# CSM_Chapters12.pdf

September 14, 2017 | Author: Clau Amaiia | Category: Trigonometric Functions, Calculus, Functions And Mappings, Analysis, Algebra

#### Description

Chapter 12

Vector-Valued Functions 12.1

Vector Functions

2 1. Since the square root function S is only defined for nonnegative values, we must have t − 9 ≥ 0. So the domain is (−∞, −3) [3, ∞).

2. Since the natural logarithm is only defined for positive values, we must have 1 − t2 > 0. So the domain is (−1, 1). 3. Since the inverse sine function is only defined for values between -1 and 1, the domain is [−1, 1]. 4. The vector function is defined for all real numbers. 5. r(t) = sin πti + cos πtj − cos2 πtk 6. r(t) = cos2 πti + 2 sin2 πtj + t2 k 7. r(t) = e−t i + e2t j + e3t k 8. r(t) = −16t2 i + 50tj + 10k 9. x = t2 ,

y = sin t,

z = cos t

10. r(t) = t sin t(i + k) = t sin ti + 0j + t sin tk so x = t sin t, 11. x = ln t,

y = 1 + t,

12. x = 5 sin t sin 3t,

z = t3

y = 5 cos 3t,

z = 5 cos t sin 3t 48

y = 0,

z = t sin t

12.1. VECTOR FUNCTIONS

13.

49

14.

z

15.

z

z

y

y

4 y

x

x

16.

x

17.

z

18.

y

y

2

y

x

2

x

x

19.

20.

z

z

y

y x

x

21. z

y x

Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper loop shown intersects the xz-plane at about (286751, 0, 286751).

50

CHAPTER 12. VECTOR-VALUED FUNCTIONS

22.

23.

z

z

y

10 10

x

10

y

x

24. z

y x

25. r(t) = h4, 0i + h0 − 4, 3 − 0it = (4 − 4t)i + 3tj, 0 ≤ t ≤ 1 y

x

26. r(t) = h0, 0, 0i + h1 − 0, 1 − 0, 1 − 0it = ti + tj + tk, 0 ≤ t ≤ 1

12.1. VECTOR FUNCTIONS 27. x = t, y = t, z = t2 + t2 = 2t2 ; r(t) = ti + tj + 2t2 k z

y

x

√ √ √ 28. x = t, y = 2t, z = ± t2 + 4t2 + 1 = ± 5t2 − 1; r(t) = ti + 2tj ± 5t2 − 1k z

y x

29. x = 3 cos t, z = 9 − 9 cos2 t = 0 sin2 t; y = 3 sin t; r(t) = 3 cos ti + 3 sin tj + 9 sin2 tk

z

y x

30. x = sin t, z = 1, y = cos t; r(t) = sin ti + cos tj + k

51

52

CHAPTER 12. VECTOR-VALUED FUNCTIONS

z

y x

31. x = t, y = t, z = 1 − 2t; r(t) = ti + tj + (1 − 2t)k

z

y x

32. x = 11,

y = t,

z = 3 + 2t;

r(t) = i + tj + (3 + 2t)k

z

y x

33. (b); Notice that the y and z values consistently increase while the x values oscillate rapidly between -1 and 1. The only vector fucntion that describes this behavior is (b). 34. (c); The trace of the graph on the xy−plane would look like a circle, while the z value oscillates between 0 and 1. The only vector function that describes this behavior is (c). 35. (d); Notice that the z value is contant. The only vector function that satisfies this constraint is (d). 36. (a); Notice that the x values consistently increase while the trace of the graph on the yz-plane would look like a circle. The only vector function that describes this behavior is (a).

12.1. VECTOR FUNCTIONS

53

37. Letting x = at cos t, y = bt sin t, and z = ct, we have c2 t 2 z2 = 2 = t2 = t2 cos2 t + t2 sin2 t 2 c c a2 t2 cos2 t b2 t2 sin2 + = a2 b2 x2 y2 = 2 + 2 a b

38. z

y

x

39. Letting x = aekt cos t, y = bekt sin t, and z = cekt , we have c2 ekt z2 = = e2kt = e2kt cos2 t + e2kt sin2 t c2 c2 a2 e2kt cos2 t b2 e2kt sin2 t = + a2 b2 2 2 x y + 2 a2 b

40. z

y x

41. x2 + y 2 + z 2 = a2 sin2 kt cos2 t + a2 sin2 kt sin2 t + a2 cos2 kt = a2 sin2 kt + a2 cos2 kt = a2

54 42.

