Cengage learning COMPLETE solutions manual for multivariable calculus 7th edition 0328693561

■ CONTENTS■ 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 1 10.1 Curves Defined by Parametric Equations 1 Laboratory Project ■ Running Circles Around Circles 15 10.2 Calculus with Parame

Trang 1

INSTRUCTOR SOLUTIONS MANUAL

Trang 2

Complete Solutions Manual

for

MULTIVARIABLE CALCULUS

SEVENTH EDITION

DAN CLEGG Palomar College BARBARA FRANK Cape Fear Community College

Australia Brazil Japan Korea Mexico Singapore Spain United Kingdom United States

Trang 3

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 107or 108of the

1976United States Copyright Act, without the prior written

permission of the publisher except as may be permitted by

the license terms below

Printed in the United States of America

1 2 3 4 5 6 7 15 14 1 3 1 2 11

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/permissions

Further permissions questions can be e-mailed to

permissionrequest@cengage.com

ISBN-13: 978-0-8400-4947-6ISBN-10: 0-8400-4947-1

Brooks/Cole

20Davis DriveBelmont, CA 94002-3098USA

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:

NOTE: UNDER NO CIRCUMSTANCES MAY THIS MATERIAL OR ANY PORTION THEREOF BE SOLD, LICENSED, AUCTIONED,

OR OTHERWISE REDISTRIBUTED EXCEPT AS MAY BE PERMITTED BY THE LICENSE TERMS HEREIN.

READ IMPORTANT LICENSE INFORMATION

Dear Professor or Other Supplement Recipient:

Cengage Learning has provided you with this product (the

“Supplement”) for your review and, to the extent that you adopt

the associated textbook for use in connection with your course

(the “Course”), you and your students who purchase the

textbook may use the Supplement as described below

Cengage Learning has established these use limitations in

response to concerns raised by authors, professors, and other

users regarding the pedagogical problems stemming from

unlimited distribution of Supplements.

Cengage Learning hereby grants you a nontransferable license

to use the Supplement in connection with the Course, subject to

the following conditions The Supplement is for your personal,

noncommercial use only and may not be reproduced, or

distributed, except that portions of the Supplement may be

provided to your students in connection with your instruction of

the Course, so long as such students are advised that they may

not copy or distribute any portion of the Supplement to any third

party Test banks, and other testing materials may be made

available in the classroom and collected at the end of each class

session, or posted electronically as described herein Any

material posted electronically must be through a protected site, with all copy and download functionality disabled, and accessible solely by your students who have purchased the associated textbook for the Course You may not sell, license, auction, or otherwise redistribute the Supplement in any form We ask that you take reasonable steps to protect the Supplement from unauthorized use, reproduction, or distribution Your use of the Supplement indicates your acceptance of the conditions set forth in this Agreement If you do not accept these conditions, you must return the Supplement unused within 30 days of receipt.

password-All rights (including without limitation, copyrights, patents, and trade secrets) in the Supplement are and will remain the sole and exclusive property of Cengage Learning and/or its licensors The Supplement is furnished by Cengage Learning on an “as is” basis without any warranties, express or implied This Agreement will be governed by and construed pursuant to the laws of the State of New York, without regard to such State’s conflict of law rules Thank you for your assistance in helping to safeguard the integrity

of the content contained in this Supplement We trust you find the Supplement a useful teaching tool.

Trang 4

This Complete Solutions Manual contains detailed solutions to all exercises in the text Multivariable

Calculus, Seventh Edition (Chapters 10–17 of Calculus, Seventh Edition, and Calculus: Early Transcendentals, Seventh Edition) by James Stewart A Student Solutions Manual is also available,

which contains solutions to the odd-numbered exercises in each chapter section, review section,True-False Quiz, and Problems Plus section as well as all solutions to the Concept Check questions.(It does not, however, include solutions to any of the projects.)

Because of differences between the regular version and the Early Transcendentals version of the text, some references are given in a dual format In these cases, users of the Early Transcendentals

text should use the references denoted by “ET.”

While we have extended every effort to ensure the accuracy of the solutions presented, we wouldappreciate correspondence regarding any errors that may exist Other suggestions or comments arealso welcome, and can be sent to dan clegg at dclegg@palomar.eduor in care of the publisher:Brooks/Cole, Cengage Learning, 20 Davis Drive, Belmont CA 94002-3098

We would like to thank James Stewart for entrusting us with the writing of this manual and ing suggestions and Kathi Townes of TECH-arts for typesetting and producing this manual as well ascreating the illustrations We also thank Richard Stratton, Liz Covello, and Elizabeth Neustaetter ofBrooks/Cole, Cengage Learning, for their trust, assistance, and patience

Trang 6

■ ABBREVIATIONS AND SYMBOLS

CD concave downward

CU concave upward

D the domain of iFDT First Derivative Test

= indicates the use of the substitution {x = cos {> gx = 3 sin { g{}

I/D Increasing/Decreasing Test

Trang 8

■ CONTENTS

■ 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES 1

10.1 Curves Defined by Parametric Equations 1

Laboratory Project ■ Running Circles Around Circles 15

10.2 Calculus with Parametric Curves 18

Laboratory Project ■ Bézier Curves 32

10.3 Polar Coordinates 33

Laboratory Project ■ Families of Polar Curves 48

10.4 Areas and Lengths in Polar Coordinates 51 10.5 Conic Sections 63

10.6 Conic Sections in Polar Coordinates 74 Review 80

11.5 Alternating Series 143 11.6 Absolute Convergence and the Ratio and Root Tests 149 11.7 Strategy for Testing Series 156

