ODDREV11

85. Let Then and Then,

xtxt ytyt ztzt r r.

2xtxt 2ytyt 2ztztrdrdt xt2 yt2 zt2

12xt2 yt2 zt212

r xti ytj ztk.r r xt2 yt2 zt2r xti ytj ztk.

87. Let where and are functions of and

(using Exercise 77)

1r3 iyz yzx jxz xzy kxy xyz 1r3 r r r

1r3

xy2 xz2 xyy xzzi x2y z2y xxy zzyj x2z y2z xxz yyzk

x2 y2 z2xi yj zk xx yy zzxi yj zk

r3

ddt

r

r rr rdrdt

r2

rr rr rrr2

r2r r rr

r3

r r.t,zy,x,r xi yj zk

89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have

Since and are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate thecoordinate system so that e lies along the positive x-axis and is the angle between e and r. Let

Then Also,

Thus,

and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse.

L2GM1 e cos r

GMr e r rr GMre cos r r r L r GMe r

rL2 L L r r L

r e r e cos re cos .e e.

rr L

r L GMrr e.

x

r

e

Sun

Planet

y

91.

Thus,

and r sweeps out area at a constant rate.

dAdt

dAd

ddt

12 r

2

ddt

12 L

A 12

r2 d

Review Exercises for Chapter 11

1.

(a) Domain: n an integer(b) Continuous except at , n an integert n

t n,

rt ti csc tk 3.

(a) Domain:(b) Continuous for all t > 0

0,

rt ln ti tj tk

68 Chapter 11 Vector-Valued Functions

5. (a)(b)(c)

(d) 2t i tt 2j 13t 3 3t2 3tk

r1 t r1 21 t 1i 1 t2j 13 1 t3k 3i j 13 k 2c 1i c 12j 13 c 13k

rc 1 2c 1 1i c 12j 13 c 13kr2 3i 4j 83 kr0 i

13.

y

x

1 21

2

3

3

2

1

z

rt ti ln tj 12 t2k 15. One possible answer is:

r3t 4 ti, 0 t 4

r2t 4i 3 tj, 0 t 3r1t 4ti 3tj, 0 t 1

17. The vector joining the points is One path isrt 2 7t, 3 4t, 8 10t.

7, 4, 10. 19.

x

y

5

1 2 33

32

z

rt ti tj 2t2kz 2t2y t,x t,

t xx y 0,z x2 y2,

21. limt2

t2i 4 t2j k 4i k

7.

1

11x

y

1 x 1

y 21 x2

x2 y2 1

yt 2 sin2 txt cos t,

rt cos ti 2 sin2 tj 9.

x

y2

1

2

1

z

z t2 z y2y t

x 1

rt i tj t2k 11.

y

x

1 212

3

2

2

3

1

z

z 1y sin t,x 1,

rt i sin tj k

t 0

x 1 1 1 1

y 0 1 0

z 1 1 1 1

1

32

2

Review Exercises for Chapter 11 69

25. and are increasing functions at and is adecreasing function at t t0.

ztt t0,ytxt 27. cos t i t cos tj dt sin t i t sin t cos tj C

29. cos t i sin tj tk dt 1 t2 dt 12t1 t2 lnt 1 t2 C

31.

rt t2 1i et 2j et 4kr0 j k C i 3j 5k C i 2j 4k

rt 2ti etj etk dt t2i etj etk C 33. 22

3t i 2t2j t3k dt 3t2

2 i 2t33 j

t4

4 k2

2

323 j

35. 2e 2i 8j 2k 20

et2i 3t2j k dt 2et2i t3j tk2

0

37.

3 cos t2 sin2 t cos2 t, 3 sin t2 cos2 t sin2 t, 0

6 sin t cos2 t 3 sin2 tsin t, 0at vt 6 cos tsin2 t 3 cos2 t cos t,

3cos2 t sin2 t 1

3cos2 t sin2 tcos2 t sin2 t 1

vt 9 cos4 t sin2 t 9 sin4 t cos2 t 9

vt rt 3 cos2 t sin t, 3 sin2 t cos t, 3

rt cos3 t, sin3 t, 3t

39.

direction numbers

Since the parametric equations are

rt0 0.1 r4.1 0.1, 16.8, 2.05z 2 12 t.y 16 8t,x t,

r4 0, 16, 2,

r4 1, 8, 12

rt 1t 3, 2t, 12

t0 4rt lnt 3, t2, 12t, 41. Range feet x v0

2

32 sin 2 752

32 sin 60 152

23.

(a)

(c)

(e)

Dtrt 10t 1

10t2 2t 1

rt 10t2 2t 1

Dtrt ut 3t2 4t

rt ut 3t2 t2t 1 t3 2t2

rt 3i j

ut ti t2j 23t3krt 3ti t 1j,

(b)

(d)

(f )

Dtrt ut 83t3 2t 2i 8t3j 9t2 2t 1k

rt ut 23t4 t3i 2t4j 3t3 t 2 tk

Dtut 2rt 5i 2t 2j 2t 2k

ut 2rt 5ti t2 2t 2j 23t3k

rt 0

43. v0 809.8sin 40 34.9 m secRange x v02

9.8 sin 2 80


45.

does not exist

does not exist

(The curve is a line.) a N

a T 0

Nt

Tt i

at 0

vt 5

vt 5i

rt 5ti 47.

a N 1

2t4t 1

a T 1

4tt4t 1

Nt i 2t j

4t 1

Tt i 12t j4t 1 2t 2t i j4t 1

at 14tt

j

vt 4t 1

2t

vt i 12t

j

rt ti tj

49.

a N 2

e2t e2t

a T e2t e2t

e2t e2t

Nt et i etj

e2t e2t

Tt eti etj

e2t e2t

at et i et jvt e2t e2t

vt eti etj rt eti etj 51.

a N 5

51 5t2

51 5t2

a T 5t

1 5t2

Nt 5t i 2j k51 5t2

Tt i 2tj tk1 5t2

at 2j kv 1 5t2

vt i 2tj tk

rt ti t2j 12t2k

53.