CHAPTER 12. VECTOR-VALUED FUNCTIONS k=1

k=2

k=3 z z

z

y

y

y

x

x

x

k=4

k = 10

k = 20

z

z

z

y

y

y

x

x

43. (a) z

y

x

(b) r1 (t) = ti + tj + (4 − t2 )k r2 (t) = ti − tj + (4 − t2 )k (c) z

y

x

44. C lies on the surface of the sphere of radius a.

x

12.2. CALCULUS OF VECTOR FUNCTIONS

55

45.

46.

k = 0.1

k = 0.2

k = 0.3 z

z

z

y y

y x

x

47.

x

k=2

k=4 z

z

y

y

x

48.

k=

x

1 10

k=1 z

z

y x

12.2

y x

Calculus of Vector Functions

1. lim [t3 i + t4 j + t5 k] = 23 i + 24 j + 25 k = 8i + 16j + 32k t→2

56

CHAPTER 12. VECTOR-VALUED FUNCTIONS 2. r(t) =

sin 2t ln t i + (t − 2)5 k + k. Using L’Hˆ opital’s Rule, t 1/t  lim+ r(t) =

t→0

 1/t 2 cos 2t k = 2i − 32j i + (t − 2)5 j + 1 −1/t2

3. Using opital’s Rule, we have  L’Hˆ    t2 − 1 5t − 1 2et−1 − 2 2t 5t − 1 2et−1 lim , , = lim = , , i = h2, 2, 2 t→1 t→1 t−1 t+1 t−1 1 t+1 1 π 4. Since lim tan−1 t = , we have t→∞ 2     e2t 1 e−1 1 −1 −1 lim , , tan t , , tan t = lim t→∞ 2e2t + t 2e−t + 5 t→∞ 2 + te−2t 2 + 5et   1 π = , 0, 2 2 The last equality follows from using L’Hˆ opital’s Rule to get lim te−2t = lim

t→∞

t→∞

1 t = lim =0 t→∞ 2e2t e2t

5. lim [−4r1 (t) + 3r2 (t)] = −4(i − 2j + k) + 3(2i + 5j + 7k) = 2i + 23j + 17k t→α

6. lim r1 (t) · r2 (t) = (i − 2j + k) = (i − 2j + k) · (2i + 5j + 7k) = −1 t→α

7. Notice that the k component ln(t − 1) is not defined at t = 1. Therefore, r(t) is not continuous at t = 1. 8. Notice that sin πt, tan πt, and cos πt are each continuous at t = 1 since the sine, cosine, and tangent function are continuous on their domains. Therefore, since each of the component functions are continuous at t = 1, we know that r(t) is continuous at t = 1. 9. r0 (t) = 3i + 8tj + (10t − 1)k so r0 (1) = 3i + 8j + 9k = h3, 8, 9i r(1.1) − r(1) h3(1.1) − 1, 4(1.1)2 , 5(1.1)2 − (1.1)i − h3(1) − 1, 4(1)2 , 5(1)2 − (1)i while = 0.1 0.1 h2.3, 4.84, 4.95i − h2, 4, 4i = 0.1 h0.3, 0.84, 0.95i = h3, 8.4, 9.5i = 0.1 −5 i + (6t + 1)j − 3(1 − t)2 k (1 + 5t)2 −5 so r0 (0) = i + j + 3k = h−5, 1, −3i 1

10. r0 (t) =

12.2. CALCULUS OF VECTOR FUNCTIONS

57

1 , 3(0.05)2 + (0.05), (1 + 0.05)3 r(0.05) − r(0) 1 + 5(0.05) while = 0.05 0.05 h0.8, 0.0575, 0.857375i − h1, 0, 1i = 0.05 h−0.2, 0.0575, −0.142625i = 0.05 = h−4, 1.15, −2.8525i 11. r0 (t) =

1 1 i − 2 j; t t

r00 (t) = −

12. r0 (t) = h−t sin t, 1 − sin ti;

1 k; 1 + t2

−

1 , 3(0)2 + (0), (1 − 0)3 1 + 5(0)

1 2 i + 3j t2 t

r00 (t) = h−t cos t − sin t, − cos ti

13. r0 (t) = h2te2t + e2t , 3t2 , 8t − 1i; 14. r0 (t) = 2ti + 3t2 j +

r00 (t) = h4te2t + 4e2t , 6t, 8i

r00 (t) = 2i + 6tj −

15. r0 (t) = −2 sin ti +√6 cos tj r0 (π/6) = −i + 3 3j

2t k (1 + t2 )2 16. r0 (t) = 3t2 i + 2tj r0 (−1) = 3i − 2j y

y

x

x

8t k (1 + t2 )2 0 r (−1) = j − 2k

17. r0 (t) = j −

18. r0 (t) = −3 sin√ ti + 3 cos√tj + 2k −3 2 3 2 r0 (π/4) = i+ j + 2k 2 2

z

z

y x

y x

58

CHAPTER 12. VECTOR-VALUED FUNCTIONS

1 1 8 19. r(t) = ti + j + t3 k; r(2) = 2i + 2j + k; r0 (t) = i + tj + t2 k; r0 (2) = i + 2j + 4k 2 3 3 Using the point (2, 2, 8/3) and the direction vector r0 (2), we have x = 2 + t, y = 2 + 2t, z = 8/3 + 4t. 20. r(t) = (t3 −t)i+