11.8 Power Series 160 11.9 Representations of Functions as Power Series 169 11.10 Taylor and Maclaurin Series 179

Laboratory Project ■ An Elusive Limit 194

11.11 Applications of Taylor Polynomials 195

Applied Project ■ Radiation from the Stars 209

Review 210

Problems Plus 223

Trang 9

viii ■ CONTENTS

■ 12 VECTORS AND THE GEOMETRY OF SPACE 235

12.1 Three-Dimensional Coordinate Systems 235 12.2 Vectors 242

12.3 The Dot Product 251 12.4 The Cross Product 260

Discovery Project ■ The Geometry of a Tetrahedron 271

12.5 Equations of Lines and Planes 273

Laboratory Project ■ Putting 3D in Perspective 285

12.6 Cylinders and Quadric Surfaces 287 Review 297

Problems Plus 307

■ 13 VECTOR FUNCTIONS 313

13.1 Vector Functions and Space Curves 313 13.2 Derivatives and Integrals of Vector Functions 324 13.3 Arc Length and Curvature 333

13.4 Motion in Space: Velocity and Acceleration 348

Applied Project ■ Kepler’s Laws 359

14.6 Directional Derivatives and the Gradient Vector 437 14.7 Maximum and Minimum Values 449

Applied Project ■ Designing a Dumpster 469Discovery Project ■ Quadratic Approximations and Critical Points 471

Trang 10

15.3 Double Integrals over General Regions 521

15.4 Double Integrals in Polar Coordinates 534

15.5 Applications of Double Integrals 542

15.6 Surface Area 553

15.7 Triple Integrals 557

Discovery Project ■ Volumes of Hyperspheres 574

15.8 Triple Integrals in Cylindrical Coordinates 575

Discovery Project ■ The Intersection of Three Cylinders 582

15.9 Triple Integrals in Spherical Coordinates 584

Applied Project ■ Roller Derby 594

15.10 Change of Variables in Multiple Integrals 595

16.5 Curl and Divergence 650

16.6 Parametric Surfaces and Their Areas 659

Trang 11

x ■ CONTENTS

■ 17 SECOND-ORDER DIFFERENTIAL EQUATIONS 711

17.1 Second-Order Linear Equations 711 17.2 Nonhomogeneous Linear Equations 715 17.3 Applications of Second-Order Differential Equations 720 17.4 Series Solutions 725

Review 729

■ APPENDIX 735

Trang 12

10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

10.1 Curves Defined by Parametric Equations

1 h31+ 11=37

h32+ 22=14

| h32+ 2

2=14

h31+ 11=37

1 h 3 11=72

h2

3 25=39

NOT FOR SALE

Trang 13

2 ¤ CHAPTER 10 PARAMETRIC EQUATIONS AND POLAR COORDINATES

6. { = 1 3 2w, | =1

2w 3 1, 32 $ w $ 4(a)

8. { = w 3 1, | = w3

+ 1, 32 $ w $ 2(a)

Trang 14

SECTION 10.1 CURVES DEFINED BY PARAMETRIC EQUATIONS ¤ 3

31 $ { $ 0 and 0 $ | $ 1 For 0 ? $ , we have 0 ? { $ 1

and 1 A |D 0 The graph is a semicircle

2 D { D 0 and 0 $ | $ 2 For @2 ? $ , we have 0 A { D 312

and 2 A |D 0 So the graph is the top half of the ellipse

(b)

13. (a) { = sin w> | = csc w, 0 ? w ?2 | = csc w = 1

sin w =

1{.For 0 ? w ?2, we have 0 ? { ? 1 and | A 1 Thus, the curve is the

portion of the hyperbola | = 1@{ with | A 1

(b)

NOT FOR SALE

Trang 15

0 ? { and 1 ? | Thus, the curve is the portion of the parabola { = |23 1

in the first quadrant As increases from3@2 to 0, the point ({> |)

approaches (0> 1) along the parabola As increases from 0 to @2, the

point ({> |) retreats from (0> 1) along the parabola

(b)

19. { = 3 + 2 cos w, | = 1 + 2 sin w, @2 $ w $ 3@2 By Example 4 with u = 2, k = 3, and n = 1, the motion of the particle

takes place on a circle centered at (3> 1) with a radius of 2 As w goes from

of the particle takes place on an ellipse centered at (0> 4) As w goes from 0 to3

2 , the particle starts at the point (0> 5) andmoves clockwise to (32> 4) [three-quarters of an ellipse]

NOT FOR SALE

Trang 16

21.{ = 5 sin w, | = 2 cos w i sin w = {5, cos w = |

2

= 1 The motion of theparticle takes place on an ellipse centered at (0> 0) As w goes from3 to 5, the particle starts at the point (0> 32) and movesclockwise around the ellipse 3 times

to (31> 0) and goes back to (0> 1) The particle repeats this motion as w goes from 0 to 2

23.We must have 1$ { $ 4 and 2 $ | $ 3 So the graph of the curve must be contained in the rectangle [1> 4] by [2> 3]

24. (a) From the first graph, we have 1$ { $ 2 From the second graph, we have 31 $ | $ 1= The only choice that satisfieseither of those conditions is III

(b) From the first graph, the values of { cycle through the values from32 to 2 four times From the second graph, the values

of | cycle through the values from32 to 2 six times Choice I satisfies these conditions

(c) From the first graph, the values of { cycle through the values from32 to 2 three times From the second graph, we have

