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r
Multivariable
Calculus
Ninth Edition
Ron Larson
The Pennsylvania State Unfrersity The Behrend College
Bruce H. Edwards
University ofFlorida
~.,. 1# BROOKS/COLE
1 • CENGAGE Learning Australia , Brazil , Japan , Korea • Mexko • Slog.pore • Spain • United Kingdom • Unite<! States
BROOKS/COLE
CfNGAGE Learning·
------ -
Multivariable Calculus, Ninth Edition Larson/Edwards
VP/Editor-in-Chief: Michelle Julet Publisher: Richard Stratton Semor Sponsoring Editor: Cathy Cantin Development Editor. Peter Galuardi Associate Editor· Jeannine Lawless Editorial Assistant Amy Haines Media Editor: Peter Galuardi Senior Content Manager: Maren Kunert Senior Marketing Manager: Jennifer Jones Marketing Communications Manager: Mary Anne Payumo Senior Content Project Manager, Editorial Production:
Tamela Ambush Art and Design Manager: JIii Haber Senior Manufacturing Coordinator: Diane Gibbons Permissions Editor: Katie Huha Text Designer: Nesbitt Graphics Art Editor: Larson Texts, Inc. Senior Photo Editor. Jennifer Meyer Dare Illustrator: Larson Texts, Inc. Cover Designer: Harold Burch
Cover Image: © Richard Edelman/Woodstock Graphics Studio
Compositor: Larson Texts, Inc.
Tl is a registered trademark ofTexas Instruments, Inc.
Mathematica is a registered trademark of Wolfram Research, Inc. Maple is a registered trademark of Waterloo Maple, Inc.
- - - - -- - -
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© 2010. 2006 Brooks/Cole, Cengage Leaming
ALL RIGHTS RESERVED. No part of this worK covered oy the copyright herein may be reproduced. transmitted, stored or used in any form or by any means graphic. electronic. or mechanical, including but not limited to photocopying, recording, scanning, digitizing, taping, Web distribution, information networks. or information storage and retrieval systems, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without the prior written permission of the publisher.
For product information and technology assistance, contact us at Cengage Learning Customer & Sales Support, 1-800-354-9706
For permission to use material from this text or product, submit all requests online at www.cengage.com/permisslons,
Further permissions quesdons can be emailed to permisslonrequest@cengage.com
Library of Congress Control Number: 2008939232 Internat ional Student Edition: ISBN-13: 978-1-4390-3032·5 ISBN-10: 1-4390-3032·4
Brooks/Cole 10 Davis Drive Belmont, CA 94002-3098 USA
Cengage learning is a leading provider of customized learning solutions with office locations around the globe. including Singapore, the United Kingdom, Australia, Mexico, Brazil. and Japan. Locate your local office at: intemational.cengage.com/reglon
Problems from the William Lowell Putnam Mathematical Competition reprinted with permission from the Mathematical Association of America, 1529 Eighteenth Street, NW. Washington, DC.
(engage learning products are represented in Canada by Nelson Education, Ltd.
For your course and learning solutions, visit www.cengage.com
Purchase any of our products at your local college store or at our preferred onllne store www.ichapters.com
Printed in the United States of America
J 1 3 4 5 6 7 12 II 10 09 08
[Clontents
A Word from the Aulhors
Textbook Features
m CHAPTER
Vectors and the Geometry of Space
11.1 Vectors in the Plane 11.2 Space Coordinates and Vectors in Space 11.3 The Dot Product of Two Vectors 11.4 The Cross Product of Two Vectors in Space 11 .5 Lines and Planes in Space SE CT ION PROJECT: Distances in Space 11.6 Surfaces in Space 11.7 Cylindrical and Spherical Coordinates
Review Exercises P.S. Problem Solving
CHAPTER lfJ
Vector-Valued Functions
12. l Vector-Valued Functions SECT ION PRO JEC T: Witch of Agnesi 12.2 Differentiation and Integration of Vector-Valued
Functions 12.3 Velocity and Acceleration 12.4 Tangent Vectors and Normal Vectors 12.5 Arc Length and Curvature
Review Exercises P.S. Problem Solving
Vl
X
763
764 775 783 792 800 811 812 822 829 831
833
834 841
842 850 859 869 881
883
iii
iv
Contents
CHAPTER IEJ
Functions of Several Variables
l3.l Introduction to Functions of Several Vruiables 13.2 Limits ru1d Continuity [3.3 PartiaJ Derivatives SECTION PROJECT: Moire Fringes 13.4 Differentials 13.5 Chain Rules for Functions of Several Variables 13.6 Directional Derivatives and Gradients 13.7 Tangent Planes and Normal Lines SECT ION PROJE CT: Wildflowers 13.8 Extrema of Functions of Two Variables 13.9 Applications of Extrema of Functions of
Two Variables SEC TIO N PROJECT: Building a Pipeline 13.10 Lagrange Multipliers
Review Exercises P.S. Problem Solving
CHAPTER eJ
Multiple Integration
14.1 Iterated Integrals and Area in the Plane 14.2 Double Integrals and Volume 14.3 Change of Variables: Polar Coordinates 14.4 Center of Mass and Moments of Inertia
SE cTI oN PROJECT: Center of Pressure on a Sail
14.5 Surface Area S EC Tl ON PROJECT: Capillary Action 14.6 Triple Integrals and Applications 14.7 Triple Integrals in Cylindrical and Spherical
Coordjnates SECTION PROJECT: Wrinkled and Bumpy Spheres 14.8 Change of Variables: JacobicU1s
Review Exercises P.S. Problem Solving
885
886 898 908 917 918 925 933 945 953 954
962 969 970 978 981
983
984 992 1004 1012 1019 l 020 1026 1027
I 038 1044 1045 1052 l 055
Acknowledgments
vii
CabAJs
5th
4th
3rd
2nd
••••••---r••••••••••••••••• ••••••
[A] cknowledgments
We would like to Lhank the many people who have helped us at various stages of this project over the la,t 35 years. T heir cncouragemenL criticisms. and suggestions have been invaluable to us.
Reviewers of the Ninth Edition
Ray Cannon. Baylor University Sadeq Elbaneh. Buffalo Srate College J. Fasteen. Pon/and Srate Unil'ersity Audrey Gillant. Binghamton Unii·ersity Sudhir Goel, Valdosta State Univer.riry Marcia Kernen. We11111·orth Jnstiwte of Teclmology Tbrahima Khalil Kaba. Embry Riddle Aeronawical University Jean-Baptiste Mcilhan. Universi1y of California Ri Ferside Catherine Moushon, Elgin Co11111w11i1y College C harles Ollion. Hous!On Community College Greg Oman. The Ohio State Uni1°ersity Dennis Pence. Western Michigan U11iversi1y Jonathan Prewett. Unirersity ofWvoming Lori Dunlop Pyle, U11i1-ersity of Central Florida Aaron Robertson. Colgare University Mauhew D. Sosa. The Pennsylrnnia Stale U11il'ersity William T. Trotter, Georgia /11stit111e ofTec/1110/ogy Dr. Draga Vidakovic. Georg ia State Unive rsity Jay Wiestling. Palomar College Jianping Zhu. Uni versiry of Texas a t /\rlington
viii
Acknowledgments
Ninth Edition Advisory Board Members
Reviewers of Previous Editions
Jim Braselton. Georgia Sowhem U11iversi1y: Sien Deng. Northern Illinois Uni11ersity; Dimirar Gramcbarov. U11i1•ersity of Texas, Arlington; Dale Hughes, Johnson Coumy C01111n1111itr College: Dr. Philippe B. Laval. Kennesaw State U11iversity; Kouok Law. Georgia Perimeter College. Clarkson Campus; Mara D. Neuse!, Texas Tech University; Charlotte Newsom, T'ide1Fater Commu11ity College. Virginia Beach Campus; Donald W. Orr, Miami Dade College, Kendall Campus; Jude Socrates, Pasadena City College: Betty Travis. University o_(Texas al San Antonio: Kuppalapalle Vajravelu. Universitr of Ce11tral Florida
Stan Adamski. O,vens Co1111111111i1v College: Alexander Arhangelskii, Ohio University; Seth G. Armstrong, Southern Utah University: Jim Ball. Indiana State University; Marcelle Bessman, Jacksonville University: Linda A. Bolte. Eastern Washington U11iversi1y: James Braselton, Georgia Southem University: Harvey Braverman. Middlesex County College; Tim Chappell, Penn Valley Community College: Oiyin Pauline Chow. Harrisburg Area Com1m111iry College: Julie M . Clark. Hollins Universitv: P.S. Crooke, Vanderbilr University; Jim Dotzler. Nassau Community College: Mun-ay Eisenberg, University of Massaclzusens ar Amherst; Donna Flint. South Dakota Stare University: l'vlichael Frantz, University ofLa Verne; Sudbir Goel, Valdosta State Un iversity; Arek Goetz, San Francisco State Universiry; Donna J. Gorton, Butler Corm(r Community College; John Gosselin. University of Georgia: Shahryar Heydari . Piedmont College; Guy Hogan, Nmfolk Staie Unirersity; Ashok Kumar, Va/dos1a Stale U11i1·ersity; Kevin J. Leith, Albuquerque Community College: Douglas B. Meade. Uniiwsity of South Carolina; Teri Murphy. University of Oklahoma: Darren Narayan. Rochesrer lns1iu11e ofTec/mology; Susan A. Natale, The Ursuline School. NY: Terence H. Perciante. Wheaton College; James Pornmersheim. Reed College; Leland E. Rogers, Pepperdi11e U11it·ersity; Paul Seeburger. Monroe Co111111w1.i1y College; Edilh A. Silver. Mercer Co1111ty Community College; Howard Speier. Cha11dler-Gilber1 Co11111umi1y College; Desmond Stephens, Florida A &M University; Jiaruhong Su. University of Texas ar Arlington: Patrick Ward, Illinois Central College: Diane Zych, Erie Co111111uni1y College
Many thanks to Roben HosteUer. The Behrend College, The Pennsylvania State University, and David Heyd, Tbe Behrend College. The Pennsylvania State University, for their significant contributions ro previous editions of tbis rext.
A special note of thanks goes to the instruclors who responded to our survey and to the over 2 million students who have used earlier editions of the text.
We would also like to thank the staff at Larson Texts, Inc.. who assisted in preparing
the manuscript. rendering the art package, typesetting, and proofreading the pages and su pplements.
On a persona[ level, we are grateful to om wives, Deanna Gilbert Larson and Consuelo Edwards, for their love. patience, and support. Also, a special note of thanks goes out to R. Scott O'Neil.
If you have suggestions for improving this text. please feel free to write to us. Over tbe years we have received many useful comments from both instructors and students. and we value these very much.
Ron Larson
Bruce H. Edwards
[yJ our Course. Your Way.
Calculus Textbook Options
The Ninth Edition of Cairn/us is a,·ai!able in a variety of textbook configurations to address the different ways instruc tors teach-and students take-their classes.
It is available in a comprehensive three-semester version or as single-variable and multivariable versions. The book can also be customized to meet your individual needs and is available through iChapters - www.ichapters.com.
TOPICS COVERED
• I. •
APPROACH
· II. I .
• •I
coverage
Late Trigonometry
3-semester
Calculus 9e
Calculus: Early Transcendental Functions 4e
Essential Calculus
ESSENTIAi. CJ\LCULUS
,..............~-- -
11,... • - ~
.
Calculus with Late Trigonometry
CALCULUS
'' "'
~l~·i-:. .,
Single Variable Only
Calculus 9e Single Variable
Calculus: Early Transcendental Functions 4e
Single Variable
Multivariable
Calculus 9e Multivanable
MULr l VARl /\ 6LE
.,..,,,... C A I, C ·u L U S
I,~
Calculus 9e Multivariable
Custom
All of these textbook choices can be customized to fit the Ind ivid ual needs of your course.
Calculus 9e
Calculus: Early Transcendental Functions 4e
Essential Calculus
Calculus with Late Trigonometry
CAlCULUS
ix
IT]extbook Features
Tools to Build Mastery
CAPSTONES )
r E'V! Capstone exercises now appear in every
section. These exercises synthesize 1.he main concepts of each section and show students how the topics relate. T hey are often multipart problems that contain conceptual and noncomputational parts, and can be used for classroom discussion or test prep.
WRITING ABOUT CONCEPTS
~ 70, I'-< di.= grnrh ot J r;.b.tV.n m the fiftllre 1U anv•ct mt f;.,llc lnr. ghc.n tiut/t1li • -l
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WRITING ABOUT CONCEPTS
These writing exercises are questions designed to test students· understanding of basic concepti. in each section. The exercises encourage students 10 verbalize and write answers. promoting technical communication ski Us that will be invaJuable in their future careers.
STUDY TIPS )
The devil is in the details. Study Tips help point out some of the troublesome common mistakes. indicate special cases that can cause confusion, or
- - Otcuu,-c mregratJon i~
ili.ually more d10icult lh:m diffen:nua\1011. )OU should al"Ar,ty-. check your 3D'\Y.cr IO
an lDlci;r.llH>n problem by differen11,ong 1'1>r instance. m ii,iimplc ➔ you ~houhJ d1ffcrentiau~\(2t - ...-~~-~ - ~ - " ~-
-----,
expand on important concepts. These tips provide students with valuable information, similar to what an instructor might comment on in class.
thtu you oblam l11c
Lnti.:r m tlus chapter,
C ,·ou wiU k.lll1 convcmcm methods fvr
~•kula11ng J(.r) clr for conunu011,
~ - - - - - - -- -- - -- - - - - - - - - - - - - - - - - - - - - - - - I_ _ _ _ _R_e_m_em_b_cr_!h_at _-,-ou c-an -.. now. }'OU must ux tbc
check )'OU? answer b) differentiating..
B EXAMPLE Evaluation of a Definite Integral
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Ji ¥-, i' l!dt-=
= td1 -l, 1..., ,Lr • ~
Solution
f, r~J,-3Jd.r 0 J.'(-x')tb-J,'J,d.r I f ( - 3td,
--f,'d.<-Jf ,d., - Jfd.r
-(1) D
+ JP) - 11?1
.,
- ;i
■
EXAMPLES Jt--- - - - - - - - - - - - - .
Throughout the text, examples are worked out step-by-step. These worked examples demonstrate Lhe procedures and techniques for solving problems, and give students an increased understanding of the concepts of calculus.
X
Textbook Features
xi
,.._E_X_ER_c_1_sE_S____J,
Practice makes perfect. Exercises are often the first place students turn to in a textbook. The auU1ors have spent a great deaJ of time analyzing and revising the exercises. and the result is a comprehensive and robust set of exercises at ilie end of every section. A variety of exercise types and levels of difficulty are included to accommodate students wiili aJl learning styles.
/11 addition to the exercises in rhe book, 3,000 algorithmic exercises appear in the WebAssign® course that acco111pa11ies Calculus.
If■ Exercises
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"When will I use this?"' The authors attempt to answer this question for students with carefully chosen appl ied exercises and examples. Applications are pulled from diverse sources, such as current events, world data industry trend!>. and more, and relate to a wide range of interests. Understanding where calculus is (or can be) used promotes fuller understanding of Lhe material.
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Review Exercises at the end of each chapter provide more practice for students. These exercise sets provide a comprehensive review of the chapter's concepts and are an excellent way for students to prepare for an exam.
m PROBLEM SOLV ING
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These sets of exercises at the end of each chapter test students' abilities with challenging, thought-provoking questions.
r
xii
Textbook Features
Classic Calculus with Contemporary Relevance
THEOREMS J
Theorems provide the conceptual framework for calculus. Theorems are clearly stated and separated from the rest of the text by boxes for quick visual reference. Key proofs often follow the lheorem. and other proofs are provided in an in-text appendix.
H IEORE~l 4.9 THE FUNDAMENTAL THEOREM OF CALCULUS
lf a function/is continuous on the closed interval [a. b] and F is an antideri\'- 11
ative offon the interval [a. b], then
11
f J(x) dx = F(b) - F(a).
DEFINITIONS 1
DEFINlTlONOF DEFJN!TE INTEGRAL
As with the theorems. definitions are clearJy
Jf/ is defined on the closed interval [a, b] and the limit of Riemann sums over
stated using precise.
partitions J
formal wording and are
separated from I.be text
by boxes for quick
exists (as described above). then/is said lo be integra ble Oil [a. b] and the
visual reference.
I l" limit is denoted by lim II J(c,) J X; =
J(x) dx.
l_ili-O i=I
11
The limit is called the definite integral off from a Lo b. lower limit of integration. and the number b is the uppe
10 complete the: chillll,';~ of Yanat,J~, m E.la.mple 5. ym.i ,,,lwd for r in tcnn~ of j j . Son1!!lirm::, 1h1:t. 1, "<f'} difficult. Foc1Unn1d; H is nol ah, a:y" 11i;4,.•cti-~J)'. ~ HlOwn 1n lht! m!M txilDlpk
Ill EXAMPLE Chanie of Variables
frind sin:! 'l1 cf.h ., 1 di.
Soluuon Bc<:a11...: 1;1cr 3.t = (sin 3tF. you ~·an lei 11 ..,... sin 1t n n:n
- , PROCEDURES j
Ju= r~3llO>tL-r ~~-. beca~ ""rK :h d.r 1~ P:lJ1 oi lh,c on~m.11 m1cgr.1l. ~ou can \\fllt"
Formal procedures are set apart from the text for easy refere!lce. The procedures provide students with stepby-step instructions that will help them solve problems quickly and efficiently.
:!3!!..= cm- -ltd,
Suh~111uting u tlll\l ,/u(', m the on.gm.ii tlllL:f!ntl yteld.c;
I ,- ' I"~J sm· .,ro),.1,I.1 ~ 'du - {frr du
- , NOTES )
- !(!::.) ~ ("
'3 •
= !9StnJ3-t ' (.'
Notes provide additional details abou1 theorems. definitions. and examples. They offer additional insigh1. or important generalizations that students might not
You C"ilD dtt::d. 1h:.. h) d11farnti.llin!'.,.
f[¾~in' :t I {!)(3H'-mJ{Fko-.'\1)0)
im mediately see. L ike the study tips. notes can be invaluable to students.
.,u,: lt cos 1,
r - - - - -- -- -- - - - - - - ' - - -- - - - - - -- - -- - - - - - , tha1 )OU have
lmli:t There are two impor1an1 points that should be made concerning th~ Trapezoidal Ruic
•
(or the Midpoint Rule). Firs!. the approximation tends 10 become more accurate a.s II iacreascs.
For instance. in Example I. if n = 16. the Trapezoidal Ruic yields an approximation of 1-99-1.
Second. alt hough you could have used the Fundamental Theorem to evalunte 1he integral in
Example I. this theorem cannot be used to evaluate an mtegral a~ simple as J;sin x 2 dx because
sin x2 has no elemental') amiderivm iw. Yet. the Trapezoidal Rule can be applied easily to
estimate this imegral.
■
Textbook Features
xiii
Expanding the Experience of Calculus
CHAPTER OPENERS )
■; ■ Differential Equations
Chapter Openers provide initial motivation for the upcoming chapter material. Along with a map of the chapter objectives. an important concept in the chapter is related to an application of the topic in the real world. Students are encouraged to see the real-life relevance of calculus.
Tht Ctmn:ru o/T1u•f>t-m! J.4 Is lhc L"Ome~ o f Thcon-m .! .! lr'll'e?1lt;it L', tf.1. f1,1ne1jon i; i.'lt...og;Jrblc, d....-.:s II h:M:: :o ~oocunuous? farlain :!,Our rca.•,.::mm~ ~ i ;W _ C~:tllliIF~·
D(;sroheshe rduuo:i~p,, a.mong rontirl:ii:~. tlitl"c:reotJ,.";tillity, a.n-d 1ruer:1bi!ity. Which 1~ 1he '-troClgcst co:0-l!Kln' W hich i; the 9-r,lk~:.t',1 W'hkh condilioru intril)" 00\C.r condttMlfij
tn11,,,cr.,p::a.,,-..,111 t.1ll.."?-dliv n•'lll~~r.11,_# ( ~
~ ,__,._,..,,, .........-:r:ca
1•, •l;Mll~M~,lll(sl:Stl'm IYl'"o l ~l"l.LUli:mo.~ tt.bl,IOOf'l'RUl,f.nl-f•lktida£.-.J ecm.-v11tTh:sJl>'•...11ta;rc7r:ie,,c: 11'<1::,nd,<""4,,o-,M~~ ~ ~ tnoni'icl~
l111 . ' " ° ~)"» '110ctl!lt.r:IIK fl,U!f"""{
1 11,...,,.,u;~s:JIIIJl('f'l'flr.idff.f
•:~~..::,:;'=,· 1 do!'l...-.l~,-.Ga,J.t ~..W,..ftl)
■ . tt,,....,,. .ei.p:anulll._.,,, .,~.,._,a-~ ,c,
■ 1~ .. _,..•f.N..ouk,i-■IJ,llc::clwltlZIXlmlal.tk;n,lli lhlb!:mllll«IIIIXIID,1¢ ~1
L p - - --
..._.., ..,ll'll. . .~ , : u ,1111t111M1•• ~
~ ·- -
-m,.,....,.....~. .m.1.1a-• • ~ c ! l l i l - -i11
-......111111!t<11l.o.i. .ll"f"'\fluta tl!IMIWll~~ l ; S -
~&,J;lnmN Ml
EXPLORATIONS )
Explorations provide students with unique challenges to study concepts that have not yet been formally
1-rzt..:linJIAnililtn•'Gtl<tt Far
, : - @ J t r i \ • a ti \'!:,d:,,.,.""TJbc:Ib: c a ~ hll\,;'lioo F
•• f1_d-= 2.T c.. Flrl - r1
b. f l .ti-.(
;i d.. F'hl::::
= e.. f1.1) ~ r. flrl - cu~.•
,...\.\bJ:l llr.11,:S)" d>d_yUIII ~'-c!tCI l1nd
covered. They allow students to learn
by discovery and introduce topics
related to ones they are presently studying.
By exploring topics in this way, students are
encouraged to think outside the box.
PUTNAM EXAM CHALLENGES )
IJ\I. lf~~
J 0 'lt'tR:dT!IIIIDrnR!wymt
; ,T - -,.: ,-o
baJ,1 R4llalll:'n:£! ~re• I.Sil, Fm:! aD lN" ~111,u,,:,,. jlDl4!Tt fr..1in-, /h !.it
ll-s.i: L.:1,'.!:lchlr
1fb,•.ts-J
f.n••.O•o
f1u1..-c.< -<i
c~ ,,._,_.,~,.,-. e-,- n.-.~)Ao,.f,.f._~...u,.,~.,.r,-.~_;,a..t.t.~<.hl~,,4._"."..-.-,,.
Putnam Exam questions appear in selected sections and are drmNn from actual Put.nam Exams. These exercises wiU push the limits of students' understanding of calculus and provide extra challenges for motivated s tuden ts.
• -i- -n A
.~. .-..b r.•...c..i ..m...e.~..O..N..i.- - • t :-M •-~-~· ~_ . .-. _-, . , _,-..-.-. . .-_ ..-c,,,;,, r-.~
wi:a #.-~11d, ~;.<J11Z.111,:;.,~1,11
l ...
HISTORICAL NOTES AND BIOGRAPH IES
Historical Notes provide students with background information on the foundations of
calculus, and Biographies help humanize calculus and teach students about the people who contributed to its formal creation.
<,[Oil{; FU[DillO I HtR.'-H.\~ (181i--lS'6l
l,e,mt::rJ!hti113!,Clf.l< i!,c:;1~:: did h i t ~ f~m:ia oa.111 ~ a,l'!1 DI ~ o : .
C ~ :-~· t1tJcqu1~i:r~U~ll~tt.!.er tt«>l ~ --nli'l!r.m·1r.-s~.ii;fi)U"
Jnd IT'!wt.~:-.;1,:h;;1I~,..~ !t!UttU!l
:i«" ecinir;._"(;~<:w.:.n1n-,~1
h1mlij~d~.Z~~~!flV'Al !P ~G-!~~· ~ll'!tl11~::tt~ a , . . t1 0 1 ' , l r i ~ r ' - . D ' "~ '-lt,t-~';z::z:;lzl ~1'1:,c,ttir=~~!'.~IJ&~;ll!(Q
e t l l ' l l(O::~.'.;!.
~- l • - ~~ .. ~ ,._
lOt - tJl
- :OCI - i!it
SECTION PROJECTS )
Projects appear in selected sections and more deeply explore applications related to the topics being studied. They provide an interesting and engaging way for students to work and investigate ideas collaboratively.
m,,,.,1,aa-111,1
Derntin.straung I.he fundamental Theorem
l'"", ~ u:il:1~ e ~ lf'I,: IJ!!:1.."'llDIII 1, • 1is;•, t"I C!1: w:~-..r ~ r< ,,- l ..,,F'.alb!~ f,;O;,..--r:il';,,o;1,n,,(,l i.-
F , " [ •111·1,b
,.1 C~.rtr1lrr,t>bl,, i.,,p114...h;dlc:•~"'ru~,..:r = :
.,,,, !~, 1;.,, t!c 11!~:;!,n:IIC.:. (C~!'011'ln • _. ~ ~ •
1~• l'""':!11:l!cte1cml,1~~~<:d. ~~ a , c ; ~
ftr• llc,on1hi~J:!Af111~1t::::.!::n:-!':m...,i,ft.,• (,11 \•..-.f) =tlh,:4:!\'it!M',:;(.v - , ·~ :, - 1£::::•.'.l-, .-,:;
(r.d. m,1.,.tilt .a >b&JI":~ b',t, . , _ . , . ~ ~ : I rr'~~u,1...t11, ru,.'l:lrt,r:a1,,
,.
xiv
Textbook Features
Integrated Technology for Today's World
B ( ; EXAMPLE Change of Variables
Find Jx✓lr - I dx.
Solution As in 1hc previous example. lei 11 = 2r - I and obiain d.\ - t/11/2.
Because the iutegrnnd contains a faclOr of x. )'OU mu, L also solve for .l in terms of 11.
as shown.
u=2t- l
X = (11 + 1)/ 2
111! of _._..
Now. using , ub~1uu1ion. you obtain
f("; Jx,'2r- ldt =
1 ),,•~(~')
~I= (u3l1 + 111") d11
¾(~;; =
T ~;;) + C
CAS INVESTIGATIONS , 1 - - - - - ~
Examples throughout the book are accompanied by CAS Investigations. These investigations are linked explorations that use a computer algebra system (e.g., Maple®) to further explore a related example in the book. They allow students to explore calculus by manipulating functions. graphs, etc. and observing the results. (Formerly called Open Explorations)
= ~ (2r - J)5, l-'- ¾121 - l) lll + C. 1
J GRAPHING TECH EXERCISES
Understanding is often enhanced by using a graph or visualization. Graphing Tech Exercises are exercises that ask students to make use of a graphing utility to help find a solution. These exercises are marked with a special icon.
Slope Fi~lds In Exercises .55 and 56. (D) use a graphing utillt)· to gn.1ph a slope field for the differential equation. (bl u.se intejlrolion and the i:,•hen poilll to find the 1>articuh1r solution of the dlfTcrcnliel equation, and (cl graph the solution and the slope fichl 111 !he same ,;ewlng \\indow.
56. ~ 2./x. (-1. 12)
TECHNOLOGY
Gl).&t. , , . . , ~ - ~I~ funtU•lb
/h,1') = 1 1 - 1:
..udllierv,eii 1J -~. :,
.. fa) L'tc: ..r. ~ Virl'a ~),11::111 Id p-..i.r,b LDC >lll'faCe ~.J~~~
,b1 ~ ~ ~ lJcm.llllC IJ.f1•,-~ ,
f.a:iimd,.._atc-=:•-t;r,1111 - teaflJ l!w•CTCf'IIO"
= al.,"';1!.."s::lt,~IO~d!tf~-lD'lca:lhtuwcn:aJ[Q, ~~
('1 -'J'9",.i.mi.= ~
Qf \.Ix f ~ U1 r,a,11l-1 .d)d
lllkrprtt.ea,.i:r.r,;lfxecaotl!llhr.~.
- -· . ... ;::J ln Liffnla'7'0-11..._•cvmputr:rllfS'rlln,)ttuitoin,pll tbt
7'. ::, -
1.,: "' ~
!iL - "'.--.r.l-
•
CL 1 .... -..1 - T-;:c=.-!,
1n l~s,t--M,.dmrlld!Wll_. #t/mt' phD,c,.ET~
............
Id! 411'ff'•IIIW!'t!E~r.t1~00\<>ldlrf11n,.11<101111r,wtJb1
.-S ~Ui3mlbc(ot1101ril 1/r ~
•11
•~• Fi!W 'TJI ◄. -1'• mJ nrlum 11, TLit111n•'1•r,.
'""' Uu.111-.enu,rurt!<II cm tf' I,.'« I e n ~ aJ;di,-.1 '-)"•l
In l~,rftb.t-t! l-.!4,mt•comp1:112rll]1,..,.brw Jt°"~ll'I 1t1RrnJ u 1t,
,.,/thc-foJ1%..bCD/»thck•i:1 dw ,,:,,;q Rl.lhelkrn"OUO ~:d>iptoClc:~~i:..
•n1l.i unil ,T"<1or urthacunul tn u •nd , .
' II.\I
15,7)
"'.' \!$.9,.l) "4.·?J-9,.
·~ ::1. u- \-, -t .tj
":a. (Hl.
:,
!-t.. a.,.. 0 7\
• • 0-'l - IHJ • G.:¼
' - 1•, - ......
- CAS EXERCISES ) '
NEW! Like the Graphing Tech Exercises, some exercises may bcsl be solved using a computer algebra system. These CAS Exercises are new to this edition and are denoted by a special icon.
Throughout the book. technology boxes give students a glimpse of how technology may be used to help solve problems and explore the concepts of calculus. They provide discussions of not onl} where technology succeeds. but also where it may fail.
~lost ~r.irh1ng uu liue-s and computer al~c-bra S)~tcn,, ha,·c bu1l1-m program..\ Lhat can be u~ 10 appmum:ue I.he \'a)ue ofa definue 1ntc:pul Tr)· u."1ni; ~h .1 program to "1ppruum.llr: 1he intfgn.J m E,;tmple I. Ht'I\I, clo\t 1\ ;our
appt01Hn3liOD".'
W~n you U5-e such a rrogr.un. ~ou ntt:d u.,, be av..m: of 1L\ hm11.t11l'll'\ Oft~n. )OO ue gi\.-.;:n noind1c.iuon of t h ~ ~ ~ f ~ · ofthearrro,u2uuon. Otha Ut~ yoa rn~ be gn en .n .1rr,oum.umn thzt is rompletel) Y.TI)(I~ For 10..,umcc. U) UllDi! 8 bwh-1n nu1nmc-al lltlCgrlUOO pn.~~ to C\.lhllllc
: I J,,- dx.
Your c-J.lc-ulator 4-ould gn't' nn cnor mi:<,sagc Does )'Oun-1
[A] ctditional Resources
Student Resources
■ Student Solutions Manual-Need a leg up on your homework or help to prepare for an exam? The Studenr Solwio11.1 Manual contains worked-out solutions for all odd-numbered exercises in the tex1. It is a great resource to help you understand how co solve those tough problems.
■ Notetaking Guide-This notebook organizer is designed to help you organize your notes. and provides section-by-section summaries of key topics and other helpful study tools. The Notetaking Guide is available for download on the book's website.
■ WebAssign®-The most widely used homework system in higher education, WebAssign offers installl feedback and repeatable problems. everything you could ask for in an online homework system. WebAssign's homework system lets you practice and submit homework via the web. It is easy to use and loaded with extra resource,. With this edition of Larson·s Calculus. there are over 3.000 algorithm ic homework exerci~es to use for practice and review.
■ DVD Lecture Series-Comprehensive, instructional lecture presentations serve a number of uses. Tbey are great if you need ro catch up after missing a class, need to supplement online or hybrid instruction. or need material for self-study or review.
■ CalcLabs with Maple® and Mathematica®- Working with Maple or M(l(l,e111a1ica in class? Be sure to pick up one of these comprehensive manuals that will help you use each program efficiently.
xv
xvi
Additional Resources
Instructor Resources
■ WebAssign®-lnstant feedback, grading precision. and ease of use arc j ust three reasons wh) WebAssign is the most widely used home~ork system in higher cducution. WebAssign's homework delivery system lets instructors deliver. collect, grade. and record assignmems via the web. With this edition or Larson·, Calm/us, there are over 3.000 algori thmic homework exercises to choose from. These algorithmic exercises are based on the section exercises from the textbook LO ensure alignment with course goals.
■ Instructor's Complete Solutions Manual-This manual contains worked-out solutions for all exercises in the text. Tt also contains solutions for the special features in the text such as Explorations, Section Projects. etc. It is available on the instructor's Reso11rc:e Ce111er at the book's website.
■ Instructor's Resource Manual- This robust manual contains an abundance of resources keyed ro the textbook by chapter and section, including chapter summaries and teaching strategies. New to this edition•~ manual are the authors' findings l'rom CalcCha1.com (see A Word from the Authors). They offer suggestions for exercises to cover in class. identify tricky exercises wi th lips on how best lo use them, and explain whar changes were made in the exercise set based on the research.
■ Power Lecture--This comprehensive CD-ROM includes the !11s1ruc10r·s Co111ple1e Solwions Ma1111a/. PowerPoint® slides. and the computerized test bank featuring algorithmicall) created questions thar can be used to create. deliver. and cu~tomiLe tests.
■ Computerized Test Bank-Create. deliver, and customize tes ts and study guides in minutes with this easy to use assessment software on CD. The thousands of algorithmic questions in the test bank are derived from the te>-.tbook exerci,es, ensuring co□sistenc) between exams and the book.
■ Joinln on TurningPoint-Enhance your swdents' interactions with you. your lectures, and each other. Cengage Leaming is now pleased to offer you book-specific content for Re~ponse Systems tailored to Larson·s Cairn/us. allowing you to Lrnnsforrn your classroom and assess your students· progress with instant in-clas~ quizzes and polls.
Vectors and the Geometry of Space
This chapter imroduces vectors and rhe d1ree-dimensio□al coordinate system. Vectors are used to represem lines and planes. and are also used to represent quantities such as force and velociry. The three-dimensional coordinate system is used to represent surfaces such as ellipsoids and elliptical cones. Much of the material in the remainjng chapteri. relies on an understanding of this system.
In this chapter, you should learn the fo]Jowing.
