAPTER 15 REVIEW1023
52. a Evaluate yy
1 rl0 r2
yy f x, y dAf a, b Dr
dA, where n is an integer and
D is the region bounded by the circles with center the
origin and radii r and R, 0rR.
b For what values of n does the integral in part a have a
limit as r l 0?
c Find 1 dV, where E is the region
lim
1
D x2y2n2
yyy
E x2 y2 z2n2
bounded by the spheres with center the origin and radii r and R, 0rR.
d For what values of n does the integral in part c have a limit as r l 0?
PROBLEMS PLUS
1. If x denotes the greatest integer in x, evaluate the integral yy xy dA
R whereRx,y1x3, 2y5.
2. Evaluate the integral
y1 y1 emaxx2, y2 dy dx 00
where maxx2, y2means the larger of the numbers x2 and y2.
3. Find the average value of the function fxxx1 cost2dt on the interval 0, 1.
4. If a, b, and c are constant vectors, r is the position vector xiyjzk, and E is given by theinequalities0ar , 0br , 0cr ,showthat
yyy arbrcr dV 2
E 8abc
5. The double integral y1 y1 1 dx dy is an improper integral and could be defined as 0 0 1xy
the limit of double integrals over the rectangle 0, t0, t as t l 1. But if we expand the integrand as a geometric series, we can express the integral as the sum of an infinite series. Show that
y1 y1 1 dx dy 1 0 01xy n1n2
6. Leonhard Euler was able to find the exact sum of the series in Problem 5. In 1736 he proved that
1 2 n1 n26
In this problem we ask you to prove this fact by evaluating the double integral in Problem 5. Start by making the change of variables
xuv yuv s2 s2
This gives a rotation about the origin through the angle 4. You will need to sketch the corresponding region in the uvplane.
Hint: If, in evaluating the integral, you encounter either of the expressions 1sin cos or cos 1sin , you might like to use the identity
cossin 2 and the corresponding identity for sin .
7. a Show that
y1 y1 y1 1 dx dy dz 1 0 0 01xyz n1n3
Nobody has ever been able to find the exact value of the sum of this series.
1024
PROBLEMS PLUS
b Show that
y1 y1 y1 1 dx dy dz 1n1 0 0 0 1xyz n1 n3
Use this equation to evaluate the triple integral correct to two decimal places. 8. Show that
y arctan xarctanx dx ln 0x2
by first expressing the integral as an iterated integral.
9. a
b
10. a
b
Show that when Laplaces equation
2u2u2u0 x2 y2 z2
is written in cylindrical coordinates, it becomes
2u1u1 2u2u0 r2 rrr22 z2
Show that when Laplaces equation is written in spherical coordinates, it becomes
2u2 ucot u1 2u1 2u0 2222 2sin2 2
A lamina has constant density and takes the shape of a disk with center the origin and radius R. Use Newtons Law of Gravitation see Section 13.4 to show that the magnitude of the force of attraction that the lamina exerts on a body with mass m located at the point 0, 0, d on the positive zaxis is 1 1
F2 Gm d dsR2d2
Hint: Divide the disk as in Figure 4 in Section 15.4 and first compute the vertical com ponent of the force exerted by the polar subrectangle Rij.
Show that the magnitude of the force of attraction of a lamina with density that occupies an entire plane on an object with mass m located at a distance d from the plane is
F2 Gm Notice that this expression does not depend on d.
11. If f is continuous, show that
yx yy yz ftdtdzdy1 yx xt2ftdt 000 20
1025
1026
16
VECTOR CALCULUS
Parametric equations enable us to plot surfaces with strange and beautiful shapes.
In this chapter we study the calculus of vector fields. These are functions that assign vectors to points in space. In particular we define line integrals which can be used to find the work done by a force field in moving an object along a curve. Then we define surface integrals which can be used to find the rate of fluid flow across a surface. The connections between these new types of integrals and the single, double, and triple integrals that we have already met are given by the higherdimensional versions of the Fundamental Theorem of Calculus: Greens Theorem, Stokes Theorem, and the Divergence Theorem.
16.1 VECTOR FIELDS
The vectors in Figure 1 are air velocity vectors that indicate the wind speed and direction at points 10 m above the surface elevation in the San Francisco Bay area. Notice that the wind patterns on consecutive days are quite different. Associated with every point in the air we can imagine a wind velocity vector. This is an example of a velocity vector field.
a 12:00 AM, February 20, 2007 b 2:00 PM, February 21, 2007 FIGURE 1 Velocity vector fields showing San Francisco Bay wind patterns
Other examples of velocity vector fields are illustrated in Figure 2: ocean currents and flow past an airfoil.
Nova Scotia
a Ocean currents off the coast of Nova Scotia b Airflow past an inclined airfoil
Werle1974
FIGURE 2 Velocity vector fields
Another type of vector field, called a force field, associates a force vector with each point in a region. An example is the gravitational force field that we will look at in Example 4.
1027
1028
CHAPTER 16 VECTOR CALCULUS
In general, a vector field is a function whose domain is a set of points in 2 or 3 and
y
0x
whose range is a set of vectors in V2 or V3.
The best way to picture a vector field is to draw the arrow representing the vector Fx, y starting at the point x, y. Of course, its impossible to do this for all points x, y, but we can gain a reasonable impression of F by doing it for a few representative points in D as in Figure 3. Since Fx, y is a twodimensional vector, we can write it in terms of its com ponent functions P and Q as follows:
Fx,yPx,yiQx,yj Px,y,Qx,y or, for short, FP iQ j
Notice that P and Q are scalar functions of two variables and are sometimes called scalar fields to distinguish them from vector fields.
A vector field F on 3 is pictured in Figure 4. We can express it in terms of its compo nent functions P, Q, and R as
Fx, y, zPx, y, z iQx, y, z jRx, y, z k
DEFINITION Let D be a set in 2 a plane region. A vector field on 2 is a function F that assigns to each point x, y in D a twodimensional vector Fx, y.
1
Fx, y x, y
FIGURE 3
Vector field on R
z Fx,y,z
x, y, z
DEFINITION Let E be a subset of 3. A vector field on 3 is a function F that assigns to each point x, y, z in E a threedimensional vector Fx, y, z.
2
0 x
FIGURE 4
Vector field on R y
F 0, 3
y
As with the vector functions in Section 13.1, we can define continuity of vector fields and show that F is continuous if and only if its component functions P, Q, and R are continuous.
We sometimes identify a point x, y, z with its position vector xx, y, z and write Fx instead of Fx, y, z. Then F becomes a function that assigns a vector Fx to a vec tor x.
V EXAMPLE1 Avectorfieldon2 isdefinedbyFx,yyixj.DescribeFby sketching some of the vectors Fx, y as in Figure 3.
SOLUTION Since F1, 0j, we draw the vector j0, 1 starting at the point 1, 0 in Figure 5. Since F0, 1i, we draw the vector 1, 0 with starting point 0, 1. Continuing in this way, we calculate several other representative values of Fx, y in the table and draw the corresponding vectors to represent the vector field in Figure 5.
F 2, 2 F 1, 0
x, y
Fx, y
x, y
Fx, y
1, 0 2, 2 3, 0 0, 1 2, 2 0, 3
0, 1 2, 2 0, 3 1, 0 2, 2 3, 0
1, 0 2, 2 3, 0 0, 1 2, 2 0, 3
0, 1 2, 2 0, 3 1, 0 2, 2 3, 0
FIGURE 5
Fx, yy ix j
It appears from Figure 5 that each arrow is tangent to a circle with center the origin.
0x
SECTION 16.1 VECTOR FIELDS1029 To confirm this, we take the dot product of the position vector xx iy j with the
vector FxFx, y:
xFxx iy jy ix jxyyx0
This shows that Fx, y is perpendicular to the position vector x, y and is therefore tangent to a circle with center the origin and radius x sx2y2 . Notice also that
Fx,ysy2 x2 sx2 y2 x
so the magnitude of the vector Fx, y is equal to the radius of the circle. M
Some computer algebra systems are capable of plotting vector fields in two or three dimensions. They give a better impression of the vector field than is possible by hand because the computer can plot a large number of representative vectors. Figure 6 shows a computer plot of the vector field in Example 1; Figures 7 and 8 show two other vector fields. Notice that the computer scales the lengths of the vectors so they are not too long and yet are proportional to their true lengths.
565
5 5 6 6 5 5
FIGURE 6
Fx, yky, xl
FIGURE 7 FIGURE 8
Fx, yky, sin xl Fx, ykln1, ln1l V EXAMPLE 2 Sketch the vector field on 3 given by Fx, y, zz k.
SOLUTION The sketch is shown in Figure 9. Notice that all vectors are vertical and point upward above the xyplane or downward below it. The magnitude increases with the distance from the xyplane.
5 6 5
z
0 x
Fx, y, zz k
simple formula. Most threedimensional vector fields, however, are virtually impossible to
FIGURE 9
We were able to draw the vector field in Example 2 by hand because of its particularly
y
M
1030
CHAPTER 16 VECTOR CALCULUS
11 z0 z0 1 1
5 z3
sketch by hand and so we need to resort to a computer algebra system. Examples are shown in Figures 10, 11, and 12. Notice that the vector fields in Figures 10 and 11 have simi lar formulas, but all the vectors in Figure 11 point in the general direction of the negative yaxis because their ycomponents are all 2. If the vector field in Figure 12 represents a velocity field, then a particle would be swept upward and would spiral around the zaxis in the clockwise direction as viewed from above.
0 1
y11x y11x y1
1 0 0 1
1 0
FIGURE 10
Fx, y, zy iz jx k
TEC In Visual 16.1 you can rotate the vector fields in Figures 1012 as well as additional fields.
z
0
y
x
FIGURE 13
Velocity field in fluid flow
FIGURE 11 FIGURE 12
Fx, y, zy i2 jx k Fx, y, zyz ixz j4z k
EXAMPLE 3 Imagine a fluid flowing steadily along a pipe and let Vx, y, z be the velocity vector at a point x, y, z. Then V assigns a vector to each point x, y, z in a certain domain E the interior of the pipe and so V is a vector field on 3 called a velocity field. A possible velocity field is illustrated in Figure 13. The speed at any given point is indicated by the length of the arrow.
Velocity fields also occur in other areas of physics. For instance, the vector field in Example 1 could be used as the velocity field describing the counterclockwise rotation of a wheel. We have seen other examples of velocity fields in Figures 1 and 2. M
EXAMPLE 4 Newtons Law of Gravitation states that the magnitude of the gravitational force between two objects with masses m and M is
F mMG r2
where r is the distance between the objects and G is the gravitational constant. This
is an example of an inverse square law. Lets assume that the object with mass M is located at the origin in 3. For instance, M could be the mass of the earth and the origin would be at its center. Let the position vector of the object with mass m be xx, y, z. Then rx, so r2x2. The gravitational force exerted on this second object acts toward the origin, and the unit vector in this direction is
x x
Therefore the gravitational force acting on the object at xx, y, z is 3 Fx mMG x
x3
Physicists often use the notation r instead of x for the position vector, so you may see
1 1
1 0 1 0 x
z
Formula 3 written in the form FmMGr3r. The function given by Equation 3 is an example of a vector field, called the gravitational field, because it associates a vector the force Fx with every point x in space.
SECTION 16.1 VECTOR FIELDS1031
Formula 3 is a compact way of writing the gravitational field, but we can also write it in terms of its component functions by using the facts that xx iy jz k and xsx2 y2 z2 :
x
FIGURE 14
y
Fx, y, zmMGx ix2 y2 z232
mMGy
x2 y2 z232
mMGz k x2 y2 z232
M
j
The gravitational field F is pictured in Figure 14.
Gravitational force field
EXAMPLE 5 Suppose an electric charge Q is located at the origin. According to Coulombs Law, the electric force Fx exerted by this charge on a charge q located at a point x, y, z with position vector xx, y, z is
FxqQ x x3
4
whereis a constant that depends on the units used. For like charges, we have qQ0 and the force is repulsive; for unlike charges, we have qQ0 and the force is attractive. Notice the similarity between Formulas 3 and 4. Both vector fields are examples of force fields.
Instead of considering the electric force F, physicists often consider the force per unit charge:
Ex 1 Fx Q x q x3
Then E is a vector field on 3 called the electric field of Q. M GRADIENT FIELDS
If f is a scalar function of two variables, recall from Section 14.6 that its gradient f or grad fis defined by
fx, yfxx, y ifyx, y j
Therefore f is really a vector field on 2 and is called a gradient vector field. Likewise,
if f is a scalar function of three variables, its gradient is a vector field on 3 given by fx, y, zfxx, y, z ifyx, y, z jfzx, y, z k
V EXAMPLE 6 Find the gradient vector field of fx, yx2yy3. Plot the gradient vector field together with a contour map of f. How are they related?
SOLUTION The gradient vector field is given by
fx,y f i f j2xyix2 3y2j
x y
Figure 15 shows a contour map of f with the gradient vector field. Notice that the gradi ent vectors are perpendicular to the level curves, as we would expect from Section 14.6.
4 4
4
FIGURE 15
4
1032
CHAPTER 16 VECTOR CALCULUS
16.1 EXERCISES
110 Sketch the vector field F by drawing a diagram like Figure 5 or Figure 9.
II 5
5 5
5 IV 5
Notice also that the gradient vectors are long where the level curves are close to each other and short where the curves are farther apart. Thats because the length of the gradi ent vector is the value of the directional derivative of f and closely spaced level curves indicate a steep graph. M
A vector field F is called a conservative vector field if it is the gradient of some scalar function, that is, if there exists a function f such that F f . In this situation f is called a potential function for F.
Not all vector fields are conservative, but such fields do arise frequently in physics. For example, the gravitational field F in Example 4 is conservative because if we define
I 3
2 33
1. Fx,y1ij 2. Fx,yixj 2
3. Fx,yyi1 j 4. Fx,yxyixj
5. Fx,y yixj sx2 y2
7. Fx,y,zk
8. Fx,y,zyk 9. Fx,y,zxk
6. Fx,y yixj sx2 y2
10. Fx,y,zji
1114 Match the vector fields F with the plots labeled IIV.
Give reasons for your choices. 11. Fx,yy,x
12. Fx, y1, sin y
13. Fx,yx2,x1 14. Fx, yy, 1x
3 35 5 3 5
1518 Match the vector fields F on 3 with the plots labeled IIV. Give reasons for your choices.
15. Fx,y,zi2j3k 16. Fx,y,zi2jzk
fx, y, zmMG sx2 y2 z2
then
Fx, y, z
In Sections 16.3 and 16.5 we will learn how to tell whether or not a given vector field is
fx,y,z f i f j f k x y z
mMGx i mMGy j mMGz k x2 y2 z232 x2 y2 z232 x2 y2 z232
conservative.
3 III 3
17. Fx,y,zxiyj3k 18. Fx,y,zxiyjzk
I II
11
SECTION 16.1 VECTOR FIELDS1033 31. fx,yxy2 32. fx,ysinsx2 y2
I 4 II 4
z0 1
III
1 z0 1
z0 1
III 4 z0 4
1 y0 1
1 0x
y 1 1 x
1 0
1 0 1 1 0 1
4 44 4 11001 4 4
IV
1 1
IV 4
44 4
0 1
y11xyx 4 4
CAS 19. If you have a CAS that plots vector fields the command is fieldplot in Maple and PlotVectorField in Mathematica, use it to plot
Fx,yy2 2xyi3xy6x2j Explain the appearance by finding the set of points x, y
such that Fx, y0.
CAS 20. LetFxr2 2rx,wherex x,y andrx.Usea CAS to plot this vector field in various domains until you can see what is happening. Describe the appearance of the plot and explain it by finding the points where Fx0.
2124 Find the gradient vector field of f.
33. A particle moves in a velocity field Vx, yx2, xy2. If it is at position 2, 1 at time t3, estimate its location at time t3.01.
34. At time t1, a particle is located at position 1, 3. If it moves in a velocity field
Fx, yxy2, y210 find its approximate location at time t1.05.
35. The flow lines or streamlines of a vector field are the paths followed by a particle whose velocity field is the given vector field. Thus the vectors in a vector field are tangent to the flow lines.
a UseasketchofthevectorfieldFx,yxiyj to draw some flow lines. From your sketches, can you guess the equations of the flow lines?
b If parametric equations of a flow line are xxt,
yyt, explain why these functions satisfy the differ ential equations dxdtx and dydty. Then solve the differential equations to find an equation of the flow line that passes through the point 1, 1.
21. fx, yxexy
23. fx,y,zsx2 y2 z2
22. fx, ytan3x4y 24. fx,y,zxcosyz
2526 Find the gradient vector field f of f and sketch it. 25. fx,yx2 y 26. fx,ysx2 y2
CAS 2728 Plot the gradient vector field of f together with a contour map of f. Explain how they are related to each other.
27. fx,ysinxsiny 28. fx,ysinxy 2932 Match the functions f with the plots of their gradient
vector fields labeled IIV. Give reasons for your choices. 29. fx,yx2 y2 30. fx,yxxy
36. a
Sketch the vector field Fx, yix j and then sketch some flow lines. What shape do these flow lines appear to have?
b If parametric equations of the flow lines are xxt,
yyt, what differential equations do these functions satisfy? Deduce that dydxx.
c If a particle starts at the origin in the velocity field given by F, find an equation of the path it follows.
1034
CHAPTER 16 VECTOR CALCULUS
16.2
LINE INTEGRALS
In this section we define an integral that is similar to a single integral except that instead of integrating over an interval a, b, we integrate over a curve C. Such integrals are called line integrals, although curve integrals would be better terminology. They were invented in the early 19th century to solve problems involving fluid flow, forces, electricity, and magnetism.
We start with a plane curve C given by the parametric equations xxt yyt atb
or, equivalently, by the vector equation rtxt iyt j, and we assume that C is a smooth curve. This means that r is continuous and rt0. See Section 13.3. If we divide the parameter interval a, b into n subintervals ti1, tiof equal width and we let xixtiand yiyti , then the corresponding points Pi xi, yidivide C into n subarcs with lengths s1, s2, . . . , sn. See Figure 1. We choose any point Pixi, yi in the ith subarc. This corresponds to a point ti in ti1, ti. Now if f is any function of two vari ables whose domain includes the curve C, we evaluate f at the point xi, yi, multiply by the length si of the subarc, and form the sum
n
fx i , y i s i
i1
which is similar to a Riemann sum. Then we take the limit of these sums and make the fol
lowing definition by analogy with a single integral.
y
P x ,yi i i
Pi1
C Pi
Pn
0x
t i
abt ti1 ti
FIGURE 1
PTM P
P
1
DEFINITION If f is defined on a smooth curve C given by Equations 1, then the line integral of f along C is
2
if this limit exists.
fx, y dslim fxi, yi si y n
C nl i1
In Section 10.2 we found that the length of C is
yb dx2 dy2
A similar type of argument can be used to show that if f is a continuous function, then the limit in Definition 2 always exists and the following formula can be used to evaluate the line integral:
The value of the line integral does not depend on the parametrization of the curve, pro vided that the curve is traversed exactly once as t increases from a to b.
L dtdtdt a
3
C fx,yds a fxt,yt dtdt dt y yb dx2 dy2
N The arc length function s is discussed in Section 13.3.
z 0
FIGURE 2
dt dtdt
So the way to remember Formula 3 is to express everything in terms of the parameter t:
SECTION 16.2 LINE INTEGRALS If st is the length of C between ra and rt, then
ds dx 2 dy 2
1035
Use the parametric equations to express x and y in terms of t and write ds as dx 2 dy 2
In the special case where C is the line segment that joins a, 0 to b, 0, using x as the parameter, we can write the parametric equations of C as follows: xx, y0, axb. Formula 3 then becomes
y f x, y dsyb f x, 0 dx Ca
and so the line integral reduces to an ordinary single integral in this case.
Just as for an ordinary single integral, we can interpret the line integral of a positive function as an area. In fact, if f x, y0, xC f x, y ds represents the area of one side of the fence or curtain in Figure 2, whose base is C and whose height above the point
ds dtdt dt
C
x, y
fx,y y
x
y
x,yis fx,y.
EXAMPLE 1 Evaluate x 2x2y ds, where C is the upper half of the unit circle
C
x2 y2 1.
SOLUTION In order to use Formula 3, we first need parametric equations to represent C.
1y 0
Recall that the unit circle can be parametrized by means of the equations
xcos t ysin t
and the upper half of the circle is described by the parameter interval 0t
See Figure 3. Therefore Formula 3 gives
.
101x yydx2dy2 2x2y ds2cos2t sin t
dt
FIGURE 3
C 0 dtdty 2cos2t sin tssin2tcos2t dt
0
y 2cos2tsintdt 2t cos3t
y030 C 2 2
M
C 3
C C
0x
FIGURE 4
A piecewisesmooth curve
CTM
Suppose now that C is a piecewisesmooth curve; that is, C is a union of a finite num ber of smooth curves C1, C2, . . . , Cn , where, as illustrated in Figure 4, the initial point of Ci1 is the terminal point of Ci . Then we define the integral of f along C as the sum of the integrals of f along each of the smooth pieces of C:
y fx,ydsy fx,ydsy fx,ydsy fx,yds C C1 C2 Cn
1036
y
1, 2 C 1,1
0,0 x FIGURE 5
CCCTM
CTM
xx yx2 0x1
y 2x dsy1 2x dxdy dxy1 2xs14×2 dx
CHAPTER 16 VECTOR CALCULUS
EXAMPLE 2 Evaluate x 2x ds, where C consists of the arc C1 of the parabola yx2
C
from 0, 0 to 1, 1 followed by the vertical line segment C2 from 1, 1 to 1, 2.
SOLUTION The curve C is shown in Figure 5. C1 is the graph of a function of x, so we can choose x as the parameter and the equations for C1 become
Therefore
2
1 214×232
2
C1 0dxdx0
1 430
5s51 6
On C2 we choose y as the parameter, so the equations of C2 are x1 yy 1y2
and Thus
2x ds21dy2 dy2 y y2 dx2 dy2 y2
C2 1dydy1 y2xdsy 2xdsy 2xds 5s5 1 2
M
C C1 C2 6
AnyphysicalinterpretationofalineintegralxC fx,ydsdependsonthephysicalinter pretation of the function f. Suppose that x, y represents the linear density at a point x, y of a thin wire shaped like a curve C. Then the mass of the part of the wire from Pi1 to Pi in Figure 1 is approximately xi , yi si and so the total mass of the wire is approx imatelyxi , yi si. By taking more and more points on the curve, we obtain the mass m of the wire as the limiting value of these approximations:
mlimn xi,yisiy x,yds nl i1 C
For example, if fx, y2x2y represents the density of a semicircular wire, then the integral in Example 1 would represent the mass of the wire. The center of mass of the wire with density function is located at the point x, y, where
x 1 yx x,yds y 1 yy x,yds mC mC
Other physical interpretations of line integrals will be discussed later in this chapter.
V EXAMPLE3 Awiretakestheshapeofthesemicirclex2 y2 1,y0,andis thicker near its base than near the top. Find the center of mass of the wire if the linear density at any point is proportional to its distance from the line y1.
SOLUTION AsinExample1weusetheparametrizationxcost,ysint,0t , and find that dsdt. The linear density is
x, yk1y
4
SECTION 16.2 LINE INTEGRALS1037 where k is a constant, and so the mass of the wire is
myk1ydsyk1sintdtktcost0 k 2 C0
From Equations 4 we have
y 1 yy x,yds 1 yyk1yds m C k 2 C
y 1
1 y sin tsin2t dt1 20 2
4
2 2
cos t1 t1 sin 2t 240
center of mass
By symmetry we see that x0, so the center of mass is
1 0 1 x FIGURE 6
0, 4
2 2
0, 0.38
See Figure 6.
M
Two other line integrals are obtained by replacingsi by eitherx ix ix i1 or yiyiyi1 in Definition 2. They are called the line integrals of f along C with respect to x and y:
fxi ,yixi fxi ,yiyi
5
fx,ydxlim
y n
6
C nl i1 fx,ydylim
y n C nl i1
When we want to distinguish the original line integral xC f x, y ds from those in Equa tions 5 and 6, we call it the line integral with respect to arc length.
The following formulas say that line integrals with respect to x and y can also be evaluated by expressing everything in terms of t: xxt, yyt, dxxt dt, dyyt dt.
It frequently happens that line integrals with respect to x and y occur together. When this happens, its customary to abbreviate by writing
yC Px,ydxyC Qx,ydyyC Px,ydxQx,ydy
When we are setting up a line integral, sometimes the most difficult thing is to think of a parametric representation for a curve whose geometric description is given. In particular, we often need to parametrize a line segment, so its useful to remember that a vector rep
7
y fx,ydxyb fxt,ytxtdt Ca
y fx,ydyyb fxt,ytytdt Ca
1038
CHAPTER 16 VECTOR CALCULUS
resentation of the line segment that starts at r0 and ends at r1 is given by
5, 3
FIGURE 7
Formulas 7 give
b Since the parabola is given as a function of y, lets take y as the parameter and write C2 as
x4y2 yy 3y2 Then dx2y dy and by Formulas 7 we have
y y2dxxdyy2 y22ydy4y2dy C2 3
y2 2y3 y2 4dy 3
y
See Equation 12.5.4.
V EXAMPLE 4 Evaluate xC y2 dxx dy, where a CC1 is the line segment from 5, 3 to 0, 2 and b CC2 is the arc of the parabola x4y2 from 5, 3 to 0, 2. See Figure 7.
SOLUTION
a A parametric representation for the line segment is
x5t5 y5t3
Use Equation 8 with r05, 3 and r10, 2. Then dx5 dt, dy5 dt, and
0,2
CTM
C
04x
x4
0t1
y y2 dxxdyy1 5t325dt5t55dt C1 0
5y1 25t2 25t4dt 0
0
25t3 25t2 1 5 5 32 4t 6
that
C1
6
8
rt1tr0 tr1 0t1
x5t
y25t 0t1
y y2dxxdy5
y4 y3 2
234y 405 M
6
Notice that we got different answers in parts a and b of Example 4 even though the two curves had the same endpoints. Thus, in general, the value of a line integral depends not just on the endpoints of the curve but also on the path. But see Section 16.3 for con ditions under which the integral is independent of the path.
Notice also that the answers in Example 4 depend on the direction, or orientation, of the curve. If C1 denotes the line segment from 0, 2 to 5, 3, you can verify, using the parametrization
3
C
A
A
FIGURE 8
B
t
In general, a given parametrization xxt, yyt, atb, determines an orien tation of a curve C, with the positive direction corresponding to increasing values of the parameter t. See Figure 8, where the initial point A corresponds to the parameter value a and the terminal point B corresponds to tb.
If C denotes the curve consisting of the same points as C but with the opposite ori entation from initial point B to terminal point A in Figure 8, then we have
SECTION 16.2 LINE INTEGRALS1039
ab
yC fx,ydxyC fx,ydx yC fx,ydyyC fx,ydy
But if we integrate with respect to arc length, the value of the line integral does not change
when we reverse the orientation of the curve:
yC fx,ydsyC fx,yds
This is because si is always positive, whereas xi and yi change sign when we reverse
the orientation of C.
LINE INTEGRALS IN SPACE
We now suppose that C is a smooth space curve given by the parametric equations xxt yyt zzt atb
or by a vector equation rtxt iyt jzt k. If f is a function of three variables that is continuous on some region containing C, then we define the line integral of f along C with respect to arc length in a manner similar to that for plane curves:
B
C
fx, y, z dslim fxi , yi, zi si y n
C nl i1
We evaluate it using a formula similar to Formula 3:
y yb dx2 dy2 dz2
C fx,y,zds a fxt,yt,zt dtdtdt dt
Observe that the integrals in both Formulas 3 and 9 can be written in the more compact vector notation
yb frtrtdt a
Forthespecialcase fx,y,z1,weget
y dsyb rtdtL
Ca
where L is the length of the curve C see Formula 13.3.3.
9
1040
CHAPTER 16 VECTOR CALCULUS
Line integrals along C with respect to x, y, and z can also be defined. For example,
fx, y, z dzlim fxi , yi, zi zi y n
C nl i1
yb fxt,yt,ztztdt
a
6 4
z
2
Therefore, as with line integrals in the plane, we evaluate integrals of the form
yC Px,y,zdxQx,y,zdyRx,y,zdz
by expressing everything x, y, z, dx, dy, dz in terms of the parameter t.
V EXAMPLE 5 Evaluate xC y sin z ds, where C is the circular helix given by the equations xcost,ysint,zt,0t2 .SeeFigure9.
SOLUTION Formula 9 gives Cysinzds 0
dx2 dy2 dz2 sintsint dtdtdt dt
C
y y2
0 2 2
1 1 y 0 0 x
1 1
y sin2tssin2tcos2t1 dts2 y 00
s2 1 2
2 t2 sin 2t0s2
11cos 2t dt 2
10
M
FIGURE 9
EXAMPLE 6 Evaluate xC y dxz dyx dz, where C consists of the line segment C1 from 2, 0, 0 to 3, 4, 5, followed by the vertical line segment C2 from 3, 4, 5 to 3, 4, 0.
SOLUTION The curve C is shown in Figure 10. Using Equation 8, we write C1 as rt1t2, 0, 0t3, 4, 52t, 4t, 5t
or, in parametric form, as
z
3, 4, 5
C CTM
2,0,0 0 y x 3,4,0
FIGURE 10
x2t y4t z5t 0t1
Thus
0 1029tdt10t29 2 Likewise, C2 can be written in the form
rt1t3, 4, 5t3, 4, 03, 4, 55t or x3 y4 z55t 0t1
y dxz dyx dzyy
1
4t dt5t4 dt2t5 dt
C1 0
y1 t
21
24.5
0
z
Fxi,yi,ziTti
Pi1
gravitational field of Example 4 in Section 16.1 or the electric force field of Example 5 in Section 16.1. A force field on 2 could be regarded as a special case where R0 and P and Q depend only on x and y. We wish to compute the work done by this force in mov ing a particle along a smooth curve C.
We divide C into subarcs Pi1Pi with lengths si by dividing the parameter interval a, b into subintervals of equal width. See Figure 1 for the twodimensional case or Figure 11 for the threedimensional case. Choose a point Pixi , yi, zi on the ith subarc corresponding to the parameter value ti. If si is small, then as the particle moves from Pi1 to Pi along the curve, it proceeds approximately in the direction of Tti, the unit tan
Then dx0dy, so
y ydxzdyxdzy1 35dt15
C2 0 Adding the values of these integrals, we obtain
yC ydxzdyxdz24.5159.5 LINE INTEGRALS OF VECTOR FIELDS
M
Recall from Section 6.4 that the work done by a variable force fx in moving a particle from a to b along the xaxis is Wxb f x dx. Then in Section 12.3 we found that the
a
work done by a constant force F in moving an object from a point P to another point Q in l
space is WFD, where DPQ is the displacement vector.
Now suppose that FP iQ jR k is a continuous force field on 3, such as the
SECTION 16.2 LINE INTEGRALS
1041
P i
0 Pixi,yi,zi
P
Pn y
gent vector at Pi . Thus the work done by the force F in moving the particle from Pi1 to Pi is approximately
Fxi , yi, zisi TtiFxi , yi, ziTti si and the total work done in moving the particle along C is approximately
n
Fxi , yi, ziTxi , yi, zi si
i1
where Tx, y, z is the unit tangent vector at the point x, y, z on C. Intuitively, we see that these approximations ought to become better as n becomes larger. Therefore we define the work W done by the force field F as the limit of the Riemann sums in 11, namely,
WyC Fx,y,zTx,y,zdsyC FTds
Equation 12 says that work is the line integral with respect to arc length of the tangential component of the force.
If the curve C is given by the vector equation rtxt iyt jzt k, then Ttrtrt, so using Equation 9 we can rewrite Equation 12 in the form
WybFrt rtrtdtybFrtrtdt a rt a
x
FIGURE 11
11
12
1042
CHAPTER 16 VECTOR CALCULUS
DEFINITION Let F be a continuous vector field defined on a smooth curve C given by a vector function rt, atb. Then the line integral of F along C is
yFdryb FrtrtdtyFTds CaC
13
N Figure 12 shows the force field and the curve in Example 7. The work done is negative because the field impedes movement along the curve.
y
1
01x FIGURE 12
N Figure 13 shows the twisted cubic C in Example 8 and some typical vectors acting at three points on C.
This integral is often abbreviated as xC Fdr and occurs in other areas of physics as well. Therefore we make the following definition for the line integral of any continuous vector field.
When using Definition 13, remember that Frt is just an abbreviation for Fxt, yt, zt, so we evaluate Frt simply by putting xxt, yyt, and zzt in the expression for Fx, y, z. Notice also that we can formally write drrt dt.
EXAMPLE 7 FindtheworkdonebytheforcefieldFx,yx2 ixyjinmovinga particle along the quartercircle rtcos t isin t j, 0t2.
SOLUTION Sincexcostandysint,wehave Frtcos2ticost sintj
rtsin t icos t j y y2y2
Even though xC FdrxC FT ds and integrals with respect to arc length are unchanged when orientation is reversed, it is still true that
yC FdryC Fdr
because the unit tangent vector T is replaced by its negative when C is replaced by C.
EXAMPLE 8 EvaluatexC Fdr,whereFx,y,zxyiyzjzxkandCisthe twisted cubic given by
and
Therefore the work done is
Frtrt dt2 cos2t sin t dt cos3t 2 2
FdrC00
233 M 0
2 1.5 z1 0.5
Fr1
1, 1, 1
Fr34 C Fr12
yt2
rtt it2 jt3 k
rti2t j3t2 k Frtt3it5jt4k
NOTE
xt
zt3
0t1
0
y221 x
FIGURE 13
0
SOLUTION We have
01
16.2 EXERCISES
116 Evaluate the line integral, where C is the given curve.
1. x y3 ds, C:xt3, yt, 0t2 C
2. x xyds, C:xt2, y2t, 0t1 C
3. xC xy4ds, Cistherighthalfofthecirclex2 y2 16
4. xC x sin y ds, C is the line segment from 0, 3 to 4, 6
5. xC x2y3 sxdy,
C is the arc of the curve ysx from 1, 1 to 4, 2
Fx, y, zy iz jx k
9. xCxyzds,
C: x2 sin t, yt, z2 cos t, 0t
10. xC xyz2 ds,
C is the line segment from 1, 5, 0 to 1, 6, 4
xC xeyz ds,
C is the line segment from 0, 0, 0 to 1, 2, 3
SECTION 16.2 LINE INTEGRALS Thus yFdry1 Frtrtdt
1043
M
Finally, we note the connection between line integrals of vector fields and line integrals of scalar fields. Suppose the vector field F on 3 is given in component form by the equa tion FP iQ jR k. We use Definition 13 to compute its line integral along C:
yFdryb Frtrtdt Ca
yb PiQjRkxtiytjztkdt a
yb Pxt, yt, ztxtQxt, yt, ztytRxt, yt, ztztdt a
But this last integral is precisely the line integral in 10. Therefore we have
For example, the integral xC y dxz dyx dz in Example 6 could be expressed as xC Fdr where
C0 y1
t3 5t6dt 0
t4 5t71 27
4 7 0 28
12.xC2x9zds, C:xt,yt2,zt3,0t1 13.xx2yszdz, C:xt3,yt,zt2,0t1
6. x xey dx,
C C232
C is the arc of the curve xey from 1, 0 to e, 1
xC xy dxxy dy, C consists of line segments from
0, 0 to 2, 0 and from 2, 0 to 3, 2
8. xC sinxdxcosydy, Cconsistsofthetophalfofthecircle
x2y21 from 1, 0 to 1, 0 and the line segment from 1, 0 to 2, 3
C: xt , yt , zt , 0t1
7.
yC FdryC PdxQdyRdz where FPiQjRk
11.
C 14.xzdxxdyydz,
15. xC xyzdx2xdyxyzdz, Cconsistsofline segments from 1, 0, 1 to 2, 3, 1 and from 2, 3, 1 to 2, 5, 2
16. x x2 dxy2 dyz2 dz, C consists of line segments from C
0, 0, 0 to 1, 2, 1 and from 1, 2, 1 to 3, 2, 0
1044CHAPTER 16 VECTOR CALCULUS Let F be the vector field shown in the figure.
a If C1 is the vertical line segment from 3, 3 to 3, 3, determine whether xC1 Fdr is positive, negative, or zero.
b If C2 is the counterclockwiseoriented circle with radius 3
xC Fdr,whereFx,y,zysinzizsinxjxsinyk and rtcos t isin t jsin 5t k, 0t
17.
and center the origin, determine whether x Fdr is posi
x x sinyz ds, where C has parametric equations xt2, C
tive, negative, or zero.
2 1
C2
123x 27. 28.
yt3,zt4, 0t5
x zexy ds, where C has parametric equations xt, yt2,
y 3
C
zet, 0t1
24. 25. 26.
CAS 2728 Use a graph of the vector field F and the curve C to guess whether the line integral of F over C is positive, negative, or zero. Then evaluate the line integral.
3 2 10 1
2 3
18. The figure shows a vector field F and two curves C1 and C2. Are the line integrals of F over C1 and C2 positive, negative, or zero? Explain.
Fx,yxyixyj,
C is the arc of the circle x2y24 traversed counter clockwise from 2, 0 to 0, 2
Fx,y x i y j, sx2 y2 sx2 y2
C is the parabola y1x2 from 1, 2 to 1, 2
a Evaluate the line integral xC Fdr, where Fx,yex1 ixyjandCisgivenby rtt2 it3 j, 0t1.
;
;
y
C
CTM
x
29.
30.
b Illustrate part a by using a graphing calculator or com puter to graph C and the vectors from the vector field corresponding to t0, 1s2 , and 1 as in Figure 13.
a Evaluate the line integral xC Fdr, where Fx, y, zx iz jy k and C is given by rt2t i3t jt2 k, 1t1.
CAS 31. 1922 Evaluate the line integral xC Fdr, where C is given by the
vector function rt. 32.
19. Fx,yxyi3y2j,
rt11t4 it3 j, 0t1
20. Fx,y,zxyiyzjz2k, CAS
Find the exact value of x x3y2zds, where C is the curve with C
33.
b Illustrate part a by using a computer to graph C and
the vectors from the vector field corresponding to t1
and 1 as in Figure 13. 2
parametric equations xet cos 4t, yet sin 4t, zet, 0t2 .
rtt2 it3 jt2 k, 0t1
21. Fx,y,zsinxicosyjxzk,
rtt3it2jtk, 0t1
22. Fx,y,zziyjxk,
rttisintjcostk, 0t
2326 Use a calculator or CAS to evaluate the line integral correct
to four decimal places.
23. xC Fdr,whereFx,yxyisinyjand
rtet iet2 j, 1t2
34.
35.
aFindtheworkdonebytheforcefieldFx,yx2ixyj on a particle that moves once around the circle x2y24 oriented in the counterclockwise direction.
b Use a computer algebra system to graph the force field and circle on the same screen. Use the graph to explain your answer to part a.
22
A thin wire is bent into the shape of a semicircle xy4,
x0. If the linear density is a constant k, find the mass and center of mass of the wire.
A thin wire has the shape of the firstquadrant part of the circle with center the origin and radius a. If the density function is x, ykxy, find the mass and center of mass of the wire.
a Write the formulas similar to Equations 4 for the center of mass x, y, z of a thin wire in the shape of a space curve C ifthewirehasdensityfunction x,y,z.
b Find the center of mass of a wire in the shape of the helix x2sint,y2cost,z3t,0t2 ,ifthedensity is a constant k.
36. Find the mass and center of mass of a wire in the shape of the helixxt,ycost,zsint,0t2 ,ifthedensityat any point is equal to the square of the distance from the origin.
37. If a wire with linear density x, y lies along a plane curve C, its moments of inertia about the x and yaxes are defined as
SECTION 16.2 LINE INTEGRALS1045 b Is this also true for a force field Fxkx, where k is a
constantandxx,y?
46. The base of a circular fence with radius 10 m is given by
x10 cos t, y10 sin t. The height of the fence at position x, y is given by the function hx, y40.01×2y2, so the height varies from 3 m to 5 m. Suppose that 1 L of paint covers 100 m2. Sketch the fence and determine how much paint you will need if you paint both sides of the fence.
Ixy2 x,yds Iyx2 x,yds y y
Find the moments of inertia for the wire in Example 3.
38. If a wire with linear density x, y, z lies along a space curve
C, its moments of inertia about the x, y, and zaxes are defined as
Ix yy2 z2 x,y,zds C
Iy yx2 z2 x,y,zds C
Iz yx2 y2 x,y,zds C
Find the moments of inertia for the wire in Exercise 35.
39. FindtheworkdonebytheforcefieldFx,yxiy2j in moving an object along an arch of the cycloid rttsinti1costj,0t2 .
40. FindtheworkdonebytheforcefieldFx,yxsinyiyj on a particle that moves along the parabola yx2 from
1, 1 to 2, 4.
41. Find the work done by the force field
Fx,y,z yz,xz,xy onaparticlethatmoves along the line segment from 1, 0, 0 to 3, 4, 2.
42. The force exerted by an electric charge at the origin on a charged particle at a point x, y, z with position vector
rx, y, z is FrKrr 3 where K is a constant. See Example 5 in Section 16.1. Find the work done as the particle moves along a straight line from 2, 0, 0 to 2, 1, 5.
A 160lb man carries a 25lb can of paint up a helical staircase that encircles a silo with a radius of 20 ft. If the silo is 90 ft high and the man makes exactly three complete revolutions, how much work is done by the man against gravity in climbing to the top?
44. Suppose there is a hole in the can of paint in Exercise 43 and 9 lb of paint leaks steadily out of the can during the mans ascent. How much work is done?
45. a Show that a constant force field does zero work on a particle that moves once uniformly around the circle x2 y2 1.
47.
An object moves along the curve C shown in the figure from 1, 2 to 9, 8. The lengths of the vectors in the force field F are measured in newtons by the scales on the axes. Estimate the work done by F on the object.
y
meters
1 01x
meters
Experiments show that a steady current I in a long wire pro duces a magnetic field B that is tangent to any circle that lies in the plane perpendicular to the wire and whose center is the axis of the wire as in the figure. Amperes Law relates the electric current to its magnetic effects and states that
yBdr 0I C
where I is the net current that passes through any surface bounded by a closed curve C, and 0 is a constant called the permeability of free space. By taking C to be a circle with radius r, show that the magnitude BBof the magnetic field at a distance r from the center of the wire is
B 0I 2r
I
CC
C
C
43.
48.
B
1046
CHAPTER 16 VECTOR CALCULUS
16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS
Recall from Section 5.3 that Part 2 of the Fundamental Theorem of Calculus can be writ
ten as
yb Fx dxFbFa a
where F is continuous on a, b. We also called Equation 1 the Net Change Theorem: The integral of a rate of change is the net change.
If we think of the gradient vector f of a function f of two or three variables as a sort of derivative of f, then the following theorem can be regarded as a version of the Funda mental Theorem for line integrals.