0 $ | $ 2 Choice IV satisfies these conditions

(d) From the first graph, the values of { cycle through the values from32 to 2 two times From the second graph, the values of

| do the same thing Choice II satisfies these conditions

25.When w =31, ({> |) = (0> 31) As w increases to 0, { decreases to 31 and |

increases to 0 As w increases from 0 to 1, { increases to 0 and | increases to 1

As w increases beyond 1, both { and | increase For w ?31, { is positive and

decreasing and | is negative and increasing We could achieve greater accuracy

by estimating {- and |-values for selected values of w from the given graphs and

plotting the corresponding points

26.For w ?31, { is positive and decreasing, while | is negative and increasing (these

points are in Quadrant IV) When w =31, ({> |) = (0> 0) and, as w increases from

31 to 0, { becomes negative and | increases from 0 to 1 At w = 0, ({> |) = (0> 1)

and, as w increases from 0 to 1, | decreases from 1 to 0 and { is positive At

w = 1> ({> |) = (0> 0) again, so the loop is completed For w A 1, { and | both

become large negative This enables us to draw a rough sketch We could achieve greater accuracy by estimating {- and

|-values for selected values of w from the given graphs and plotting the corresponding points

27.When w = 0 we see that { = 0 and | = 0, so the curve starts at the origin As w

to 0 while { repeats its pattern, and we arrive back at the origin We could achieve greater accuracy by estimating {- and

|-values for selected values of w from the given graphs and plotting the corresponding points

NOT FOR SALE

Trang 17

28. (a) { = w4

3 w + 1 = (w4

+ 1) 3 w A 0 [think of the graphs of | = w4+ 1 and | = w] and | = w2

D 0, so these equationsare matched with graph V

(b) | =I

w D 0 { = w23 2w = w(w 3 2) is negative for 0 ? w ? 2, so these equations are matched with graph I

(c) { = sin 2w has period 2@2 = Note that

|(w + 2) = sin[w + 2 + sin 2(w + 2)] = sin(w + 2 + sin 2w) = sin(w + sin 2w) = |(w), so | has period 2.These equations match graph II since { cycles through the values31 to 1 twice as | cycles through those values once.(d) { = cos 5w has period 2@5 and | = sin 2w has period , so { will take on the values31 to 1, and then 1 to 31, before |takes on the values31 to 1 Note that when w = 0, ({> |) = (1> 0) These equations are matched with graph VI=

(e) { = w + sin 4w, | = w2+ cos 3w As w becomes large, w and w2become the dominant terms in the expressions for { and

|, so the graph will look like the graph of | = {2, but with oscillations These equations are matched with graph IV

(f) { = sin 2w

4 + w2, | = cos 2w

4 + w2 As w< ", { and | both approach 0 These equations are matched with graph III

29. Use | = w and { = w3 2 sin w with a w-interval of [3> ]

30. Use {1= w, |1= w3

3 4w and {2= w3

3 4w, |2= w with a w-interval of[33> 3] There are 9 points of intersection; (0> 0) is fairly obvious The point

in quadrant I is approximately (2=2> 2=2), and by symmetry, the point in

quadrant III is approximately (32=2> 32=2) The other six points are

approximately (~1=9> ±0=5), (~1=7> ±1=7), and (~0=5> ±1=9)

31. (a) { = {1+ ({23 {1)w, | = |1+ (|23 |1)w, 0 $ w $ 1 Clearly the curve passes through S1({1> |1) when w = 0 andthrough S2({2> |2) when w = 1 For 0 ? w ? 1, { is strictly between {1and {2and | is strictly between |1and |2 Forevery value of w, { and | satisfy the relation |3 |1= |23 |1

{23 {1({ 3 {1), which is the equation of the line through

Trang 18

32.For the side of the triangle from D to E, use ({1> |1) = (1> 1) and ({2> |2) = (4> 2)

Hence, the equations are

{ = {1+ ({23 {1) w = 1 + (4 3 1) w = 1 + 3w,

| = |1+ (|23 |1) w = 1 + (2 3 1) w = 1 + w

Graphing { = 1 + 3w and | = 1 + w with 0$ w $ 1 gives us the side of the

triangle from D to E Similarly, for the side EF we use { = 43 3w and | = 2 + 3w, and for the side DF we use { = 1and | = 1 + 4w

33.The circle {2

+ (| 3 1)2 = 4 has center (0> 1) and radius 2, so by Example 4 it can be represented by { = 2 cos w,

| = 1 + 2 sin w, 0 $ w $ 2 This representation gives us the circle with a counterclockwise orientation starting at (2> 1).(a) To get a clockwise orientation, we could change the equations to { = 2 cos w, | = 13 2 sin w, 0 $ w $ 2

(b) To get three times around in the counterclockwise direction, we use the original equations { = 2 cos w, | = 1 + 2 sin w withthe domain expanded to 0$ w $ 6

(c) To start at (0> 3) using the original equations, we must have {1= 0; that is, 2 cos w = 0 Hence, w =

2 So we use{ = 2 cos w, | = 1 + 2 sin w,

2 $ w $3

2 Alternatively, if we want w to start at 0, we could change the equations of the curve For example, we could use

{ = 32 sin w, | = 1 + 2 cos w, 0 $ w $

34. (a) Let {2@d2= sin2w and |2@e2= cos2w to obtain { = d sin w and

| = e cos w with 0 $ w $ 2 as possible parametric equations for the ellipse

{2@d2+ |2@e2= 1

(b) The equations are { = 3 sin w and | = e cos w for eM {1> 2> 4> 8}

(c) As e increases, the ellipse stretches vertically

35.Big circle: It’s centered at (2> 2) with a radius of 2, so by Example 4, parametric equations are

{ = 2 + 2 cos w> | = 2 + 2 sin w> 0 $ w $ 2

Small circles: They are centered at (1> 3) and (3> 3) with a radius of 0=1 By Example 4, parametric equations are

(left) { = 1 + 0=1 cos w> | = 3 + 0=1 sin w> 0 $ w $ 2

and (right) { = 3 + 0=1 cos w> | = 3 + 0=1 sin w> 0 $ w $ 2

Semicircle: It’s the lower half of a circle centered at (2> 2) with radius 1 By Example 4, parametric equations are