When

Direction numbers when

z t 34y

2t 2,x 2t 2,

c 1b 2,a 2,t 34 ,

rt 2 sin ti 2 cos tj k

z 34 .y

2,x 2,t 34 ,

z ty 2 sin t,x 2 cos t,rt 2 cos ti 2 sin tj tk,

55. v 9.56 1044600 4.56 mi sec

57.

13t5

0 513

s ba

rt dt 504 9 dt

rt 2i 3j224

4

2

6810121416

4 6 8 10 12 14x

y

(0, 0)

(10, 15)

rt 2ti 3tj, 0 t 5

Review Exercises for Chapter 11 71

59.

s 420

30 cos t sin t dt 120 sin2 t

2 2

0 60

30cos t sin t rt 30cos4 t sin2 t sin4 t cos2 t

rt 30 cos2 t sin ti 30 sin2 t cos tj

10

10

2

10 102 2x

y rt 10 cos3 ti 10 sin3 tj

61.

s ba

rt dt 309 4 16 dt 3

029 dt 329

rt 3i 2j 4k

xy6 8

2 410

22

64

81012

z

(0, 0, 0)

(9, 6, 12) rt 3ti 2tj 4tk, 0 t 3

63.

x

y

z

4

468

68

(8, 0, 0)

(0, 8, )22

s ba

rt dt 20

65 dt 652

rt < 8 sin t, 8 cos t, 1, rt 65

rt 8 cos t, 8 sin t, t, 0 t 2 65.

52

0 dt 52 t

0

52

014 cos2 t sin2 t dt

s 0

rt dt

rt 12 i cos tj sin tk

0 t rt 12ti sin tj cos tk,

67.

Line

k 0

rt 3ti 2tj 69.

K r rr3

20

5t2 432 25

4 5t232

r r i20 jt1 k2t2 4j 2k, r r 20 rt j 2k rt 2i tj 2tk, r 5t2 4

rt 2ti 12t2j t2k

71.

At and r 1732 1717.x 4, K 11732

K y1 y232 1

1 x232

y 1

y x

y 12x

2 273.

At and r 22.x 1, K 1232

1

22

24

K y1 y232 1x2

1 1x2]32

y 1x, y

1x2

y ln x

75. The curvature changes abruptly from zero to a nonzeroconstant at the points B and C.


Problem Solving for Chapter 11

1.

(a)

(b)

At

(c) K a lengtht a, K a.

K t cos2t2

2 t sin2t2

2 1 t

xt t sint2

2 , yt t cost2

2

s a0xt2 yt2 dt a

0dt a

xt cost2

2 , yt sint2

2

xt t0cosu

2

2 du, yt t

0sinu

2

2 du 3. Bomb:Projectile:At 1600 feet: Bomb:

seconds.

Projectile will travel 5 seconds:

Horizontal position:

At bomb is at

At projectile is at Thus,

Combining,

v0 200

cos 447.2 ftsec

v0 sin v0 cos

400200 tan 2 63.4.

v0 cos 200.

5v0 cos .t 5,

5000 40010 1000.t 10,

v0 sin 400.

5v0 sin 1625 1600

3200 16t2 1600 t 10

r2t v0 cos t, v0 sin t 16t2

r1t 5000 400t, 3200 16t2

5.

Thus, and

s2 2 16 cos2 t2 16 sin2t2 16.

1K 4 sin

t2

1

4 sin 2

K 11 cos cos sin sin 2 sin 2

3 1 cos 1

8 sin3 2

x sin , y cos

st t

2 sin 2 d 4 cos

2t

4 cos t2

2 2 cos 4 sin2 2

x2 y2 1 cos 2 sin2

x 1 cos , y sin , 0 2

7.

ddt rt

rt rtrt rt rt rt rt

ddt rt

2 2rt ddt rt

r2t rt rt

Problem Solving for Chapter 11 73

9.

x

y

z

3

123

4 4

6 T

T

B

B

N

N

B2 35i

45k

N2 j

At t 2, T

2 45i

35k

B T N 35 sin ti 35 cos tj

45k

N cos ti sin tj

T 45 cos ti 45 sin tj

T 45 sin ti 45 cos tj

35k

rt 4 cos ti 4 sin tj rt 4 sin ti 4 cos tj 3k, rt 5

rt 4 cos ti 4 sin tj 3tk, t 2 11. (a)

Hence, and

for some scalar

(b) Using Exercise 10.3, number 64,

Now,

Finally,

KT B.

B KN N T

Ns dds B T B T B T

KN dTds TsTs Ts dTds . N.

T NT T TN

B T T N T T T N

T

N NT N TN

B N T N N N T N

B T N.

.

dBds T

dBds N

dBds B

T T N T T TT 0

T dBds T T N T T N

dBds

dds T N T N T N

constant length dBds BB T N 1

13.

(a)

(c)

(e) limt

K 0

K2 0.51

K1 2 2

2 132 1.04

K0 2

K 2t2 2

2t2 132

3 3

2

2

0 t 2rt t cos t, t sin t,(b)

(graphing utility)

(d)

(f ) As the graph spirals outward and the curvaturedecreases.

t,

0 50

5

202t2 1 dt 6.766

Length 20

rt dt


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