THEOREM Let C be a smooth curve given by the vector function rt, atb. Let f be a differentiable function of two or three variables whose gradient vector f is continuous on C. Then
yC fdrfrbfra
2
NOTE Theorem 2 says that we can evaluate the line integral of a conservative vector field the gradient vector field of the potential function fsimply by knowing the value of f at the endpoints of C. In fact, Theorem 2 says that the line integral of f is the net change in f. If f is a function of two variables and C is a plane curve with initial point
Ax1, y1and terminal point Bx2, y2 , as in Figure 1, then Theorem 2 becomes yC fdrfx2,y2fx1,y1
If f is a function of three variables and C is a space curve joining the point Ax1, y1, z1to the point Bx2, y2, z2 , then we have
yC fdrfx2,y2,z2fx1,y1,z1 Lets prove Theorem 2 for this case.
Ax, y
yz
BxTM, yTM
Ax, y, z 0Cx0
x
C
FIGURE 1
1
BxTM, yTM, zTM
y
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS1047 PROOF OF THEOREM 2 Using Definition 16.2.13, we have
y fdryb frtrt dt
yb f dxf dyf dz dt
Ca a x dt y dt z dt
yb d f rt dt a dt
by the Chain Rule
f rbf ra
The last step follows from the Fundamental Theorem of Calculus Equation 1.
M
Although we have proved Theorem 2 for smooth curves, it is also true for piecewise smooth curves. This can be seen by subdividing C into a finite number of smooth curves and adding the resulting integrals.
EXAMPLE 1 Find the work done by the gravitational field Fx mMG x
x3
in moving a particle with mass m from the point 3, 4, 12 to the point 2, 2, 0 along a
piecewisesmooth curve C. See Example 4 in Section 16.1.
SOLUTION From Section 16.1 we know that F is a conservative vector field and, in fact,
Ff, where
fx, y, zmMG sx2 y2 z2
Therefore, by Theorem 2, the work done is
WyC FdryC fdr
f 2, 2, 0f 3, 4, 12 mMG mMG mMG11 M
s22 22 s32 42 122 INDEPENDENCE OF PATH
2s2 13
Suppose C1 and C2 are two piecewisesmooth curves which are called paths that have the same initial point A and terminal point B. We know from Example 4 in Section 16.2 that, in general, xC1 FdrxC2 Fdr. But one implication of Theorem 2 is that
y fdry fdr C1 C2
whenever f is continuous. In other words, the line integral of a conservative vector field depends only on the initial point and terminal point of a curve.
In general, if F is a continuous vector field with domain D, we say that the line integral xC FdrisindependentofpathifxC1 FdrxC2 FdrforanytwopathsC1 andC2 in
1048
CHAPTER 16 VECTOR CALCULUS
FIGURE 2
A closed curve
CTM
yFdry Fdry Fdry Fdry Fdr0 C C1 C2 C1 C2
A C
since C1 and C2 have the same initial and terminal points.
Conversely, if it is true that xC Fdr0 whenever C is a closed path in D, then we
demonstrate independence of path as follows. Take any two paths C1 and C2 from A to B in D and define C to be the curve consisting of C1 followed by C2. Then
0yFdry Fdry Fdry Fdry Fdr C C1 C2 C1 C2
and so xC1 FdrxC2 Fdr. Thus we have proved the following theorem.
Since we know that the line integral of any conservative vector field F is independent of path, it follows that xC Fdr0 for any closed path. The physical interpretation is that the work done by a conservative force field such as the gravitational or electric field in Section 16.1 as it moves an object around a closed path is 0.
The following theorem says that the only vector fields that are independent of path are conservative. It is stated and proved for plane curves, but there is a similar version for space curves. We assume that D is open, which means that for every point P in D there is a disk with center P that lies entirely in D. So D doesnt contain any of its boundary points. In addition, we assume that D is connected. This means that any two points in D can be joined by a path that lies in D.
PROOF Let Aa, b be a fixed point in D. We construct the desired potential function f by defining
fx, yyx, y Fdr a, b
for any point x, y in D. Since xC Fdr is independent of path, it does not matter which path C from a, b to x, y is used to evaluate f x, y. Since D is open, there exists a disk contained in D with center x, y. Choose any point x1, y in the disk with x1x and let C consist of any path C1 from a, b to x1, y followed by the horizontal line segment C2 from x1, y to x, y. See Figure 4. Then
fx,yy Fdry Fdryx1,y Fdry Fdr C1 C2 a, b C2
FIGURE 3
y
0x
FIGURE 4
x, y D
C
B
D that have the same initial and terminal points. With this terminology we can say that line integrals of conservative vector fields are independent of path.
A curve is called closed if its terminal point coincides with its initial point, that is, rbra. See Figure 2. If xC Fdr is independent of path in D and C is any closed path in D, we can choose any two points A and B on C and regard C as being composed of the path C1 from A to B followed by the path C2 from B to A. See Figure 3. Then
THEOREM xC FdrisindependentofpathinDifandonlyifxC Fdr0for every closed path C in D.
3
THEOREM Suppose F is a vector field that is continuous on an open connected region D. If xC Fdr is independent of path in D, then F is a conservative vector field on D; that is, there exists a function f such that fF.
4
x, y C
a, b
CTM
y
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS Notice that the first of these integrals does not depend on x, so
fx,y0y Fdr x x C2
If we write FP iQ j, then
y Fdry PdxQdy
C2 C2
On C2, y is constant, so dy0. Using t as the parameter, where x1tx, we have
1049
x,y CTM C
a, b
fx,yy PdxQdyyx Pt,ydtPx,y
0x
ment, using a vertical line segment see Figure 5, shows that
Thus
which says that F is conservative.
x, y D
x x C2
x x1
by Part 1 of the Fundamental Theorem of Calculus see Section 5.3. A similar argu
fx,yy PdxQdyyy Qx,tdtQx,y
FIGURE 5
y y1 x y
y y C2
FPiQj f i f jf
The question remains: How is it possible to determine whether or not a vector field F is conservative? Suppose it is known that FP iQ j is conservative, where P and Q have continuous firstorder partial derivatives. Then there is a function f such that Ff, that is,
Pf and Qf x y
Therefore, by Clairauts Theorem,
P 2f2f Q y yx xy x
M
THEOREM If Fx, yPx, y iQx, y j is a conservative vector field, where P and Q have continuous firstorder partial derivatives on a domain D, then throughout D we have
PQ y x
5
simple, not closed
simple, closed
FIGURE 6
Types of curves
not simple, not closed
not simple, closed
The converse of Theorem 5 is true only for a special type of region. To explain this, we first need the concept of a simple curve, which is a curve that doesnt intersect itself any where between its endpoints. See Figure 6; rarb for a simple closed curve, but rt1rt2whenat1 t2 b.
In Theorem 4 we needed an open connected region. For the next theorem we need a stronger condition. A simplyconnected region in the plane is a connected region D such
1050CHAPTER 16 VECTOR CALCULUS
simplyconnected region
regions that are not simplyconnected
that every simple closed curve in D encloses only points that are in D. Notice from Figure 7 that, intuitively speaking, a simplyconnected region contains no hole and cant consist of two separate pieces.
In terms of simplyconnected regions we can now state a partial converse to Theorem 5 that gives a convenient method for verifying that a vector field on 2 is conservative. The proof will be sketched in the next section as a consequence of Greens Theorem.
6 THEOREM Let FP iQ j be a vector field on an open simplyconnected region D. Suppose that P and Q have continuous firstorder derivatives and
FIGURE 7
PQ y x
throughout D
10
10
is conservative.
FIGURE 8
10
C
10
Then F is conservative.
V EXAMPLE 2 Determine whether or not the vector field
N Figures 8 and 9 show the vector fields in
Examples 2 and 3, respectively. The vectors in
Figure 8 that start on the closed curve C all
appear to point in roughly the same direction
asC.Soitlooksasifx Fdr0andthere C
Fx,y32xyix2 3y2j SOLUTION LetPx,y32xyandQx,yx2 3y2.Then
P2xQ y x
Also, the domain of F is the entire plane D2, which is open and simply connected. Therefore we can apply Theorem 6 and conclude that F is conservative. M
In Example 3, Theorem 6 told us that F is conservative, but it did not tell us how to find the potential function f such that F f . The proof of Theorem 4 gives us a clue as to how to find f. We use partial integration as in the following example.
EXAMPLE 4
a IfFx,y32xyix2 3y2j,findafunction f suchthatFf. b Evaluate the line integral xC Fdr, where C is the curve given by
rtet sin t iet cos t j 0t
fore F is not conservative. The calculation in Example 2 confirms this impression. Some of the vectors near the curves C1 and C2 in Figure 9 point in approximately the same direction as the curves, whereas others point in the opposite direction. So it appears plausible that line inte grals around all closed paths are 0. Example 3 shows that F is indeed conservative.
is conservative.
2
2 2
2
C
FIGURE 9
CTM
Fx, yxy ix2 j SOLUTION LetPx,yxyandQx,yx2.Then
P1 Q1 y x
Since PyQx, F is not conservative by Theorem 5. M V EXAMPLE 3 Determine whether or not the vector field
7
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS1051
SOLUTION
a From Example 3 we know that F is conservative and so there exists a function f with fF, that is,
fxx, y32xy
fyx,yx2 3y2 Integrating 7 with respect to x, we obtain
fx, y3xx2yty
Notice that the constant of integration is a constant with respect to x, that is, a function
of y, which we have called ty. Next we differentiate both sides of 9 with respect to y: fyx, yx2ty
Comparing 8 and 10, we see that
ty3y2 Integrating with respect to y, we have
tyy3 K where K is a constant. Putting this in 9, we have
fx,y3xx2yy3 K as the desired potential function.
b To use Theorem 2 all we have to know are the initial and terminal points of C, namely, r00, 1 and r 0, e . In the expression for f x, y in part a, any value of the constant K will do, so lets choose K0. Then we have
y Fdry fdrf0, e f0, 1e31e31 CC
This method is much shorter than the straightforward method for evaluating line inte grals that we learned in Section 16.2. M
A criterion for determining whether or not a vector field F on3 is conservative is given in Section 16.5. Meanwhile, the next example shows that the technique for finding the potential function is much the same as for vector fields on 2.
V EXAMPLE5 IfFx,y,zy2i2xye3zj3ye3zk,findafunctionfsuch that fF.
SOLUTION If there is such a function f, then
fxx, y, zy2
fyx, y, z2xye3z fzx, y, z3ye3z
8
9
10
11
12
13
1052
CHAPTER 16 VECTOR CALCULUS
Integrating 11 with respect to x, we get
fx,y,zxy2 ty,z
where ty, z is a constant with respect to x. Then differentiating 14 with respect to y, we have
fyx, y, z2xytyy, z and comparison with 12 gives
tyy, ze3z Thus ty, zye3zhz and we rewrite 14 as
fx,y,zxy2 ye3z hz
Finally, differentiating with respect to z and comparing with 13, we obtain hz0
and therefore hzK, a constant. The desired function is fx,y,zxy2 ye3z K
It is easily verified that fF. M CONSERVATION OF ENERGY
Lets apply the ideas of this chapter to a continuous force field F that moves an object along a path C given by rt, atb, where raA is the initial point and rbB is the terminal point of C. According to Newtons Second Law of Motion see Sec tion 13.4, the force Frt at a point on C is related to the acceleration atrt by the equation
Frtmrt So the work done by the force on the object is
WyFdryb Frtrtdt yb mrtrtdt Caa
14
m yb d rtrt dt 2 a dt
2 a dt rt 2 dt2rt 2a mybd mb
m rb2 ra2 2
Therefore
Theorem 13.2.3, Formula 4 Fundamental Theorem of Calculus
15
W1mvb2 1mva2 22
where vr is the velocity.
The quantity 1 m vt 2, that is, half the mass times the square of the speed, is called the
kinetic energy of the object. Therefore we can rewrite Equation 15 as WKBKA
16
2
whose gradient is continuous. Find xC fdr. y
SECTION 16.3 THE FUNDAMENTAL THEOREM FOR LINE INTEGRALS1053
which says that the work done by the force field along C is equal to the change in kinetic energy at the endpoints of C.
Now lets further assume that F is a conservative force field; that is, we can write Ff. In physics, the potential energy of an object at the point x, y, z is defined as Px, y, zfx, y, z, so we have FP. Then by Theorem 2 we have
WyC FdryC PdrPrbPraPAPB
Comparing this equation with Equation 16, we see that
PAKAPBKB
which says that if an object moves from one point A to another point B under the influence of a conservative force field, then the sum of its potential energy and its kinetic energy remains constant. This is called the Law of Conservation of Energy and it is the reason the vector field is called conservative.
5. Fx,yexsinyiexcosyj
6. Fx,y3x2 2y2i4xy3j
Fx,yyex sinyiex xcosyj
8. Fx,yxycosxysinxyix2 cosxyj 9. Fx,ylny2xy3i3x2y2 xyj
10. Fx,yxycoshxysinhxyix2 coshxyj
The figure shows the vector field Fx, y2xy, x2and three curves that start at 1, 2 and end at 3, 2.
a Explain why xC Fdr has the same value for all three
16.3 EXERCISES
1. ThefigureshowsacurveCandacontourmapofafunction f
C
20
0x
7.
30
40
60
50
10
2. A table of values of a function f with continuous gradient is given. Find x fdr, where C has parametric equations
curves.
b What is this common value?
C
xt2 1 yt3 t 0t1
y 3
2 1
0
11.
x
y
0
1
2
0
1
6
4
1
3
5
7
2
8
2
9
310 Determine whether or not F is a conservative vector field. If it is, find a function f such that Ff.
3. Fx,y2x3yi3x4y8j 4. Fx,yexcosyiexsinyj
123
x
1054CHAPTER 16 VECTOR CALCULUS
1218 a Find a function f such that Ff and b use
26.
Let Ff, where fx, ysinx2y. Find curves C1 and C2 that are not closed and satisfy the equation.
a y Fdr0 b y Fdr1 C1 C2
ShowthatifthevectorfieldFPiQjRkisconser vative and P, Q, R have continuous firstorder partial deriva tives, then
PQ PR QR y x z x z y
Use Exercise 27 to show that the line integral
xC ydxxdyxyzdzisnotindependentofpath.
part a to evaluate xC Fdr along the given curve C. 12. Fx, yx2 iy2 j,
C is the arc of the parabola y2×2 from 1, 2 to 2, 8
13. Fx,yxy2ix2yj,
C: rttsin1 t,tcos1 t, 0t1 22
y2
14. Fx,y 1×2 i2yarctanxj,
C: rtt2i2tj, 0t1 Fx,y,zyzixzjxy2zk,
C is the line segment from 1, 0, 2 to 4, 6, 3
16. Fx,y,z2xzy2i2xyjx2 3z2k,
C: xt2, yt1, z2t1, 0t1
17. Fx,y,zy2 coszi2xycoszjxy2 sinzk,
C:rtt2isintjtk, 0t
18. Fx,y,zeyixeyjz1ezk,
C: rttit2jt3k, 0t1
1920 Show that the line integral is independent of path and
28.
27.
15.
33.
evaluate the integral.
19. x tan y dxx sec2 y dy,
2932 Determine whether or not the given set is a open, b connected, and c simplyconnected.
29. x,yx0, y0 30 x,yx0 31. x,y1x2 y2 4
32. x,yx2 y2 1 or 4×2 y2 9
LetFx,yyixj. x2 y2
C
C is any path from 1, 0 to 2, 4
a Show that PyQx.
b Show that x Fdr is not independent of path.
20. x1yexdxexdy, C
C is any path from 0, 1 to 1, 2
2122 Find the work done by the force field F in moving an
object from P to Q.
21. Fx, y2y32 i3xsy j; P1, 1, Q2, 4
22. Fx, yey ixey j; P0, 1, Q2, 0
2324 Is the vector field shown in the figure conservative?
Explain.
23. y 24. y
xx
CAS 25. IfFx,ysinyi1xcosyj,useaplottoguess whether F is conservative. Then determine whether your guess is correct.
34. a
C
Hint: Compute xC1 Fdr and xC2 Fdr, where C1
and C2 are the upper and lower halves of the circle
x2y21 from 1, 0 to 1, 0. Does this contradict Theorem 6?
Suppose that F is an inverse square force field, that is, Frcr
for some constant c, where rx iy jz k. Find the work done by F in moving an object from a point P1 along a path to a point P2 in terms of the distances d1 and d2 from these points to the origin.
r3
b An example of an inverse square field is the gravita tional field FmMGrr3 discussed in Example 4 in Section 16.1. Use part a to find the work done by the gravitational field when the earth moves from aph elion at a maximum distance of 1.52108 km from the sun to perihelion at a minimum distance of
1.47108 km. Use the values m5.971024 kg, M1.991030 kg, and G6.671011 Nm2kg2.
c Another example of an inverse square field is the electric force field FqQrr 3 discussed in Example 5 in Section 16.1. Suppose that an electron with a charge of 1.61019 C is located at the origin. A positive unit charge is positioned a distance 1012 m from the electron and moves to a position half that distance from the elec tron. Use part a to find the work done by the electric force field. Use the value 8.98510 9.
y
16.4
GREENS THEOREM
Greens Theorem gives the relationship between a line integral around a simple closed curve C and a double integral over the plane region D bounded by C. See Figure 1. We assume that D consists of all points inside C as well as all points on C. In stating Greens Theorem we use the convention that the positive orientation of a simple closed curve C refers to a single counterclockwise traversal of C. Thus if C is given by the vector func tion rt, atb, then the region D is always on the left as the point rt traverses C. See Figure 2.
SECTION 16.4 GREENS THEOREM1055
D
C 0x
FIGURE 1
FIGURE 2
N Recall that the left side of this equation is another way of writing xC Fdr, where FP iQ j.
yy
DD
C
1
C 0x0x
a Positive orientation
b Negative orientation
GREENS THEOREM Let C be a positively oriented, piecewisesmooth, simple closed curve in the plane and let D be the region bounded by C. If P and Q have continuous partial derivatives on an open region that contains D, then
y P dxQ dyyyQPdA C x y
D
NOTE
The notation
y PdxQdy or gPdxQdy CC
is sometimes used to indicate that the line integral is calculated using the positive orienta tion of the closed curve C. Another notation for the positively oriented boundary curve of D is D, so the equation in Greens Theorem can be written as
yyQPdAy PdxQdy
xy D D
Greens Theorem should be regarded as the counterpart of the Fundamental Theorem of Calculus for double integrals. Compare Equation 1 with the statement of the Fundamental Theorem of Calculus, Part 2, in the following equation:
yb Fx dxFbFa a
In both cases there is an integral involving derivatives F, Qx, and Py on the left side of the equation. And in both cases the right side involves the values of the original functions F, Q, and P only on the boundary of the domain. In the onedimensional case, the domain is an interval a, b whose boundary consists of just two points, a and b.
1056CHAPTER 16 VECTOR CALCULUS
N Greens Theorem is named after the selftaught English scientist George Green 17931841. He worked fulltime in his fathers bakery from the age of nine and taught himself mathematics from library books. In 1828 he published privately An Essay on the Application of Mathematical Analysis to the Theories of Electricity and Magnetism, but only 100 copies were printed and most of those went to his friends. This pamphlet contained a theorem that is equivalent to what we know as Greens Theorem, but it didnt become widely known
at that time. Finally, at age 40, Green entered Cambridge University as an undergraduate
but died four years after graduation. In 1846 William Thomson Lord Kelvin located a copy of Greens essay, realized its significance, and had it reprinted. Green was the first person to try to formulate a mathematical theory of elec tricity and magnetism. His work was the basis for the subsequent electromagnetic theories of Thomson, Stokes, Rayleigh, and Maxwell.
Greens Theorem is not easy to prove in general, but we can give a proof for the special case where the region is both of type I and of type II see Section 15.3. Lets call such regions simple regions.
PROOF OF GREENS THEOREM FOR THE CASE IN WHICH IS A SIMPLE REGION Notice that Greens Theorem will be proved if we can show that
2
y y
yy P D
yy Q D x
C P dx y dA
and
3
4
y
C D
ygx
0abx
FIGURE 3
We prove Equation 2 by expressing D as a type I region: Dx,yaxb, t1xyt2x
where t1 and t2 are continuous functions. This enables us to compute the double integral on the right side of Equation 2 as follows:
yyPdAybyt2x Px,ydydxybPx,t2xPx,t1xdx D y a t1x y a
where the last step follows from the Fundamental Theorem of Calculus.
Now we compute the left side of Equation 2 by breaking up C as the union of the
four curves C1, C2, C3, and C4 shown in Figure 3. On C1 we take x as the parameter and write the parametric equations as xx, yt1x, axb. Thus
ygTMx C
C
Q dy
dA
CTM
y Px, y dxyb Px, t1x dx C C1 a
Observe that C3 goes from right to left but C3 goes from left to right, so we can write the parametric equations of C3 as xx, yt2x, axb. Therefore
y Px,ydxy Px,ydxyb Px,t2xdx C3 C3 a
On C2 or C4 either of which might reduce to just a single point, x is constant, so dx0 and
y Px,ydx0y Px,ydx C2 C4
Hence
yPx,ydxy Px,ydxy Px,ydxy Px,ydxy Px,ydx
C C1 C2 C3 C4
yb Px, t1x dxyb Px, t2x dx aa
D
y 0, 1
0,0
FIGURE 4
y1x
SOLUTION Although the given line integral could be evaluated as usual by the methods of Section 16.2, that would involve setting up three separate integrals along the three sides of the triangle, so lets use Greens Theorem instead. Notice that the region D enclosed by C is simple and C has positive orientation see Figure 4. If we let Px, yx4 and Qx, yxy, then we have
SECTION 16.4 GREENS THEOREM1057 Comparing this expression with the one in Equation 4, we see that
y Px, y dxyy P dA C Dy
Equation 3 can be proved in much the same way by expressing D as a type II region see Exercise 28. Then, by adding Equations 2 and 3, we obtain Greens Theorem. M
EXAMPLE 1 Evaluate x x4 dxxy dy, where C is the triangular curve consisting of the C
line segments from 0, 0 to 1, 0, from 1, 0 to 0, 1, and from 0, 1 to 0, 0.
C
D
yx4dxxydyyyQPdA 1 1xy0dydx
1,0 x
yy
C xy 00 D
y2 dx1x2 dx 2y0 2
y11y1x 1y1
00
1
11×3 1 M 606
N Instead of using polar coordinates, we could simply use the fact that D is a disk of radius 3 and write
yy4dA4 3236 D
7xsy4 13yesinx dA yy
V EXAMPLE 2 Evaluatex 3yesinxdx7xsy4 1dy,whereCisthe
C
circle x2y29.
SOLUTION The region D bounded by C is the disk x2y29, so lets change to polar
coordinates after applying Greens Theorem:
y 3yesinxdx7xsy4 1dy C
y2 y3 73rdrd 4y2 d y3 rdr36 M
x y 0000
D
In Examples 1 and 2 we found that the double integral was easier to evaluate than the line integral. Try setting up the line integral in Example 2 and youll soon be convinced! But sometimes its easier to evaluate the line integral, and Greens Theorem is used in the reverse direction. For instance, if it is known that Px, yQx, y0 on the curve C, then Greens Theorem gives
yyQPdAy P dxQ dy0
xy C D
no matter what values P and Q assume in the region D.
Another application of the reverse direction of Greens Theorem is in computing areas.
SincetheareaofDisxxD 1dA,wewishtochoosePandQsothat
QP1 x y
1058
CHAPTER 16 VECTOR CALCULUS
There are several possibilities:
Px, y0 Qx, yx
Px, yy Qx, y0
Px, y1 y 2
Qx, y1 x 2
Then Greens Theorem gives the following formulas for the area of D:
x2 y2 EXAMPLE 3 Find the area enclosed by the ellipse a2b21.
SOLUTION The ellipse has parametric equations xa cos t and yb sin t, where
5
Ay xdyy ydx1 y xdyydx CC2C
0t2
. Using the third formula in Equation 5, we have
A1 yxdyydx 2C
1 y2 acostbcostdtbsintasintdt 20
aby2 dt ab M 20
C
FIGURE 5
Although we have proved Greens Theorem only for the case where D is simple, we can now extend it to the case where D is a finite union of simple regions. For example, if D is the region shown in Figure 5, then we can write DD1D2 , where D1 and D2 are both simple. The boundary of D1 is C1C3 and the boundary of D2 is C2C3 so, apply ing Greens Theorem to D1 and D2 separately, we get
QP dA
D
DTM
C
CTM
C
C
y
yy
C1C3
x y
yC2C3
D2
x y
P dxQ dyPdxQdyyyQP dA
D1
If we add these two equations, the line integrals along C3 and C3 cancel, so we get
PdxQdyyyQPdA
yC1C2 x y D
FIGURE 6
of nonoverlapping simple regions see Figure 6.
V EXAMPLE 4 Evaluate x y2 dx3xy dy, where C is the boundary of the semiannular
which is Greens Theorem for DD1D2, since its boundary is CC1C2.
The same sort of argument allows us to establish Greens Theorem for any finite union
C
region D in the upper halfplane between the circles x2y21 and x2y24.
SOLUTION Notice that although D is not simple, the yaxis divides it into two simple regions see Figure 7. In polar coordinates we can write
Dr, 1r2,0
D
FIGURE 7
4 CC
x y D
y
Therefore Greens Theorem gives
0 1
x
ysin d y2r2drcos1r3214 031
y y2dx3xydyyy 3xyy2dA
SECTION 16.4 GREENS THEOREM1059
yyydAyy2rsin rdrd
D
01
M
Greens Theorem can be extended to apply to regions with holes, that is, regions that are not simplyconnected. Observe that the boundary C of the region D in Figure 8 con sists of two simple closed curves C1 and C2. We assume that these boundary curves are oriented so that the region D is always on the left as the curve C is traversed. Thus the positive direction is counterclockwise for the outer curve C1 but clockwise for the inner curve C2. If we divide D into two regions D and D by means of the lines shown in Figure 9 and then apply Greens Theorem to each of D and D, we get
CTM
D
FIGURE 8
Da
Daa
013
C
yy QP dAyy QP dAyy QP dA
x y x y x y D D D
FIGURE 9
yD PdxQdyyD PdxQdy
Since the line integrals along the common boundary lines are in opposite directions, they
cancel and we get
yyQPdAy PdxQdyy PdxQdyyPdxQdy
xy C1 C2 C D
which is Greens Theorem for the region D.
V EXAMPLE 5 IfFx,yyixjx2 y2,showthatx Fdr2 forevery
y
Ca
C
SOLUTION Since C is an arbitrary closed path that encloses the origin, its difficult to compute the given integral directly. So lets consider a counterclockwiseoriented circle
C with center the origin and radius a, where a is chosen to be small enough that C lies inside C. See Figure 10. Let D be the region bounded by C and C. Then its positively oriented boundary is CC and so the general version of Greens Theorem gives
yPdxQdyy PdxQdyyyQPdA C C x y
Therefore thatis,
positively oriented simple closed path that encloses the origin.
C
D
FIGURE 10
x
yy y2x2 y2x2 D
222222
xy xydA0
D
yC PdxQdyyC PdxQdy
yC FdryC Fdr
1060
CHAPTER 16 VECTOR CALCULUS
We now easily compute this last integral using the parametrization given by
rtacostiasintj,0t2 .Thus yFdry Fdry2 Frtrtdt
C C 0
y2 asintasintacostacost dt y2 dt2 M
0 a2 cos2ta2 sin2t 0
We end this section by using Greens Theorem to discuss a result that was stated in the
preceding section.
SKETCH OF PROOF OF THEOREM 16.3.6 Were assuming that FP iQ j is a vector field on an open simplyconnected region D, that P and Q have continuous firstorder partial derivatives, and that
PQ throughout D y x
If C is any simple closed path in D and R is the region that C encloses, then Greens Theorem gives
16.4 EXERCISES
14 Evaluate the line integral by two methods: a directly and
b using Greens Theorem.
1. xC x ydxxydy,
C is the circle with center the origin and radius 2
2. xCxydxx2dy,
C is the rectangle with vertices 0, 0, 3, 0, 3, 1, and 0, 1
6. x cos y dxx2 sin y dy, C
y Fdry PdxQdyyyQPdAyy0dA0
CC xy R
R
A curve that is not simple crosses itself at one or more points and can be broken up into
a number of simple curves. We have shown that the line integrals of F around these
simple curves are all 0 and, adding these integrals, we see that x Fdr0 for any C
closed curve C. Therefore x Fdr is independent of path in D by Theorem 16.3.3. It C
follows that F is a conservative vector field. M
3.
23 xxydxxy dy,
8. x xe2x dxx42x2y2 dy, C
4. C
C
C is the triangle with vertices 0, 0, 1, 0, and 1, 2
C is the boundary of the region between the circles x2 y2 1andx2 y2 4
x x dxy dy, C consists of the line segments from 0, 1 to 0, 0 and from 0, 0 to 1, 0 and the parabola y1x2 from 1, 0 to 0, 1
510 Use Greens Theorem to evaluate the line integral along the given positively oriented curve.
9. x y3 dxx3 dy, Cisthecirclex2 y2 4 C
10. x sinydxxcosydy, Cistheellipsex2 xyy2 1 C
C is the rectangle with vertices 0, 0, 5, 0, 5, 2, and 0, 2
7. xC yesxdx2xcosy2dy,
C is the boundary of the region enclosed by the parabolas yx2 andxy2
5. x xy2 dx2x2y dy, C
1114 Use Greens Theorem to evaluate xC Fdr. Check the orientation of the curve before applying the theorem.
11. Fx, ysxy 3, x 2sy ,
C consists of the arc of the curve ysin x from 0, 0 toand the line segment from, 0 to 0, 0
, 0
C is the triangle with vertices 0, 0, 2, 2, and 2, 4
12. Fx, yy2 cos x, x22y sin x,
C is the triangle from 0, 0 to 2, 6 to 2, 0 to 0, 0
13. Fx,yexx2y,eyxy2,
C is the circle x 2y 225 oriented clockwise
14. Fx, yylnx2y2, 2 tan1yx, C is the circle x22y321 oriented counterclockwise
CAS 1516 Verify Greens Theorem by using a computer algebra sys tem to evaluate both the line integral and the double integral.
15. Px, yy2ex, Qx, yx2ey,
C consists of the line segment from 1, 1 to 1, 1 followed by the arc of the parabola y2x2 from 1, 1 to 1, 1
16. Px, y2xx3y5, Qx, yx3y8, Cistheellipse4x2 y2 4
Use Greens Theorem to find the work done by the force Fx,yxxyixy2 jinmovingaparticlefromthe origin along the xaxis to 1, 0, then along the line segment to 0, 1, and then back to the origin along the yaxis.
18. A particle starts at the point 2, 0, moves along the xaxis to 2, 0, and then along the semicircle ys4x2 to the starting point. Use Greens Theorem to find the work done on this particle by the force field Fx, yx, x33xy2 .
19. Use one of the formulas in 5 to find the area under one arch of the cycloid xtsin t, y1cos t.
SECTION 16.5 CURL AND DIVERGENCE1061 c Find the area of the pentagon with vertices 0, 0, 2, 1,
1, 3, 0, 2, and 1, 1.
22. Let D be a region bounded by a simple closed path C in the xyplane. Use Greens Theorem to prove that the coordinates of the centroid x, y of D are
x1 y x 2 d y 2A C
y 1 y y 2 d x 2A C
; 20.
If a circle C with radius 1 rolls along the outside of the circlex2 y2 16,afixedpointPonCtracesouta curve called an epicycloid, with parametric equations
x5 cos tcos 5t, y5 sin tsin 5t. Graph the epi cycloid and use 5 to find the area it encloses.
a If C is the line segment connecting the point x1, y1 to the point x2, y2, show that
yC xdyydxx1y2 x2y1
b If the vertices of a polygon, in counterclockwise order, are x1, y1 , x2, y2 , . . . , xn , yn , show that the area of the polygon is
A1x1y2x2y1x2y3x3y2 2
where A is the area of D.
23. Use Exercise 22 to find the centroid of a quartercircular
region of radius a.
24. Use Exercise 22 to find the centroid of the triangle with
vertices 0, 0, a, 0, and a, b, where a0 and b0.
25. A plane lamina with constant density x, yoccupies a region in the xyplane bounded by a simple closed path C. Show that its moments of inertia about the axes are
Ix y y3dx Iy y x3dy 3C 3C
26. Use Exercise 25 to find the moment of inertia of a circular disk of radius a with constant density about a diameter. Compare with Example 4 in Section 15.5.
IfFisthevectorfieldofExample5,showthatxC Fdr0 for every simple closed path that does not pass through or enclose the origin.
28. Complete the proof of the special case of Greens Theorem by proving Equation 3.
29. Use Greens Theorem to prove the change of variables formula for a double integral Formula 15.9.9 for the case where fx,y1:
yy dx dyyy x, y du dv
R S u, v
Here R is the region in the xyplane that corresponds to the region S in the uvplane under the transformation given by xtu, v, yhu, v.
Hint: Note that the left side is AR and apply the first part of Equation 5. Convert the line integral over R to a line inte gral over S and apply Greens Theorem in the uvplane.
17.
21.
27.
A
xn1ynxnyn1xny1x1yn
16.5 CURL AND DIVERGENCE
In this section we define two operations that can be performed on vector fields and that play a basic role in the applications of vector calculus to fluid flow and electricity and mag netism. Each operation resembles differentiation, but one produces a vector field whereas the other produces a scalar field.
1062
CHAPTER 16 VECTOR CALCULUS CURL
IfFPiQjRkisavectorfieldon3 andthepartialderivativesofP,Q,andR all exist, then the curl of F is the vector field on 3 defined by
As an aid to our memory, lets rewrite Equation 1 using operator notation. We introduce the vector differential operatordel as
ijk x y z
It has meaning when it operates on a scalar function to produce the gradient of f : fi fj fk ff if jf k
1
curlFRQiPRjQPk y z z x x y
x y z x y z
If we think ofas a vector with components x, y, and z, we can also consider
the formal cross product ofwith the vector field F as follows: ijk
Fx y z
PQR
RQiPRjQPk
y z z x x ycurl F
Thus the easiest way to remember Definition 1 is by means of the symbolic expression
EXAMPLE1 IfFx,y,zxzixyzjy2k,findcurlF. SOLUTION Using Equation 2, we have
2
ijk curlFF
N Most computer algebra systems have commands that compute the curl and divergence of vector fields. If you have access to a CAS, use these commands to check the answers to the examples and exercises in this section.
2
y2 xyzi y2 xzj
y2x ix jyz k
M
curl FF
x y z xz xyz y
y z
xyzxzk
x z
x y
2yxy i0x jyz0 k
PROOF We have
ijk
SECTION 16.5 CURL AND DIVERGENCE1063
Recall that the gradient of a function f of three variables is a vector field on 3 and so we can compute its curl. The following theorem says that the curl of a gradient vector field is 0.
THEOREM If f is a function of three variables that has continuous second order partial derivatives, then
curl f 0
3
f f f
x y z
2f2f i2f2f j2f2f k
yz zy zx xz xy yx 0i0j0k0
by Clairauts Theorem. M Since a conservative vector field is one for which Ff, Theorem 3 can be rephrased
N Notice the similarity to what we know from Section 12.4: aa0 for every threedimensional vector a.
curlff x y z
N Compare this with Exercise 27 in Section 16.3.
as follows:
This gives us a way of verifying that a vector field is not conservative.
V EXAMPLE2 ShowthatthevectorfieldFx,y,zxzixyzjy2kisnot conservative.
SOLUTION In Example 1 we showed that
curl Fy2x ix jyz k
This shows that curl F0 and so, by Theorem 3, F is not conservative. M
The converse of Theorem 3 is not true in general, but the following theorem says the converse is true if F is defined everywhere. More generally it is true if the domain is simplyconnected, that is, has no hole. Theorem 4 is the threedimensional version of Theorem 16.3.6. Its proof requires Stokes Theorem and is sketched at the end of Section 16.8.
If F is conservative, then curl F0.
THEOREM If F is a vector field defined on all of 3 whose component func tions have continuous partial derivatives and curl F0, then F is a conservative vector field.
4
1064
CHAPTER 16 VECTOR CALCULUS
V EXAMPLE 3
a Show that
is a conservative vector field.
Fx, y, zy2z3 i2xyz3 j3xy2z2 k b Find a function f such that Ff.
SOLUTION
a We compute the curl of F:
curl FF
ijk x y z
curl Fx, y, z
23322 y z 2xyz 3xy z
6xyz26xyz2i3y2z23y2z2j2yz32yz3k
0
Since curl F0 and the domain of F is 3, F is a conservative vector field by
Theorem 4.
b The technique for finding f was given in Section 16.3. We have
fxx, y, zy2z3
fyx, y, z2xyz3
fzx, y, z3xy2z2 Integrating 5 with respect to x, we obtain
fx,y,zxy2z3 ty,z
Differentiating 8 with respect to y, we get fyx, y, z2xyz3tyy, z, so comparison
with 6 gives tyy, z0. Thus ty, zhz and
fzx, y, z3xy2z2hz
Then 7 gives hz0. Therefore
fx,y,zxy2z3 K M
The reason for the name curl is that the curl vector is associated with rotations. One connection is explained in Exercise 37. Another occurs when F represents the velocity field in fluid flow see Example 3 in Section 16.1. Particles near x, y, z in the fluid tend to rotate about the axis that points in the direction of curl Fx, y, z and the length of this curl vector is a measure of how quickly the particles move around the axis see Figure 1. If curl F0 at a point P, then the fluid is free from rotations at P and F is called irro tational at P. In other words, there is no whirlpool or eddy at P. If curl F0, then a tiny paddle wheel moves with the fluid but doesnt rotate about its axis. If curl F0, the paddle wheel rotates about its axis. We give a more detailed explanation in Section 16.8 as a consequence of Stokes Theorem.
5
6
7
8
x,y,z
FIGURE 1
SECTION 16.5 CURL AND DIVERGENCE1065
DIVERGENCE
IfFPiQjRkisavectorfieldon3 andPx,Qy,andRzexist,then the divergence of F is the function of three variables defined by
Observe that curl F is a vector field but div F is a scalar field. In terms of the gradient oper ator x iy jz k, the divergence of F can be written symbolically as the dot product ofand F:
EXAMPLE4 IfFx,y,zxzixyzjy2k,finddivF.
SOLUTION By the definition of divergence Equation 9 or 10 we have
divFFxzxyzy2zxz M x y z
33
If F is a vector field on, then curl F is also a vector field on. As such, we can
compute its divergence. The next theorem shows that the result is 0.
9
div FPQR x y z
10
div FF
THEOREM IfFPiQjRkisavectorfieldon3 andP,Q,andR have continuous secondorder partial derivatives, then
div curl F0
11
N Note the analogy with the scalar triple product: aab0.
PROOF
Using the definitions of divergence and curl, we have
div curl FF
RQPRQP
x y z y z x z x y2R2Q2P2R2Q2P
xy xz yz yx zx zy 0
because the terms cancel in pairs by Clairauts Theorem. M V EXAMPLE5 ShowthatthevectorfieldFx,y,zxzixyzjy2kcantbe
written as the curl of another vector field, that is, Fcurl G. SOLUTION In Example 4 we showed that
div Fzxz
1066CHAPTER 16 VECTOR CALCULUS
N The reason for this interpretation of div F will be explained at the end of Section 16.9 as a consequence of the Divergence Theorem.
and therefore div F0. If it were true that Fcurl G, then Theorem 11 would give div Fdiv curl G0
which contradicts div F0. Therefore F is not the curl of another vector field. M Again, the reason for the name divergence can be understood in the context of fluid
flow. If Fx, y, z is the velocity of a fluid or gas, then div Fx, y, z represents the net rate of change with respect to time of the mass of fluid or gas flowing from the point x, y, z per unit volume. In other words, div Fx, y, z measures the tendency of the fluid to diverge from the point x, y, z. If div F0, then F is said to be incompressible.
Another differential operator occurs when we compute the divergence of a gradient vec tor field f. If f is a function of three variables, we have
divff 2f2f2f x2 y2 z2
and this expression occurs so often that we abbreviate it as 2 f. The operator 2
is called the Laplace operator because of its relation to Laplaces equation 2 f2f2f2f0
x2 y2 z2
We can also apply the Laplace operator 2 to a vector field
in terms of its components:
FPiQjRk
2 F 2P i 2Q j 2R k
VECTOR FORMS OF GREENS THEOREM
The curl and divergence operators allow us to rewrite Greens Theorem in versions that will be useful in our later work. We suppose that the plane region D, its boundary curve C, and the functions P and Q satisfy the hypotheses of Greens Theorem. Then we con sider the vector field FP iQ j. Its line integral is
y Fdry PdxQdy CC
and, regarding F as a vector field on 3 with third component 0, we have ijk
curlF QP k x y z x y
Px, y Qx, y 0
Therefore
curlFkQPkk QP x y x y
SECTION 16.5 CURL AND DIVERGENCE1067
y
If C is given by the vector equation
rtxt iyt j
then the unit tangent vector see Section 13.2 is
Ttxt irt
atb yt j
and we can now rewrite the equation in Greens Theorem in the vector form
Equation 12 expresses the line integral of the tangential component of F along C as the double integral of the vertical component of curl F over the region D enclosed by C. We now derive a similar formula involving the normal component of F.
12
D
C
ntyt irt
xt j rt
Tt rt nt
rt
You can verify that the outward unit normal vector to C is given by
0x
FIGURE 2
See Figure 2. Then, from Equation 16.2.3, we have
y Fn dsyb Fntrtdt Ca
y FdryycurlFkdA
C
D
yb Pxt, yt ytQxt, yt xtrt dt
yb Pxt, yt yt dtQxt, yt xt dt
a rt rt a
y P dyQ dxyyPQdA C x y
D
by Greens Theorem. But the integrand in this double integral is just the divergence of F. So we have a second vector form of Greens Theorem.
This version says that the line integral of the normal component of F along C is equal to the double integral of the divergence of F over the region D enclosed by C.
13
y FndsyydivFx,ydA
C
D
1068CHAPTER 16 VECTOR CALCULUS
16.5 EXERCISES
18 Find a the curl and b the divergence of the vector field.
Fx, y, zxyz ix2y k
2. Fx,y,zx2yzixy2zjxyz2 k
3. Fx,y,zixyzjxyszk
4. Fx,y,zcosxzjsinxyk
5. Fx,y,z 1 xiyjzk sx2 y2 z2
6. Fx,y,zexysinzjytan1xzk
7. Fx, y, zln x, lnxy, lnxyz
8. Fx,y,zex,exy,exyz
911 The vector field F is shown in the xyplane and looks the same in all other horizontal planes. In other words, F is indepen dent of z and its zcomponent is 0.
1318 Determine whether or not the vector field is conservative. If it is conservative, find a function f such that Ff.
13. Fx,y,zy2z3 i2xyz3 j3xy2z2 k 14. Fx,y,zxyz2 ix2yz2 j x2y2zk
Fx,y,z2xyix2 2yzjy2 k
16. Fx,y,zez ijxez k
17. Fx,y,zyex iex j2zk
18. Fx,y,zycosxyixcosxyjsinzk
Is there a vector field G on 3 such that curl Gx sin y, cos y, zxy? Explain.