• How to write vectors, perform basic vector operations. and represent vectors graphically. (11.1)
■ Ho~ to plot_points in a three-dimensional coordinate system ai1d analyze vectors in space. ( 11 .2)
■ How to find die dot product of two vectors (in die plane or in space). (11.3)
■ How to find the cross product of two vectors (in space). (11.4)
■ How to find equations of lines and planes in space, and how ro sketch their graphs. ( 11.5)
■ Hm, to recognize and write equations of cylindrical and quadric surfaces and of surfaces. of revolution. (11.6 )
■ How to use cylindrical and spherical coordinates to represent surfaces in space. ( 11.7)
Mark Huni/Hunr Stock
Two tugboats are pushing an ocean liner, as shown above. Each boat is exerting a force of 400 pounds. What is the resultant force on the ocean liner? (See Section 11.1 , Erample 7.)
/ 7 ---::__ I
Vectms indicate quantities that involve both magnitude and direction. In Chapter 11, you will study operations of vectors 1n the plane and in space. You will also learn how to represent vector operations geometrically. For example, the graphs shown above represent vector addition in the plane.
763
764
Chapter 11 Vectors and the Geometry of Space
- Vectors in the Plane
■ Write the component form of a vector. ■ Perform vector operations and interpret the results geometrically. ■ Write a vector as a linear combi nation of standard unit vectors. ■ Use vectors to solve problems involving force or velocity.
Q
p / lniti:il
point
~emiinal
point
A directed line segment Figure 11.1
Equivalent directed line segments Figure 11.2
y
4
(4. -1) s
3
J
The vectors u and v are equivalcn1. Fig ure 11.3
Component Form of a Vector
Many quantities in geometry and physics, such as area. volume, temperature. mass.
and time, can be cbarac1erized by a single real number scaled to appropriate units of measure. These are called scalar quantities, and the real number associated with each is called a scalar.
Other quantities, such a!; force, velocity, and acceleration. involve both magnirude and direction and cannot be characterized completely by a single real number. A di rected line segment is used to represenr such a quantity, as ~hown in Figure I I .I.
The directed line segment PQ has injtjaJ point P and terminal point Q. and its length
(or magnitude) is denoted by IIPQ 11-Directed line segments that ha\·e the same length
and direction arc equivalent, as shown in Figure I J.2. The set of all directed line
segments tJ1a1 are equivalent to a given directed line segment PQ is a vector in the plane and is denoted by \' = PQ. In typeset material, vectors are usually denoted by
lowercase. boldface leucrs such ai; u. v, and w. When written by hand. however,
u. vector!, are often denoted by letters with arrows above them, such as I". and 1?.
Be sure you understand that a vector represents a set of directed line segments (each having the same length and direction). In practice, however. it is common not co distinguish between a vector and one of its representative!>.
D EXAMPLE Vector Representation by Directed Line Segments
Let v be represented by the directed line segment from (0. 0) lO (3. 2). and let u be represented by the directed line segment from (I. 2) to (4. 4). Show that v and u are equivale nt.
Solution Let P(O. 0) and Q(3, 2) be the initial and terminal points of v, and le l
R( I. 2) and S(4. 4) be the initial and lerminal points of u. as shown in Figure 11.3. You
RS can use the Distance Formula to show that PQ and have the same length.
JIPQII = ✓(3 - 0)2, (2 - 0)2 = J13
11 7?sll = -✓(4 - 1)2 - H - 2)2 = .Ji3
L~ngth of PQ Lt:ngth c,i R.~·
Both line segments have the same direction, because they both are directed toward the upper right on lines having the same slope.
~ 2-0 2
Slope of PQ = 3 _ 0 = 3
and
Slope
-of RS
=
..i-2 4_ I
=
2
3
RS Because PQ and have the same length and direction. you can conclude that the two
vectors are equivalent. That is, v and u are equivalent.
■
11.1 Vectors in the Plane
765
(0. 0)
PJ
2
A vector in standard position
Figure 11.4
The directed line segment whose initial point is rhe origin is often the most convenient representative of a set of equivalent directed line segments such as those shown in Figure 11.3. This representation of v is said to be in standard position. A directed line segment whose ini tial point is the origin can be uniquely represented by the coordi nates of its tenninaJ point Q(111, v2), as shown in F igure 11 .4.
OEFfNITION OF COMPONENT FORM OF A VECTOR IN THE PLANE
If vis a vector in the plane whose initial point is the origin and whose temlinal
point is (v1. v2 ). then the component form or vis given by
= V {v1. 1•2) .
The coordinates 111 and 112 are called the components of v. If both the initial point and the terminal point lie at the origin, then v is called the zero vector
and is denoted by O = (O, 0) .
_,Q (-2. 5) 6
•
--~-~---+--~ ,-4-► <
-6 -I
4 6
? (3. - 7)
Component form of,,: 1· = (- 5. 12) Figure 11.5
= This definitio n implies that two vectors u (u1, 112 ) and v = (v1, 112) are equal if and
only if u1 = 1•1 and u2 = v2. The fo llowing procedures can be used to convert directed line segments to
component form or vice versa.
= 1.
lf P( p1, p2 ) and Q(q1• q2) are segment, the component fom1
the initial and tem1inal poi111s of a directed line of the vector v represented by -P-Q" is (v1, 112 )
(q1 - p 1, q2 - p ) . Moreover, from the D istance Fonnula you can see tha1 the
length (or m agnitude) of vis
!l vll = J(q 1- P1P + (q2 - P2F
= ✓v?- + vf.
Lcngrh of a vecmr
2. If v = (v1, v2), v can be represented by the direc1ed line segment. in standard
position. from P(O. 0) to Q(v1, v2 ).
= The length of v is also called the norm of v. If II v II I , v is a unit vector.
Moreover, II v 11 = 0 if and only if v is 1he zero vecwr 0.
f l EXAMPLE Finding the Component Form and Length of a Vector
F ind the component fom1 and length of the vector v rhat has initial point (3, - 7) and terminal point ( - 2, 5).
Solution Let P(3, -7) = (p1.p2 ) and Q(- 2. 5) = (q1,q2 ) . Tben the components
o f v = (1•1. v2) are
= = " 1 C/1 - Pr -2 - 3 = - 5
= = 1'2 fJ2 - P2 = 5 - (- 7) 12.
So. as shown in Figure I 1.5. v = (- 5. 12). and the length of v is
ll vll = ✓(- s)2 + 122
✓169
13.
■
766
Chapter 11 Vectors and the Geometry of Space
I - - -- - V
- - -· -
-
v
- -3 v 2
-- _ __ 1 _ _ _/ _
- --
- - - - --
--·-
- - - - - ---
Thescalar multiplication or v Figure I 1.6
Vector Operations
DEFINITIONS OF VECTOR ADDITION ANDSCALAR MULTIPLICATION
Let u = (11 1, 112) and v = (v1, 112) be vectors and let c be a scalar.
1. The vector sum or u and vis the vector u + v = (u + v u +
,
1•
)
.
1
1
2
2
2. The scalar multiple of c and u is the vector c u = (c11 1, cu2).
3. The negative of v is the vector
= -v = (- l)v ( - 111. -v2 ).
4. The difference of u and v is
Geometrically, the scalar multi ple of a vecwr v and a scalar c is the vector th at is
kl times as long as v. as shown in Figure J1.6. If c is positive, cv has the same
direction as v. lf c is negative, c v has Lhe opposite direction.
The sum of two vectors can be represented geometrically by positioning the
vectors (wilhoul changing their magnitudes or cLirections) so that the initial point of one coincides wi th the term inal point of the other. as shown in Figure l] .7. The
vector u + v, called the resultant vector, is the diagonal of a parallelogram having u
and v as its adjacent sides.
a + v~ _ _
-=:::::::::::
To lind u + I',
Figure 11.7
(I) move the initial point ofv (2) mo\"e the initial point or u to the terminal point of u, or to the terminal point or\'.
WILLlAMl!OWAN l:iAMLLTON (1805-1865)
Some of the earliest worlt with vectors was done by the Irish mathematician William Rowan Hamilton. Hamilton spent many years developing a system of vector-like quantities called quatem,ons. Although Hamilton was convinced of the benefits of quaternions, the operations he defined did not produce good models !or physical phenomena. II wasn't until thelatter half of the n111eteenth century that the Scottish physicist James Maxwell (1831- 1879) restructured Hamilton's quaternions in a form useful for representing physical quantities such as force, velocity.
_ J and acceleration._ _ _ _ __
Figure 11.8 shows the equivalence of the geometric and algebraic definitions of vector addirion and scalar multiplication. and presents (at far rigbt) a geometric interpretation of u - v.
Cul + 1'1· u2 + "1)
(111'112!_ _
Vector addition Figure I 1.8
Scalar multiplication
Vector subtraction
11.1 Vectors in the Plane
767
IJ EXAMPLE Vector Operations
G iven v = ( - 2, 5) and ,, = (3, ..i), fi nd each ofrhe vec1or~.
b. \\ - V c. '" + 2 \1
Solution
a. ½v = (½(- 2).½(5)) = (- 1. ~)
= = = b. w - V ( 11· 1 - l' 1, ll'2 - 1•2> (3 - ( -2).-+ - 5) (5. - 1)
c. Using 2w = (6. 8). you ha,e
, + 2\\ = (- 2.5) + (6. 8)
= (- 2 -'- 6. 5 + 8)
= (4. 13) .
•
Vec1or adLlition and scalar multiplication share many properties o f ordinary ,u-ith me1ic. a'> shown in the following theorem.
THEOHE\! 11 .1 PROPERTIES OF \'ECTOR OPERATIONS
Let u, v. and w be l'ectors in rhc plane. and let c and d be ~calars.
L.u +v = v + u
2. (u + v) + w = u + (v + w)
3. u + 0 = u
4.u +(- u) = O
5. c(du) = (cd )u
6. (c + d )u = c u , d u 7. d u - v) = c u + cv
8. l{u) = u, O(u ) = 0
Curnmulat1,·e Property ..\'.'rl,oc1auvc Prupcrt~
AJJuivc In, er,,;• Propt'rty
D1,1ribu1i1e Pr.•pcrr~ D1-tribuu, c Propcrr:
~ The proof of the Associmi,1e Pmp err:, of vector addition uses the A ssocia1ive Property of addition of rea l numbers.
= ( U + Y ) + W + [ ( 11 1. 111 ) ( 11 1. 1•2) ] -f- ( 11•1. 1112)
= + ( 11 1 -f- 1' 1.111
r 2) -f- (11·1. 11',)
= ((11 1 + 1·1) + 11·1. (112 + ,·2 l + ,r 2)
= + ( 11 , -f- (1·1 + 11'1).112 -f- ( 1·2 11·2))
= + = ( 11 1. 111)
( v1 + 1111. v1 + 1111)
11 -
(v + w )
Similarly. the proof of the D i.1tri b111i ve Property o f vectorn depends on the Di~Lributive P ropeny of real numbers.
+ (c d )u = (c , d)(111. 112) ""((c + d ) u 1. (c + c/) 112) = (cu1 + du 1• rn 2 + d u1) = (cul . rn2) + (du1. d11 2) = ru + d u
The m hcr properties can be proved in a similar manner.
■
768
Chapter 11 Vectors and the Geometry of Space
Any set of vectors (\~ ith an accompanying set of scalars) that satisfies the eight properties given in Theorem l l. l is a vector sp ace.* The eight properties are the vecror space axioms. So. this theorem states that the set of vectors in the plane (with the set of real numbers) forms a vector space.
THEORlDI 11.2 LENGTH OF A SCALAR MULTIPLE
'
Let v be a vector and let c be a scalar. Then
llcvll = kl llvll-
( PROOF ) Because cv = (n•1. cv2). jt follows that
DL\1\ N0ETI IER (1882-193-5-')- ~
One person who contnbuled to our knowledge of axiomatic systems was lhe German
llnll
=
ll(n-1, c112)l1
=
J(C\'Y
+
2 (c1~)
= ✓c'lPP, + c1,,f
mathematician Emmy Noether. Noelher is
Jc1h~ + rfl
generally recognized as the leading woman mathemallc1an 1n recent history.
icl Jv~ + ''i
!cl llvll-
■
FOR Fl,lffHER rKFQR,\JATIOJli For more infom1ation on Emmy Noelher. see the article "Emmy Noether. Gremest Woman Mathematician" by Clark Kimberling in The /lla1/wmarics Teacher. To view thh article. go 10 the website
w1,·11·.111,11har1ic:le.~.co111.
In man) applications of vectors. it is useful to find a unit vector chat has the same direction as a given \'ector. The following theorem gives a procedure for doing this.
THEO RDI 11.3 UNIT\ ECTOR JN THE DIRECTIONOF,,
If v is a nonzero vector in the plane, then the vcc.:Lor
'"
I
u= w=rvr
has lenglh 1 and the same direction as"·
( PROOF) Because 1/llvll is positive and u = (I /llvll)v. you can conclude that u has the same direction as v. To see Lhat II u II == 1. note that
!lull= IIC1:ll)vll
== 111~111llvll
I
= w llvll
= I.
So, u has length I and the !'.::IJJ1C direction as , .
■
ln Theorem 11.3, u is called a unit vector in the direction of v. The process of
multiplying vb) l / II vii lo get a unit vector is called normalization of,•.
"' For mor<! i11fomw1io11 a/10111 1•ec1or spac<!s, see Elementary Linear /'I. lgebra, SiXlh Edi1io11, by
wrso11 am/ Fa/l•o (Bos1011: H oughum Mi.fl1i11 Harcourt P11blishi11g Co111pa11y. 2009}.
Triangle inequality Figure 11.9
I Tl j = (0. I )
I -
--'---
-i =•(-'I--.
0) -
-
-'-------• .\
1
2
Standard unit vectors i and j Figure 11.10
11.1 Vectors in the Plane
769
IJ EXAMPLE Finding a Unit Vector
Find a un.it vector in the direction of v = (- 2, 5) and verify that it has length I.
Solution From Theorem 11 .3, lhe unit vector in the direction of Y is
~\w 11:11= ✓(~;)~,
~ = <- 2• 5> = (~- ~)-
This. vector has length I. because
J( m )1 ( )2 -?. + ✓'519 = ✓~ 29 +29 = ✓/2T99= 1.
■
Generally. the lengrh of' the sum of two vectors is not equal to the sum or
their lengths. To sec this, consider the vectors a and v as shown in Figure 11.9. By considering u and v as rwo sides of a triangle, you can see that rhc length of the third
side is IIu + v II. and you have
llu + vii s llull + llvll-
Equaliry occurs only if the vectors u and v have the same direction. This n:sult is called the triangle inequality for vectors. (You are asked to prove this in Exercise 91. Section 11.3.)
Standard Unit Vectors
The unit vecrors (I, 0) and (0. 1) are called the standard unit vectors in the plane and arc denoted by
i = (1,0) and j = (0. 1)
S1andarJ unit WCIOf~
as shown in Figure I 1.10. These vectors can be used to represent any vector uniquely, as follows.
T he vector v = 111 i + ,·2 j is called a linear combination of i and j. The scalars v1
and 1•2 are called the horizontal and vertical components of v.
1:1 EXAMPLE Writing a Linear Combination of Unit Vectors
Let u be the vector with initial point (2. - 5) and terminal point (- I. 3). and let
v = 2i - j. Write each vector as a linear combination of i and j .
a. u
b. w = '.!u - 3v
Solution
a. u = (q1 - /J1,l/2 - P1)
= (- I - 2, 3 - ( -5))
= ( - 3_8) =-Ji + 8j
b. w = '.!u - 3v = 2( - 3i + 8j ) - 3(2i - j )
= - 6i + I6j - 6i + 3j
= -12i + 19j
■
770
Chapter 11 Vectors and the Geometry of Space
I
. . 1 _- • I I ---_(cos e. ,in OJ ' a. --
,,,
"''
IJ
\ sin 0
-1
cos e
, .. ... fl ., ,,. ,'
---1 - - -
The angle 0 li-0111 the positive x-axis w the 1ecror u Figure 11.ll
lf u is a unit vector and O is the angle (measured counterclockwise) from I.be posirive x-axis to 11. then the terminal point of u lies on the unit circle. and you have
= u = (cos 0, sin 8) cos 0i + .~in Oj
1_ nit vcclor
as shown in Figu re 1 1.11. Moreover. it follows that any other nonzero vector v making
an angle (} with the posiri,e x-axis has the same direction as u. and you can write
v = II v ll(cos 0, sin 8) = I vii cos 0i + II v I sin 0j.
II EXAMPLE Writing a Vector of Given Magnitude and Direction
The vector v has a magnitude uf 3 and makes an angle of 30° = 1r/ 6 with I.be positive
x-a'lis. Write v as a linear combination of the unit vectors i and j .
Solution Because the angle between v and the positive x-axis is 8 = 11/ 6. you can
write the following.
v = Iv II cos Oi + II,.I sin Oj
= + •3
COS•
7T ,
-6 1
3
.
Sill
-16TJ.
=
3.J3.
--1
+ -3J.
2
2
■
400 co,(- 20") -100 ,in(-'.!O' /
Applications of Vectors
Vectors have many applications in physics and engineering. One example is force. A
vector can be used to represent force, because force has both magnirude and direction. If two or more forces are acLing on an object, then the rcsuJt.ant force on the object is the vector sum of the vector force,,.
II EXAMPLE Finding the Resultant Force
Two tugboa!S are pushing an ocean liner. as sbown in Figure 11.12. Each boat is exerting a force of 400 pounds. What is the resuJtanc force on the ocean liner?
Solution lJsing F igure 11.12. you can represent the forces exerted by the first and second tugboats as
F1 = 400(cos 20', sin 20°)
= -100 cos(20°) i + -+00 sin{20°)j F1 = -I-OO(cos(-20°). sin(-20°))
= 400 cos(20°)i - -100 sin(20°)j .
The resultant force on the ocean liner is
F = F1 + F2 = [400 cos(20°)i + 400 sin(20°)j] + [400 cos(20°)i - 400 sin(20°)j]
= 800 cos(20°)i = 752i.
So. the resulrant force on the ocean liner is approximately 752 pounds in the direction
o f tl1e positive x-axis.
■
The resultant rorcc on the ocean liner ihat is exerted b) the two tugboats
Figure 11.12
In surveying and navigation, a bearing is a direction that measures I.be acute angle that a path or line of sight makes with a fixed north-south line. In air navigation, bearings arc measured in degrees clockwise from north.
11.1 Vectors in the Plane
771
y
N
W+E
s
El (.; EXAMPLE Finding a Velocity
An airplane is Lraveling at a fi xed altitude with a negligible wind factor. The airplane is traveling at a speed of 500 mile, per hour with a bearing of 330 . as shown in Figure l I. l3(a). As the airplane reaches a certa in point. it encounters wind with a velocicy of 70 miles per hour in the direction N 45° E (-+Sc east of north). as shown in Figure l I. I3(b). What are the resultant speed and direction of the airplane?
(a) Din.'Clion Wllhout \\ ind
Solution Using Figure I 1. l 3(a). repre~enc the velocity of the airplane (alone) as
v1 = 500 cos( I20°}i + 500 sin( 120°)j .
The velocity of the wind i~ represented by the vector
v2 = 70 cos(45°)i + 70 sin(45°)j .
The resultant velocit) of the airplane (in the wind) is
, = v1 + v2 = 500cos( l20°) i + 500 sin(l20°)j + 70cos(45°)i + 70sin(45°)j
= -200.:i i + 482.5j.
To find the resulcam speed and direction. write Y = II Y/lh:os 8 i + sin 0 j ). Because
= l\ v\\ = J( - 200.5)2 + (482.5)2 522.5. you can write
= j)= v 522.5 ( --~~O/ i + ~~~-)S-
522.5 [cos( 1l2.6°)i + sin( l 12.6°)j).
j -·
) ___
- - - - - - ~-.......1..---~ x
(bl Dir~ction with \\ ind Figure 11.13
The new speed of the airpla ne. as altered by the wind, is approximately 522.5 miles per
hour in a pat11 that makes an angle of 112.6° with the po~itive x-axis.
■
Ill Exercises
Se:> .1M1 CalcChat.com for 1•,or,ed-out solu: ons to odd-numbered e,erc1ses.
In Exercises 1- 4, (a l find the component form of the wctor v and (b) sketch the vector with its initial point al the origin.
I.
2.
-t- t l 3 -I 5
3.
-'-
"+ (--t - 3 1 (2. -3)
-6-
V
-I I 2 -I ; 6
t3. -2)
4.
(- 1. 3) -1 -~
~ ft v (2.I J
-2 - I
,I •
I 2
In Exercises 5-8, find the vectors u and v whose initial and terminal points are given. Show that u a nd v are equivalent.
5. u: (3. 2). (5. 6) v: ( I. 4). (3. 8)
7. u: (0. 3). (6. - 2) v· (3. 10). (9. 5)
6. u: (-4. O). (l. 8) v: (2. - I). (7. 7)
8. u: (--l.- 1) . (11.-4) v: ( I0, I 3). (25. !O)
In Exercises 9-1 6, the initial and terminal points of a vecl<lr v are gi"en. (a) Sketch the given directed line segmen t. (b) ,1rite the vector in component form. (c) write the vector as the Linear combination of the s tandard unit vectors i and j , and (d ) s ketch the vector with its initial point at the origin.
/11i110/ Poi111
9. (2. 0) 11. (8. 3)
T em1inal Pui11r
(5. 5) (6. - I )
lnirial Poim
10. (4. - 6) 12. (0. - 4)
Tem1i11al Poim
(3.6) (- 5. - 1)
TT1e icon ~ i11dicares rhat you ll'i/1 find a CAS l m·e!itigmicm 011 the book 's websi1e. The CAS
lm•eHigmio11 is a cnllabomril'e explora1io11 of1/,is example 11si11g rhe compurer algebra sys1e111s
Maple and Mathematica.
772
Chapter 11 Vectors and the Geometry of Space
Tnitial Point
13. (6. 2) 1:~,. ~'~ ~j )
Ter111i11af Poim
(6, 6)
(½. 3)
Initial Poinr
14. (7, -1) 16. (0.12, 0.60)
Terminal Poim
(-3. - 1)
(O.~. 1.25)
In Exercises 17 and 18, sketch each scalar multiple of v.
17. \' = (3. 5 )
(a) 2\' (b) - 3,·
18. r = (- 2, 3)
(a) -I\' (b) - }I V
(C) ; V
(c) o,
(d) fv
(d) -6v
In Exercises 19-22. use the figure to sketch a graph of the vector. To print an enlarged copy of the graph. go to the website www.ma1'1graph s.co111 .
\' !
I
---,--- - - -- - - - . ,
19. - u 21. U - V
20. 2u 22. u - 2v
In Exercises 23 and 24, find (a) 3u. (b), - u, and (cl 2u + 5\'.
23. u = (-1, 9) , = (2. - 5)
24. u = (-3. -8) V = (8. 25)
In Exercises 25-28, find the vector v where u = (2, - 1) and w = (1. 2). Illustrate the vector operations geometrically.
25. = V ~U
27. \'=UT 2w
26. V = U + \\" 28. v = Su - 3w
In Exercises 29 and 30, lhe \'ector v and its initial 11oint arc given. Find the terminal point.
29. \' = (- I. 3): Initial po1m: (-1. 2) 30. v = (4, - 9): lnilial point: (5, 3)
In Exercises 31-36, find the magnitude of v.
31. v = 7i 33. V = (4, 3)
35. , = 6i - 5j
32. v = -3i
34. v = ( l2. -5)
36. v = - lOi + 3j
[n Exercises 37-40, find the unH vector in lhe direction of v and verif) that it has lenglh 1.
37, V = (3, 12) 39. ~ = (i. i)
38. V = (- 5. 15) -10. V = (-6.2,3.-1)
In Exercises 41-44, find the following.
(a) II ull
( b) II vii
(c} II u + vii
(d) II 11 : d
(e) ll 11 :1, II (f) II II : : : II II
-U. u = (l. - I) V = (- [. 2)
43. u = (J. ½) V = (2. 3)
42. u= (O, 1) = V (3. -3)
4-t. u= (2. --1)
V = (5. 5)
[n Exercises 45 and 46, sketch a graph of u, v. and u + v. Then
demonstrate the triangle inequality using the vectors u and v.
= -15. LI = (2. I). V = (5, -1) -16. U = (-3. 2). V (I, - 2)
In Exercises 47-50, find the vector v with the given magnitude and the same direction as u.
Mngni111de
47. llvll = 6 -I8. ll vll =-+ 49. llvll = s 5o. ll vll == 2
Direction
u = (0, 3)
u = (I, I)
u = (- 1, 2) u = (-/3, 3)
In Exercises 51-54, find lhe component form of v given its magnilude and the angle it makes with the positive x-axis.
51. II v II = 3. 0 = 0° 53. llvll = 2. o= 1so0
52. ll vll = s. 0 = 120° 54. I vII = 4. 0 = 3.5°
In Exercises 55-58, find the component form of u + v given the
lengths of u and v and the angles that LI and ,, make with the positive x-axis.
55_ llu11 = 1.
llvll = 3. 57. /l ull= 2.
11 vii = t.
Bu= 00 8,. = 45° Ou= 4 8, =2
56. 1 u11 = -1.
ll vll = 258. llull = 5.
ll vll = s.
Ou= 00
8, = 60°
Ou= - 0.5
8, = 0-5
WRITING ABOUT CONCEPTS
59. In your own words. state the difference between a scalar and a vector. Gi1·e examples of each.
60. Give geometric descriptions of rhe operations of addiLion of vecLor.. and multiplicaLion of a vector b) a scalar.
61. ldcmify the quantity as a scalar or as a vccmr. Explain your reasoning. (al The muuJe velocity of a gun (b) The price of a company's stock
62. Identify the quantity as a scalar or as a vecmr. Explain your reasoning.
(a) The air temperature in a room (h) The weight of a car
11.1 Vectors in the Plane
773
In Exercises 63-68, find a and b such that v = au + bw, where tit ln Exercises 79 and 80, use a graphing utility to find the
u = (1. 2) and w = (1. - 1).
magnitude a nd direction of the resultant or the vectors.
63. ..- = (2. l ) 65. " = (3. 0) 67. , = ( I. I)
64. v = (0. 3) 66. v = (3. 3) 68. ,. = ( - I, 7)
79.
.r
80 .
In Exercises 69-74. !ind a unit Yector (a) [larallcl to and
(b) perpendicular to the graph off at the given point. Then
sketch the graph or/ and sketch the \"CCIors at the given point.
F1111ctio11
69. / (x) = xl 70. J (x ) = 5 -x! -'71. J(x) = x3
72. f (x ) = .r '
73. J(x) = J2s - .r!
Point (3. 9) ( l. 4)
( I. I ) (-2. -8) (3. 4)
74. f(x) = tan x
81. Resultant Force Forces with magnirudes of 500 pounds nnd 200 pounds act on a machine pan al angles of 30° and --1-5°, respectively. with the x-axis (see figure). Find the direction and magnitude of lhe resultant force.
fn Exercises 75 and 76, find the component form of v given the
magnitudes of u and u + v and the angles that u and u + v
make with the positive x-axis.
75. !l ull= I. 0 = -I-Se
11 u + ,.11 = , 2. 0 = 90°
76. !l ull= -1-. 0 = 30° II u + \' II = 6. () = I 20°
77. Programming You are given the magnitudes of u and v and the angles that u and v make with the positive x-ax.is. Write a program for a graphing utility in which the outpm is the following.
(a) u + v (b) II u + vII
(c) The angle that u -r v makes with the positive x-axis
(d) Use the program ro find tl1c magnitude and direction of the resul!am o f the vectors sbO\\ n.
Figure for 81
Figure for 82
82. Numerical and Graphical Analysis Forces with mngnirudcs of 180 newwns and 275 newtons act on a hook (see figure). The angle between the two forces is 0 degrees.
= (a) If 0 30°, find the dircclioa and magrritutlc of the resu ltant
force.
(b) Write the magnitude M and direction a of the rcsullant force as functions of 0, where 0° 5. 0 5. 180°.
(c) Use a graphing utility 10 complete the table.
I (} 0° 30° 60° 90° 120° 150° 1so0
M
_I _ ____
j
CAPSTONE
78. The initial and terminal points of vector v are (3. --1-) and (9. I). rehpecrivcly. (a) Wri1e ,· in component form. (b) Write v as the linear combination or the standard uni1 \'ector.. i and j. (c) Sketch v wi th its inilial point a1 the origin. (dl Find the magnitude of,•.
(d) Use a graphing utility to graph the two functions Mand a .
(c ) Explain why one of the ft.ms_1ions decreases for increasing values of Owhereas the other does not.
83. Resultant Farce Three forces \\ 11h magnitudes of 75 pounds, I00 pounds, and J25 pounds act on an object at angles of 30°, -15°. and 120°. respectively. with the positive x-axis. Find the direction and magnitude of the resultant force.
84. Resultant Force Three force, wi th magnitudes of 400 newtons, 280 newtons. and 350 newtons act on an object al angles of -30°. -1-5°. and 135°. respectively. with the positive x-axis. Find the direction and magnitude of the resultant force.
85. Think About It Consider two forces of equal magnitude acting on a point.
(al [f the magnirudc of the resultant i~ the sum of lhe m.ignitutlcs of the 1wo forces, make a conjecture about the angle between the fon:es.
774
Chapter 11 Vectors and the Geometry of Space
(b) ff the resultant of the forces is 0. make a rnnjeclure abou t the angle between the forces.
(c) Can the magnirndc of the resullanr be greawr th an 1he su m of the magn1tudei. of the 1wo forces? Explain.
e. 86. Graphical Reasoning Con,ider two forces F 1 = ('.!0. 0) and
F : = I0(cos sin 0).
(a) Find IIF, + F2)1.
Cb) De1er111ine thl' magnitude of the resultam as a function of 0. Use a graphing utility to graph the function for 0~0<~;;.
{c) Use the graph in part (h) to dctennine the range of the functilltl. \\11aL is its ma'-imum and for\\ hat q1lue of 0 doc~ it occur? What is it<, minimum and for\\ hat va lue of {I does 11 occur'?
(d) Explain wh} the magnitude of the re~ultant b never 0.
87. Thn:c vertice~ of a parallelogram are ( I, 2), (3. I ). and (8, 4J.
Find the three pos,ible fourth vertices (sec figure).
j
b
5
.j
• (8.4)
3 2
.
• n. 2>
I
• (3. I J
- I - + - ; - -t-+--.-t--+-+ +-t-t--- .r
- .J - 3-! - I
I ~ 3 -I 5 6 7 s 9 IO
88. Use vector, to find the points of trisection of the line segment with endpoints (I. 2) and (7. 5).
Cable Te11sio11 In Exercises 89 and 90, use the figure to determine the tension in each cable supporting the given load.
89.
90.
8
~ .,_._.::2:.:.:0_i.c:.n·c.._..,
1 A
B
~4 in.
L _---
91. Projectile Motion A gun with a munJe velocity o r 1200 feel per ~ecoad 1s fired at an angle of 6° above the hori/onml. Find the ,crtical and honzontal components of the velocity.
92. Shared Loatl To CaIT} a IO0-pountl cylindrical weight two worker~ lift on the ends of shon ropes tied 10 an eyelet on the top center or the cylinder. One rope makes a 20° :rngle awa) from the vertical and the other makes n 30° .inglc (see figure).
(a) Find each rope', tension if the resultant force is vertical.
(b) Find the venic,1J componc111 of each worker's force.
✓ -10•~ '' 30
~;J•\;/
N
W+ E
~
s
, , 100 km/hr
- ------- -
\ -15"
Figure for 92
Figure for 93
93. Nm•igation A plane i\ fl ying with a bearing of 302°. ll~ speed with respect to the air i~ 900 kilometers per hour. The I\ ind al the plane·s alwude i~ from the ,ourhwesr at I00 ki lometer, per hour (see ligureJ. Whal is 1hc true direction of the plane. and what is its ~peed wit.h respect to the ground?
9-t Nm•igatio11 A plane flies at a conqant groundspeed of .WO
miles per hour due east and encounters a 50-milc-per-hour wind from the northwest. Find the airspeed and compass direction !l1nc will allo\\ the plane ro maintain its grouodspeed and eastward direction.
Tnie or False? In Exercises 95-100, determine whether the statement is true or fal~e. If it is false, e:q1lain why o r give an example lhut shows it is false.
95. If u and v ha1·e the same magnitude and direction. then II and v arc equivalent.
96. If u i, a unit vector 111 the direction of v. then v = ll vllu.
97. If u = a i + bj is a unit 1·cccor. then a ' - b = I.
98. lfv = a i + bj = 0. then a= - b.
99. ff a= b. then ll a i + bj = .,.,:!a.
100. If u and v have the same magnitude but opposite directions.
then u + ,. = 0.
101. Pro1e that u = (cos (Jli - (sin O)j and , = (sin 0)i - (co~ O)j
are un11 vectnrs for any angle 0. 102. Geometry Ltsing vector-.. prm-c tha1 the line seg:memjoining
the midpoinls of two sides of a triangle is parallel to. and onehalf the length or, th!! thinl side.
103. Geometry Usi ng 1eccors, prove that t.he diagonaJs of a parallelogram bisect each other.
104. Prove that 1he vector w = ll ull v + ll viu bisects the angle
between u and ,•.
!05. Consider the vector u = (x. v). Describe the set of all points
11.y) ,uch th,11 lull= 5.
PUTNAM EXAM CHALLENGE
L06. A coast artillery gun can fire at any angle of elevation bef\1 t'en 0° illld 90' in a fJ.).ed , cnical plane. If air resistam.:c is ncglectcu and the munJe veloc ity is com.tam (= 1·0). determine the set Hof points in the plane and abo1e the hori10111al which t:an be hit.
Thi, pmhlem \\a., coinpo,rd h} !ht: Comm111ee on the Pu1nam Pnze Compcuuon C: lltc 1\1.uhcm;mt.-al ,\~sociut10n of Americn, ,\ll right:-- reserved
11.2 Space Coordinates and Vectors in Space
775
IIIJ Space Coordinates and Vectors in Space
■ Understand the three-dimensional rectangular coordinate system. ■ Analyze vectors in space. ■ Use three-dimensional vectors to solve real-life problems.
..:-plane
y;:-plane
."
.I
.T)·-plane
1
The three-dimensional coordinate sys1em
Figure 11.14
Coordinates in Space
Up to this poinr in the text. you have been primari ly concerned with the two-dimensional coordinate system. Much of the remaining part of your study of calculus w ill involve the three-dimensional coordinate system .
Before extending the concept of a vector to three dimensions. you must be able to identify points in the three-dimens ional coordinate system. You can conscruct
this system by passing a :-axis perpendicular to both the x- and y-axes at the origin.
Figure 1 1. 14 shows the positive porlion o f each coordinate axis. Taken as pairs. the axes de1ermine three coordinate planes: the xy -plane, rhe xz-plane, and the yz-plane. These three t:oordinatc planes separate three-space into eight octants . The first octant is the one for which all three coordi nates are positive. In this 1hreedimcnsional system , a poi nt P in space is determined by an ordered triple (x. y, z) where x. y, and .: are as fol lows.