{ = 2 + 1 cos w> | = 2 + 1 sin w> $ w $ 2

To get all four graphs on the same screen with a typical graphing calculator, we need to change the last w-interval to[0> 2] inorder to match the others We can do this by changing w to 0=5w This change gives us the upper half There are several ways toget the lower half—one is to change the “+” to a “3” in the |-assignment, giving us

{ = 2 + 1 cos(0=5w)> | = 2 3 1 sin(0=5w)> 0 $ w $ 2

NOT FOR SALE

Trang 19

36. If you are using a calculator or computer that can overlay graphs (using multiple w-intervals), the following is appropriate

Left side: { = 1 and | goes from 1=5 to 4, so use

To get the x-assignment, think of creating a linear function such that when w = 0, { = 1 and when w = 2=5,

{ = 10 We can use the point-slope form of a line with (w1> {1) = (0> 1) and (w2> {2) = (2=5> 10)

{ 3 1 = 10 3 1

2=5 3 0(w 3 0) i { = 1 + 3=6w.

Handle: It starts at (10> 4) and ends at (13> 7), so use

{ = 10 + 1=2w> | = 4 + 1=2w> 0 $ w $ 2=5(w1> {1) = (0> 10) and (w2> {2) = (2=5> 13) gives us { 3 10 = 13 3 10

Trang 20

Left wheel: It’s centered at (3> 1), has a radius of 1, and appears to go about 30above the horizontal, so use

2>136 gives us 356 =

13

6 356 5

(c) { = h33w= (h3w)3 [so h3w= {1@3],

| = h32w= (h3w)2= ({1@3)2= {2@3

If w ? 0, then { and | are both larger than 1 If w A 0, then { and |

are between 0 and 1 Since { A 0 and | A 0, the curve never quite

reaches the origin

38. (a) { = w, so | = w32= {32 We get the entire curve | = 1@{2traversed in a

the curve when { A 0, that is, cos w A 0, and we get the second quadrant

portion of the curve when { ? 0, that is, cos w ? 0

(c) { = hw, | = h32w= (hw)32= {32 Since hwand h32ware both positive, we

only get the first quadrant portion of the curve | = 1@{2

NOT FOR SALE

Trang 21

39. The case

2 ? ? is illustrated F has coordinates (u> u) as in Example 7,

and T has coordinates (u> u + u cos(3 )) = (u> u(1 3 cos ))

[since cos(3 ) = cos cos + sin sin = 3 cos ], so S has

coordinates (u3 u sin( 3 )> u(1 3 cos )) = (u( 3 sin )> u(1 3 cos ))

[since sin(3 ) = sin cos 3 cos sin = sin ] Again we have the

parametric equations { = u(3 sin ), | = u(1 3 cos )

40. The first two diagrams depict the case ? ? 32 , g ? u As in Example 7, F has coordinates (u> u) Now T (in the seconddiagram) has coordinates (u> u + g cos(3 )) = (u> u 3 g cos ), so a typical point S of the trochoid has coordinates(u + g sin( 3 )> u 3 g cos ) That is, S has coordinates ({> |), where { = u 3 g sin and | = u 3 g cos When

g = u, these equations agree with those of the cycloid

41. It is apparent that { =|RT| and | = |TS | = |VW | From the diagram,

{ = |RT| = d cos and | = |VW | = e sin Thus, the parametric equations are

{ = d cos and | = e sin To eliminate we rearrange: sin = |@e i

sin2 = (|@e)2and cos = {@d i cos2 = ({@d)2 Adding the two

equations: sin2 + cos2 = 1 = {2@d2+ |2@e2 Thus, we have an ellipse

42. D has coordinates (d cos > d sin ) Since RD is perpendicular to DE, {RDE is a right triangle and E has coordinates(d sec > 0) It follows that S has coordinates (d sec > e sin ) Thus, the parametric equations are { = d sec , | = e sin

43. F = (2d cot > 2d), so the {-coordinate of S is { = 2d cot Let E = (0> 2d)

Then_RDE is a right angle and _RED = , so |RD| = 2d sin and

D = ((2d sin ) cos > (2d sin ) sin ) Thus, the |-coordinate of S

is | = 2d sin2

44. (a) Let be the angle of inclination of segment RS Then|RE| = cos 2d

Let F = (2d> 0) Then by use of right triangle RDF we see that|RD| = 2d cos

Now

|RS | = |DE| = |RE| 3 |RD|

= 2d

1cos 3 cos

= 2d1 3 cos2cos = 2d

sin2cos = 2d sin tan

So S has coordinates { = 2d sin tan · cos = 2d sin2

and | = 2d sin tan · sin = 2d sin2 tan

(b)

NOT FOR SALE

Trang 22

45. (a) There are 2 points of intersection:

2 satisfies (1) and (2) but w =

2 does not So the only collision pointoccurs when w =32, and this gives the point (33> 0) [We could check our work by graphing {1and {2together asfunctions of w and, on another plot, |1and |2as functions of w If we do so, we see that the only value of w for which both

pairs of graphs intersect is w =32.]