20. Is there a vector field G on 3 such that curl Gxyz, y2z, yz2? Explain.
Show that any vector field of the form
Fx, y, zfx ity jhz k
where f , t, h are differentiable functions, is irrotational. 22. Show that any vector field of the form
Fx, y, zfy, z itx, z jhx, y k is incompressible.
2329 Prove the identity, assuming that the appropriate partial derivatives exist and are continuous. If f is a scalar field and F, G arevectorfields,then fF,FG,andFGaredefinedby
f Fx, y, zfx, y, z Fx, y, z
FGx, y, zFx, y, zGx, y, z FGx, y, zFx, y, zGx, y, z
23. divFGdivFdivG
24. curlFGcurlFcurlG
25. divfFf divFFf
26. curlfFf curlFfF
27. divFGGcurlFFcurlG 28. divft0
29. curlcurl Fgraddiv F2F
3032 Letrxiyjzkandr r .
1.
a b
Is div F positive, negative, or zero? Explain.
Determine whether curl F0. If not, in which direction does curl F point?
9. y
10. y
11.
12.
0x0x y
0x
Let f be a scalar field and F a vector field. State whether each expression is meaningful. If not, explain why. If so, state whether it is a scalar field or a vector field.
a curl f
c div F
e grad F
g divgrad f
i curlcurl F
k grad f div F
b gradf
d curlgradf
f graddiv F
h graddivf
j divdivF
l divcurlgrad f
30. Verify each identity. a r3
c 2r312r
b rr4r
15.
19.
21.
31.
Verify each identity. arrr c1rrr3
b r0 d lnrrr2
w
SECTION 16.5 CURL AND DIVERGENCE z
1069
32. IfFrrp,finddivF.Isthereavalueofpforwhich div F0?
33. Use Greens Theorem in the form of Equation 13 to prove Greens first identity:
yyf2tdAy ftndsyyftdA C
DD
where D and C satisfy the hypotheses of Greens Theorem
and the appropriate partial derivatives of f and t exist and are continuous. The quantity tnDn t occurs in the line inte gral. This is the directional derivative in the direction of the normal vector n and is called the normal derivative of t.
34. Use Greens first identity Exercise 33 to prove Greens second identity:
yyf2tt2fdAy fttfnds
C
where D and C satisfy the hypotheses of Greens Theorem
and the appropriate partial derivatives of f and t exist and are continuous.
dv P
B
D
div E0
curl E 1 H
following: aE1 2E
div H0
curl H1 E
38.
Maxwells equations relating the electric field E and magnetic field H as they vary with time in a region containing no charge and no current can be stated as follows:
x
0
y
35. Recall from Section 14.3 that a function t is called harmonic onDifitsatisfiesLaplacesequation,thatis,2t0onD. Use Greens first identity with the same hypotheses as in Exer
c2 t2 2
c t
where c is the speed of light. Use these equations to prove the
c t
cise 33 to show that if t is harmonic on D, then xC
Dntds0.
bH1 H c2 t2
Here Dn t is the normal derivative of t defined in Exercise 33.
36. Use Greens first identity to show that if f is harmonic
on D, and if fx, y0 on the boundary curve C, then xx f 2 dA0. Assume the same hypotheses as in
Exercise 33.
37. This exercise demonstrates a connection between the curl
vector and rotations. Let B be a rigid body rotating about the zaxis. The rotation can be described by the vector wk, where is the angular speed of B, that is, the tangential speed of any point P in B divided by the distance d from the axis of rotation. Let rx, y, z be the position vector of P.
a By considering the angle in the figure, show that the velocity field of B is given by vwr.
bShowthatv yi xj. c Show that curl v2w.
1 2E c 2Ec2 t2
2
Hint: Use Exercise 29.
D
d2H 1 H c2 t2
39.
We have seen that all vector fields of the form Ft
satisfy the equation curl F0 and that all vector fields of the form Fcurl G satisfy the equation div F0 assuming continuity of the appropriate partial derivatives. This suggests the question: Are there any equations that all functions of the form fdiv G must satisfy? Show that the answer to this question is No by proving that every continuous function f on 3 is the divergence of some vector field. Hint: Let
Gx, y, ztx, y, z, 0, 0,where tx, y, zx0x ft, y, z dt.
1070
CHAPTER 16 VECTOR CALCULUS
16.6
PARAMETRIC SURFACES AND THEIR AREAS
So far we have considered special types of surfaces: cylinders, quadric surfaces, graphs of functions of two variables, and level surfaces of functions of three variables. Here we use vector functions to describe more general surfaces, called parametric surfaces, and com pute their areas. Then we take the general surface area formula and see how it applies to special surfaces.
PARAMETRIC SURFACES
In much the same way that we describe a space curve by a vector function rt of a single parameter t, we can describe a surface by a vector function ru, v of two parameters u and v. We suppose that
ru, vxu, v iyu, v jzu, v k
is a vectorvalued function defined on a region D in the uvplane. So x, y, and z, the com ponent functions of r, are functions of the two variables u and v with domain D. The set of all points x, y, z in 3 such that
xxu, v yyu, v zzu, v
and u, v varies throughout D, is called a parametric surface S and Equations 2 are called parametric equations of S. Each choice of u and v gives a point on S; by making all choices, we get all of S. In other words, the surface S is traced out by the tip of the posi tion vector ru, v as u, v moves throughout the region D. See Figure 1.
1
2
z
D
u,
r
S
z
0,0,2 0
x
2, 0, 0
FIGURE 2
y
EXAMPLE 1 Identify and sketch the surface with vector equation ru, v2 cos u iv j2 sin u k
SOLUTION The parametric equations for this surface are
x2 cos u yv z2 sin u
So for any point x, y, z on the surface, we have
x2z24 cos2u4 sin2u4
FIGURE 1
A parametric surface
x
y
0u
ru, 0
This means that vertical crosssections parallel to the xzplane that is, with y constant are all circles with radius 2. Since yv and no restriction is placed on v, the surface is a circular cylinder with radius 2 whose axis is the yaxis. See Figure 2. M
x
FIGURE 3
z
0, 3, 2 0
y
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS1071 In Example 1 we placed no restrictions on the parameters u and v and so we obtained
the entire cylinder. If, for instance, we restrict u and v by writing the parameter domain as
0u 2 0v3
then x0, z0, 0y3, and we get the quartercylinder with length 3 illustrated in Figure 3.
If a parametric surface S is given by a vector function ru, v, then there are two useful families of curves that lie on S, one family with u constant and the other with v constant. These families correspond to vertical and horizontal lines in the uvplane. If we keep u constant by putting uu0, then ru0, v becomes a vector function of the single parame ter v and defines a curve C1 lying on S. See Figure 4.
TEC Visual 16.6 shows animated versions u , of Figures 4 and 5, with moving grid curves,
r
for several parametric surfaces.
z
constant
C
Similarly, if we keep v constant by putting vv0 , we get a curve C2 given by ru, v0that lies on S. We call these curves grid curves. In Example 1, for instance, the grid curves obtained by letting u be constant are horizontal lines whereas the grid curves with v constant are circles. In fact, when a computer graphs a parametric surface, it usually depicts the surface by plotting these grid curves, as we see in the following example.
EXAMPLE 2 Use a computer algebra system to graph the surface ru,v 2sinvcosu,2sinvsinu,ucosv
Which grid curves have u constant? Which have v constant?
SOLUTION We graph the portion of the surface with parameter domain 0u4 , 0v2 in Figure 5. It has the appearance of a spiral tube. To identify the grid curves, we write the corresponding parametric equations:
x2sin v cos u y2sin v sin u zucos v
If v is constant, then sin v and cos v are constant, so the parametric equations resemble those of the helix in Example 4 in Section 13.1. So the grid curves with v constant are the spiral curves in Figure 5. We deduce that the grid curves with u constant must be the curves that look like circles in the figure. Further evidence for this assertion is that if u
is kept constant, uu0, then the equation zu0cos v shows that the zvalues vary fromu0 1tou0 1. M
In Examples 1 and 2 we were given a vector equation and asked to graph the corre sponding parametric surface. In the following examples, however, we are given the more challenging problem of finding a vector function to represent a given surface. In the rest of this chapter we will often need to do exactly that.
x
u constant
y
FIGURE 5
z
uu0u
D
0 FIGURE 4 x
CTM
y
1072
a
CHAPTER 16 VECTOR CALCULUS
EXAMPLE 3 Find a vector function that represents the plane that passes through the
P
b
b
point P0 with position vector r0 and that contains two nonparallel vectors a and b.
P SOLUTION If P is any point in the plane, we can get from P0 to P by moving a certain distance in the direction of a and another distance in the direction of b. So there are scalars u and v such that PA0 Puavb. Figure 6 illustrates how this works, by means of the Parallelogram Law, for the case where u and v are positive. See also Exercise 40 in Section 12.2. If r is the position vector of P, then
rOAP 0PA0 Pr 0u av b So the vector equation of the plane can be written as
ru,vr0 uavb
where u and v are real numbers.
Ifwewriter x,y,z,r0x0,y0,z0,a a1,a2,a3,andb b1,b2,b3,
then we can write the parametric equations of the plane through the point x0, y0, z0as follows:
xx0 ua1 vb1 yy0 ua2 vb2 zz0 ua3 vb3 M V EXAMPLE 4 Find a parametric representation of the sphere
x2 y2 z2 a2
SOLUTION The sphere has a simple representationa in spherical coordinates, so lets choose the angles and in spherical coordinates as the parameters see Section 15.8. Then, puttinga in the equations for conversion from spherical to rectangular coordi nates Equations 15.8.1, we obtain
ua FIGURE 6
N One of the uses of parametric surfaces is in computer graphics. Figure 7 shows the result of tryingtographthespherex2 y2 z2 1 by solving the equation for z and graphing the top and bottom hemispheres separately. Part of the sphere appears to be missing because of the rectangular grid system used by the computer. The much better picture in Figure 8 was pro duced by a computer using the parametric equations found in Example 4.
xa sin cos ya sin sin za cos
as the parametric equations of the sphere. The corresponding vector equation is
r , asin cos iasin sin jacos k
Wehave0and0 2 ,sotheparameterdomainistherectangle
D0, 0, 2 . The grid curves with constant are the circles of constant lati
tude including the equator. The grid curves with circles, which connect the north and south poles.
constant are the meridians semi
M
FIGURE 7
FIGURE 8
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS1073
EXAMPLE 5 Find a parametric representation for the cylinder x2 y2 4 0z1
SOLUTION The cylinder has a simple representation r2 in cylindrical coordinates, so we choose as parameters and z in cylindrical coordinates. Then the parametric equations of the cylinder are
x2 cos y2 sin zz where0 2 and0z1.
M
V EXAMPLE 6 Find a vector function that represents the elliptic paraboloid zx22y2. SOLUTION If we regard x and y as parameters, then the parametric equations are simply
TEC In Module 16.6 you can investigate several families of parametric surfaces.
In general, a surface given as the graph of a function of x and y, that is, with an equa tion of the form zf x, y, can always be regarded as a parametric surface by taking x and y as parameters and writing the parametric equations as
xx yy zfx, y
Parametric representations also called parametrizations of surfaces are not unique. The
next example shows two ways to parametrize a cone.
EXAMPLE 7 Find a parametric representation for the surface z2sx2y2 , that is, the
tophalfoftheconez2 4×2 4y2.
SOLUTION 1 One possible representation is obtained by choosing x and y as parameters:
and the vector equation is
xx yy zx2 2y2
rx,yxiyjx2 2y2k M
N For some purposes the parametric represen tations in Solutions 1 and 2 are equally good, but Solution 2 might be preferable in certain situations. If we are interested only in the part of the cone that lies below the plane z1, for instance, all we have to do in Solution 2 is change the parameter domain to
0r1 0 2 2
SOLUTION 2 Another representation results from choosing as parameters the polar coor dinatesrand .Apointx,y,zontheconesatisfiesxrcos ,yrsin ,and z2sx2 y2 2r.Soavectorequationfortheconeis
rr, rcos irsin j2rk
wherer0and0 2 . M
SURFACES OF REVOLUTION
Surfaces of revolution can be represented parametrically and thus graphed using a com puter. For instance, lets consider the surface S obtained by rotating the curve yf x, axb, about the xaxis, where f x0. Let be the angle of rotation as shown in
So the vector equation is
xx yy z2sx2 y2 rx,yxiyj2sx2 y2 k
1074
CHAPTER 16 VECTOR CALCULUS
z
FIGURE 10
y
z
0
y
y
x z
Figure 9. If x, y, z is a point on S, then
3 xx yf x cos zf x sin
Therefore we take x and as parameters and regard Equations 3 as parametric equations ofS.Theparameterdomainisgivenbyaxb,0 2 .
EXAMPLE 8 Find parametric equations for the surface generated by rotating the curve ysin x, 0x2 , about the xaxis. Use these equations to graph the surface of revolution.
SOLUTION From Equations 3, the parametric equations are
xx ysinx cos zsinx sin
andtheparameterdomainis0x2 ,0 2 .Usingacomputertoplotthese equations and rotate the image, we obtain the graph in Figure 10. M
We can adapt Equations 3 to represent a surface obtained through revolution about the y or zaxis. See Exercise 30.
TANGENT PLANES
We now find the tangent plane to a parametric surface S traced out by a vector function ru, vxu, v iyu, v jzu, v k
at a point P0 with position vector ru0 , v0 . If we keep u constant by putting uu0 , then ru0, v becomes a vector function of the single parameter v and defines a grid curve C1 lying on S. See Figure 11. The tangent vector to C1 at P0 is obtained by taking the partial derivative of r with respect to v:
x, y, z x
FIGURE 9
FIGURE 11
0 CTM
y
x
4
rvx u0,v0i y u0,v0j z u0,v0k v v v
z u ,
r D uu
0u
x
Similarly, if we keep v constant by putting vv0, we get a grid curve C2 given by ru, v0that lies on S, and its tangent vector at P0 is
5 rux u0,v0i y u0,v0j z u0,v0k u u u
r C
P
ru
N Figure 12 shows the selfintersecting surface in Example 9 and its tangent plane at 1, 1, 3.
If rurv is not 0, then the surface S is called smooth it has no corners. For a smooth surface, the tangent plane is the plane that contains the tangent vectors ru and rv, and the vector rurv is a normal vector to the tangent plane.
V EXAMPLE 9 Find the tangent plane to the surface with parametric equations xu2, yv2, zu2v at the point 1, 1, 3.
z
1, 1, 3
x
FIGURE 12
SOLUTION We first compute the tangent vectors:
rux iy jz k2u ik
The image of the subrectangle Rij is the patch Sij.
0
u
x
y
FIGURE 13
ui,j
0
y
u u u
rvx iy jz k2v j2 k
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS1075
v v v Thus a normal vector to the tangent plane is
ijk
ru rv2u 0 1 2vi4uj4uvk
0 2v 2
Notice that the point 1, 1, 3 corresponds to the parameter values u1 and v1, so
the normal vector there is
Therefore an equation of the tangent plane at 1, 1, 3 is
2 i4 j4 k
2x14y14z30
or x2y2z30 M SURFACE AREA
Now we define the surface area of a general parametric surface given by Equation 1. For simplicity we start by considering a surface whose parameter domain D is a rectangle, and we divide it into subrectangles Rij. Lets choose ui, vj to be the lower left corner of Rij. See Figure 13. The part Sij of the surface S that corresponds to Rij is called a patch and has the point Pij with position vector rui, vj as one of its corners. Let
ruruui, vj and rvrvui, vj be the tangent vectors at Pij as given by Equations 5 and 4.
z
Rij Iu
I
r
Pij Sij
1076
CHAPTER 16 VECTOR CALCULUS
Sij
a
Ir
I u r u
b
Approximating a patch by a parallelogram
Figure 14a shows how the two edges of the patch that meet at Pij can be approximated by vectors. These vectors, in turn, can be approximated by the vectors u ru and v rv because partial derivatives can be approximated by difference quotients. So we approxi mate Sij by the parallelogram determined by the vectors u ru and v rv . This parallelo gram is shown in Figure 14b and lies in the tangent plane to S at Pij. The area of this parallelogram is
u ruv rvrurv u v and so an approximation to the area of S is
mn
ru rvuv
i1 j1
Our intuition tells us that this approximation gets better as we increase the number of sub rectangles, and we recognize the double sum as a Riemann sum for the double integral xxD rurv du dv. This motivates the following definition.
Pij
DEFINITION If a smooth parametric surface S is given by the equation ru, vxu, v iyu, v jzu, v k u, vD
and S is covered just once as u, v ranges throughout the parameter domain D, then the surface area of S is
ASyyru rvdA D
6
where rux iy jz k u u u
rvx iy jz k v v v
FIGURE 14
EXAMPLE 10 Find the surface area of a sphere of radius a. SOLUTION In Example 4 we found the parametric representation
xa sin cos ya sin sin za cos where the parameter domain is
D , 0, 0 2We first compute the cross product of the tangent vectors:
ijk
xyz i j k
rr acos cos acos sin asin x y z a sin sin a sin cos 0
a2 sin2 cos ia2 sin2 sin ja2 sin cos k
Thus
so and
Thus we have
x ijk
y
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS1077
since sin
r r sa4sin4 cos2 a4sin4 sin2 a4sin2 cos2sa4 sin4a4 sin2 cos2a2ssin2a2 sin
0 for 0 . Therefore, by Definition 6, the area of the sphere is AyyrrdAy2 ya2sin d d
00
D
a2y2 d ysin d a2224a2 M
00
SURFACE AREA OF THE GRAPH OF A FUNCTION
For the special case of a surface S with equation zf x, y, where x, y lies in D and f has continuous partial derivatives, we take x and y as parameters. The parametric equations are
xx
rx ifk
f f f rxry i
yy
zfx, y
ry jfk
1 0 x x y
jk
01
f2
x
f y
z2
V EXAMPLE 11 Find the area of the part of the paraboloid zx2y2 that lies under
the plane z9.
N Noticethesimilaritybetweenthesurfacearea formula in Equation 9 and the arc length formula
rxry
and the surface area formula in Definition 6 becomes
f2
y1
z2 1xy
L 1 dx ybdy2
a dx from Section 8.1.
z
9
SOLUTION The plane intersects the paraboloid in the circle x2y29, z9. There fore the given surface lies above the disk D with center the origin and radius 3. See Figure 15. Using Formula 9, we have
D
A 1 xy dA yy z2 z2
yy
s12x22y2 dA
y xDD
3
FIGURE 15 yys14x2 y2dA D
7
8
AS 1 xy dA yy z2 z2
D
9
1078
CHAPTER 16 VECTOR CALCULUS
Converting to polar coordinates, we obtain
Ay2 y3 s14r2 rdrd y2 d y3 rs14r2 dr 0000
1 2 2 32 3
2 8314r 0637s371 M
The question remains whether our definition of surface area 6 is consistent with the surface area formula from singlevariable calculus 8.2.4.
We consider the surface S obtained by rotating the curve yfx, axb, about the xaxis, where f x0 and fis continuous. From Equations 3 we know that para metric equations of S are
xx yf x cos zf x sin axb 0 2 To compute the surface area of S we need the tangent vectors
Thus
rx ifxcos jfxsin k r fxsin jfxcos k
ijk
0 fx sin fx cos
rxr 1 fxcos fxsin
fxfxifxcos jfxsin k
rxrsfx2fx2fx2 cos2fx2 sin2
and so
because f x0. Therefore the area of S is
sfx21fx2fxs1fx2 Ayyrx r dAy2 yb fxs1fx2 dxd
0a 2 ybfxs1fx2 dx
a
This is precisely the formula that was used to define the area of a surface of revolution in singlevariable calculus 8.2.4.
D
16.6 EXERCISES
12 Determine whether the points P and Q lie on the given surface.
1. ru, v2u3v, 15uv, 2uv P7, 10, 4, Q5, 22, 5
2. ru,v uv,u2 v,uv2 P3, 1, 5, Q1, 3, 4
36 Identifythesurfacewiththegivenvectorequation. ru,vuvi3vj14u5vk
4. ru, v2 sin u i3 cos u jv k, 0v2
5. rs, ts, t, t2s2
6. rs, ts sin 2t, s2, s cos 2t
; 712 Use a computer to graph the parametric surface. Get a printout and indicate on it which grid curves have u constant and which have v constant.
7. ru,vu2 1,v3 1,uv, 1u1, 1v1 8. ru,vuv,u2,v2, 1u1, 1v1
9. ru,vucosv,usinv,u5, 1u1, 0v2
3.
10. ru, vcos u sin v, sin u sin v, cos vln tanv2, 0u2 , 0.1v6.2
11. xsin v, ycos u sin 4v, zsin 2u sin 4v, 0u2,2v2
12. xu sinu cosv, yu cosu cosv, zu sinv
1318 Match the equations with the graphs labeled IVI and give reasons for your answers. Determine which families of grid curves have u constant and which have v constant.
13. ru,vucosviusinvjvk
14. ru,vucosviusinvjsinuk,u
15. ru, vsin v icos u sin 2v jsin u sin 2v k
16. x1u3cos v cos 4 u, y1u3cos v sin 4 u, z3u1usinv
Iz II
x
SECTION 16.6 PARAMETRIC SURFACES AND THEIR AREAS1079 1926 Find a parametric representation for the surface.
19. The plane that passes through the point 1, 2, 3 and containsthevectorsijkandijk
20. The lower half of the ellipsoid 2×24y2z21
21. The part of the hyperboloid x2y2z21 that lies to the
right of the xzplane
22. The part of the elliptic paraboloid xy22z24 that lies
infrontoftheplanex0
23. Thepartofthespherex2 y2 z2 4thatliesabovethe
conezsx2 y2
24. The part of the sphere x2y2z216 that lies between
the planes z2 and z2
25. The part of the cylinder y2z216 that lies between the
planes x0 and x5
26. The part of the plane zx3 that lies inside the cylinder
x2 y2 1
CAS 2728 Use a computer algebra system to produce a graph that
17. xcos3u cos3v, ysin3u cos3v,
18. x1ucosv, y1usinv, zu
zsin3v z
looks like the given one.
27.
28.
x y y 3 0110 01 IIIzIVz 3y05xy11x
x
Vz VIz
y
; 29. ; 30. ; 31.
; 32.
Find parametric equations for the surface obtained by rotating the curve yex, 0x3, about the xaxis and use them to graph the surface.
Find parametric equations for the surface obtained by rotating thecurvex4y2 y4,2y2,abouttheyaxisand use them to graph the surface.
a What happens to the spiral tube in Example 2 see Fig ure 5 if we replace cos u by sin u and sin u by cos u? b What happens if we replace cos u by cos 2u and sin u
by sin 2u?
The surface with parametric equations
x2 cosr cos 2 y2 sinr cos 2 zrsin 2
where1 r1 and0 2 ,iscalledaMobius 22
strip. Graph this surface with several viewpoints. What is unusual about it?
x
y
xy
y
x
3
z0 z0
1080CHAPTER 16 VECTOR CALCULUS
3336 Find an equation of the tangent plane to the given para 51. metric surface at the specified point. If you have software that
graphs parametric surfaces, use a computer to graph the surface
and the tangent plane. CAS
a Use the Midpoint Rule for double integrals see Sec tion 15.1 with six squares to estimate the area of the surfacez11x2 y2,0x6,0y4.
b Use a computer algebra system to approximate the surface area in part a to four decimal places. Compare with the answer to part a.
Find the area of the surface with vector equation
ru, vcos3u cos3v, sin3u cos3v, sin3v, 0u, 0v2 . State your answer correct to four decimal places.
Findtheexactareaofthesurfacez12x3y4y2, 1×4, 0y1.
Set up, but do not evaluate, a double integral for the area of the surface with parametric equations xau cos v, ybusinv,zu2,0u2,0v2 .
b Eliminate the parameters to show that the surface is an elliptic paraboloid and set up another double integral for the surface area.
c Use the parametric equations in part a with a2 and b3 to graph the surface.
d For the case a2, b3, use a computer algebra system to find the surface area correct to four decimal places.
a Show that the parametric equations xa sin u cos v, ybsinusinv,zccosu,0u ,0v2 , represent an ellipsoid.
b Use the parametric equations in part a to graph the ellip soidforthecasea1,b2,c3.
c Set up, but do not evaluate, a double integral for the sur face area of the ellipsoid in part b.
xuv, y3u2, zuv; 2, 3, 0
34. xu2, yv2, zuv; u1,v1
35. ru,vu2 i2usinvjucosvk; u1,v0
36. ru,vuviusinvjvcosuk; u0,v
3747 Find the area of the surface.
The part of the plane 3x2yz6 that lies in the
first octant
38. Thepartoftheplane2x5yz10thatliesinsidethe
cylinder x2y29
39. Thesurfacez2x32 y32, 0x1, 0y1
40. The part of the plane with vector equation
ru, v1v, u2v, 35uv that is given by 0u1, 0v1
The part of the surface zxy that lies within the cylinder x2 y2 1
CAS 52.
33.
37.
CAS 53.
54. a
41.
3
;
CAS
43. The part of the hyperbolic paraboloid zy2x2 that lies ; betweenthecylindersx2 y2 1andx2 y2 4
42. Thepartofthesurfacez13x2y2 thatliesabovethe triangle with vertices 0, 0, 0, 1, and 2, 1
44. The part of the paraboloid xy2z2 that lies inside the cylindery2 z2 9
45. The part of the surface y4xz2 that lies between the planes x0, x1, z0, and z1
;
56. a
b Use the parametric equations in part a to graph the
46. The helicoid or spiral ramp with vector equation
ru, vu cos v iu sin v jv k, 0u1, 0v
The surface with parametric equations xu2, yuv, z1v2,0u1,0v2
4849 Find the area of the surface correct to four decimal places by expressing the area in terms of a single integral and using your calculator to estimate the integral.
48. The part of the surface zcosx2y2 that lies inside the cylinder x2y21
49. The part of the surface zex2y2 that lies above the disk x2 y2 4
CAS 50. Find, to four decimal places, the area of the part of the surface z1x21y2 that lies above the square
xy 1. Illustrate by graphing this part of the surface.
hyperboloid for the case a1, b2, c3.
c Set up, but do not evaluate, a double integral for the sur
face area of the part of the hyperboloid in part b that lies betweentheplanesz3andz3.
Findtheareaofthepartofthespherex2 y2 z2 4zthat lies inside the paraboloid zx2y2.
58. The figure shows the surface created when the cylinder
y2z21 intersects the cylinder x2z21. Find the area of this surface.
47.
2
55.
Show that the parametric equations xa cosh u cos v, yb cosh u sin v, zc sinh u, represent a hyperboloid of one sheet.
57.
z
x
y
;
b Use the parametric equations found in part a to graph the torus for several values of a and b.
c Use the parametric representation from part a to find the surface area of the torus.
a b, 0, 0
59. Findtheareaofthepartofthespherex2 y2 z2 a2 that lies inside the cylinder x2y2ax.
60. a Find a parametric representation for the torus obtained
by rotating about the zaxis the circle in the xzplane with center b, 0, 0 and radius ab. Hint: Take as parame ters the angles and shown in the figure.
SECTION 16.7 SURFACE INTEGRALS1081 z
D
0u
PARAMETRIC SURFACES
Suppose that a surface S has a vector equation
ru, vxu, v iyu, v jzu, v k
u, vD
r
z
S
0 x
FIGURE 1
of f over the surface S as
Rij
I Iu
We first assume that the parameter domain D is a rectangle and we divide it into subrect angles Rij with dimensions u and v. Then the surface S is divided into corresponding patches Sij as in Figure 1. We evaluate f at a point Pij in each patch, multiply by the area Sij of the patch, and form the Riemann sum
mn
fP i j S i j
i1 j1
Then we take the limit as the number of patches increases and define the surface integral
P i j Sij
yy
S
m n
fPijSij
16.7
SURFACE INTEGRALS
y
Notice the analogy with the definition of a line integral 16.2.2 and also the analogy with the definition of a double integral 15.1.5.
To evaluate the surface integral in Equation 1 we approximate the patch area Sij by the area of an approximating parallelogram in the tangent plane. In our discussion of surface area in Section 16.6 we made the approximation
Sij ru rvuv
x
The relationship between surface integrals and surface area is much the same as the rela tionship between line integrals and arc length. Suppose f is a function of three variables whose domain includes a surface S. We will define the surface integral of f over S in such a way that, in the case where f x, y, z1, the value of the surface integral is equal to the surface area of S. We start with parametric surfaces and then deal with the special case where S is the graph of a function of two variables.
1
fx,y,zdS lim
m,nl i1 j1
0
x, y, z
y
1082
CHAPTER 16 VECTOR CALCULUS
where rux iy jz k rvx iy jz k
N We assume that the surface is covered only once as u, v ranges throughout D. The value of the surface integral does not depend on the parametrization that is used.
are the tangent vectors at a corner of Sij. If the components are continuous and ru and rv are nonzero and nonparallel in the interior of D, it can be shown from Definition 1, even when D is not a rectangle, that
This should be compared with the formula for a line integral:
N Here we use the identities
cos21 1cos 2
sin21cos2
Instead, we could use Formulas 64 and 67 in the Table of Integrals.
2
Observe also that
y fx, y, z dsyb frtrtdt Ca
yy1dSyyru rvdAAS SD
u u u v v v
yyfx,y,zdSyyfru,vru rvdA SD
Formula 2 allows us to compute a surface integral by converting it into a double inte gral over the parameter domain D. When using this formula, remember that f ru, v is evaluated by writing xxu, v, yyu, v, and zzu, v in the formula for f x, y, z.
EXAMPLE 1 Compute the surface integral xx x2 dS, where S is the unit sphere
S
x2 y2 z2 1.
SOLUTION As in Example 4 in Section 16.6, we use the parametric representation
xsin cos ysin sin zcos 0 0
thatis, r , sin cos isin sin jcos k As in Example 10 in Section 16.6, we can compute that
2
r rsin yyx2dSyysin cos 2r r dA
SD
y2 ysin2 cos2 sin d d y2 cos2 d ysin3 d 0000
d
Therefore, by Formula 2,
2
y2 1 1cos 2d y 020
sinsin cos2 112134
2 2 sin 2 0 cos3 cos 03
M
Surface integrals have applications similar to those for the integrals we have previously considered. For example, if a thin sheet say, of aluminum foil has the shape of a surface
S and the density mass per unit area at the point x, y, z is of the sheet is
myy x,y,zdS S
and the center of mass is x, y, z, where
x 1 yyx x,y,zdS y 1 yyy x,y,zdS
x, y, z, then the total mass
andsowehave Thus
x
y
3
xx
rx itk
yy
SECTION 16.7 SURFACE INTEGRALS1083
z 1 yyz x,y,zdS mSmSmS
Moments of inertia can also be defined as before see Exercise 39.
GRAPHS
Any surface S with equation ztx, y can be regarded as a parametric surface with para metric equations
ztx, y
ry jtk
rxry xy1 z2 z2
rxry t it jk x y
and
Therefore, in this case, Formula 2 becomes
4
f x, y, z dSf x, y, tx, y xy1 dA yy yy z2 z2
SD
z
FIGURE 2
y
Similar formulas apply when it is more convenient to project S onto the yzplane or xzplane. For instance, if S is a surface with equation yhx, z and D is its projection on the xzplane, then
f x, y, z dSf x, hx, z, z xz1 dA yy yy y2 y2
SD
EXAMPLE2 EvaluatexxydS,whereSisthesurfacezxy2,0x1,0y2.
See Figure 2. SOLUTION Since
S
x
z 1 and z 2y x y
1084
CHAPTER 16 VECTOR CALCULUS
Formula 4 gives
yy
ydS y 1 xy dA yyz2 z2
z y
0 STM
FIGURE3
S z1x
S 1
SD
y1 y2 ys114y2 dydx
00
y1 dxs2 y2 ys12y2 dy 00
1 2 2 32 2 13s2
s24312y 0 3 M
If S is a piecewisesmooth surface, that is, a finite union of smooth surfaces S1 , S2, . . . , Sn that intersect only along their boundaries, then the surface integral of f over S is defined by
yy f x, y, z dSyy f x, y, z dSyy f x, y, z dS SS1 Sn
V EXAMPLE 3 Evaluate xxS z dS, where S is the surface whose sides S1 are given by the cylinderx2 y2 1,whosebottomS2 isthediskx2 y2 1intheplanez0,and whosetopS3 isthepartoftheplanez1xthatliesaboveS2.
SOLUTION The surface S is shown in Figure 3. We have changed the usual position of the axes to get a better look at S. For S1 we use and z as parameters see Example 5 in Section 16.6 and write its parametric equations as
xcos
2 and
ysin
0z1x1cos
zz
where
x
0
Therefore
and
Thus the surface integral over S1 is
ijk
rrsin cos 0 cos isin j
z
001
r rzscos2 sin2 1 yyzdSyyzr rzdA
S1 D
y2 y1cos zdzd y2 11cos 2d
1 y2 12 cos1 1cos 2d 202
00 02
1 32 sin1 sin 22240
2
23
Since S2 lies in the plane z0, we have
yy z dSyy 0 dS0
S2 S2
ThetopsurfaceS3 liesabovetheunitdiskDandispartoftheplanez1x.So,
taking tx, y1x in Formula 4 and converting to polar coordinates, we have zdS 1x 1 xy dA
SECTION 16.7 SURFACE INTEGRALS1085
yy yyz2 z2 S3 D
y2 y1 1rcos s110 rdrd 00
s2y2 y1rr2cos drd 00
s2y2 11cosd
0
sin 2
23
s223 s2 0
Therefore
yy z dSyy z dSyy z dSyy z dS S S1 S2 S3
3 0s2 3s2 M
2
ORIENTED SURFACES
2
P
To define surface integrals of vector fields, we need to rule out nonorientable surfaces such as the Mobius strip shown in Figure 4. It is named after the German geometer August Mobius 17901868. You can construct one for yourself by taking a long rectangular strip of paper, giving it a halftwist, and taping the short edges together as in Figure 5. If an ant were to crawl along the Mobius strip starting at a point P, it would end up on the other side of the strip that is, with its upper side pointing in the opposite direction. Then, if the ant continued to crawl in the same direction, it would end up back at the same point P without ever having crossed an edge. If you have constructed a Mobius strip, try drawing a pencil line down the middle. Therefore a Mobius strip really has only one side. You can graph the Mobius strip using the parametric equations in Exercise 32 in Section 16.6.
FIGURE 4
A Mobius strip
TEC Visual 16.7 shows a Mobius strip with a normal vector that can be moved along the surface.
B C AD
B D
FIGURE 5
Constructing a Mobius strip
AC
1086
CHAPTER 16 VECTOR CALCULUS
z
n
From now on we consider only orientable twosided surfaces. We start with a surface S that has a tangent plane at every point x, y, z on S except at any boundary point. There are two unit normal vectors n1 and n2n1 at x, y, z. See Figure 6.
If it is possible to choose a unit normal vector n at every such point x, y, z so that n varies continuously over S, then S is called an oriented surface and the given choice of n provides S with an orientation. There are two possible orientations for any orientable sur face see Figure 7.
nTM 0nn
n
n
xyn
nn nn
n
FIGURE 6
FIGURE 7
The two orientations
of an orientable surface
For a surface ztx, y given as the graph of t, we use Equation 3 to associate with the surface a natural orientation given by the unit normal vector
t it jk x y
nt2 t2 1 xy
z
0
Since the kcomponent is positive, this gives the upward orientation of the surface.
If S is a smooth orientable surface given in parametric form by a vector function
ru, v, then it is automatically supplied with the orientation of the unit normal vector n rurv
rurv
and the opposite orientation is given by n. For instance, in Example 4 in Section 16.6 we
found the parametric representation
r , asin cos iasin sin jacos k
for the sphere x2y2z2a2. Then in Example 10 in Section 16.6 we found that
y
x
FIGURE 8
Positive orientation
z
Negative orientation
r r a2sin2 cos ia2sin2 sin ja2sin cos k r ra2sin
and
So the orientation induced by r
,is defined by the unit normal vector
n r r sin cos isin sin jcos k1r ,
x
FIGURE 9
5
6
r r a
y
Observe that n points in the same direction as the position vector, that is, outward from the sphere see Figure 8. The opposite inward orientation would have been obtained see Figure 9 if we had reversed the order of the parameters because rrrr .
For a closed surface, that is, a surface that is the boundary of a solid region E, the convention is that the positive orientation is the one for which the normal vectors point outward from E, and inwardpointing normals give the negative orientation see Figures 8 and 9.
z
0
FIGURE 10
n
Sij
S
Fv
y
SECTION 16.7 SURFACE INTEGRALS1087 SURFACE INTEGRALS OF VECTOR FIELDS
Suppose that S is an oriented surface with unit normal vector n, and imagine a fluid with density x, y, z and velocity field vx, y, z flowing through S. Think of S as an imagi nary surface that doesnt impede the fluid flow, like a fishing net across a stream. Then the rate of flow mass per unit time per unit area is v. If we divide S into small patches Sij, as in Figure 10 compare with Figure 1, then Sij is nearly planar and so we can approxi mate the mass of fluid crossing Sij in the direction of the normal n per unit time by the quantity
vnASij
x
where , v, and n are evaluated at some point on Sij. Recall that the component of the vec tor v in the direction of the unit vector n is vn. By summing these quantities and tak ing the limit we get, according to Definition 1, the surface integral of the function vn over S:
yy vndSyy x,y,zvx,y,znx,y,zdS SS
and this is interpreted physically as the rate of flow through S.
If we write Fv, then F is also a vector field on 3 and the integral in Equation 7
7
becomes
A surface integral of this form occurs frequently in physics, even when F is not v, and is
called the surface integral or flux integral of F over S.
yy Fn dS S
DEFINITION If F is a continuous vector field defined on an oriented surface S with unit normal vector n, then the surface integral of F over S is
yy FdSyy Fn dS SS
This integral is also called the flux of F across S.
8
In words, Definition 8 says that the surface integral of a vector field over S is equal to the surface integral of its normal component over S as previously defined.
If S is given by a vector function ru, v, then n is given by Equation 6, and from Def inition 8 and Equation 2 we have
yyFdSyyF rurv dS
S S
rurv
yyFru,v ru rv ru rv dA
N Compare Equation 9 to the similar expres sion for evaluating line integrals of vector fields in Definition 16.2.13:
y Fdryb Frtrt dt Ca
D rurvwhere D is the parameter domain. Thus we have
9
yyFdSyyFru rvdA SD
1088CHAPTER 16 VECTOR CALCULUS
EXAMPLE 4 FindthefluxofthevectorfieldFx,y,zziyjxkacrosstheunit
N Figure 11 shows the vector field F in Example 4 at points on the unit sphere.
z
x
FIGURE 11
spherex2 y2 z2 1.
SOLUTION Using the parametric representation
r , sin cos isin sin jcos k 00 2
wehave Fr , cos isin sin jsin cos k and, from Example 10 in Section 16.6,
y
r r sin2 cos isin2 sin jsin cos k
Fr , r r cos sin2 cos sin3 sin2 sin2 cos cos
Therefore
and, by Formula 9, the flux is
yyFdSyyFr rdA SD
y2 y 2sin2 cos cos sin3 sin2 d d 00
2ysin2 cos d y2 cos d ysin3 d y2 sin2 d 0000
0y sin3 d y2 sin2 d sincey2 cos d 0 000
4 3
by the same calculation as in Example 1. M
If, for instance, the vector field in Example 4 is a velocity field describing the flow of a fluid with density 1, then the answer, 4 3, represents the rate of flow through the unit sphere in units of mass per unit time.
In the case of a surface S given by a graph ztx, y, we can think of x and y as param eters and use Equation 3 to writet t
Frx ryPiQjRk x i y jk Thus Formula 9 becomes
10 yy FdSyyP tQ tRdA S Dxy
This formula assumes the upward orientation of S; for a downward orientation we multi ply by 1. Similar formulas can be worked out if S is given by yhx, z or xky, z. See Exercises 35 and 36.
z
S
x
FIGURE 12
V EXAMPLE 5 EvaluatexxS FdS,whereFx,y,zyixjzkandSisthe boundary of the solid region E enclosed by the paraboloid z1x2y2 and the plane z0.
SOLUTION S consists of a parabolic top surface S1 and a circular bottom surface S2. See Figure 12. Since S is a closed surface, we use the convention of positive outward orientation. This means that S1 is oriented upward and we can use Equation 10 with D being the projection of S1 on the xyplane, namely, the disk x2y21. Since
t yyFdSyyPxQyR dA
SECTION 16.7 SURFACE INTEGRALS1089
STM
y
Px,y,zy onS1 and
we have
Qx,y,zx t 2x
x
Rx,y,zz1x2 y2 t 2y
y
t
S1 D
yyy2xx2y1x2 y2dA
D
yy14xyx2 y2dA D
y2 y1 14r2cos sin r2rdrd 00
y2 y1rr34r3cos sindrd 00
y2 1cos sind120 44
02
The disk S2 is oriented downward, so its unit normal vector is nk and we have
yy FdSyy Fk dSyy z dAyy 0 dA0 S2 S2 D D
since z0 on S2. Finally, we compute, by definition, xxS FdS as the sum of the sur face integrals of F over the pieces S1 and S2:
yyFdSyyFdSyyFdS 2 0 2 M S S1 S2
Although we motivated the surface integral of a vector field using the example of fluid flow, this concept also arises in other physical situations. For instance, if E is an electric field see Example 5 in Section 16.1, then the surface integral
yy EdS S
is called the electric flux of E through the surface S. One of the important laws of electro
1090
CHAPTER 16 VECTOR CALCULUS
statics is Gausss Law, which says that the net charge enclosed by a closed surface S is
Q0 yy EdS S
11
where 0 is a constant called the permittivity of free space that depends on the units used. 122 2
In the SI system, 08.854210 C Nm. Therefore, if the vector field F in Example 4 represents an electric field, we can conclude that the charge enclosed by S is Q4 03.
Another application of surface integrals occurs in the study of heat flow. Suppose the temperature at a point x, y, z in a body is ux, y, z. Then the heat flow is defined as the vector field
FK u
where K is an experimentally determined constant called the conductivity of the sub stance. The rate of heat flow across the surface S in the body is then given by the surface integral
yy FdSK yy udS SS
V EXAMPLE 6 The temperature u in a metal ball is proportional to the square of the distance from the center of the ball. Find the rate of heat flow across a sphere S of radius a with center at the center of the ball.
SOLUTION Taking the center of the ball to be at the origin, we have ux,y,zCx2 y2 z2
where C is the proportionality constant. Then the heat flow is
Fx, y, zK uKC2x i2y j2z k
where K is the conductivity of the metal. Instead of using the usual parametrization of the sphere as in Example 4, we observe that the outward unit normal to the sphere
x2 y2 z2 a2 atthepointx,y,zis
n1 x iy jz k a
andso Fn2KCx2 y2 z2 a
ButonSwehavex2 y2 z2 a2,soFn2aKC.Thereforetherateofheat flow across S is
yy FdSyy Fn dS2aKC yy dS SSS
2aKCAS2aKC4 a2 8KC a3 M
16.7 EXERCISES
1. Let S be the boundary surface of the box enclosed by the
13. xxS y dS,
S is the part of the paraboloid yx2z2 that lies inside the cylinderx2z24
planesx0,x2,y0,y4,z0,andz6.Approx
imate xx e0.1xyz dS by using a Riemann sum as in Defini S
tion 1, taking the patches Sij to be the rectangles that are the faces of the box S and the points Pij to be the centers of the rectangles.