.r = directed d istance from _,·:-plane to P
y = d irected distance l'rom .r;:-plane to P .: = d irected distance l'rom .1:r-plane to P
Several points are shown in Figure 11. 15.
(2. - 5. 3)
•
' ,
(-2.5. 4)
•
Right-banded system Figure 11.1 6
• \3. 3. -2)
Points in the three-dimensional coordinate system arc reprcsemcd b) ordered triples. Figure J1.15
1rt
X
\'
..,
Left-handed system
A three-dimensional coordinare system can have either a left-handed or a rightbanded orientation. To determine the orientation of a system. imagine 1bat you are sranding at the origin. with your arms pointing in the direction of the positive x- and y-axes, and with rhe ::-axis pointing up. as shown in Figure 11. I6. T he system is right-handed or left-handed dependi ng on which hand points along 1.he .r-axis. In this text, you will wo rk exclusively with the right-handed system.
CilD T he three-dimensional rotatable graph, that are a\'ailablc in the premium cBook for
Lhi~ text will help you visualize points or objects in a three-dimen,ional t:oordinate system. ■
776
Chapter 11 Vectors and the Geometry of Space
The distance between two points in space
F igure 11.17
Figure 11.1 8
Many of the formulas established for the two-dimensional coordinate system can
be extended Lo three dimensions. For example, to find the distance between two points
in space, you can use the Pythagorean T heorem twice, as shown in Figure I 1.17. By
doi ng this, you will obtain the formula for the distance between lhe points (x . y • :: )
and (x2, y2. .:2 ).
1 11
IJ1s1anc,· Fonnuta
D EXAMPLE Finding the Distance Between Two Points in Space
The distance between the points (2, - I. 3) and ( I, 0, - 2) is
d= ✓(I -2)2+(0+ 1)2+ (-2-3)2
= ✓ 1 -1- I + 25 = Jn
Distance Formula
=3J3.
•
A sphere with center at (x0 , y0 . 2o) and radius r is defined to be the se1 of all points (x, y, .::) sucb tha t the distance between (x, y, z) and (x0, y0 , z0 ) is r. You can use the Distance Formula to find the standard equation of a sphere of radius r. cemered al
(.r0 , y0, .::0). If (x. y, z) is an arbitrary point on che sphere. the equation of the sphere 1s
Equaoon ,11 sphcr,·
ai, shown in Figure 11.18. Moreover, the rnidpoim of the line segment joining the points (x 1• y 1. ::1) and (x2• Ji, z2) has coordinates
:\liJpt1mt fonnula
f l EXAMPLE Finding the Equation of a Sphere
Find the standard equation of the sphere that has the points (5. - 2, 3) and (0, 4, - 3) as endpoints of a diameter.
Solution Using the Midpoint Formula. the center of the ~pbere is
(l ( 5
+ 2
o -2 + • 2
4 '
3
2
3)
=
2'
I '
o) •
;\lidpoi11t Formula
By the Dist.ance Fom1ula. the radius i~
J( s) m 2
,. =
0 - 2 + (-l - 1)?- + (- 3 - 0)2 = ✓ 4 = 2.ff·i
~r T herefore, U1e standard equation of the sphere is
(x -
-1- ( y - I )2 + z2 = ~ .
•
11.2 Space Coordinates and Vectors in Space
777
(0. 0. I) k y
(I, 0. 0)
X
The standard unit vectors in space Figure 11.19
.t
Figure 11.20
Vectors in Space
In space. vectors arc denoted by ordered triples v = (v1, v2. 1·1). The zero vector is denoted by O = (0,0,0). Using the unjt vectors i = ( i,O,O). j = (0. l,O),
= and k (O, 0 , I) in the direction of the positi,·e :-axjs_ the standard unit vector
notation for v is
= + + V 11, i
V2 j
\13 k
as shown in Figure l l. l9. If v is represented by the directed line segment from P( p 1. p 2. p3) to Q(q1• q2. q3), as shown in Figure 11.20. the component form of v is given by subtracting the coordinates of the initial point from the coordinates of the
terminal poin1. a<; follows.
v = (v1, 1'2• 1•_;) = (q1 - P 1· q2 - P2· q3 - p3)
VECTORS IN SPACE
Ii
= = Let u (111, 112 , u3) and v (v1, v2, 1•3) be veccors in space and let c be a
scalar.
= = 1. Equality of Vectors: u v if and only if 111 = v1• 11~ v2 • and 113 = v3•
2. Component Form: ff v is represented by the direcced line segment from
P( P1, P2· P3) to Q(q1 . C/2• q3), then
= = v (v1, 1'2· 1'3) ( q1 - P1• C/2 - P2, qJ - p ~).
3. Length: llvll = ✓v? + vf + vl
4. Unit Vector in the Direction ofv: ll: II = (11 : 11) (11 1• v2• v3).
5. \lec1or Addition: v + u = + (v1 11 1, 112 + + 112, v3 u3)
6. S('{llar Multiplication: cv = (n•1. cv2, cv3 )
v =I=- 0
CID The properties of vector addition and scalar multiplication given in Theorem 11.1 are
a l~o valid for vecIOrs in space.
■
D ( ; EXAMPLE Finding the Component Form of a Vector in Space
Find the component form and magnitude of the vector v having initial poinr (- 2, 3, I) and terminal point (0. - 4, 4). Then find a unit vector in the di rection or v.
Solution The component form of v is
V = (q, -p1,q1 - p2,q3 -p3) = (0 - (-2), - 4 - 3, 4- I)
=(2, - 7,3)
which implies that its magnitude is
ll vll = ✓22 + ,-1F + Y = J6i.
The unit vector in the di rection or vis
N V
I
/ 2 -7 3 \
u = = JG2 <2. - 1. 3) = \ v®· -162' v®l
■
778
Chapter 11 Vectors and the Geometry of Space
Recall from the definition of scalar multiplication that positive scalar multiples of a nonzero vector v have the same direction as v, whereas negative multiples have the direction opposice of"· In geneml. two nonzero vectors u and ,. are parallel if there
is some scalar c such that u = cv.
DEFINITIO~ OF PARALLEL VECTORS Two nomero vecror~ a and v ::ire para llel if there is some scalar c such that
= U CV.
P:irallcl ,ertors Figure 11.2 1
The points P. Q. and R lie on the same line.
Figure 11.22
For example. in Figure 11.21. the vec tors a. v. and ware parallel because u = 2v and \\' = -v.
II EXAMPLE Parallel Vectors
Vector w has initial point (2. -1, 3) and terminal point (---k 7 . 5). Which of 1he followi ng vectors is parallel tow?
a. II = (3.- ➔,- I }
b. v = (12. - 16.-4)
Solution Begin by writing win component form.
= w (-.i - 2, 7 - t- t). s - 3) = (-6. 8, 2) = a. Because u (3, -➔. -1 ) = -½(-6. 8, 2) = - ½w. )OU can conclude tha1 u is
parallel to w.
b. ln th is case, you want to find a scalar c such that
( 12. - 16. 4) = c(-6, 8. 2).
12 = - 6c ➔ C = - 2
= - l6
8c ➔ C = - 2
4 = 2c ➔ C = 2
Because there is no c for which the equation has a solution. the vectors are not parallel.
II EXAMPLE Using Vectors to Determine Collinear Points
Determine whether the point!- P( I. - 2. 3). Q(2. I. 0). and R(-l. 7. -6) are collinear.
Solution The componem fom1s of -P-Q-' and -P-'R> are
PQ = (2 - I. l - (-21. 0 - 3) = (I. 3, -3)
,Uld
PR= <-i - 1. 1 - (-2). -6 - 3) = (3. 9. -9).
These two vectors have a common inilial point. So. P. Q, and R lie on the same line
~
~
~
~
if and only if PQ and Pl? are parallcl-wl1ich they arc because PR = 3 PQ, as shown
in Figure 11.22.
■
QI (0. - I. 0) , . - ,
/ ~,(1H,
.t
Figure 11.23
11.2 Space Coord inates and Vectors in Space
779
II EXAMPLE Standard Unit Vector Notation
a. Write the vector v = ..J-i - 5k in component fonn. b. Find Lhc terminal poim o f the vcccor v = 7i - j + 3k. given that the initial point
is P( - 2, 3. 5).
Solution
a. Because j is missing. its componem is Oand
v = 4i - 5k = (4. 0, - 5).
b. You need to find Q(q1. q2, r13) such that v = PQ = 7i - j + 3k. This implies that
q , - t- 2) = 7. q2 - 3 = - I. and q3 - 5 = 3. The solution of these Lhree
equacions is q 1 = 5. q2 = 2. and q3 = 8. Therefore. Q is (5. 2. 8).
■
Application
IJ EXAMPLE Measuring Force
A television camera weighing 120 pounds is supponed by a tripod. as shown in Figure 11.23. Represent the force exened on each leg of the tripod as a veccor.
Solution Let the vectors F 1. F2. and F 3 repre;,ent the forces exened on the three legs. From Figure 11.23. you can detem1inc the directions of F 1. F 2. and F3 to be as fo llo w s .
PQ1 = (0 - 0. -1 - 0. 0 - 4) = (0. - I. -4)
½- l PQ2 = ('; - o. o.o - 4) = (';. -➔)
PQ.
·'
=
I -
\
J3 2
-
o' J2.. -
o • cJ -
➔) = / - A J. - 4) \ 2 ' 2·
Because each leg has the same length, and the total force is distributed equally among
the three legs. you know that IIF1JI = IJ F2II = IJ F3II-So. there exists a constant c such
that
c(-;3.} F1 = c(O. - I. --t). F2 =
- ➔). -4) . and F3 = c\- ' ;-½,
Let the total force exerted by the object be given by F = (0. 0. - 120). Then, using
the fact that
F = F1 + F2 + F,
you can conclude that F 1, F2. and F3 all have a vertical component of -40. This
implies that d- 4) = - -tO and c = IO. Therefore. the forces exened on the legs can
be represented by
F1 = (0. - JO. --W)
F ~ = (s J3, 5. -40)
F_1 = (-5../3, 5. --to).
•
780
Chapter 11 Vectors and the Geometry of Space
IIIJ Exercises
See ,•1Mv.CalcCnat.com ior wor~ed-0111 solutions to odd-nu'llbered t-xerc1ses
In Exercises 1 and 2. approximate the coordinates of I.he points.
1.
2.
In Exercises 3-6. plot the points on the same three-dimensional coordinate system.
3. (a) n, I. 3)
4. (a) (3. - 2, 5) 5. (a) (5, - 2, 2) 6. (al (0. 4. -5)
(b) (-1. 2, Ll
(b) (½. 4. - 2)
(b) (5, - 2, - '.!) (b) (4. 0. 5)
In Exercises 7-10, find I.he coordinates of the point.
7. The point is located three units behind the y::-plane, four units to the right of the x;:-plane, and live units above the xy-plane.
8. The point is located seven llllits in front of the yz-plane. two units ro the left of the xz-plane. and one unit below the xy-plane.
9. The poim is located on the x-axis, 12 units in from of the y::-plane.
10. The point is located in the y::-plane. three units to the right of the .c-plane. and two units above the .\}'-plane.
11. Think About It .t)·-plane'?
12. Think About II y::-plane?
What is the ::-coordinate of any point in the What is the x-coordmate of any poim in the
In Exercises 13-24, determfoe th e location of a point (x,y,z) that satisfies the condition(s).
13. :: = 6
JS. x = - 3
14. y = 2
16. z = -i
17. y < 0
19. Lvl 5: J
21. X)' > 0. :,: = - J
18. X > 0
20. lxl > 4
22. XJ < 0, : = 4
23. xy: < 0
24. .l}";: > 0
In Exercises 25- 28, find the distan ce between the poiuls.
25. (0, 0. 0). (-4. 2. 7) 26. (-2. 3. 2), (2. - 5, -2) 27. (1. -2,4), (6, - 2.-2) 28. (2. 2. 3). (4, - 5. 6)
In Exercises 29-32, find lhe lengths of the sides of the triangle \\ilh the indicated vertices, and determfoe whether the triangle is a right triangle, an isosceles triangle, or neither.
29. (0. 0. 4), (2. 6. 7), (6, 4, - 8) 30. (3, 4, I), (O, 6, 2), (3. 5, 6)
31. (- I. 0, - 2). (- 1. 5, 2), (-3, - I. Il
32. (4, - 1. - 1). (2. 0. -4). (3, 5. - 1)
33. Think About It The triangle in Exercise 29 is translated five uniIS upward along the ::-axis. Detennine the coordinates of U1e translated triangle.
34. Think About Jr The triangle in Exercise 30 b rranslated three un its to the righ t along the y-axis. Detennine the coordinates of the translated triangle.
In Exercises 35 and 36, find the coorclinates of the midpoint of I.he line segment joining the points.
35. (5. -9, 7), (- 2. 3. 3)
36. (4, 0. - 6), (8, 8. 20)
In Exercises 37-40, find lhe standard equation of the sphere.
37. Center: (0. 2. 5)
38. Center: (4, - 1, I)
Radius: 2
Radius: 5
39. Endpoints of a diameter: (2, 0, 0). (0, 6. 0)
40. Center: (- 3. 2, 4). tangent lo the yz-plane
In Exercises 41- 44, complete the square to write the equation of the sphere in standard form. Find the center and radius.
41. x2 + y 2 + : 2 - 2x + 6y + 8:: + I = 0 42. x2 + y1 + ~2 + 9x - 2y + I0:: + l 9 = 0 43. 9x2 + 9y1 + 9::2 - 6x, l8y -r I = 0
44. 4x2 + 4y2 + 4::2 - 24.1 - 4y + 8z - 23 = 0
In Exercises 45-48, describe U1c solid satisfying I.he condition.
45. x2 + y2 + ;:2 '5 36
46. r2 + y2 + ::2 > 4
47. x2 + y 2 + ::2 < 4x - 6y + 8z - 13
..J8..r2 + ,.2 + ::~ > -4x + 6y - 8:: - 13
In Exercises 49-52, (a) find the component form of I.he ,·ector v, (b) write the vector using standard unil vector notation, and (c} skelch the veclor with its initial point at the origin.
50.
11 .2 Space Coordinates and Vectors in Space
781
52.
6'
l-I ('.!. 3. -1)
2
,
2 -- ~ 4 (2. 3. 0)
_,C,,.
rn Exercises 53-56, find the component form and magnitude or Lhe vector v with the given initial and terminal points. Then find a unit vector in the direction or v.
l11i1ial Point 53. (3. 2. 0) 54. {4. - 5. 2) 55. (-4. 3.1 ) 56. (I. - 2. 4)
Tem1i11al Poim (4. I. 6) (- 1.7. - 3) (- 5. 3, 0) (2. 4. - 2)
ln Exercises 57 and 58, lhe initial and terminal poinL~ or a vector v are given. ta) S ketch the directed line segment, (bl find the component form of the vector, (c) write the vector us ing st:indard unit vector notation, a nd (d ) sketch t:he vector with its initial point at the origin.
57. lni1ial poim: (- 1, 2. 3) Te rminal poim: (3. 3, 4)
58. lni1ial poim: (2. - I. - 2) Terminal poim: (--l. 3. 7)
Jn Exercises 59 and 60, the vector v and its initial point are given. Find the terminal point.
59. " = (3. - 5. 6) Initial point: (0. 6. 2)
60. ,. = ( !. -½- ½) Initial poinc: (o. 2. {)
In Exercises 61 and 62, find each scalar multiple of v and sketch its graph.
6). V = ( 1. 2. 2) (a) 2, (b l - v (c) ! Y (d ) Ov
62. v = (2. - 2. I) (a ) - ,- (b) 2v
(c) ! v (d) ~v
In Exercises 63- 68, find the vector z, given that u = (1, 2, 3),
v = (2, 2, -1), and w = (4, 0, -4).
63. z = u - \' 65. z = 2u -'- -lv - w 67. 2z - 3u = w
64. z = u - v + 2w
66. z = 5u - 3v - ! w 68. 2u + v - w + 3z = 0
.In Exercises 69-72, determine which of the vectors is (are) pa ra llel to z. Use a graphing utility lo co11firm your results.
69. z = (3. 2. - 5)
ii 70. z = i i - + ¾k
(a) (- 6. - ➔ , 10)
(b)
(?-·
:! 3•
_.ill) 3
(c) (6. 4, 10)
(d ) ( I. - 4. 2)
(a) 6i - 4j + 9k
h - (bl - i +
fk
(c) l2i + 9k
(<l) ii - j + ~k
7J. 'I. has initial point ( I, - I. 3) and 1em1inal poinl (- 2, 3, 5).
{a) -6i + 8j + -lk
(b) 4j + 2k
72. z has initial point (5. -l. I) and terminal point (- 2. - 4. -I).
(a) (7, 6. 2)
(b) (14, 16, - 6)
In Exercises 73-76, use vectors to determine whether the points are collinear.
73. (0. -2. -5). (3. 4. 4). (2. 2. I) 74. (4.-2.7), (- 2.0, 3).(7, -3.9) 75. ( I. 2. 4), (2, 5. 0). (0. I. S) 76. (O. 0. 0), ( I, 3, -2), (2. - 6, 4)
1n Exercises 77 and 78, use vectors to s how that the points form the vertices of a parallelogram.
77. (2. 9. I), (3. 11, 4), {O. I0. 2). ( I, 12 5) 78. ( 1. I. 3), (9. -1. - 2). (11. 2. -9), (3, 4, - 4)
In Exercises 79-84, find lhe magnitude of v.
79. v = (O, 0, O) 8 1. V = 3j - 5k
83. v = i - 2j - 3k
80. v = ( I, 0, 3)
82. v = 2i + Sj - k
84. v = - 4i + 3j + 7k
In Exercises 85-88, find a unit vector (a) in the direction of v and (b ) in the direction opposite of v.
85. V = (2. - J. 2) 87. v =(3. 2, - 5)
86. V = (6. 0, 8) 88. v = (8. 0, O)
89. Programming You are given the componem forms of the
veclOrs u and v. Wri1c a program for a gr.iphing utility in which
1he ou1pm is (a) the componem fonn of u + v, (b) ll u + vii,
(c) !l u ll, and (d) llvll, (c) Ruo the program for the vectors
u = ( - 1, 3, -l) and v = (5. 4.5, - 6).
90. Consider the two nonzero vectors u and v. and let s and, be real numbers. Describe lhe geometric figure generated by 1he terminal poims of lhe three vectors 1v. u + TV, .ind
HI + IV.
In Exercises 91 and 92, determine the values of c tha t satisfy the
equa tion. Let u = -i + 2j + 3k an d ,. = 2i + 2j - k.
91. llcvll = 1
92. lieu11= 4
In Exercises 93-96, lind the vector v with the given magnitude and direction u.
Magnitude
93. 10
94. 3
95.
l
'.!
96. 7
Direcrio11 u = (O. 3. 3) u = ( l. 1. 1) u = (2. -2. I) u =(-4. 6,2)
782
Chapter 11 Vectors and the Geometry of Space
In Exercises 97 and 98, sketch the vector v and write its component form.
97. v lies in 1hc y::-plane. has magnirude 1. and makes an angle of 30' with 1hc positive v-axb.
98. v lies in the x:-plane. has magnitude 5. and makes an angle or
..J5° with 1hc positive :-axis.
In Exercises 99 and 100, use vector s to find the point that lies
Cwo-thirds of the way from P to Q.
99. P(-1-. 3. 0). Q( I. - 3. 3) 100. P( J. 2. 5). Q(6 , 8. 2)
IOI. Ler u = i + j , ,. = j + k. and w = au + bv.
(a) Sketd1 u ::ind v.
(b) ff w = 0, show that a and b must both be zero. = (c) Find 11 and b such thai I\ i + 2j + k . (d) Show !hat no choice of Cl and b yields \\ = i + 2j -,. 3k.
102. fl'riring The initial and tcm1inal poin ts of the vector ,. :ire (x1. y 1. ~ 1) and (x.v. z). Describe the set or aU poin1s (.1.y. ::)
sucb Lhm l\vl\ = 4.
WRITING ABOUT CONCEPTS
103. A point in 1hc three-dimensional coordinate system has
coord1m1tc, (x0. y0, ~il, Describe whut each coordinate
measures.
JO.i. Give the formula for the <lis1ance between the poim~ (.r1. Y1· :::1) and Cx2. Y2· :::21-
LOS. Give the standard equation of a ~pherc of radius r,
cemercd at (x0 . y0, :::ol.
106. State the definition of parallel vec1ors.
I07. Leu\. 8. and C be vertices ofa criangle. Find AB + BC + 01.
JOS.
Let r poims
= (.r,
(.r. _I'.
\', ;:)
;s:u) cahndthart0llr=
(
-
I. I.
roll
I)
=
. Desi:ribe 2.
c.he
set
of
all
109. Numerical, Graphical, and AJw(rtic Analysis The lighb in
an auditorium are 24-pound discs of radius IS inches. Each
disc is supported by three equally spai:cJ cables 1ha1 are L inches long (,ee figure).
r_r l 50 2_0ij_lbS_- 1_30_' 3.5_ _ 4_0 _ 45
(c) Use a graphing mility m graph the function in par! (a). Determine the asymptotes of the graph.
(d) Confirm the asympto1cs or the graph in pan (cJ analyticall}.
(e) De1ermine the minimum length of each cable if a cable is
designed IO carry a m11ximum load of IOpounds.
l JO. Think About ft Suppose the length ofeach cable inExercise
= I09 has a fi,cu length L 11. and the radius or each disc is r0
inches. Make a conjecture about che limit lim T and give u
reason for your answer.
r,,- a
111. Diagonal of a Cube Find the component form of the uni1 vector v in 1he direction of 1hc diagonal of the cube shown in the figure.
Figure for lJ1
Figure for 1 12
112. Tower Guy ll'ire Tbe guy wire supporting a 100-foot tower has a 1ension of 550 pounds. Using the distances shown in 1he figure. write the component form of rhe vei:tor F representing the te1L,. ion in the wire.
113. toad Supports Find the tension in each of the supporting cables in the figure if the weight of the crate is 500 newtons.
18 in.
(a) Wri1c the tension T in each cabll! as a funcuon of l. Determine the domain of the function.
tb) Use a graphing utili1y and the function in pan (a) to complet.: the table.
Figure for 113
Figure for .1U
1U. Constmctio11 A precas1concrere wall is temporarily kept in
its vertical posi1ion by ropes (see figure). Find the torn] force exerted on the pin at position ,\. The tension~ in AB and AC are -1-:!0 pound, und 650 pound~.
115. Wri1e an equation whose graph consists of the set of poi111s P(.r, y. ::) th,11 are twice as for from A(O. - I. I ) as f'rom B( I. 2. 0).
IIJ The Dot Product of Two Vectors
11.3 The Dot Product of Two Vectors
783
■ Use properties of the dot product of two vectors. ■ Find the angle between two vectors using the dot product. ■ Find the direction cosines of a vector in space. ■ Find the projection of a vector onto another vector. ■ Use vectors to find the work done lly a constant force.
EXPLORATION
Interpreting a Dot Product Se1 era] vecror-, are shown below on rhc unir circle. Fino the dot products of ~cveral pairs of vee!On,. Then find rhe angle berwcen each pair thul you used. Make a conjccrure about the relationship between the dot pro<lurt of 1wo vecwr~ and the angle berwecn the vectors.
The Dot Product
So far you have studic<l 1wo operations with vector;,-vcctor adclilion and 111ultiplica1ion by a !>calar-each of which yields another vector. In this section you will swdy a third vector operation. called the dot product. This product yie lds a scalar. raLher than a vector.
DEHNITION OF DOT PRODL'CT
mJ:I Because the dot product of two vectors yields a scalar. it is also callc:<l rhe scalar
product (or inner produr t Jof the two vector,.
■
THEORE,t I U PROPERTIES OF THE DOT PRODUCT
Lei u. v. and w be ve(;tors in the plane o r in space and let c be a scalar.
l.u·v =v· u
= + + 2. U ' ( \' w) LI • V U • W
3. c(u • v) = cu • v = u • cv
4. 0 • v = 0
= 5. y. V 11 ,,11!
CornrnulJtt\C Pri,pcn:,
= + U • V 1111'1 -'- 1111'2 1131"3
= + + l'1ll1 \"2ll1 \ '31/_\
=" . u.
For Lhc fifth property. let v = (1•1. v2. ,-). Then
+ Y • v = v,1 + l 'i \'·l
= (Jv,2+ i} + P])2
= llvlf.
Proofs or Lhc othe r properties are left co you.
•
784
Chapter 11 Vectors and the Geometry of Space
~
~/
Origin
The angle between two wctors Figure 11.24
D EXAMPLE Finding Dot Products
Given u = (2. -2). v = (5. 8). and w = (- -I. 3). find each ofrbe following.
a. a· v
b. (u • v)w
c. u • (2vl d. II\\ 112
Solution
a. U • V = (2. -2) ' (5. 8) = 2(5) + (- 2)(8) == - 6
b. (u • v)" = - 6(-4. 3) = (2-t - 18)
c. u· (2v) = 2(u· v) = 2(-6) = - I2
rh~or<'lll 11 ..J
d. ll wll2 = \\ • w
Theorcn l 1.4
= ( "L J) • ( - -l. 1)
= (- ➔)(-4) + (3)(3)
',ubs1nu1c ( - 4, 3} fur w. Defimlll>ll ul d nt product
= 25
Simplil)
Notice that the rcsu lL of part (b) is a vector quantity. whereas the results of tbe other
three parts are srnlar quantities.
■
Angle Between Two Vectors
The angle between two nonzero vectors is tbe angle 0, 0 s 0 s 'TT', between their respecti ve standard position vectors. as shown in Figure I l.24. The next cbeorem shows how ro find this angle using the dot product. (Note that the angle between the zero vector and anmher vecror is not defined here.)
THIWRE\l I J.5 ANGLE BETWEEN TWO VECTORS If 0 is the angle between two nonzero vectors u and v. then
u·v
cos O= !lull lh·II'
( PROOF ) Consider the triangle determined by vecrors u. v. and v - u. ru, shown in Figure 11. 2➔. By the Law of Cosines. you can wri te
llv - ui12 = l\ulP + llv1l2 - 2llull llvllcos e.
Using the properties or the dot product. the left side can be rewritten ru,
ll v - ull2 = (v - u) • (v - u)
= (v - u) • v - (, - u) • u =v ·v - o·v - v· u-u ·u
= llvll2 - 2u • v + llu[F
and substitution back into the La,, of Cosine~ yields
l!v\12 - 2u • v + llull2 = llu ll2 + llv\F - 2llull ll vll cos 8 - 2u • v = -2llu\l llvll cos 0
u·v
cos B = llull llvll"
•
11.3 The Dot Product of Two Vectors
785
If the angle between two vectors is known. rewriting Theorem l l .5 in the form
u • V = II u 1111VII cos 8
,\11~ma11ve form ui dvl produ.:1
produces an alternative way 10 calculate the dot product. From this form. you can see
that because llu11 and llvII are always positive. u • v and cos 8 will always have the
same sign. Figure 11.15 shows 1he possible orientations of two vectors.
Opposi1c dircc1ion
afJ
u
I
0= " cos 0 = - I
F igure 11.25
U • V <0
½_
\
,-/ 1<8<7r
- I < cos 8 < 0
r··".
I'
0 = ,-j'l cos 0 = 0
u • \' > 0
L \
0 < 0 < 1r/2
0 < cos O < I
Same direc1ion
u
\'
0=0
cos 0 = I
From Theorem I 1.5. you can see that two nonzero vectors meet at a right angle if and only if their dot product is zero. Two such vectors are said to be orthogonal.
DEFINITION OF ORTHOGONAL VECTORS The vectors u and v are orthogonal if u • v = 0.
C 9 The 1cm1s ··perpendicular," "orthogonal," anJ " nomal" all mean essentially the same
thing- meeting at right angles. Ilowever. it is common Lo say that rwo vectors arc onhogmwl.
two Jines or plant:s are perpentlirnlar. and a vector is normal w a given line or plane.
■
From this definition. it follows thar rhe zero vector is orrhogonal 10 every vector
u, because O • u = 0. Moreover. for O 5: 8 ::; 1r, you know that cos O= 0 if and only
= if 0 1r/1. So. you can use Theorem 11.5 to conclude thar rwo 11011.;;ero vector~ are
orthogonal if and only if the angle between them is n/2.
f l ( ; EXAMPLE Finding the Angle Between Two Vectors
For u = (3. -1. 2}. v = (--k 0. 2}. w = {I. - 1. -2). and z = (2. 0. -1). find the
angle between each pair of vectors.
a. u and v b. u and w c. v and z
Solution
a.
cos
8
=
u·v
~
=
-
11 + 0+4
F4Jw
-8 1ffiJ5
-4 fio
Bccau~e u • v < 0, 0 = arccos .-./740 = 2.069 radians.
U•W 3+1 - -1-
Q
b. cos e = !lull llwll = ffi.)6 = ffi = O
Because u • w = 0, u anti ware orthogo11al. So. 8 = 11/ 1.
MW v·z - 8 +0- 2 - 10
C, cos O=
= JwJs = .Jioo = - I
Consequently. 0 = 11. Note that v and z are parallel, with v = -1z.
■
786
Chapter 11 Vectors and the Geometry of Space
k
; \
Direction a11gles Figure 11.26
ex = angle between v and i
fJ = angle bctwcc11 v ~nd j
y = angle between ,. antl k
41 .1.
2 '/ , = 2i.,. 3j + -lk
47
,._
A',,./3
2 3
,
J
X
The c.lirecrion angles of v Figure 11.27
Direction Cosines
For a vector m the plane. you have seen that it is coll\·enieot to measure direction in tenm of the angle, measured couoterclockwise. ji·om the positive x-axis ro the vector. In space it is more convenient to measure direction in Icrms of lhe angles berween the nonzero vector v and the three unit vectors i. j . and k. :is shown in Figure 11.26. The angles a. /3. and y are the direction angles of v, and cos a. cos (3. and cos y are the direction cosines of v. Because
v • i = II v 1111 i II cos a = II v II cos a
and
V • j = ( 1•1,1'2.1,3) • ( J.Q.Q) = 1•1
it follows that cos a= 1'1/llvll- 8 ) similar reasoning with the unit vec1ors j and k. you have
HI'
COSO' =
I',
cos 13 = llv-11
cos 'Y = ~1\ TI'
/1 i, 1hc angle hct11ecn, and j
Consequently. any non1ero vector v in space has the normalized form
114 iNjV
\'
=N
i
-,-
1'2
llv ll j
-,-
I'
k = cos a i + cos /3.i
f- cos y k
and because , /ll v[I is a unit vec tor. it follows rhat
cos2 a + cos2 f3 + co~2 y = l.
II EXAMPLE Finding Direction Angles
= Find the direction cosines and angles for 1he vector Y 2i - 3j + 4k, and show 1har
cos 2 a+ cos1 f3 + cos2 y = I .
llvll Solution Because = ./22 + Y + 42 = ,/29. you can write the following.
viiI, COSO' = -1'1 =✓-2= 29
F'9 cos {3 =
\'
11 ,211
=
= O' 68.2° = /3 56.1°
\11glc between , .,111I j
GIiI'•
4
cos 'Y =
= -./29
'Y = 42.0°
Furthcm1ore. the sum of Ihe squares of the direc1ion cosines is
,
cos-
a
+
,
cos-
(3
+ co~1-
y
=
-1-
+
9
+
16
29 29 29
29
29
= I.
See Figure I I .'27.
•
The force due 10 grarity pulls 1he boat agairu.1 1he ramp and down the ramp. Figure J1.28
11.3 The Dot Product of Two Vectors
787
Projections and Vector Components
You have already seen applications in which two vec1ors are added to produce a resultant vector. Many applications in physics and engineering pose the reverse problem- decomposing a gi ven vector into the sum of two vector components. T he following physical example enables you to see the usefulness of this procedure.
Com,ider a boat on an inclined ramp. as shown in Figure 11.28. Tue force F due
ro grnviry pulls the boat down the ramp and against the ramp. These two forces, w 1 and w2. are orthogonal- they arc called the vector components of F.
The forces w 1 and w2 help you analyze Lhe effect of gravity on the boat. For example, w1 indicates the force necessary to keep Lhc boat from rolling down rhe ramp. whereas w2 indicates the force that Lhc tires must withstand.
DEFINITIO\S OF PROJECTION ,\ ND \ECTOR COMPONE\TS
= Lei u and v be nonzero vectors. Moreo,·er. let u w 1 + w2. where w 1 is
parallel Lo,·, and \\'2 is ortJ1ogonal to v. as shown in Figure _11.29.
1. w I i~ called tile projection of' u onto v or the vector component of u along
v, and is denoted by w 1 = proj,.u.
2. w1 = u - w I is called the vector component of u orthogonal to v.
Ii
0 i, acute.
0 " obtuse.
u
u
(-3. -I J ,
•
4 -
"'
+ U = II I
11'2
Figure I 1.30
(5. IO)
n s
'•
11·1 = proj,u = projection of u onto r = vector component of u along 1·
11·2 = vect0r component of uorthogonal lo v
figure 11.29
IJ EXAMPLE Finding a Vector Component of u Orthogonal to v
Find the vector component of u = (5. I 0) that is orthogonal Lo v = (4. 3), gi\en that
w1 = proj ,,u = (8, 6) and U = (5, 10) = W I + W 2.
Solution Because u = w1 + w2. where w I is parallel to v. it folJO\vs thar w J i~ lhe
vecror component of u orthogonal 10 v. So. you have
= (5. 10) - ( 8. 6) = ( - 3.4),
Check to sec that w2 is orthogonal 10 v. as shown in Figure J 1.30.
■
788
Chapter 11 Vectors and the Geometry of Space
G:a Note the distinction berween
the term.~ "component" and "vector com-
ponem." For example, using the srandard
unit vectors with u = u1i T u2j. 111 is
the component of u in rhe direction of i
and u 1i is Lhe vecror compone111 in rhe direction of i.
From Example 4, you can see that it is easy to find the vector component w once 2
you have found the projection. w 1, of u onto v. To find this projection. use the dor product given in the theorem below, wbjch you will prove in Exercise 92.
THEOREM 11.6 PROJECTION USL'IG THE DOT PRODUCT If u and v are nonzero vectors. then che projection of u onto v is given by
proj,.u = ( 1I1IY•ll2V) v.
U = w1 + W2
Figure 11.31
Figure 11.32
The projection of u onto v can be written as a scalar multiple of a unit vector in
the direction of v. That is.
Tvf N e u • V) ( u • V) V
V
( llvll2 V =
M = (k)M
k = o· v = llu ll cos 0.
The scalar k is called the component of u in the direction of v.