(c) The circle is centered at (3> 1) instead of (33> 1) There are still 2 intersection points: (3> 0) and (2=1> 1=4), but there are

no collision points, since (B) in part (b) becomes 5 cos w = 6 i cos w = 6

= 34=9k

w23250 4=9w +125

4=9

2l+12524=9 = 34=9

w 3125 4=9

2

+12524=9 $ 12524=9

with equality when w =1254=9 s, so the maximum height attained is1254=92 E 3189 m

(b) As (0? ? 90) increases up to 45, the projectile attains a

greater height and a greater range As increases past 45, theprojectile attains a greater height, but its range decreases

y0cos

2

= (tan ){ 3

j2y2cos2

{2

,which is the equation of a parabola (quadratic in {)

NOT FOR SALE

Trang 23

47. { = w2> | = w3

3 fw We use a graphing device to produce the graphs for various values of f with 3 $ w $ Note that allthe members of the family are symmetric about the {-axis For f ? 0, the graph does not cross itself, but for f = 0 it has acusp at (0> 0) and for f A 0 the graph crosses itself at { = f, so the loop grows larger as f increases

49. { = w + d cos w> | = w + d sin w> d A 0 From the first figure, we see that

curves roughly follow the line | = {, and they start having loops when d

is between 1=4 and 1=6 The loops increase in size as d increases

While not required, the following is a solution to determine the exact values for which the curve has a loop,

that is, we seek the values of d for which there exist parameter values w and x such that w ? x and

(w + d cos w> w + d sin w) = (x + d cos x> x + d sin x)

In the diagram at the left, W denotes the point (w> w), X the point (x> x),and S the point (w + d cos w> w + d sin w) = (x + d cos x> x + d sin x)

Since S W = S X = d, the triangle S W X is isosceles Therefore its baseangles, =_SWX and = _SXW are equal Since = w 3

Trang 24

Since W X = distance((w> w)> (x> x)) =s

2(x 3 w)2=I

2 (x 3 w), we see thatcos =

(2) Now cos

w 3 4

= sin

2 3

w 3 4

= sin3

4 3 w,

so we can rewrite (2) as x3 w =I2 d sin3

by w + 2 merely increases { and | by 2 The curve’s basic shape repeats every time we change w by 2.] Solving for d in

(3), we get d =

I

23

4 3 wsin3

4 3 w Write } =3

4 3 w Then d =

I

2 }sin }, where } A 0 Now sin } ? } for } A 0, so d A

I2.k

As }< 0+, that is, as w<3

4

3, d<I2l

50.Consider the curves { = sin w + sin qw, | = cos w + cos qw, where q is a positive integer For q = 1, we get a circle ofradius 2 centered at the origin For q A 1, we get a curve lying on or inside that circle that traces out q3 1 loops as wranges from 0 to 2

Note: {2+ |2= (sin w + sin qw)2+ (cos w + cos qw)2

= sin2w + 2 sin w sin qw + sin2qw + cos2w + 2 cos w cos qw + cos2qw

= (sin2w + cos2w) + (sin2qw + cos2qw) + 2(cos w cos qw + sin w sin qw)

= 1 + 1 + 2 cos(w 3 qw) = 2 + 2 cos((1 3 q)w) $ 4 = 22,with equality for q = 1 This shows that each curve lies on or inside the curve for q = 1, which is a circle of radius 2 centered

Trang 25

itself at the origin and there are loops above and below the {-axis In general, the figures have q3 1 points of intersection,all of which are on the |-axis, and a total of q closed loops

52. { = cos w, | = sin w 3 sin fw If f = 1, then | = 0, and the curve is simply the line segment from (31> 0) to (1> 0) The

graphs are shown for f = 2> 3> 4 and 5

It is easy to see that all the curves lie in the rectangle [31> 1] by [32> 2] When f is an integer, {(w + 2) = {(w) and

|(w + 2) = |(w), so the curve is closed When f is a positive integer greater than 1, the curve intersects the x-axis f + 1 times

and has f loops (one of which degenerates to a tangency at the origin when f is an odd integer of the form 4n + 1)

As f increases, the curve’s loops become thinner, but stay in the region bounded by the semicircles | =±

1 +I

1 3 {2and the line segments from (31> 31) to (31> 1) and from (1> 31) to (1> 1) This is true because

||| = |sin w 3 sin fw| $ |sin w| + |sin fw| $I1 3 {2+ 1 This curve appears to fill the entire region when f is very large, asshown in the figure for f = 1000

NOT FOR SALE

Trang 26

LABORATORY PROJECT RUNNING CIRCLES AROUND CIRCLES ¤ 15

When f is a fraction, we get a variety of shapes with multiple loops, but always within the same region For some fractionalvalues, such as f = 2=359, the curve again appears to fill the region

LABORATORY PROJECT Running Circles Around Circles

1.The center T of the smaller circle has coordinates ((d3 e)cos > (d 3 e)sin )

Arc S V on circle F has length d since it is equal in length to arc DV

(the smaller circle rolls without slipping against the larger.)

Thus,_STV =d

e and_STW = d

e 3 , so S has coordinates{ = (d 3 e)cos + e cos(_S TW ) = (d 3 e)cos + e cos

2.With e = 1 and d a positive integer greater than 2, we obtain a hypocycloid of d

cusps Shown in the figure is the graph for d = 4 Let d = 4 and e = 1 Using the

sum identities to expand cos 3 and sin 3, we obtain

{ = 3 cos + cos 3 = 3 cos +

4 cos3

3 3 cos

= 4 cos3and | = 3 sin 3 sin 3 = 3 sin 3

3 sin 3 4 sin3

= 4 sin3

3.The graphs at the right are obtained with e = 1 and

d = 12, 13, 14, and101 with32 $ $ 2 We

conclude that as the denominator g increases, the graph

gets smaller, but maintains the basic shape shown

Trang 27

Letting g = 2 and q = 3, 5, and 7 with32 $ $ 2 gives us the following:

So if g is held constant and q varies, we get a graph with q cusps (assuming q@g is in lowest form) When q = g + 1, weobtain a hypocycloid of q cusps As q increases, we must expand the range of in order to get a closed curve The followinggraphs have d =32, 54, and1110