14. xx y2 dS,
S 222
2. AsurfaceSconsistsofthecylinderx2 y2 1,1z1, together with its top and bottom disks. Suppose you know that
f is a continuous function with
f 1, 0, 02 f 0, 1, 03 f 0, 0, 14
Estimate the value of xx f x, y, z dS by using a Riemann sum, taking the patches Sij toSbe four quartercylinders and the top and bottom disks.
3. LetHbethehemispherex2 y2 z2 50,z0,and suppose f is a continuous function with f 3, 4, 57,
f 3, 4, 58, f 3, 4, 59, and f 3, 4, 512. By dividing H into four patches, estimate the value of
xxH fx,y,zdS.
4. Supposethat fx,y,ztsx2 y2 z2 ,wheretisa function of one variable such that t25. Evaluate
xxS fx,y,zdS,whereSisthespherex2 y2 z2 4.
518 Evaluate the surface integral. 5. xxx2yzdS,
Sisthepartofthespherex y z 4thatlies inside the cylinder x 2y 21 and above the x yplane
SECTION 16.7 SURFACE INTEGRALS1091
xxSx2zy2zdS,
Sisthehemispherex2 y2 z2 4,z0
16. xxSxzdS,
S is the boundary of the region enclosed by the cylinder y2 z2 9andtheplanesx0andxy5
17. xxzx2ydS, S
S is the part of the cylinder y2z21 that lies between the planesx0andx3inthefirstoctant
18. xxSx2y2z2dS,
S is the part of the cylinder x2y29 between the planes z0 and z2, together with its top and bottom disks
1930 Evaluate the surface integral xxS FdS for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive outward orientation.
Fx,y,zxyiyzjzxk, Sisthepartofthe S22
Sisthepartoftheplanez12x3ythatliesabovethe rectangle 0, 30, 2
S is the triangular region with vertices 1, 0, 0, 0, 2, 0, and 0, 0, 2
7. xxSyzdS, Sisthepartoftheplanexyz1thatliesinthe first octant
8. xxS y dS,
Sisthesurfacez2x32 y32,0x1,0y1
9. xxS yzdS,
S is the surface with parametric equations xu 2, yu sin v, zu cos v, 0u1, 0v2
10. xxSs1x2y2dS,
S is the helicoid with vector equation
ru, vu cos v iu sin v jv k, 0u1, 0v
paraboloid z4xy that lies above the square 0x1,0y1,andhasupwardorientation
20. Fx,y,zyixjz2 k,
S is the helicoid of Exercise 10 with upward orientation
21. Fx,y,zxzey ixzey jzk, Sisthepartoftheplanexyz1inthefirstoctantand has downward orientation
22. Fx,y,zxiyjz4 k,
Sisthepartoftheconezsx2 y2 beneaththeplanez1 with downward orientation
23. Fx,y,zxizjyk,
Sisthepartofthespherex2 y2 z2 4inthefirstoctant, with orientation toward the origin
24. Fx,y,zxzixjyk,
Sisthehemispherex2 y2 z2 25,y0,orientedinthe direction of the positive yaxis
6. xxxydS, S
3
11. xxx2z2dS,
Sisthepartoftheconez x y thatliesbetweenthe planesz1andz3
Fx,y,zyjzk, S22222
12. xxzdS, S
Sconsistsoftheparaboloidyx z,0y1, andthediskx2 z2 1,y1
26. Fx,y,zxyi4x2 jyzk, Sisthesurfacezxey, 0x1, 0y1, with upward orientation
S is the surface xy2z2, 0y1, 0z1
15.
19.
25.
1092CHAPTER 16 VECTOR CALCULUS
27. Fx,y,zxi2yj3zk,
S is the cube with vertices 1, 1, 1
28. Fx,y,zxiyj5k, Sistheboundaryoftheregion enclosed by the cylinder x2z21 and the planes y0 and xy2
29. Fx,y,zx2 iy2jz2 k, Sistheboundaryofthe solidhalfcylinder0zs1y2,0x2
30. Fx,y,zyizyjxk,
S is the surface of the tetrahedron with vertices 0, 0, 0, 1, 0, 0, 0, 1, 0, and 0, 0, 1
39. a Give an integral expression for the moment of inertia Iz about the zaxis of a thin sheet in the shape of a surface S if the density function is .
b Find the moment of inertia about the zaxis of the funnel in Exercise 38.
40. LetSbethepartofthespherex2 y2 z2 25thatlies above the plane z4. If S has constant density k, find
a the center of mass and b the moment of inertia about the zaxis.
41. A fluid has density 870 kgm3 and flows with velocity vziy2 jx2 k,wherex,y,andzaremeasuredin meters and the components of v in meters per second. Find the rate of flow outward through the cylinder x2y24, 0z1.
CAS 31. CAS 32. CAS 33.
CAS 34.
Evaluate xxS xyz dS correct to four decimal places, where S is the surface zxy, 0x1, 0y1.
Find the exact value of xx x 2 yz dS, where S is the surface in 3
S
and flows in a velocity field
Exercise 31. Findthevalueofxxx2y2z2dScorrecttofourdecimalplaces,
S whereSisthepartoftheparaboloidz32x2 y2 that
lies above the xyplane. Find the flux of
Fx,y,zsinxyzix2yjz2ex5 k
across the part of the cylinder 4y2z24 that lies above the xyplane and between the planes x2 and x2 with upward orientation. Illustrate by using a computer algebra system to draw the cylinder and the vector field on the same screen.
43. Use Gausss Law to find the charge contained in the solid hemispherex2 y2 z2 a2,z0,iftheelectricfieldis
Ex, y, zx iy j2z k
44. Use Gausss Law to find the charge enclosed by the cube
with vertices 1, 1, 1 if the electric field is Ex, y, zx iy jz k
The temperature at the point x, y, z in a substance with con ductivity K6.5 is ux, y, z2y22z2. Find the rate of heat flow inward across the cylindrical surface y2z26, 0x4.
46. The temperature at a point in a ball with conductivity K is inversely proportional to the distance from the center of the ball. Find the rate of heat flow across a sphere S of radius a with center at the center of the ball.
47. Let F be an inverse square field, that is, Frcrr 3 for someconstantc,whererxiyjzk.Showthatthe flux of F across a sphere S with center the origin is inde pendent of the radius of S.
42. Seawater has density 1025 kgm
vy ix j, where x, y, and z are measured in meters and the components of v in meters per second. Find the rate of flow outward through the hemisphere x2y2z29,
z0.
35. Find a formula for xxS FdS similar to Formula 10 for the case where S is given by yhx, z and n is the unit normal that points toward the left.
36. Find a formula for xx FdS similar to Formula 10 for the S
37.
case where S is given by xky, z and n is the unit normal that points forward that is, toward the viewer when the axes are drawn in the usual way.
Find the center of mass of the hemisphere x2y2z2a2, z0, if it has constant density.
38. Find the mass of a thin funnel in the shape of a cone zsx2 y2,1z4,ifitsdensityfunctionis
45.
x, y, z10z.
16.8 STOKES THEOREM
Stokes Theorem can be regarded as a higherdimensional version of Greens Theorem. Whereas Greens Theorem relates a double integral over a plane region D to a line integral around its plane boundary curve, Stokes Theorem relates a surface integral over a surface S to a line integral around the boundary curve of S which is a space curve. Figure 1 shows
z
an oriented surface with unit normal vector n. The orientation of S induces the positive orientation of the boundary curve C shown in the figure. This means that if you walk in the positive direction around C with your head pointing in the direction of n, then the sur face will always be on your left.
n
n
C 0
S
1
SECTION 16.8 STOKES THEOREM1093
STOKES THEOREM Let S be an oriented piecewisesmooth surface that is bounded by a simple, closed, piecewisesmooth boundary curve C with positive orientation. Let F be a vector field whose components have continuous partial derivatives on an open region in 3 that contains S. Then
yC FdryycurlFdS S
xy
FIGURE 1
N Stokes Theorem is named after the Irish mathematical physicist Sir George Stokes
1819 1903. Stokes was a professor at Cam bridge University in fact he held the same position as Newton, Lucasian Professor of Mathematics and was especially noted for his studies of fluid flow and light. What we call Stokes Theorem was actually discovered by
the Scottish physicist Sir William Thomson 18241907, known as Lord Kelvin. Stokes learned of this theorem in a letter from Thomson in 1850 and asked students to prove it on an examination at Cambridge University in 1854. We dont know if any of those students was able to do so.
Since
curl Fn dS
Stokes Theorem says that the line integral around the boundary curve of S of the tangen tial component of F is equal to the surface integral of the normal component of the curl of F.
The positively oriented boundary curve of the oriented surface S is often written as S, so Stokes Theorem can be expressed as
yycurlFdSyS Fdr S
There is an analogy among Stokes Theorem, Greens Theorem, and the Fundamental Theorem of Calculus. As before, there is an integral involving derivatives on the left side of Equation 1 recall that curl F is a sort of derivative of F and the right side involves the values of F only on the boundary of S.
In fact, in the special case where the surface S is flat and lies in the xyplane with upward orientation, the unit normal is k, the surface integral becomes a double integral, and Stokes Theorem becomes
yC FdryycurlFdSyycurlFkdA SS
This is precisely the vector form of Greens Theorem given in Equation 16.5.12. Thus we see that Greens Theorem is really a special case of Stokes Theorem.
Although Stokes Theorem is too difficult for us to prove in its full generality, we can give a proof when S is a graph and F, S, and C are well behaved.
PROOF OF A SPECIAL CASE OF STOKES THEOREM We assume that the equation of S is ztx, y, x, yD, where t has continuous secondorder partial derivatives and D is a simple plane region whose boundary curve C1 corresponds to C. If the orientation of S is upward, then the positive orientation of C corresponds to the positive orientation of C1. See Figure 2. We are also given that FP iQ jR k, where the partial deriva tives of P, Q, and R are continuous.
z
n
zgx, y C
FdrFT ds and curl FdS
y y yy yy
SS
CC
S
xD C
FIGURE 2
0
y
1094
CHAPTER 16 VECTOR CALCULUS
Since S is a graph of a function, we can apply Formula 16.7.10 with F replaced by
curl F. The result is yy curl FdS
2
S
yyRQzPRzQPdA
y z x z x y x y
D
where the partial derivatives of P, Q, and R are evaluated at x, y, tx, y. If
xxt yyt atb
is a parametric representation of C1, then a parametric representation of C is
xxt yyt ztxt, yt atb
This allows us, with the aid of the Chain Rule, to evaluate the line integral as follows:
y Fdryb P dxQ dyR dzdt C a dt dt dt
yb P dxQ dyRz dxz dydt a dt dt xdtydt
yb PR zdx QR zdydt a x dt y dt
yPR zdxQR zdy yyQRzPRz dA
C1 xy
x y y x D
where we have used Greens Theorem in the last step. Then, using the Chain Rule again and remembering that P, Q, and R are functions of x, y, and z and that z is itself a func tion of x and y, we get
yFdryyQQ zR zR z zR 2zC x z x x y z x y xy
D
PP zR zR z zR 2z dA y z y y x z y x yx
Four of the terms in this double integral cancel and the remaining six terms can be arranged to coincide with the right side of Equation 2. Therefore
yC Fdryy curl FdS M S
z
C
yz2
be counterclockwise when viewed from above.
SOLUTION The curve C an ellipse is shown in Figure 3. Although xC Fdr could be evaluated directly, its easier to use Stokes Theorem. We first compute
SECTION 16.8 STOKES THEOREM1095 V EXAMPLE1 EvaluatexFdr,whereFx,y,zy2ixjz2kandCisthe
C
curve of intersection of the plane yz2 and the cylinder x2y21. Orient C to
ijk
curlF 12yk
Although there are many surfaces with boundary C, the most convenient choice is the elliptical region S in the plane yz2 that is bounded by C. If we orient S upward, then C has the induced positive orientation. The projection D of S on the xyplane is the diskx2 y2 1andsousingEquation16.7.10withztx,y2y,wehave
yC FdryycurlFdSyy12ydA SD
x y z 22
y x z
x
y
FIGURE 3
y2 y1 12rsin rdrd 00
y 2
231
r2rsin d2 12sind
y2 02300
12 0 M
23
z
S C
0
z4
V EXAMPLE 2 Use Stokes Theorem to compute the integral xxS curl FdS, where Fx,y,zxziyzjxykandSisthepartofthespherex2 y2 z2 4that lies inside the cylinder x2y21 and above the xyplane. See Figure 4.
SOLUTION To find the boundary curve C we solve the equations x2y2z24 and x2 y2 1.Subtracting,wegetz2 3andsozs3 sincez0.ThusCisthe circle given by the equations x2y21, zs3. A vector equation of C is
x
FIGURE 4
1
y
rtcostisintjs3 k so rtsin t icos t j
Also, we have
Therefore, by Stokes Theorem,
0t2
S
D0
Frts3 cos t is3 sin t jcos t sin t k yycurlFdSyFdry2 Frtrtdt
S
C0
y2 s3 costsints3 sintcostdt
0
s3y2 0dt0 M 0
1096
CHAPTER 16 VECTOR CALCULUS
T Cv
a jC vdr0, positive circulation T
C
v
b jC vdr0, negative circulation FIGURE 5
N Imagine a tiny paddle wheel placed in the fluid at a point P, as in Figure 6; the paddle wheel rotates fastest when its axis is parallel to curl v.
Note that in Example 2 we computed a surface integral simply by knowing the values of F on the boundary curve C. This means that if we have another oriented surface with the same boundary curve C, then we get exactly the same value for the surface integral!
In general, if S1 and S2 are oriented surfaces with the same oriented boundary curve C and both satisfy the hypotheses of Stokes Theorem, then
yycurlFdSyC FdryycurlFdS S1 S2
This fact is useful when it is difficult to integrate over one surface but easy to integrate over the other.
We now use Stokes Theorem to throw some light on the meaning of the curl vector. Suppose that C is an oriented closed curve and v represents the velocity field in fluid flow. Consider the line integral
yC vdryC vT ds
and recall that vT is the component of v in the direction of the unit tangent vector T. This means that the closer the direction of v is to the direction of T, the larger the value of vT. Thus xC vdr is a measure of the tendency of the fluid to move around C and is called the circulation of v around C. See Figure 5.
Now let P0x0, y0, z0be a point in the fluid and let Sa be a small disk with radius a and center P0. Then curl FPcurl FP0 for all points P on Sa because curl F is con tinuous. Thus, by Stokes Theorem, we get the following approximation to the circulation around the boundary circle Ca:
yCa vdryycurlvdSyycurlvndS Sa Sa
yycurl vP0nP0dScurl vP0nP0 a2 Sa
This approximation becomes better as a l 0 and we have
1 curlvP0nP0lim 2 y vdr
Equation 4 gives the relationship between the curl and the circulation. It shows that curl vn is a measure of the rotating effect of the fluid about the axis n. The curling effect is greatest about the axis parallel to curl v.
Finally, we mention that Stokes Theorem can be used to prove Theorem 16.5.4 which states that if curl F0 on all of 3, then F is conservative. From our previous work Theorems 16.3.3 and 16.3.4, we know that F is conservative if xC Fdr0 for every closed path C. Given C, suppose we can find an orientable surface S whose boundary is C. This can be done, but the proof requires advanced techniques. Then Stokes Theorem gives
yC FdryycurlFdSyy0dS0 SS
A curve that is not simple can be broken into a number of simple curves, and the integrals around these simple curves are all 0. Adding these integrals, we obtain xC Fdr0 for any closed curve C.
al0 a Ca
curl v
FIGURE 6
3
4
16.8 EXERCISES
1. A hemisphere H and a portion P of a paraboloid are shown. Suppose F is a vector field on 3 whose components have con tinuous partial derivatives. Explain why
yycurlFdSyycurlFdS HP
zz
10.
11.
Fx,y,zxyi2zj3yk, Cisthecurveofintersection oftheplanexz5andthecylinderx2 y2 9
4 H
4
P
a
b
c a
b c
field F and surface S.
13. Fx,y,zy2 ixjz2 k,
S is the part of the paraboloid zx2y2 that lies below the plane z1, oriented upward
14. Fx,y,zxiyjxyzk,
S is the part of the plane 2xyz2 that lies in the first octant, oriented upward
15. Fx,y,zyizjxk,
Sisthehemispherex2 y2 z2 1,y0,orientedinthe direction of the positive yaxis
16. Let C be a simple closed smooth curve that lies in the plane xyz1. Show that the line integral
xC zdx2xdy3ydz
depends only on the area of the region enclosed by C and not
on the shape of C or its location in the plane.
17. A particle moves along line segments from the origin to the points 1, 0, 0, 1, 2, 1, 0, 2, 1, and back to the origin under the influence of the force field
Fx,y,zz2 i2xyj4y2 k Find the work done.
; ;
26 Use Stokes Theorem to evaluate xxS curl FdS. ;
2. Fx,y,z2ycosziexsinzjxey k,
Sisthehemispherex2 y2 z2 9,z0,oriented ; upward
3. Fx, y, zx2z2 iy2z2 jxyz k,
S is the part of the paraboloid zx2y2 that lies inside the cylinder x2y24, oriented upward
4. Fx,y,zx2y3zisinxyzjxyzk, Sisthepartoftheconey2 x2 z2 thatliesbetweenthe planes y0 and y3, oriented in the direction of the positive yaxis
5. Fx,y,zxyzixyjx2yzk,
S consists of the top and the four sides but not the bottom of the cube with vertices 1, 1, 1, oriented outward Hint: Use Equation 3.
6. Fx,y,zexy coszix2zjxyk,
S is the hemisphere xs1y2z2 , oriented in the direc tion of the positive xaxis Hint: Use Equation 3.
710 Use Stokes Theorem to evaluate xC Fdr. In each case C is oriented counterclockwise as viewed from above.
7. Fx,y,zxy2iyz2jzx2k,
C is the triangle with vertices 1, 0, 0, 0, 1, 0, and 0, 0, 1
8. Fx, y, zex iex jez k,
C is the boundary of the part of the plane 2xy2z2 in the first octant
9. Fx,y,zyzi2xzjexy k, Cisthecirclex2 y2 16,z5
12.
x2 2yx2 2y
SECTION 16.8 STOKES THEOREM1097
UseStokesTheoremtoevaluatexC Fdr,where Fx, y, zx2z ixy2 jz2 k
and C is the curve of intersection of the plane xyz1andthecylinderx2 y2 9oriented counterclockwise as viewed from above.
Graph both the plane and the cylinder with domains chosen so that you can see the curve C and the surface that you used in part a.
Find parametric equations for C and use them to graph C. Use Stokes Theorem to evaluate xC Fdr, where
Fx, y, zx2 y i1 x3 jxy k and C is the curve of 3
intersection of the hyperbolic paraboloid zy 2x 2 and the cylinder x2y21 oriented counterclockwise as viewed from above.
Graph both the hyperbolic paraboloid and the cylinder with domains chosen so that you can see the curve C and the surface that you used in part a.
Find parametric equations for C and use them to graph C. 1315 Verify that Stokes Theorem is true for the given vector
1098CHAPTER 16 VECTOR CALCULUS 18. Evaluate
xC ysin x dxz2cos y dyx3 dz
where C is the curve rtsin t, cos t, sin 2t, 0t2 .
20.
Suppose S and C satisfy the hypotheses of Stokes Theorem and f , t have continuous secondorder partial derivatives. Use Exercises 24 and 26 in Section 16.5 to show the following.
a xCf tdrxxS ftdS bxC ffdr0
c xCf ttf dr0
19.
Hint:ObservethatCliesonthesurfacez2xy.
If S is a sphere and F satisfies the hypotheses of Stokes
Theorem, show that xxS curl FdS0.
THREE MEN AND TWO THEOREMS
WRITING PROJECT
N The photograph shows a stainedglass window at Cambridge University in honor of George Green.
Courtesy of the Masters and Fellows of Gonville and Caius College, University of Cambridge, England
www.stewartcalculus.com
The Internet is another source of infor mation for this project. Click on History of Mathematics. Follow the links to the St. Andrews site and that of the British Society for the History of Mathematics.
Although two of the most important theorems in vector calculus are named after George Green and George Stokes, a third man, William Thomson also known as Lord Kelvin, played a large role in the formulation, dissemination, and application of both of these results. All three men were interested in how the two theorems could help to explain and predict physical phenomena in electricity and magnetism and fluid flow. The basic facts of the story are given in the margin notes on pages 1056 and 1093.
Write a report on the historical origins of Greens Theorem and Stokes Theorem. Explain the similarities and relationship between the theorems. Discuss the roles that Green, Thomson, and Stokes played in discovering these theorems and making them widely known. Show how both theorems arose from the investigation of electricity and magnetism and were later used to study a variety of physical problems.
The dictionary edited by Gillispie 2 is a good source for both biographical and scientific information. The book by Hutchinson 5 gives an account of Stokes life and the book by Thompson 8 is a biography of Lord Kelvin. The articles by GrattanGuinness 3 and Gray 4 and the book by Cannell 1 give background on the extraordinary life and works of Green. Additional historical and mathematical information is found in the books by Katz 6 and
Kline 7.
1. D. M. Cannell, George Green, Mathematician and Physicist 17931841: The Background to
His Life and Work Philadelphia: Society for Industrial and Applied Mathematics, 2001.
2. C. C. Gillispie, ed., Dictionary of Scientific Biography New York: Scribners, 1974. See the article on Green by P. J. Wallis in Volume XV and the articles on Thomson by Jed Buchwald and on Stokes by E. M. Parkinson in Volume XIII.
3. I. GrattanGuinness, Why did George Green write his essay of 1828 on electricity and magnetism? Amer. Math. Monthly, Vol. 102 1995, pp. 387396.
4. J. Gray, There was a jolly miller. The New Scientist, Vol. 139 1993, pp. 2427.
5. G. E. Hutchinson, The Enchanted Voyage and Other Studies Westport, CT : Greenwood
Press, 1978.
6. Victor Katz, A History of Mathematics: An Introduction New York: HarperCollins, 1993, pp. 678680.
7. Morris Kline, Mathematical Thought from Ancient to Modern Times New York: Oxford University Press, 1972, pp. 683685.
8. Sylvanus P. Thompson, The Life of Lord Kelvin New York: Chelsea, 1976.
1
SECTION 16.9 THE DIVERGENCE THEOREM1099
16.9 THE DIVERGENCE THEOREM
In Section 16.5 we rewrote Greens Theorem in a vector version as
yC FndsyydivFx,ydA D
where C is the positively oriented boundary curve of the plane region D. If we were seek ing to extend this theorem to vector fields on 3, we might make the guess that
yy Fn dSyyy div Fx, y, z dV SE
where S is the boundary surface of the solid region E. It turns out that Equation 1 is true, under appropriate hypotheses, and is called the Divergence Theorem. Notice its similarity to Greens Theorem and Stokes Theorem in that it relates the integral of a derivative of a function div F in this case over a region to the integral of the original function F over the boundary of the region.
At this stage you may wish to review the various types of regions over which we were able to evaluate triple integrals in Section 15.6. We state and prove the Divergence Theo rem for regions E that are simultaneously of types 1, 2, and 3 and we call such regions simple solid regions. For instance, regions bounded by ellipsoids or rectangular boxes are simple solid regions. The boundary of E is a closed surface, and we use the convention, introduced in Section 16.7, that the positive orientation is outward; that is, the unit normal vector n is directed outward from E.
THE DIVERGENCE THEOREM Let E be a simple solid region and let S be the bound ary surface of E, given with positive outward orientation. Let F be a vector field whose component functions have continuous partial derivatives on an open region that contains E. Then
yy FdSyyy div F dV SE
N The Divergence Theorem is sometimes called Gausss Theorem after the great German mathe matician Karl Friedrich Gauss 17771855, who discovered this theorem during his investigation of electrostatics. In Eastern Europe the Diver gence Theorem is known as Ostrogradskys Theorem after the Russian mathematician Mikhail Ostrogradsky 18011862, who pub lished this result in 1826.
Thus the Divergence Theorem states that, under the given conditions, the flux of F across the boundary surface of E is equal to the triple integral of the divergence of F over E.
PROOF LetFPiQjRk.Then
div FPQR
x y z
so yyydivFdVyyyP dVyyyQ dVyyyR dV
E E x E y E z
If n is the unit outward normal of S, then the surface integral on the left side of the
1100
CHAPTER 16 VECTOR CALCULUS
Divergence Theorem is
yy FdSyy Fn dSyy P iQ jR kn dS SSS
yy P in dSyy Q jn dSyy R kn dS SSS
Therefore, to prove the Divergence Theorem, it suffices to prove the following three equations:
yyPindSyyyP dV S Ex
yyQjndSyyyQ dV S Ey
yyRkndSyyyR dV S Ez
To prove Equation 4 we use the fact that E is a type 1 region:
Ex, y, z x, yD, u1x, yzu2x, y
where D is the projection of E onto the xyplane. By Equation15.6.6, we have yyyRdVyy u2x,y Rx,y,zdz dA
y
z
STM zuTMx, y
yyyR dVyyRx,y,u2x,yRx,y,u1x,ydA E z D
The boundary surface S consists of three pieces: the bottom surface S1, the top surface S2, and possibly a vertical surface S3, which lies above the boundary curve of D. See Figure 1. It might happen that S3 doesnt appear, as in the case of a sphere. Notice that on S3 we have kn0, because k is vertical and n is horizontal, and so
E z D u1x, y z
and therefore, by the Fundamental Theorem of Calculus,
S
E
0
xD FIGURE 1
S zux, y y
yy R kn dSyy 0 dS0 S3 S3
Thus, regardless of whether there is a vertical surface, we can write
yy R kn dSyy R kn dSyy R kn dS S S1 S2
The equation of S2 is zu2x, y, x, yD, and the outward normal n points upward, so from Equation 16.7.10 with F replaced by R k we have
yy R kn dSyy Rx, y, u2x, y dA S2 D
On S1 we have zu1x, y, but here the outward normal n points downward, so
2
3
4
5
6
N The solution in Example 1 should be yy comparedwiththesolutioninExample4
FdS divFdV
yyy yyy
B B
4 1dV VB3
3 4
13 M
in Section 16.7.
S
z
0,0,1
1,0,0
x
FIGURE 2
y2z
0,2,0 y z1
V EXAMPLE 2 Evaluate yy FdS, where S
we multiply by 1:
Divergence Theorem gives the flux as
SECTION 16.9 THE DIVERGENCE THEOREM
1101
N Notice that the method of proof of the Divergence Theorem is very similar to that of Greens Theorem.
yy R kn dSyy Rx, y, u1x, y dA S1 D
Therefore Equation 6 gives
yy R kn dSyy Rx, y, u2x, yRx, y, u1x, ydA SD
Comparison with Equation 5 shows that
yyRkndSyyyR dV S Ez
Equations 2 and 3 are proved in a similar manner using the expressions for E as a type 2 or type 3 region, respectively. M
V EXAMPLE 1 FindthefluxofthevectorfieldFx,y,zziyjxkovertheunit spherex2 y2 z2 1.
SOLUTION First we compute the divergence of F:
divFzyx1
x y z TheunitsphereSistheboundaryoftheunitballBgivenbyx2 y2 z2 1.Thusthe
2 Fx,y,zxyiy2 exz jsinxyk
0
and S is the surface of the region E bounded by the parabolic cylinder z1x2 and the planes z0, y0, and yz2. See Figure 2.
SOLUTION It would be extremely difficult to evaluate the given surface integral directly. We would have to evaluate four surface integrals corresponding to the four pieces of S. Furthermore, the divergence of F is much less complicated than F itself:
divFxyy2 exz2sinxyy2y3y x y z
Therefore we use the Divergence Theorem to transform the given surface integral into a triple integral. The easiest way to evaluate the triple integral is to express E as a type 3 region:
Ex,y,z1x1, 0z1x2, 0y2z
1102
CHAPTER 16 VECTOR CALCULUS
Then we have
yy FdSyyy div F dVyyy 3y dV SEE
y1 y1x2 2z2 31 2z3 11 2 3
y1 y1x2 y2z
3 1 0 0 ydydzdx3 1 0 2 dzdx
y1×2
3 dx 1x18dx
2
y1 x6 3×4 3×2 7dx184 M
1 0
0 35
y
2
nTM
FIGURE 3
Although we have proved the Divergence Theorem only for simple solid regions, it can be proved for regions that are finite unions of simple solid regions. The procedure is similar to the one we used in Section 16.4 to extend Greens Theorem.
For example, lets consider the region E that lies between the closed surfaces S1 and S2, where S1 lies inside S2. Let n1 and n2 be outward normals of S1 and S2. Then the boundary surfaceofEisSS1 S2 anditsnormalnisgivenbynn1 onS1 andnn2 onS2. See Figure 3. Applying the Divergence Theorem to S, we get
yyy div F dVyy FdSyy Fn dS ESS
yyFn1dSyyFn2 dS S1 S2
yy FdSyy FdS S1 S2
Lets apply this to the electric field see Example 5 in Section 16.1:
ExQ x x3
where S1 is a small sphere with radius a and center the origin. You can verify that div E0. See Exercise 23. Therefore Equation 7 gives
yyEdSyyEdSyyydivEdV yyEdSyyEndS S2 S1 E S1 S1
The point of this calculation is that we can compute the surface integral over S1 because S1 is a sphere. The normal vector at x is xx . Therefore
En Q xxQ xx Q Q x3 x x4 x2 a2
since the equation of S1 is x a. Thus we have yyEdSyyEndSQ yydS
7
a2
S2 S1 S1
QAS1Q4 a24 Q a2 a2
This shows that the electric flux of E is 4 Q through any closed surface S2 that contains
y
PTM
FIGURE 4
the origin. This is a special case of Gausss Law Equation 16.7.11 for a single charge. The relationship betweenand 0 is 14 0 .
Another application of the Divergence Theorem occurs in fluid flow. Let vx, y, z be the velocity field of a fluid with constant density . Then Fv is the rate of flow per unit area. If P0x0, y0, z0is a point in the fluid and Ba is a ball with center P0 and very small radius a, then div FPdiv FP0for all points in Ba since div F is continuous. We approximate the flux over the boundary sphere Sa as follows:
yyFdSyyydiv F dVyyydiv FP0dVdiv FP0VBa Sa Ba Ba
This approximation becomes better as a l 0 and suggests that div FP0 lim 1 yy FdS
SECTION 16.9 THE DIVERGENCE THEOREM1103
P
x
al0 VBa Sa
The vector field F i j 16.9 EXERCISES
Equation 8 says that div FP0is the net rate of outward flux per unit volume at P0. This is the reason for the name divergence. If div FP0, the net flow is outward near P and P is called a source. If div FP0, the net flow is inward near P and P is called a sink.
For the vector field in Figure 4, it appears that the vectors that end near P1 are shorter than the vectors that start near P1. Thus the net flow is outward near P1, so div FP10 and P1 is a source. Near P2, on the other hand, the incoming arrows are longer than the outgoing arrows. Here the net flow is inward, so div FP2 0 and P2 is a sink. We can use the formula for F to confirm this impression. Since Fx2 iy2 j, we have divF2x2y,whichispositivewhenyx.Sothepointsabovethelineyx are sources and those below are sinks.
1 4 Verify that the Divergence Theorem is true for the vector field F on the region E.
Fx, y, z3x ixy j2xz k, Eisthecubeboundedbytheplanesx0,x1,y0, y1,z0,andz1
2. Fx,y,zx2ixyjzk, Eisthesolidboundedbytheparaboloidz4x2 y2 and the xyplane
3. Fx,y,zxyiyzjzxk, Eisthesolidcylinderx2 y2 1,0z1
4. Fx,y,zxiyjzk, Eistheunitballx2 y2 z2 1
515 Use the Divergence Theorem to calculate the surface integral xxS FdS; that is, calculate the flux of F across S.
5. Fx,y,zex sinyiex cosyjyz2k,
S is the surface of the box bounded by the planes x0, x1,y0,y1,z0,andz2
6. Fx,y,zx2z3 i2xyz3jxz4 k,
S is the surface of the box with vertices 1, 2, 3
Fx,y,z3xy2 ixez jz3 k,
S is the surface of the solid bounded by the cylinder y2 z2 1andtheplanesx1andx2
8. Fx,y,zx3yix2y2 jx2yzk,
S is the surface of the solid bounded by the hyperboloid x2 y2 z2 1andtheplanesz2andz2
Fx, y, zxy sin z icosxz jy cos z k, Sistheellipsoidx2a2 y2b2 z2c2 1
10. Fx,y,zx2yixy2j2xyzk,
S is the surface of the tetrahedron bounded by the planes x0, y0, z0, and x2yz2
11. Fx,y,zcoszxy2ixezjsinyx2zk, S is the surface of the solid bounded by the paraboloid zx2 y2 andtheplanez4
12. Fx,y,zx4ix3z2j4xy2zk,
S is the surface of the solid bounded by the cylinder x2 y2 1andtheplaneszx2andz0
13. Fx,y,z4x3zi4y3zj3z4 k,
S is the sphere with radius R and center the origin
1.
8
7.
9.
1104
CHAPTER 16 VECTOR CALCULUS
14. CAS 15.
CAS 16.
17.
18.
Frr,whererxiyjzk,
S consists of the hemisphere zs1x2y2 and the diskx2y21inthexyplane
Fx,y,zey tanziys3x2 jxsinyk,
S is the surface of the solid that lies above the xyplane
4 4 andbelowthesurfacez2x y ,1×1,
1y1
Use a computer algebra system to plot the vector field
Fx, y, zsin x cos2 y isin3y cos4z jsin5z cos6x k in the cube cut from the first octant by the planes x2, y2, and z2. Then compute the flux across the surface of the cube.
UsetheDivergenceTheoremtoevaluatexxS FdS,where Fx,y,zz2xi1 y3 tanzjx2zy2k
22. Fx,yx2,y2
23. VerifythatdivE0fortheelectricfieldEx x3 x.
19.
Q
24. Use the Divergence Theorem to evaluate xx 2x2yz2 dS
25.
whereSisthespherex2 y2 z2 1. S
2530 Prove each identity, assuming that S and E satisfy the con ditions of the Divergence Theorem and the scalar functions and components of the vector fields have continuous secondorder partial derivatives.
yyandS0, whereaisaconstantvector S
yy
26. VE1 3
FdS, whereFx,y,zxiyjzk curl FdS0
3222S andSisthetophalfofthespherex y z 1.
Hint: Note that S is not a closed surface. First compute integrals over S1 and S2, where S1 is the disk x2y21, oriented downward, and S2SS1.
LetFx,y,zztan1y2iz3 lnx2 1jzk. Find the flux of F across the part of the paraboloid
x2 y2 z2thatliesabovetheplanez1andis oriented upward.
A vector field F is shown. Use the interpretation of diver gence derived in this section to determine whether div F is positive or negative at P1 and at P2.
2
P
2 2
PTM
2
27. 28. 29. 30.
31.
yy
S
yyDn fdSyyy2fdV SE
f tn dS
yy yyy
f 2tft dV yyfttfndSyyyf2tt2fdV
SE
Suppose S and E satisfy the conditions of the Divergence The orem and f is a scalar function with continuous partial deriva tives. Prove that
yy f n dSyyyf dV SE
These surface and triple integrals of vector functions are vectors defined by integrating each component function. Hint: Start by applying the Divergence Theorem to Ff c, where c is an arbitrary constant vector.
A solid occupies a region E with surface S and is immersed in a liquid with constant density . We set up a coordinate system so that the xyplane coincides with the surface of the liquid and positive values of z are measured downward into the liquid. Then the pressure at depth z is ptz, where t is the acceleration due to gravity see Section 6.5. The total buoyant force on the solid due to the pressure distribution is given by the surface integral
Fyy pn dS S
where n is the outer unit normal. Use the result of Exercise 31 to show that FWk, where W is the weight of the liquid displaced by the solid. Note that F is directed upward because z is directed downward. The result is Archimedes principle: The buoyant force on an object equals the weight of the displaced liquid.
SE
20. a Are the points P1 and P2 sources or sinks for the vector field F shown in the figure? Give an explanation based solely on the picture.
b Given that Fx, yx, y2, use the definition of diver gence to verify your answer to part a.
2
2 2
PTM
CAS 2122 Plot the vector field and guess where div F0 and where div F0. Then calculate div F to check your guess.
21. Fx,yxy,xy2
32.
P
2
SECTION 16.10 SUMMARY1105
16.10 SUMMARY
The main results of this chapter are all higherdimensional versions of the Fundamental Theorem of Calculus. To help you remember them, we collect them together here with out hypotheses so that you can see more easily their essential similarity. Notice that in each case we have an integral of a derivative over a region on the left side, and the right side involves the values of the original function only on the boundary of the region.
Fundamental Theorem of Calculus
Fundamental Theorem for Line Integrals
GreensTheorem
StokesTheorem
yb Fx dxFbFa a
yCfdrf rbf ra
ab
rb C
C
yyQ P
D
dAyC PdxQdy
xy
D
ra
yycurlFdSyC Fdr S
yyy div F dVyy FdS ES
C
n
S
n
S
n
Divergence Theorem
E
1106CHAPTER 16 VECTOR CALCULUS
16 REVIEW
1. What is a vector field? Give three examples that have physical
meaning.
2. a What is a conservative vector field? b What is a potential function?
3. a Write the definition of the line integral of a scalar function f along a smooth curve C with respect to arc length.
b How do you evaluate such a line integral?
c Write expressions for the mass and center of mass of a thin
wire shaped like a curve C if the wire has linear density
function x, y.
d Write the definitions of the line integrals along C of a
scalar function f with respect to x, y, and z. e How do you evaluate these line integrals?
4. a Define the line integral of a vector field F along a smooth curve C given by a vector function rt.
b If F is a force field, what does this line integral represent? c If FP, Q, R, what is the connection between the line integral of F and the line integrals of the component func
tions P, Q, and R?
5. State the Fundamental Theorem for Line Integrals.
CONCEPT CHECK
6. a What does it mean to say that xC Fdr is independent of path?
c If F is a velocity field in fluid flow, what are the physical interpretations of curl F and div F?
10. IfFPiQj,howdoyoutesttodeterminewhetherFis conservative? What if F is a vector field on 3?
11. a What is a parametric surface? What are its grid curves? b Write an expression for the area of a parametric surface. c What is the area of a surface given by an equation
ztx, y?
12. a Write the definition of the surface integral of a scalar func tion f over a surface S.
b How do you evaluate such an integral if S is a parametric surface given by a vector function ru, v?
c What if S is given by an equation ztx, y?
d If a thin sheet has the shape of a surface S, and the density
at x, y, z is x, y, z, write expressions for the mass and center of mass of the sheet.
13. a What is an oriented surface? Give an example of a non orientable surface.
b Define the surface integral or flux of a vector field F over an oriented surface S with unit normal vector n.
c How do you evaluate such an integral if S is a parametric surface given by a vector function ru, v?
d What if S is given by an equation ztx, y? 14. State Stokes Theorem.
15. State the Divergence Theorem.
16. In what ways are the Fundamental Theorem for Line Integrals, Greens Theorem, Stokes Theorem, and the Divergence Theorem similar?
5. If FP iQ j and PyQx in an open region D, then F is conservative.
6. xC fx,ydsxC fx,yds
7. If S is a sphere and F is a constant vector field, then
8. There is a vector field F such that curlFxiyjzk
b If you know that xC Fdr is independent of path, what can you say about F?
7. State Greens Theorem.
8. Write expressions for the area enclosed by a curve C in terms
of line integrals around C.
9. Suppose F is a vector field on 3. a Define curl F.
b Define div F.
TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. If F is a vector field, then div F is a vector field.
2. If F is a vector field, then curl F is a vector field.
3. If f has continuous partial derivatives of all orders on 3, then divcurlf 0.
4. If f has continuous partial derivatives on 3 and C is any circle,thenxC fdr0.
xxFdS0. S
Cisthearcoftheparabolayx2 from0,0to1,1 3. x yz cos x ds,
where C is the circle x2y24 with counterclockwise orientation.
C
C: xt, y3 cos t, z3 sin t, 0t
18. Find curl F and div F if
Fx,y,zex sinyiey sinzjez sinxk
19. Show that there is no vector field G such that curlG2xi3yzjxz2k.
20. Show that, under conditions to be stated on the vector fields F and G,
curlFGFdivGGdivFGFFG 21. If C is any piecewisesmooth simple closed plane curve
and f and t are differentiable functions, show that xC fxdxtydy0.
22. If f and t are twice differentiable functions, show that 2ftf2tt2f2f t
23. If f is a harmonic function, that is,2 f0, show that the line integral x fy dxfx dy is independent of path in any simple region D.
24. a Sketch the curve C with parametric equations
xcos t ysin t zsin t 0t2
b Find x 2xe2y dx2x2e2y2y cot z dyy2 csc2zdz. C
25. Find the area of the part of the surface zx22y that lies above the triangle with vertices 0, 0, 1, 0, and 1, 2.
26. a Find an equation of the tangent plane at the point 4, 2, 1 to the parametric surface S given by
ru,vv2 iuvju2k 0u3,3v3
4. x ydxxy2dy, Cistheellipse4x2 9y2 36with C
counterclockwise orientation
5. xC y3 dxx2 dy, Cisthearcoftheparabolax1y2
from 0, 1 to 0, 1
6. x sxy dxey dyxz dz,
C 423 Cisgivenbyrtt it jt k,0t1
7. x xydxy2 dyyzdz, C
C is the line segment from 1, 0, 1, to 3, 4, 2
8. x Fdr, where Fx, yxy ix2 j and C is given by
C
rtsin t i1t j, 0t
9. x Fdr, whereFx,y,zez ixzjxykand C
Cisgivenbyrtt2 it3jtk,0t1
17. Use Greens Theorem to evaluate x x2 y dxxy2 dy, C
CHAPTER 16 REVIEW1107
EXERCISES
1. A vector field F, a curve C, and a point P are shown. a Is xC Fdr positive, negative, or zero? Explain. b Is div FP positive, negative, or zero? Explain.
y
C
29 Evaluate the line integral. 2. xC x ds,
1314 Show that F is conservative and use this fact to evaluate xC Fdr along the given curve.
13. Fx, y4x3y22xy3 i2x4y3x2y24y3j, C:rttsin ti2tcos tj, 0t1
14. Fx, y, zey ixeyez jyez k,
C is the line segment from 0, 2, 0 to 4, 0, 3
15. Verify that Greens Theorem is true for the line integral
x xy2 dxx2 ydy, where C consists of the parabola yx2
C
from 1, 1 to 1, 1 and the line segment from 1, 1 to 1, 1.