El EXAMPLE Decomposing a Vector into Vector Components
Find the projection of u onto v and the vector component of u orthogonal to v for the
)'
vecrors u = 3i - 5j + 2k and v = 7i + j - 2k shown in Figure 11.31.
Solution The pr~jection of u onto v is
(11;1,n" W1 =
= G!)(7i + .i - 2k) = 19\ + i j - ik.
The vector component of u orthogonal to v is the vector
(~i w = u - w = (:Ii - 5J" + ?k) -
2
I
-
-
9
+ I9J· -
i9k) = Q9 i -
47 9
J·
+
22
9k ·
II EXAMPLE Finding a Force
A 600-poWld boat sits on a ramp inclined al 30°. as shown in Figure l L.32. What force is required to keep the boac from rolling down the ramp?
Solution Because the force due to gravity is veitical and downward, you can
represent the gravitational force by the vector F = - 600j. To find the force required
to keep the boat from rolling down the ramp. project F onto a unit vecror v in the
direction of the ramp, as follows.
2 2 V
=
+ COS
"oo
.J
I •
SU, l ."}0oJ• =
./31. +
IJ.
L"nic vcc cor aJung. ramp
Therefore. the projection of F omo v is given by
f w 1 = proj,.F = (riv· :)v = (F • v)v = (- 600)(½)v = -300( i + ½j )11
Tbe magnitude of this force is 300, and therefore a force of 300 pounds is required to
keep the boat from rolling down the ramp.
■
11.3 The Dot Product of Two Vectors
789
p
Work= 111-"IIII PQII (a) Force acb along the line of motion.
Work
The work W done by the constant force F acting along rhe l ine of motion of an objecI
is given by
Q
W = (magnitude of force)(distance) = IIFIIIIPQI
as shown in Figure l l .33(a). If the constant force F is not Jirecced along the line of motion. you can ~ee from Figure l l .33(b) Lhat the work IV done by the force is
W = llprojPQFllll7>QII = (cos 8)IIFllll7>QII = F • PQ.
This notion of work is summarized in the foll owing definition.
p
Q
Work= llprojiiQ F IIII PQII
(b) Force acts mangle 6 "i!h the line of motion.
Figure 11.33
DEFrNITION OF WORK
The work W done by a constant force F a.~ its point of applicacion moves
along the vector PQ is given by either of the following.
1. IV = llprojl'Q Fii llPQII
Proje~1ion form
2. 11/ = F • PQ
Dot product form
Figure Ll.34
mExercises
II EXAMPLE Finding Work
To close a sl iding door, a perso11 pulls on a rope with a constant force of 50 pou11ds at a constant angle of 60°, as shown in Figure 11 .34. Find the work. done in moving the door 12 feet Lo its closed position.
Solution Using a projection, you can calculate the work as follows.
w = llproj,,Q Flllll5QII = cos(60°) IIFII IIPQII
Pmjecuon form for work
= ½(50)(12)
= 300 foot-pounds
■
See v.....,.CalcChaLCom for worked -out solutions to odd-numbered exercises.
In Exercises 1-8. find (a) u • ,·. (b) u • u. (c) 'l ull\ (d) (u • \')v.
and (e) u • (2v).
1. U = (3, 4). Y = (-1,5) 2. U = (4. 10), V = (-2. 3)
3. U = (6. -4). Y = (- 3. 2) 4. U = ( - 4, 8). V = (7, 5)
5. U = (2, - 3, 4). V = (Q. 6. 5} 6. ti = i, V = j
7. u = 2i - j + k
"= i - k
8. u = 2i + j - 2k , = i - 3j + 2k
In Exercises 9 a nd 10. find u • ,·.
9. !lull = 8. jJ vll = 5, and the angle between u and v is -.r/3.
10. IIu II = 4O. JI vi\ = 25. and Lhe ang le between u and vis 5 7T/6.
In E xercises 11-1 8, find the angle Obetween the vectors.
11. u =(I.!). ,.= (2. - 2)
12. u = (3. I). v = (2. - I )
13. u = 3i + j ," = - 2i + 4j
i) i)j 14. u = cos( i -r ,in(
4 4 V
=
COS(
3,;r)
.
I
+
S.i ll( 37T) J.
15. u = ( I. I. I )
,. = (2. I. - I )
17. u = 3i T 4j
v = -2j 1 3k
16. u = 3i + 2j + k
v = 2i - 3j
18. u = 2i - 3j + k v = i - 2j + k
In Exercises 19-26, determine whether u and v are orthogonal, parallel, or neil'her.
19. U = (-l. 0). v = ( I. I)
20. U = (2. 18), V = (½. -i)
790
Chapter 11 Vectors and the Geometry of Space
o. 21. u = (4. 3) V = - f)
23. u = j + 6k v=i -2j - k
25. u = (2. - 3. I) v= ( -1 , - 1. - 1)
22. u = - i (i - 2j)
V = 2i - 4j 24. u = - 2i + 3j - k
v = 2i + j - k
26. u = (cos 0. sin 0. - 1) V = (sin tJ. - COS 0, 0)
In Exercises 27- 30, the vertices of a triangle are given. Determine whether the t r iangle is an acute triangle. an obtuse triangle, or a r ight t riangle. Explain yom· reasoning.
27. ( l.2.0),(0,0,0). (-2. 1. 0) 28. ( - 3. 0, 0), (0, 0, 0), ( I. 2, 3) 29. (2. 0. I). (0, l . 2). (- 0.5, 1.5. 0) 30. (2. - 7. 3), (- 1, 5. 8). (4. 6. - l)
ln Exercises 31-34, find the direction cosines of u and demonstrate that t he sum of tlle sq uares of the direction cosines is I.
31. u = i + 2j + 2k 32. u = Si + 3j - k 33. u = (0. 6, - 4) 34. u = (a, b. c)
Ln Exercises 35-38, find the direction a ngles of the vector.
35. a = 3i + 2j - 2k
37.u= (-l. 5, 2)
36. u = - 4i + 3j + 5k
38. u = (- 2. 6. I)
I i'... l.n Exercises 39 and 40, use a graphing utility lo find the magnJtude a nd d irection an gles of the resultant of forces F I and F; with initial points at t he origi n. The magnitude and terminal
point of each vector are given.
In Exercises 43- 50. (a) find the proj ection of u onto v, and (b) find the vector component of u orthogonal to v.
43. u = (6, 7). v = ( l, 4) 44. U = (9, 7). V = ( 1, 3)
45. u = 2i + 3j . ,, = Si + j
46. U = 2i - 3j . V = 3i + 2j
47. u = (0,3,3), V = ( - 1. l. l )
48. u = (8, 2, 0), v = (2, 1, - I) 49. u = 2i + j + 2k, v = 3j + 4k 50. u = i + 4k. v = 3i + 2k
WRITING ABOUT CONCEPTS
51. Defme the dot product of vectors u and ,.
52. Srate the de fi nilion of orthogonal V<'!Crors. If vecwrs are neilher parallel nor onhogonal, how do you find the angle becween them? Explain.
53. Determine which of the following are defined for nonzero
vecrors u , v, and w. Explain yom reasoning.
+ (a) u • (y w)
(b) (u • v)w
(c) u • , , + w
(dJ llull • (v + w)
54. Describe direction cosines and direccion angles of a vector ,,.
55. Give a geometric description of the-projeclion of u onto v.
56. What can be said abolll !he vectors u and v if (a) !he
projection of u on to v eq1mls u and (b) the projection of u
onto V equals o~
57. If rhc projection of u onto v has the same magnitude as !he
projection of vonto u, can you conclude that ll ull = llvll?
E x plai n.
\lec1or 39. F 1
F2 40. Fl
F2
Magniw de 50 16 80 lb 300 N IOON
Tem rinal Point (10. 5,3) (12. 7. - 5) (-20, -10. 5) (5, 15. 0)
41. Load-S upporting Cables A load i.s supported by three cables. as show n in the figure. Find the direction ang les of the loadsupporting cable OA.
l- 4. - 6, 10) 11 •
(4, - 6. 10)
(0, IO, !OJ
/ ~ / •;1
0 300 lb
y
X
42. Load-S11pporri11g Cables T he tension in lhe cable OA in Exercise 4 1 is 200 newtons. Detennine the weight of lhe load.
58. What is known about 0. ihe angle between cwo nonzero vectors u and , . if
(a) u • v = O? (b) u • v > O? (c) u • ,. < O?
59. Reven ue T he vector u = (3240, 1450, 2235) gives the
numbers of hamburgers. chicken sandwiches. and cheeseburgers. respectlvely. sold at a fast-food restaurant in one week. The
vector v = (1.35. 2.65. 1.85) gives the prices (in dollars) per unit
for the three food irems. Find the dot product u • v, and explain what i..nfon11ation it gives. 60. R evenue Repeat Exercise 59 afler increasing prices by 4Sr. Identify the vector operation used to increase prices by 4%.
61. Programming G iven vectors u and v in component form. write a program for a graphing utility in which the output is
(a) ll ull,(b) llvll,and (c) the angle between uand v.
H 62. Programming Use the program you wrote in Exercise 6 1 10
ri nd the angle between rhe vectors u = (8, - 4. 2) and V = (2. 5. 2).
11.3 The Dot Product of Two Vectors
791
63. Programming Given vectors u and v in componem form, write a program for a graphing utility in which lhe output is the component form of the projecti on of u onto v.
64. Progrm11mi11g Use the program you wrote in Exercise 63 to find the projection of 11 onto v for u = (S, 6, 2) and v=( - 1.3,4).
Think About It In Exercises 65 and 66, use the figure to determine mentally the projection of u onto v. (The coordinates of the terminal points of the vectors in standard position are given.) Verify your results analytically.
65.
,I
T ,. 6 I
•
(4, 6)
4-
21
-+-----'f--,_--1--!-----!--+- X
2 ➔ 6
66.
.r
V • (4, 6)
-+--t----t--1--i'-H-► X
-2
4 6
- 2 u • (3, - 2)
ln E.xercises 67-70, find two vectors in opposite directions tl1a1 are orthogonal to the vector u. (The answers are not unique.)
67. ll = - ¼i + 1j
69. u = (3, I, - 2)
68. u = 9i - 4j 70. u = (4, - 3, 6)
71. Braking Load A 48,000-pound rruck is parked on a I0° slope (see figure). Assume the only force to overcome is that due to gravity. Find (a) the force required to keep the trnck from rolling down the hill and (b) the force perpendicular to the hi ll.
(5, - 5. 20) .B
C•i i (-5. - 5, 20)
A•
{1({20) 12--
Figure for 71
1000 kg
y
Figure for 72
72. Load-Supporting Cables Find the magnitude of the projection of the load-supporting cable 0.4 onto the positive z-axis as shown in rhe figure.
73. Work An object is pulled 10 feet across a floor. using a force of 85 pounds. The direction of the force is 60° above the horizontal (see figure). Find the work done.
-I I
~~ :~~~
Figure for 73
Figure for 74
74. Work A toy wagon is pulled by exerting a force of 25 pounds on a handle that makes a 20° angle with the horizontal (see figure in left column). Find the work <lone in pulling the wagon SO feet.
75. Work A car is towed usi ng a force or 1600 newtons. The chain used to pull the car makes a 25° angle with the horizontal. Find the work done in towing the car 2 ki lometers.
76. Work A sled is pulled by exerting a force of 100 newtons on a rope that makes a 25° angle with the horizontal. Find the work done in pulling the sled 40 meters.
True or False? fn Exercises 77 and 78, determine wbether the statement is true or false. If it is false, explain why or give an example that shows it is false.
* 77. lf u • v = u • wand u 0, then v = w.
78. lf u and v are orthogonal tow. then u + vis orthogonal cow.
79. Find the angle between a cube"s diagonal and one of its edges.
80. Find the angle between the diagonal of a cube and the diagonal of one of its sides.
In Exercises 81-84, (a) fi nd all points of intersection of the graphs of the two equations, (b) find the uni t tangent vectors to each curve at their points of intersection, and (c) find the angles (0° :S 0 s 90°) between the curves at their points of intersection.
SL y = x2, y = x 113
= 82. \I x3. y = xl / J 83. y = 1 - x2. y = x1 84. ( y + 1)2 = x. y = x 3 - I
85. Use vectors ro prove that the diagonals of a rhombus are perpendicular.
86. Use vecLors to prove that a parallelogram is a rectangle if and only i f its diagonals are equal in length.
87. Bond Angle Consider a regular tetrahedron with vertices (0, 0, 0), (k. k, 0). (k, 0. k), and (0, k. k), where k is a positive real number.
(a) Sketch the graph of the tetrahedron.
(b) Find Lhe length of each edge.
(c) Find the angle between any rwo edges.
(d) Find the angle between the line segments from the centroid (k/2, k/2, k/2) to rwo vertices. This is the bond angle for a molecule such as CI-1 4 or PbC1 4, where the structure of the molecule is a tetrahedron.
88. Consider the vectors u = (cos a, sin a, O) and
v = (cos /3, sin {3, 0), where a > (3. Find the dot product of the
vectors and use the result to prove the identity
cos(a - (3) = cos ll' cos f3 + sin a sin (3.
89. Prove that llu - vii" = llu ii2 + llvll1 - 2u • v. I 90. Prove the Cauchy-Schwarz Inequality u • vi :s !l u ll ll v ll-
91. Prove the triangle inequality Ji u+ v/1 :S )l u ll + jjv)I. 92. Prove Theorem I 1.6.
792
Chapter 11 Vectors and the Geometry of Space
1111 The Cross Product of Two Vectors in Space
■ Find the cross product of two vectors in space. ■ Use the triple scalar product of three vectors in space.
EXPLORATION
Geometric Property ofthe Cross Product Three pairs of vectors are shown below. Use the definition to find the cross product of each pair. Sketch all three vectors in a three-dimensional system. Describe any relationshi ps among the three vectors. Use your description to write a conjecrure about u. v, and u x v.
a. U = (3,0,3), V = (3, 0. - 3)
The Cross Product
Many applications in physics, engineering, and geometry involve .finding a vector in space that is orthogonal to two given vectors. In this section you will study a product that will yield such a vector. lt is called the cross product, and it is most conveniently defined and calculated using the standard uni t vector form. Because tbe cross product yields a vector, it is also called tbe vector product
DEFINITION OF CROSS PRODUCT OF TWO VECTORS IN SPACE
Let
= u = 11 1i + u2 j + 113k and v v1i + F2 j + v3k
be vectors in space. The cross product of u and v is the vector
+ U X V= (1121'3 - l13ll2)i - (11 1\13 - !l31'1)j (u 1V2 - U2V1)k.
b. u = (0. 3, 3), v = (0, -3, 3)
2
>_r
-2
)'
-3
c. u = (3. 3. O). v = (3, - 3, O)
tmJa Be sure you see LhaL this definition applies only to three-dimensional vectors. The
cross product is not defined for two-dimensional vectors.
■
A convenient way to calculate u x vis to use the following determi11amform with cofactor expansion. (This 3 x 3 de1ennjnant form is used simply co help remember the formula for the cross product-it is technically not a determinant because the entries of the correspondi_ng matrix are not all real numbers.)
j k
u xv= 11 I
ll2
It,
·'
v, \12 \13
------- Put --u•· in Row 1. Pm "v" in Row l
U.2
~ 113 i -
I
2
3 j + u,
.i
"2
"2
\13
V I ,1.~
V 3
V I
"2
Il/2 Vz
I /.I3 1i - u I
V3
11 1
I+ ll31 j
u I
\/3
11 1
I11 2
k
"2
= + (112\13 - U3l'2)i - (11,v3 - U3V 1) j (u,112 - Ll2V1)k
~
f;
k
Note the minus sign in front of the j-component. Each of the three 2 x 2 detem1inants can be evaluated by us ing the fo!]owing diagonal panern.
Here are a couple of examples.
\~ -~I= = (2)(- 1) - (4)(3) = -2 - 12 -14
1-: ~I = (4)(3) - (0)(-6) = 12
NOTATIOt'-< FOR DOT AND CROSS PRODUCTS
The notation for the dot product and cross product of vectors was iirst introduced by !he American physicist Josiah Willard Gibbs (1839-1903). In the early 1880s, Gibbs built a system io represent physical quantities called "vector analysis." The system was a departure from Hamilton's theo1y of quaternions.
11.4 The Cross Product of Two Vectors in Space
793
D EXAMPLE Finding the Cross Product
Given u = i - 2j + k and v = 3i + j - 2k, !ind each of the following.
a. U XV
b. V X LI
C. V X V
Solution
j a. U X V = l -2
l l k
= 1-~
-2' I·I -
1 3
11
-21J· + 3
-2I1 k
3
-2
= (4 - 1) i - (- 2 - 3)j + (I + 6) k
= 3i + 5j + 7k
j k
b. V X u = 3
-2 = 1-~
-~I i -
I~
- 2I
1, J
+
13l
- 2I l k
-2
= ( I - 4)i - (3 + 2)j + (- 6 - I }k
= -3i - 5j - 7k
Note that this result is the negative of that in part (a).
j k
c. vx v = 3 I - 2 = O
•
3
-'>
The results obtained i_n Example l suggest some interesting algebraic properties of the cross product. For instance. u x v = -(v x u), and v x v = 0. These properties, and several others. are summarized in the foll owing theorem.
THEOREM 11.7 ALGEBRAIC PROPERTIBS OF THE CROSS PRODUCT
Let u. v. and w be vectors in space, uncl let c be a scalar.
1.u x v =-(v x u)
2. u x (v + w) = (u x v) + (n x w)
3. c(u x v) = (c u) x v = u x (cv) 4. U X O = 0 X LI = 0 5. U X U = 0 6. u • (v x w) = (u x v) · w
~ To prove Propeny I, let o = 11 1i + u2 j + 113k and ,, = v1i + v2 j + v3k .
Then.
and
v x u = ( v2113 - 11311~ i - (v,113 - \1311 1)j -I (1•1112 - v 211.)k
which implies that u xv = -(v x u). Proofs of Properties 2. 3. 5. and 6 are left as
exercises (see Exercises 59- 62).
■
794
Chapter 11 Vectors and the Geometry of Space
l:mzl It follows from Propcrues l
and 2 in Theorem 11.8 that if n is a unit vector orthogonal to both u and v. then
a x v = ±(l!ull llv/l~inl:J)n.
Note that Property l of Theorem I 1.7 indic.:aces that U1e cross product is not commllfative. In particular. this propeny indicates that the vectors u x ,. and v x u have equal lengths but opposite direccions. The following theorem lists some other geometric properties of the c.:ross product of two vectors.
THEORE\ l 11.8 GEmtETRTC PROPERTIES OF THE CROSS PRODUCT
Let u and v be nonzero vectors in space. and let ebe the angle between u and v.
1. u x vis orthogonal to both u and v.
= 2. ll u x v ]I ll u lJ llvll sin 0
3. u x v = 0 if and only if u and v are scalar mulei pies of each other. 4. II u x vii = area of parallelogram having u and v as adjacent sides.
u
The vectors u and v form adjacent sides of a parallelogram. Figure 11.35
= CPROOF To prove Property 2, note because cos 0 (u • v)/(ll ull IJv JI), it follows that
!Jul! ll vllsine = llull llvll ✓ l - cos2 0 ✓ (u • v)2
= ll ull llvll I l!u ll211vll2
JJJu !!2 !lv !J1- (u • v)2
v (11f + u} + ul)(v?, vf + 1}) - + (11 11•1 1111'~--'- u3v3)1
= + ( + J(11~1•1 - 113112>2
11 1v3 - 11 3111F (u1\12 - 11 2l'1P
= llu x vii.
To prove Prope11y 4, refer lo Figure 11.35. which is a parallelogram having v and u as adjacent sides. Because the height of the parallelogram i s l!vll sin O. the area is
Area = (base)(height) = !lu ll !vii sine = !lu x vii-
Proofs of Properlies I and 3 are left as exercises (see Exercise:, 63 and 64).
■
Both u x v and v x u are perpendicular to the pl ane determined by u and v. One
way to remember the orientations of the vectors u. v. and u x v is to compare them
with the unit vectors i, j. and k = i x j . as shown in Figure I 1.36. The three vectorl.
u. v. and u x ,, form a right-handed .!>}'Stem. whereas the three vectors u. v, and v u
form a left-handed system.
k =ixj
j uxv
I
Right-handed ~ystcrns Figure 1l.36
Plane determined by o and,,
11.4 The Cross Product of Two Vectors in Space
795
fl ( ; EXAMPLE Using the Cross Product
• (-3. '.1. 11)
IJ t
IO t
sr
!I -0. I I u
] /;
-~
~
~,
4
4-,(/ /
~
( 2~3.0)
The \ec!Or u x , i.s orthogonal 10 both u and 1. Figu re 11 .37
Find a unit vecror that is orlhogonal to both
u = i - 4j + k and v = 2i + 3j .
Solution T he cross product u x v, as sh own in Figure 11.37, is orthogonal to both u and v.
j k
u x v= - 4 l
2 30
= - 3i + 2j + 11k
Crn,, pn du,·1
Because
llu x vii = ✓(-3) 2 + 2 2 + 11 2 = , 1134
a unit vector orthogonal ro borh u and v is
U X V _ _ 3_ i _,_ __2_j + _ I_I_ k
llu x vii JT34 fiTI ✓n+ •
■
G D In Example 2. note that you cou ld have used the cross product v x u LO form a unit
vector that is orthogonal to bolh u and v. With that choice. you would have obtained the
neguti ve of the unir vector fou nd in the example.
■
II EXAMPLE Geometric Application of the Cross Product
C= (2. ~. 71
_,
/
6/ -;of A -- C•l, ,__OJ
8= 12. 6. I I
The area of the parallelogram is approximately 32. 19. Figure 11.38
Show that the quadri lateral witJ1 vertices at the following points is a parallelogram. and fi nd its area.
A = (5. 2 . O)
C = (2, 4. 7)
B = (2. 6, 1) D = (5. 0. 6)
Solution F rom Figure 1 1.38 you can see that the sides or the quadrilateral cone-
spond to the fol lowing four vecrors.
AB = - 3i + 4j + k AD = Oi - 2j + 6k
- CD= 3i - 4j - k = - AB
= C--B' = Oi + 2j - 6k - AD
So. AB is parallel io CD and AD is parallel to CB . and you can conclude that the quadrila1eral is a parallelogram with AB and AD as adjacent ~ides. .Moreover. because
j k
XB x Al>= -3 4 I
0 -2 6
= 26i + l8j + 6k
thl! area of the parallelogram is
108 X ADIi = Jio36 = 32.19.
Is the p,u·allelogram a r~ can~ You rnn determine whether it i\ by find ing the angle
between the vectors A B and AD .
■
796
Chapter 11 Vectors and the Geometry of Space
l\l
/
The moment of Fabout P Figure 11.39
-\
Avertical force of 50 pounds is applied at poim Q. Figure 11.40
lo physics, the cross product can be used to measure torque-the moment M of a force F about a point P, as shown in Figure 11.39. ff Lhe poim of application of the force is Q, the moment of F about P is given by
M = PQ X F.
\}llJllL'lll n1 I· .ihout P
The magn itude of d1e moment M measures the Lendency of the vector PQ to rotate
counterclockwise (using die right-hand rule) about an axis directed along lhe vector M.
IJ EXAMPLE An Application of the Cross Product
A vertical force of 50 pounds is appl ied to the end of a one-foot lever that is arrached Lo an axJe at point P. as shown in Figure 11 .40. Find the moment of this force about
the poim P when 0 = 60°.
Solution If you represent the SO-pound force as F = - SOk and the lever as
PQ = cos(60°)j + sin(60°)k = ½j + " ; k
the moment of F abom P is given by
j k
M = PQ x F = 0
2
,'3
2
0 0 -so
-2Si.
\ lom,·111 ,,, t· abou1 r>
The magnitude of this moment is 25 foot-pounds.
•
G D ln Example 4. note that the moment (the tendency of the lever 10 rmatc about its axle)
i& dependent on the angle 0. When 0 = 'rr/2. the 1110111cm is 0. The moment is greatest when
8 = 0.
■
The Triple Scalar Product
For vectors u, v, and win space. the dot product of u and v x w u • (v x w)
i~ called the triple scalar product, as defined in Theorem 11.9. The proof of this theorem is left as an exercise (see Exercise 67).
FOR FURTHER IXFOR~I \TIO/\ To
see ho11 the cross product i, used 10 model the lOrquc of the robot ann of a space shuttle. sec the a11ic:le ..The Long Ann of Calculus" b~ Ethan Bcrkove and Rich 1'vlan::hand in The College Mathematics Journal. To , ,cw this article. go to the 11eb,i1e
,,ww.111athar1icles.cc>n1.
THEORE:\I 11.9 THE TRrPLE SCALA R PRODLCT
= For u u1i + u2j + u3 k." = v1i + i>::d -r 1·.1k. and\\ = 11·1i , w1j + w3k,
the triple scalar product is given by
ll1
111
113
u • (v x ,v) = v1 v2 ''.i
11'1 11'2 H'3
c m The value of a deterrninam is multiplicc.l by - I if two rows arc interchanged. After
two such interchanges. the \alue of the determjnant will bc unchanged. So. the following triple
scalar products are equivalent.
u • (v x w) = v • (w x u ) = w • (u x v)
■
11.4 The Cross Product of Two Vectors in Space
797
l' X W
If 1he vectors 11. v. and w do not lie in the same plane. the triple scalar product
u • (v x w) can be used to determine the volume of the parallelepiped (a polyhedron.
all of whose faces are parallelograms) with u. v. and was adjacent edges. as shown in
Figure I 1.41. This is escablished in the following theorem.
I
ll proj, " ~u ,
Arca of base = II,· x wll Volume ofparallelepiped = Ju • (v x w)I
Figure 11.41
•tO. 2. - 2)
The parallelepiped has a volume of 36. Figure l lA2
THEOREM 11.10 GEOMETRIC PROPERTY OFTHE TRTPLE SCALAR PRODUCT
The volume \/ of a parallelepiped with vectors u. v. and w as adjacent edges is gi,en by
I' = Ju • (v x w) I.
~ In Figure J 1.41. note that
ll vx wll = area of base
and
llproj, , ,.uil = height of parallelepiped.
Therefore. the volume is
\I = (hcight)(area of base) = \lproj, , "u\l llv x wll
=
ju
• (v
IJv x
x w)I
wjj llv
:<
wll
= ju • (v x w)I.
•
I I EXAMPLE Volume by the Triple Scalar Product
Find the volume of the parallelepipecl shown in Figure 11.42 having
u = 3i - 5j + k. v = 2j - 2k. and w = 3i + j + k as adjacent edges.
Solution By Theorem 11.10, you have
\I = lu • (v x w)j
3 -5
= 0 2 _ ')
3 l l
Tnpk ,calm pn1Juct
= 31~ -~, - (- 5)1~ -~1 + (I),~
= 3(4) + 5(6) + 1(- 6)
= 36.
•
A natural consequence of Theorem 11.10 is rhat the rnlume of the parallelepiped
is Oif and only if the three vectors are coplanar. That is. if the ,·ectors u = (u 1. 112 . 113) .
= = v ( 1· 1. 1•2. 1·3) . and w ( 11· 1. 11·2. 11·3) have the same initial point. they lie in Lhe same
plane if and only if
II l 112 IIJ
= u • (v x w) = v1 1'2 1'3 0.
11 '1 11'2 11 '3
798
Chapter 11 Vectors and the Geometry of Space
IIIJ Exercises
See 1N1w.CalcChatcom tor worked -out silul1ons to odd-numberedeY:rc,ses
ln Exercises 1-6, and tbe cross product of the unit vectors :rnd sketcl1 your result..
1. j X j
3. .i x k
5. i x k
2. i X j 4. k x j 6. k x i
In Exercises 7-10, find (a) u x v, (b) \' x u, and (c) v x v.
7. u = - 2i + 4j
v = 3i + 2j + 5k 9. u = (7. 3. 2)
Y = (I, -1. 5)
8. u = 3i + 5k v = 2i + 3j - 2k
10. u = (3, - 2. - 2)
v= ( 1. 5.1 )
Ln Exercises 11-16, find u x v and show that it is orthogonal to both u and v.
11. u = (12. - 3, O) v = (- 2, 5, 0)
13. u = (2. -3, l )
v = (I, - 2. I ) 15. u = i + j + k
v = 2i + .i - k
12. u = ( - 1, 1. 2) V = (0. 1, 0)
14. u = ( - 10, 0, 6)
v = (5. - 3. 0) 16. u = i + 6j
\' = - 2i + j + k
Think About It In Exercises 17-20, use the vectors u and v shown in the figure to sketch a vector in tbe direction of lhe indicated cross product in a right-banded system.
Area In Exercises 27-30, find tbe area of tbe parallelogram that has the given vectors as adjacent sides. Use a computer algebra system or a graphing utility to verify your result.
27. u = j
v= j + k 29. u = (3, 2, - I)
v = ( l. 2,3)
28. u = i + j + k V =j + k
30. u = (2, - I, 0) v = ( -L 2,0)
Area In Exercises 31 and 32, verify that tbe points are the vertices of a parallelogram, and find its area.
31. A(0. 3. 2). 8(1. 5. 5). C(6 . 9. 5). D(S, 7. 2) 32. A(2. - 3, IJ. 8(6. 5. - I). C(7. 2, 2). D(3. - 6. 4)
Area In Exercises 33-36, find the area of the triangle with the
given vertices. (Hi11t: ½II ux v II is the area of the triangle having
u and v as adjacent sides.) 33. A(O. 0. 0), B(I, 0. 3), C( - 3, 2. 0) 34. A(2. -3,4). B(0.1,2). C(-1.2. 0)
35. A(2, -7, 3). 8(- 1. 5, 8). C(4. 6, -1 )
36. A(l.1.0). 8( - 2. 1. 0), C(0,0.0)
37. Torque A child applies the brakes on a bicycle by applying a downward force of 20 pounds OLl the pedal when the crank makes a 40° angle wi th the horizontal (see figure). TI1e crank is 6 inches in length. Find the torque at P.
17, U X V 19. ( - v)x u
18. \' X U 20. u x (u x v)
C!D In Exercises 21-24, use a computer algebra system to find u x v
and a unit vector orthogonal to u and v.
21. u = (4. - 3.5. 7)
v = (2.5. 9, 3) 23. u = - 3i + 2j - 5k
v = 0.4i - 0.8j + 0.2k
22. u = ( -8. - 6, 4)
v = (10, - l2, - 2) 24. u = 0.7k
v = I.Si + 6.2k
r.., 25. Program111i11g G iven the vecrors u and v in componem form,
write a program for a g raphing mility in which the ourpur is u x v and llu x vii-
,.-, 26. Programming Use the program you wrote in Exercise 25 to find u x v and llu x vii for u = (-2, 6, 10) and v = (3, 8, 5).
Figure for 37
Figure for 38
38. Torque Both the magnitude and the direction of the force on a crankshafl change as the crankshaft rotates. Find the torque on the crankshaft using the position and data shown in the figure.
39. Optimization A force of 56 pounds acts on the pipe wrench shown in the figure on the next page.
(a) F~ the magnitude of the momem about O by evaluating
IIOA x Fil- Use a graphing utility LO graph the resulting
runction of 0.
(bl Use the result of pan (a) to determine the magnitude of the moment when (} = 45°.
(c) Use the result of pari (a) to determine the angle /J when the magn itude of the moment is maximum. ls the answer what you expected? Why or why not?
11.4 The Cross Product of Two Vectors in Space
799
~rB % 180 lb
12 in.
l
Figure for 39
Figure for 40
40. Optimization A force of 180 pounds acts on the bracket shown in the figure.
AB (a) Detennine the vector and che vector F representing the
force. (F will be in terms of 0. )
(bl F~ the magnitude of the moment about A by evaluating
I\ AB X F ll-
(c) Use the result of pan (b) to determine the magnitude of the momenr when 0 = 30°.
(d) Use the result of pan (b ) to detenn ine the angle fJ when the magnitude of the moment is maximum. At th~angle. what is the relmionsh ip between the vectors F and AB? l s it what you expected? W hy or why nor'l
t9 (e) Use a graphing utility to graph the fu nction for the
magnitude of die moment about A for 0° s 0 s 180°. Find the zero of the function in the given domain. Interpret the meaning of rhe zero in the context of die problem .
In Exer cises 41--M, find u • (v x w).
41. u = i V =j w =k
43. u = (2. 0, 1)
V = (Q, 3. 0)
w = (O. 0. I )
42. u = ( !. I . I)
" = (2. I. 0)
w = (0. 0, I)
44. ll = (2. 0. 0) v = (l. l. l)
w = (0. 2. 2)
Volume In Exercises 45 and ~6, use the triple scalar product to find the volume of the parallelepiped having adjacent edges u, v, and w.
45. U = i T j V = j .,. k w = i + k
46. u = ( l. 3, I) v = (O. 6. 6) w = (- 4, 0, - 4)
...
Volume In Exercises 47 and 48, find the volume of the parallelepiped with the given vertices.
47. (0. 0. 0), (3, 0. 0). (0. 5, I). (2. 0. 5) (3. 5, 1). (5. 0. 5), (2. 5. 6). (5. 5. 6)
48. (0. 0. 0). (0. 4. 0), (- 3. 0, 0). (- l. I. 5) (- 3, 4 , 0), ( - 1, 5. 5). ( - 4, l. 5). (- -+. 5. 5)
49. If u x v = 0 and u • v = 0. what can you conclude about u
and v?
50. Identify the dot products that are equal. Explain your reasoning. (A ssume u. v. and w are nonzero vectors.)
(a) u • (v x w )
(b) (v x w) • u
(CJ (u x v) • w
(d ) (u x - w) • v
{e) u • (w x v)
(f) w • (v x u)
(g) (- u x v) • w
(hJ (w x u) • v
WRITING ABOUT CONCEPTS
51. Define the cross product of vecwrs u and v. 52. State the geometric propenics of the cross product. 53. lf the magnimdes of two vectors are doubled, how wi ll the
magnitude of the cross product of lhe Yecrorschange?Explain.
CAPSTONE
54. The vertices or a u·iangle i 11 space are (.r1. _Y1. ::i). (x2 • y2, .:2) ,
and (x3, y3. ::,). Explain how to fi.nd a vector perpendi cular to d1e triangle.
True or False? fn Exercises 55- 58, determine whether the statement is true or false. If it is fa lse, explain why or give an example that shows it is false.
55. It is possible ro find the cross product of two vec10rs in a two-dimensional coordinate system.
56. ff u and v arc Yectors in space that arc nonzero and nonparallel. then u X V = V X ll .
57. Ir u i= 0 and u x v = u x w. then v = w.
58. I f u i= 0. u · v = u · w. and u x v = u x w. then ,. = w.
Ill Exercises 59-66, prove the property of t he cross product.