4. If e = 1, the equations for the hypocycloid are

{ = (d 3 1) cos + cos ((d 3 1) ) | = (d 3 1) sin 3 sin ((d 3 1) )which is a hypocycloid of d cusps (from Problem 2) In general, if d A 1, we get a figure with cusps on the “outside ring” and

if d ? 1, the cusps are on the “inside ring” In any case, as the values of get larger, we get a figure that looks more and morelike a washer If we were to graph the hypocycloid for all values of , every point on the washer would eventually be arbitrarilyclose to a point on the curve

d =I

2, 310 $ $ 10 d = h 3 2, 0 $ $ 446

5. The center T of the smaller circle has coordinates ((d + e) cos > (d + e) sin )

Arc S V has length d (as in Problem 1), so that_STV =d

e ,_STU = 3d

e,and_STW = 3 d

e 3 = 3

d + ee

NOT FOR SALE

Trang 28

LABORATORY PROJECT RUNNING CIRCLES AROUND CIRCLES ¤ 17

6.Let e = 1 and the equations become

{ = (d + 1) cos 3 cos((d + 1)) | = (d + 1) sin 3 sin((d + 1))

If d = 1, we have a cardioid If d is a positive

integer greater than 1, we get the graph of an

“d-leafed clover”, with cusps that are d units

from the origin (Some of the pairs of figures are

not to scale.)

d = 3, 32 $ $ 2 d = 10, 32 $ $ 2

If d = q@g with q = 1, we obtain a figure that

does not increase in size and requires

3g $ $ g to be a closed curve traced

exactly once

d = 1

4,34 $ $ 4 d =1

7,37 $ $ 7

Next, we keep g constant and let q vary As q

increases, so does the size of the figure There is

an q-pointed star in the middle

d = 2

5,35 $ $ 5 d =7

5,35 $ $ 5

Now if q = g + 1 we obtain figures similar to the

previous ones, but the size of the figure does not

Trang 29

10.2 Calculus with Parametric Curves

1. { = w sin w, | = w2+ w i g|gw = 2w + 1,g{

gw = w cos w + sin w, andg|

g{= g|@gwg{@gw= 2w + 1

I

3and g|@g{ =3 cot(@6) = 3I3,

so an equation of the tangent line to the curve at the point corresponding to = @6 is |33

8

I

3 = 3I3

{ 31 8

,

or | =3I3 { +1

2

I3

Trang 30

SECTION 10.2 CALCULUS WITH PARAMETRIC CURVES ¤ 19

2 The interval [0> 2] gives the complete curve, so we need only find

the values of w in this interval Thus, w = 2 or w = 23 or w =43 Checking w = 2,32, 23 , and43 in the equation for |,

we find that w =2 corresponds to (31> 1) The slope of the tangent at (31> 1) with w =2 is 0 3 2

31 3 0= 2 An equation

of the tangent is therefore |3 1 = 2({ + 1), or | = 2{ + 3

11.{ = w2+ 1, | = w2+ w i g|

g{ =g|@gwg{@gw =2w + 1

2w = 1 +

12w i g

2

|g{2 =

ggw

g|

g{

g{@gw = 31@(2w

2

)2w = 3 1

4w3.The curve is CU when g

2|g{2 A 0, that is, when w ? 0

12.{ = w3+ 1, | = w2

3 w i g|

g{ = g|@gwg{@gw = 2w 3 1

3w2 = 2

3w3 13w2 i

3w3

3w2 =

2 3 2w3w3

3w2 = 2(1 3 w)

9w5 The curve is CU wheng

2

|g{2 A 0, that is, when 0 ? w ? 1

ggw

g|

g{

g{@gw =

2whw3 hw· 2(2w)2

2w =

2hw

(w 3 1)(2w)3 =h

w

(w 3 1)4w3 The curve is CU when g

2|g{2 A 0, that is, when w ? 0 or w A 1

NOT FOR SALE

Trang 31

ggw

g|

g{

g{@gw = 33

ggw

g|

g{

g{@gw =

gw = 3w2

3 3 = 3(w + 1)(w 3 1), sog{

gw = 0 C

w = 31 or 1 C ({> |) = (2> 32) or (32> 32) The curve has a horizontal

tangent at (0>33) and vertical tangents at (2> 32) and (32> 32)

31

2> 1, or (31> 31)

g{

g = 3 sin , sog{

g = 0 C sin = 0 C = 0 or C({> |) = (1> 1) or (31> 31) Bothg|gand g{

g equal 0 when = 0 and .

To find the slope when = 0, we find lim

31

2> 1, and there are no vertical tangents

NOT FOR SALE

Trang 32

20.{ = hsin , | = hcos The whole curve is traced out for 0$ ? 2

g|

g = 3 sin hcos , sog|

g = C sin = 0 C = 0 or C({> |) = (1> h) or (1> 1@h) g{

g = cos hsin , sog{

g = 0 C cos = 0 C

=

2 or 3

2 C ({> |) = (h> 1) or (1@h> 1) The curve has horizontal tangents

at (1> h) and (1> 1@h), and vertical tangents at (h> 1) and (1@h> 1)

21.From the graph, it appears that the rightmost point on the curve { = w3 w6

, | = hw

is about (0=6> 2) To find the exact coordinates, we find the value of w for which the

graph has a vertical tangent, that is, 0 = g{@gw = 13 6w5

3 2w, | = w + w4are (1=5>30=5) and (31=2> 1=2), respectively To find the

exact coordinates, we solve g|@gw = 0 (horizontal tangents) and g{@gw = 0

=

9

E (31=19> 1=19)

23.We graph the curve { = w43 2w33 2w2, | = w33 w in the viewing rectangle [32> 1=1] by [30=5> 0=5] This rectanglecorresponds approximately to wM [31> 0=8]

We estimate that the curve has horizontal tangents at about (31> 30=4) and (30=17> 0=39) and vertical tangents at

about (0> 0) and (30=19> 0=37) We calculateg|g{= g|@gw

g{@gw= 3w

2

3 14w33 6w23 4w The horizontal tangents occur wheng|@gw = 3w2

3 1 = 0 C w = ± 1

I

3, so both horizontal tangents are shown in our graph The vertical tangents occur when