16. Use Greens Theorem to evaluate xC s1x3 dx2xydy, where C is the triangle with vertices 0, 0, 1, 0, and 1, 3.
x P
10. Find the work done by the force field
Fx, y, zz ix jy k in moving a particle from the point 3, 0, 0 to the point 0, 2, 3 along
a a straight line
bthehelixx3cost, yt, z3sint
1112 Show that F is a conservative vector field. Then find a func tion f such that Ff.
11. Fx,y1xyexy iey x2exyj 12. Fx,y,zsinyixcosyjsinzk
1108
;
CAS
CHAPTER 16 VECTOR CALCULUS
b Use a computer to graph the surface S and the tangent
plane found in part a.
c Set up, but do not evaluate, an integral for the surface area
of S. d If
z2 x2 y2 Fx,y,z 1×2 i 1y2 j 1z2 k
find xxS FdS correct to four decimal places. 2730 Evaluate the surface integral.
Evaluate xC Fdr, where C is the curve with initial point 0, 0, 2 and terminal point 0, 3, 0 shown in the figure.
27. xxzdS, whereSisthepartoftheparaboloidzx2 y2 S
3,0,0
x
Let Fx,y
that lies under the plane z4
28. xx x2zy2zdS, where S is the part of the plane
38.
S22 z4xythatliesinsidethecylinderx y 4
2×3 2xy2 2yi2y3 2x2y2xj x2 y2
Fdr, where C is shown in the figure. y
C
0x
z
0,0,2
0
0, 3, 0 1, 1, 0
y
29. xxFdS, whereFx,y,zxzi2yj3xkandSis
Evaluate x C
S
the sphere xyz4 with outward orientation
2 2 2
30. xxFdS, whereFx,y,zx2ixyjzkandSisthe
S
partoftheparaboloidzx2 y2 belowtheplanez1
with upward orientation
31. Verify that Stokes Theorem is true for the vector field Fx,y,zx2 iy2 jz2 k,whereSisthepartofthe paraboloid z1x2y2 that lies above the xyplane and S has upward orientation.
32. Use Stokes Theorem to evaluate xxS curl FdS, where Fx,y,zx2yziyz2 jz3exy k,Sisthepartofthe spherex2 y2 z2 5thatliesabovetheplanez1,and S is oriented upward.
33. Use Stokes Theorem to evaluate xC Fdr, where
Fx, y, zxy iyz jzx k, and C is the triangle with vertices 1, 0, 0, 0, 1, 0, and 0, 0, 1, oriented counter clockwise as viewed from above.
39.
FindxxFndS,whereFx,y,zxiyjzkandSis S
the outwardly oriented surface shown in the figure the boundary surface of a cube with a unit corner cube removed.
z
2,0,2
0, 2, 2
11
34. Use the Divergence Theorem to calculate the surface integral xx FdS, where Fx, y, zx3 iy3 jz3 k and S is the
1
S
surface of the solid bounded by the cylinder x2y21 and
the planes z0 and z2.
35. Verify that the Divergence Theorem is true for the vector
field Fx, y, zx iy jz k, where E is the unit ball x2 y2 z2 1.
y
36. Compute the outward flux of
Fx,y,z xiyjzk
40. 41.
If the components of F have continuous second partial deriva tives and S is the boundary surface of a simple solid region, showthatxxS curlFdS0.
Ifaisaconstantvector,rxiyjzk,andS isanori ented, smooth surface with a simple, closed, smooth, positively oriented boundary curve C, show that
yy2adSyardr S C
x
S 2, 2, 0
x2 y2 z232 through the ellipsoid 4×29y26z236.
37. Let
Fx, y, z3×2 yz3y ix3z3x jx3 y2z k
FIGURE FOR PROBLEM 1
where r is the radius vector from P to any point on S, rr , and the unit normal vector n is directed away from P.
This shows that the definition of the measure of a solid angle is independent of the radius a of the sphere. Thus the measure of the solid angle is equal to the area subtended on a unit sphere. Note the analogy with the definition of radian measure. The total solid angle sub tended by a sphere at its center is thus 4 steradians.
2. Find the positively oriented simple closed curve C for which the value of the line integral xy3 ydx2x3dyisamaximum.
C
3. Let C be a simple closed piecewisesmooth space curve that lies in a plane with unit normal
Pa
S
PROBLEMS PLUS
1. Let S be a smooth parametric surface and let P be a point such that each line that starts
at P intersects S at most once. The solid angle S subtended by S at P is the set of lines starting at P and passing through S. Let Sa be the intersection of S with the surface of the sphere with center P and radius a. Then the measure of the solid angle in steradians is defined to be
S area of Sa a2
Apply the Divergence Theorem to the part of S between Sa and S to show that Syyrn dS
Sa
S r3
vector na, b, c and has positive orientation with respect to n. Show that the plane area
Intake Compression
Explosion Exhaustion
Water
Flywheel
P
!
0
Crankshaft Connecting rod
zsinuv. Start by graphing the surface from several points of view. Explain the
appearance of the graphs by determining the traces in the horizontal planes z0, z1,
and z 1. 2
5. Prove the following identity:
FGFGGFFcurl GGcurl F
6. The figure depicts the sequence of events in each cylinder of a fourcylinder internal combus tion engine. Each piston moves up and down and is connected by a pivoted arm to a rotating crankshaft. Let Pt and Vt be the pressure and volume within a cylinder at time t, where
atb gives the time required for a complete cycle. The graph shows how P and V vary through one cycle of a fourstroke engine. During the intake stroke from x to y a mixture of air and gasoline at atmospheric pressure is drawn into a cylinder through the intake valve as the piston moves downward. Then the piston rapidly compresses the mix with the valves closed in the compression stroke from y to z during which the pressure rises and the vol ume decreases. At z the sparkplug ignites the fuel, raising the temperature and pressure at almost constant volume to . Then, with valves closed, the rapid expansion forces the piston downward during the power stroke fromto . The exhaust valve opens, temperature and pressure drop, and mechanical energy stored in a rotating flywheel pushes the piston upward, forcing the waste products out of the exhaust valve in the exhaust stroke. The exhaust valve closes and the intake valve opens. Were now back at x and the cycle starts again.
a Show that the work done on the piston during one cycle of a fourstroke engine is WxC P dV, where C is the curve in the PVplane shown in the figure.
Hint: Let xt be the distance from the piston to the top of the cylinder and note that the force on the piston is FAPt i, where A is the area of the top of the piston. Then WxC1 Fdr, where C1 is given by rtxt i, atb. An alternative approach is
to work directly with Riemann sums.
b Use Formula 16.4.5 to show that the work is the difference of the areas enclosed by the two loops of C.
C
FIGURE FOR PROBLEM 6
V
enclosedbyCis 1 x bzcydxcxazdyaybxdz. 2C
;4. Investigatetheshapeofthesurfacewithparametricequationsxsinu,ysinv,
1109
17
SECONDORDER DIFFERENTIAL EQUATIONS
y
x
Most of the solutions of the differential equation y4ye3x resemble sine functions when x is negative but they all look like exponential functions when x is large.
The basic ideas of differential equations were explained in Chapter 9; there we concen trated on firstorder equations. In this chapter we study secondorder linear differential equations and learn how they can be applied to solve problems concerning the vibrations of springs and the analysis of electric circuits. We will also see how infinite series can be used to solve differential equations.
1110
17.1 SECONDORDER LINEAR EQUATIONS
A secondorder linear differential equation has the form
Px d2yQx dyRxyGx dx2 dx
where P, Q, R, and G are continuous functions. We saw in Section 9.1 that equations of this type arise in the study of the motion of a spring. In Section 17.3 we will further pur sue this application as well as the application to electric circuits.
In this section we study the case where Gx0, for all x, in Equation 1. Such equa tions are called homogeneous linear equations. Thus the form of a secondorder linear homo geneous differential equation is
Px d2y Qx dy Rxy0 dx2 dx
If Gx0 for some x, Equation 1 is nonhomogeneous and is discussed in Section 17.2. Two basic facts enable us to solve homogeneous linear equations. The first of these says that if we know two solutions y1 and y2 of such an equation, then the linear combination
yc1y1c2y2 is also a solution.
1
2
THEOREM If y1x and y2x are both solutions of the linear homogeneous equation 2 and c1 and c2 are any constants, then the function
yxc1y1xc2y2x is also a solution of Equation 2.
3
PROOF Since y1 and y2 are solutions of Equation 2, we have Pxy1Qxy1Rxy10
and Pxy2Qxy2Rxy20 Therefore, using the basic rules for differentiation, we have
PxyQxyRxy
Pxc1y1c2y2Qxc1y1c2y2Rxc1y1c2y2
Pxc1y1c2y2Qxc1y1c2y2Rxc1y1c2y2
c1Pxy1Qxy1Rxy1c2Pxy2Qxy2Rxy2c10c200
Thus yc1y1c2y2 is a solution of Equation 2. M 1111
1112
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
The other fact we need is given by the following theorem, which is proved in more advanced courses. It says that the general solution is a linear combination of two linearly independent solutions y1 and y2. This means that neither y1 nor y2 is a constant multiple of the other. For instance, the functions fxx2 and tx5×2 are linearly dependent, but fxex and txxex are linearly independent.
Theorem 4 is very useful because it says that if we know two particular linearly inde pendent solutions, then we know every solution.
In general, it is not easy to discover particular solutions to a secondorder linear equa tion. But it is always possible to do so if the coefficient functions P, Q, and R are constant functions, that is, if the differential equation has the form
where a, b, and c are constants and a0.
Its not hard to think of some likely candidates for particular solutions of Equation 5 if
we state the equation verbally. We are looking for a function y such that a constant times its second derivative y plus another constant times y plus a third constant times y is equal to 0. We know that the exponential function yerx where r is a constant has the prop erty that its derivative is a constant multiple of itself: yrerx. Furthermore, yr2erx. If we substitute these expressions into Equation 5, we see that yerx is a solution if
ar2erxbrerxcerx0 or ar2 brcerx 0
But erx is never 0. Thus yerx is a solution of Equation 5 if r is a root of the equation
Equation 6 is called the auxiliary equation or characteristic equation of the differen tial equation aybycy0. Notice that it is an algebraic equation that is obtained from the differential equation by replacing y by r2, y by r, and y by 1.
Sometimes the roots r1 and r2 of the auxiliary equation can be found by factoring. In other cases they are found by using the quadratic formula:
bsb2 4ac bsb2 4ac r12a r22a
We distinguish three cases according to the sign of the discriminant b24ac.
THEOREM If y1 and y2 are linearly independent solutions of Equation 2, and Px is never 0, then the general solution is given by
yxc1y1xc2y2x where c1 and c2 are arbitrary constants.
4
5
6
aybycy0
ar2 brc0
7
If the roots r1 and r2 of the auxiliary equation ar2brc0 are real and unequal, then the general solution of aybycy0 is
yc1er1x c2er2x
8
N In Figure 1 the graphs of the basic solutions fxe2x andtxe3x ofthedifferential equation in Example 1 are shown in blue and red, respectively. Some of the other solutions, linear combinations of f and t, are shown
in black.
1f fgg1
N CASE I b24ac0
In this case the roots r1 and r2 of the auxiliary equation are real and distinct, so y1er1x and y2er2 x are two linearly independent solutions of Equation 5. Note that er2 x is not a constant multiple of er1x. Therefore, by Theorem 4, we have the following fact.
EXAMPLE 1 Solve the equation yy6y0. SOLUTION The auxiliary equation is
r2 r6r2r30
whose roots are r2, 3. Therefore, by 8, the general solution of the given differen
8 5fg
f5g
tial equation is
We could verify that this is indeed a solution by differentiating and substituting into the
SECTION 17.1 SECONDORDER LINEAR EQUATIONS1113
fg
5
FIGURE 1
gf
yc1e2xc2e3x
differential equation. M
d2y dy EXAMPLE2Solve3dx2 dxy0.
SOLUTION To solve the auxiliary equation 3r2r10, we use the quadratic formula: r 1s13
6
Since the roots are real and distinct, the general solution is
yc1e1s13 x6c2 e1s13 x6 M In this case r1r2; that is, the roots of the auxiliary equation are real and equal. Lets
N CASE II b24ac0
denote by r the common value of r1 and r2. Then, from Equations 7, we have
rb so 2arb0 2a
We know that y1erx is one solution of Equation 5. We now verify that y2xerx is also a solution:
ay2by2cy2a2rerxr2xerxberxrxerxcxerx 2arberx ar2 brcxerx
0erx0xerx0
9
1114CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
If the auxiliary equation ar2brc0 has only one real root r, then the general solution of aybycy0 is
yc1erx c2xerx
10
N Figure 2 shows the basic solutions f xe3x2 and txxe3x2 in
Example 3 and some other members of the family of solutions. Notice that all of them approach 0 as x l .
The first term is 0 by Equations 9; the second term is 0 because r is a root of the auxiliary equation. Since y1erx and y2xerx are linearly independent solutions, Theorem 4 pro vides us with the general solution.
V EXAMPLE 3 Solve the equation 4y12y9y0.
SOLUTION The auxiliary equation 4r212r90 can be factored as
2r32 0
so the only root is r 3 . By 10, the general solution is
2 fg
g
5fg f5g gf 2
yc1e3x2c2 xe3x2 M In this case the roots r1 and r2 of the auxiliary equation are complex numbers. See Appen
FIGURE 2
fg 8 f2
5
N CASE III b24ac0
dix H for information about complex numbers. We can write
r1 i r2 i
where and are real numbers. In fact,b2a,s4acb22a. Then,
using Eulers equation
from Appendix H, we write the solution of the differential equation as
yC1er1x C2er2x C1e i x C2e i x
C1e xcos xisin xC2e xcos xisin x
e xC1 C2cos xiC1 C2sin x e xc1cos xc2sin x
where c1C1C2, c2iC1C2. This gives all solutions real or complex of the dif ferential equation. The solutions are real when the constants c1 and c2 are real. We summa rize the discussion as follows.
ei cos isin
If the roots of the auxiliary equation ar2brc0 are the complex num bersr1i ,r2i ,thenthegeneralsolutionofaybycy0 is
ye xc1cos xc2sin x
11
N Figure 3 shows the graphs of the solu tions in Example 4, fxe 3 x cos 2 x and txe3x sin 2x, together with some linear combinations. All solutions approach 0
asx l .
3 2 3
FIGURE 3
SECTION 17.1 SECONDORDER LINEAR EQUATIONS1115 V EXAMPLE 4 Solve the equation y6y13y0.
SOLUTION The auxiliary equation is r26r130. By the quadratic formula, the roots
are
fg
r 6s36526s16 32i 322
g fg f
By 11, the general solution of the differential equation is
ye3xc1 cos 2xc2 sin 2x M INITIALVALUE AND BOUNDARYVALUE PROBLEMS
An initialvalue problem for the secondorder Equation 1 or 2 consists of finding a solu tion y of the differential equation that also satisfies initial conditions of the form
yx0y0 yx0y1
where y0 and y1 are given constants. If P, Q, R, and G are continuous on an interval and Px0 there, then a theorem found in more advanced books guarantees the existence and uniqueness of a solution to this initialvalue problem. Examples 5 and 6 illustrate the technique for solving such a problem.
N Figure 4 shows the graph of the solution of the initialvalue problem in Example 5. Compare with Figure 1.
20
2 0 2
FIGURE 4
EXAMPLE 5 Solve the initialvalue problem
yy6y0 y01 y00
SOLUTION From Example 1 we know that the general solution of the differential equa tion is
yxc1e2x c2e3x Differentiating this solution, we get
yx2c1e2x3c2 e3x To satisfy the initial conditions we require that
y0c1 c2 1 y02c1 3c2 0
From 13, we have c22 c1 and so 12 gives 3
c12 c11 c13 c22 355
Thus the required solution of the initialvalue problem is y3e2x 2e3x
55
EXAMPLE 6 Solve the initialvalue problem
yy0 y02
M
12
13
y03
SOLUTION The auxiliary equation is r210, or r21, whose roots are i. Thus
0,1, and since e0x1, the general solution is yxc1 cos xc2 sin x
Since yxc1 sin xc2 cos x
1116CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
N The solution to Example 6 is graphed in Figure 5. It appears to be a shifted sine curve and, indeed, you can verify that another way of writing the solution is
the initial conditions become
ys13 sinx
2
where tan
2 3
y0c2 3
yx2 cos x3 sin x M
FIGURE 5
5
y0c1 2
Therefore the solution of the initialvalue problem is
5
2
A boundaryvalue problem for Equation 1 or 2 consists of finding a solution y of the differential equation that also satisfies boundary conditions of the form
yx0y0 yx1y1
In contrast with the situation for initialvalue problems, a boundaryvalue problem does
not always have a solution. The method is illustrated in Example 7.
V EXAMPLE 7 Solve the boundaryvalue problem
y2yy0 y01 y13
SOLUTION The auxiliary equation is
r2 2r10 or r12 0
whose only root is r1. Therefore the general solution is yxc1exc2 xex
The boundary conditions are satisfied if
y0c1 1
y1c1e1 c2e1 3
The first condition gives c11, so the second condition becomes
e1 c2e1 3
Solving this equation for c2 by first multiplying through by e, we get
1c2 3e so c2 3e1 Thus the solution of the boundaryvalue problem is
yex 3e1xex M SUMMARY: SOLUTIONS OF aybyc0
N Figure 6 shows the graph of the solution of the boundaryvalue problem in Example 7.
5
1 5
5
FIGURE 6
Rootsofar2 brc0
General solution
r1, r2 real and distinct r1 r2 r
r1, r2 complex:i
yc1er1x c2er2x
yc1erx c2xerx
ye xc1 cos xc2 sin x
1.
SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS
1117
17.1 EXERCISES
113 Solve the differential equation. yy6y0
3. y16y0
5. 9y12y4y0 7. y2y
2. y4y4y0 4. y8y12y0 6. 25y9y0
8. y4yy0
19. 4y4yy0, y01, y01.5 20. 2y5y3y0, y01, y04
y16y0, y 43, y 44 22. y2y5y0, y 0, y 2
y2y2y0, y02, y01 24. y12y36y0, y10, y11
2532 Solve the boundaryvalue problem, if possible. 25. 4yy0, y03, y 4
26. y2y0, y01, y12
27. y3y2y0, y01, y30
28. y100y0, y02, y 5
29. y6y25y0, y01, y 2
y6y9y0, y01, y10
31. y4y13y0, y02, y 21 32. 9y18y10y0, y00, y 1
33. Let L be a nonzero real number.
a Show that the boundaryvalue problem y
y4y13y0
2 d2y2 dyy0
10. y3y0
21.
23.
9.
11.
dt2 dt
12. 8 d2y 12 dy 5y0
dt2 dt
13. 100 d2P 200 dP 101P0
dt2 dt
; 14 16 Graph the two basic solutions of the differential equation and several other solutions. What features do the solutions have in common?
14. d2y4dy20y0 dx2 dx
15. 5 d2y2 dy3y0 dx2 dx
d2y dy 16.9dx2 6dxy0
1724 Solve the initialvalue problem. 2y5y3y0, y03, y04
18. y3y0, y01, y03
y0, y00, yL0 has only the trivial solution y0 for
30.
thecases 0and 0.
b For the case0, find the values of for which this
problem has a nontrivial solution and give the corre sponding solution.
34. If a, b, and c are all positive constants and yx is a solution of the differential equation aybycy0, show that limxl yx0.
17.
17.2 NONHOMOGENEOUS LINEAR EQUATIONS
In this section we learn how to solve secondorder nonhomogeneous linear differential equa
tions with constant coefficients, that is, equations of the form
aybycyGx
where a, b, and c are constants and G is a continuous function. The related homogeneous
equation
aybycy0
is called the complementary equation and plays an important role in the solution of the original nonhomogeneous equation 1.
1
2
1118
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
THEOREM The general solution of the nonhomogeneous differential equation 1 can be written as
yxypxycx
where yp is a particular solution of Equation 1 and yc is the general solution of the complementary Equation 2.
3
PROOF All we have to do is verify that if y is any solution of Equation 1, then yyp is a solution of the complementary Equation 2. Indeed
ayypbyypcyypayaypbybyp cycyp aybycyaypbyp cyp
txtx0 M
We know from Section 17.1 how to solve the complementary equation. Recall that the solution is ycc1y1c2y2, where y1 and y2 are linearly independent solutions of Equa tion 2. Therefore Theorem 3 says that we know the general solution of the nonhomoge neous equation as soon as we know a particular solution yp. There are two methods for finding a particular solution: The method of undetermined coefficients is straightforward but works only for a restricted class of functions G. The method of variation of parameters works for every function G but is usually more difficult to apply in practice.
THE METHOD OF UNDETERMINED COEFFICIENTS
We first illustrate the method of undetermined coefficients for the equation
aybycyGx
where Gx is a polynomial. It is reasonable to guess that there is a particular solution yp that is a polynomial of the same degree as G because if y is a polynomial, then aybycy is also a polynomial. We therefore substitute ypxa polynomial of the same degree as G into the differential equation and determine the coefficients.
V EXAMPLE 1 Solve the equation yy2yx2. SOLUTION The auxiliary equation of yy2y0 is
r2 r2r1r20
with roots r1, 2. So the solution of the complementary equation is
ycc1exc2e2x
Since Gxx2 is a polynomial of degree 2, we seek a particular solution of the form
ypxAx2 BxC
Then yp2AxB and yp2A so, substituting into the given differential equation, we have
2A2AxB2Ax2 BxCx2
N Figure 1 shows four solutions of the differen tial equation in Example 1 in terms of the partic ular solution yp and the functions fxe x and txe2x.
8 yp2f3g
yp3g
3 3
5
FIGURE 1
N Figure 2 shows solutions of the differential equation in Example 2 in terms of yp and the functions f xcos 2x and txsin 2x. Notice that all solutions approachas x land all solutions except yp resemble sine functions when x is negative.
4
or 2Ax2 2A2Bx2AB2Cx2 Polynomials are equal when their coefficients are equal. Thus
SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS
1119
2A1 2A2B0 The solution of this system of equations is
2AB2C0 C3
A1 B1 224
A particular solution is therefore
yp2f
yp 224
ypx1x2 1×3 and, by Theorem 3, the general solution is
yyc yp c1ex c2e2x 1×2 1×3 224
M
If Gx the right side of Equation 1 is of the form Cek x, where C and k are constants, then we take as a trial solution a function of the same form, ypxAe k x, because the derivatives of e k x are constant multiples of e k x.
EXAMPLE 2 Solve y4ye3x.
SOLUTION The auxiliary equation is r240 with roots 2i, so the solution of the com
plementary equation is
ycxc1 cos 2xc2 sin 2x
For a particular solution we try ypxAe3x. Then yp3Ae3x and yp9Ae3x. Substi
tuting into the differential equation, we have
9Ae3x4Ae3x e3x so 13Ae3xe3x and A1 . Thus a particular solution is
ypx1 e3x 13
and the general solution is
yxc1 cos 2xc2 sin 2x
If Gx is either C cos kx or C sin kx, then, because of the rules for differentiating the
sine and cosine functions, we take as a trial particular solution a function of the form ypxA cos kxB sin kx
V EXAMPLE 3 Solveyy2ysinx. SOLUTION We try a particular solution
ypxA cos xB sin x
Then yp AsinxBcosx ypAcosxBsinx
ypfg ypg
4 p2 13
y
ypf
2
FIGURE 2
1 e3x M 13
1120
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
so substitution in the differential equation gives
A cos xB sin xA sin xB cos x2A cos xB sin xsin x or 3AB cos xA3B sin xsin x This is true if
3AB0 and A3B1 The solution of this system is
so a particular solution is
A1 B3 10 10
ypx1 cosx 3 sinx 10 10
In Example 1 we determined that the solution of the complementary equation is ycc1exc2e2x. Thus the general solution of the given equation is
yxc1ex c2e2x1 cosx3sinx M 10
If Gx is a product of functions of the preceding types, then we take the trial solu tion to be a product of functions of the same type. For instance, in solving the differential equation
we would try
y2y4yx cos 3x
ypxAxB cos 3xCxD sin 3x
If Gx is a sum of functions of these types, we use the easily verified principle of super position, which says that if yp1 and yp2 are solutions of
aybycyG1x aybycyG2x respectively, then yp1yp2 is a solution of
aybycyG1xG2x V EXAMPLE 4 Solve y4yxexcos 2x.
SOLUTION The auxiliary equation is r240 with roots 2, so the solution of the com plementary equation is ycxc1e2xc2e2x. For the equation y4yxex we try
yp xAxBex 1
Then ypAxABex, ypAx2ABex, so substitution in the equation
11 gives
Ax2ABex 4AxBex xex or 3Ax2A3Bexxex
N In Figure 3 we show the particular solution
ypyp1yp 2 of the differential equation in Example 4. The other solutions are given in terms of fxe2x andtxe2x.
5 yp2fg
ypg
ypf
4 yp 1
1
For the equation y4ycos 2x, we try
yp2xC cos 2xD sin 2x
Substitution gives
4C cos 2x4D sin 2x4C cos 2xD sin 2xcos 2x or 8C cos 2x8D sin 2xcos 2x Therefore 8C1, 8D0, and
yp x1 cos 2x 28
By the superposition principle, the general solution is
yycypypc1e2xc2 e2x1 x2 ex1 cos 2x
SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS Thus 3A1 and 2A3B0, so A1, B2, and
1121
39 yp x1 x2 ex
39
FIGURE 3
2
M
Finally we note that the recommended trial solution yp sometimes turns out to be a solu tion of the complementary equation and therefore cant be a solution of the nonhomoge neous equation. In such cases we multiply the recommended trial solution by x or by x2 if necessary so that no term in ypx is a solution of the complementary equation.
EXAMPLE 5 Solveyysinx.
SOLUTION The auxiliary equation is r210 with roots i, so the solution of the com
plementary equation is
ycxc1 cos xc2 sin x Ordinarily, we would use the trial solution
ypxA cos xB sin x
but we observe that it is a solution of the complementary equation, so instead we try
ypxAxcos xBxsin x
Then ypxA cos xAx sin xB sin xBx cos x
ypx2A sin xAx cos x2B cos xBx sin x
Substitution in the differential equation gives
ypyp 2Asinx2Bcosxsinx
12 398
1122CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
N The graphs of four solutions of the differential equation in Example 5 are shown in Figure 4.
4
2 2 yp
so A1, B0, and 2
The general solution is
ypx1 x cos x 2
yxc1 cos xc2 sin x1 x cos x M 2
We summarize the method of undetermined coefficients as follows:
SUMMARY OF THE METHOD OF UNDETERMINED COEFFICIENTS
1. If GxekxPx, where P is a polynomial of degree n, then try ypxekxQx, where Qx is an nthdegree polynomial whose coefficients are determined by substituting in the differential equation.
2. If GxekxPx cos mx or GxekxPx sin mx, where P is an nthdegree polynomial, then try
ypxekxQx cos mxekxRx sin mx where Q and R are nthdegree polynomials.
Modification: If any term of yp is a solution of the complementary equation, multiply yp by x or by x2 if necessary.
FIGURE 4
4
EXAMPLE 6 Determine the form of the trial solution for the differential equation y4y13ye2x cos 3x.
SOLUTION Here Gx has the form of part 2 of the summary, where k2, m3, and Px1. So, at first glance, the form of the trial solution would be
ypxe2xA cos 3xB sin 3x
But the auxiliary equation is r24r130, with roots r23i, so the solution
of the complementary equation is
ycxe2xc1 cos 3xc2 sin 3x
This means that we have to multiply the suggested trial solution by x. So, instead, we
use
ypxxe2xA cos 3xB sin 3x M THE METHOD OF VARIATION OF PARAMETERS
Suppose we have already solved the homogeneous equation aybycy0 and writ ten the solution as
yxc1y1xc2y2x
where y1 and y2 are linearly independent solutions. Lets replace the constants or parame ters c1 and c2 in Equation 4 by arbitrary functions u1x and u2x. We look for a particu
4
SECTION 17.2 NONHOMOGENEOUS LINEAR EQUATIONS1123 lar solution of the nonhomogeneous equation aybycyGx of the form
ypxu1x y1xu2x y2x
This method is called variation of parameters because we have varied the parameters c1
and c2 to make them functions. Differentiating Equation 5, we get yp u1y1 u2y2u1y1 u2y2
Since u1 and u2 are arbitrary functions, we can impose two conditions on them. One con dition is that yp is a solution of the differential equation; we can choose the other condition so as to simplify our calculations. In view of the expression in Equation 6, lets impose the condition that
u1y1 u2y2 0
Then ypu1 y1u2 y2u1y1u2y2 Substituting in the differential equation, we get
au1y1 u2y2 u1y1u2y2bu1y1 u2y2cu1y1 u2y2G or
u1ay1by1 cy1u2ay2by2 cy2au1y1 u2y2G But y1 and y2 are solutions of the complementary equation, so
ay1by1 cy1 0 and ay2by2 cy2 0 and Equation 8 simplifies to
au 1 y 1u 2 y 2 G
Equations 7 and 9 form a system of two equations in the unknown functions u1 and u2. After solving this system we may be able to integrate to find u1 and u2 and then the partic ular solution is given by Equation 5.
EXAMPLE 7 Solvetheequationyytanx,0x 2.
SOLUTION The auxiliary equation is r210 with roots i, so the solution of
yy0 is c1 sin xc2 cos x. Using variation of parameters, we seek a solution of the form
ypxu1x sin xu2x cos x
Then yp u1sinxu2cosxu1cosxu2sinx
Set
5
6
7
8
9
10
u1 sin xu2 cos x0
1124
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
N Figure 5 shows four solutions of the differential equation in Example 7.
Then ypu1 cos xu2 sin xu1 sin xu2 cos x For yp to be a solution we must have
ypyp u1cosxu2sinxtanx Solving Equations 10 and 11, we get
u1sin2xcos2xcos x tan x u1 sinx u1xcosx
We seek a particular solution, so we dont need a constant of integration here. Then, from Equation 10, we obtain
u2 sinx u1 sin2xcos2x1 cosxsecx cos x cos x cos x
So u2xsin xlnsec xtan x Notethatsecxtanx0for0x 2.Therefore
2.5
0 1
2
ypxcosx sinxsinxlnsecxtanxcosxcos x lnsec xtan x
yp
and the general solution is
yxc1 sin xc2 cos xcos x lnsec xtan x M
FIGURE 5
17.2 EXERCISES
110 Solve the differential equation or initialvalue problem
using the method of undetermined coefficients.
1. y3y2yx2
2. y9ye3x
3. y2ysin 4x
4. y6y9y1x
5. y4y5yex
6. y2yyxex
7. yyex x3, y02, y00
8. y4yex cosx, y01, y02
yyxex, y02, y01
10. yy2yxsin 2x, y01, y00
; 1112 Graph the particular solution and several other solutions. What characteristics do these solutions have in common?
11. y3y2ycosx 12. y4yex 1318 Write a trial solution for the method of undetermined
coefficients. Do not determine the coefficients.
13. y9ye2x x2 sinx 14. y9yxex cos x 15. y9y1xe9x
16. y3y4yx3 xex 17. y2y10yx2ex cos 3x 18. y4ye3xx sin 2x
9.
11
1922 Solve the differential equation using a undetermined coef ficients and b variation of parameters.
19. 4yycosx 20. y2y3yx2 y2yye2x
22. yyex
2328 Solve the differential equation using the method of
24. yysec3x, 0x2 1
21.
SECTION 17.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS
1125
y3y2y 1ex
variation of parameters.
23. yysec2x, 0x 2
26. y3y2ysinex
27. y2yy ex
1×2
28. y4y4ye2x x3
17.3
APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS
Secondorder linear differential equations have a variety of applications in science and engineering. In this section we explore two of them: the vibration of springs and electric circuits.
VIBRATING SPRINGS
We consider the motion of an object with mass m at the end of a spring that is either ver tical as in Figure 1 or horizontal on a level surface as in Figure 2.
In Section 6.4 we discussed Hookes Law, which says that if the spring is stretched or compressed x units from its natural length, then it exerts a force that is proportional to x:
restoring forcekx
where k is a positive constant called the spring constant. If we ignore any external resist ing forces due to air resistance or friction then, by Newtons Second Law force equals mass times acceleration, we have
m
FIGURE 1
FIGURE 2
equilibrium position
0 xm
x d2x d2x
m dt2 kx or m dt2 kx0
equilibrium position
m
x
This is a secondorder linear differential equation. Its auxiliary equation is mr2k0
with roots r
i, where
skm . Thus the general solution is xtc1cos tc2sin t
xtAcos tfrequency
0x
which can also be written as
whereskm
25.
1
Asc12c22 amplitude
cosc1 sin c2is the phase angle
See Exercise 17. This type of motion is called simple harmonic motion.
AA
1126
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
FIGURE 3
m
V EXAMPLE 1 A spring with a mass of 2 kg has natural length 0.5 m. A force of 25.6 N is required to maintain it stretched to a length of 0.7 m. If the spring is stretched to a length of 0.7 m and then released with initial velocity 0, find the position of the mass at any time t.
SOLUTION From Hookes Law, the force required to stretch the spring is k0.225.6
so k25.60.2128. Using this value of the spring constant k, together with m2 in Equation 1, we have
2 d2x128x0 dt2
As in the earlier general discussion, the solution of this equation is
xtc1 cos 8tc2 sin 8t
We are given the initial condition that x00.2. But, from Equation 2, x0c1.
Therefore c10.2. Differentiating Equation 2, we get
xt8c1 sin 8t8c2 cos 8t
Since the initial velocity is given as x00, we have c20 and so the solution is
xt1 cos 8t M
DAMPED VIBRATIONS
We next consider the motion of a spring that is subject to a frictional force in the case of the horizontal spring of Figure 2 or a damping force in the case where a vertical spring moves through a fluid as in Figure 3. An example is the damping force supplied by a shock absorber in a car or a bicycle.
We assume that the damping force is proportional to the velocity of the mass and acts in the direction opposite to the motion. This has been confirmed, at least approximately, by some physical experiments. Thus
damping forcec dx dt
where c is a positive constant, called the damping constant. Thus, in this case, Newtons Second Law gives
2
5
m d2xrestoring forcedamping forcekxc dx dt2 dt
or
3
m d2x c dx kx0 dt2 dt
Schwinn Cycling and Fitness
SECTION 17.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS1127 Equation 3 is a secondorder linear differential equation and its auxiliary equation is
mr2 crk0.Therootsare
csc2 4mk csc2 4mk
r12m r22m According to Section 17.1 we need to discuss three cases.
4
x 0t
x 0t
FIGURE 4
Overdamping
N CASE I c24mk0 overdamping
In this case r1 and r2 are distinct real roots and
xc1er1tc2er2t
Since c, m, and k are all positive, we have sc24mkc, so the roots r1 and r2 given by Equations 4 must both be negative. This shows that x l 0 as t l . Typical graphs of x as a function of t are shown in Figure 4. Notice that oscillations do not occur. Its pos sible for the mass to pass through the equilibrium position once, but only once. This is because c24mk means that there is a strong damping force highviscosity oil or grease compared with a weak spring or small mass.
N CASE II c24mk0 critical damping This case corresponds to equal roots
x
xAec2mt
0t xAec2mt
FIGURE 5
Underdamping
It is similar to Case I, and typical graphs resemble those in Figure 4 see Exercise 12, but the damping is just sufficient to suppress vibrations. Any decrease in the viscosity of the fluid leads to the vibrations of the following case.
N CASE III c24mk0 underdamping Here the roots are complex:
r1 ci r2 2m
s4mkc2 2m
xec2mtc1 cos tc2 sin t
We see that there are oscillations that are damped by the factor ec2mt. Since c0 and m0,wehavec2m0soec2mt l0astl.Thisimpliesthatxl0astl; that is, the motion decays to 0 as time increases. A typical graph is shown in Figure 5.
and the solution is given by
r1r2 c 2m
xc1c2tec2mt
whereThe solution is given by
1128
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
N Figure 6 shows the graph of the position func tion for the overdamped motion in Example 2.
V EXAMPLE 2 Suppose that the spring of Example 1 is immersed in a fluid with damping constant c40. Find the position of the mass at any time t if it starts from the equilibrium position and is given a push to start it with an initial velocity of 0.6 ms.
SOLUTION From Example 1, the mass is m2 and the spring constant is k128, so the differential equation 3 becomes
2 d2x40 dx128x0 dt2 dt
or d2x20 dx64x0 dt2 dt
Theauxiliaryequationisr2 20r64r4r160withroots4 and 16, so the motion is overdamped and the solution is
xtc1e4tc2 e16t
We are given that x00, so c1c20. Differentiating, we get xt4c1e4t16c2e16t
so x04c116c20.6 Since c2c1, this gives 12c10.6 or c10.05. Therefore
0.03
0
FIGURE 6
1.5
x0.05e4te16tM Suppose that, in addition to the restoring force and the damping force, the motion of the
FORCED VIBRATIONS
spring is affected by an external force Ft. Then Newtons Second Law gives m d 2 xrestoring forcedamping forceexternal force
dt2
kxc dxFt
dt
Thus, instead of the homogeneous equation 3, the motion of the spring is now governed by the following nonhomogeneous differential equation:
The motion of the spring can be determined by the methods of Section 17.2.
5
m d2xc dxkxFt dt2 dt
6
switch
E
FIGURE 7
R
C
L
A commonly occurring type of external force is a periodic force function
FtF0 cos 0t where 0 skm
In this case, and in the absence of a damping force c0, you are asked in Exercise 9 to
use the method of undetermined coefficients to show that
xtc1cos tc2sin t F0 cos 0t m202
If 0, then the applied frequency reinforces the natural frequency and the result is vibrations of large amplitude. This is the phenomenon of resonance see Exercise 10.
ELECTRIC CIRCUITS
In Sections 9.3 and 9.5 we were able to use firstorder separable and linear equations to analyze electric circuits that contain a resistor and inductor see Figure 5 on page 582 or Figure 4 on page 605 or a resistor and capacitor see Exercise 29 on page 607. Now that we know how to solve secondorder linear equations, we are in a position to analyze the circuit shown in Figure 7. It contains an electromotive force E supplied by a battery or generator, a resistor R, an inductor L, and a capacitor C, in series. If the charge on the capacitor at time t is QQt, then the current is the rate of change of Q with respect to t: IdQdt. As in Section 9.5, it is known from physics that the voltage drops across the resistor, inductor, and capacitor are
RI LdI Q dt C
respectively. Kirchhoffs voltage law says that the sum of these voltage drops is equal to the supplied voltage:
L dI RI Q Et dt C
Since IdQdt, this equation becomes
which is a secondorder linear differential equation with constant coefficients. If the charge Q0 and the current I0 are known at time 0, then we have the initial conditions
Q0Q0 Q0I0I0
and the initialvalue problem can be solved by the methods of Section 17.2.
SECTION 17.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS1129
7
L d2QR dQ1 QEt dt2 dtC
1130
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
A differential equation for the current can be obtained by differentiating Equation 7
with respect to t and remembering that IdQdt:
L d 2IR dI1 IEt
8
dt2 dtC
V EXAMPLE 3 Find the charge and current at time t in the circuit of Figure 7 if R40,L1H,C16104 F,Et100cos10t,andtheinitialchargeand current are both 0.
SOLUTION With the given values of L, R, C, and Et, Equation 7 becomes d 2Q40 dQ625Q100 cos 10t
dt2 dt
The auxiliary equation is r240r6250 with roots
r 40s900 2015i 2
so the solution of the complementary equation is
Qcte20tc1 cos15tc2 sin15t
For the method of undetermined coefficients we try the particular solution
QptA cos10tB sin10t
Then Qpt10Asin10t10Bcos10t
Qpt100 A cos 10t100B sin 10t Substituting into Equation 8, we have
100Acos10t100Bsin10t4010Asin10t10Bcos10t
625A cos10tB sin10t100 cos10t
or 525A400Bcos10t400A525Bsin10t100cos10t Equating coefficients, we have
525A400B100 21A16B4
or
or 16A21B0
and B64 , so a particular solution is
400A525B0 The solution of this system is A84
697
697
Qpt1 84cos10t64sin10t 697
and the general solution is
QtQctQpt
e20tc1 cos15tc2 sin15t4 21cos10t16sin10t 697
0.2
Qp
0Q 1.2
0.2
FIGURE 8
md2xcdx kxFt dt2 dt
L d2QR dQ1 QEt dt2 dtC
SECTION 17.3 APPLICATIONS OF SECONDORDER DIFFERENTIAL EQUATIONS1131 Imposing the initial condition Q00, we get
Q0c184 0 c1 84 697 697
To impose the other initial condition, we first differentiate to find the current:
IdQe20t20c115c2cos15t15c120c2sin15t dt
40 21sin10t16cos10t 697
I020c1 15c2 640 0 c2 464 697 2091
Thus the formula for the charge is
Qt4 e20t 63cos15t116sin15t21cos10t16sin10t 697 3
and the expression for the current is
It1 e20t1920 cos 15t13,060 sin 15t12021 sin 10t16 cos 10t 2091
M
NOTE 1 In Example 3 the solution for Qt consists of two parts. Since e20t l 0 as t land both cos 15t and sin 15t are bounded functions,
Qct4 e20t63cos15t116sin15t l 0 2091
So, for large values of t,
QtQpt4 21cos10t16sin10t
as t l
5
697
and, for this reason, Qpt is called the steady state solution. Figure 8 shows how the graph of the steady state solution compares with the graph of Q in this case.
NOTE 2 Comparing Equations 5 and 7, we see that mathematically they are identical. This suggests the analogies given in the following chart between physical situations that, at first glance, are very different.
7
Spring system
Electric circuit
x displacement dxdt velocity
m mass
c damping constant k spring constant Ft external force
Q
IdQdt L
R
1C
Et
charge
current
inductance resistance elastance electromotive force
We can also transfer other ideas from one situation to the other. For instance, the steady state solution discussed in Note 1 makes sense in the spring system. And the phenomenon of resonance in the spring system can be usefully carried over to electric circuits as elec trical resonance.
1132CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
17.3 EXERCISES
1. A spring has natural length 0.75 m and a 5kg mass. A force of 25 N is needed to keep the spring stretched to a length of 1 m. If the spring is stretched to a length of 1.1 m and then released with velocity 0, find the position of the mass after t seconds.
2. A spring with an 8kg mass is kept stretched 0.4 m beyond its natural length by a force of 32 N. The spring starts at its equi librium position and is given an initial velocity of 1 ms. Find the position of the mass at any time t.
A spring with a mass of 2 kg has damping constant 14, and
a force of 6 N is required to keep the spring stretched 0.5 m beyond its natural length. The spring is stretched 1 m beyond its natural length and then released with zero velocity. Find the position of the mass at any time t.
4. A force of 13 N is needed to keep a spring with a 2kg mass stretched 0.25 m beyond its natural length. The damping con
stant of the spring is c8.
a If the mass starts at the equilibrium position with a ;
velocity of 0.5 ms, find its position at time t. b Graph the position function of the mass.