59. u x (v + w) = (u x ,·) + (u x w) 60. c(u x v) = (cu) x v = u x (cv)
61. U X ll = () 62. u • (v x w) = (u x v) • w 63. u x v is orthogonal to both u and \'. 64. u x v = 0 if and onl y if u and v are scalar multiples of each
oll1er.
65. Prove that llu x vii = llull llvl if u and v are orthogonal.
66. Prove that u x (v x w) = (u • w)v - (u • v) w.
67. Pro\·e Theorem 1l.9.
800
Chapter 11 Vectors and the Geometry of Space
IIIJ Lines and Planes in Space
■ Write a set of parametric equations for a line in space. ■ Write a linear equation to represent a plane in space. ■ Sketch the plane given by a linear equation. ■ Find the distances between points, planes, and lines in space.
Q(x,y,::)
' L
b. \'= (a. c>
' '' _v
X
Line Land its direction vector v Figure 11.43
Lines in Space
In the plane, slope is used to determine an equation of a line. Tn space. it is more
convenient to use vec10rs to determine the equaLion of a line.
the
In Figure 11.43,
vec.;tor v = (c1, b.
c.;onsider tbe line L through the point c). The vector v is a direction vector
P(xi. y1. ::1) and parallel to for the line Land a, b, and
care direction numbers. One way of des_c_r_i_b, ing the line L is to say that it cons_is_t_s,, of all points Q(x. y. z) for which the vector PQ is parallel to v. Th.is means tbat PQ is
a scalar multiple of v, and you can write PQ = tv, where t is a scalar (a real
number).
PQ = (x - x1,_v - Yi· z - ::i) = (al, bt. ct) = tv
By equating corresponding components. you can obtain parametric equations of a Line in space.
THEORE.\111. l I PARAMETRIC EQUATIONS OF A LlNE IN SPACE
A line L parallel to the vector v = (a. b, c) and passing through the point
P(x,, Yi , Zi) is represented by the parametric equations
x = Xi + at, y = Yt + bt, and :: = :: 1 + ct.
(1.-2, 4)
--l
.j .t
-4 y
\ V = ( 2. 4. -➔)
\ The vector ,, is parallel to the line L. Figure 11.44
Ii' the directio n numbers a, b. and c are all nonzero, you can eliminate the parameter t lO obtain symmetric equations of the line.
= ,\; - X1 = )' - Y1 l. - l.1
a
b
c
D EXAMPLE Finding Parametric and Symmetric Equations
Find parametric and symmetric equations of the line L that passes through the point (I. - 2, 4) and is parallel LO,, = (2, 4, - 4).
Solution To find a set of parametlic equations of the line, use the coordinates Xi = I, y 1 = -2, and z1 = 4 and direction numbers a = 2, b =-+,and c = -4 (see Figure I 1.44).
x = I + 2,, y = -2 + 41, z = 4 - 4,
Because a. b, and care all nonzero, a set of symmetric equations is
■
11.5 Lines and Planes in Space
801
Neither paramellic equations nor symmetric equations of a given Jjne are uruque.
= For instance. in Example 1, by letting 1 l in the parametric equations you would
obtain the point (3, 2, 0). Using this point with the direction num bers a = 2, b = 4.
and c = - 4 would produce a different set of parametiic equations x = 3 + 2r, y = 2 + 4t. and z = - 4 1.
U fl EXAMPLE Parametric Equations of a Line Through Two Points
Find a set of paramet1i c equations of the line lhat passes through the points (- 2. l, 0) and (1, 3. 5).
Solution Begin by using the points P(-2, I, O) and Q(l, 3. 5) to find a direction vector for the line passing through P and Q, given by
v = PQ = ( 1 - (- 2), 3 - I, 5 - O) = (3, 2. 5) = (a. b, c).
Using I.he direction numbers a = 3. b = 2. and c = 5 with the poin t P (- 2, I. 0), you
can obtain the parametric equations
v x = - 2 + 3t, = I + 21, and ::; = 51.
■
m g As I varies over all real numbers, the parametric equations in Example 2 detem1ine the
n
poinrs (x. y, z) on the line. ln particular, note tbat 1 = 0 and 1 = 1 give the original points
(- 2. 1.0) and (J.3. 5).
■
> r
The aormal vector n is orthogonal to each
l'ector PQ in the plane. ~
Figure 11.45
Planes in Space
You have see.n how an equation of a l ine in space can be obtained from a point on the line and a vector parallel to it. You will now see that an equation of a plane in space can be obtained from a point in the plane and a vector normal (perpendicular) to the plane.
Consider the plane containing the point P(x i. y 1. :::1) having a nonzero normal vector n = (a. b, c), as shown in Figure 11 .45. This plane consists of all points Q(x, y. .::) for which vector PQ is orthogonal to n . Using the dot product, you can write the following.
n· PQ =O
= (a, b, c) • (x - X1, y - Yi , ::; - Z1) 0
a (x - x1) + /J(y - y1) + c(.:: - Zi) = 0
T he third equation of the plane is said to be in standard form.
THEOREI\I I l.12 STAN DARD EQUATlON OF APLANE INSPACE
T he plane contaiujng the point (x i, y 1, Zi) and having normal vector n = (a, b. c) can be represented by the standard form of the equation of a plane
a(x - x i) + b( y - y.) + c(z - = Zi) 0.
By regroupi11g terms, you obtain the general form of the equation of a plane in space.
ax + by + cz + d = 0
General fonn of equ ati,,n of plane
802
Chapter 11 Vectors and the Geomet ry of Space
(-2. l. 4)
!'
Aplane determined by uand v Figure 11.46
Given the general form of the equation of a plane. it is easy to lind a normal
vector to the plane. Simply use the coefficients of x, y, and z and w1ite n = (a, b. c).
I I EXAMPLE Finding an Equation of a Plane in Three-Space
Find the general equation of the plane containi ng the points (2. l. 1). (0. 4 , 1), and (-2. 1,4).
Solution To apply Theorem 11 . L2 you need a point in the p lane and a vecror that is normal to the plane. There are three choices for the point. but no normal vector is given. To obtain a normal vector. use the cross product of vectors u and v extending from t he point (2. I, l) to the points (0. ➔, I) and (- 2, 1. 4), as shown in Figure l l .--l6. The component forms of u and Y are
u = (0 - 2.4 - l. l - 1) = ( - 2. 3,0) v = ( - 2 - 2, I -1,4- 1) = ( - 4.0.3)
and it fo llows that
n=u x v
i
k
-2 3 0
0 3
= 9i + 6j + 12k
= (a. b, c)
is normal to the given plane. Using the direction num bers for n and the poim
(x1• y 1• z1) = (2, l. J), you can determine an equation of the plane to be
a(x - x 1) + b( y - y1} + c(.: - ::1) = 0
9(x - 2) + 6(y - 1) + 12(:: - l ) = 0
Standard form
9x + 6y + 12;:: - 36 = 0
General tom1
3x + 2y + 4;:: - l2 = 0 .
Simplified gcncr~l torm
•
C B ln Example 3, check to sec !.hat each of the three original points satisfies the equatio n
+ + = 3x 2y 4~ - 12 0 .
■
Two disrinct planes in th ree-space either arc parallel or intersect in a line. If they
intersect. you can detem1ine the angle (0 :s 0 ::; w/ 2) between them from the angle
between their normal vectors, as shown in Figure 11.47, Specifically, if vectors n 1 and n1 are nom1al to two intersecting planes. the an.gle Obetween the normal vectors is equal to the angle between the two planes and is given by
The angle 0 between two planes Figtu-e 11.47
\ngk between two plane,
Consequently, two planes with normal vectors n 1 and n2 are
I. perpendiculnr if 11 1 • n1 = 0.
2. parallel if n 1 LS a scalar m ultiple of n2.
Line of
Pla ne I
1 intersection
Plane 2 ,
e/
---
Figure 11.~8
11.5 Lines and Planes in Space
803
II EXAMPLE Finding the Line of Intersection of Two Planes
Find the angle between the two planes given by
X - 2y +:; = Q
2x + 3y - 2:: = 0
Equauon of plane I Equation or plane :?.
and find parametric equations of 1heir line of imersection (see Figure l l.--l8).
Solution Normal vectors for the planes are n 1 = ( I. - 2, 1) and n1 = (2, 3. - 2).
Co nsequentl y. the angle between the two planes is determined as follows .
, cos
0
_ -
Jn 1 • n2J
lintIIlln2ll
l- 61 .)6 fi7
Cusine nf ,mglc bet\\ e~n n .md n.
6
✓ 1 02
= 0.59409
This impl ies that the angle between the two planes is 0 = 53.55°. You can li nd the line
of intersection of the two planes by simultaneously solving the two linear equations representing rhe planes. One way to do this is to multiply the first equation by - 2 and add the result ro the second equation.
= X - 2_\' + Z Q
2l - 3r - 2:; == Q
- 2x + 4y - 2:; = 0
2x + 3y - 2:: = 0
7y - 4:; = 0
4:: .\' = -7
Substituting -" = -1-::./7 back into one of the original equations. you can determine that
x = :;/ 7. Finally. by !erring r = ::/7, you obtain 1he parametric equations
.r = 1, y == 4r, and ;:: = 7t
Line uf lllt~rsection
which indicate tha t l. 4. and 7 are di rection numbers for the line of intersection.
■
Note that the direction numbers in Example 4 can be obtained from the cross product of the two normal vectors as follows.
j k
n1 x n2 =
-2 I
2 3 -2
= i + ➔j + 7k
I·1
1
! -2 .I + 2
T hi~ means that the line u f' intersection of 1he two planes is parallel to the cross product o f their normal vecto rs.
G m The lhrec-dimensional rotatable graphs chat arc arnilable in the premium eBook for
this 1c:1.1 can help you visualize surfaces such as those shown in Figure I IA8. If you have access
to 1hc~e graphs. you should use them to he lp your spatiul intuition when Mudying tJ1is section
and other sections in the 1ex1 that deal with vectors. curves. or surfaces in space.
■
804
Chapter 11 Vectors and the Geometry of Space
Sketching Planes in Space
If a plane in space intersects one of the coordinate planes, the line of intersection is called the trace of the given plane in the coordinate plane. To sketch a plane in space. it is helpful to find its points of intersection wit h the coordinate axes and ics traces in the coordinate planes. For example, consider the plane given by
3.r + 2_1 + -l-;: == 12.
You can find the .\)'-trace by letting::: = 0 and sketching the line 3x + 2y = 12
in the xv-plane. This line intersects the x-axis at (·l 0, 0) and I.he y-axis at (0, 6, 0). [n Figure 1J.49, this process is continued by finding the y:::-rrnce and the x:::-trace. and then shading the triangular region lying in the first octant.
(0. 0. 3)
(0. 6. 0)
~''.
0) I
~
\
,, (-+, 0. 0)
·'
, (4. 0. 0)
I
.r
.i:r-trace (: = 0): Jx + ly = 12
_r:-trace (x = 0): ly + 4: = 12
Tracesof1heplane3x + 2_r + .t: = 12
Figure 11.49
x:-trace (y = 0): 3x + 4: = 12
l, Plane: 2r + ~ = I (0. 0. I l
_ j_ ,
If an equation of a plane has a missing variable, such as 2,-, .: = I. the plane
must be parallel 10 1he axis represented by the missing variable, as shown in Figure JJ.50. If two variables are missing from au equation of a plane, it is parallel ro 1he coordi11n1e pfo11e represented by I.he missing \'ruiables, as shown in Figure 11.51.
Plane lr + : = l is parallel to tbe y-axis.
Figure 11.50
-__,
Plane ax + d = 0 is parallel
to 1he y:-plane
Figure 11.51
Plane ~r + d = 0 is parallel
to thex:-plane
Plane c - d = 0 is parallel
to 1he .1:r-plane
proj PQ
0
p
D = IIprojo PQII
The distance between a point and a plane
Figure 11.52
11.5 Lines and Planes in Space
805
Distances Between Points, Planes, and Lines
This section is concluded with the follow ing discussion of two basic types of problems invoh ing distance in space.
l. Finding the distance between a point and a plane 2. Finding the distance between a point and ri line
The solutions of these problems illustrate the versatiliry and usefulness of vectors in coordinate geometry: the first problem uses the dot product of two vectors. and the second problem uses the cross product.
The distance D between a point Q and a plane is the length of the shortest line segment connecting Q to the plane. as shown in Figure 11.52. If Pis c111y point in the
plane. you can line! this disrance by projecting the vector PQ onto the nonual vector
n. The length or this projection is the desired distance.
THEORE.\I 11.13 DISTANCE BETWEEN A PON TA\lD ,\ PLANE
The distance between a plane and a point Q (not in the plane) is
_ D -
. _,. _
llproJ11PQ II -
175Qll n•ll nl
where Pis a point in the plane and n is nonnal to the plane.
To find a point in the plane given by ax + bv + c:: + d = 0 (a i= 0). let y = 0
and :: = 0. Then, from the equation ax + d = 0. you can conclude that the point
(-d/ a. 0, 0) lies in the plane.
El EXAMPLE Finding the Distance Between a Point and a Plane
Find the distance between the point Q(l. 5. - 4} and the plane given by
3.r - y + 1;: = 6.
Solution You know that n = (3. - J. 2) is nonnaJ to the given plane. To frnd a point
in the plane, let y = 0 and :: = 0. and obtain the point P(2. 0. 0). The vector from P
to Q is given by
PQ = (l - 2.5-0, - 4 - 0)
= (- I. 5. -~).
Using the Distance Fonnula given in Theorem 11.13 produces
D=
175Q • nl llnll
-
l<- 1.5, - 4>
J9 +
• (3. I + 4
1. 2>1
I >ist.rncc betwo~n ,l point anJ ~ plane
_ J- 3 - 5 -8I
- -IN
■
~ The choice of the point Pin Example 5 is arbitrary. Try choosing a different point in
the plane to vcrif"y that you obtain the same distance.
■
806
Chapter 11 Vectors and the Geometry of Space
From TI1eorem 11.13. you can detenni11e that che d istance between che point
Q(x0. y0. .:a) and the plane g iven by ax + by + c::. + d = 0 is D = )a(x0 - .r1) + b(y0 - y1) + d =o - ;:1)) Val + IJ1 + cl
or
D = lax0 + by0 + c::0 + d)
✓al+ b2 + c1
J
f
~ D
•
6x - 21·+ 4: + -t = 0
The distance between the parallel planes is approximately 2. 14.
Figure 1 1.53
I I EXAMPLE Finding the Distance Between Two Parallel Planes
Find the distance between che two parallel planes given by
3x - y + 2:: - 6 = 0 and 6x - 2y + 4:: , 4 = 0.
Solution The rwo plane!> are shown in Figure 11.53. To find the distance between the
planes. choose a point in the first plane, say (.r0, y0, -=a) = (2. 0, 0). Then, from the
second plane. you can determine that a = 6, b = - 2, c = 4, and d = -+, and conclude
that the distance is
_ D-
)a.\ 0 ✓+a'-b+y0
+ b,-
c;i +
+ c?-
d
)
)6(2) + (- 2)(0) + (4)(0) + ➔,
✓62 + (-2)2 + 42
=
16 tu -v56
= -,/8I-4
=
2. 14.
•
The formula for the distance between a poi nt ant.I a line in space resembles that
ror the distance between a point and a plane--except that you replace the dot prot.lucr
with che length of the cross product and the normal vector n with a direccion vector for the line.
lHEORE\I 11.1 .J DISTANCE BETWEEN A POINT Al\D A Lll\'E I~ SP,\CE
The distance between a point Q and a line in space is given by
D = IIPQ x u)I
1/ ull
where u is a direction vector for the line and Pis a poim on the line.
Point Q
D = II PQII sin 0
p 0 u
r, Line
Ihe distance between a paint and a line
Figure 11.5~
: PROOF ) In Figure 11.54. let D be the distance between the point Q and che given
line. Then D = 1175QII sin 0. where 0 is the angle between u and PQ. By Property 2 of
Theorem I I.8. you have
)lull )IPO II sin 0 = )l u x PQ II = )IPQ x u!I.
Con~equently,
D
=
IIPO II
sin
0
=
IIPQ x Hull
u!I
•
11.5 Lines and Planes in Space
807
D EXAMPLE Finding the Distance Between a Point and a Line
Find the di stance between the point Q(3, - I , 4) and Lhe line given by
x = - 2+31, y= - 21. and .:: = 1 + 4i.
Solution U~ing the direction numbers 3. - 2. ancJ 4. you know that a direction
vector for the Line is
u = (3. -2. -t).
Dir~ctHH1 ,ector 11.,r hnc
= T◊ tint! a point on the line, let 1 0 and obtain
P = ( - 2, 0. I ).
Point nn the luw
So,
D
PQ = (3 - ( - 2). - I - 0, 4 - I) = (5, - 1. 3)
•
Q = 13. - l.4)
and you can form the cross product
j -P--Q--' x u = 5 - I
3 -2
k 3 = 2i - 1Jj - 7k = ( 2. - 11. - 7). 4
Finally. using Theorem 11. 14, you can fi nd the distance to be
X
D = IIPQ x ull llull
JTTT
The distance between the point Qand the
= lineis ,/ 6 2.-45.
= v'29
Figure 11.55
= ,16 = 2.-+5.
■
mExercises
Se" w1w, CalcCha1 1m for wor►ed-~ul solu:1ons lo odd-n r•1oered iemses
In Exercises 1 and 2. lhe figure shows the graph of a line given by the parametric equations. (a) Draw an arrow on the line to indicate its orientation. To print an enlarged copy of the graph. go to the '"·ebsite www.mathgraphs.com. (b) Find the coordinates of two points, P and Q, on the line. Determine the vector PQ. Wltat is the relationship between the components of the vector and the coefficients oft in the parametri c equations? Why is this true? (c) Determine the coordinates of any points of iJ1tcrsection rnth the coordinate planes. If the line does not intersect a coordinate plane, explain why.
1. x = I + 31
y = 2-t
: = 2 + 51
2. X = 2 - 31 y= 2
ln Exercises 3 and 4. determine whclher each point lies on the line.
3. X = - 2 + /, _\' = 31,: = ➔ -"- I
(a) (0. 6. 6)
(b) (2, 3, 5)
x- 3 y- 7
.i. ~
= · -8 - = : - 2
(al (7. 23. 0)
(bl ( I. - J. - 3)
In Exercises 5- 10, find sets of (a) parametric equations and (b) symmetric equations of lhe line lhrough the point parallel to the given vector or line (if possible). (For each line, ,\Tile tbe direction numbers as integers.)
Poi111 5. (0. 0. OJ 6. (0. 0. 0) 7. (-2.0.3 ) 8. (- 3. 0, 2) 9. II. 0. l )
JO. ( - 3. 5, 4)
Parallel IV
\' = (3, l. 5)
V = (- 2, f 1)
v = 2i ..- 4j - 2k
v = 6j + 3k X = J + ] /. _I' = ) - 2t. : = - 7 , I
.I - I \' + l
3 - -2 - ~
3
808
Chapter 11 Vectors and the Geometry of Space
I n Exercises I l-1-t fi nd sets of (a) pa ra metric equation!. and (b) symmetric ef)ua tions of the line through the two points tif possible ). (For each line. \H ite the direction n umbers as integers.)
11. (s. - 3. -:n.(-rl 1)
13. (7, -2. 6). {- 3. 0, 6)
12. (0. -l-. 3). ( - I.~- 5) 14. (0, 0. 25), ( I0 , JO. OJ
30 '
L
I:
x-3
~
=
y-2 - 1-
=
:+ 2
~
L,:
.1- l --
= ·-y
--
1 = :+3
➔
2
➔
L, :
x +2
- ,-
=
Q,. -
5 I
=
: -
-3 I -
In E...xercises 15-22, find a set of param et ric equations of the line.
L· x- 3 =L±....!_ = : - 2
•1• 2
➔
-1
15. The line passes 1hrough the poim (2, 3. 4) and is parnll.:1 to Lhc x:-plane anc.J the y:-planc.
16. The li ne passes 1hrough Lhe pnint (- ➔. 5, 2) and is paralle l 10 the .1)·-planc antl Lhe y:-plane.
17. The line passes through 1he point (2. 3. 4) and i~ perpendicular
10 the plane given by 3x + 2y - ;; = 6.
J.n Exe rcis es 31-34, determine whether the li nes intersect, and if so, find the point of intersection and t he cosine of the a ngle of intersection.
31. .I = ➔I + 2. y = 3. ;; = - £ +- I X = 2-I + 2, _\' = 2s + 3, ;; = J +
18. The line p;1sses through the poi nt ( - ➔. 5. 2) and is perpcnc.Jicular
10 the plane gfren by -.1 + 2y + : = 5.
19. The line passes Lhrough the point (5. - 3, - ➔land is parallel 10
, = (2. - I. 3) .
32. .I = - 3£ + I. y = 41 -'- I. : = 21 + 4
x = 3J + I. y = ls + 4. : = - J + I
33.
x -3
=
v- 2 '---I_
=
:
+
I.
x- 1 - 4-
= •1·
+
2
=
:.,..3 --3-
20. The line passeS through Lhe point (-I. 4, - 3) anc.J is parallel to
v = Si - j .
6 34.
.~ r - 1
= r-2
= :
-
~
·'·
x- 3
: +2
- 2-
=1••+
5
=-➔
21. The lme passes Lhrough 1he point (2. I. 2) and is parallel to Lhc
line .1 = - 1, " = I + = £, ;: - 2 + ,.
m , In Exercises 35 and 36, use a comp uter aJgebra system to g raph
22. The line pass•es 1hmugh Lhe poim ( -6. 0, 8) and i;. parallel 10
the pa ir or intersectin g lines a nd fi nd the point ofinte rsection.
!he line x = 5 - 21. y = - 4 + 21. .: = 0.
35. .r = 21 + 3. y = 51 - 2. .: = - t + l
In Exercises 23-26, find the coord inates of a point P on the line a nd a w clor \ parallel to t he line.
x = - 2s + 7. y = s + 8. ~ = '.!I - l 36. .\ = 21 - I._,.= -41 + 10. .: = I
23. .r = 3 - I, _I' = - I + 2r. : = - 2
x = - 5.1· - 12.y = 3s + 11. : = - 2s - -I
= 2.J. X = 4£. _1' = 5 - I. .: 4 + 31
.-,.~. -
-X - -7
-I
= ._\'-+2
-6
=
-
~
+
-.,
Cross Product In Exercises 37 and 38, (a ) find the coordina tes
of three points P. Q. and R in the plane, and deter mine the
vectors PQ and PR. (b) Find PQ x PR. ·w ha t is the relation-
In Exercises 27-30, deter mine if any of the lines a re parallel or identical.
27. L1: x = 6 - 3r. y = - 2 + 21. : = 5 + 41
L2: r = 61. y = 2 - ➔I. : = 13 - 8r
= L.: .1 10 - 6r. y = 3 + 41. .: = 7 + 81
L.:\: .1 = - -l- + 61, y = 3 + -l-£, .: = 5 - 6r
28. L1: x = 3 + 2r. y = -6£. : = I - 21
l..2: x = I + 2r. _\' = - I - I. = .: 31
L3: x = - I T 2,. .r = 3 - !Or. : = I - -11 L,: ., = 5 + 2£. y = I - £. ;; = 8 - Jr
29.
Li:
.I -
- 4-
8
=
V + 5
• -2
=
: +9
~
L-,·.
X
+ 2
7
=
)'
I
4
= :
+ 5
6
L ,: -.I +- -+ = y- - - I = - : +- I 8
• -8
4
-6
shi p between the components or the cross product and the coefficien ts of the eq ua tion of the plane? Why is this true?
37. 4.r - 3y - 6.: = 6
38. 2x + 3y + 4z = 4
....,,
X
X
---...__.... _\'
In Exercises 39 and 40. determine whether the pla ne passes through each point.
39. .l -r 2.r - 4: - I = 0
(a) ( - 7. 2, - I l
40. 21 + y + 3:; - 6 = 0
(a) (3. 6. - 2)
(b) (5. 2. 2) (bl (-1. 5. - I)
L: -.\ -- = 2 -,,-+ -3 = -;; - -4
- -2
I
1.5
l l .5 Lines and Planes in Space
809
In Exercises -11-46. find an equation of the plane passing through the point perpendicular to the given YCctor or Linc.
In Exercises 65-70. dcterrninc whether the planes are parallel, orthogonal. or neither. If they are neilher parallel nor
Poi111 41. ( 1.3. - 7) -t2. (0. - I. 4) -t3. (3. 2. 2) -t4. (0. 0. 0) 45. (- 1.4.0)
Perpendicular tn
= 11 j
n = k
n = 2i + 3j - k
a = -3i + 2k
x = - I + 21. y = 5 - /. .: = 3 - 21
x- 1
, : +3
- 4-== y+ _ =-=T
orthogonal, lind the angle of intersection.
65. 5x - 3y + :: = 4
x + 4y+7.:. == l
67. X - Jy + 6;: = ➔ 5x + y - : = -I
69. X - 5y - : = I
5x - 25y - 5:: = - 3
66. 3x + y - -\;: = 3
- 9.\ - 3y + 12: = 4
68. Jr + 2y - .: = 7 X - ➔}' + 2.: = ()
70. 2l· - :: == I
4.\· + .r ..,. 8.: = I0
ln Exercises 71-78, sketch a graph of the plane and lal.>el any
In Exercise.~ -t7-58, find an equation of the plane.
intercepts.
47. The plane passes through (0. O. 0). (2. 0. 3 ). and ( - 3 , - I. 5). -18. The plane passe~ through (3. - I. 2). (1. I. 5). and ( I. - 2. - 2). -19. The plane passes through l I. 2, 3). (3. 2, I). and (- I. - 2. 2). 50. The plane passes through the point ( I. 2. 3) and is parallel to
71 4x + 2y + 6:: = 12 73. 2r - y + 3:: = 4 75. X + ;: = 6
77. X = 5
72. 3.r + 6y + 2:: = 6 74. h - v + .: = 4 76. 2r + y = 8 78. : = 8
the •v:-plane.
( m ln Exerc.ises 79- 82. USC a cornputer algebra system to graph the
51. The plane passes Lhroug.h the poin1 ( I. 2. 3) and is parallel 10
plane.
Lhe xy-plane.
52. The plane comains 1he y-axis and makes an angle of .r/ 6 with the positive .r-axis.
= 79. 2\' -'- y - :: 6 8 I. - 5x + 4y - 6:: = - 8
80. .r - 3:: = 3 82. 2. lx - 4.7y - : = -3
53. The plane contains the lines gi~en by
x-1 --
-2
=
" •
4 =:
and
x- 2 --
-3
=
y-)
-·- 4
=
-: --2.
-I
In Exercises 83- 86, d etermine if any of the planes are parallel or identical.
54. The plane pas~es Lhrnugb 1he pnin1 (2, 2. I) and comains the line gi\'cn by
55. The plane passes Lhrough the points (2. 2. I) and (- I. I. - I)
and is perpendicular to the plane 2.r - 3y + :; = 3.
56. The plane pa,scs through the points (3. 2, I) and (3, I. - 5) and is perpendicular 10 the plane 6.r + 7y + 2: = I0.
57. The plane passes 1hrough lhe points ( I. - 2. - I) and (2. 5. 6) and is parallel to the x-axis.
58. The plane passes through the points (4. 2. I) and (- 3. 5. 7) and is parallel to the .::-axis.
P~: -5x + 2y - 8:: = 6 = P1: 6.1 - 4.\' + 4:, 9 P4: 3,1 - 2\' - 2.:: = 4 85. P1: 3x - 2y + 5:: = 10
P~: - 6x -'- 4y - l0:=5
P3: - 3x + 1y + 5:: = 8
PJ: 75.r - 50y + 125.:: = 250
86. P1: -60.r-'- 90y ~ 30:: = 27
P~: 6x - 9y - 3.: = 2 P,: -20.r - 30y + 10:: = 9
PJ: 12x - 18y + 6:=5
P2: 3.r - 5y - 2:: = 6
P3: 8x - 4y + 12.: = 5
P.: - 4.r - 2y - 6: = 11
ln Exercises 59 and 60. sketch a graph of the line and find the points (if any) where the line intersects the xy-, xz-, and yz-planes.
59. X = 1 - 11.
l. _ -,
60.•_.1_- = .,. -
= y -1 -1- 31. I = -- ---..,3
;: = -4 + I
ln Exercises 61-64, and an equation of the plane lhat contains all tJ1e points that arc equidistant from the giYen points.
61. (1. 2. 0). (0, 2. 2) 63. (- 3, I, .2). (6. - 2. 4)
62. ( I. 0. 2). (2. 0. IJ 64. (-5. 1. - 3). (2. - 1. 6)
In Exercises 87-90, describe the family of planes represen ted by the equation, where c is any real number.
87. X + .\' + ;: = C 89. cy +:: = 0
88. X + y = C 90. X + C;: = 0
In Exercises 91 and 92. (a) find the angle between the two planes, and (b) find a set of parametric equations for the line of intersection of the planes.
9 l. 3x + ly - : = 7
.1 - 4y + 2: = 0
92. 6.r - 3y + :: = 5
- _r T )' + 5;: = 5
810
Chapter 11 Vectors and the Geometry of Space
In Exercises 93-96, lind the point(s) of interseclion (if any) of lhe plane and the Line. Also determine whether the line lies in the plane.
93. ,_,,. _ ,-,, + -_ = I'->,
,- _ _! = y + (3/2) = ::: + I
., 2
-l
2
94. 2x + 3y = - 5.
x-l=I = .:-3 4 26
95. 2x + 3y = I0.
96. 5x + Jy = 17•
= x - 1 y+l
~
- 3-
~ = .: - .::i
- .I -2-4 --
- ·y -+3-I
.: + 2
- - ~
In Exercises 97-100. find the distance between the point and the plnne.
97. (0. 0. 0) lr ... 3y +.: = 12
99. (2. 8...i) 2x---y-'-:::=5
98. (0, 0. 0)
5.\ + y - ;: = 9
100. (1.3. -1 )
= J\ - 4y .,. 5: 6
In Exercises Hll -104, verify that the two planes arc parallel, and find the dislance between the plnnes.
101. ~ - 3y + 4: = 10 x - 3y + -k = 6
103. - 3_, .c. 6y.,;. 7: -
6x - 12.1· 14: = 25
102. -h - 4_r + 9: = 7
➔X - -+_\' _,_ 9: = ]8
I04. lT - 4: = 4 2r - 4.: = 10
In Exercises 105-108. find the distance between the point and tbc line given by the set of parametric equations.
l05. 11. 5. - 2): x=41 -2. _r=3. := - 1 106. ( 1. - 2. -+): X = 21, I = I - 3. ;: = 21 + 2 107. (-2.. 1.3): .\ = 1 - I. _I'= 2 + /, ;: = -21
108. (4. - 1. 5): x = 3. 1 = I + 31. : = I - r
[n Exercises 109 and lJ0, verify that the Lines are paraUel. and find the distance between them.
l09. L1: r = 2 - 1. _I'= 3 + 21. ;: = ➔ + I = L1: .t 3r. _,. = I - 6r. : = ..i - 3r
110. L1: .t = J + 61. y = -1 + 9r. : = I - 111 L.: .t = - I + 4r, J = 3 + 61. : = - 81
WRITING ABOUT CONCEPTS
J 11. Give the parametric equations aml the symmetric equations of a line in space. Descri be what is reqaired to find these equations.
112. Give the ~landard equation of a plane in space. Describe what i~ required to find this equation.
1J 3. Describe a method of finding the line of intersection of mo plane~.
114. Describe each ;,urfa1:e given by tl1e equations ., = a. r = b. and : = c.
WRITING ABOUT CONCEPTS (continued>
115. Describe a method for detennin.ing when rwo planes a1x + b1y + c 1: + d1 = 0 and
a:x + h:i.Y + c2;: + d2 = 0
are (al parallel and (b) perpendicular. Explain your reasoning. 116. Let l 1 and L::_ be nonparallel line~ 1ha1 do not intersec1. 1s it possible to find a nonzero vector , such that v is perpend icular IO both L1 and ~'? Explain your reasoning. 117. Find an equation of the plane with x-intercept (a. 0. 0). y-imercept (0. b, 0), and ;:-intercept 10. 0. c). (Assume a. h. and c are nonzero.)
CAPSTONE
118. Match the equation or set of equation~ with the description it represent,. (a) Set of parametric equations of a line (b) Set of symmetric equations of a line (Cl Standard equation of a plane in space (d) General form of an equauon of a plane in space
= (i) (x - 6)/ 2 (_r, l)/-3 = : / I
(ii) 2, - 7y -,- 5:: -,- IO = 0
= (iii) r -1- + 71. _r = 3 + r,::. = 3 - 31
= (i,) 1tx - I) + (y + 3J - 4(:: - 5) 0
I 19. Describe and find an equation for 1hc surface generated by all points (x. y, ;:) that are four units from the point (3. -2, 5).
120. Describe and find an equation for the surface generated bj all points (x. y. ;:) that arc four units from the plane
= 4X - 3y .L :: I 0.
121. Modeling DaJa Per capita consumptions (in gallons) of di.ffcrcm types of milk in the United States from 1999 through 2005 are shown in the table. ConsumptioIL~ of flavored milk. plain reduced-fat milk. nod plain light and skim milk:, are represented by the ,·ariables x. y. and :, respectively. (Source: U.S. Department ofAgric11/r11re)
-- - ----Yeur I 1999 I 2000 2001 2002 2003 2004 2005
- ~-~- r - -- ---~-- - -
x _ I 1.4 . 1.4 1.4 1.6 [ 1.6 1.7 J.7
cry - - -7.3
-
7. 1 7.0 7.0
- -r::-:-:---
6.9 6.9 6.9
--~- - -
6.2 _ _!_,1_1_ 5 ~ 5 ~ 5.6 5.5 5.6
A model for the data is given by 0.92t - = I .03y J.. ; 0.02.
(a l Complete a founh fll\l in the table using the model to approximate: for the given value~ ofx and y. Compare the approximation~ with the actual values of:.
!bl According to this mode l. ,my increases in consumption of 1wo type~ of milk will have what e ffect on the consumption of the third t}'J)l~?
11.5 Lines and Planes in Space
811
122. ,\,Jechanical Design The figure shows a chuce at che cop of a grain elevator of a combine 1h::11 funnels the grain into a bin. Find the angle becween m·o adjacenc sides.
8 in.
6 in. 123. Distance Two insects are crawling along differem lines in
three-space. At ti me 1 (in minutes), the fi rst insec t is at the
poim (x. y, ;;) on the line x = 6 + = t. _,· 8 - 1, :: = 3 + ,.
Also, at time 1. the second insect is ai the point (x_y. ;;) on the
= = line x l -1- 1, y = 2 + 1, z 2T.