6t j+(2t+1)2 k; t+1

r(1) = 3j+9k;

r0 (t) = (3t2 −1)i+

6 j+(8t+4)k; (t + 1)2

3 r0 (1) = 2i + j + 12k. Using the point (0, 3, 9) and the direction vector r0 (1), we have x = 2 2t, y = 3 + 23 , z = 9 + 12t. p √ 2 = 6 21. r0 (t) = het + tet , 2t + 2, 3t2 − 1i so r0 (0) = h1, 2, −1i and |r0 (0)| = 12 + 22 + (−1)  r0 (0) 1 2 −1 h1, 2, −1i √ The unit tangent vector at t = 0 is given by 0 = √ ,√ ,√ = |r (0)| 6 6 6 6 To find the parametric equations of the tangent line at t = 0, we first compute r(0) = 1 2 −1 h0, 0, 0i. The tangent line is then given in vector form as p(t) = h0, 0, 0i + t √ , √ , √ = 6 6 6   1 2 −1 1 2 −1 √ t, √ t, √ t or in parametric form as x = √ t, y = √ t, z = √ t. 6 6 6 6 6 6 p √ 2 + (2)2 + (1)2 = 22. r0 (t) = h3 cos 3t, 2 sec2 2t, 1i so r0 (π) = h−3, 2, 1i and |r0 (π)| = (−3) 14.   −3 2 1 r0 (π) h−3, 2, 1i = √ ,√ ,√ The unit tangent vector at t = π is given by 0 = √ |r (π)| 14 14 14 14 To find the parametric equations of the tangent line at t = π, we first compute r0 (π) = h1, 0, πi. The tangent line is then given in vector  form as −3 2 1 p(t) = h1, 0, πi + t √ , √ , √ 14 14 14   −3 2 1 = 1 − √ t, √ t, π + √ t 14 14 14 2 1 −3 or in parametric form as x = 1 − √ t, y = √ t, z = π + √ t 14 14 14 * √ + 1 3 π 23. r(π/3) = , , 2 2 3 r0 (t) = h− * sin t, cos t, 1i+ √ 3 1 r0 (π/3) = − , ,1 2 2 so the tangent * √ line+is given * by + √ 1 3 π 3 1 p(t) = , , +t − , ,1 2 2 3 2 2 * + √ √ 1 3 3 1 π = − t, + t, + t 2 2 2 2 3 24. r(0) = h6, 1, 1i r0 (t) = h−3e−t/2 , 2e2t , 3e3t i

12.2. CALCULUS OF VECTOR FUNCTIONS

59

r0 (0) = h−3, 2, 3i So the tangent line is given by r(t) = h6, 1, 1i + th−3, 2, 2i = h6 − 3t, 1 + 2t, 1 + 3ti d [r(t) × r0 (t)] = r(t) × r00 (t) + r0 (t) × r0 (t) = r(t) × r00 (t) dt d d [r(t) · (tr(t))] = r(t) · (tr(t))+ = r(t) · (tr0 (t) + r(t)) + r0 (t) · (tr(t)) 26. dt dt = r(t) · (tr0 (t)) + r(t) · r(t) + r0 (t) · (tr(t)) = 2t(r(t) · r0 (t)) + r(t) · r(t)

25.

27.

d d [r(t) · (r0 (t) × r00 (t))] = r(t) · (r0 (t) × r00 (t)) + r0 (t) · (r0 (t) × r00 (t)) dt dt = r(t) · (r0 (t) × r000 (t) + r00 (t) × r00 (t)) + r0 (t) · (r0 (t) × r00 (t)) = r(t) · (r0 (t) × r000 (t))

28.

d d [r1 (t) × (r2 (t) × r3 (t))] = r1 (t) × (r2 (t) × r3 (t)) + r0 (t) × (r2 (t) × r3 (t)) dt dt = r1 (t) × (r2 (t) × r03 (t) + r02 (t) × r3 (t) + r01 (t) × (r2 (t) × r3 (t)) = r1 (t) × (r2 (t) × r03 (t)) + r1 (t) × (r02 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t))

d 1 [r1 (2t) + r2 ( 1t )] = 2r0 (2t) − 2 r02 ( 1t ) dt t d 3 2 30. [t r(t )] = t3 (2t)r0 (t2 ) + 3t2 r(t2 ) = 2t4 r0 (t2 ) + 3t2 r(t2 ) dt 2 Z 2  Z 2  Z 2  Z 2 2 2 3 1 2 2 3 31. r(t)dt = tdt i + 3t dt j + 4t dt k = t i + t3 −1 j + t4 −1 k = i + 9j + 15k 2 −1 2 −1 −1 −1 −1  Z 4  Z 4  Z 4 Z 4 √ √ 2t + 1dt i + 32. − tdt j + sin πtdt k r(t)dt = 29.