5. For the spring in Exercise 3, find the mass that would produce critical damping.
6. For the spring in Exercise 4, find the damping constant that would produce critical damping.
12.
Consider a spring subject to a frictional or damping force. a In the critically damped case, the motion is given by
xc1ertc2tert. Show that the graph of x crosses the
taxis whenever c1 and c2 have opposite signs. b In the overdamped case, the motion is given by
xc1er1 tc2er2 t, where r1r2. Determine a condition on the relative magnitudes of c1 and c2 under which the graph of x crosses the taxis at a positive value of t.
A series circuit consists of a resistor with R20 , an inductor with L1 H, a capacitor with C0.002 F, and a 12V battery. If the initial charge and current are both 0, find the charge and current at time t.
A series circuit contains a resistor with R24 , an induc tor with L2 H, a capacitor with C0.005 F, and a 12V battery. The initial charge is Q0.001 C and the initial cur rent is 0.
a Find the charge and current at time t. b Graph the charge and current functions.
The battery in Exercise 13 is replaced by a generator produc ing a voltage of Et12 sin 10t. Find the charge at time t.
The battery in Exercise 14 is replaced by a generator pro ducing a voltage of Et12 sin 10t.
a Find the charge at time t.
b Graph the charge function.
Verify that the solution to Equation 1 can be written in the formxtAcos t .
The figure shows a pendulum with length L and the angle from the vertical to the pendulum. It can be shown that , as a function of time, satisfies the nonlinear differential equation
d2 t
dt2 Lsin 0
where t is the acceleration due to gravity. For small values of we can use the linear approximation sinand then the
differential equation becomes linear.
a Find the equation of motion of a pendulum with length
1 m if is initially 0.2 rad and the initial angular velocity
is d dt1 rads.
b What is the maximum angle from the vertical?
c What is the period of the pendulum that is, the time to
complete one backandforth swing?
d When will the pendulum first be vertical?
e What is the angular velocity when the pendulum is
;
14.
15. 16.
18.
; 7.
; 8.
A spring has a mass of 1 kg and its spring constant is k100. The spring is released at a point 0.1 m above its equilibrium position. Graph the position function for the following values of the damping constant c: 10, 15, 20, 25, 30. What type of damping occurs in each case?
A spring has a mass of 1 kg and its damping constant is
c10. The spring starts from its equilibrium position with a velocity of 1 ms. Graph the position function for the follow ing values of the spring constant k: 10, 20, 25, 30, 40. What type of damping occurs in each case?
Suppose a spring has mass m and spring constant k and letskm . Suppose that the damping constant is so small
that the damping force is negligible. If an external force FtF0 cos 0t is applied, where 0, use the method of undetermined coefficients to show that the motion of the mass is described by Equation 6.
3.
;
13.
17.
9.
10. As in Exercise 9, consider a spring with mass m, spring con stant k, and damping constant c0, and letskm .
If an external force FtF0 cos t is applied the applied frequency equals the natural frequency, use the method of undetermined coefficients to show that the motion of the mass isgivenbyxtc1 cos tc2 sin tF02m tsin t.
11. Show that if 0, but0 is a rational number, then the motion described by Equation 6 is periodic.
vertical?
L
1
SECTION 17.4 SERIES SOLUTIONS1133
17.4
SERIES SOLUTIONS
Many differential equations cant be solved explicitly in terms of finite combinations of simple familiar functions. This is true even for a simplelooking equation like
y2xyy0
But it is important to be able to solve equations such as Equation 1 because they arise from physical problems and, in particular, in connection with the Schrodinger equation in quan tum mechanics. In such a case we use the method of power series; that is, we look for a solution of the form
yfx cnxn c0 c1xc2x2 c3x3
n0
The method is to substitute this expression into the differential equation and determine the values of the coefficients c0, c1, c2, . . . . This technique resembles the method of undeter mined coefficients discussed in Section 17.2.
Before using power series to solve Equation 1, we illustrate the method on the simpler equation yy0 in Example 1. Its true that we already know how to solve this equa tion by the techniques of Section 17.1, but its easier to understand the power series method when it is applied to this simpler equation.
V EXAMPLE 1 Use power series to solve the equation yy0. SOLUTION We assume there is a solution of the form
yc0 c1xc2x2 c3x3cnxn
n0
We can differentiate power series term by term, so
yc1 2c2x3c3x2ncnxn1
n1
y2c2 23c3x nn1cnxn2
n2
In order to compare the expressions for y and y more easily, we rewrite y as follows:
y n2n1cn2xn
n0
Substituting the expressions in Equations 2 and 4 into the differential equation, we
N By writing out the first few terms of 4, you can see that it is the same as 3. To obtain 4, we replaced n by n2 and began the sum mation at 0 instead of 2.
obtain
or
n2n1cn2xncnxn 0
2
3
4
5
n0
n0
n2n1cn2 cnxn 0 n0
1134
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS
If two power series are equal, then the corresponding coefficients must be equal. There
fore the coefficients of xn in Equation 5 must be 0: n2n1cn2 cn 0
cn2 cn n0, 1, 2, 3, . . . n1n2
Equation 6 is called a recursion relation. If c0 and c1 are known, this equation allows us to determine the remaining coefficients recursively by putting n0, 1, 2, 3, . . . in succession.
6
Putn0: c2 c0 12
Putn1: c3 c1 23
Putn2: c4 c2 34
Putn3: c5 c3 45
Putn4: c6 c4 56
Putn5: c7 c5 67
By now we see the pattern:
c0 c0 1234 4!
c1 c1 2345 5!
c0
4! 56
c1
5! 67
c0 6!
c1 7!
For the even coefficients, c2n1n c0 2n!
For the odd coefficients, c2n11n c1
2n1!
Putting these values back into Equation 2, we write the solution as
yc0 c1xc2x2 c3x3 c4x4 c5x5
c0
x2x4x6 1
2! 4! 6!
x2n 1n
x3
x5 5!
x7 7!
2n!1n
x2n1 2n1!
c1 x
3!
c1
n0
1n
x 2 n 2n!
n0
1n
x 2 n1 2n1!
c0
Notice that there are two arbitrary constants, c0 and c1.
M
2ncnxn
n1 n0
n0
n2n1cn2 2n1cn 0
cn22n1 cn n0, 1, 2, 3, . . .
2ncnxn
n2n1cn2 2n1cnxn 0 This equation is true if the coefficient of xn is 0:
as in Example 1. Substituting in the differential equation, we get
n2n1cn2xn 2x ncnxn1cnxn 0
n0
7
n1 n0
SECTION 17.4 SERIES SOLUTIONS1135
NOTE 1 We recognize the series obtained in Example 1 as being the Maclaurin series for cos x and sin x. See Equations 11.10.16 and 11.10.15. Therefore we could write the solution as
yxc0 cos xc1 sin x
But we are not usually able to express power series solutions of differential equations in
terms of known functions.
V EXAMPLE 2 Solvey2xyy0.
SOLUTION We assume there is a solution of the form
y cnxn n0
Then y ncnxn1
n1
and y nn1cnxn2n2n1cn2xn n2 n0
n2n1cn2xn2ncnxncnxn 0 n0 n1 n0
n1n2
We solve this recursion relation by putting n0, 1, 2, 3, . . . successively in Equation 7:
Put n0: Put n1: Put n2: Put n3:
c21 c0 12
c3c4c5
1 c1 23
3 c2 3 c0 3 c0 34 1234 4!
5 c315 c115 c1 45 2345 5!
1136
CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS Putn4:
Put n5:
Putn6:
Putn7:
In general, the even coefficients are given
c67 56
c79 67
c811 78
c437 c0 37 c0 4! 56 6!
c5159c1159c1 5! 67 7!
c6 3711 c0 8!
c715913 c1 9!
by
c913 89
c2n 37114n5 c0 2n!
and the odd coefficients are given by
c2n11594n3 c1 2n1!
yc0 c1xc2x2 c3x3 c4x4
c01 1 x23 x437 x63711 x8
or
The solution is
8
2! 4! 6! 8!
c1x 1 x315 x5159 x715913 x93! 5! 7! 9!
1 yc0 1 x2
374n5
1594n3
c1x x2n1 M
2 !
n22 n!
x2n
n1 2n1!
In Example 2 we had to assume that the differential equation had a series solu tion. But now we could verify directly that the function given by Equation 8 is indeed a solution.
NOTE 3 Unlike the situation of Example 1, the power series that arise in the solution of Example 2 do not define elementary functions. The functions
y1x11x2 374n5x2n 2! n2 2n!
and y2xx1594n3 x2n1 n1 2n1!
NOTE 2
2 T
2 2
are perfectly good functions but they cant be expressed in terms of familiar functions. We can use these power series expressions for y1 and y2 to compute approximate values of the functions and even to graph them. Figure 1 shows the first few partial sums T0, T2, T4, . . . Taylor polynomials for y1x, and we see how they converge to y1. In this way we can graph both y1 and y2 in Figure 2.
NOTE 4 If we were asked to solve the initialvalue problem
y2xyy0 y00 y01
we would observe from Theorem 11.10.5 that
c0 y00 c1 y01
This would simplify the calculations in Example 2, since all of the even coefficients would
be 0. The solution to the initialvalue problem is
yxx1594n3 x2n1
n1 2n1!
11. yx2yxy0, y00, y01 12. The solution of the initialvalue problem
x 2 y x y x 2 y0 y0 1 y0 0
is called a Bessel function of order 0.
a Solve the initialvalue problem to find a power series
expansion for the Bessel function.
b Graph several Taylor polynomials until you reach one that
looks like a good approximation to the Bessel function on the interval 5, 5.
b What is the complementary equation? How does it help solve the original differential equation?
c Explain how the method of undetermined coefficients works.
d Explain how the method of variation of parameters works.
4. Discuss two applications of secondorder linear differential equations.
5. How do you use power series to solve a differential equation?
CHAPTER 17 REVIEW1137
FIGURE 1
2.5
FIGURE 2
T8
15
15
fi
2.5
17.4 EXERCISES
111 Use power series to solve the differential equation.
1. yy0 yx2y
2. yxy
4. x3y2y0 6. yy
3.
5. yxyy0
7. x1yy0
8. yxy
9. yxyy0, y01, y00
;
10. yx2y0, y01, y00
CONCEPT CHECK
1. a
Write the general form of a secondorder homogeneous
linear differential equation with constant coefficients.
b Write the auxiliary equation.
c How do you use the roots of the auxiliary equation to solve
the differential equation? Write the form of the solution for each of the three cases that can occur.
17 REVIEW
What is an initialvalue problem for a secondorder differ ential equation?
2. a
b What is a boundaryvalue problem for such an equation?
3. a
Write the general form of a secondorder nonhomogeneous linear differential equation with constant coefficients.
1138CHAPTER 17 SECONDORDER DIFFERENTIAL EQUATIONS TRUEFALSE QUIZ
Determine whether the statement is true or false. If it is true, explain why. If it is false, explain why or give an example that disproves the statement.
1. Ify1 andy2 aresolutionsofyy0,theny1 y2 isalso a solution of the equation.
2. Ify1 andy2 aresolutionsofy6y5yx,then c1 y1c2 y2 is also a solution of the equation.
EXERCISES
110 Solve the differential equation.
1.
y2y15y0
3. The general solution of yy0 can be written as yc1 coshxc2 sinhx
4. The equation yyex has a particular solution of the form yp Aex
16. Use power series to solve the equation yxy2y0
17. A series circuit contains a resistor with R40 , an inductor with L2 H, a capacitor with C0.0025 F, and a 12V bat tery. The initial charge is Q0.01 C and the initial current
is 0. Find the charge at time t.
18. Aspringwithamassof2kghasdampingconstant16,anda force of 12.8 N keeps the spring stretched 0.2 m beyond its natural length. Find the position of the mass at time t if it starts at the equilibrium position with a velocity of 2.4 ms.
19. Assume that the earth is a solid sphere of uniform density with mass M and radius R3960 mi. For a particle of mass m within the earth at a distance r from the earths center, the gravitational force attracting the particle to the center is
FrGMrm r2
where G is the gravitational constant and Mr is the mass of the earth within the sphere of radius r.
a Show that FrGMm r. R3
b Suppose a hole is drilled through the earth along a diame ter. Show that if a particle of mass m is dropped from rest at the surface, into the hole, then the distance yyt of the particle from the center of the earth at time t is given by
ytk2yt
where k2GMR3tR.
c Conclude from part b that the particle undergoes simple
harmonic motion. Find the period T.
d With what speed does the particle pass through the center
of the earth?
2. y4y13y0
3. y3y0
4. 4y4yy0
5. d2y
dx2 6. d2y
dx2 7. d2y
4 dy 5ye2x dx
dy2yx2 dx
2 dyyx cos x dx
dx2 d2y
4ysin 2x dy
8. dx2 d2y
dx 6y1e2x ycscx, 0x 2
9. dx2 d2y
10. dx2
1114 Solve the initialvalue problem.
11. y6y0, y13, y112
12. y6y25y0, y02, y01 13. y5y4y0, y00, y01 14. 9yy3xex, y01, y02
15. Use power series to solve the initialvalue problem yxyy0 y00 y01
APPENDIXES
F Proofs of Theorems
H ComplexNumbers
I Answers to OddNumbered Exercises
A1
A2
APPENDIX F PROOFS OF THEOREMS
F PROOFS OF THEOREMS
In this appendix we present proofs of several theorems that are stated in the main body of
SECTION 11.8
the text. The sections in which they occur are indicated in the margin. In order to prove Theorem 11.8.3, we first need the following results.
THEOREM
1. If a power seriescn xn converges when xb where b0, then it converges whenever xb .
2. If a power seriescn xn diverges when xd where d0, then it diverges
whenever xd .
PROOF OF 1 Suppose thatcnbn converges. Then, by Theorem 11.2.6, we have
limn lcnbn0. According to Definition 11.1.2 with 1, there is a positive integer Nsuchthat cnbn1whenevernN.Thus,fornN,wehave
n cnbnxn nxn xn cnxbncnb bb
If xb , then xb1, soxb n is a convergent geometric series. Therefore,
by the Comparison Test, the series nn
nN cn x is convergent. Thus the seriescn x is absolutely convergent and therefore convergent.
M
PROOF OF 2 Suppose thatcndn diverges. If x is any number such that xd , then
cn xn cannot converge because, by part 1, the convergence ofcn xn would imply the convergence ofcndn. Thereforecn xn diverges whenever xd . M
PROOF Suppose that neither case 1 nor case 2 is true. Then there are nonzero numbers b and d such that cnxn converges for xb and diverges for xd. Therefore the set
Sxcnxn converges is not empty. By the preceding theorem, the series diverges if
xd,so xd forallxS.Thissaysthat d isanupperboundforthesetS.
Thus, by the Completeness Axiom see Section 11.1, S has a least upper bound R. If
x R, then xS, so cnxn diverges. If x R, then xis not an upper bound for
SandsothereexistsbSsuchthatb x .SincebS, cnbn converges,sobythe
preceding theoremcn xn converges.
THEOREM For a power seriescn x n there are only three possibilities: 1. The series converges only when x0.
2. The series converges for all x.
3. There is a positive number R such that the series converges if x R and diverges if x R.
M
SECTION 14.3
PROOF If we make the change of variable uxa, then the power series becomescnun and we can apply the preceding theorem to this series. In case 3 we have con vergence for u R and divergence for u R. Thus we have convergence for
xa R and divergence for xa R.
PROOF For small values of h, h0, consider the difference
h f ah, bhf ah, b f a, bhf a, b
Notice that if we let txfx, bhfx, b, then htahta
By the Mean Value Theorem, there is a number c between a and ah such that tahtatchhfxc, bhfxc, b
Applying the Mean Value Theorem again, this time to fx , we get a number d between b and bh such that
fxc, bhfxc, bfxyc, dh Combining these equations, we obtain
hh2fxyc, d Ifhl0,thenc,dla,b,sothecontinuityof fxy ata,bgives
lim hlim fxyc, dfxya, b hl0 h2 c, dla, b
Similarly, by writing
h f ah, bhf a, bh f ah, bf a, b
and using the Mean Value Theorem twice and the continuity of fyx at a, b, we obtain
M
APPENDIX F PROOFS OF THEOREMSA3
THEOREM For a power seriescnxan there are only three possibilities: 1. The series converges only when xa.
2. The series converges for all x.
3. There is a positive number R such that the series converges if xa R and diverges if xa R.
3
CLAIRAUTS THEOREM Suppose f is defined on a disk D that contains the point
a, b. If the functions fx y and fyx are both continuous on D, then fx ya, bfyxa, b.
It follows that fxya, bfyxa, b.
M
lim hfyxa, b hl0 h2
A4
APPENDIX F PROOFS OF THEOREMS
y
PROOF Let
zfax, byfa, b
According to 14.4.7, to prove that f is differentiable at a, b we have to show that we
can write z in the form
zfxa, b xfya, b y1 x2 y
where 1 and 2 l 0 as x, y l 0, 0. Referring to Figure 1, we write
z f ax, byf a, by f a, byf a, b Observe that the function of a single variable
txfx, by
is defined on the interval a, ax and txfxx, by. If we apply the Mean
Value Theorem to t, we get
taxtatu x
where u is some number between a and ax. In terms of f , this equation becomes
f ax, byf a, byfxu, by x
This gives us an expression for the first part of the right side of Equation 1. For the second part we let h yf a, y. Then h is a function of a single variable defined on the interval b, by and hyfya, y. A second application of the Mean Value Theorem then gives
hbyhbhv y
where v is some number between b and by. In terms of f, this becomes
f a, byf a, bfya, v y We now substitute these expressions into Equation 1 and obtain
zfxu, byxfya, vy
fxa, b x fxu, byfxa, b xfya, b yfya, vfya, b y fxa,bxfya,by1x2 y
SECTION 14.4
THEOREM If the partial derivatives fx and fy exist near a, b and are continu ous at a, b, then f is differentiable at a, b.
8
aIx, bIy u, bIy
a, bIy
0x
FIGURE 1
1
a,
a,b
R
where 1 fxu,byfxa,b 2fya, vfya, b
Since u, by l a, b and a, v l a, b as x, y l 0, 0 and since fx and fy are continuous at a, b, we see that 1 l 0 and 2 l 0 as x, y l 0, 0.
Therefore f is differentiable ata, b .
COMPLEX NUMBERS
M
APPENDIX H COMPLEX NUMBERSA5
H
Im
A complex number can be represented by an expression of the form abi, where a and b are real numbers and i is a symbol with the property that i 21. The complex num ber abi can also be represented by the ordered pair a, b and plotted as a point in a plane called the Argand plane as in Figure 1. Thus the complex number i01i is identified with the point 0, 1.
The real part of the complex number abi is the real number a and the imaginary part is the real number b. Thus the real part of 43i is 4 and the imaginary part is 3. Two complex numbers abi and cdi are equal if ac and bd; that is, their real parts are equal and their imaginary parts are equal. In the Argand plane the horizontal axis is called the real axis and the vertical axis is called the imaginary axis.
The sum and difference of two complex numbers are defined by adding or subtracting their real parts and their imaginary parts:
abicdiacbdi abicdiacbdi
For instance,
1i47i1417i56i
The product of complex numbers is defined so that the usual commutative and distributive
laws hold:
abicdiacdibicdiacadibcibdi 2
Since i 21, this becomes
abicdiacbdadbci
EXAMPLE 1
13i25i125i3i25i
25i6i1511311i M
Division of complex numbers is much like rationalizing the denominator of a rational expression. For the complex number zabi, we define its complex conjugate to be zabi. To find the quotient of two complex numbers we multiply numerator and denominator by the complex conjugate of the denominator.
23i
42i
22i
FIGURE 1
i
01 Re i
32i
Complex numbers as points in the Argand plane
A6
APPENDIX H COMPLEX NUMBERS
Im
i
0 i
FIGURE 2
Im
bi
zabi
zabi
zabi b
Re
EXAMPLE 2 Express the number 13i in the form abi. 25i
SOLUTION We multiply numerator and denominator by the complex conjugate of 25i, namely 25i, and we take advantage of the result of Example 1:
13i13i25i1311i1311 i M 25i 25i 25i 2252 29 29
The geometric interpretation of the complex conjugate is shown in Figure 2: z is the reflection of z in the real axis. We list some of the properties of the complex conjugate in the following box. The proofs follow from the definition and are requested in Exercise 18.
The modulus, or absolute value, zof a complex number zabi is its distance from the origin. From Figure 3 we see that if zabi, then
Notice that
zzabiabia2 abiabib2i2 a2 b2
and so
This explains why the division procedure in Example 2 works in general:
zzw zw w ww w2
Since i 21, we can think of i as a square root of 1. But notice that we also have i2 i2 1andsoiisalsoasquarerootof1.Wesaythati istheprincipal square root of 1 and write s1i. In general, if c is any positive number, we write
sc sci
With this convention, the usual derivation and formula for the roots of the quadratic equa
tionax2 bxc0arevalidevenwhenb2 4ac0:
SOLUTION Using the quadratic formula, we have
1s12 41 1s3 1s3i x222M
PROPERTIES OF CONJUGATES
zwzw zwzw zn zn
zsa2 b2
0 a Re FIGURE 3
zzz2
x
EXAMPLE 3 Findtherootsoftheequationx2 x10.
bsb2 4ac 2a
z a b
Im
We observe that the solutions of the equation in Example 3 are complex conjugates of each other. In general, the solutions of any quadratic equation ax2bxc0 with real coefficients a, b, and c are always complex conjugates. If z is real, zz, so z is its own conjugate.
We have seen that if we allow complex numbers as solutions, then every quadratic equa tion has a solution. More generally, it is true that every polynomial equation
anxn an1xn1 a1xa0 0
of degree at least one has a solution among the complex numbers. This fact is known as
the Fundamental Theorem of Algebra and was proved by Gauss.
POLAR FORM
We know that any complex number zabi can be considered as a point a, b and that
zabircos rsin i Thus we can write any complex number z in the form
where rzsa2 b2 and tan
The angle is called the argument of z and we writeargz. Note that argz is not
unique; any two arguments of z differ by an integer multiple of 2 . EXAMPLE 4 Write the following numbers in polar form.
a z1i b ws3 i
SOLUTION
a Wehaverzs12 12 s2 andtan 1,sowecantake4. Therefore the polar form is
APPENDIX H COMPLEX NUMBERSA7
0a
FIGURE 4
Re
r
b
ar cos as in Figure 4. Therefore we have
br sin
abi
any such point can be represented by polar coordinates r,
with r0. In fact,
zrcos isin
b a
Im
0
FIGURE 5
cos 4 isin 4
1i 2
zs2
b Herewehaverws312andtan 1s3.Sincewliesinthe
4
6
2
Re
3i
fourth quadrant, we take 6 and
w2 cos 6 isin 6
The numbers z and w are shown in Figure 5. M
A8
APPENDIX H COMPLEX NUMBERS
zTM
TM
zzTM
FIGURE 6
Im
Im
TM
z
The polar form of complex numbers gives insight into multiplication and division. Let
z1 r1cos 1 isin 1 z2 r2cos 2 isin 2 be two complex numbers written in polar form. Then
z1z2 r1r2cos 1 isin 1cos 2 isin 2
r1r2cos 1 cos 2 sin 1 sin 2isin 1 cos 2 cos 1 sin 2
Therefore, using the addition formulas for cosine and sine, we have
This formula says that to multiply two complex numbers we multiply the moduli and add the arguments. See Figure 6.
A similar argument using the subtraction formulas for sine and cosine shows that to divide two complex numbers we divide the moduli and subtract the arguments.
Inparticular,takingz1 1andz2 zandtherefore 1 0and 2,wehavethefol lowing, which is illustrated in Figure 7.
EXAMPLE 5 Find the product of the complex numbers 1i and s3i in polar form. SOLUTION From Example 4 we have
1is2 cos 4 isin 4
and s3 i2cos6isin6
r
0Re
1 r
FIGURE 7
1 z
z
Re
Im
2
z1i
zw
22
So, by Equation 1,
cosisin
1is3 i2s2
0Re 4646
FIGURE 8
This is illustrated in Figure 8.
M
12
1
z1z2 r1r2cos 12isin 12
z1 r1 cos1 2isin1 2 z20 z2 r2
If zrcos isin , then 11 cos isin . zr
2
2s2 cos 12 isin 12
w3i
Repeated use of Formula 1 shows how to compute powers of a complex number. If
zrcos isinthen z2 r2cos2 isin2
and z3zz2r3cos 3i sin 3
In general, we obtain the following result, which is named after the French mathematician
Abraham De Moivre 16671754.
This says that to take the nth power of a complex number we take the nth power of the modulus and multiply the argument by n.
EXAMPLE 6 Find 11 i10. 22
SOLUTION Since 11i11i, it follows from Example 4a that 11i has the polar 222 22
form
11 is2 cosi sin22244
So by De Moivres Theorem,
APPENDIX H COMPLEX NUMBERSA9
DE MOIVRES THEOREM If zrcosi sinand n is a positive integer, then
zn rcos isin n rncosn isinn
2
1 1 10 s210 10
22i2 cos 4 isin 4
10
25cos5 isin51i M
210 2 2 32
De Moivres Theorem can also be used to find the nth roots of complex numbers. An nth root of the complex number z is a complex number w such that
wnz Writing these two numbers in trigonometric form as
wscos isinand zrcos isinand using De Moivres Theorem, we get
sncosn isinn rcos isinThe equality of these two complex numbers shows that
sn r or sr1n
and cos ncos and sin nsin
A10
APPENDIX H COMPLEX NUMBERS
From the fact that sine and cosine have period 2 it follows that
nn
Since this expression gives a different value of w for k0, 1, 2, . . . , n1, we have the following.
n2k or
Thus wr1ncos2kisin2k
2k n
ROOTSOFACOMPLEXNUMBER Letzrcos isin andletnbeaposi tive integer. Then z has the n distinct nth roots
wk r1ncos2k isin2knn
where k0, 1, 2, . . . , n1.
3
Notice that each of the nth roots of z has modulus wk r1n. Thus all the nth roots of z lie on the circle of radius r1n in the complex plane. Also, since the argument of each successive nth root exceeds the argument of the previous root by 2 n, we see that the nth roots of z are equally spaced on this circle.
EXAMPLE 7 Find the six sixth roots of z8 and graph these roots in the complex plane.
SOLUTION In trigonometric form, z8cosi sin . Applying Equation 3 with n6,
we get
wk 816cos 2k isin 2k66
We get the six sixth roots of 8 by taking k0, 1, 2, 3, 4, 5 in this formula:
w0816cos
w1 816cos
w2 816cos
w3 816cos
w4 816cos
isin s2s31i 6622
2 isin 2s2 i
5 isin 5 s2 s31
6622
7 isin 7 s2 s31 6622
3 isin 3 s2 i 22
i i
i
M
w5 816cos
All these points lie on the circle of radius s2 as shown in Figure 9.
11 isin 11 s2 s31 6622
COMPLEX EXPONENTIALS
We also need to give a meaning to the expression ez when zxiy is a complex num ber. The theory of infinite series as developed in Chapter 11 can be extended to the case where the terms are complex numbers. Using the Taylor series for ex 11.10.11 as our guide, we define
1z
and it turns out that this complex exponential function has the same properties as the real
exponential function. In particular, it is true that
ez1z2ez1ez2
If we put ziy, where y is a real number, in Equation 4, and use the facts that i2 1, i3 i2ii, i4 1, i5 i, …
e
n0n!
2! 3!
APPENDIX H COMPLEX NUMBERSA11
z zn
z2 z3
4
weget
iy iy2 iy3 iy4 iy5
e 1iy 2!3!4!5!
y2 y3 y4 y5 1iy 2! i 3!4! i 5!
y2 y4 y6 y3 y5 12!4!6! i y3!5!
5
N We could write the result of Example 8a as
ei 10
This equation relates the five most famous num
cos yi sin y
Here we have used the Taylor series for cos y and sin y Equations 11.10.16 and 11.10.15.
The result is a famous formula called Eulers formula: eiy cosyisiny
Combining Eulers formula with Equation 5, we get
exiy exeiy excosyisiny EXAMPLE 8 Evaluate: a ei b e1i 2
SOLUTION
a From Eulers equation 6 we have
ei cos isin 1i01 b Using Equation 7 we get
1i 2 11 i
ee cos 2i sin 2e 0i1e M
Finally, we note that Eulers equation provides us with an easier method of proving De Moivres Theorem:
bers in all of mathematics: 0, 1, e, i, and
.
6
7
rcos isin n rei n rnein rncosn isinn
A12APPENDIX H COMPLEX NUMBERS
H EXERCISES
114 Evaluate the expression and write your answer in the formabi.
3336 Find the indicated power using De Moivres Theorem.
41i95i 2 2
12i83i 2i1 i
2
32i 14i
3
43i
15. 125i 16. 12s2i
17. 4i
18. Prove the following properties of complex numbers. a zwzw b zwzw
c znzn, where n is a positive integer
Hint: Write zabi, wcdi.
1924 Find all solutions of the equation.
33. 1i20
35. 2s32i5
34. 1s3i5 36. 1i8
1. 3. 5.
7.
9.
11. 13.
number.
56i32i 25i4i 127i
1i i3
2. 4. 6.
8. 1 10.
3740 Find the indicated roots. Sketch the roots in the complex plane.
14i 32i
37. The eighth roots of 1 39. Thecuberootsofi
38. The fifth roots of 32 40. Thecuberootsof1i
s25
1517 Find the complex conjugate and the modulus of the
41. ei 2 43. ei 3 45. e2i
42. e2 i 44. ei 46. e i
12. 14.
i100
s3 s12
4146 Write the number in the form abi.
47. Use De Moivres Theorem with n3 to express cos 3 and sin3 intermsofcos andsin .
48. Use Eulers formula to prove the following formulas for cos x and sin x:
eixeix 2
eixeix 2i
cos x
49. If uxf xitx is a complexvalued function of a real
variable x and the real and imaginary parts f x and tx are differentiable functions of x, then the derivative of u is defined to be uxf xitx. Use this together with Equation 7 toprovethatifFxerx,thenFxrerx whenrabi is a complex number.
sin x
19. 4×2 90
21. x2 2×50 23. z2 z20
20. x4 1
22. 2×2 2×10
24. z2 1z1 0 24
2528 Write the number in polar form with argument between 0 and2 .
50. a
If u is a complexvalued function of a real variable, its indefinite integral x ux dx is an antiderivative of u. Evaluate
25. 33i 27. 34i
26. 1s3i 28. 8i
ye1ix dx
b By considering the real and imaginary parts of the integral
in part a, evaluate the real integrals
yex cosxdx and yex sinxdx
c Compare with the method used in Example 4 in Sec tion 7.1.
2932 Find polar forms for zw, zw, and 1z by first putting z and w into polar form.
29. zs3 i, w1s3i 30.z4s34i, w8i 31.z2s32i, w1i 32. z4s3 i, w33i
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA13
CHAPTER 10
EXERCISES 10.1 N
PAGE 626
I ANSWERS TO ODDNUMBERED EXERCISES
15. a y1lnx1 b
2
17. a y2 x2 1,y1 b
y
1 01x
y
1 0x
1.y 3.y t5
t
5
0,
15, 5
t4 3, 0
5. a
7. a
t0 1, 0
5, 1 t0
8, 1 t1
y
x
2, 3 t1
y
t00,0 5x
b y2x13 33
b x1y52 2, 4
3y11
19. Moves counterclockwise along the circle
x32y124 from 3, 3 to 3, 1
21. Moves 3 times clockwise around the ellipse
x 225 y 241, starting and ending at 0, 2
23. It is contained in the rectangle described by 1×4 and 2y3.
1, 5 t2
0x
25.
29.
y 27. y
1 t12
7 , 11 t3
1,0 t0
0, 1
0, 1
3
t1
t1
x
t0 1 x
2, 5 t0
15
x
14, 3 t4
4 4, 0
t2
9. a y 0x
2,3 t4
11. a x2 y2 1,×0
b y1x2,x0
13. a y1x,y1 b
3 3
3
b x25t,y78t,0t1 a x2 cos t, y12 sin t, 0t2 x2 cos t, y12 sin t, 0t6
b
y
x0.
41. xacos ,ybsin ;x2a2y2b21,ellipse 43. y
2a
Ox
0,1 t0 1,0 t1
31.
33.
b
c
37. The curve yx23 is generated in a. In b, only the portion with x0 is generated, and in c we get only the portion with
x2 cos t, y12 sin t, 2t3
2
0,1 y
0 x
0,1
1, 1 0x
A14APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES 45. a Two points of intersection
4
6 6
4
b One collision point at 3, 0 when t3 2
c There are still two intersection points, but no collision point. 47. Forc0,thereisacusp;forc0,thereisaloopwhosesize
27. a dsin rdcos29. 16,29,2,4 27 9
31. ab 33.3e 35.2r2 d2 37. x2s14t2 dt3.1678
1
39. x2 s32 sin t2 cos t dt10.0367 41. 4s22 0
43. s103ln3s10 s2ln1s2
increases as c increases.
e3 11e8
0
3
1.5 0
1
1.5
49. As n increases, the number of oscillations increases; a and b determine the width and height.
612.3053 a
15
t0, 4
45.s2e 1
8
0
21
25
1
51. 6s2, s2
15
15
15
x1 2 t21etse2tt12t22t2 dt103.5999
47.
49. 55.
b
57. 59. 65.
EXERCISES 10.3 N 1. a
21
1
2.5
31
1
1
1 2
0
EXERCISES 10.2 N PAGE 636
1. 2t1
t cos tsin t
5. y2ex3 9. y1 x
6
10
11. 13 t, 34t, t0 2
3. yx
7. y2x1
20
2
294 0
2 1215
247s1364
61. 6 a2 5
63. 59.101
24 54
71. 1 PAGE 647
b 2, 3
10
949s261
13. et, et1et, t0
15. 3 tant,3 sec3t, 2t3 2
243O
17. Horizontal at 6, 16, vertical at 10, 0
19. Horizontal at s2, 1 four points, vertical at 2, 0 21. 0.6, 2; 5665, e615
23.
O
3, 2, 4 3
2
O
2, 1, 5 2
3 1,4
34
7.5
1
25. yx,yx y
2, 7
1, 5
4, 1, 4
c
8 . 5
3
0 x
1 , 2
1, 3
3. a
b
37.
39. 3
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
A15
52
8
1,
1, 0
O
O O61
c
2 3
2,2334
1, s3
3 4
41.
45.
49.
43.
5 66
O
3 2, 4
O
s2, s2
5. a i 2s2, 7 4
b i 2, 2 3 7.
ii 2s2, 3 4 ii 2, 5 3
9.
47.2
33
r1
r2
O
O
2
3,
3, 0
r2
7 3
r3
5 3
6
11.
r4
2 1 O
O
51.
O
2,0 6,0
1
15. Circle, center O, radius 2 Circle, center 0, 3 , radius 3
2s3
Horizontal line, 1 unit above the xaxis
13.
17.
19.
21.
27.a6 bx3 29.
r3 sec 23. rcot 31.
csc
25. r2c cos 1, 2
O
55. a For c1, the inner loop begins atsin1 1c and ends at sin1 1c;
for c1, it begins at
22
53.
63.
vertical at 3, 0, 0, 2
65. Horizontal at 3 , 3, 0,the pole, and 3 , 5 3;
vertical at 2, 0, 1, 2 3, 1, 4 3 22
67. Horizontal at 3, 2, 1, 3 2; vertical at 31 s3, ,
O
O
6
59.
Horizontal at 3s2, 4, 3s2, 3 4;
O
ends at 61. 1
sin1 1c and
2sin1 1c.
57. s3
33.
35.
22
3 1s3,where sin11 1s3 22 22
69. Center b2, a2, radius sa2b22
22
A16
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
71.
75.
77.
or
79.
2.6 73.
3.5
3 3
2.5
43. Intersection at0.89, 2.25; area3.46 45.
7 7
47. 821321 3
53. 16 3
0.75
49. 29.0653 51. 9.6884 55. b 2 2s2
1.25
3. 0, 0, 0,1 , y1 88 1616
3.4
1.8
1
1
2.6
7
EXERCISES 10.5 N
1. 0, 0, 1 , 0, x1
PAGE 660
y
1 x 8
y
1
8,0 16
y 1 0,1
16
7
By counterclockwise rotation through angle
x
about the origin
a A rose with n loops if n is odd and 2n loops if n is even
b Number of loops is always 2n
81. For 0a1, the curve is an oval, which develops a dimple as a l 1. When a1, the curve splits into two parts, one of which has a loop.
5. 2, 3, 2, 5, y1 yy
EXERCISES 10.4 N
PAGE 653
0x
x1
6, 3,
x
7.
2, 1, 5, 1, x1
7. 41 4x
5, 1 2, 1
121 s3
11. 4
y1
2, 5
1. 9. 9
13.
510,240
3.
5. 2 84
O
O
3
9. xy2, focus 1, 0, directrix x1 44
11. 3, 0, 2, 0 13. 0, 4, 0, 2s3 yy
5
4
6
15. 3
3
303x 202x 5
3
3
1 s3
15. 1,3,1,s5
4
x2 y2
17. 49 1,foci0,s5
17. 1
820232 1,3
19. 9
4s34 27. 29. 5 1s3 31. 1 1
21.3 s3
3 2442
23. 1
y
1 0
25.
33.1 1 35.1 3s3
844
37. 3, 6, 3, 5 6, and the pole 22
3 x
39. 1,where12, 5 12, 13 12, 17 12
and 1,where7 12, 11 12, 19 12, 23 12
41. 1 s3, 3, 1 s3, 2 3, and the pole 22
1,3
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
A17
19. 12, 0, 13, 0, y5 x
12
21. 0, 2, 0, 2s2, yx
11. a 1 4
d
b Ellipse
4, 2
3, O 3,0
c y12
yy y5 x
12
2
12, 3052
12 0x 2
yx
x
23. 4, 2, 2, 2; 3s5, 2;
y22x3
y12
13. a 1 b Ellipse c x9 y32
d
15.
d
x92
0x
3 2, 2
99 4, 8,0
2, 2
4, 2
35, 2
35, 2
O
3 3 2, 2
a 2
34 , 0
b Hyperbola
c x3 8
25. 29. 33.
37.
41.
45.
49.
51.
55.
Parabola, 0, 1, 0, 3 4
27. Ellipse, s2, 1, 1, 1
Hyperbola, 0, 1, 0, 3; 0, 1s5
31. x28y
41 ,
O
x 38
y2 12×1 x2 y2 25211
35. y32x22
x2 39. 12
y42
16 1
x12 12
y12 25
y42 16
x32 39
11
x2 y2
43. 9161
x2 y2
47. 9361
y2 3,763,6003,753,1961
17.
a2,y1 1 2
2 2
y12
x2
121x 2 121y 2
a 1,500,6253,339,3751
a Ellipse b Hyperbola c No curve
b 248 mi
59. 9.69 61. b2c ablna wherec2 a2 b2
EXERCISES 10.6 N
br 1 12sin 3 4
2
2 2
2
e1.0 e0.6
e0.8
a bc PAGE 668
15
43 cos
4
3
1. r5. r
42
47 sin
8
1sin
3. r
7. r
9. a 1 d
b Parabola
1 2, 2
O
2cos c y1
y1
19. The ellipse is nearly circular when e is close to 0 and becomes more elongated as e l 1. At
e1, the curve becomes a parabola.
e0.4
A18APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
2.26108 10.093 cos
2 23.1
1sint, 1costsint 1cos t 1cos t3
25. r
27. 35.64AU 29. 7.0107 km
21. 25.
Vertical tangent at 3 a, 1 s3 a, 3a, 0;
horizontal tangent at a, 0, 1 a, 3 s3a
27. 11,3
3a, 0
35.1 1 2
CHAPTER 10 REVIEW N TrueFalse Quiz
PAGE 669
5. True
22
5, 1, t1
1, 1,0
37.
39.
2 s4 21
31. 3.6108 km
8
4
29.
y
a, 0 0x
1. False Exercises
3. False
7. False
9. True
2
2
1. xy2 8y12 yy
3. y1x
0, 6, t4
33.2,3 25s51
471,295 1024
curve bulges to the right. At c1, the curve is the line x1.
For 1c0, it bulges to the left. At c0 there is a cusp at 0, 0. For c0, there is a loop.
31.18
5. xt,yst;xt4,yt2; xtan2t,ytant,0t 2
41.
43.
xx
ln s 21
All curves have the vertical asymptote x1. For c1, the
2s 21s4 21 2
7. a
4, 2 3
b 3s2, 3 3s2, 7 4
2 3
4,
45. 1, 0, 3, 0 47.25 , 3, 1, 3 O 24
yy 22
9.
13.
2, 2s3
2,
11.
15.
1, 0
3 0 3 x22
x2 y2 25 9 1
1, 3
0 x
1, 2
3 1, 2
6
1, 0
y2
51. 725851
49.
53. 57.
PROBLEMS PLUS N PAGE 672 1. ln 2
3. 3 s3, 3 s31, 2 44
x2
O
x2
8y3992 160,801 1
4 55. r 3cos
25
xacot sin cos ,ya1sin2
3, 32
1
2, 2,0 3
O y2 1
5. a At0,0and3,3 O22
1, 2b Horizontal tangents at 0, 0 and s3 2, s3 4 ; vertical tangents at 0, 0 and s3 4, s3 2
2 dyg3
17.rcos sin
19. 0.75
2
0.3
rsin
0.75
1.2
x
yx1
CHAPTER 11
EXERCISES 11.1 N
Abbreviations: C, convergent; D, divergent
1. a A sequence is an ordered list of numbers. It can also be defined as a function whose domain is the set of positive integers. b The terms an approach 8 as n becomes large.
c The terms an become large as n becomes large.
9. a C 19. D 31. D 41. 2
9
b D 21. D
11. 9 23. D
13. D
25. 5 2
15. 60 27. D
17. 1 7
29. D 39. e1
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
A19
PAGE 684
35. 3
45. 50633300
33. ee1 43. 1138333
37. 11 26
47. 3×3;
51. Allx; 2 53.1
x 3x
49. 1 x1; 1
4 4 14x
2cos x 9. an 12n1 11. an 5n3 2
3. 0.8, 0.96, 0.992, 0.9984, 0.99968 5. 3, 3, 1, 1,1 2 28 40
7. 3,5,9,17,33 13. an 2n1
3
15. 1,2,3,4, 5 , 6 ;yes;1
55. a1 0,annn1 forn1,sum1
57. a Sn1
63. nn1
71. snis bounded and increasing.