Assume 1hac dis1anccs arc given in inches.
(a} Find the di,cance between the rwo insec[s at Lime 1 = 0.
(b) Use a graphi ng utility 10 graph the distance between the
insecrs from , = 0 10 r = I0 .
(cJ Using che graph from pan (b). whar can you conclude abom the distance between the insects'!
(d) How close to each other do the insects gel?
124. Find the standard equation of the sphere with cellter ( - 3, 2, 4)
tha1 is mngenc to the plane given by 2.x + -ly - 3.: = 8. 125. Find the point of intersection of the plane 3.r - y - -+: = 7
and the line through (5. 4. - 3) that is perpe ndicular 10 this plane.
126. Show that the plane lx - y - 3.: =-+ i, parallel 10 the line
x = - 2 + 21, y = - l + 41. i = 4. ,md fi nd the distance
berween them.
127. Find the point of intersection of the line rhrough (1. -3. I)
and (3. - 4. 2). and the plane given by .1· - y + :: = 2.
128. Find a set of parametric equations for the line passing through
the point ( I. 0. 2) that is parallel to the plane given by
x + y + z = 5. and perpendicular to the line x = t. v = I + = I,: I + r.
Tr11e or False? l n E xercises 129- 13-l. deter mine ,1helher the
rr statement is t rue or false. it is false. explain why or give an
example that s hows it is false.
129. If v = ai i + b,j + c 1k is any vector in £he plane given by = = a1x + b"Jy - c2: - d" 0. then ail½ + bib1 + c 1c: 0.
130. Every two lines in space are either imer~ecting or parallel.
131. Two planes in space are either intersecting or parallel.
132. lf two lines L1 and Lz arc parallel LO a plane P. then L1 and Lz
are parallel.
133. Two planes perpendicular lO a ihird plane in space are parallel.
134 . A plane and a line in space are either imersecting or parallel.
Distances in Space
You have learned two distance formulas in this section-the
discance between a point and a plane, and the disrance between a
point and a line. In th.is project you will study a third distance problem- the distance between rwo skew lines. Two lines in space are skei,· if the~ are neither parallel nor intersectin)! (see figure).
(a) Consider the following two line, in space.
Li: X = 4 + St. y = 5 + 51. z = 1 - 41 Lz: x = ➔ T s. _\' = -6 - Ss, : = 7 - 3s
(i) Show that these lines are noi parallel.
(ii) Show that these lines do aot intersect. and therefore arc skew lines.
(iii) Sha,~ that the rwo lines lie in parallel planes.
(iv) Find the distance between the parallel planes from part (iii). This is the distance hetwecn rhe original skew lines.
(b) Use the prm;edure in pan (a) to find the distance between the lines.
= = = L1: X 21, y ➔I. : 61
-I + s
(c) Use the procedure in part (al to find the distance between the lines.
L1: X = 31. y = 2 - /. .: = - I + / = = L2: x I + -+S. y - 2 + .1, : = - 3 - 3s
(d) Develop a formula for finding rhc distance between the skew lines.
= = L 1: X X1 + {Ill. ,I' = .1'1 + bit, Z. .:1 + C1l + l 2: x = x2 a 2s, y === y2 + b1s. .:: = .::? + c1J
~
812
Chapter 11 Vectors and the Geometry of Space
Im Surfaces in Space
■ Recognize and write equations of cylindrical surfaces. ■ Recognize and write equations of quadric surfaces. ■ Recognize and write equations of surfaces of revolution.
Right circular cylinder. x2 + y2 = a2
Rulingsare parallel 10 :-axis. Figure 11.56
Cylindrical Surfaces
The first five sections of this chapter contained the vector portion of the preliminary work necessary to study vector calculus and the calcu lus of space. ln thls and the next section, you \\ ill study surfaces in space and alternative coordinate systems for space. You have already sc udied two special types of surfaces.
1. Spheres: (.\ - x0l2 + (y - y0f + k - .::0F = r 2 2. Planes: ax T by + c::. + d = 0
Secuon 11 ~
A third type of surface in space is called a cylindrical surface, or simply a
cylinder. To define a cylinder, consider the fami liar right circular cylinder shown in
Figure 11 .56. You can imagine that this cylinder is generated by a vertical line moving
around the circle x2 + y 2 = a 2 in the XY-planc. This circle is called a generating
curve for the cylinder, as indicated in the fo llowing definition.
0EFINITIO~ OF A CYLINDER
n
Let C be a curve in a plane and let L be a Line not in a parallel plane. The set
of all lines parallel to Land intersecting C is called a qlinder. C is called the
generating curve (or directrix) of the cylinder, and the par:illel lines are
II
called rulings.
II
Ruling intersec1iog C
Generating curve C
G a WithoUI loss of gcneraliry. you can assume that C lies in one of che chree coordinare
plane~. \lloreoYer. Lhis text rcstncts the discussion ro right cylindcrs--cylinders whose rulings
are pcrpemlicular to the coordina1e plane containing C. as shown in Figure 11.57.
■
For the right circular cylinder shown in Figure 11.56, the equation of the generating curve is
Cylinder: Rulings intersect C and arc parallel to the given line. Figure J 1.57
To find an equation of the cylinder, note that you can generate any one of the rulings by fixing the rnlues of.randy and then allowing ;: to rake on all real values. ln this sense. the value of::: is arbitrary and is. therefore. not included in the equation. In other words, the equation of this cylinder is simply the equation of its generarjng curve.
EQUATIONS OF CYLINDERS
The equation of a cylinder whose rulings are parallel to one of the coordinate axes contain!> only the variables corresponding to the other two axes.
11.6 Surfaces in Space
813
D EXAMPLE Sketching a Cylinder
Sketch the surface represented by each equation.
a. :: = y2
b. :: = siru. 0 ~ x ~ 2-rr
Solution
= _,· a. The graph is a cylinder whose generating curve. :z: 2 • is a parabola in the
y::-plane. The rulings of the cylinder are parallel to the .r-axis. as shown in Figure I l .58(a).
b. The graph is a cylinder generated by the sine curve in the x::-plane. The rulings are parallel 10 the y-axis. as shown in Figure I l .58(b}.
Generating curve C ltc, in y:::-plane
Generating cun·e C lies in -~-plane
X
Cylinder: : = y2
(a) Ruling, are parallel 10 .\-axis.
Figure 11.58
Cylinder: : = ~in x
(bl Ruling, are parallel to y-axis.
■
EPIID■m• ln the table on pages 8 14 and 8 I5. on!} one of several orientations of each quadric surface is shown. If the surface i, oriented along a different axis, irs ~1andard cqua1ion will change accordingly. as illustrn1ed in Examples 2 and 3. The fact that the two types of paraboloids have one variable raised to the fir~! power can be helpful in classifying quadric surface~. The OEhcr four types of basic quadric surfaces ha\"e equation~ that are of second dei1ree in all three variables.
Quadric Surfaces
The founh basic type of surface in space is a quadric surface. Quadric surfaces are the three-dimensional analogs of conic sectioll5.
QUADRIC SURFACE
The equation of a quadric surface in space is a second-degree equation in
th ree variables. The general form of the equation is
'
I
+ + + + + + Ax2
B y2
C z1
Dxy
Ex~
+ F y:.
Gx
0. Hy+ I :.+ .I=
T here are six basic Lypes of quadric surfaces : ellipsoid, hyperboloid of one sheet, hy perboloid of two sheets, elliptic cone, efliptic paraboloid, and hyp erbolic paraboloid.
The intersection of a surface with a plane is called the trace of the surface in the plane. To visuali1e a su1face in space. it is helpful to determine its traces in some wellchosen planes. The trace!-. of quadric surfaces are conic~. These traces, together with the standard form of the equation of each quadric surface, are shown in the table on pages 81-l and 815.
814
Chapter 11 Vectors and the Geometry of Space
Ellipsoid
Trace
Plmre
Ellipse
Parallel to .,}·-plane
Ellipse
Par::tllel to x:-plane
Elli pse
ParaJlel to y:-plane
X
The surface is a sphere if
a= b = c =/= 0.
Hyperboloid of One Sheet
Trace
Plane
E llipse H y pe rb o l a Hyperbola
Parallel to xy-plane Parallel ro xz-plane Parallel to yz-plane
The axis of the hyperboloid co1Tesponds ro rhe variable whose coefficient is negative.
y y
x:•trat:e
Hy perboloid of Two Sheets
y:-trnce
.rz-trace
Trace
P/al!e
Ellipse
Parallel to .\)'-plane
Hyperbola Parallel to x:-plane
Hyperbola Parallel to y:-plane
I ,
The axis of the hypcrboloi<l
I
corresponds to the , ariable whose
coefficient is posiuve. There is
no trace in the coordinate plane
perpendicular to this axis.
11.6 Surfaces in Space
815
E lliptic Con e
Trace
Ellipse Hyperbola Hyperbola
-Pla-ne - - -- -
Parallel to xy-plane Parallel to .r;:-plane Parallel 10 y;:-plane
The axis of the cone corresponds to the variable whose coefficient is negative. T he traces in the coordinate planes parallel to this axis are incersecting lines.
.r.:-trace
Elliptic Paraboloid
Trace
Plane
Ellipse Parabola Parabola
Parallel to xy-planc Parallel to x.:-plane Parallel to y.:-pJane
T he axis or the paraboloid corresponds to the variable raised to the first power.
y~-tracc
Hyperbolic Paraboloid
\'2 x2
-==.!..---
tr'
_\\.:-trace
Trace
Hyperbola Parallel 10 .rr-plane
y
Parnbolu Parallel to .r;:-piane
Parabol11 Parallel 10 y:::-plane
The axis of 1he paraboloid corresponds 10 rhe variable raised to the Iirs1 power.
c -trace
parallel to .')·-plane
816
Chapter 11 Vectors and the Geometry of Space
To classify a quadric surface. begin by wricj ng the surface in standard fonn. Then.
determine several traces taken in the coordinate planes or taken in plane;, that are
parallel lo the coordinate planes.
fl EXAMPLE Sketching a Quadric Surface
= Classify and sketch the surface gh en by -lx2 - 3y2 + 12::2 + 12 0.
Solution Begin by writing the equation in standard form.
= 4x2 - 3y2 +- J2z2 + 12 0
\\ m~ ,,riginal cqun11on
y
Hyperboloid of two sheet,,:
4v2
-
x-2
3
-
:
.,
-
=
I
Figure 11.59
Elliptic parnboloid: x=~·~+4:2
Figure 11.60
r2 x z .:2
'---- - -=! 43 I
From the table on pages 8 1-1 and 8 15, you can conclude 1hat the surface is a hyperboloid of two sheets with the y-axis as its axis. To sketch the graph of this surface. it helps lO find the traces in the coordinate planes.
-"':''-trace (:: = 0):
x::-trace (y = 0): .-r2 +z-2 = - I
3 I
y;:-trace (.r = 0):
H~r~rhola
The graph is shown in Figure 11.59.
II EXAMPLE Sketching a Quadric Surface
Classify and sketch the surface given by .r - y 2 - 4::2 = 0.
Solution Because .r is raised only to the first power. the surface is a paraboloid. The axis of the paraboloid is the x-axis. In the standard form, the equation is
X = _r2 + 4;:2.
St;111Jard 11' 11111
Some convenient traces are a.s follows.
.\y-trace (:: = 0):
x.:-trace (y = 0):
= X y-1
= X 4;:2
Par.,ho!,1
parallel to y;:-planetx = -+):
The surface is an elliptic paraboloid, a'i shown in Figure 11.60,
■
Some second-degree equation!<. in x. y, and ;: do not represent any of the basic l)'pes of quadric ~urfaces. Here are two examples.
x= - _r2 + ::" = 0 Xl + _\-" =
Single p,1rn1 R1gh1 drrn lnr .:ylln,kr
11.6 Surfaces in Space
817
I 2. -1. 11
•
An ellipsoid centered at (2. - I. I) Figure 11.61
For a quadric ~urface not centered at the ongrn. you can form the surndard equation by completing the square, as demonstrated in Example 4.
II (.; EXAMPLE A Quadric Surface Not Centered at the Origin
C la~sify and sketch the surface given by
x2 + 1y2 1· z2 - -l-x + 4y - 2z + 3 = 0.
Solution Completing the square for each variable produces the following.
(x2 - -l-x + ) + 2( J 2 + 2y + ) + (:2 - 2;; + ) = - 3
(x2 4.r + 4) + 2( y 2 + 2,· + l ) + (z2 2;; + 1) = - 3 + 4 + 2 +
-
-
Lr - 2)2 ~ 2(y + I)2 + (;; - If = 4
-(x -- 2-)2 + ~lv-+-'-I )-2 - -(z -- -'I-)2- = I
4
2
4
From this equation. you can see that the quadric surface is an ellipsoid that is centered
at (2. - 1. I). Its graph is shown in F igure 11.61.
■
.~IIf:IJ::I[IlfJ A computer algebra system can help you visualize a surface in space.* Most or these computer algebra systems create three-dimensional illusions by sketching several traces of the surface and then applying a '·hidden-line·· routine that blocks out portions of the surface that lie behind other portions of the surface. Two examples of figures that were generated by Mmhemarica are shown below.
H) perbolic paraboloid
ri xi -=:.- - -
• 16 16
Using a graphing urility to graph a surface in space requires practice. For one thing, you must know enough about the surface 10 be able to specify a 11ie1ri11g wi11do11 that gi,es a representative \'iew of the surface. Also, you can often improve the view of a surface by rotating the axes. For instance. note Lhat the elliptic p,u-aboloid in the ligure is seen from a line of sight that is ..higher·' than the line or sight used 10 view the hyperbolic paraboloid.
*Some 3 -D graphing wilirieJ require s111faces 10 be e111ered with pam111erric eq11ario11s. For
a discussion of rhis 1ech11iq111'. see Senion 15.5.
818
Chapter 11 Vectors and the Geometry of Space
Circular cross \ection
Genera11ng .:un e
t r= r(:)
:,\-(O.O. :)
• • ::.;...-. (0, r (: J. :)
I
Surfaces of Revolution
The fifth special type of i,urface you wiU smdy is called a surface of revolu tion. ln Section 7.-+. you studied a method for finding rhe area of such a surface . You will no\\ look at a procedure for finding il~ equation. Consider the graph of the radius funct ion
y = r(.::)
in the y;:-p lane. lf this graph is revolved about r.be ;:-axis. it forms a surface of revolution. as shown in Figure 11.62. The trace of the surface in the plane.: = .::0 is a circle whose radius is r(¼i) and whose equation is
Figure 11.62
Replacing : 0 11ith " produces an equation that is valid for all values of;:. ln a similar manner, you can obtain eq11ations for surfaces ofrevolution for the other two axes. and the resu lts are summarized as folio\\~-
SURFACE OF REVOLUTION
If the graph of a radius hmction r is re\'Ol\'ed about one of the coordinate axes. the equation of the resulting surface of revolution has one of the following forms.
1. Revol\'ed about the x-axis: y1 -'- .::1 = [,{r)P
= 2. Revolved about the y-axis: .r1 ,- ;:2 (dy)F
3. Revolved aboUL rhe :-axis : .r~ + y 2 = [r(::ff
Surface:
El EXAMPLE Finding an Equation for a Surface of Revolution
a. An equation for rhe surface of revolution formed by revolving the graph of
y= about the ::-axi~ is
R,,dms 1unu!(ln
Generating curve 9x~ = y 3
Figure 11.63
b. To find an equ,1tion for the surface formed b) re,·olving the grap h of 9x2 = y-'
about Lhc y-axis. solve for x in 1em1:, of y to obtain
= = X 'IiY"1 -, r ( _\'}•.
Ratli" f,111cti,.
So, the equation for 1his surface is
= x .:? + :::1 [r(yff .\.:i + ::2 = (h3, 2)2
= , .;J ,r- _.;
A
~
!9H,• J •
S11h,111u1c I tn1 r I I
The graph is shown i.11 Figure 11.63.
•
11.6 Surfaces in Space
819
The :o:,eneratineo cun·e for a surface of revolution is not uniq ue . For i.nstance, the surface
can be formed by revol\'ing either the graph of x = e-) about rhe y-axis or the graph of z. = e-.1 about rhe y-axis, as shown in Figure I l .64.
Surface:
,.2 + ::.1 = e-2\·
l . Generating curve in y:-plane I ' ,.,-•
Generating curve
in xy-plane x= ✓9-3y2
Generating curve
in y:-plane : = ✓9-3y~
Surface: t'~ + 3v1 + : ~ = 9
Figure 11.65
Gcneraung curve
in .\)'"plane x = e-,-
Figure 1L.6-t
/~ .\
-
-)
II EXAMPLE Finding a Generating Curve for a Surface of Revolution
Find a generming curve and the axis of revolution for the surface given by
x1 + 3r" + :::: = 9.
Solution You now know thar the equatio n has one of the following forms.
x 2 + Y2 = [r(::IF y1 + ::" = [r(.r)]2 = + x l .::2 [ r(y)F
Because the coefficients of .,..2 and :::2 are equal, you ~hould choose rhe lhird form and write
.r1 + :::' = 9 - ]y".
The y-,v:i~ is the ai;is of revolution. You can choose a generating curve from eithe r of the following traces.
x2 = 9 - 3y1
::2 = 9 - 3y::'
For example, using the fi rst trace. the generating curve is the semie llipse given by
The graph of thi~surface i~ shown in Figure 1 1.65.
•
mExercises 820
Chapter 11 Vectors and the Geometry of Space
Se'= v,~t1CalcC0 a; com 101 ,•inned-out solut1ons to odo-1111mbe.ecf exercises
Ln Exercises 1-6, match the equation with its graph. [The
(c)
(d)
graphs are labeled Ca). (b), (c), (d), (e). and (f}.]
(a)
(b)
Figures for 17
(m 18. Use a computer algebra system to graph a vie,\ of the cylinder
(c)
(d)
y1 + :::2 = 4 from each point.
(a) (I 0. 0, 0)
(b) (0. 10. OJ
(c) (10, 10, IO) -5
In Exercises 19-32. identify and sketch the quadric surface. Use a computer algebra system to confirm your sketch.
(e)
(f )
19.
x2
+
y2 ~
-'-
:2
=
I
-1
= 21. l6x2 - y 2 T 16:::'.' 4
23. -+X2 - y2 - :.2 = I
20.
-x-'+
\'"' :-'
~..!.- =
16 25 25
22. -8x2 + l8y2 , 18:::2 = 2
= y2
24 •
.2
~
-
.\''.' -
'-4--
I
= 25. X'.' - )' T ;:' 0
= 27. .r2 - y 2 ' . : 0
26. :: = x 2 + -1y2
28. 3:: = -y1 + x1
1 x-' , y-' - ~' = 1
• 9 16 9
= 3. 4.r'.' - y1 + ,k:2 -I
= 5. 4x1 - -1_,. - :::2 0
2. 15x2 - -1"y 1 + 15.:1 = --1
3 1. 16:t'.' T 9_r2 T 16::2 - J2t - 36}' + ]6 = 0
= + 4. y2 4.r'.' 9:'.'
32. 9x1 + y1 - 9z2 - 54x - 4y - 54:: + 4 = 0
6. -h2 - \'2 + ·k = 0
C!D In Exercises 33-42, use a computer algebra system to graph the
In Exercises 7-16, describe and sketch the s urface.
surface. (Hint: It may be necessary to solve for ;: and acquire
7. y = 5
8. :: = 2
two equations to graph the surface.)
9. y'.' + ::2 = 9 11. x'.' - ,. = 0
JO. r2 + ::'.' = 25 12. ,·2 + :: = 6
33. : = 2 cosx
= 35. : 2 x 2 + 1.5y2
34. ;: = x2 - 0.5y1 36. 3.25 \' = .\2 + ::2
13. -1x 2 .1- y'.' = -+
15. ;:-smy=0
= 14. y 2 -::2 16 16. :: - = p.l 0
17. Think About It The four figures are graphs of the quadric
surface :: = x1 + y1. M:1tch each of the four graphs with the
poim in space from which the paraboloid is viewed. The four
poim~ are (0, 0, 20). (0. 20. 0). (20. 0. 0), and (10. 10. 20)
(al
Cb)
M 39_ : = 10-
41. 6x1 - .i_,.2 + 6:.2 = - 36
-10. ~
-x 8 + x2 -,- y2
-12. 9x2 "t' -ly1 - 8:2 = 72
In Exercises 43--16, sketch lbe region bounded IJy the graphs of the equations.
43. : = 2 Jx2 + \'2• : = 2
44. : = J 4 - x 2• y = ../4 - x2. x = 0. _r = O. :: = 0 45. X~ -,- _I'" = I, X + ;:: = '.!. : = 0
46. :: = ../4 - x 2 - y2. y = 2::. : = 0
11.6 Surfaces in Space
821
ln Exercises 47-52, find an equation for the surface of revolution generated by revolving the curve in the indicated coordinate plane about the gh·en axis.
Equation of Cun•e
47. :2 = -ly
48. : = 3y
49. : = 2y 50. 2: = '-4 - x1
51. xy = 2 52. ~= lny
Coordinate Plane _v:-plane yz-plane _v:-plane x:-pl anc .\}'·plane yz-planc
Axis ofRe1•0/111io11 y-axi~ y-axis :-axis
.l·UXI:.
x~nxi s ::-axi s
In Exercises 53 and 54. find an equa lion of a generating curve given the equation of its surface of revolution.
53. x2 + y1 - 2z =0
54. x 2 + : 2 = cos2 .r
WRITING ABOUT CONCEPTS
55. State the definition of a cylinder.
56. What is meant by the trace of a surface? How do you lind a trace'!
57. Identify the six quadric surfaces and give the standard form of each.
58. What does the equation : = x 2 represent in the xz-plane'?
What does ll represent in three-space'!
In Exercises 59 and 60, use the s hell method to find the \"Olu me of the soUd below the s urface of revolution and above the xy-plane. 59. The curve : = -tr - x' in the .,~-plane is revol\ed about the
.:-axis.
60. The curve : = sin y (O :5 y s. 'lT) in the _1·z-plane is revolved
about the ::-axis.
In Exercises 61 and 62, analyze the trace when the surface
64. The set of all points equidistam from the point (0. 0, 4) and the .\')'- pla n e
65. Geography Because of the forces caused by its rotation, Eanh is an oblate ellipsoid rather than a sphere. The equatorial radius is 3963 miles and the polar radius is 3950 miles. Find an equation of the cllip~oid. (A~sume that the center of Earth is
at the origin and that the trace formed by the plane :: = 0
corresponds to the equator.) 66. Machine Design The iop of a rubber bushing designed to
absorb vibrations in an automobile is the surface of revolution
½.i generated by revolving the curve z = + l (O S. -" :5 2) in
the _r:::-plru1e about the .:-axis.
(a) Find an equation for the surface of revolution.
(bJ All measurements are in centimeters and the bushing is set on the xy-plane. Use the shell method to find its volume.
(c) The bushing has a hole ofdia meter l centimeter through iL~ center ttnd parallel to the axis of revolution. Find the \'Ol ume of lhe rubber bushing.
67. Determine the intersection of the hyperbolic paraboloid
= ~ y1/ b2 - .r-/a2 with the plane bx+ ay - .:: = 0. (As~ume
ll. b > 0 .) 68. Explain why the curve of intersection of the surfaces
x" + 3y1 - 2:~ + 2y = 4 and 2,-~ + 6_r" - 4::~ - 3x = 2
lies in a plane.
True or False? ln Exercises 69-72, determine whether the s tatement is true or false. If it is false. explain why or give a n example that shows it is false.
69. A sphere i~ an e llipsoid.
70. The generating curve for a surface of revolution is unique.
71. All craces of an ellip~oid arc ellipses.
72. All u-.1ces of a hyperboloid of one ,heel arc hyperboloids.
73. Think About it Three types or cla~sic '·topological" ~urfaccs are shown below. The sphere and torus have both an ..inside'· and an ·•outside." Does the Klein bottle have both an inside and an ou1side'? Explain.
is intersected b) t he indkated planes.
61. Find rhc lengths or rhe major and minor a,e~ and the coordinates of the foci of the ellipse generated when the ~urface is imcrsected by the planes given by
(al : = 2 and (b) : = 8.
62. Find the coordinates of the focus of the parabola formed when the surface is intersected by the planes givcn by
(al -" = .i and Cb) x = 2.
In Exercises 63 and 64, find an equation of the surface satisfying lhe conditions. and identify the surface.
63. The set of all points equidistant from 1he point (0. 2. OJ and the
planc y = -'>
Sphere Klein bottle
Torus Klein bottle
822
Chapter 11 Vectors and the Geometry of Space
11D Cylindrical and Spherical Coordinates
■ Use cylindrical coordinates to represent surfaces in space. ■ Use spherical coordinates to represent surfaces in space.
Rectangular
coordina1e,:
r = r co , 8
r = r ~in e
Cylindrical <:oordinal~s:
r'· = x,· + y-'
lan 0=::.\..' X
(.\, y, :) P' (r8.:I
_I'
Cylindrical Coordinates
You have already seen that some two-dimensional graphs are easier to represent in polar coordinates than in rectangular coordinates. A similar situation exists for surfaces in space. l.n rhis secrion. you will swdy two alternative space-coordinate systems. The first, the cylindrical coordinate system , is an extension of polar coordinates in the plane to three-dimensional space.
THE CYLf'IDRICAL COORDl1'ATE SYSTEM
In a cylindrical coordinate system, a point Pin space is represented by an
ordered triple (,-, e. ;:).
1. (r. 0) is a polar represemation of the rrojection of Pin 1he .1y-plane. 2. z is the directed distance from (r. 0) to P.
Figure 11.66
5;. ( r. e. : )= (.1. .1)
Figure J 1.67
To convert from rectangular 10 cylimlrical coordinates (or ,·ice \'ersa). use the following con\'ersion guidelines for polar coordinates. as illustrated in Figure 11.66. Cyli11drical to rectangular:
x = rcos 8. y = r sin 0.
Rectangular to cylindrical:
ran0=y-. X
The point (0. 0. 0) is called the pole. Moreover. because the representation of a poim in the polar coordinale system is not unique. it follows that the repre~entation in the cylindrical coordinate ~ystem is also not unique.
D EXAMPLE Converting from Cylindrical to Rectangular Coordinates
5J:.3) Convert the point (r. e. ::) = (4.
to recrangular coordinates.
Solution Using the cylindrical-LO-rectangular conversion cqualions produces
= 5
X ..J. COS 7T =
4 (- '-'}) =
-
2 _/3
6
y
=
4
sin
5 ;
= ➔ (½) = 2
;: = 3.
So, in rectangular coordinates. the point is (x, y. ::) = (- 2-/3. 2. 3). al. shown in
Figure 11.67.
■
2 3
/
Figure 11.68
11.7 Cylindrical and Spherical Coordinates
823
El EXAMPLE Converting from Rectangular to Cylindrical Coordinates
Convert the point (x . y. z:) = (I. fi. 2) to cylintlrical coordinates.
Solution Use the rectangular-to-cylindrical conversion equations.
r = ±fiTI= =2
tan O= fi
f e = arctan (.J3) + 1177 = + 1177
: = 2
You have rwo choices for r and infinitely many choices for 0. As shown in Figure 11.68. two conven ient representations of the point are
r > () ;ind II 111 Qu~J rnnl I
r < ll ,111J ti m Qu3drant II I
■
Cylindrical coordinates arc especially convenient for representing cyLindricaJ surfaces and surfaces of revolution with the :-ax is as the axis of symmetry, as shown in Figure 11.69.
x~+y 1 =9 r =3
x1 + v 2 = 4: r=2·.../:.
.r2+J~ =.:1
r=:
x1 +y~ - .:~=I ,: = :2 + I
r - -1
Cylinder
Paraboloid
Cone
Figure 11.69
Hyperboloid
Venical planes containing the :-axis and horizontal planes also have simple cylindrical coordinate equations. as shown in Figure 11.70.
Vertical plane: 8=c
Horizonllll pl3ne: := c
Figure J1.70
824
Chapter 11 Vectors and the Geometry of Space
Rectangular.
r+ -"~= ➔:~
Cylimlric-JJ:
r1= ~.::?
Figure 11.71
Rectangular: y2=x ~
Cylindrical: r : CSC fi COi fi
Figure 11.72
II EXAMPLE Rectangular-to-Cylindrical Conversion
Find an equation in cylindrical coordinates for the surface represented by each rectangular equation.
a. x2 + v~ = •k:1
b. Y2 = X
Solution
a. From the preceding section. you know that the graph x 1 + y 2 = -k:2 is an el!jptic
cone with its axis along the ::-axis. as shown in Figure 11.7 1. If you replace
x" + y: with r2, the equation in cylindrical coordinates is
x2 + y2 = 4z2
,.2 = 4;,".
R, ct.111gul.ir ,,-111a1io11
C tndricct. cquauon
b. The graph of the surface y2 = .r is a parabolic cylinder with nilings parallel to the
r ::-axis. as shown in Figure 11.72. By replacing y2 wilh sin2 8 and .r with r cos 0.
you obtain the folio\, ing equation in cylindrical coordinates.
= _\'2 X
r 2 sin2 f} = r cos 0
r(r sin2 0 - cos 0) = 0 r sin2 0 - cos 0 = 0 r= -co~- e
sin2 0
r = csc Ocot 0
ffrc1angula1 ,·qu~1ion
C,,llec t tcnn, .,nd tacl01
Snl\'C fllr, ( • 'mdncal cquanon
Note that this equation includes a point for which r = 0, so nothing was lost by
dividing each side b) the factor r.
■
Cylindrical: r1 co~ 29 + : 2 + I = 0
3
Rectangular: y2-x2-_:2 = I
Figure 11.73
Converting from cylindrical coordinates to rectangular coordinates is les, straightforv,1ard than converting from rect::1ngular coordinates LO cylindrical coordinates. as demonstrated in Example ..J..
II EXAMPLE Cylindrical-to-Rectangular Conversion
Find an equation in rectangular coordin.ites for the surface represented b) the cylindrical equation
+ ,.: CO\ 20 T .:2 l = 0.
Solution
r~ cos ~ 0 + ::~ + I = 0
r 2(cos2 0 - sin2 0) + ::1 + I = 0
r 2 cos~ 0 - r 2 sin~ 0 + ::2 = - I
Ci 1,rdricctl ·u,mor
Tru.:unomelTJl 1dcntt1y
R,·pl.,c:e r cc" fl w11h ., ,111d r ,rn O11,lh , .
This is a hyperboloid of two sheets whose axis lies along the y-axis, as shown 111
Figure 11.73.
■
11.7 Cylindrical and Spherical Coordinates
825
Prime mcriilian
so~w
40~ N
;r
Spherical Coordinates
ln lhe s pherical coordinate system, each point is represented by an ordered triple: the first coordi.nate is a distance. and the second and third coordinates are angles. This system is similar to the latitude-longitude system used to identify poiml> on the surface of Eai1h. For example. the point on the surface of Earth whose latitude is -1-0° North (of the equator) and whose longitude is 80° West (of the prime meridian) is shown in Figure l l .74. Assuming that the EarLh is spherical and has a radius of -1-000 miles. you would label this point as
/
R.,d,u,
(-1-000, - 80°, 50°).
' /
~
S!f d,,d.,11,~ from
50 dll\l 11 from
pn m~ meridian
North ['.,le
Figure 11.74
TH ESPH ERICAL COORDINATE SYSTEM
ln a spherical coordinate system, a point P in space is represented by an ordered aiple (p. 0. <f>).
I. pis the distance between P and the origin, p 2:: 0. 2. 0 is the same angle used in cylindrical coordinates for r ~ 0.
3. cl> is the angle between the positive : -ax is and the line segment OP ,
0'5</>:5 11.
Note that the first and third coordinates, p and c/). are nonnegative. p is the lowercase Greek letter rho, and ¢ is the lowercase Greek letter phi.
Spherical coordina1es Figure 11.75
The relationship between rectangular and spherical coordinate~ is illustrated in Figure I l.75. To convert from one system to the other, use the following. Spherical to rectangular:
X = p sin q> COS 0. y = p sin </> sin 0. z = p cos <I>
Rectangular to spherical:
can0= I,
X
( ) <f, = arccos J x2 +~-yi + ,:2
To change coordinates between the cylindrical and spherical systems. use the following. Spherical to cylindrical (r ?:. 0):
= r2 = p2 sin2 </>, 0 0. .;: = pcos </>
Cylindrical to spherical (r ~ 0):
0 = 0,
</>
=
arccos(J
2
r-,., T
z-?)
826
Chapter 11 Vectors and the Geometry of Space
The spherical coordinate system :is useful primarily for surfaces in space that ha\·e a point or center of symmelry. For example, Figure 11.76 shows Lhree surfaces with simple spherical equations.
Rcc1angular:
I 2 + y2 + ~l - 4.:= Q
Figure lJ.77
Sphere: p=c
Figure 11.76
.t
Vertical half-plane: 8=t·
(o OH. =alfc-cone: <c<i1-r)
El ( ; EXAMPLE Rectangular-to-Spherical Conversion
Find an equation in spherical coordinate~ for the surface represented by each recranguJar equation,
a. Cone: x2 + _r2 = :2 b. Sphere: x 1 - y 1 + ::2 - -k = 0
Solution
o. Making 1he approprimc replacements for x. y. and:: in the given equation yields the following.
Spherical· p=4co, ,P
e = p1 sin2 </J cos2 FJ + p 2 sin2 </> sin2
p2 cos2 <b
p2 s:in1 <f; (cos2 0 + sin1 e) = p2 cos1 <b
p2 sin2 <b = p2 cos2 <b
sin2 cos2
<b
cp
=
tan2 <b =
p 2'. (1
The eq uation <I> == n-/4 represents the upper half-cone. and the equation¢ = 377/4
represents the lower half-cone.
b. Because p 2 = x2 + y2 + ::2 and ;: = p cos cp. the given equation has ilie following
spherical form.
p2 - 4pcos <b = 0
p(p - 4cos <f;) = 0
Temporarily discarding the possibility that p = 0. you have the spherical equation
= p - 4 cos c/J 0 or
Nore that the ~olution seL for this equaLion include!> a poinr for which p = 0. so
nothing is lost by discarding the factor p. The sphere represented by the equation
p = -+ cos <bi~ shown in Figure I 1.77.
■
11.7 Cylindrical and Spherical Coordinates
827
11D Exercises
Sec 1·r.vN CalcCnat.com !or worked-out solulions to odd-numbered exercises
In Exercises 1-6, con~erl the point from cylindrical coordina tes lo rectangular coorilinatcs.
I. (-7.0.5) 3. (3. -rr/ -+. I) 5. (-l. h / 6. 3)
2. (:>.. - rr. - -l) 4. (6. - 11/ -l. 2) 6. (- 0.:'i. J, rr/ 3. 8)
In E.xercises 7-U, convert the point from rectangular coordinates to cylindrical coordinates.