0

0

0

0

4 4 4 26 1 2 3/2 1 16 3/2 = (2t + 1) i − t j − cos πt k = i− j 3 3 π 3 3 0 0 0 Z  Z  Z  Z 2 33. r(t)dt = tet dt i + −e−2t dt j + tet dt k     1 t2 1 2 1 −2t 1 t t + c2 j + e + d3 k = et (t − 1)i + e−2t j + et k + c, = [te − e + c1 ]i + e 2 2 2 2 where c = c1 i + c2 j + c3 k. Z  Z  Z  Z 1 t t2 34. r(t)dt = dt i + dt j + dt k 1 + t2 1 + t2 1 + t2   Z   1 1 −1 2 k = [tan t + c1 ]i + ln(1 + t ) + c2 j + 1− 2 1 + t2   1 = [tan−1 t + c1 ]i + ln(1 + t2 ) + c2 j + [t − tan−1 t + c3 ]k 2 1 = tan−1 ti + ln(1 + t2 )j + (t − tan−1 t)k + c, 2

60

CHAPTER 12. VECTOR-VALUED FUNCTIONS

where c = c1 i + c2 j + c3 k.   R  R R R 6dt i + 6tdt j + 3t2 dt k = [6t + c1 ]i + [3t2 + c2 ]j + [t3 + c3 ]k 35. r(t) = r0 (t)dt = Since r(0) = i + 2j + k = c1 i + c2 j + c3 k, c1 − 1, c2 = −2, and c3 = 1. Thus, r(t) = (6t + 1)i + (3t2 − 2)j + (t3 + 1)k    R R    R t sin t2 dt i + − cos 2tdt j = − 21 cos t2 + c1 i + − 21 sin 2t + c2 j 36. r(t) = r0 (t)dt = Since r(0) = 32 = (− 12 + c1 )i + c2 j, c1 = 2, and c2 = 0. Thus,  r(t) =

 1 1 2 − cos t + 2 i − sin 2tj. 2 2

R  R  R  R 37. r0 (t) = r00 (t)dt = 12tdt i + −3t−1/2 dt j + 2dt k = [6t2 + c1 ]i + [−6t1/2 + c2 ]j + [2t + c3 ]k Since r0 (1) = j = (6 + c1 )i + (−6 + c2 )j + (2 + c3 )k, c1 = −6, c2 = 7, and c3 = −2. Thus, r0 (t) = (6t2 − 6)i + (−6t1/2 + 7)j + (2t − 2)k. Z r(t) =

0

Z

r (t)dt =

 Z  Z  1/2 (6t − 6)dt i + (−6t + 7)dt j + (2t − 2)dt k 2

= [2t3 − 6t + c4 ]i + [−4t3/2 + 7t + c5 ]j + [t2 − 2t + c6 ]k. Since r(1) = 2i − k = (−4 + c4 )i + (3 + c5 )j + (−1 + c6 )k, c4 = 6, c5 = −3, and c6 = 0. Thus, r(t) = (2t3 − 6t + 6)i + (−4t3/2 + 7t − 3)j + (t2 − 2t)k. 38. r0 (t) =

Z

r00 (t)dt =

Z

 Z  Z  sec2 tdt i + cos tdt j + − sin tdt k

= [tan t + c1 ]i + [sin t + c2 ]j + [cos t + c3 ]k Since r0 (0) = i + j + k = c1 i + c2 j + c3 k, c1 = 1, c2 = 1, and c3 = 0. Thus, r0 (t) = (tan t + 1)i + (sin t + 1)j + cos tk. Z r(t) =

r0 (t)dt =

Z

 Z  Z  (tan t + 1)dt i + (sin t + 1)dt j + cos tdt k.