3 5 7 9 11 13
2
27. D 39. e2
b D
59. Convergent by the Monotonic Sequence Theorem; 5L8
17. 1
29. 0
41. ln 2
51. D
55. a 1060, 1123.60, 1191.02, 1262.48, 1338.23 57. 1r1
19. 5 31. 0
21. 1 33. 0
23. 1 35. 0
25. 0 37. 1
D1c n1c
1
59. 2s31
43. D 53. 0
45. D
47. 1
49. 1 2
b 5
65. The series is divergent.
61. Decreasing; yes
65. Decreasing; yes
71. b 11s5 2
63. Not monotonic; no
73. a 75. a
0, 1 , 2 , 1 , 2 , 7 , 8 , 1 993399
1 5 23 119 n1!1
67. 2 73. a 0
69. 1 3s52
b 9,11
2,6,24,120;
n1! PAGE 703
c 1
EXERCISES 11.2 N PAGE 694
EXERCISES 11.3 N 1.C y
1. a A sequence is an ordered list of numbers whereas a series is the sum of a list of numbers.
b A series is convergent if the sequence of partial sums is a con vergent sequence. A series is divergent if it is not convergent.
y 1 x 1.3
a … 01234 x
3. 2.40000, 1.92000, 2.01600, 1.99680, 2.00064, 1.99987, 2.00003, 1.99999, 2.00000, 2.00000; convergent, sum2
5. 1.55741, 0.62763, 0.77018, 0.38764, 2.99287, 3.28388, 2.41243, 9.21214, 9.66446, 9.01610; divergent
7. 0.29289, 0.42265, 0.50000, 0.55279, 0.59175, 0.62204, 0.64645, 0.66667, 0.68377, 0.69849; convergent, sum1
1
aTM
0
sand
ssn d
10
10
a
3. D
17.C 19.C 21.D 23.C 25.C 27.p1 29. p1 31. 1,
33. a 1.54977, error0.1
c n1000
35. 0.00145 41. b1e
5. C 7. C 9. D
a
11. C
13. D
15. C
9. C 21. C
PAGE 713
3
2
10
1
b 1.64522, error0.005
0
EXERCISES 11.4 N 1. a Nothing
PAGE 709
27. C
san d
11. C 13. C
23. C 25. D
33. 1.249, error0.1 45. Yes
7. C 19. D
b C 15. C
3. C
17. D
5. D
31. D
ssn d
29. C
35. 0.76352, error0.001
sn
an
1. a A series whose terms are alternately positive and negative b 0bn1bn and limnl bn0,
EXERCISES 11.5 N
0 11
where bnan3. C 5. C 15. C 17. C
c Rnbn1
7. D 9. C 11. C 13. D
19. D
Copyright 2008 Thomson Learning, Inc. All Rights Reserved. May not be copied, scanned, or duplicated, in whole or in part.
A20APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
21. 1.0000, 0.6464,
0.8389, 0.7139, 0.8033,
0.7353, 0.7893, 0.7451, 0.7821, 0.7505; error0.0275
1
EXERCISES 11.9 N PAGE 733
ssnd 1
EXERCISES 11.6
N PAGE 719
13.
b c
15. 19.
25. 4
27. 0.9721
33. p is not a negative integer
x n, 1, 1
1. 10 3.1nx n, 1, 1 5. 2n0 n0 3
x n, 3, 3 7.1n x2n1, 3, 3 9. 12x n, 1, 1
0
10
a 1nn1xn,R1
n0 1
1nn2n1x n, R1 2 n0
n1
sand 1
23. 5
31. An underestimate
35. bnis not decreasing
29. 0.0676 11.1n1
1
n0 9n1 1
n1
n0
2
n1
Abbreviations: AC, absolutely convergent; CC, conditionally convergent
12 n2
1nnn1xn, R1
1. a D 3. AC 15. AC 25. AC
b C 5. CC
17. CC
c May converge or diverge
7. AC 19. AC
9. D 11. AC
13. AC
xn
ln 5 n , R5
n2
17.n1 xn, R2
21. AC
31. aandd
23. D
n1 n5
11n
n3 2 x2n1, R4
0.25
27. D
35. a 6610.68854, error0.00521
29. D
960
b n11, 0.693109
n0 16n1
EXERCISES 11.7 N
PAGE 722
s s
s
f s
1. C 15. C 27. C
3. D 17. D 29. C
5. C 19. C
31. D PAGE 727
7. D
21. C
9. C
23. D
13. C 25. C
37. C
11. C 35. C
sTM
4 4
sTM f
33. C
s s
EXERCISES 11.8 N
s s
0.25
1. A series of the form n0 cnxan, where x is a variable and a and the cns are constants
3. 1, 1, 1 9. 2, 2, 2
5. 1, 1, 1 11. 1 ,1 , 1
7. , ,13. 4, 4, 4
21.
2x2n1
,R1
222
s sTM
n0 2n1
15. 21.
27. 31. 35.
37.
1, 1, 3
b, ab, ab 23. 0, 1
17. 1, 13, 11 333
19. , ,25. 1, 1, 0
3 s f
, ,
kk 33. No
b No
2 2
3
29. a Yes a , b,c
2 ssTM s
242
J 8 8
2 s s s 1, 1, fx12×1×2 41. 2
23.
25.
27. 33.
t8n2 n0 8n2
, R1 x2n1
C
C 1n1
, R1
29. 0.000983 31. 0.09531
n1 0.199989 b 0.920
4n21
37. 1, 1, 1, 1, 1, 1
EXERCISES 11.10 N 1. b8 f858!
PAGE 746
3. n1xn,R1
n0
41.
n1
1n1 n1!
xn,R
2 n2
n1n2
53.
61.
65. 1s2
3 120 2 24
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
A21
6
5. n1xn,R1
T
T T
n0
2n1 34
x2n1,R
x2n1
11.,R n0 2n1!
13. 12x13x12 4×13 x14, R
f
TTM TT
7. 1n n0
T
TTM
T
T T T
2n1! 9.xn,R
5n n0 n!
f
6
15.
17.
19.
25.
27.
29. 31.
33. 35. 37. 39.
e3 n0 n!
x3n, R
x 2n,R
43. 45.
b
47.
49.
0.81873 a 1
n1
1352n1 x2n 2nn!
1 1n1
1352n1 x x2n1
n0 2n!
n1 2n12nn!x6n2
C1n ,R n0 6n22n!
C 1n
1352n1
1n
n0 23n!
x9n, R9
1 x1n1 1352n3 xn,R1
n 2n1
n
2 n!
xn, R2 x2n1,R
x2n,R 51. 0.440 0.40102 55. 1 57. 1 59. 13×2 25×4
11x27x4 63.ex4 6 360
3
67. e1
1
n1 2n 2n!
1n 1n
2 n4 2n1
2n1!
n0
n0
2n1
EXERCISES 11.11
N
PAGE 755
xn , R
12
1. a T0x1T1x,T2x12x T3x,
n0 n!
1n n0
1x1n
T4x12x 24x T5x, Tx11x21x41 x6
11 24
1
2 2 n 2n!
6
2 24 720
x4n1 , R 1352n1
x2n1,R2
TT
T
2
2 TTMT
T T
2
2
1n1 n1
n!2 x2n,R
3n1
n1
f
22n1 2n!
2
1n
n0
1
2n!
x4n, R 1.5
1.5
x
f
T0T1
T2T3
T4T5
T6
4 2
0.7071
0 1
1
1 1
0.6916
0.2337 3.9348
0.7074
0.0200 0.1239
0.7071
0.0009 1.2114
b
1.5
T TTTMT
1.5
TT T T f
TTTT
c As n increases, Tnx is a good approximation to f x on a larger and larger interval.
A22APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
3. 1 1x21x221 x23
1 1 2 5 13. a 24x464x4 b 1.562510
2 4
8 16
2
0
15. a 12x11x124 x13 3 9 81
17. a 11×2 b 0.0015 19. a 1×2 f2
21. a x21 x4 b 0.042 23. 0.17365 T 6
b 0.000097 b 0.00006
25. Four 27. 1.037×1.037 29. 0.86×0.86
4
31. 21 m, no 37. c They differ by about 8109 km.
1.1
CHAPTER 11 REVIEW N TrueFalse Quiz
PAGE 759
5. False 13. True
13 5.x26 x2
1. False 9. False 17. True
Exercises
3. True 11. True
19. True
7. False 15. False
7. x1x3 6
T
f
0
2
1.6
1. 1 2
13. 25. 37. 41.
45.
47.
51.
53.
5 5 . 57.
b
5. 0
17. C
7. e12
19. C
11. C 21. C
f
27. 1 11
29. 4
3. D 15. D
9. 2 31. ee
T
C
AC
0.18976224, error6.4107 4, 6, 2 43. 0.5, 2.5, 3.5
23. CC 35. 0.9721
1.1
1
12n s3 2n1 1n xx
1 1
2n! 1nxn2,R1
6
2n1! 6xn
49.,R1 n1 n
2 n0
T f
1
T f
x 8 n4
1n ,R
n0
n1
Cl nxx n
1.6
3
n0 1
2n1!
1594n3
2 n!2 xn,R16
9. x2x2 2×3
6 n1
T
283 11.T5x12x4 2×4 3 x4 f
n1 nn!
a 11x11x121 x13
1.5
2 8 16
1.5
c 0.000006
4
10 4 64 5 3×4 15×4
0
59. 1 6
2
T T
5 T
PROBLEMS PLUS N
1. 15!5!10,897,286,400
3. b 0ifx0,1xcotxifxk ,kaninteger
f TTM
PAGE 762
2 02
5. a s 34n,l 13n,p 4n3n1 c 2 3
T
TTM n n n 5s
4 f T T
9. 1, 1,
13. a 250 101
x3 4×2 x
1×4 11.
en1 5en 5
1 ln 2
b 250 101
2
CHAPTER 12
EXERCISES 12.1
1. 4, 0, 3
5. A vertical plane that
intersects the xyplane in the line y2x, z0 see graph at right
7. a 4,2 y
B2, 1
N PAGE 769 3. Q; R
A2, 3
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA23
9. a 3,1 y
A1, 3
B2, 2
y2x a0x 0ax y2x, z0
z
0
2 y xz
2
11. a2, 0, 2 A0, 3, 1
13. 5, 2 y
k6, 2l k1, 4l
k5, 2l 0x
7. PQ 6, QR 2s10, RP 6; isosceles triangle 9. a No b Yes
11. x12 y44 z32 25;
x12 z32 9,y0acircle
13. x32 y82 z12 30
15. 3, 2, 1, 5
17. 2, 0, 6, 9s2 19. b 5, 1 s94, 1 s85 222
21. a x22 y32 z62 36 b x22 y32 z62 4
c x22 y32 z62 9
23. A plane parallel to the xzplane and 4 units to the left of it
25. A halfspace consisting of all points in front of the plane x3 27. All points on or between the horizontal planes z0 and z6 29. All points on or inside a sphere with radius s3 and center O
31. All points on or inside a circular cylinder of radius 3 with axis the yaxis
0
a
y
x
15.
B2, 3, 1
0, 1, 1 z
k0, 1, 2l
k0, 0, 3l
k0, 1, 1l
17. 2, 18, 1, 42, 13, 10
x
y
19. ij2k,4ij9k,s14,s82 21. 3 i 7 j 23.8i1j4k
33. 0x5 37. a 2,1,4
35. r2 x2 y2 z2 R2
s58 s58 9 9 9
25. 2, 2s3 27. 45.96 fts, 38.57 fts
29. 100s7264.6 N, 139.1
31. s49322.2 mih, N8W
33. T1 196i3.92j,T2 196i3.92j
b L C
x
P
z
0
LTM Ay
B
d s9,t11 77
35. i4 js17 39. a,b
b
41. A sphere with radius 1, centered at x0, y0, z0EXERCISES 12.3 N PAGE 784
1. b, c, d are meaningful
3.14 5.19 7.32 9.15 11. uv1,uw1
37. 0
39. 14x6y10z9, a plane perpendicular to AB EXERCISES 12.2 N PAGE 777
1. a Scalar b Vector c Vector d Scalar
llllllll
3. ABDC,DACB,DEEB,EACE
5. a u
b
uv
uv
v
u
c
vw d v 15. cos1 94s795 17. cos1 5 v22
81
w v w wvu
20 s1015
u
19. cos11101 21. 45, 45, 90 2s7
y
0
sa
a
c
x
tb
A24APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES 23. a Neither b Orthogonal
c Orthogonal d Parallel
25. Yes 27. ijks3 orijks3
39.
z 41. z 0, 0, 10
0, 2, 0
29. 3 , 4 , 1 ; 65, 56, 45 5s2 5s2 s2
31. 2, 3,6; 73, 65, 149 777
33. 1s3, 1s3, 1s3; 55, 55, 55 35. 3,9,12 37. 9,27,54,18
39.1s21,2 i1 j4 k 21 21 21
43. 0, 0, 2 s10or any vector of the form s, t, 3s2s10, s, t
45. 144 J 47. 2400 cos401839 ftlb 51. cos11s355
0, 0, 320
1,0,0
0
0, 2, 0
x
5 5 7 49 49 49
5,0,0
x
49. 13 5
45.
47.
2, 3, 5 Perpendicular
a x1,yt,zt x1,y2z
43. 49.
55. 57.
59. 63.
65. 67.
EXERCISES 12.6 N 1. a Parabola
2, 3, 1
1, 0, 1 51. Neither, 70.5
53. Parallel b cos15 15.8
EXERCISES 12.4 N 1. 16i48k
7. t4i2t3jt2k 9. 0
13. a Scalar b Meaningless
d Meaningless e Meaningless f Scalar
15. 24; into the page 17. 5, 3, 1, 5, 3, 1
19. 2s6, 1s6, 1s6, 2s6, 1s6, 1s6 27. 16 29. a 6, 3, 2 b 7
31. a 13, 14, 5 33. 82 35. 3
3s3
61. xaybzc1
PAGE 792
3. 15i3j3k
5. 1 ij3k 22
x2yz5 x3t,y1t,z22t
P1 and P3 are parallel, P2 and P4 are identical
75. 1s6
11. ijk c Vector
s6114 69. 18 7
2 2
71. 52s14 PAGE 810
b 1 s390
39. 10.8 sin 8010.6 Nm
41. 417 N 49. a No
43. b s973 b No c Yes
b Parabolic cylinder with rulings parallel to the zaxis c Parabolic cylinder with rulings parallel to the xaxis
EXERCISES 12.5 N PAGE 802
1. a True b False c True
f True g False h True
k True
3. r2i2.4j3.5kt3i2jk; x23t, y2.42t, z3.5t
5. ri6kti3jk;
x1t, y3t, z6t
3. Elliptic cylinder z
x
7. Cylindrical surface
x
5. Parabolic cylinder z
d False
i True
e False j False
y
z
y
7. x15t,y3,z22t; x1z2,y3 1 52
9. x22t,y12t,
z34t;
x222y2z34
11. x1t,y12t,z1t; x1y12z1
13. Yes
15. a x11y52z63 b 1, 1, 0, 3 , 0, 3, 0, 3, 3
17. rt2ij4kt2i7j3k,0t1 19. Parallel 21. Skew
y
9. a xk,y2 z2 1k2,hyperbolak1; yk,x2 z2 1k2,hyperbolak1;
zk,x2 z2 1k2,circle
b Thehyperboloidisrotatedsothatithasaxistheyaxis c Thehyperboloidisshiftedoneunitinthenegative ydirection
22
23. 2xy5z1
27. 2xy3z0
31. xyz2
35. 33x10y4z190 37. x2y4z1
25. xyz1 29. 3x7z9
33. 13x17y7z42
x
y
y
11. Elliptic paraboloid with axis the xaxis z
x y2 z2 31. 632
Elliptic paraboloid with vertex 0, 0, 0 and axis the xaxis
z
x
y
13. Elliptic cone with axis the xaxis z
y
z32 1 Ellipsoid with center 0, 2, 3
0, 4, 3
x
15. Hyperboloid of two sheets
17. Ellipsoid
19. Hyperbolic paraboloid
21. VII
x2 y2 z2
z
z
35. y12 x22 z12 Circular cone with vertex 2, 1, 1
2,1,1 z
x
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
A25
y22 33. x24
z 0, 0, 3
and axis parallel to the yaxis
37.
y
0
xy
y
x
23. II
29.94361
x
27. VIII
Hyperboloid of two sheets with axis the zaxis
0, 0, 6
x
y
2 z1 0
1
25. VI
b Circle
c Ellipse
y0 0x 11
x
y
0, 0, 1
z
41. z z2
z
0
xy
43. yx2 z2 45. 4xy2 z2,paraboloid
x2 y2 z2
47. a 6378.13726378.13726356.52321
z
51.
0, 0, 6
1
y
4
z0 z0
4 4 2
40 0x 20220x
2
0, 6, 0
x 1, 0, 0 y y 4 4 y
39.
2
A26APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
CHAPTER 12 REVIEW N TrueFalse Quiz
PAGE 812
5. True 7. True 9. True
7.
y
9.
z
1. True 11. False
3. True 13. False
15. False
17. True
1
0, 1, 0
y , 1, 0
Exercises
1. a x12 y22 z12 69
b y22 z12 68,×0
c Center 4, 1, 3, radius 5
3. uv3s2;uv3s2;outofthepage
5. 2,4 7. a 2 b 2 c 2 d 0
11.
z
x
13. z
b x2 y2 t2 1,zt CHAPTER 13
EXERCISES 13.1 N PAGE 822 1. 1, 2 3. 1, 0, 0
c 4 3
5. ijk
10
x
9. cos11 71 11. a 4, 3, 4 3
13. 166N,114N
15. x43t,y12t,z23t 17. x22t,y2t,z45t 19. 4x3yz14 21. 1,4,4
b s412
23. Skew 27. 22s26
25. xyz4 31. Cone
rtt, 2t, 3t, 0t1; xt, y2t, z3t, 0t1
17. rt3t1, 2t1, 5t2, 0t1; x3t1, y2t1, z5t2, 0t1
15.
x
y 1x
y
29. Plane zz
0
27.
0, 0, 0, 1, 0, 1
19. VI 25.
21. IV z
23. V
x
y
x
y
0, 2, 0
0
33. Hyperboloid of two sheets zz
y
1, 1, 0
x
29.
1
z0
1 1
10
z0
10
35. Ellipsoid
0, 1, 2
y
0, 2, 0
x
37. 4×2 y2 z2 16
1 0x
1
0 10 10 0 y x
xy
0, 1, 2
PROBLEMS PLUS N
PAGE 815
y
01
1. s31.5 m
3. a x12cycc2 1zcc2 1
31.
10
y
33.
2
z
rtt cos tsin t, 2t, cos 2t2t sin 2t
rt4e4t k 13. rt2tet2 i313t k
rtb2tc 17. 1,2,2 19. 3 j4 k 33355
1, 2t, 3t2, 1s14, 2s14, 3s14, 0, 2, 6t, 6t2, 6t, 2 x3t,y2t,z24t
x1t, yt, z1t
xt,y1t,z2t
0 2
9.
11.
15.
21.
23.
25.
27.
29.
31.
37.
y0
22
2
37. rtti1t2 1j1t2 1k
39. x2 cos t, y2 sin t, z4 cos2t
EXERCISES 13.2 N 1.a y
b, d
PAGE 828
h l 0 3. a, c
3, 2
ra1
5. a, c y
h
r4
b rt1, 2t
y
1
CR r4.5
Q
T4 P
29
19. a 1 s2et, e2t, 1, e2t 1
1 1e2t, s2et, s2et e2t 1
r a40x
r4
b rtcos t i2 sin t j
ra0 1, 1
22
r4.5r4 1Q
r4.5 r4.2
EXERCISES 13.3 N
PAGE 836
r4 01x
13.rts 2 si 1 3 s j 5 4 s k
r4.5r4 0.5
r4.2
21. 24t1 23. 25 25. 7 s14
27. 24×28×532 29. 15sx1100×332
r4 01x
31. 1 ln 2, 1s2; approaches 0 2
0
x
c r4 lim r4hr4;T4 r4
ykx
r4.2r4 P
1. 20s29 9. 1.2780
3. ee1 11. 42
5. 1 1332 8 27
7. 15.3841
r4.2r4 0.2
s29 s29 s29 15. 3 sin 1, 4, 3 cos 1
17. a 2s29cos t, 5s29, 2s29sin t, sin t, 0, cos t b 2
y
4 4 1
41.
Yes
x t,y t,z t
t2it3j2t32 2k CR 33
45. 2t cos t2 sin t2 cos t sin t
r1
0x kt
7. a, c y
2 ,22
r0 1 01x
333333333
b rtet i3e3t j
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA27
39.
66 33. 4i3j5k
et it2 jtlnttkC
35. ijk
b s2e2te2t 12
232 4 119
33. a P
b 1.3,0.7 35.
4
37. aisyfx,bisy x
6s4 cos2t12 cos t13 1712 cos t32
integer multiples of 2
0 2 4 6 t
41. 1s2et 43. 2,2,1,1,2,2,2,1,2
45. y6x ,x6y6
39. t
yx
A28APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
47. x52 y2 81,×2 y52 16 19. t4 21. rttitj5t2 k,vts25t2 2 2439 2
7.5
2.5
b23.6 upstream 12
0 40
12
etet, s2 41. t1
5 49. 1, 3, 1 57. 2t44t21
59. 2.071010 A2 m EXERCISES 13.4 N PAGE 846
1. a 1.8i3.8j0.7k, 2.0i2.4j0.6k, 2.8i1.8j0.3k, 2.8i0.8j0.4k
b 2.4i0.8j0.5k, 2.58
3. vtt, 1
CHAPTER 13 REVIEW N TrueFalse Quiz
PAGE 850
5. False
at 1, 0vt st21
5. vt3 sin t i2 cos t j at3 cos t i2 sin t j vt s5 sin2 t4
7. vti2tj z at2j
vt s14t2
x
9. 2t, 3t2, 2t, 2, 6t, 2, ts9t28
11. s2iet jet k,et jet k,et et
1. True 9. False
Exercises 1. a
x
3. False 11. True
7. True
0, 1, 0 y
2, 1, 0
0x
y
15. vtti2tjk,rt1t2 1it2 jtk 2
b rti sin tj cos tk, rt 2cos tj 2sin tk
13. etcostsintisintcostjt1k,
et2sinti2costjt2k,etst2 2t3
17. a rt1t3titsin t1 j11 344
cos 2tk
b
0.6 z 0.4 0.2 0
200 0 x
10
200
10
0y
5
23. a 22 km b 3.2 km c 500 ms
1, 1, 2
a t2, t, 1st4t21
2t, 1t4, 2t3tst84t62t45t2
v2 y 2, 2
a2
0, 2
0x
a1 v1
3. 5.
11.
rt4 cos t i4 sin t j54 cos tk, 0t2 1i22j2k 7.86.631 9. 2
3
y
v3
32 , 3
a33, 0
25. 30 ms 29. 13.031. a 16 m
20
0 40
4
33. 6t, 6 35. 0, 1 37. 39. 4.5 cms2, 9.0 cms2
27. 10.2, 79.8
36.0, 55.4 85.5
z
b
c st 4t 2t 5tt t 1
13. 121732 15. x2y2 0
17. vt1lntijet k, vts22lntlnt2 e2t,at1tiet k 19. a About 3.8 ft above the ground, 60.8 ft from the athlete b 21.4 ft c 64.2 ft from the athlete
21. c 2et vdet R
PROBLEMS PLUS N PAGE 852
1. a v Rsin ticos tj c a 2r
3. a 90, v022t
5. a 0.94 ft to the right of the tables edge, 15 fts b 7.6 c 2.13 ft to the right of the tables edge 7. 56
8642422
CHAPTER 14
EXERCISES 14.1 N
21. z3,horizontalplane z
23. 4x5yz10,plane z
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA29
PAGE 865
0, 0, 10
1. a 27; a temperature of 15C with wind blowing at 40 kmh feels equivalent to about 27C without wind.
b When the temperature is 20C, what wind speed gives a wind chill of 30C? 20 kmh
c With a wind speed of 20 kmh, what temperature gives a wind chill of 49C? 35C
d A function of wind speed that gives windchill values when the temperature is 5C
e A function of temperature that gives windchill values when the wind speed is 50 kmh
3. Yes
5. a 25; a 40knot wind blowing in the open sea for 15 h will cre ate waves about 25 ft high.
b f 30, t is a function of t giving the wave heights produced by 30knot winds blowing for t hours.
c f v, 30 is a function of v giving the wave heights produced by winds of speed v blowing for 30 hours.
0 x
y
0
0, 2, 0
25. zy21, parabolic cylinder
2.5, 0, 0
x
x
z
7. a 4 b 2 c 0,
9. a e b x,y,zzx2 y2
c 1, y
0x
y
19 1 0x
11. x,yyx
13. x,y1x2 y2 1 9
yx
27. z4x2 y2 1 elliptic paraboloid
29. zsx2 y2, top half of cone
15. x,y1x1,1y1
y
33. Steep; nearly flat 35. z 37. z
x
y
10 1 x
41. ylnxk
zz
0, 0, 1
x
x
y
0
y
31. 56, 35 15
0
0
17. x,yyx2,x1 y
1 1x
1
14
x y
39. y2x2 k yy
x
y
19. x,y,zx2 y2 z2 1
z
0
2 1 0 1
y
4321 0 1234
x0x 2
y
y
A30APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
43. ykex 45. y2 x2 k2 yy
123 3 2
local maximum points but no local minimum point.
0
0z0
47.
x2 9y2 k
2 2 0y
49.
51.
y
0x
3. 5 2
z
2
1
0x0x 1
2 2 102 13 3 0
x
2
0 3
35. x,y,zy0,ysx2 z2
3
yz 1 234
z4 z3
z2 z1
y
The function values approach 0 as x, y become large; as x, y approaches the origin, f approachesor 0, depending on the direction of approach.
71. If c0, the graph is a cylindrical surface. For c0, the level curves are ellipses. The graph curves upward as we leave the ori gin, and the steepness increases as c increases. For c0, the level curves are hyperbolas. The graph curves upward in the ydirection and downward, approaching the xyplane, in the xdirection giving a saddleshaped appearance near 0, 0, 1.
73. c2, 0, 2 75. b y0.75×0.01 EXERCISES 14.2 N PAGE 877
13. 0 15. Does not exist 17. 2 19. 1 21. Does not exist
23. The graph shows that the function approaches different num bers along different lines.
25. hx,y2x3y62 s2x3y6; x, y2x3y6
27. Along the line yx 29. x, yyex2 31. x,y y0 33. x,y x2 y2 4
0x
2 0 22 0 2 00xyx
d Reflect the graph of f about the xyplane and then shift it
y33
b II 57. a F b VI
b I
a C
a B
Family of parallel planes
Family of hyperboloids of one or two sheets with axis
55.
59.
61.
63.
the yaxis
65. a Shift the graph of f upward 2 units
b Stretch the graph of f vertically by a factor of 2 c Reflect the graph of f about the xyplane
37. x, yx, y0, 0 43.
2 1 0 1
39. 0
2
41. 1
upward 2 units
f is continuous on 2 EXERCISES 14.3 N PAGE 888
67.
20 0 20
405
1. a The rate of change of temperature as longitude varies, with latitude and time fixed; the rate of change as only latitude varies; the rate of change as only time varies.
b Positive, negative, positive
z
0 5 y055x
f appears to have a maximum value of about 15. There are two
53.
z0
x
69.
10
5
z
5
1. Nothing; if f is continuous, f3, 16
5. 1 7. 2 9. Does not exist 11. Does not exist
7
20 0x y22
3. a fT 15, 301.3; for a temperature of 15C and wind speed of 30 kmh, the windchill index rises by 1.3C for each degree the temperature increases. fv15, 300.15; for a temperature of 15C and wind speed of 30 kmh, the windchill index decreases by 0.15C for each kmh the wind speed increases.
33. uxy sin1yz, uyx sin1yzxyzs1y2z2, uzxy2s1y2z2
35. fxyz2 tanyt, fyxyz2t sec2ytxz2 tanyt,
fz2xyz tanyt, ftxy2z2 sec2yt
37. uxixisx12x2xn2
39. 1 41. 1 54
43. fxx,yy2 3x2y, fyx,y2xyx3 45. z3yz2x, z3xz2y
x 2z3xy y 2z3xy
b Positive, negative
5. a Positive b Negative 7. a Positive b Negative 9. cf,bfx,afy
11.
13.
16 16
1, 2, 8
1, 2, 8
25.
2r2
utewt1wt, uwewt
2rs
27.
29.
31.
wz3x2y3z
23.
25.
27.
29.dR 2cos d 2 cos d31. z0.9225, dz0.9 33. 5.4 cm2
c 0
fx1, 28slope of C1, fy1, 24slope of C2 zz
z
x
C
0 4 0CTM4
2y2y
65. er 2 sincosr sin67. 69. 12.2,16.8,23.25 81. R2R21
4y2z3, 0
1, 2 x fx 2x2xy, fy 2yx2
10 f z
0
2
10 z0
10 2
z0
1, 2
002 1 x 2 2 y
y0 01 11x
fx
,fyx,y
e No, since fxy and fyx are not continuous.
002 x 2 2 y
fy
2 1
002 x 2 2 y
z200 z0 1
fxx,y3y,fyx,y5y4 3x
15.
17.
19.
21.
23. wcos cos , w
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA31
z 1y2z2 z
47. x1yy2z2, y1yy2z2
49. a fx, ty b fxy, fxy
51. fxx6xy524x2y, fxy15x2y48×3fyx, fyy20x3y3 53. wuuv2u2v232, wuvuvu2v232wvu,
wvvu2u2v232
55. zxx 2x1x22,zxy 0zyx,zyy 2y1y22 61. 12xy, 72xy
63. 24 sin4x3y2z, 12 sin4x3y2z
87. No 93. 2
95. a
0.2 z0 0.2
89. x1t,y2,z22t
x4y4x2y3 y5 b fxx,y 2 2 2
x5 4x3y2 xy4 2 2 2
c 0, 0
EXERCISES 14.4 N
1. z8x2y 3. xy2z0 5. zy
7.
400
0
9 9 1005y22x
fxx,t etsin x,ftx,tetcos x zx202x3y , zy302x3y
fxx, y2yxy2, fyx, y2xxy2
00
sin
frr,s r2 s2 lnr2 s2,fsr,s r2 s2
11. 19.
x 10 5 0y
2x1y1 13. 1x2y2 15. 1 y 4 993
2x7y20;2.846 21. 3x2y6z;6.9914 333 777
fxz10xy3z4, fy15x2y2z4, fzx20x2y3z3 wx1x2y3z,wy2x2y3z,
2sin d
35. 16 cm3
sin
xy
xy
PAGE 899
4TH329; 129F
dz3×2 lny2dx2x3y dy dm5p4q3 dp3p5q2 dq
9.
A32APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
37. 150 39. 10.05917
43. 1 x,2 y EXERCISES 14.5 N PAGE 907
1. 2xycost2yxet
3. xtysints1x2 y2
41. 2.3
43. a xyz1 45.
2
b x1yz
47. 2, 3, 2x3y12 y
I
yz2 z1 5. e 2txz2xyz
2x3y12
7. zs2xy3cos t3x2y2 sin t,
zt2sxy3sin t3sx2y2cos t 00x 9.zst2cos cos 2stsin sin ,
zt2st cos coss2 sin sin
11.zertcos s sin, x12 1y2 1
s ss2t2 z er scost sin
53. No 59. x110t,y116t,z212t
63. Ifu a,b andv c,d,thenaf bf andcf df are
t ss2 t2
13. 62 15. 7, 2
17. uu xu y,uu xu y,
xyxy known, so we solve linear equations for fx and fy.
r x r y r s uu xu y
x s
y s
EXERCISES 14.7 N PAGE 930
1. a f has a local minimum at 1, 1.
b f has a saddle point at 1, 1.
3. Local minimum at 1, 1, saddle point at 0, 0
5. Maximumf1,111 2
7. Minima f1, 10, f1, 10, saddle point at 0, 0
9. Saddle points at 1, 1, 1, 1
11. Minimum f 2, 18, saddle point at 0, 0
13. None 15. Minimum f 0, 00, saddle points at 1, 0 17. Minima f 0, 1f, 1f 2 , 11,
saddle points at2, 0, 3 2, 0
21. Minima f1, 13, f1, 13
23. Maximum f3, 33 s32,
minimum f 5 3, 5 33 s32, saddle point at,
25. Minima f 1.714, 09.200, f 1.402, 00.242, saddle point 0.312, 0, lowest point 1.714, 0, 9.200
27. Maxima f 1.267, 01.310, f 1.629, 1.0638.105, saddle points 0.259, 0, 1.526, 0,
highest points 1.629, 1.063, 8.105
29. Maximum f 2, 09, minimum f 0, 314
31. Maximum f1, 17, minimum f0, 04
33. Maximum f3, 083, minimum f1, 10
35. Maximum f 1, 02, minimum f 1, 02
t x t y t
19. ww rw sw t,
w y 21.
27. 31.
33.
x r x s x t x w r w s w t
r ys yt y
85, 178, 54
4xy32y
x2x2sxy
23. 9 , 9 77
3yz2x 3xz2y ,
2z3xy 2z3xy
1y2z2 z 1yy2z2 ,1yy2z2
11. 2310
19. 25
13. 8s10 21. 4s2, 1, 1
17. 92s5 23. 1, 0, 1
2
3 2y
1 0 1 x
36, 24, 30 sinxyey 29. sinxyxey
2Cs 37.0.33 ms per minute a 6 m3s b 10 m2s c 0 ms
35.
39.
41.
45.
zzxr sinzyr cos
51. 4rs 2zx24r24s2 2zx y4rs 2zy22 zy
EXERCISES 14.6 N PAGE 920
1. 0.08 mbkm 3. 0.778 5. 2s32
0.27 Ls 43. 112s3rads a zrzx coszy sin
,
7. a fx, y2 cos2x3y, 3 cos2x3y b 2, 3 c s33
37. 1, 0, 0 0
1 z
1, 2, 0
2
9. a e2yz, 2xze2yz, 2xye2yz
b 1, 12, 0 15. 4s30
c 22 3
25. 1, 3, 6, 2 27. b 12, 92
29. All points on the line yx1 31. a 403s3
4 2
25.
47. 4 49. Cube, edge length c12 41. a4x5yz4 b 3
x2 y1 z1
4 5 1
39. s3 41. 2, 1, s5, 2, 1, s5 43. 100, 100, 100 13 333
33. a 32s3 b 38, 6, 12 c 2s406 35. 327
39. a xyz11 b x3y3z5
51. Square base of side 40 cm, height 20 cm 53. L 33s3
45. 8r33s3
xy6
3, 2
f 3, 2
EXERCISES 14.8 N PAGE 940
1. 59, 30
3. No maximum, minima f1, 1f1, 12
5. Maxima f2, 14, minima f2, 14
7. Maximum f 1, 3, 570, minimum f 1, 3, 570
9. Maximum 2s3, minimum 2s3
11. Maximum s3, minimum 1
15. tutan1v,tvu1v2
17. Tp lnqer,Tq pqer,Trperqer 19. fxx 24x, fxy 2yfyx, fyy 2x
21. fxxkk1xk2 ylzm, fxyklxk1yl1zmfyx, fxzkmxk1ylzm1fzx, fyyll1xk yl2zm,
fyzlmxk yl1zm1fzy, fzzmm1xk ylzm2
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA33
13. Maximum f1,1,1,12,
2222 84
25. a z8x4y1 b x1y2 1z x2 y1 z1
b x38t,y42t,z14t 31. 2, 1, 1, 2, 1, 1
33. 60×24 y32 z120; 38.656 55
35. 2xy316p3x2y2pepep4z3p cos psin p
minimum f1,1,1,12 2222
15. Maximum f1,s2,s212s2,
minimum f1,s2,s212s2
17. Maximum 3 , minimum 1 22
19. Maxima f 1s2, 12 s2 e 14, minima f 1s2, 12 s2 e14
2737. See Exercises 3949 in Section 14.7.
39. L33s3
41. Nearest 1 , 1 , 1 , farthest 1, 1, 2 222
43. Maximum 9.7938, minimum 5.3506 45. a cn b Whenx1 x2 xn
27. a 2x2y3z3 b 446 29. a 4x y2z6
22
37. 47, 108 43. 47. s1452, 4, 9
zex sy zsy, xz2sy, 2 49. 5 knotmi
45. 43 5
CHAPTER 14 REVIEW N TrueFalse Quiz
PAGE 944
5. False
53. Maximum f 1, 11; saddle points 0, 0, 0, 3, 3, 0
55. Maximum f 1, 24, minimum f 2, 464
57. Maximum f 1, 02, minima f 1, 13, saddle points 1, 1, 1, 0
59. Maximum f s23, 1s3 23 s3 , minimum fs23,1s323s3
61. Maximum 1, minimum 1
63. 314, 314s2, 314 , 314, 314s2, 31465. P2s3, P3s36, P2s333
1. True 11. True
Exercises
3. False
7. True
9. False
51. Minimum f 4, 111
28
1.
x,yyx1 3. z y
1
1 xy
1 x 1
yx1
y 7.y 2
345 12
01 x
PROBLEMS PLUS N
1. L2W2,1L2W2 4
7. s62, 3s22 CHAPTER 15
EXERCISES 15.1 N
PAGE 948
3. a xw3,basew3
PAGE 958
b Yes
5.
2
9.
11. a 3.5Cm, 3.0Cm
Equation 14.6.9 Definition 14.6.2 gives 1.1Cm. c 0.25
13. fx 1s2xy2, fy ys2xy2
012x
1. a 288 b 144
3. a 224.935 b 0
5. a 6 b 3.5
7. UVL
9. a 248 b 15.5
11. 60 13. 3
15. 1.141606, 1.143191, 1.143535, 1.143617, 1.143637, 1.143642
3
EXERCISES 15.2 N 1. 500y3, 3×2
PAGE 964
3. 10 15. 2
b0.35Cm by
7. 261,63245 17. 9 ln 2
9. 21 ln 2 2
11. 0 13.
19. 1s31 1
21. 1e2 3
5. 2
21
2 12 2
A3423. z
4
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
01 1y
45. 1e9 1 47. 1 ln9 49. 12s21 51. 1 633
53.16e116 xx ex2y22 dA 16 55. 3 Q4
59. 8 61. 2 3 EXERCISES 15.4 N PAGE 978
1. x3 2 x4 frcos ,rsin rdrd 3. x1 xx12 fx,ydydx 00 10
x
5. y
0
33 2
25. 47.5 27. 166 33. 21e57
31. 64 27 3
2
z
29. 2
47
x
R
0 1432
0
7. 0 9. 2
sin 9 11.21e13. 64 17.1 2 19.16 21.4
y
0
15. 12
11x
Fubinis Theorem does not apply. The integrand has an infinite
EXERCISES 15.3 N PAGE 972 1. 32 3. 3 5. e1
833
35. 5
37.
discontinuity at the origin.
23. 4 a3 63
25. 2 311s2 27. 8 36424s3
7. 4 9.
10 3 22
29. 1 1cos 9 2
33.1800 ft3 EXERCISES 15.5 N
31. 2s23
35.15 37.as 4 bs 2
16
PAGE 988
11. 1e16 17
13. 11cos1 15. 147 17. 0 19. 7 21. 31 2 20 188
644433
3. 3 , 3 , 0 5. 6, 4 , 2e2 1 4e3 1
23. 33.
6
1,0,0
x
z
0
4 2e219e21
25. 128 27. 1 29. 0, 1.213, 0.713 31. 64 153 3
1. 3 C
7. 1e2 1, ,
0, 0, 1
9. L4, L2, 16911. 3 , 3 16 13. 0, 45148
15. 2a5, 2a5 if vertex is 0, 0 and sides are along positive axes 17. 1 e4 1,1e2 1, 1 e4 2e2 3
19. 7ka6180, 7ka6180, 7ka690 if vertex is 0, 0 and sides are along positive axes 2 1 16
21.m28,x,y 3 ,9 ,Ix3264, Iy1 43 2,I0 4169 264
16
23. bh33, b3h3; bs3, hs3
25. a4 16, a4 16; a2, a2
27. a 1 b 0.375 c 50.1042 2 48
29. b i e0.20.8187
ii 1e1.8 e0.8 e1 0.3481 c 2,5 31. a 0.500 b 0.632
33. a xx k2020sxx02 yy 2dA,whereDis D0
the disk with radius 10 mi centered at the center of the city
0, 1, 0
16 8 16
13,984,735,61614,549,535
37. 2 3 s9x2
y
3 0y03x
35. 39. y
2
x4 0 y0 4 x
43. xln2 x2 fx,ydxdy 0 ey
x2x4
yx
41. x3 x0
3
fx,ydydx 9
0 y2 fx,ydxdy
k3209k, 200
28 k136k, on the edge 9
PAGE 998
y
ln 2
ylnx or xe
b 200 EXERCISES 15.6
N
1. 27 3. 1
7. 1
9. 4
5. 1e3 1
1 16
11. 65 4 3 3 28
x2
y0
13. 83e 15. 60 17. 16 3 19. 3
21. 36
23. a x1 xx xs1y2 dzdydx b 1 1
012x
00043 25. 60.533
y
27.
29.
1 31. x2
z
1
1.a z b
z
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES EXERCISES 15.7 N PAGE 1004
A35
2
4,3,5
0 1y
x
x2 2
2, 4 , 10
x4x2 xs4x2y2 0 s4x 2y2
5
21 40
f x, y, z dz dy dx f x, y, z dz dx dy f x, y, z dx dy dz
f x, y, z dx dz dy x4x24z2 f x, y, z dy dz dx
4 x3
x4 xs4y xs4x2y2
y
0 x1
s4y s4x 2y2 x44z2 xs4y4z2
y
1
x4 xs4y2
s4y4 z 2 xs4y4z2
x
2, 2s3, 5 3, 2
0
0 s4y2
s2, s2, 1 3. a s2, 7
s4y4 z 2 2 s4x22 0
4, 4
5. Vertical halfplane through the zaxis
2, 4
x2 xs4x22
b b r2sin
x1
9. a zr2 11.
xs44z2 x4x24z2 f x, y, z dy dx dz s44z2 0
7. Circular paraboloid
x4 x2y2 fx, y, z dz dy dx 2 x2 0
z
1 z1
x4 xsy x2y2 fx,y,zdzdxdy 0 sy 0
x2 x42z xsy 00 sy
x4 x2y2 xsy 00 sy
fx,y,zdxdydz fx,y,zdxdzdy
2y
x2 x2x22 x42z fx, y, z dy dz dx 20 x2
x2 xs42z x42z fx, y, z dy dx dz 0 s42z x2
33. x1 x1 x1y fx,y,zdzdydx 0 sx 0
x1 xy2x1y fx,y,zdzdxdy 000
x1 x1z xy2 fx,y,zdxdydz 000
x1 x1y xy2 fx,y,zdxdzdy 000
x1 x1sx x1z fx,y,zdydzdx 00 sx
x1 x1z2 x1z fx, y, z dy dx dz 00 sx
13. 15.
17. 23.