7. (0. 5. I) 9. (2. - 2. -J,) 11. (1. -./3.J,)
8. (2-/2. -2./2, .1,) 10. (3. - 3. 7) 12. (2 .fi_ - 2. 6)
ln Exercises 49-56. find an equation in rectangular coordinates ror the equation gh en in s pherical coordinates. and sketch its graph.
49. p == 5
51. 1, = !!.
6
53. p = ➔ cos cb 55. p = csc </1
50. 0 = 37i
-+ 52. d, = -r:,r
54. p = 1 sec <h
56. (1 = -+ csc cD sec e
Ln Exercises 57-6-l. convert the point from cylindrical coorilinates to sphericaJ coordinates.
In Exercises 13-20, find an equation in cylindrical coordinates for the equation given in rectangular coordinates.
13. :=J, 15. x 2 -'- ye + ::2 = I7 17. y = .re 19. y2 = 10 - ::2
]4. .\" = 9 16. : = x2 -ry2 - l l 18. x 2 + y 2 = Sx
20. .r2 + y1 + ::2 - 3: = 0
57. (-l, rr/ J,, 0) 59. (-+... '2. J, ) 61. (4. - ../ 6. 6) 63. (] 2. 7T. 5)
58. (3. - 1r/4. 0) 60. (2. 2 rr/3. - 2) 62. (-4. r,/3. J,) 64. (4. rr/ 2. 3)
In Exercises 65-72, comert the 1>oint from spherical coordinates to cylindrical coordinates.
ln Exercises 21-28. find an equation in rectangular coordinates
65. ( I0. rr/6. 1r/ 2)
66. (4. 11/18. rr/ 2)
for the equa tion given in cylindric:1J coordina1cs. and s ketch its graph.
67. (36. Tr. 11/ 2 ) 69. (6. - rr/ 6. 11/ 3)
68. ( 18. TT/ 3. rr/ 3) 70. (5. - 57i/ 6, TT)
21. r = 3
22. : = 2
71. (8. 7../ 6, rr/ 6)
72. (7. 11/ 4. 3;;/ J,)
23. ti= 1r/6 25. ,.1 -;- ,:2 = 5
24. r = 2I : 26. i: = r2 cos" 0
(m In Exercises 73-88, tLSe a computer algebra sys tem or g raphing
utility to convert t he point from one system to another among
27. r = 2 ~in 0
28. r=2cosll
the rectangular. cylindrical. and spherical coordinate systems .
1n Exercises 29-3➔ , con\'erl the point from rectangular coordinates to spherical coordinates.
29. 14. 0, 0)
o. 30. (- ➔. 0)
31. (-1. 2../3. 4)
33. (JI I. 2J3)
32. (2, 2. J,,/2) 34. (- 1.2. 1)
In Exercises 35-40, convert the point from spherical coordinates to rectangular coordinates.
35. (J,. '71/' 6. r,/J,) 37. (12. -1r/ 4,0) 39. (5, 1114. 31r/ 4)
36. ( 12. 37r/4. rr/ 9) 38. (9. 11/ 4. -rr) 40. (6. II, 11/2)
In Exercises 41- 48, find an equation in spherical coordinates for the equation gi,en in rectangular coordinates.
.n. ,. = 2
42. :: = 6
43. .r2 + 1'2 + : 2 = 49
44. X 2 + y 2 - 3:2 = 0
46. X = 13
48. x2 + .,.2 l- : 2 - 9: = O
Recwngular
73. (4. 6. 3) 74. (6. -2. - 3) 75 . 76. 77. 78. 79. (3. - 2, 2)
80. (3 ../ 1. 3./2. -3)
81. (5/ 2, 4/3. - 3/ 2) 82. (0. - 5,-+) 83. 84 . 85. 86. 87. 88.
Cylindrirnl
(5. ;;/9. 8) (J O, -0.75. 6J
(5. 3../4. - 5) (- 2. 1l rr/6. 3) (- 3.5. 2.5. 6) (8.25. 1.3. - 4)
Spherical
(20. 27i/3.../ ~) (7.5. 0.25. I)
(3. 37i 4. 7T/3) (8. - r./ 6. 77)
828
Chapter 11 Vectors and the Geometry of Space
In Exercises 89-9-t match lhe cqu.ltion (written in terms of cylindrical or spherical coor dinates) with its graph. [The grap hs
are labeled (a), (bJ, (c), (d ), (e), a nd (f}.l
(a}
(bl
I 1I
(d)
.\
(f}
89. r = 5
91. p = 5 93. r 1 =;;:
90. 0 = !!.
4
92. qJ = !!.
-I
94. p = 4 sec cb
WRITING ABOUT CONCEPTS 95. Gi,e the equations for the coordinate conversion from
rectangular to cylindrical coordinates and vice versa.
96. Explain why in spherical coordinates the graph of O= c is a half-plane and not an e111ire plane.
97. Gi,e the e4uations for the coordinate conversion from rectangular to spherical coordinates and vice versa.
98. (a) For constants a. b. and c. describe the graphs of the
equation, r = rt, 0 = b, and ;;: = c in cylindrical
coordinates.
(b) For constants a, b. and c. describe the graphs of the
equations p = a, I} = b, a nd cb = c in spherical
coordinates.
In Exercises 99-106, convert the rectangular equation to an
equation in (a) cylindrical coordinates and (b) spherical
coordinates.
... 99. \"2 y2 + 1 = 25
... 100. -l(x2 y2) = 1
101. - \':! + y2 + ' - 2: = 0
= - 102. x.2 ..L y:! -
103. r'· + y2 ::c -ly
104. x·' + y'· = 36
= 105. - .\:! ."2 9
106. y = -I
In Exercises 107- 110, s ketch lhe solid that bas the ghen description in cylindrical coordinates.
107. 0 s O s 7r/2, 0 $ r S 2, 0 s : s 4 108. - ,,/ '!. s /J $ ,,/'2. 0 $ r 5 3, 0 :S : S r cos 0 109. 0 s II s 2,,. 0 s rs a. r s : s a
110. 0 s OS 2,,. 2 s rs -1. : 2 s -r!..,.. 6r - 8
fn Exercises lll-ll4, sketch the solid that has the given d escription in spherical coordinates.
lJ I. 0 s (J s 2,,. 0 s cp s r,/6. 0 s p s a ~ec <i> 112. 0 s O s 2,,. ,,/4 s cb s r,/2. 0 s p s I
113. o s n s ,,f2_ o s cb s ,,/2. o s p s '2
114. 0 s O s "· 0 s cb s 1r/2. I s p s 3
Think About It In Exercises 115-120, find inequalities tha1 describe the solid, and state the coordinate system used. Position the solid on the coordinate system s uch that the inequalities arc as simple as possible.
115. A cube wirh each edge 10 cemimeter, long
116. A cylindrical ,hell 8 meters long with an inside diameter of 0.75 meter and an outside diameter of 1.25 merers
117. A ,pherical shell with inside nnd our~ide radii of4 inches and 6 inches, respectively
118. The solid that remains after a hole I inch m diameter is drilled through the center of a $phere 6 inches in diameLer
119. The solid inside boLh x2 + y2 -;- z2 = 9 and
120.
The
2x .L
solid y2 +
between
;:2 = 9. and
the spheres x inside the cone
2 + v2 + ;:2
;:2 = .r2 -r- y
1 =
4
and
Tnie or False? fn Exercises 121-124. determine \\ hether the statement is true or false. If it is false, explain \\ by or give an example that shows it L~ false.
Ul. In cylindrical coordinate,. the equation r = ;: is a cylinder.
122. The equations p = 2 and .r2 + y 2 + ;:2 = 4 represent the:
same ~urface. 123. The cylindrical coordinates of a point (x, r, ;:) are unique. 12-t The spherical coordinates of a poim (.,. y. :) are unique.
125. Identify the curve of inwrsection of the surface, (in cylindrical coordinates);;: = sin Band r = I
126. Idemif) the cur,e of intersection of the surface~ (in spherical
cuordinme~) p = 2 sec <I> and p = -L
Review Exercises
829
I I REVIEW EXERCISES
Stt ,w,v, CalcChat.corn for wml•d -out solutions to o~d-numbered exercises
In Exercises 1 and 2, let u = PQ and v = PR, and (a) write u
and v in component form, (b) write u as the linear combination of the standard unit vectors i a nd j , (c) find the magnitude of v,
and (d) find 2u + v.
I. P = ( I. 2). Q = (-k I), R = (5. 4)
2. P = (- 2, - I). Q = (5, - I). R = (2. 4)
In Exercises 3 and 4, find the component form of v giYen its magnitude and the angle it makes with the pos itiYe x-axis.
t 4. llvll = o = 225°
5. Find the coordinates of the pt)int in the .\)"•plane four units to the right of the x::-plane and five units behind the y;:-plane.
6. Find the coordinates of the point located o n the y-axis and seven units 10 the lefl of the xz-plane.
ln Exer cises 7 and 8, determine !he location of a point (x, y, z) that satisfies the condition.
7. Y:. > 0
8. xy < 0
In Exercises 9 a nd 10. find the s ta ndard equation of the s phere.
9. Ce111er: (3. -2. 6): Diameter: 15 LO. Endpoints of a diarnc1cr: (0. O. 4). (4. 6. OJ
In Exercises 11 and ll, complete the square to write the equation of the sphere in standard form. Find the center a nd radius.
I J. x 2 + y 2 + ;::1 - 4x - 6y + 4 = 0 12. x~ ... y 2 -r- ;:1 - IOx + 6y - 4:: -,- 34 = 0
In Exercises 13 and 14. t he initial and ter mi nal points of a vector are giYen. (a) Sketch the directed Line segment, (b) find I.he component form of the vector, (c) write the vector us ing standard unit vector notation, a nd (d) sketch the vector with it.~ initial point a t the orig in.
13. Initial point: (2. - I. 3)
14. lni1ial poin1: (6, 2, 0)
Terminal point: (4, 4, - 7)
Terminal point: (3, - 3. 8)
ln Exercises 15 and 16. use vectors to deter m ine whether t he points are collinear.
15. (3. 4. - I). (-1, 6, 9). (5, 3, -6) 16. (5. -+, 7). (8. -5. 5). (l I, 6. 3)
17. Find a uni1 vector in the dim:tion of u = (2. 3. 5).
18. Find the vector v o r magnitude 8 in the direction (6, -3. 2).
In Exercises 19 and 20, let u = PQ and v = PR, and find (a) the
component forms of u a nd v, (b) u • v, and (c) v • v.
19. P = {5. 0, 0). Q = (4. -+. O), R = (2, 0, 6) 20. P = (2. - I, 3), Q = (0. 5. I). R = (5, 5, O)
In Exercises 21 a nd 22, determi ne whether u and v a re orthogonal, parallel, or neither.
21. u = (7. -2. 3)
v= (- 1.4,5)
22. u = (-4. 3. -6}
V = ( 16, -12, 24)
1n Exercises 23- 26, find the a ngle Obetween the vectors.
23. u = 5(cos(31r/4)i + sin(37T/4)j ] v = 2(cos(2r./3)i + sin(2rr/3)j]
24. u = 6i + 2j - 3k. v = - i + 5,i
25. u = (10, - 5, 15). ,. = (- 2. I, -3)
26. u = ( I. 0, - 3). v = (2, -2, I)
27. Find two vectors in opposite direclion~ 1ha1 are orthogonal to Lhe vec!Or u = (5. 6. - 3).
28. \York An object is pulled 8 fee1 across a noor using a force of •75 pounds. The di rection of the force is 30° above the horizontal. Find the work done.
In Exer cises 29- 38, let u = (3, - 2, 1). v = (2, - 4, - 3), and w = (-1. 2, 2).
29. Show 1ha1 u • u = l! u W.
30. Find 1he angle between u and v. 3 1. De1em1ine rhe projection of w 01110 u. 32. Find the work done in moving an object along 1he vec1or u if
1he applied force is w. 33. De1ermine a uni t vecmr perpendicular to the plane containing
,. and w. 3-'. Sho\\ that u " ,, = -(v x u). 35. Find the volume of the solid whose edges are u, v. and w.
36. Show 1ha1 u x (v + w) = (u :< v) + {u x w).
37 . Find the area or 1he para llelogram with adjacent sides u and v. 38. Find the area of the triangle with adjacent sides v and w.
i 39. Torque The ,pecifications for a trac1or state lhat the torque on a boll with head size inch cannot exceed 200 foot-pounds. Determine the maximum force IIFI thal can be applied to the
wrench in the figure.
830
Chapter 11 Vectors and the Geometry of Space
40. l'olumt• Use the triple scalar product to find the volume of the
parallelepiped having adjacent edges u = 2i + j, v = 2j + k, and w = - j + 2k.
In Exercises -H and 42. find sets of (a) parametric equation.~ and (b) symmetric equations of the line through the two points. (For each line. write the direction numbers as integers.)
.n. (3, 0, 2), (9, 11, 6)
42. (-1.4.3), (8, 10,5)
In Exercises 43-46. (a) find a set of parametric equations for the line. (b) find a set of symmetric equations for the line. and (c) sketch a graph of the line.
43. The line passes through the point (I, 2, 3) and is perpendicular to the x;:-plane.
44. The line passe~ through the point ( I. 2, 3) and is parallel to the
line given by x = y = z.
45. The intersection of the planes 3x - 3y - 7z = -4 and
= X - )' + 2: 3.
46. The line passes through the point (0. l. 4) and is perpendicular
to u = (2, - 5. I) and Y = (- 3, I, 4).
In Exerci<,es 47-50. find an e<1uatio11 of the plane and sketch its graph.
47. The plane passes through (- 3. -4, 2), (- 3, 4, I), and ( I. 1. - 2).
48. The plane passes through the point (- 2, 3. I) and is perpendicular to n = 3i - j + k.
49. The plane contains the lines given by
X- l -=-,=y=;:+
and x+I
-2 =y-1=--'
50. The plane passes through the points (5, I, 3) and (2, - 2, I) and
= is perpendicular to the plane 2t° + y - ;: 4.
51. Find the distance between the point (I, O. 2) and the plane
2x - 3y + 6:. = 6.
52. Find the distance between the point (3. - 2, 4) aml the plane 2x - Sy + .:: = I0.
53. Find the distam:e between the planes 5x - 3y + ..: = 2 and
5x - 3y +.:: = -3.
54. Find the distance between the point (-5, I, 3) and the line
given by x = 1 + 1, y = 3 - 2t. and;: = 5 - t.
In Exercises 55-64. describe and sketch the surface.
55. X + 2y + 3;: = 6
56. y = .-: 2
57. y = ½z
58. y =cos:.
~9 x·' + y·' + "2 =
- • 16 9 ~
60. l6.r2 + 16y2 - 9.:: 2 = 0 9 r: y:
61. .16 - + .-:2 = - I
x2 v2
~2
6' -+~--=
-· 25 4 100
63. x 2 + .-:2 = 4
64. y 2 + :.: = 16
65. Find an equation of a generating curve of the surface of
= revolution y 1 + ;: 2 - 4.r 0.
66. Find an equation of a generating curve of the surface of
revolution .t-2 + 1y2 + :.1 = 3y.
67. Find an equation for the surface of revolution generated by revolving the curve z2 = 2y in the y::-plane about the y-axis.
68. Find an equation for the surface of revolution generated by
revolving the curve 2x + 3;: = I in the x:.-plane about the
x-axis.
In Exercises 69 and 70, comert the point from rectangular coordinates to (a) cylindrical coordinates and (b) spherical coordinates.
69. (-2,.12, 2v'2. 2)
In Exercises 71 and 72. com·ert the point from cylindrical coordinates to spherical coordinates.
- f 71. ( JOO, 50)
In Exercises 73 and 74, comert the point from spherical coordinates to cJlindrical coordinates.
rr) 73•
(2-s,
-~_4·,'
3 4
In Exercises 75 and 76, com·ert the rectangular equation to an equation in (a) cylindrical coordinates and (b) spherical coordinates.
75. x 1 - r1 = 2;:
76. x 2 + y 2 + ;: 1 = 16
In Exercises 77 and 78. find an equation in rectangular coordinates fur the equation given in cylindrical coordinates. and sketch its graph.
77. r = 5 cos fJ
78.;: = 4
In Exercises 79 and 80. find an equation in rectangular coordinates for the equation giwn in spherical coordinates, and sketch its graph.
f 79. 0 =
80. p = 3 cos cp
P.S. Problem Solving
831
•
PROBLEM SOLVING
l. Using \·ector~. pro\"e the La\\ of Sines: If a. b . and care the three sides of I.he triangle shown in the figure, then
~in r\ sin B sin C
Tai = hi! = Tcf·
a
C b
Ju~ 2. Consider the functionJ(x) = ('
dr.
f±! (a) Use a graphing utility to graph the !"unct ion on the inter\'al
-2 ~ X ~ 2.
(b l Find a unit vector parallel 10 I.he graph of/at the poim fO. 0).
(cJ Find a unit vecwr perpendicular to the graph ofjar the point (0. 0).
td) Find I.he parametric equations of the tangem line 10 the graph off at the point (0, 0).
3. Using vectors, prove that the line scgmellls joining the midpui 111s of the side~ of a parallelogram fonn a parallelogram (see figure).
-'· Using ,ectors. prove that the diagonals of a rhombus are perpendicular (see ligure ).
5. (al Find the short..:st distance between the point Q('.!. 0. 0) and the line dctennined by the points P 1(0, 0. 11and P~(0. I. 2).
(b) Find the shortest disrnnc:e between the point Q('.!. 0. 0) and the line segment joi ning the points P 1(0. 0. l ) and P~(0. I. 2 ).
6. Let P0 be a point in ll1c plane with normal vector n. Dcsc:rib~ the
set of~1ms Pin the plane for which l n + Pft) is onhogonal to
(n - PP,-.).
7. (a) Find the ,·olume of the solid bounded bclm1 by the parab-
oloid ;: = x ~ -"- y ~ and abo1·e by the plane :: = I .
(b) Find the rnlume of I.he solid bounded below by the elliptic-
paraboloid
;:
=
:aX..·:,
-L
)'2
~1, ,-
and
above
by
the
plane
;:
=
k.
,,here k > 0.
(c) Show that the volume of the solid in part (b) b equal 10 one-half I.he product of the area of the base time~ the ahirude, as sh01vn in the figure.
8. (a) Use the di~k method lO find the 1·olume of the \phere
+ = x :? y ! + ::2 ,.2.
(b)
Find
the
volume
of
the
r0ellipsoid:a..-,
+
·b,•,--'
+
z-' c---;-;-
=
I.
9. Sketch ll1c graph of each equation given in spherical coordinate~. (a) {) = 2 sin ,t,
(b) p = 2 cos</;
I0. Sketch the graph of each equation given in cylindrical coordinates.
(a l r = 2 cos O !bl ;: = ,~co~ 20 11. Prove the following property or ll1e cross product.
( u x v) x ( w
= x z )
( u x Y • z)w -
(u x v • w )z
12. Consider the line given by the parametric equations
.\ = - r -+ 3. y = ½r + I. ~ = :.r - I
and the point (4. 3. s) for any real num bers.
(a) Write the distance between the point and the line as a function of s.
(bl Use a graphing uti lity 10 graph the f'unc1ion in pan {a). Use the graph to find the value of s ~uch that the distance berweea the poim and the line is minimum.
(c) Use the :00111 feawre of a graphing u tility to zoom our several times on the graph in pan (b). Does tl appear that the graph ha~ slant asymptote,'? Explain. lf it appears to have slant as) mptoks. li nd them.
832
Chapter 11 Vectors and the Geometry of Space
- 13. A 1e1herball weighing I pound is pulleJ outward from the pole by a hori,mntal force u until the rope makes an angle of fJ degrees with the pole bee figure).
(a) Determine the resul ting tension in the rope and the magni-
tude of u II hen 0 = 30°.
(b) Write the tension Tin the rope and the magni111de of u as functions of 0. Determine the domains of the functions.
ccJ Use a graphing utility 10 complete the 1ahle.
16. L os Angeles is located at 34.05° North latiwde and 118.'.!4" West longi1t1de. and Rio de Janei ro, Bra,jJ is located at 22.90° South latitude and -B.23° Wes, longitude (sec figure). Assume th,11 Eartll is spherical and has a radius of 4000 miles.
(d) Use a graphing 111ility IO graph the 1wo functions for 0 5 (I :S 60°.
(e) Compare T and IIu II as fl increases.
(fJ Find (i f' possible) Jim,_ T and . lim, 11 u 11- Are the
tJ---rtr·-
~;:-/~
re;ults what you expected'/ Explain.
u
Figu re for 13
F igure for 14
14. A loaded barge is being towed by two mgboats. and Lhc magnitude of the resultant i, 6000 pound, directed along the a,is of the barge (see figure). Each towline makes an angle of 8 degree, with the axis of the barge.
(a) Find the tension in the towlines if 8 = 20°.
{b) Write the tension T of each line as a function of 1:1. Determine the domain of the f unction.
(e) Use a grnph111g u1iliry to complete the table.
(d) Use a graphing uLiliry 10 graph 1he 1ension function.
(eJ Explain why the tension increases ru. (I increru,es.
15. Consider 1he vectors u = (cos a.. ~in a. 0) and v =
(rns {3, sin (3, 0), where a > (3. Find the cro,s product of the vectors and use the result 10 prove the idenrity
sin( a: - f3) = sin a co~ f3 - co~ a sin {3.
(al Find the ~pherical coordinates for the location of each city.
(b) Find the rectangular coordinares for the location of each city.
(c) Find the angle (in radians) between the vectors from the center of EMth 10 the rwo cities.
(d) Find the great-circle distance s berween the cities. (HiJ11: s = r0)
(e) Repeat parts (a)-(d) for the cities of Boston. located at 42.36° North latitude and 71.06° West longitude, and Honolulu, located at 2 1.31 ° North latitude and l 57.86° West longitude.
17. Consider the plane that pa,~es through the poims P.R. and S. Show that the distance from a poim Q 10 this plane is
= .
Distance
i
u
• ( ll u
v
x
xviwi )I
where u = PR, v = P.~. and w = PQ.
18. ShO\I thac the distance between the parallel planes
= cu + by + er. -r d 1 = 0 and ax + by + c: - dz 0 is
= .
Distance
✓
I,d1 -
d,I , -
,
rr + b- + c-
19. Show chat tht' curve of interxcction of the plane;: = '.!y and the
+ cylinder .i2 y 1 = I b an ellipse.
20. Read the article "Tooth Tables: Solution of a Dental Problem by Vector Algebra" by Gary Hosler Meisters in Mmhemarics /lfaga;:i11e. (To view this article, go to the website 1nn,:ma1Jiarric/es.cn111.) Then write a paragraph explaining how vectors and vector algebra can be used in the construction of dental inlayh.
12 Vector-Valued Functions
This chapter inrroduces the concept of vector-valued functions. Vector-valued functions can be used Lo study curves in Lhe plane and in pace. These funcLions can also be used to study the motion of an object along a curve.
In this chapLer, you should learn the following.
■ Hm, to analyze and sketch a space curve represented by a vector-valued function. Hou. co apply the concepts of limits and continuity to vecLor-valued functions. (12.1)
■ Ho,, to differentiate and inLegrate vector-valued functions. (12.2)
■ How to describe Lhe velocity and acceleration associated with a vectorvalued function and how to use a vector-valued function Lo analyze projectile motion. (12.3)
■ How ro find tangent veccors and nonnal vectors. (12.4)
■ l-fow to find the arc length and curvature ofa curve. (12.5)
.
Jerry Driebdl/Geny Image.<
A ferris wheel is constructed using the basic principles of a bicycle wheel. You can use a vector-valued function to analyze the motion of a Ferris wheel, including its position and velocity. (See P.S. Problem Solvrng, Exercise 14.)
, •(0)
I
a CO)
a (Ol
a(Ol
a (O}
A vector-valued funct ion maps real numbers to vectors. You can use a vector-valued function to represent the motion of a particle along a curve. In Section 12.3 , you will use the first and second derivatives of a position vector to find a particle's velocity and acceleration.
833
834
Chapter 12 Vector-Valued Functions
Im Vector-Valued Functions
■ Analyze and sketch a space curve given by a vector-valued function. ■ Extend the concepts of limits and continuity to vector-valued functions.
Space Curves and Vector-Valued Functions
ln Section I0.1, a plane curve was defined as the set of ordered pairs U(r), g(r))
together with their defining parametric equations
X = f(t) and y = g(t)
where f and g are continuous functions of r on an interval /. This definition can be extended naturally ro three-dimensional space as follows. A space curve C is the set of all ordered triples (f(r). g(r), h(r)) together with their defini ng parametric equations
X = .f(T), Y = g(r), and :: = h(r)
where_{, g, and hare continuous functions of r on an interval /. Before looking at examples of space curves. a new type of funccion. called a
vector-va lued function, is inu·oduced. This type of function maps real numbers 10 vectors.
Cune in a plane
Cune in ,pace
r (l~)
rtr 1J
/
:
r ! '11>: ~ ' '\ _I / /
; C
.
'
I
I I
I
I
, /
-;I,:.
.t
~,
•
Curre C is m1ced oui b) the terminal point of position vector r(t). Figure 12.1
DEFINITION OF VECTOR-VALUED FUNCTION
A function of the form
r (t) = f(t) i + g(r)j
Plane
or
r(t) = J(t) i + g(1)j + Jz(t)k
Spa~c
is a vector-valued function, wbere the component functions/. g. and hare real-valued functions of the parameter t. Vector-valued function~ are sometimes denoted as r(,) = (/(r), g(i)) or r (r) = (J(r), g(r). h(r) ).
Technicall). a curve in the plane or in space consists of a collection of points and the defining parametric equations. Two different curves can have the same graph. For instance, each of the curves given by
r (r) = sin r i + cos r j and r(t) = sin r1 i + cos t2 j
has the unit circle as its graph, but these equations do not represent the same curvebecause Lhe circle is traced out in different ways on the graphs.
Be sure you see the distinction between the vector-valued funccion r and the real-Yalued functions f, R- and /1. All arc func tions of the real variable r. but r (r) is u vector. whereas/(r). g(t). and h(t) are real numbers (for each specific value oft).
Vector-valued fum;tions serve dua.l roles in the representation of curves. By leuing the parameter I represent time, you can use u vector-valued function to represent 11101io11 along a curve. Or. in the more general case. you can use a ,eccorvalued function to trace rhe graph of a curve. In either case. the terminal point of the position vector r (r) coincides with the point (x, y) or (x, y, ;::) on the curve given by the parametric equations. as shown in Figure 12.1. The arrowhead on rhe curve indicates the curve's orientation by pointing in the direction of increasing values of/.
12.1 Vector-Valued Functions
835
Unless stated mherwise, the d omain of a vec1or-rnlued func tion r is considered
to be the intersection of the domains of the component functions f g. and h. For
instance. the domain of r (t) = Int i + Jl=,j + 1k is the interval (0. I ].
D EXAMPLE Sketching a Plane Curve
Sketch the plane curve represented by the vector-valued function
r (t) = 2 cos t i - 3 sin 1j, 0 s I s 2n.
Vecmr-valu~d function
Solution From the position vector r (t), you can write 1be parametric equations
.r = 2 cos I and y = - 3 sin 1. Solving for cos r and sin r and using the identity
T
cos 2 r + sin" r = I produces rhe rectangu lar equation
r (r) = 2 cos 1i - 3 sin rj
The ellipse is traced clockwise as r increases
Rectungular equation
from Oto ln. Figure 12.2
The graph of this rectangular equation is the ellipse shown in Figure 12.2. The c urve has a doc/.:wise orientation. That is, as r increases from O to 21r, the position \'ecror
Cylinder:
r (r) moves clockwise, and i1s terminal point rraces the e llipse.
xi+ y2 = 16
El ( ; EXAMPLE Sketching a Space Curve
r (1) = 4 cos 1i +-l sin rj + rk
As I increases from O10 -hr. two spirals on 1he helix are rraced out. Figure 12.3
Sketch rhe space curve represented by the vector-valued function
r (/) = -l cos ri + +sin rj --- rk , 0 :s E :s -l1r.
Vecwr-valued funcuon
Solution From the first two paramc:u·ic equations x = 4 cos f and y = -+ sin £. you
can obtain
Rectangular equation
This means that the curve lies on a right circular cylinder of radius 4. cemered about
the : -axis. To locate the curve on this cylinder. you can use the third parametric equa-
tion : = r. In Figure l'.U, note that as r increases from O Lo 41r. the point (x, y. z)
spirals up rhe cylinder to produce a helix. A real-life example of a helix is shown in
the drawing at the lower left.
■
In 1953 Francis Crick and James D. Wa1sun discO\cred the double helix structure of DNA.
In Examples I and 2. you were given a vector-valued function and were asked to sketch rhe corresponding curYe. T he next two examples address the reverse problem-fi nding a vector-Yal ued function 10 represenr a given graph. Of course. if the graph is described parametrically. representation by a vector-valued function is straightforward. For instance, to represent the line in space given by
.\ = 2 + I. = .,· 31. and ;: = 4 - /
you can simpl y use 1he vector-valued function given b)
r (t) = (2 + r) i + 3tj + (4 - 1) k .
If a set of parametric equations for the graph is 1101 given. the problem of representing the graph by a ,,ector-valued function boils down to finding a set of parametric equations.
The icon ( ; inrfic11res thll/ you ll'ill find a CAS lnl'esrigmion m1 £he book \ ll ebsire. The CAS Inves1igario11 is a co/labomm·e e..rplorarion ofthir example using 1he compwer algebra systems Mapk a11d Mathematica
836
Chapter 12 Vector-Valued Functions
t=-'.!
I= 1
/= 0
y=x2 + I
--'-------1----i--- + - ~ X
-! - I
2
There are many ways to parametrize this
graph. One way is to let x = r.
Figure 12A
Ill EXAMPLE Representing a Graph by a Vector-Valued Function
Represent 1l1e parabola given b)' y = x2 + I by a vector-valued function.
Solution Although there are many ways Lo choose the parameter t, a natural choice
is 10 let x = t. Then _v = t 2 + I and you have
r(t) =ti+ (t 2 + l)j .
Vc.:10H,1lurd f unc11on
Note in Figure J2.4 lhe orientation produced by this particular choice of parameter. Had you chosen x = - t as the parameter, lhe curve would have been oriented in the opposite d irection.
II EXAMPLE Representing a Graph by a Vector-Valued Function
Sketch the ~pace curve C represented by rhe intersection of the semicllipsoid
C D Curves in space can be specified
in various ways. For instance, the curve
in Example -I is described as the intersection of two surfaces in space.
= and the parabolic cylinder y x 2. Then, find a vector-valued function to represent the
graph.
Solution The intersection of the two surfaces is shown in Figure 12.5. As 1n Example 3. a natural choice of parameter is x = r. For this choice. you can use the
= given equation y = x2 to obtain y t2. Then, it follows that
.-:2::__=I_:_r_2
_
_v,2__=
1
-12-
-t
4 -
=24--
-2-t 2
- -
-l~
(6 + 12)(4 - 12)
4
12 24
12 24
24
24
Because the curve lies above the .\)'-plane, you should choose the positive square root
for ;: and obtain the following parametric equations.
✓(6 = x = I. Y = ,2. and z
+ ,2~4 - r2)
The resulling vecLOr-valued function is
✓( r (I) = t i + r2j +
6
+
12 4
~
-
12
) k.
- 2 $ I $ 2.
\ ·e,tor- ,Jlu.c<l fonrllon
(Note 1hat the k -component of r (1) implies -2 s; r :5 2.) From the points ( - 2. 4, 0)
and (2, -+. 0) shown in Figure 12.5, you can see that the curve is traced as t increases
from -2 lo 2.
Parabol.ic cylinder Ellipsoid
(0, o. 2)
C: .t=I
),• = 12
_ j,--<6_+_1--2-=-)(-➔ ---11::--)
~-
6
Curve in space
y
The curve Cis the intersection of the scrniellipsoid and the parabolic cylinder.
Figure 12.5
•
12.1 Vector-Valued Functions
837
Limits and Continuity
Many techniques and definitions used in the calcul us of real- valued functions can be appl ied tO vector-valued functions. For i nstance, you can add and subtract vectorrnlued functions. multiply a vector-valued function by a scalar. take the limit of a vec tor-val ued function, differentiate n vector-valued function, and so on. The basic approach is to capitalize on the linearity of vector operations by extending the defi nitions on a component-by-component basis. For example, 10 add or subtract two \'ector-val ued functions (in the plane), you can write
1)1) +
r = [/ + + [J + 2(/)
1(r) i
gi(Ilj ]
2( r)i
g
2(r)j]
= [Jl(t) + f/r)] i + [g1(t) + g2(t)].j
l)l) - r 2(t) = Lf1(l) i + g,(t)j ] - [J~(t) i + g1(t)j ]
= U1(1l - f2(rl Ji + [g1(rl - gz(r)Jj .
.Sum Differ~nce
Similarly. to multiply and di vide a vccwr-valued function by a scalar. you can write
cr (r) = + 1 c[J (t) i
g
1( r )j ]
= cf1(t)i + cg 1(1)j
r (d
[I + 1(1)i
g 1( r ). i J
C
C
C"F-0
Scalar multipl ica1 iun Scalar divi,10n
This componem-by-component extension of operations with real -\'a)ued functions lO vector-valued functions i$ further i l lustrated in the followi ng definition of the limit of
a vector-val uec..l function.
DEFINITION OF THE LIMIT OF A VECTOR-VAL UED FUNCTION
Ii
= 1. If r is a vector-valued function such that r (t) + J(r )i then g(t ) j ,
Ii
Ii
Lim r (t) = [lim f(r)Ji + [ 1im g(t) jj
f - n.J
f -.tl
t -tl
Plw1e
provided/and g have limits as 1 ➔ a.
= + + 2 . If r is a vector-mlued function such that r (1) J(r) i
g(r)j
h(t)k
.
then
Ii
r lri➔m" r (t)
==
[1im .f(r)]i
t - > tl
+
[ Jim g(r)-,j ,-,ti
+
Lim h(r)]k
J 1--,a
Space
provided); g, and/, have limits as r ➔ a.
0
As r approaches u. r(1) approaches the limit L. For the limi1 L to exist. it is not neccssar~ that r (a) be defined or that r(a) beequal to L. Figure 12.6
lf r (t) approaches the vector L as r ➔ a, the length of rhe vector r(r) - L approaches 0. T hat is.
ll r (r) - Lil ➔ 0
as
I ➔ a.