= [ln | sec t| + c4 ]i + [− cos t + t + c5 ]j + [sin t + c6 ]k Since r(0) = −j + 5k = (−1 + c5 )j + (c6 )k, c4 = 0, c5 = 0, and c6 = 5. Thus, r(t) = (ln | sec t| + t)i + (− cos t + t)j + (sin t + 5)k. p √ 39. r0 (t) = −a sin ti + a cos tj + ck; |r0 (t)| = (−a sin t)2 + (a cos t)2 + c2 = a2 + c2 2π √ √ R 2π √ a2 + c2 dt = a2 + c2 t 0 = 2π a2 + c2 s= 0

12.2. CALCULUS OF VECTOR FUNCTIONS

61

40. r0 (t) = i + p(cos t − t sin t)j + (sin t + t cos t)k √ |r0 (t)| = 12 + (cos t − t sin t)2 + (sin t + t cos t)2 = 2 + t2 √ √ √ √ √  Rπ√ π s = 0 2 + t2 dt = 2t 2 + t2 + ln |t + 2 + t2 | 0 = π2 2 + π 2 + ln(π + 2 + π 2 ) − ln 2 t t t 41. r0 (t) = (−2e + et sin 2t)j + et k p sin 2t + e cos 2t)i + (2e cos 2t √ √ 2 0 2t 2 2t 2t |r (t)| = 5e cos 2t + 5e sin 2t + e = 6e2t = 6et √ 3π √ R 3π √ t s= 0 6e dt = 6et = 6(e3π − 1) 0

q √ √ √ 42. r0 (t) = 3i + 2 3tj + 2t2 k; |r0 (t)| = 32 + (2 3t)2 + (2t2 )2 = 9 + 12t2 + 4t4 = 3 + 2t2  1 R1 s = 0 (3 + 2t2 )dt = 3t + 23 t3 = 3 + 23 = 11 3 0

Rt s 43. From r0 (t) = h9 cos t, −9 sin ti, we find |r0 (t)| = 9. Therefore, s = 0 9du = 9t so that t = . By 9 E D D s s sE s . Note that r0 (s) = sin , cos substituting for t in r(t), we obtain r(s) = 9 sin , 9 cos 9 9 9 9 r s s so that r0 (s) = sin2 + cos = 1. 9 9 √ Rt 44. From r(t) = h−5 sin t, 12, 5 cos ti, we find |r0 (t)| = 169 = 13. Therefore, s = 0 13du = 13t s s 12 5 s . By substituting for t in r(t), we obtain r(s) = h5 cos 13 , 13 s, 13 cos 13 i. so that t = 13

5 s 12 5 s 0 Note that r (t) = − 13 sin 13 , 13 , 13 cos 13 so that r 25 144 25 25 s 0 sin + 169 + cos2 13 |r (s)| = =1 169 13 169 √ √ Rt√ 45. From r0 (t) = h2, −3, 4i, we find |r0 (t)| = 29. Therefore,  s = 0 29du = 20t so that  2 3 4 s t = √ . By substituting for t in r(t), we obtain r(s) = 1 + √ s, 5 − √ s, 2 + √ s . 29 29 29 29 r D E 9 16 4 2 3 4 Note that r0 (s) = √29 , − √29 , √29 so that r0 (s) = + + = 1. 29 29 29 t t t t 46. From r0 (t) p= he cos t − e sin t, e sin t + e cos t, 0i we find √ √ |r0 (t)| = e2t cost −2e2t cos t sin t + e2t sin2 t + e2t sin2 t + 2e2t sin t cos t + e2t cos2 t = 2e2t = et 2.   √ Rt √ Therefore, s = 0 eu 2du = 2(et − 1) so that t = ln √s2 + 1 . By substituting for t in r(t), we obtain        E D √s + 1 cos(ln √s + 1 √s + 1 sin ln √s + 1 r(s) = , , 1 Note that 2  2  D 2    2       E r0 (s) = √12 cos ln √s2 + 1 − √12 sin ln √s2 + 1 , √12 sin ln √s2 + 1 + √12 cos ln √s2 + 1 , 0 so that v             u u 1 cos2 ln √s + 1 − cos ln √s + 1 sin ln √s + 1 + 1 sin2 ln √s + 1 u 2  2   2   2  2  2  |r0 (s)| = t + 12 sin2 ln √s2 + 1 + sin ln √s2 + 1 cos ln √s2 + 1 + 21 cos2 ln √s2 + 1 s       s s = cos2 ln √ + 1 + sin2 ln √ + 1 =1 2 2

62

CHAPTER 12. VECTOR-VALUED FUNCTIONS d d d (r · r) = |r|2 = c2 = 0 and dt0 dt dt Thus, r is perpendicular to r.

47. Since

d dt (r

· r) = r · r0 + r0 · r = 2r · r0 , we have r · r0 = 0.