NOT FOR SALE

Trang 33

g{@gw = 2w(2w2

3 3w 3 2) = 0 C 2w(2w + 1)(w 3 2) = 0 C w = 0, 31

2 or 2 It seems that we have missed one verticaltangent, and indeed if we plot the curve on the w-interval [31=2> 2=2] we see that there is another vertical tangent at (38> 6)

24. We graph the curve { = w4+ 4w33 8w2, | = 2w23 w in the viewing rectangle [33=7> 0=2] by [30=2> 1=4] It appears that there

is a horizontal tangent at about (30=4> 30=1), and vertical tangents at about (33> 1) and (0> 0)

+ 3w 3 4) = 0 C 4w(w + 4)(w 3 1) = 0 We have missed one vertical tangent corresponding to w = 34, and if weplot the graph for wM [35> 3], we see that the curve has another vertical tangent line at approximately (3128> 36)

25. { = cos w, | = sin w cos w g{@gw = 3 sin w, g|@gw = 3 sin2w + cos2w = cos 2w

({> |) = (0> 0) C cos w = 0 C w is an odd multiple of

2 When w =

2,g{@gw = 31 and g|@gw = 31, so g|@g{ = 1 When w =3

2 , g{@gw = 1 andg|@gw = 31 So g|@g{ = 31 Thus, | = { and | = 3{ are both tangent to the

curve at (0> 0)

26. From the graph, we discover that the graph of the curve { = cos w + 2 cos 2w,

| = sin w + 2 sin 2w crosses itself at the point (32> 0) To find w at (32> 0),solve | = 0 C sin w + 2 sin 2w = 0 C sin w + 4 sin w cos w = 0 Csin w (1 + 4 cos w) = 0 C sin w = 0 or cos w = 31

4 We find that

w = ± arccos

314

corresponds to (32> 0)

31 4

, cos w =31

4, sin w =

I15

4 3I15 2

= 3 3

15 4

3I15 4

= 3I15 By symmetry, w = 3 arccos

31 4

so the trochoid can have no vertical tangent if g ? u

NOT FOR SALE

Trang 34

28.{ = d cos3, | = d sin3

(a) g{

g = 33d cos2 sin ,g|

g = 3d sin2 cos , sog|

g{ = 3cos sin = 3 tan

(b) The tangent is horizontal C g|@g{ = 0 C tan = 0 C = q C ({> |) = (±d> 0)

The tangent is vertical C cos = 0 C is an odd multiple of

2 C ({> |) = (0> ±d) =(c) g|@g{ =±1 C tan = ±1 C is an odd multiple of

4 C ({> |) =

±I2

4 d> ±I2

4 d[All sign choices are valid.]

29.{ = 2w3, | = 1 + 4w3 w2 i g|g{ =g|@gw

g{@gw =4 3 2w

6w2 Now solveg|

g{= 1 C 4 3 2w6w2 = 1 C6w2+ 2w 3 4 = 0 C 2(3w 3 2)(w + 1) = 0 C w =23 or w =31 If w = 23, the point is16

27>299, and if w =31,the point is (32> 34)

If this line is to pass through (4> 3), we must have 33 (2w3+ 1) = w[4 3 (3w2+ 1)] C 2w33 2 = 3w33 3w C

w33 3w + 2 = 0 C (w 3 1)2(w + 2) = 0 C w = 1 or 32 Hence, the desired equations are | 3 3 = { 3 4, or

| = { 3 1, tangent to the curve at (4> 3), and | 3 (315) = 32({ 3 13), or | = 32{ + 11, tangent to the curve at (13> 315)

31.By symmetry of the ellipse about the {- and |-axes,

] 2 0

(w23 2w)

1

2I

wgw

= 3U2 0

= 3I2

38 15

= 8 15

I2

33.The curve { = 1 + hw, | = w3 w2

= w(1 3 w) intersects the {-axis when | = 0,that is, when w = 0 and w = 1 The corresponding values of { are 2 and 1 + h

The shaded area is given by

NOT FOR SALE

Trang 35

] {=1+h

{=2

(|W3 |E) g{ =

] w=1 w=0

U(1 3 cos 2) sin22 g

=1 8

0 (u 3 g cos )(u 3 g cos ) g =U2

0 (u23 2gu cos + g2cos2) g

36. (a) By symmetry, the area ofR is twice the area inside R above the {-axis The top half of the loop is described by

{ = w2, | = w33 3w, 3I3 $ w $ 0, so, using the Substitution Rule with | = w33 3w and g{ = 2w gw, we find that

of the centroid of the top half ofR, the area of which is1

U3

0 {| g{ = 5

12 I 3

U3 3

0 w2(w3

3 3w)2w gw = 5

6 I 3

1

7w7

33

5w53 3 0

= 6I53

k

1

7(331@2)7335(331@2)5l

= 6I53

37. { = w + h3w, | = w3 h3w, 0$ w $ 2 g{@gw = 1 3 h3wand g|@gw = 1 + h3w, so

(g{@gw)2+ (g|@gw)2

= (1 3 h3w)2+ (1 + h3w)2

= 1 3 2h3w+ h32w+ 1 + 2h3w+ h32w= 2 + 2h32w.Thus, O =Ue

d

s(g{@gw)2+ (g|@gw)2gw =U2

Trang 36

39.{ = w 3 2 sin w, | = 1 3 2 cos w, 0 $ w $ 4 g{@gw = 1 3 2 cos w and g|@gw = 2 sin w, so

(g{@gw)2+ (g|@gw)2= (1 3 2 cos w)2+ (2 sin w)2= 1 3 4 cos w + 4 cos2w + 4 sin2w = 5 3 4 cos w