25. 29. b
x2
Cylindricalcoordinates:6r7,0 2 ,0z20
z 4
x4 4y
384 19. e6 e5 a 162 b 0, 0, 15
64 3
35. x01 xy1 x0y fx,y,zdzdxdyx01 x0x x0y fx,y,zdzdydx x1 x1 x1 fx,y,zdxdydzx1 xy x1 fx,y,zdxdzdy
21. 2 5 a xxxC hPtP dV, where C is the cone
0zy 00y
x01 x0x xzx fx,y,zdydzdxx01 xz1 xzx fx,y,zdydxdz
Ka 28, 0, 0, 2a3 27. 0 3.11019 ftlb
37. 79 , 358 , 33 , 57139. a 5, 7a12, 7a12, 7a12 30 553 79 553
41. Ix Iy Iz 2kL5 43. 1 kha4 32
45. a mx3 xs9x2 x5y sx2 y2 dzdydx 3 s9x2 1
EXERCISES 15.8 N
PAGE 1010
1. a
z
1 0
b x, y, z, where
x1mx3 xs9x2 x5y xsx2 y2 dzdydx
1, 0, 0
0, 0, 1
1s2,1s6,s2 2 2
3 s9x2 1
y1mx3 xs9x2 x5y ysx2 y2 dzdydx
3 s9x2 1
z1mx3 xs9x2 x5y zsx2 y2 dzdydx
3 s9x2 1
c x3 xs9x2 x5y x2 y232 dzdydx
x
y
3 s9x2 1 47. a 311
32
b x,y,z
c 1 6815
240
49. a 1 b 1 8 64
51. L38
53. The region bounded by the ellipsoid x22y23z21
24
z
9 44 45 220 135 660
b
28 30 128 45 208 , ,
2,3,4 0
3
4
2
c 1
5760
x
y
A36APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
3. 5. 7. 9.
11.
13.
a 4, 3, 6 b s2, 3 2, 3 4
CHAPTER 15 REVIEW N TrueFalse Quiz
PAGE 1021
5. True
15. 17.
19. 21. 27. 31. 33.
35. 37. 41.
PAGE 1032
Halfcone
Sphere, radius 1, center 0, 1, 0 22
1. True Exercises
1. 64.0
3. True
7. False
5. 1 sin 1 7. 2 23
a cos2
sin2 z
2
b
2
2
2sin2
cos2
cos2
9
2
2224
xy
z
21. 81 5 23. 40.5 25. 96
31. 2 33. 2ma39 3
35. a1 b1,8 4 315
c Ix1 , Iy1 ; y1s3, x1s6 12 24
37. 0, 0, h4
39. 97.2 41. 0.0512
43. a 1 b 1 c 1 15 3 45
1 1z sy
45. x0 x0 xsy fx,y,zdxdydz 47. ln2
15
29. 176
49. 0
xy3
0
x
4, 0z
3
6
cos
9 42s3
PAGE 1024
7. b0.90
x2 x3 x2 frcos ,rsin ,zrdzdrd
000 312,500 7
s31 a33 0, 525 , 0
296
23. 15 16 29. a 10
25. 1562 15
b 0, 0, 2.1
a 0, 0, 3 a 8
4 1
PROBLEMS PLUS N 1. 30 3. 1 sin1
2
CHAPTER 16
EXERCISES 16.1 N 1. y
1
2 1 0 1 x 1
2
3. y 5. y 2
y
b 4K a515
2 311s2, 0, 0, 382s2
5 6
39. 4s2515 43. 136 99
EXERCISES 15.9 N PAGE 1020
1. 16 3. sin2cos2 5. 0
7. The parallelogram with vertices 0, 0, 6, 3, 12, 1, 6, 2
9. The region bounded by the line y1, the yaxis, and ysx
11. 3 13. 6 15. 2 ln 3
17. a 4 abc b 1.0831012 km3
x
3
21. 3 sin 1 23. ee1
19. 8 ln 8 52
the first quadrant
3. 4e24e3
9. x0 x24 frcos ,rsin rdrd
11. The region inside the loop of the fourleaved rose rsin 2 in
13. 1 sin 1 15. 1 e67 17. 1 ln 2
19. 8 27. 64
02x0x 2
7.z 9.z
y
y
x
11. II 19.
4.5
13. I
15. IV
17. III
The line y2x
4.5
27.3 2 3
2.5
2.5
27.
6
29. III 35. a
y
33. 2.04, 1.03
b y1x,x0
6 6
yCx EXERCISES 16.2 N
x2y2a2 in the first quadrant EXERCISES 16.5 N PAGE 1068
6 31. II
b 77 b 0
25. Conservative
4.5
4.5
21. fx,yxy1exy ix2exy j
x 23.fx,y,zsx2y2z2 i
sx2 y2 z2
25. fx,y2xij
y 2
6 4 2 0 2
b 1.6
0
0.2 s21e14
Fr 12
y j z k
Fr0
sx2 y2 z2
1
1.6
4 6 x
z1mxC z x,y,zds,wheremxC
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
A37
29. a 111e 8
31.
172,704
5,632,705
33. 2
k, 4
, 0
2.5
2
.5
Fr1
35. a x1mxC x x,y,zds,
y1mxC y x,y,zds,
x,y,zds
39. 2 2 41. 26 43. 1.67104 ftlb 45. b Yes
47. 22 J
EXERCISES 16.3 N PAGE 1053
1. 40 3. fx,yx2 3xy2y2 8yK
5. fx,yexsinyK 7. fx,yyex xsinyK
9. fx,yxlnyx2y3 K
11. b 16 13. a fx,y1x2y2 b 2 2
15. a fx,y,zxyzz2 17.afx,y,zxy2cosz
19. 2 21. 30 23. No
29. a Yes b Yes c Yes 31. a Yes b Yes c No
b 0, 0, 3
37. Ix k1
23 23
0x 2 12 2 23. 4a3 , 4a3if the region is the portion of the disk
PAGE 1043
1. a
3. a
5. a 0 b 2sx2 y2 z2
1. 1 14532 1
17. a Positive b Negative
19. 45 21. 6 cos1sin1 23. 1.9633 5
5. 243
54 83
3. 1638.4
11. 1 s14e61 13. 1 15. 97
9. s5
25. 15.0074
7. a
9. a
11. a Zero b curl F points in the negative zdirection 13. fx,y,zxy2z3 K 15. fx,y,zx2yy2zK 17. Not conservative 19. No
12 53
7. 17
1y, 1x, 1x b 1x1y1z Negative b curl F0
4,Iy k1 2
EXERCISES 16.4 N PAGE 1060
1. 8 3. 2 5. 12 7. 1 9. 24 11. 42
333
13. 625 15. 8e48e1 17. 1 19. 3 21. c 9
x2 i3xyjxzk b yz xyiyjk b z12sz
A38APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES
EXERCISES 16.6 N PAGE 1078
1. P: no; Q: yes
3. Plane through 0, 3, 1 containing vectors 1, 0, 4, 1, 1, 5 5. Hyperbolic paraboloid
7.
51. a 24.2055 b 24.2476
53. 45 s1415 ln11s53s70 3s5s70
55. b
9.
u constant
00
57.4 59.2a2 2
11.
where Dprojection of S on xzplane
37. 0, 0, a2
39.aIzxxSx2y2x,y,zdS
41. 0 kgs 43. 8 a30 45. 1248 3
13. IV
u constant 17. III
z0 5
x3 cos t, y3 sin t,
z0
2 0
u constant
1
2
7. s324 3
1
z0
1
constant 1
constant
EXERCISES 16.7 N PAGE 1091
1. 49.09 3. 900 5. 171s14
9. 5s5481240 11. 364 s2
1
00x y11
xxS FdSxxD PhxQRhz dA,
constant 2z0
1
z0
b4329s2 5
33. 3xy3z3 39. 4 352 272 1
15
35. x2z1 37. 3s14 41. 2 32s21
15. 341s26081 arcsins33
1
15. II
1
7. 1 b
9. 80
2 0 2 2 0x y
1 1.5 x
y y21x
22
0 y
0 11×5
19. x1uv,y2uv,z3uv
21. xx,zz,ys1x2z2
23. x2sin cos ,y2sin sin ,
z2 cos , 0 4, 0 2
2222 orxx,yy,zs4x y,x y 2
2
25. xx, y4 cos , z4 sin x
, 0x5, 0
2
z13cos tsin t,
29.xx,ye cos, zex sin ,0x3, 0 2
31. a Direction reverses
0t2 4
43.617s175s5
45. 1s2117ln2s21lns17
49. 13.9783
17. 13 20
21. divF0inquadrantsI,II;divF0inquadrantsIII,IV
24
47. 4
19. Negative at P1, positive at P2
1
z0
1 1
2
0
2
PAGE 1103
19. 1 29. 35.
180 6 3
010 2 yx
17. 3
EXERCISES 16.9 N
b Number of coils doubles
5. 2
9.0
7. 9 2
11.32 3 13.0
8 16
EXERCISES 16.8 N 3. 0 5. 0
11. a 81 2
PAGE 1097
c
2
2 0 01
c x2 x s36 sin4u cos2v9 sin4u sin2v4 cos2u sin2u du dv
13.
60391s171 15. 16
713 21. 1 23. 4 25. 0
8
23 31. 0.1642 33. 3.4895
17. 12 27. 48
20
z
20 02 y22x
CHAPTER 16 REVIEW N TrueFalse Quiz
1. False 3. True
Exercises
1. a Negative b
7. 110 9. 11 4e 3 12
PAGE 1106
5. False
17. ypxex Ax2BxC cos 3xDx2ExF sin 3x 19. yc1cos1xc2 sin1x1 cosx
APPENDIX I ANSWERS TO ODDNUMBERED EXERCISESA39
7. True
223 21. yc1ex c2xex e2x
Positive
5. 4 15
23. yc1 sinxc2 cosxsinxlnsecxtanx1
25. yc1 ln1exex c2 ex ln1exe2x
27. yexc1 c2x1 ln1x2xtan1x 2
3. 6 s10
11. fx,yey xexy
13. 0
17. 8 25. 1 275s5 6
27.60391s171 29. 64 3
39. 21
EXERCISES 17.3 N
PAGE 1132
33. 1 2
1. 7.
x0.35 cos2s5 t c10
3. x1 e6t6 et 5. 49 kg 5 5 12
1.4
37. 4 CHAPTER 17
0.02 0
0.11
c15
c20 c25
c30
EXERCISES 17.1 N
1. yc1e3xc2e2x 3. yc1 cos 4xc2
5. yc1e2x3c2xe2x3 7. yc1c2ex2 9. ye2xc1 cos 3xc2 sin 3x
PAGE 1117
sin 4x
13. Qte10t2506 cos 20t3 sin 20t3 ,
It3 e10t sin 20t 5
15. Qte10t 3 250
125
yc1es31t2c2es31t2
Petc1 cos1 tc2 sin1 t
11. 13.
15.
17. 21.
25.
29. 31. 33.
EXERCISES 17.2 N PAGE 1124
1. yc1e2x c2ex 1×2 3×7
3. yc1 c2e2x1 cos4x 1 sin4x 40 20
5. ye2xc1 cosxc2 sinx 1 ex 10
7. y3 cosx11 sinx1ex x3 6x 222
cos 20t3 500
sin 20t
c0ex33 x2n1
10
10
3 cos 10t3 250 125
EXERCISES 17.4 Nxn
sin 10t PAGE 1137
10
All solutions approach either 0 oras x l .
f
3 3
10
y2e3x2 ex
y3 cos 4xsin 4x
1. c0c0ex n0 n!
x3n 3. c0n
g
1n 5.c0 n
2nn! x2nc1
n0 3n!
19. yex2 2xex2
23. yex2 cos x3 sin x
n0 2n!
xn
n0 2n1!
7. c0 c1c0 c1 ln1xforx1
x3 2x y3cos1x4sin1x 27. y ee
n1 n
2
2
e31 1e3 b n L,napositiveinteger;yCsinn xL
x2n 9.n
ex22
1n22523n12
No solution
n0 2n!
ye2x2 cos 3xe 222
sin 3x
11. x n1
3n1!
x3n1
CHAPTER 17 REVIEW N TrueFalse Quiz
PAGE 1138
224
1. True Exercises
3. True
9. yex1x2 x2 2
1. yc1e5xc2e3x
11.
13. 15.
3
38 yp
3
The solutions are all asymptotic toyp1 cosx 3 sinxas
xl. Except for yp, all solutions approach eitheroras xl.
3. yc1 coss3xc2 sins3x 2 2
3x 2x 1 1 2x 9. yc1e c2e 6 5xe
11. y52e6x1 13. ye4x ex3 15.2nn! x2n1
n0 2n1!
17. Qt0.02e10tcos 10tsin 10t0.03 19. c 2 k85 min d 17,600 mih
5. ye2xc1 cosxc2 sinx1
10 10
7. yc1ex c2xex 1 cosx1x1sinx
yp Ae2x Bx2 CxDcosxEx2 FxGsinx yp AxBxCe9x
A40APPENDIX I ANSWERS TO ODDNUMBERED EXERCISES APPENDIXES
31. 4 s2 cos7 12i sin7 12, 2s2cos13 12i sin13 12, 4
EXERCISES H N
PAGE A12
1 cos 6i sin 6 5. 127i s321 i, i
1. 84i 11 10i
7. 13 13
15. 125i, 13 17.
3. 1318i
37. 1, i, 1s21i 39. 2 3 Im Im
i
01Re 0Re
i
2
47. cos3 cos3 3cos sin2 ,sin3 3cos2 sin sin3
33. 1024 35. 512s3512i
1 1i 9. 2 2
11. i 19.
13. 5i 2 i
4i, 4
1 s72i
21. 12i 23. 2
25. 3 s2 cos3 4i sin3 4
27. 5costan14isintan14 33
29. 4cos 2i sin 2, cos 6i sin 6,
1 cos 6i sin 6 2
41. i
1s32i 43. 2
45. e
INDEX
RP denotes Reference Page numbers.
absolutely convergent series, 714
absolute maximum and minimum, 923, 928 absolute maximum and minimum
values, 923, 928 absolute value, A6
acceleration of a particle, 839 components of, 842
as a vector, 839
addition of vectors, 770, 773
Airy, Sir George, 728
Airy function, 728
alternating harmonic series, 711
alternating series, 710
Alternating Series Estimation Theorem, 712 Alternating Series Test, 710
angle between vectors, 779 angular momentum, 848 angular speed, 840 aphelion, 667
apolune, 661 approximation
by Taylor polynomials, 749
by Taylors Inequality, 737 approximation, linear, 894, 898
to a tangent plane, 894 Archimedes Principle, 1104 arc length
of a parametric curve, 633 of a polar curve, 652
of a space curve, 830, 831
arc length function, 831 area
by Greens Theorem, 1058
enclosed by a parametric curve, 632 in polar coordinates, 650
of a sector of a circle, 650
area, surface, 635, 1075, 1077 argument of a complex number, A7 arithmeticgeometric mean, 686 astroid, 629
asymptote of a hyperbola, 658 auxiliary equation, 1112
complex roots of, 1114
real roots of, 1113
average value of a function, 956, 1000 average velocity, 838
axes, coordinate, 765
axis of a parabola, 655
basis vectors, 774, 775 Bernoulli, John, 625, 736 Bessel, Friedrich, 724 Bessel function, 724, 728 Bezier, Pierre, 639
Bezier curves, 624, 639 binomial coefficients, 742 binomial series, 742, 748
discovery by Newton, 748 binormal vector, 834 blackbody radiation, 757 boundary curve, 1093 boundaryvalue problem, 1116 bounded sequence, 682 bounded set, 928
Brache, Tycho, 844 brachistochrone problem, 625 branches of a hyperbola, 658
C1 transformation, 1013 Cantor, Georg, 696 Cantor set, 696 cardioid, 643
CAS. See computer algebra system Cassini, Giovanni, 649
Cauchy, AugustinLouis, 961 CauchySchwarz Inequality, 786 center of mass
of a lamina, 981 of a solid, 996
of a surface, 1083 of a wire, 1036
centripetal force, 852
centroid of a solid, 996
Chain Rule for several variables, 901,
903, 904 change of variables
in a double integral, 975, 1013, 1016
in a triple integral, 1002, 1007, 1019 characteristic equation, 1112
charge, electric, 980, 996
in a circuit, 1129
charge density, 980, 996
circle of curvature, 835
circular paraboloid, 810
circulation of a vector field, 1096 cissoid of Diocles, 628, 648
Clairaut, Alexis, 885
Clairauts Theorem, 885, A3
clipping planes, 804
closed curve, 1048
Closed Interval Method for a function
of two variables, 929 closed set, 928
closed surface, 1086
Cobb, Charles, 856
CobbDouglas production function, 857,
887, 940 cochleoid, 670
A41
A42INDEX
coefficients
binomial, 742
of a power series, 723 static friction, 815
comets, orbits of, 668
common ratio, 689 Comparison Test, 705 comparison tests for series, 705 complementary equation, 1117 Completeness Axiom, 682 complex conjugate, A5 complex exponentials, A11 complex numbers, A5
argument of, A7
division of, A5, A8 equality of, A5 imaginary part of, A5 modulus of, A6 multiplication of, A5, A8 polar form, A7
powers of, A9
principal square root of, A6 real part of, A5
roots of, A10
component function, 817, 1028
components of acceleration, 842 components of a vector, 772, 782 composition of functions, continuity of, 875 computer algebra system, graphing with
for creating a threedimensional scene, 804
function of two variables, 859 level curves, 864
parametric equations, 624 parametric surface, 1072 partial derivatives, 885
polar curve, 646 sequence, 680 space curve, 820 vector field, 1029
computer algebra system, integrating with, 732
conchoid, 626, 648
conditionally convergent series, 715 conductivity of a substance, 1090 cone, 808
parametrization of, 1073 conic section, 654, 662
directrix, 662 eccentricity, 662
focus foci, 662
polar equations for, 664 shifted, 659
vertex vertices, 655 conjugates, properties of, A6 connected region, 1048
conservation of energy, 1052 conservative vector field, 1032, 1053 constraint, 934, 938
continued fraction expansion, 686 continuity
of a function of three variables, 876
ovals of Cassini, 649 parametric, 621, 818 piecewisesmooth, 1035 polar, 641
simple, 1049
smooth, 831
space, 818
strophoid, 653, 671 swallowtail catastrophe, 629 toroidal spiral, 820
trefoil knot, 820
trochoid, 628
twisted cubic, 820
witch of Maria Agnesi, 628
cusp, 626 cycloid, 624 cylinder
parabolic, 805
parametrization of, 1073 cylindrical coordinate system, 1001
conversion equations for, 1001 triple integrals in, 1002
damped vibration, 1126 damping constant, 1126 decreasing sequence, 681 definite integral, 951
of a vectorvalued function, 827 del , 913
De Moivre, Abraham, A9 De Moivres Theorem, A9 density
ofa lamina, 980
ofa solid, 996
dependent variable, 855, 903 derivatives
directional, 910, 911, 914 higher partial, 884 normal, 1069
notation for partial, 880 partial, 879, 880
of a power series, 729 second partial, 826
of a vector function, 824
determinant, 786 differentiable function, 895 differential, 896, 898 differential equation
homogeneous, 1111
linearly independent solutions, 1112 logistic, 687
nonhomogeneous, 1111, 1117 partial, 886
secondorder, 1111
differentiation formulas for, RP5
of a function of two variables,
of a vector function, 818 contour curves, 860
contour map, 860 convergence
absolute, 714 conditional, 715 interval of, 725 radius of, 725
of a sequence, 677 of a series, 688
convergent sequence, 677 convergent series, 688
properties of, 693
cooling tower, hyperbolic, 810 coordinate axes, 765 coordinate planes, 765 coordinate system
cylindrical, 1001
polar, 639
spherical, 1005 threedimensional rectangular,
coplanar vectors, 791
Coriolis acceleration, 851 Cornus spiral, 637
cosine function, power series for, critically damped vibration, 1127 critical points, 923, 933
cross product, 786
direction of, 788
geometric characterization of, magnitude of, 789
properties of, 790
crosssection of a surface, 804 curl of a vector field, 1062 curvature, 638, 832
curves
Bezier, 624, 639
boundary, 1093
cissoid of Diocles, 628, 648 closed, 1048
dog saddle, 868
epicycloid, 630 equipotential, 868
grid, 1071
helix, 818
length of, 830
level, 854, 860
monkey saddle, 868 orientation of, 1039, 1055
874
766
740
789
formulas for vector functions, 826 implicit, 883, 905
partial, 878, 883, 884
of a power series, 729
term by term, 729
of a vector function, 824 directional derivative, 910, 911, 914
maximum value of, 915
of a temperature function, 910, 911 direction angles, 781
direction cosines, 781
direction numbers, 795
directrix, 655, 662 displacement vector, 770, 783 distance
between planes, 801
between point and line in space, 793 between point and plane, 793, 800 between points in space, 767
distance formula in three dimensions, 767
divergence
of an infinite series, 688 of a sequence, 677
of a vector field, 1065
Divergence, Test for, 692 Divergence Theorem, 1099 divergent sequence, 677 divergent series, 688 division of power series, 745 DNA, helical shape of, 819 dog saddle, 868
domain of a function, 855 Doppler effect, 909
dot product, 779
in component form, 779
properties of, 779 double integral, 951, 953
change of variable in, 1013, 1016 Midpoint Rule for, 955
over general regions, 965, 966
in polar coordinates, 974, 975 properties of, 958, 970
over rectangles, 951 double Riemann sum, 954 Douglas, Paul, 856
dumpster design, minimizing
construction cost, 933
e the number as a sum of an infinite series, 739
eccentricity, 662
electric charge, 980, 996
electric circuit, analysis of, 1129
electric field force per unit charge, 1031 electric force, 1031
electric flux, 1089 ellipse, 656, 662 directrix, 662
eccentricity, 662
foci, 656, 662
major axis, 657
polar equation, 664, 667 reflection property, 658 vertices, 657
ellipsoid, 806, 808
elliptic paraboloid, 806, 808 energy
conservation of, 1052 kinetic, 1052 potential, 1053
epicycloid, 630 equations
differential. See differential equation of an ellipse, 657, 664
heat conduction, 890
of a hyperbola, 658, 659, 664 Laplaces, 886, 1066
linear, 798
of a line in space, 794, 795, 796 of a parabola, 655, 664 parametric, 641, 795, 818, 1070 of a plane, 798
polar, 641, 664
of a space curve, 818
of a sphere, 768
symmetric, 795
vector, 794
wave, 886
equipotential curves, 868
equivalent vectors, 770
error estimate for alternating series, 712 error in Taylor approximation, 750 estimate of the sum of a series, 700, 708,
712, 717
Euler, Leonard, 698, 739
Eulers formula, A11 expected values, 987 exponential function,
integraion of, 743, 744
power series for, 736 Extreme Value Theorem, 928
family of hypocycloids, 629 Fibonacci, 686
Fibonacci sequence, 676 field
conservative, 1032 electric, 1031 force, 1031 gradient, 919, 1031 gravitational, 1031
incompressible, 1066 irrotational, 1064 scalar, 1029
vector, 1027, 1028 velocity, 1027, 1030
first octant, 765
firstorder optics, 754
flow lines, 1033
fluid flow, 1030, 1066, 1088 flux, 1087, 1089
flux integral, 1087
focus
of a conic section, 662 of an ellipse, 656, 662 of a hyperbola, 658
of a parabola, 655
folium of Descartes, 672 force
centripetal, 852 constant, 783 resultant, 776 torque, 791
forced vibrations, 1128 force field, 1027, 1031 fourleaved rose, 643 FrenetSerret formulas, 838 Fubini, Guido, 961
Fubinis Theorem, 961, 991 functions
Airy function, 728
arc length, 830, 831
average value of, 956, 1000 Bessel, 724, 728 CobbDouglas production, 857,
887, 940
component, 817, 1028 composite, 875
continuity of, 818, 874, 876 differentiability of, 895
domain of, 855
gradient of, 913, 915
graph of, 858
harmonic, 886
homogeneous, 909
integrable, 953
joint density, 985, 996
limit of, 871, 876
linear, 858
maximum and minimum values
of, 922, 923
of n variables, 865 polynomial, 874 potential, 1032 probability density, 985 range of, 855
rational, 874
INDEXA43
A44INDEX
functions continued
representation as a power series, 728 of several variables, 855, 864
of three variables, 864
of two variables, 855
vector, 817
vectorvalued, 817
Fundamental Theorem of Calculus, higherdimensional versions, 1105 for line integrals, 1046
for vector functions, 828
Galileo, 625, 633
Gauss, Karl Friedrich, 1099 Gaussian optics, 754
Gausss Law, 1090
Gausss Theorem, 1099 geometric series, 688 geometry of a tetrahedron, 794 gradient, 913, 915
gradient vector, 913, 915
interpretations of, 919, 920 gradient vector field, 919, 1031 graph
of a function of two variables, 858 of a parametric curve, 622
polar, 641
of a parametric surface, 1083
of a sequence, 680
graphing device. See computer algebra system gravitational field, 1031
great circle, 1011
Green, George, 1056, 1098
Greens identities, 1069
Greens Theorem, 1055, 1098
vector forms, 1066, 1067 Gregory, James, 732, 736 Gregorys series, 732
grid curve, 1071
ground speed, 778
halfspace, 864
harmonic function, 886 harmonic series, 691
harmonic series, alternating , 711 heat conductivity, 1090
heat conduction equation, 890 heat flow, 1090
heat index, 878
Hecht, Eugene, 754
helix, 818
higher partial derivatives, 884 homogeneous differential
equation, 1111 homogeneous function, 909 Hookes Law, 1125
horizontal plane, equation of, 766 Huygens, Christiaan, 625 hydroturbine optimization, 943 hyperbola, 658, 662
asymptotes, 658 branches, 658
directrix, 662 eccentricity, 662 equation, 658, 659, 664 foci, 658, 662
polar equation, 664 reflection property, 662 vertices, 658
hyperbolic paraboloid, 807, 808 hyperboloid, 808, 810 hypersphere, 1000
hypocycloid, 629
i imaginary number, A5
i standard basis vector, 774
ideal gas law, 891
image of a point, 1013
image of a region, 1013
implicit differentiation, 883, 905 Implicit Function Theorem, 906 incompressible velocity field, 1066 increasing sequence, 681
increment, 898
independence of path, 1047 independent random variable, 986 independent variable, 855, 903 inertia moment of , 983, 996, 1045 infinite sequence. See sequence infinite series. See series
initial point
of a parametric curve, 622
of a vector, 770 initialvalue problem, 1115 inner product, 779 integrable function, 953 integrals
change of variables in, 1011, 1016, 1019 conversion to cylindrical
coordinates, 1002
conversion to polar coordinates, 975 conversion to spherical
coordinates, 1007
definite, 827, 951
double, 951, 953. See also double
integral
iterated, 959, 960
line, 1034. See also line integral surface, 1081, 1087
table of, RP610
triple, 990. See also triple integral
Integral Test, 697, 699
integration
formulas, RP610
partial, 960
of a power series, 729 reversing order of, 962, 970 term by term, 729
of a vector function, 827
intermediate variable, 903
intersection of planes, 799
intersection of polar graphs, area of, 651 intersection of three cylinders, 1005 interval of convergence, 725
inverse transformation, 1013
irrotational vector field, 1064
isobars, 854, 861
isothermals, 861, 868
iterated integral, 959, 960
j standard basis vector, 774
Jacobi, Carl, 1015
Jacobian of a transformation, 1015, 1019 joint density function, 985, 996
k standard basis vector, 774 Kepler, Johannes, 844, 848 Keplers Laws, 844, 848 kinetic energy, 1052 Kirchhoffs Laws, 1129 Kondo, Shigeru, 739
Lagrange, JosephLouis, 935 Lagrange multiplier, 934, 935 lamina, 980
Laplace, Pierre, 886, 1066 Laplace operator, 1066 Laplaces equation, 886, 1066 Law of Conservation of Angular
Momentum, 848
Law of Conservation of Energy, 1053 least squares method, 932
least upper bound, 682
Leibniz, Gottfried Wilhelm, 748 length
of a parametric curve, 633 of a polar curve, 652
of a space curve, 830
of a vector, 773
level curves, 854, 860
of barometric pressure, 854 of temperatures, 861
level surface, 865 tangent plane to, 917
limacon, 647 limit
of a function of three variables, 876 of a function of two variables, 871
of a sequence, 677
of a vector function, 817 Limit Comparison Test, 707 Limit Laws
for functions of two variables, 873
for sequences, 678
linear approximation, 894, 898 linear combination, 1111
linear differential equation, 1111 linear equation of a plane, 798 linear function, 858
linearity of an integral, 958 linearization, 894
linearly independent solutions, 1112 line in space
normal, 918
parametric equations of, 795 skew, 797
symmetric equations of, 795 tangent, 824
vector equation of, 794, 795
line integral, 1034
Fundamental Theorem for, 1046 for a plane curve, 1034
with respect to arc length, 1037 for a space curve, 1039
of vector fields, 1041, 1042 work defined as, 1041
Lissajous figure, 629
lithotripsy, 658
local maximum and minimum values, 923 logistic difference equation, 687
logistic sequence, 687
LORAN system, 661
Maclaurin, Colin, 736 Maclaurin series, 734, 736
table of, 743
magnitude of a vector, 773
major axis of ellipse, 657
marginal productivity, 887 marginal propensity to consume or
save, 695 mass
of a lamina, 980 of a solid, 996
of a surface, 1083 of a wire, 1036
mass, center of. See center of mass mathematical model
CobbDouglas, for production costs, 857, 887, 940
for vibration of membrane, 724 maximum and minimum values, 922, 923 Mean Value Theorem for
double integrals, 1023
method of Lagrange multipliers, 934, 935, 938
method of least squares, 932 method of undetermined
coefficients, 1118, 1122 Midpoint Rule
for double integrals, 955
for triple integrals, 998 Mobius, August, 1085 Mobius strip, 1079, 1085 modulus, A6
moment
about an axis, 981
of inertia, 983, 996, 1045 of a lamina, 981
about a plane, 996
polar, 983
second, 983
of a solid, 995
monkey saddle, 868
monotonic sequence, 681 Monotonic Sequence Theorem, 683 motion in space, 838
motion of a spring, force affecting
restoring, 1125 damping, 1126 resonance, 1129
multiple integrals. See double integral; triple integral
multiplication, scalar, of vectors, 771, 773 multiplication of power series, 745 multiplier Lagrange, 934, 935, 938 multiplier effect, 695
natural exponential function, power series for, 736
ndimensional vector, 774
Newton, Sir Isaac, 748, 844, 848 Newtons Law of Gravitation, 844, 1030 Newtons Second Law of Motion, 840,
844, 1125 Nicomedes, 626
nonhomogeneous differential equation, 1111, 1117
nonparallel planes, 799
normal component of acceleration, 842 normal derivative, 1069
normal line, 918
normal plane, 835
normal vector, 797, 834
nthdegree Taylor polynomial, 737 number, complex, A5
octant, 765
onetoone transformation, 1013 open region, 1048
optics
firstorder, 754 Gaussian, 754 thirdorder, 755
orbits of planets, 844, 848
order of integration, reversed, 962, 970 ordered triple, 765
Oresme, Nicole, 692
orientation
of a curve, 1039, 1055
of a surface, 1086 oriented surface, 1085, 1086 orthogonal projection, 785 orthogonal surfaces, 922 orthogonal vectors, 781 osculating circle, 835 osculating plane, 835 Ostrogradsky, Mikhail, 1099 ovals of Cassini, 649 overdamped vibration, 1127
parabola, 655, 662 axis, 655
directrix, 655, 662 equation, 655, 656 focus, 655, 662 polar equation, 664 vertex, 655
parabolic cylinder, 805 paraboloid, 806, 810 parallelepiped, volume of, 791 Parallelogram Law, 771, 786 parallel planes, 799
parallel vectors, 771 parameter, 621, 795, 818 parametric curve, 621, 818
arc length of, 633
area under, 632
slope of tangent line to, 630
parametric equations, 621 of a line, 795
of a space curve, 818 of a surface, 1070
of a trajectory, 841 parametric surface, 1070
graph of, 1083
surface area of, 1075, 1076 surface integral over, 1081 tangent plane to, 1974
parametrization of a space curve, 820 smooth, 831
with respect to arc length, 831 partial derivatives, 879, 880
of a function of more than three variables, 883
interpretations of, 881
INDEXA45
A46INDEX
partial derivatives continued notations for, 880
as rates of change, 880
rules for finding, 880 second, 884
as slopes of tangent lines, 881 partial differential equation, 886 partial integration, 960
partial sum of a series, 688 particle, motion of, 838
path, 1047
perihelion, 667
perilune, 661
perpendicular vectors, 781 piecewisesmooth curve, 1035 Plancks Law, 757
planes, 797
coordinate, 765
equations of, 797, 798
horizontal, 766
normal, 835
osculating, 835
parallel, 799
tangent to a surface, 892, 917, 1074 vertical, 766
plane region of type I, 966 plane region of type II, 967 planetary motion, 844 points in space
coordinates of, 765 projection of, 766 distance between, 767
polar axis, 639
polar coordinate system, 639
area in, 650
conversion equations for Cartesian
coordinates, 640,641 conic sections in, 662 conversion of double integral
to, 974, 975 polar curve, 641
arc length of, 652 graph of, 641 symmetry in, 644 tangent line to, 644
polar equation, graph of, 641
polar equation of a conic, 664
polar form of a complex number, A7 polar graph, 641
polar moment of inertia, 983
polar rectangle, 974
polar region, area of, 650
pole, 639
polynomial function of two variables, 874 position vector, 773
positive orientation
of a boundary curve, 1093 of a closed curve, 1055
of a surface, 1086
potential energy, 1053 potential function, 1032 power series, 723
coefficients of, 723 for cosine, 740 differentiation of, 729 division of, 745
for exponenial function, 740 integration of, 729
interval of convergence, 725 multiplication of, 745
radius of convergence, 725 representations of functions as, 728 for sine, 740
principal square root of a complex number, A6
principal unit normal vector, 834 principle of superposition, 1120 probability, 985
probability density function, 985 product
cross, 786. See also cross product dot, 779. See also dot product scalar, 779
scalar triple, 790
triple, 790
projectile, path of, 629, 841 projection, 766, 782, 783, 785 pseries, 699
quadratic approximation, 933 quadric surfaces, 805
cone, 808
cylinder, 805 ellipsoid, 806, 808 hyperboloid, 808, 810 paraboloid, 806, 810 table of graphs, 808
radiation from stars, 757 radius of convergence, 725 radius of gyration, 984
range of a function, 855 rational function, 874
Ratio Test, 716 RayleighJeans Law, 757 rearrangement of a series, 719 rectangular coordinate system
conversion to cylindrical coordinates, 1001
conversion to spherical coordinates, 1006
threedimensional, 766 recursion relation, 1134 reflection property
of an ellipse, 658
of a hyperbola, 662 region
connected, 1048
open, 1048
plane, of type I or II, 966, 967 simple plane, 1056
simple solid, 1099 simplyconnected, 1049
solid of type 1, 2, or 3, 991, 993
remainder estimates
for the Alternating Series Test, 712 for the Integral Test, 701
remainder of the Taylor series, 737 representation of a function as a
power series, 728 resonance, 1129
restoring force, 1125
resultant force, 776
reversing order of integration, 962, 970 Riemann sums for multiple
integrals, 954, 990 righthand rule, 765, 788
Roberval, Gilles de, 633
rocket science, 941
roller derby, 1012
roots of a complex number, A10 Root Test, 718
rubber membrane, vibration of, 724 ruled surface, 812
ruling of a surface, 804
saddle point, 924
sample point, 952
satellite dish, parabolic, 810 scalar, 771
scalar equation of a plane, 798 scalar field, 1028
scalar multiple of a vector, 771 scalar product, 779
scalar projection, 782, 783 scalar triple product, 790
geometric characterization of, 791 secant vector, 824
second derivative of a vector
function, 826
Second Derivatives Test, 924 second moment of inertia, 983 secondorder differential equation,
solutions of, 1111, 1116 second partial derivative, 884
sector of a circle, area of, 650 sequence, 675
bounded, 682 convergent, 677 decreasing, 681 divergent, 677 Fibonacci, 676 graph of, 680 increasing, 681 limit of, 677
limit laws for, 678 monotonic, 681
of partial sums, 688
terms of, 675 series, 687
absolutely convergent, 714 alternating, 710
alternating harmonic, 711, 715 binomial, 742, 748 coefficients of, 723 conditionally convergent, 715 convergent, 688
divergent, 688 geometric, 688 Gregorys, 732 harmonic, 691 infinite, 687 Maclaurin, 734, 736 p, 699
partial sum of, 688 power, 723 rearrangement of, 719 strategy for testing, 721 sum of, 688
Taylor, 734, 736 terms of, 687 trigonometric, 723
series solution of a differential equation, 1133
set, bounded or closed, 928 shifted conics, 659
shock absorber, 1126 Sierpinski carpet, 696 simple curve, 1049
simple harmonic motion, 1125 simple plane region, 1056 simple solid region, 1099 simplyconnected region, 1049 Simpson, Thomas, 949
sine function, power series for, 740 sink, 1103
skew lines, 797
smooth curve, 831
smooth parametrization, 831 smooth surface, 1075
snowflake curve, 761
solid, volume of, 991, 992 solid angle, 1109
solid region, 1099
source, 1103
space, threedimensional, 765 space curve, 818
arc length of, 830, 831 speed of a particle, 839 sphere
equation of, 768
flux across, 1088 parametrization of, 1072 surface area of, 1076
spherical coordinate system, 1005 conversion equations for, 1006 triple integrals in, 1006
spherical wedge, 1007
spring constant, 1125
Squeeze Theorem for sequences, 679 standard basis vectors, 774, 775 static friction, coefficient of, 815 stationary points, 923
steady state solution, 1131
Stokes, Sir George, 1093, 1098 Stokes Theorem, 1092, 1093 strategy for testing series, 721 streamlines, 1033
strophoid, 653, 671
sum
of a geometric series, 689 of an infinite series, 688 telescoping, 691
of vectors, 770, 773
surfaces, 766 closed, 1086
graph of, 1083
level, 865
oriented, 1086
parametric, 1070
positive orientation of, 1086
quadric, 805. See also quadric surface smooth, 1075
surface area
of a parametric surface, 635, 1075 of a sphere, 1076
of a surface zfx, y, 1077
surface integral, 1081
over a parametric surface, 1081 of a vector field, 1087
surface of revolution, parametric representation of, 1073
swallowtail catastrophe curve, 629 symmetric equations of a line, 795 symmetry in polar graphs, 644
T and T 1 transformations, 1013
table of differentiation formulas, RP5 tables of integrals, RP610
tangential component of acceleration, 842 tangent line
to a parametric curve, 630 to a polar curve, 644
to a space curve, 824
tangent plane
to a level surface, 917
to a parametric surface, 1074 to a surface Fx, y, zk, 917 to a surface zfx, y, 892
tangent plane approximation, 894 tangent vector, 824
tautochrone problem, 625
Taylor, Brook, 736
Taylor polynomial, 737, 933 applications of, 749
Taylor series, 734, 736
Taylors Inequality, 737
telescoping sum, 691 temperaturehumidity index, 866, 878 termbyterm differentiation and
integration, 729 terminal point
of a parametric curve, 622
of a vector, 770
term of a sequence, 675
term of a series, 687
Test for Divergence, 692
tests for convergence and divergence
of series
Alternating Series Test, 710 Comparison Test, 705 Integral Test, 697, 699 Limit Comparison Test, 707 Ratio Test, 716
Root Test, 718
summary of tests, 721
Test for Divergence, 692
tetrahedron, 794
Thomson, Sir William Lord Kelvin, 1056,
1093, 1098
threedimensional coordinate system, 766 TNB frame, 835
toroidal spiral, 820
torque, 791, 848
Torricelli, Evangelista, 633
torsion of a space curve, 838
torus, 1081
total differential, 896
total electric charge, 980, 996
trace of a surface, 804
trajectory, parametric equations for, 841
INDEXA47
A48INDEX
transformation, 1013 inverse, 1013
Jacobian of, 1015, 1019 onetoone, 1013
tree diagram, 903
trefoil knot, 820
Triangle Inequality for vectors, 786 Triangle Law, 771
trigonometric series, 723
triple integrals, 990
applications of, 995
in cylindrical coordinates, 1002 over a general bounded region, 991 Midpoint Rule for, 998
in spherical coordinates, 1007, 1008
triple product, 790
triple Riemann sum, 990
trochoid, 628
twisted cubic, 820
type I or type II plane region, 966, 967 type 1, 2, or 3 solid region, 991, 993
ultraviolet catastrophe, 757 underdamped vibration, 1127 undetermined coefficients, method of,
1118, 1122
unit normal vector, 834 unit tangent vector, 824
unit vector, 775
variable
dependent, 855, 903 independent, 855, 903 independent random, 986 intermediate, 903
variables, change of. See change of variables variation of parameters, method of,
1122, 1123 vectors, 770
acceleration as, 839 addition of, 770, 773 algebraic, 772
angle between, 779 basis, 774, 775 binormal, 834 components of, 772 coplanar, 791
cross product of, 786 difference of, 771 displacement, 770, 783 dot product, 779, 780 equivalent, 770
force, 1030
geometric representations of, 772 gradient, 913, 915
i, j, and k, 775
initial point of, 770
length of, 773
magnitude of, 773 multiplication of, 771, 773 ndimensional, 774 negative, 771
normal, 797, 834 orthogonal, 781
parallel, 771 perpendicular, 781 position, 773
principal unit normal, 834 projection of, 782, 783 properties of, 774
scalar multiple of, 771 standard basis, 775 subtraction of, 773
sum of, 770, 773
tangent, 824
terminal point of, 770 threedimensional, 772 triple product, 790, 791 twodimensional, 772 unit, 772
unit normal, 834
unit tangent, 824
velocity, 831
wind velocity, 764
zero, 770
vector equation
of a line, 794, 795
of a line segment, 797 of a plane, 798
of a plane curve, 818
vector field, 1027, 1028 conservative, 1032
curl of, 1062
divergence of, 1065 electric flux of, 1089
force, 1027, 1031
flux of, 1087
gradient, 1031 gravitational, 1031 incompressible, 1066 irrotational, 1064
line integral of, 1041, 1042 surface integral of, 1087 velocity, 1027
vector function, 817 continuity of, 818 derivative of, 824
differentiation formulas for, 826 integration of, 827
limit of, 817
parametric equations of, 818
vector product, 786 properties of, 790
vector projection, 782, 783
vector triple product, 791
vectorvalued function. See vector function velocity field, 1030
air flow, 1027
ocean currents, 1027 wind patterns, 1027
velocity vector, 831
velocity vector field, 1027 vertex of a parabola, 655 vertical plane, equation of, 766 vertices
of an ellipse, 657
of a hyperbola, 658
vibrating spring, 1125
vibration of a rubber membrane, 724 vibrations, 1125, 1126, 1128
volume
by double integrals, 951 of a hypersphere, 1000 of a solid, 953
by triple integrals, 995
wave equation, 886
windchill index, 856
wind patterns in San Francisco Bay
area, 1027
witch of Maria Agnesi, 628 work, 783,
defined as a line integral, 1041 Wren, Sir Christopher, 635
xaxis, 765 xcoordinate, 765 Xmean, 987 xyplane, 766 xzplane, 766
yaxis, 765 ycoordinate, 765 Ymean, 987 yzplane, 766
zaxis, 765 zcoordinate, 765 zero vector, 770
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