48. Let v = ai + bj and r(t) = x(t)i + y(t)j. Then Z

b

Z v · r(t)dt =

a

b

Z

b

Z

Z

b

r(t)dt. a

a

a

a

b

y(t)dt = v ·

x(t)dt + b

[ax(t) + by(t)]dt = a

Rt 49. From r(t) = r0 + tv, we get r0 (t) = v so that |r0 (t)| = |v|. Therfore s = 0 |r0 (t)|du = Rt s |v|du = |v|t which gives t = . Substituting for t in r(t), we have r0 (s) = r0 + 0 |v| v |v| s 0 0 v . Note that r (s) = |v| so that |r (s)| = |v| = 1. |v|v = r0 + s |v|   p s 3 −4 , 50. (a) |h3, −4i| = 32 + (−4)2 = 5 so r(s) = h1, 2i + h3, −4i = h1, 2i + s 5 5 5 √ √ (b) r(t) + th1, 2, −1i and |h1, 2, −1i| = 1 + 4 + 1 = 6 so r(s) = h1, 1, 10i +  = h1, 1, 10i  2 −1 1 s √ ,√ ,√ 6 6 6

12.3

Motion on a Curve

1. v(t) = 2ti + t3 j; v(1) = 2i + j; |v(1)| = a(t) = 2i + 3t2 j; a(1) = 2i + 3j

4+1=

5;

y a

v x

√ √ 2 j; v(1) = 2i − 2j; |v(1)| = 4 + 4 = 2 2; 3 t 6 a(t) = 2i + 4 j; a(1) = 2i + 6j t

2. v(t) = 2ti −

y a

v

x

12.3. MOTION ON A CURVE

63

3. v(t) = −2 sinh 2ti + 2 cosh 2tj; v(1) = 2j; |v(0)| = 2; a(t) = −4 cosh 2ti + +4 sinh 2tj; a(0) = −4i

y

v

a

x

√ 1 4. v(t) = −2 sin ti + cos tj; v(π/3) = − 3i + j; 2 p √ 1/4 = 13/2; a(t) = −2 cos ti − sin tj; |v(π/3)| = 3 + √ 3 a(π/3) = −i − j 2

y v a x

5. v(t) = (2t − 2)i + k; v(2) = 2j + k; |v(2)| = a(t) = 2j; a(2) = 2j

4+1=

5;

z

v a y x

6. v(t) = i + j; v(2) = i + j + 12k; |v(2)| = a(t) = 6tk; a(2) = 12k

1 + 1 + 144 =

146;

z a v

y x

64

CHAPTER 12. VECTOR-VALUED FUNCTIONS 7. v(t) = i + 2tj + 3t2 k; √ √ mathbf v(1) = i + 2j + 3k; |v(1)| = 1 + 1 + 9 = 14; a(t) = 2j + 6tk; a(1) = 2j + 6k

z

a

v y

x

8. v(t) = i + 3t2 j + k; √ √ v(1) = i + 3j + k; |v(1)| = 1 + 9 + 1 = 11; a(t) = 6tj; a(1) = 6j

z

v a y x

9. The particle passes through the xy-plane when z(t) = t2 −5t = 0 or t = 0, 5 which gives us the points (0, 0, 0) and (25, 115, 0). v(t) = 2ti + (3t2 − 2)j + (2t − 5)k; v(0) = −2j − 5k, v(5) = 10i + 73j + 5k; a(t) = 2i + 6tj + 2k; a(0) = 2i + 30j + 2k 10. If a(t) = 0, then v(t) = c1 and r(t) = c1 t + c2 . The graph of this equation is a straight line. √ 11. Initially we are given s0 = 0 and v0 = (480 cos 30◦ )i + (480 cos 30◦ )j = 240 3i + 240j. Using a(t) = −32j we find Z v(t) =

a(t)dt = −32tj + c

√ 240 3i + 240j = v(0) = c √ √ v(t) = −32tj + 240 3i + 240j = 240 3i + (240 − 32t)j Z √ r(t) = v(t)dt = 240 3ti + (240t − 16t2 )j + b 0 = r(0) = b. √ √ (a) The shell’s trajectory is given by r(t) = 240 3ti + (240t − 16t2 )j or x = 240 3t, y = 240 − 16t2 . (b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum altitude is y(15/2) = 900 ft. (c) Solving y(t) = 240t − 16t2 = 16t(15 − t) = 0, we see that the √ shell is at ground level when t = 0 and t = 15. The range of the shell is s(15) = 3600 3 ≈ 6235 ft.

12.3. MOTION ON A CURVE

65

(d) From (c), impact is when t = 15. The speed at impact is p √ |v(15)| = |240 3i + (240 − 32 · 15)j| = 2402 · 3 + (−240)2 = 480 ft/s.