2

= 1 +I1

w+ 14w+ 1 3I1

w+ 14w = 2 +

12w.Thus, O =

] e

d

s(g{@gw)2+ (g|@gw)2gw =

] 1 0

u

2 + 12wgw = limw <0 +

]1 w

u

2 + 12wgw E 2=0915

]1 0

6ws

1 + w2gw = 6

] 2 1

I

x1

2gx[x = 1 + w 2 , gx = 2w gw]

3 2 + h32w+ 4 = h2w+ 2 + h32w= (hw+ h3w)2

.Thus, O =U3

= w2cos2w + 2w sin w cos w + sin2w + w2sin2

w 3 2w sin w cos w + cos2w

= w2(cos2w + sin2w) + sin2w + cos2w = w2+ 1

44.{ = 3 cos w 3 cos 3w, | = 3 sin w 3 sin 3w, 0 $ w $ . g{

gw = 33 sin w + 3 sin 3w andg|gw = 3 cos w 3 3 cos 3w, so

gw

2

= 9 sin2

w 3 18 sin w sin 3w + 9 sin2(3w) + 9 cos2

w 3 18 cos w cos 3w + 9 cos2(3w)

= 9(cos2w + sin2

w) 3 18(cos w cos 3w + sin w sin 3w) + 9[cos2(3w) + sin2(3w)]

= 9(1) 3 18 cos(w 3 3w) + 9(1) = 18 3 18 cos(32w) = 18(1 3 cos 2w)

Trang 37

(cos w 3 sin w)]2+ [hw(sin w + cos w)]2

= (hw)2(cos2w 3 2 cos w sin w + sin2w)

+ (hw)2(sin2w + 2 sin w cos w + cos2w

= h2w(2 cos2w + 2 sin2w) = 2h2wThus, O =U

0

I2h2wgw =U

47. The figure shows the curve { = sin w + sin 1=5w, | = cos w for 0$ w $ 4

g{@gw = cos w + 1=5 cos 1=5w and g|@gw = 3 sin w, so(g{@gw)2+ (g|@gw)2= cos2w + 3 cos w cos 1=5w + 2=25 cos21=5w + sin2w.Thus, O =U4

2 + 2h2w Then by Simpson’s Rule with q = 6 and {w = 63(36)6 = 2, we get

O E2

3[i (36) + 4i(34) + 2i(32) + 4i(0) + 2i(2) + 4i(4) + i(6)] E 612=3053

Trang 38

50.{ = 2d cot i g{@gw = 32d csc2 and | = 2d sin2 i g|@gw = 4d sin cos = 2d sin 2

16

+ 2i3

8

+ 4i7

16

+ i

2, because the curve is the segment of { + | = 1 that lies in the first quadrant(since {, |D 0), and this segment is completely traversed as w goes from 0 to

31

I4x2+ 1 gx

= 8U1 0

I4x2+ 1 gx = 8Utan12

= (33d cos2 sin )2+ (3d sin2 cos )2

= 9d2cos4 sin2 + 9d2sin4 cos2

= 9d2sin2 cos2(cos2 + sin2) = 9d2sin2 cos2

The graph has four-fold symmetry and the curve in the first quadrant corresponds

Trang 39

55. (a) { = 11 cos w3 4 cos(11w@2), | = 11 sin w 3 4 sin(11w@2)

Notice that 0$ w $ 2 does not give the complete curve because

{(0) 6= {(2) In fact, we must take w M [0> 4] in order to obtain the

complete curve, since the first term in each of the parametric equations has

period 2 and the second has period 2

11@2 = 4

11, and the least commoninteger multiple of these two numbers is 4

(b) We use the CAS to find the derivatives g{@gw and g|@gw, and then use Theorem 6 to find the arc length Recent versions

of Maple express the integralU4

0

s(g{@gw)2+ (g|@gw)2gw as 88H

2I

2 l, where H({) is the elliptic integral] 1

56. (a) It appears that as w< ", ({> |) <1

2>1 2

, and as w< 3", ({> |) <

31

2> 31 2

.(b) By the Fundamental Theorem of Calculus, g{@gw = cos

2w2andg|@gw = sin

2w2, so by Formula 4, the length of the curve from the origin

to the point with parameter value w is

tcos2

2x2+ sin2

2x2gx

=Uw

01 gx = w [or 3w if w ? 0]

We have used x as the dummy variable so as not to confuse it with the upper limit of integration

57. { = w sin w, | = w cos w, 0 $ w $ @2 g{@gw = w cos w + sin w and g|@gw = 3w sin w + cos w, so

(g{@gw)2+ (g|@gw)2= w2cos2w + 2w sin w cos w + sin2w + w2sin2w 3 2w sin w cos w + cos2w

= w2(cos2w + sin2w) + sin2w + cos2w = w2+ 1

Trang 40

2w2s9w4+ 4w2gw = 2

] 1 0

w2s

w2(9w2+ 4) gw

= 2

] 13 4

x 3 49

I

k3x5@2

= (32 sin + 2 sin 2)2+ (2 cos 3 2 cos 2)2

= 4[(sin2 3 2 sin sin 2 + sin22) + (cos2 3 2 cos cos 2 + cos22)]

= 4[1 + 1 3 2(cos 2 cos + sin 2 sin )] = 8[1 3 cos(2 3 )] = 8(1 3 cos )

We plot the graph with parameter interval [0> 2], and see that we should only

integrate between 0 and (If the interval [0> 2] were taken, the surface of

revolution would be generated twice.) Also note that

| = 2 sin 3 sin 2 = 2 sin (1 3 cos ) So

I2(25@2) =128

Định dạng
Số trang	751
Dung lượng	34,87 MB