T hi~ is i llustrated graphicalJy in Figure 12.6. Wi th this definition of the limir of a vector-valued fu nction, you can develop vector versions of most of the limit theorems given i n Chapter l . For example, the limit of the sum of two vector-valued functions is the sum of thei r individual limits. A lso. you can use rhe orientation of the curve r(t) to defrne one-sided limits of vector-val ued functi ons. T he next definition ex1cnds Ihe notion of cominui1y to vcctor-\'alucd functions.
838
Chapter 12 Vector-Valued Functions
II>
(l=-4
,~
I~ 10
~
6
DEFI NITIONOF CONTINUITY OF AVECTOR-VALUED FUNCTION
A vector-valued fu nction r is continuous at the point given by , = a if the limit of r (,l exists a~ 1 ~ a and
lim r (t) = r (a).
I ·~a
A vector-,·alucd function r is continuous on an interva l / if it is continuous II
at every point in the interval.
11
1:
From this definition. it t'ollows 1hat a \'ector-valued func1ion is continuous at
t = a if und only if each of its component functions is continuous at r = a.
El EXAMPLE Continuity of Vector-Valued Functions
Discuss the cominuity of the vecror-valued function given by
= r (r) ti ..1. a j + (aJ - ,J)k
a1 t = 0.
Solution As 1 approaches 0, 1hc lim it is
,l-i--m-,o r (t) = [1,-im,o ,J i - [1,-i,mo a]j + [l,i~mo (a 2 - ,~)] k = Oi + aj -,- a2k
= " .i + a 2k.
Because
,1 =-!
r (O) = (O)i + (a)j + (a2)k
=aj +a1 k
you can conclude thai r is continuous al , = 0. By similar reasoning. you can
conclude 1ha1 the vector-valued func tion r is continuous at all real-number value:-
of 1.
■
For each value of a. 1he cur;e represenred by the vector-valued function in Example 5.
= r (t) t i - a j - (a2 - 1J)k
a 1," constant
is a parabola. You can think of each parabola as the intersection of rhe vertical plane
y = a and 1hc hyperbolic: paraboloid
a=O
Cl=-2
G= 2
For each 1aluc of"· 1he cu rve represented
by rhc 1ecror-1alued function
r(1) = ti -'- aj + (u! - 11)k i, a parabola.
Figure 12.7
as :.hown in Figure 12.7.
___.~,:.'f,,f/lj'r'i3"l!!l: Almost any type of rbree-dimensional sketch ii. difficult co do by
hand. but ske1c:hing curves in space is especially difficulL The problem is in trying
10 create the illusion of three dimcnsio111>. Graphing util ities use a varie1y of
techniques to add "three-dimcnsionali1y" 10 graphs of space curves: one wa) is co sho\\ the curve on a surface, as in Figure l '.?..7.
12.1 Vector-Valued Functions
839
mExercises
See ,•1-.•1w.CalcCilal.com ror workea iut sol11t1on1 to odd-numbered e~erc1s.;s_
In Exercises 1-8, find lhe domain of the vector-valued function.
1. r(1) = - 1- i --'-.!,j - 31k
I- l
2
2. r (1) = ~
i + r 1j - 61k
3. r (I) = In Ii - e' j - rk
4. r(tl = sin ti + -l cos rj ..,. 1k
= 5. r(1) F(1) + G (1) where
F(rJ = cos1i - sintj - ./rk.
G(t) = co~t i + sin1j
6. r (i ) = F(1) - G (t) where
F(1) = In ti -;- 51j - 311 k. G(I) = i + 41j - 3r2k 7. r{I) = F (rJ "- G(r) where
F(r) = sin ri + cos rj, G(r) = sin rj + cos tk
8. r(tl = F(r) x G(r) where
F(t) = 1'i - rj -r 1k.
1
G(t)
= ~1;
i +
-
I
-r -I
j
-
(1 + 2)k
In Exercises 9-12. evaluate (if possible) the vector-valued fw1ction al each given value oft.
9. r(r) = 111 i - (1 - l)j (aJ r ( I l (b) r ((J) (c) r (~ + I) td l r(2 + ..irl - r(2)
IO. r(r) = co, Ii - 2 sin 1j
(a) r (0) (b ) r (r./ 4 ) (c) r ( 0 - r.l Id) r(r./ 6 + ..ir) - r(ri/ 6)
11. r (1) = lnri --"-- .!1_·I. + 31k
(a) r \2) (b ) r ( - 3 ) (C) r{i - 4) (d ) dl+ ..iil- r( I)
12. r(r) = .._f, i , r311 j + e 11" k
(aJ r (0) (b l r(-l) (c) r(c , 2) (d ) r (9 _._ ..it) - r(9)
In Exercises 13 a nd 14, find 11r(t)[I.
13. r (r) = ._/1 i ..,.. 31j - 41 k 14. r(r) = sin 31i + cos 3ij , ik
Jn Exercis es 15-18, represent the line segment from P to Q by a vector-valued function and by a set of parametric equations.
15. P(0. 0. 0). Q(3. I, 2) 17. P(-2.5. - 3). Q(- l. 4. 9) 18. P( I, -6. 8). Q(- 3. - 2. 5)
16. P(0. 2. - 1). Q(4. 7. 2)
Think .--\how Ir In Exercis es 19 a nd 20. find r(t) • u(t). Is the result a vector-valued function? Explnin.
19. r(I) = (31 - l )i + ½i·1 j -'- -lk. u(rl = i 2 i - SJ + r' k 20. r (r) = (3 cos 1. 2 ,inf. r - 2). u(r) = (-l sin 1. - 6 co, 1. 1 2}
In Excrdscs 21- 24, match the equation with its graph. [The graphs a re labeled (a). (b ). (cJ, and (d ).]
(a )
(b)
-2
( C)
(d)
.r
= 2L r (f) ri T 2rj + F:, k , -2 s Is 2 22. r(r) = cos( r.1)i + sin(r,t)j + i 1 k. - I s r s 23. r(f) = ti + 11 j + e'1' 51 k, - 2 s Is 2
24.
r (I)
=
Ii
+
ln 1 j
+
'l I
t
k,
0.1 s r s 5
25. Think About It The four figures below are graphs of the
, ector-l'alued function r(i} = 4 cos Ii - -l sin , j -,- (i/ 4)k .
Match each ol' the four graphs with the poilll in space from
which the helix is viewed. The four points are (0. 0, 20).
(20, 0, 0 ). (- 20. 0. 0). and ( IO, 20. 10).
(a)
(bl
<irn~rnuJ In \fahrmotiro
le )
,.
Gr11rruud b\• ,\lmlrcmutu..·a
(d l
G~nuauJ b\' Mo.1Jirmmic,1
GrMra:Ji'd /Jy Matlirm.irica
26. Sketch the three graphs o f the vector-valued fu ncrio n r(r) = ti - rj + 2k as ,·iewed from each point.
(:.1) (0. 0. 20) (b ) ( 10. 0. 0) (c l (5. 5. 5)
840
Chapter 12 Vector-Valued Functions
In Exercises 27-42, sketch the curve reprcsen1ed l>y the vectorvalued function and give the orientation of the curve.
27. r(r) = ± i + V - I)j
28. r(,) = (5 - 1)i + Jtj
29. r l1) = 1-1i + r1j
31. r(8) = cos Oi + 3 sin 8j
30. r(t) = (,~ + r)i + (12 - i)j
32. r(1) = 2 cos ti + 2 sin rj
33. r (0} = 3 sec 8i + 2 1an 8,i
34. r(r) = 2 cos.1 ti + 2 sin3 rj
35. r (1) = (-, + l )i + (41 + 2)j + (2, -i- 3)k
36. r(/) = ti + (21 - 5)j + 31k
37. r (r) = 2 cos ti + 2 sin ,j + 1k
38. r (i) = ii + 3 cos 1j + 3 sin rk
39. r (1) = 2 s in Ii + 2 cos rj + e I k
40. r (i) = 12i + 2tj + iik
f 41. r (r\ = (,. 12. 1 '}
42. r (r) = (cos 1 + 1 sin 1. sin, - , cost. 1)
GD In Exercises 43-46, use a computer nlgebra sys1e01 to graph the
vector-valued function and identify the common curve.
2 2 4"3, r(/) = - lT ' 't + IJ' - ./} 1'- k.
44 .
r (/)
=
.
ti
-
l ,
/
3
'.
1-J
+
l I
'k
1-
2 2 2✓3) k• 4:->. r(_/) = SI•D II• + (,]/3 COS I - If).)· + ( l COS I +
46. r (,l = - fi sin ti + 2 cos 1j + fi sin tk
( m Think About It ln Exercises ,n and 48, use a computer al gebra
system Lo graph the vector-v.llued funclion r(t). For each u(t), make a conjecture about the transformation (if any) of the gr:1pb of r(t). Use a computer algebra system to verify your
con,ieclure.
47. r{t) = 2 cos ti + 2 sin ij + ½1k (a) 0 (1) = 2(c:os 1 - 1)i + 2 sin ,j + !rk
(b) u(,) = 2 cos Ii + 2 sin , j + 21k
h- (cl u(t) = 2 co,(- i)i + 2 sin(- r)j + t) k
(d) 0 (1) = ½i i .1.. 2 sin 1j ~ 2 cos 1k
(el u(i) = 6cos ti + 6 ~in,j + ir k
~8. r(i) = ri + r2j + ½r'k
½, (a) 0 (1) = ri + (,~ - 2J j + ' k
(b) 0 (1) = r2i + 1j + ½r3k (c) u (I) = ti + r:?j + {½13 + 4)k
(d) u(i) = ri + 12j + ! 13k (e) o [/) = (- l)i + (-r)2j + 1(-1)·1k
In Exercises 49-56, represent the plane curYe by a , ectorvalued function. (There are many correct answers.)
49. _I"= X-,- 5 51. J = (x - 2)2
50. '.!x - 3y + S = 0
52. y = ➔ - x~
53. x2 .,. y 2 = 25
~,.. x1 y1
:,:,, 16-4= I
54. (x - 2F + y2 =-+
In Exercises 57 and 58, find vector-valued functions forming the boundaries of the region in the figure. State the interval for the parameter of each function.
57. Y
58. y
121 10 +
;±
21
- - .' - - - - -- ' - - .t
Jn Exercises 59-66, sketch the space cun·e represented by the intersection of the surfaces. Then represent the curve b) a ,·eclor-valued function using the ginn parameter.
S11,faces
59. .; = x2 + y 2, X + )' = Q 60. ;: = Xe + y1• Z = 4 61. x2 + y:? = 4. z "' x2 62. 4.r2 + 4_v2 + : 1 = 16, X = : 2
63. X1 + y 2 + : 2 = 4. X + : = 2
64. x1. + y 2 + : 1 = JO. x + y =-+
65. x 1 + :" = 4, y2 + : 2 = -166_ x 2 + y 2 + ;:2 = 16. .1y = -1-
Parameter
x=1
X = 2 COS I x = 2 sin, .: = / x = l t- sin 1
x = 2 + sin 1 x = 1 (first octant) x = r (first octant)
67. Show Ihm 1he vector-valued function
r(t) = ii + 2r cos 1j -"- 21 sin 1k
i lies on the cone 4x2 = + .::2. Sketch the curve.
68. Show chat the ,ector-valued function
r (1) = e-, cos 1i + e-1 sin 1j + e ' k
lies on t.he cone ;:2 = x 2 + v2. Sketch the curve.
In Exercises 69- 74, find the limit (if it exists).
69. ,li_m" (ii .,_ cos 1j + sin 1k )
70 .
h•m (3n• T
I· ,2
-,-2 - J• r- - l
+ -l k) /
12.1 Vector-Valued Functions
841
In Exercises 75- 80, determine the interval(s) on which the vector-valued fu nction is continuous.
= 1:-i. r(r) I •I T -IJ• 1
76. r (r) = Jr i + J;-=--rj
77. r (r) = r i + arcsin rj + (t - l ) k
78. r (r) = 2e-1 i -r 1,-,j + In(, - 1) k
79. r (1) = (e-1. 11. 1an 1)
80. r(r) = (8. Jr. -Yr)
WRITING ABOUT CONCEPTS
81 . Consider the vector-valued function
r(1) = r1i + (1 - 3).i + 1k.
Write a vector-valued fu□c1 ion s(1) that is the specified cransforma1ion of r.
(a) A vertical rranslmion 1hree units upward
(b) A horizon1al rranslarion two units in the direction of the nega1i,·e x-axis
(c) A horizomal 1ransla1ion ti,·e uni!~ in the direction of the positive y-axis
82. Stace !he definition of cominuity of a ,·ector-valued funccion. Give an example of a vector-valued func1ion Lhm
is defined but not continuous al t = 2.
C?l) 83. The oucer edge of a playground slide is in the shape of a helix of
radius 1.5 meters. The slide has a height of 2 meters and makes one complete revolution from top Lo bottom. Find a vecmrvalued fllllction for the helix. Use a computer a lgebra S) stem 10 graph your function . (There are many correct answers.)
CAPSTONE
8-t. Which of the following vector-valued functions represent the same graph?
(al r (r) = (-3cost + l)i + (5 sin, + 2)j + 4k (b) r (r) = 4i + (- 3 cos I + I)j + (5 sin r + 2)k
(c) r (r) = (3 cos I - I)i + (- 5 sin, - 2)j + 4k
(d) r (1) = (- 3 cos 21 + I)i + (5 sin 11 + 2)j + 4k
85. Let r (1) and u(,) be vccmr-valued funcuons whose limits exist ~ t ➔ c. Pro\·e that
!im [r(t) x u(rJ] = lim r{r) x lim u (r).
: - ; .,
1 - t( '
1--,,
86. Let r(,) and u(,) be veccor-valued function~ whose limits exist a!> t ➔ c. Prove that
li111 [rll) • ult)]= Jim r t1) • lim u (1).
i-•
1-.:
1- t:
87. Prove lhat if r is a vector-valued function that i!> continuous at
c. then II rll is continuous al c.
88. Verify that the converse or Exercise 87 is nc11 true by findi ng a vector-valued func tion r such that II r II is contim1om at c bllt r is not cominuous at c.
In Exercises 89 and 90, l\10 particles travel along lhe space curves r (r) and u(t). A collision will occur at the point of intersection P if both particles arc at Pal the same time. Do the particles collide? Do their paths intersect?
89. r(1) = t"i + (9t - 20)j + t2k u(r) = (31 + 4)i + i2j + (51 - 4)k
90. r (r) = ri + rlj + r'k
u (n = (-21 + 3)i + 81j + (121 + 1)k
Think Abolll It ln Exercises 91 and 92, two particles travel along the space cu rves r(I) a nd u(1).
91. If r(1) and u(t) intcrsecL. w ill the particles collide?
92. If the particles collide. do their path, r (1) and u (1) intersecc?
Tnie or False? In Exercises 93--96. determine whether the statement is true or false. lf it is fa lse. explain why or give an example that s hows it is false.
93. If f. g. and h are first-degree polynomial function~. then the
curve given by x =f(t) , y = g( r). and z = h(r) is a line. 94. If the curve given by .r = f(t) . y = g(r). and.: = h(r) is a line,
t11en/ g. and hare firsc-degree polynomi al iunc1ion, of,.
95. 1\vo pani cles travel along the space curves r(tJ and u(1). The intersec tion of their paths depends only on the curves rraced Olli b, r (t) and u(1). while collision depends on the parameteriLarions.
96. The vector-\'alucd function r(i) = r1 i , 1 sin, j -'- 1co~ 1 k lies on the paraboloid .r = v2 + ,:1_
Witch of Agnesi
In Section 3.5. you studied a famous curve called lhe Witch o r Agnesi. ln this project you will take a closer look at this function .
Consider a circle of radi us a centered on the y-ax:is at (0. a). Let A be a point on the horizontal line y = 2a. lee O be the origin. and Jet B be the point where the segmcm OA imersects the circle. A poim P is on the Witch of Agnesi if P lies on the horizontal line through Band on the ,·ertical line th rough A .
(a) Show that the point A is Lraced out by Lhe vector- \'alued function
r ,(O) = 2a cot f:li -'- 2aj . 0 < B < 1r
\\ here Ois the angle that OA make\ with the positive x-axis. (b) Show that the point Bis traced our by me vector-valued function
rn(O) = a sin 20i -r a( I - cos 20).i, 0 < fl< -rr.
(c) Combine !he results of' parts (a) and (bl to find the vcctor-
valu.:d function r (fJ) for the Witch of Agnesi. Use a graphing
utility to graph this curve for a = I.
(d) Describe the limits lim r (fJ) and lim r l0).
11➔0 '
d- ,,.--
(C) Eliminate the parameter O and deccrmine the rectangular
equacion of the Witch of Agnesi. Use a graphi11g utility to graph
this function for a = I and compare your graph wit11 that
obtained in pan (cl.
842
Chapter 12 Vector-Valued Functions
lflJ Differentiation and Integration of Vector-Valued Functions
■ Differentiate a vector-valued function. ■ Integrate a vector-valued function.
Differentiation of Vector-Valued Functions
In Section.1, 12.3-12.5. you will sLUd) several importam applications in\"O]\ing the calculus of vector-valued functions. ln preparaLion for that study. this section is devoted 10 the mechanics of d ifferenriation and imegration o f vector-valued functions.
The definiLion of the derivative of a vector-valued function parallels the definit ion given for real-valued functions.
DEFINITIONOF THE DERIVATIVE OF AVECTOR-VALUED FU~CTI0:-1
The derivatfre of a vector-valued function r is defined by
r
't ) 1
=
11. m r( r
..'lt->O
+
.lr) .lt
r (r)
for all r for which the limit exists. l f r '(r) exists. then r is differentiable at I. if r ' (r) exists for all I in an open interval J. then r is differentiable on the interval I. Differentiability of vecror-valued fu nction~ can be exLended to closed imervals by considering one-sided li mits.
Figure 12.8
r (1 + ..'>t) - rtn I
r 'tr)
G D In addiuon to r '(r). other 11mations for t.he derivative of a vector-valued function arc
d
dr
D.[r(r)]. -d[r (r)]. am! -d.
■
•
f
f
Differentiation of vector- valued functions can be done on a componem-byco111pone11t basis. To sec why l11is is true. consider the function given by
r (r) =f(r)i + g(r)j .
Applying the definition of the derivative produces the following.
= .J-~11 '( ) I" r (1 i. .11) - r (I)
r I
ur
=
. /(r I1m
.l,......o
..L
.lr)i
+
g(r
+ ur)j
.lr
-
f(1)i -
g(r) j
I = li.m {r..r(, + .11) - j(,)li + lg(, ..;.. ~ ,) - g(,)J ·}
.lr-+o
.lr
.lr
J
= r {
li!TI
. l1 - 0
. (
(r
+
.lr) .lT
•f{r)l} 1· +
{ .l1,•L-il-,ol[g~ (r --.- .lrl)I-- -g{-1)]}J.
= f'{r)i + g '(llj
T his important result is listed in the theorem on the next page. Note that the derivative o f 1hc vector-valued function r is its~lf a vector-valued function. You can see from Figure 12.8 that r '(r) is a vector tangent LO the curve given by r (,) and pointing in the direction of increasing 1-rnlucs.
12.2 Differentiat ion and Integration of Vector-Valued Functions
843
1· r (I} =t i + (12 + 2)j
rlfEORDI 12.l OIFFERE'.'lTIATION OF VECTOR-VALUED FUNCTIONS
l. If r (r) = f( t)i + g(t)j, where.fand g are differentiable functions of r, then
r 1(1) = f'(t) i + g 1(1) j.
= 2. lf r(1) J(t) i + g(/)j + h(t)k, where/ g. and Ii are differentiable functions
of 1. then
r '(,) = f'(,) i J.. g'(t)j + h'(t)k.
Sp,1c~
G D EXAMPLE Differentiation of Vector-Valued Functions
= For the vector-valued function giYen by r(, ) i i + (12 + 2)j. find r '(,). Then sketch
the plane curve represented by r (,l. and the graphs of r(l) and r '( I ).
Solution Differentiate on a component-by-component basi1, ro obtain
r '(1) = i + 21j .
Derhalivc
- --3---_, --_...l.._l_y,____.__...2.__3_ .< Figure 12.9
From the position vector r (1). you can write the parametric equations x = 1 and
y = 12 -r 2. The corresponding rectangular equation is y = x2 .,_ 2. When c = I.
r ( I ) = i + 3j and r 1(1) = i + 2j . In Figure 12.9, r ( I ) is drawn starting at the origin.
and r '(I) is drawn ~tarting at the terminal point of r ( I ).
■
Higher-order derivatives of vector- valued functions arc obtained by successi ve diffcrenciarjon of each component function.
fl EXAMPLE Higher-Order Differentiation
For the vector-val ued function given by r (i) = cos Ii T sin i j , 21k, lind each of the following.
a. r '(i) c. r '(i) • r"(t)
b. r "{t) d. r '( t) x r "(t)
Solution
a. r '(i) = -sin ti +cos tj + 2k
b. r "(/) = -cos Ii - sin 1j + Ok
= -cos Ii - sin ,j
c. r '(t) · r "(,) = ~in r cos, - sin, cos,= 0
j k
d. r '(r) x r "(i) = - sin t cos 1 2
-CO\, 1 -sin 1 0
=I c~~ , 2I i -I - sint
-s111 t O
- cos r
= 2 sin ri - 2 cos t.i + k
FiN Jeri,a1i,c
Doi produci Cm" prnduc1
21. I-sin t cos 1 \ k 0 J + -cos , -s1111
Note that the dot product in part (c) is a r eal-1•alued function. nol a vec1or-valued
function.
■
844
Chapter 12 Vector-Valued Functions
-6-'-
= 1'(1) (S cos 1 - cos 5tli + (5 sin t - sin 5/)j
The epic)cloid is not smooth at the points where it intersects the axes. Figure 12.10
The parametrization of the curve represented by the vector-valued funcrion
r(t) = f(t) i + g(t)j -'- /z(t)k
is s mooth on an open interval / iff', g'. and /,' are continuous on / and r '(,) if: 0 for any value oft in the interval /.
IJ EXAMPLE Finding Intervals on Which a Curve Is Smooth
Find the intervals on which the epicycloid C given by
r (t) = (5 cos r - cos 5r)i + (5 sin 1 - sin St)j , 0 $ r $ 2,.
is smooth.
Solution The derivative of r is
r ' (r) = (- 5 sin t + 5 sin Sr)i + (5 cos t - 5 cos 5/)j.
In the inter\'al [O, 2 rr]. the only values of I for which
r '(t) = Qi + Oj
are 1 = 0. rr/'2. rr. 3r./'2. and 2,r. Therefore. you can conclude that C is smooth in the
intervals
(0, 27T), (2". ) ( 7T, • 7T, 2 , 37T)
and
(
3,. 2 '
21T)
as shown in Figure 12. 10.
■
m . In Figure 12.10. note that the curve is 001 ~mooth at points at which the curve makes
abrupt change, in direction. Such points are called cusps or nodes.
■
Most of the differentiation rules jn Chapter 2 have counterparts for vector-valued functions. and several are listed in the folio\\ ing theorem. Note that the theorem contains three versions of "product rule!>." Property 3 gives the derivative of the product of a reaJ-valuctl function 11· and a vector-valued function r. Property .i g ives the derivative of the dot product of 1wo vector-valu ed functions. and Property 5 gives the derivative of the cross product of two vector-valued functions (in space). Note that Property 5 applit!s only to three-dimensional vector-valued functions, because the cross product is not defined for two-dimensional vectors.
THEORE.\I 12.2 PROPERTIESOF THE DERIVATIVE
Let rand u be differenriable vector-valued functions of r, let 11· be a differentiable real-vaJuetl function of ,. and let c be a scalar.
1. Di[cr(r)] = cr'(t) 2. D,[r(r) ± u(t)] = r ' (T) ± u '(E)
3. D,[w(r)r (il] = w(r)r '(t) + w'(T)r (t) -t D,[r (r) • u (T)] = r (r) • u '(t) + r '(r) • u(T)
= 5. D1[ r (T) x u (t)] r (I) x u '(t) ... r '(r) x u (r) 6. D1[r(111(t))] = r1 (11'(1))w'(t) 7. If r (t) • r (r) = c. then r (r) • r '(E) = 0.
EXPLORATION
Let r(1) = cos ti + sin rj. Sketch
the graph of r (t). Explain why
the graph is a circle of radius I
centered m the origin. Calculate r (,./4) and r '(n-/4). Position the vector r '( r,/ 4) so that its initial point is at the terminal point of r('lT/4). What do you observe" Show that r (r) • r {r) is constant
and that r(r) • r '(1) = 0 for all r.
How does this example relate to Propeny 7 ofTheorem J2.2?
12.2 Differentiation and Integration of Vector-Va lued Functions
845
( PROOF ) To prove Property -t let
r(t) = / 1(r)i + gt(1)j and u(t) = fi(r)i + gir)j
whcre/1.f2. g 1. and g2 are differentiable functio ns of 1. The n.
r(t) • u (1) = f, (1)Ji(t) t- g 1(1)g2(1)
and it follows that
D,[r (t) · u(r)] =f,(r)j/(1) + f 1'(t)f2(t) + g 1(1)g2' (t) + g,'(t)g2(r)
= [.r,(nJ/(1) + 81 (1)g2'(1)J + [J,'(1)!2(1) + g,'(1)gi1)J
= r (r) • u'(t) + r '(t) • u (1).
Proofs of the other properties are left as exercises (see Exercises 77-81 and
Exe rcise 84).
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El EXAMPLE Using Properties of the Derivative
For tbe vector-value-el functions given by
7 r (t) = i - j + In 1k and a (r) = r2i - 2rj + k
find
a. D,[r(t) • u(r)] and b. D,[u(1) x u '(1)].
-fi 7 Solution
a. Because r'(t) =
i + k (tnd u'(t) = 2ti - 2j. you hm e
D,[r(t) • u(t)] = r(t) • u '(t) + r '(t) • u(t)
+ = ( i - J + In tk ) • (2ti - 2j )
(-~ i +
+ .!. k) · (t2 i - 21j + k)
r·
I
= 2 + 2 + (-
1
1) + -
I
= 3 .... .!._ I
b. Because u '(t) = 2ti - 2 j anti u"(1) = 2 i. you have D,[u(1) x u'(tl] = [u(r) x u"(r)] + [u '(t) x u'(r)]
i j k
= rJ -2r l + 0
2 0 0
= Oi - (-2)j + 4tk
= 2j + 4rk.
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G D Try reworking parts (al and (bl in Example 4 by first forming the dot and cross
product~ and then differentiating to see that you obtain the same results.
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846
Chapter 12 Vector-Valued Functions
Integration of Vector-Valued Functions
The following definition is a rational consequence of the definition of the derivative of a vector-valued functjon.
DEFINITIO~ OF INTEGRATION OF \'ECTOR-VALUED FC~CTIONS
I!
J. u· r(i) = /(1)i - g(t)j . where / and g arc continuous on [a, b]. then the
indefinite integral (antiderirntive) of r i~
Plane
and its definite integral over the interval a :s r $ b is
f r(r) cir= [fr(,) dr} + [ fg(1J i11 jJ.
2. If r(I) = /( r)i + g(1)j ..1. h(t)k. where f g. and h are continuous on [a. b]. II
then the indefinite integral (antiderivative) of r is
S pnce
II
and its defi nite int egral over the interval a ::; 1 s b is
f [f r (1) d1 = l f r(r) d1 Ii+ [ J,"g(r) drl j + h(1) dr] k .
The amiderivnt ive of u vector-valuell function is a family or vector-valued functions all differing by a constant vector C. For instance. if r (T) is a three-dimensional vector-valued function. then for 1he imlefi □ite imegral f r(t) di. you obtain three constants of inregration
Jh(1) dr = H(r) + C3
where F'(r) = /(1). G'(1) = g(t) , and /-/ '(r) = h (r). These three scalar constants
produce one ,·ecror constant of imegration.
I r(1 ) di = [F(!) ..L C1Ji + [G(I) + C2]j + [H(!) + C3]k
= [F(!)i + C(dj T H (l)k ] + [C1 i + C2j - C,k ] = R (1) + C
where R '(r) = r (r).
El EXAMPLE Integrating a Vector-Valued Function
Find the indefinite imegral
J(,j • 3j ) cit.
Solution Integrating on a component-by-component basis produces
J f= (r i ..1. 3j) dt i + 3rj + C.
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12.2 Differentiation and Integration of Vector-Valued Functions
847
Example 6 shows how to evaluare the definite integral of a vector-valued function.
II EXAMPLE Definite Integral of a Vector-Valued Function
Evaluate the integral
L f (~ r(r) dr =
~ 11i +
j + e-1 k) dt.
Solution
■
As with real-valued functions. you can narrow the family of antiderivaLives of a vector-va lued function r ' down to a single antiderivative by imposing an initial condition on !he vector-valued function r. This is demonstrated in the next example.
II EXAMPLE The Antiderivative of a Vector-Valued Function
Find the antiderivative of
r '(t) = cos 2ri - 2 ~in tj + -1-+1- ,-1 k
that satis11es the iniria l condition r (O) = 3i - 2.i -'- k.
Solution
f= r (t)
r '(r) dr
(f (f (J = cos2rdr} + -2si111d1) j + 1 : 12 dr)k
G = sin 21 + C1} + (2 cos, + C2)j + (arctan t + C,)k
Letting r = 0 and using the fact that r (O) = 3i - 2j + k. you h,l\C
r (O) = (0 - C 1)i -'- (2 + C2)j -1- (0 + C)k
= + 3i ...!.. {-'.!.)j k.
Equating corresponding components produces
and C., = I.
So, tbc antiderirntive that satisfie~ the given initial rnndilion is
r (r) = (~ sin 2r + 3 ) i ..,_ (2 cos T - -t)j - (arctan I - l)k.
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848
Chapter 12 Vector-Valued Functions
rm Exercises
S~ ,...,.,., Cal Chat.corn •or ,,or~td· u: solutions ;u c-dd-numbnd e~erc1~es
In Exercises 1-8. sketch the plane cun-e represented by the
vector-valued fu nction, and sketch the vectors r(/0 ) and r '{t0) for the given value of lo- Position the vectors s uch that the initia l
point of r (t0) is at the origin and the initial point of r '(t0) is al the terrninal point of r{t0). What is the relation.ship between r'(t0) and the curve?
]. r (,) = 12i + tj , to = 2
2. r(, ) = ri + (r1 - l)j . I r, =
7 3.
r (t)
=
,.
1· 1
+
I J..
in = ,-
4. r(1) = (I , 1)i + r3j . r0 = I
5. r (t) = cos ti + sin rj,
7T
lo = 2
6. r (!) = 3 sin ti + --!- cos tj.
7. r(,) = (e'. e1'). r0 = 0 8. r(1) = (e-1. e'). 10 = 0
In Exercises 9 a nd 10, (a) sketch the space curve represented by the vector-valued function, and (bJ sketch the vectors r(t0) and r '(/0) for the ghen ,·alue of 10.
9. r (1) = 2 co, ti + 1 s i □ tj + 1k. 10 = -3;,,--
10. r (t) = ti + r2j + ! k. 10 = 2
[n Exercises 11-22, find r '(t).
11. r(1) = t3i - 3rj
12. r(1) = J, i T ( I - r3 )j
13. r (t) = ( 2 cos r, 5 sin r)
14. r (t) = ( 1 cos 1. - 2 sin 1)
15. r (1) = 6ri - 711j + 13k
16.
r (1)
=
1 -i
+
l61j
,2
+ -k
I
2
17. r(, ) = acos3 1i ..L asin1 1j + k
18. r(r) = 4-J, i + r2 J, j + In 12k
19. r (t) = e-1 i + 4-j + 5te' k
20. r (r) = (1·1, cos 31. sin 31)
2 1. r (t) = (1 sin r. t cos 1. 1)
22. r (r) = (arcsin r. arccos 1. 0)
In Exercis es 23-30. find (a ) r'(t), (b) r"(t). and (c) r' (t) • r"(t).
23. r(t) = 1·' i + ½11j
24. r (t) = (12 + !)i + (12 - r)j 25. r(t) = 4 cos Ii ..L 4 sin tj 26. r(r) = 8 cos 1i + 3 sin rj
!r 27. r(t) = 2i - rj + ~13k
28. r(1) = ti + (21 + 3)j + (3r - 5)k 29. r(1) = (cos t + 1 s in 1. sin 1 - 1co~ t . 1) 30. r(1) = (e '. 12. tan 1)
In Exercises 31 and 32, a veclor·Yalucd function and its graph
are and
g r
iven. The "(10)/Jl r"(10
gr
)JI.
aph also s hows tbe Find these two unjt
unit vectors r'(t0 Mlr '(t0)1! ,·ectors and identify them
on the graph.
-± 31. r (i) = cos(rr rli + sin(r.r)j + r2k . r0 =
32. r (r) = i ri + t 2j + e-1 k. 10 = ¾
1
Figure for 31
Figure for 32
In Exercises 33- :12. find the open inter,al(s) on which the curve given by the vector-valued function i.s smooth.
34• r(r) = ~ I .• - 3IJ.
= 35. r ( 0) 2 cos 1 0i + 3 sin3 0j
36. r (O) = (fJ + sin O)i T ( I - co~ 0)j
37. r (0) = (0 - 2 sin O)i + (I - 2 cos 0}.i
! 21 . 2,2 .
38. dt) : 8 + /3 I -I 8 + ,, J
39. r (1) = (1 - I li + j - ,~k
I
4-0. dr) = e'i - e-rj ...._ 3rk
41. r (1) = ri - 31j + tan i k
4-2. r (r) = J, i -'- (12 - J)j + 1tk
In Exercises -'3 and 4--'. use the properties of the derivative lo find the following.
(a ) r '{t)
(b) r"(t)
(c) D,[r(t} • u(t)J
(d) D, [3r(t) - u(t)) (e) D,[r(t) x u(t)) (f) D,Ulr(t):□. t > 0
-'3. r (r) = Ii +- 31j + ; 2k. u(t) : 4-ti - 11j T 13k .W. r(,) = ti + 2 sin tj - 2 cos ,k.
u(r) = -I 1. + 2s.in rJ. + -? cos 1k
I
In Exercises 45 and :16. find ( a) D,[r (t) • u(r)] and (b) D, [r(I) x n(t)J in two different ways.
(i) Find the product first. then differentiate. (ii) Apply the properties of Theorem 12.2.
4-5. r (,) = ti + 2r2j + r'k. u(t) = ,'k
46. r(t) = cos ii + ~in 1j + rk. u(rl = j + 1k
ln Exercises .i7 and :18, find the angle 0 between r(t) and r '(t) as a function of I. Use a graph.ing utility lo graph 8 (1). Use the graph to find a ny extrema of the function. Find any values oft at whlch the nclors arc orthogonal.
47. r(r) = ., sin Ii + 4- cos rj 48. rft) = l"i ..,. tj