√ 12. Initially we are given s0 = 1600j and v0 = (480 cos 30◦ )i + (480 sin 30◦ )j = 240 3i + 240j. Using a(t) = −32j we find Z v(t) = a(t)dt = −32tj + c √ 240 3i + 240j = v(0) = c √ √ v(t) = −32tj + 240 3i + 240j = 240 3i + (240 − 32t)j Z √ r(t) = v(t)dt = 240 3ti + (240t − 16t2 )j + b 1600j = r(0) = b. √ √ (a) The shell’s trajectory is given by r(t) = 240 3ti+(240t−16t2 +1600)j or s = 240 3t, y = 240t − 16t2 + 1600. (b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum altitude is y(15/2) = 2400 ft. (c) Solving y(t) = −16t2 + 240t + 1600 = −16(t − 20)(t + 5) = 0, we √ see that the shell hits the ground when t = 20. The range of the shell is x(20) = 4800 3 ≈ 8314 ft. (d) From (c), impact is when t = 20. The speed at impact is p √ √ |v(20)| = |240 3i + (240 − 32 · 20)j| = 2402 · 3 + (−400)2 = 160 13 ≈ 577 ft/s.

13. We are given s0 = 81j and v0 = 4i. Using a(t) = −32j, we have Z v(t) =

a(t)dt = −32tj + c 4i = v(0) = c

v(t) = 4i − 32tj Z r(t) =

v(t)dt = 4ti − 16t2 j + b 81j = r(0) = b

r(t) = 4ti + (81 − 16t2 )j. Solving y(t) = 81 − 16t2 = 0, we see that the car hits the water when t = 9/4. Then p √ |v(9/4)| = |4i − 32(9/4)j| = 42 + 722 = 20 13 ≈ 72.11ft/s.

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CHAPTER 12. VECTOR-VALUED FUNCTIONS

14. Let θ be the angle of elevation. Then v(0) = 98 cos θi + 98 sin θj. Using a(t) = −9.8j, we have Z v(t) = a(t)dt = −9.8tj + c 98 cos θi + 98 sin θj = v(0) = c v(t) = 98 cos θi + (98 sin θ − 9.8t)j r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j + b. Since r(0) = 0, b = 0 and r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j. Setting y(t) = 98t sin θ − 4.9t2 = t(98 sin θ − 4.9t) = 0, we see that the projectile hits the ground when t = 20 sin θ. Thus, using x(t) = 98t cos θ, 490 = s(t) = 98(20 sin θ) cos θ or sin 2θ = 0.5. Then 2θ = 30◦ or 150◦ . The angles of elevation are 15◦ and 75◦ . √ √ s 2 s 2 i+ j. Using a(t) = 15. Let s be the initial speed. Then v(0) = s cos 45◦ i + s sin 45◦ j = 2 2 −32j, we have Z v(t) = a(t)dt = −32j + c √ √ s 2 s 2 i+ j = v(0) = c 2 2 ! √ √ s 2 s 2 i+ − 32t j v(t) = 2 2 ! √ √ s 2 s 2 r(t) = ti + t − 16t2 j + b. 2 2 Since r(0) = 0, b = 0 and √ s 2 r(t) = ti + 2

! √ s 2 2 t − 16t j. 2

√ √ 2 Setting √ y(t) = s 2t/2 − 16t = t(2 √2/2 − 16t) = 0 we see that the ball hits the ground when t√ = 2s/32. Thus, using x(t) = s 2t/2 and the fact that 100 yd = 300 ft, 300 = x(t) = √ s 2 √ s2 ( 2s/32) = and s = 9600 ≈ 97.98 ft/s. 2 32 16. Let s be the initial speed and θ the initial angle. Then v() = s cos θi + s sin θj. Using a(t) = −32j, we have Z v(t) = a(t)dt = −32tj + c s cos θi + s sin θj = v(0) = c v(t) = s cos θi + (s sin θ − 32t)j r(t) = st cos θi + (st sin θ − 16t2 )j + b.

12.3. MOTION ON A CURVE

67

Since r(0) = 0, b = 0 and r(t) = st cos θi + (st sin θ − 16t2 )j. Setting y(t) = st sin θ − 16t2 = t(s sin θ − 16t) =, we see that the ball hits the ground when t = (s sin θ)/16. Using x(t) = st cos θi, we see that the range of the ball is   s sin θ s2 sin θ cos θ s2 sin 2θ x = = . 16 16 32 √ 2 ◦ 2 ◦ ◦ 2 ◦ For √ θ2 = 30 , the range is s sin 60 /32 = ◦3s /64 and for θ = 60 the range is s sin 120 /32 = 3s /64. In general, when the angle is 90 − θ then range is [s2 sin 2(90◦ − θ)]/32 = s2 [sin(180◦ − 2θ)]/32 = s2 (sin 2θ)/32. Thus, for angles θ and 90◦ − θ, the range is the same. 17. r0 (t) = v(t) = −r0 ω sin ωti + r0 ω cos ωtj; v = |v(t)| = 00 2 2 ω = v/r0 ; a(t) q = r (t) = −r0 ω cos ωti − r0 ω sin ωtj

q

r02 ω 2 sin2 ωt + r02 ω 2 cos2 ωt = r0 ω

r02 ω 4 cos2 ωt + r02 ω 4 sin2 ωt = r0 ω 2 = r0 (v/r0 )2 = v 2 /r0 . p √ 18. (a) v(t) = −b sin ti + b cos tj + ck; |v(t)| = b2 sin2 t + b2 cos2 t + c2 = b2 + c2 √ Rt Rt√ ds √ 2 (b) s = 0 |v(t)|du = 0 b2 + c2 du = t b2 + c2 ; = b + c2 dt p d2 s = 0; a(t) = −b cos ti−b sin tj; |a(t)| = b2 cos2 t + b2 sin2 t = |b|. Thus, d2 s/dt2 = (c) 6 2 dt |a(t)|. a = |a(t)| =

19. Let the initial speed of the projectile be s and let the target be at (x0 , y0 ). Then vp (0) = s cos θi + s sin θj and vt (0) = 0. Using a(t) = −32j, we have R vp (t) = a dt = −32tj + c s cos θi + s sin θj = vp (0) = c vp (t) = s cos θi + (s sin θ − 32t)j rp (t) = st cos θi + (st sin θ − 16t2 )j + b.

y (x0,y0)

x0 tan θ θ x0

x

Since rp (0) = 0, b = 0 and rp (t) = st cos θi + (st sin θ − 16t2 )j. Also, vt (t) = −32tj + c and since vt (0) = 0, c = 0 and vt (t) = −32tj. Then rt (t) = −16t2 tj + b. Since rt (0) = x0 i + y0 j, bx0 i + y0 j and rt (t) = x0 i+(y0 −16t2 )j. Now, the horizontal component of rp (t) will be x0 when t = x0 /s cos θ at which time the vertical component of rp (t) will be (sx0 /s cos θ) sin θ − 16(x0 /s cos θ)2 = x0 tan θ − 16(x0 /s cos θ)2 = y−) − 16(x0 /s cos θ)2 . Thus, rp (x0 /s cos θ) = rt (x0 /s cos θ) and the projectile will strike the target as it falls. 20. The initial angle is θ = 0, the initial height is 1024 ft, and the initial speed is s = 180(5280)/3600 = 264 ft/s. Then x(t) = 264t and y(t) = −16t2 + 1024. Solving y(t) = 0 we see that the pack hits the ground at t = 8 seconds. The horizontal distance tranvelled is x(8) = 2112 feet. From the figure in the text, tan α = 1024/2112 = 16/33 and α ≈ 0.45 radian or 25.87◦ . 21. By Problem 17, a = v 2 /v0 = 15302 /(4000 · 5280) ≈ 0.1108. We are given mg = 192, so m = 192/32 and we = 1192 − (192/32)(0.1108) ≈ 191.33 lb.

68

CHAPTER 12. VECTOR-VALUED FUNCTIONS

22. By problem 17, the centripetal acceleration is v 2 /r0 . Then the horizontal force is mv 2 /r0 . The vertical force is 32m. The resultant force is U = (mv 2 /r0 )i + 32mj. From the figure, we see that tan φ = (mv 2 /r0 )/32m = v 2 /32r0 . Using r0 = 60 and v = 44 we obtain tan φ = 442 /32(60) ≈ 1.0083 and φ ≈ 45.24◦ .

< 0, 32m>

φ

23. Solving x(t) = (v0 cos θ)t for t and substituting into y(t) − 21 gt2 + (v0 sin θ)t + s0 we obtain 1 y=− g 2



x v0 cos θ

2 + (v0 sin θ)

x g x2 + (tan θ)x + s0 , + s) = − 2 v0 cos θ 2v0 cos2 θ

which is the equation of a parabola. 24. Since the projectile is launched from ground level, s0 = 0. To find the maximum height we maximize y(t) = − 21 gt2 + (v0 sin θ)t. Solving y 0 (t) = −gt + v0 sin θ = 0, we see that t = (v0 /g) sin θ is a critical point. Since y 00 (t) = −g ≤ 0,  H=y

v0 sin θ g

 =

1 v02 sin2 θ v0 sin θ v02 sin2 θ g + v sin θ = 0 2 g2 g 2g

is the maximum height. To find the range we solve y(t) = − 21 gt2 + (v0 sin θ)t = t(v0 sin θ − 1 2 gt) = 0. The positive solution to this equation is t = (2v0 sin θ)/g. The range is thus x(t) = (v0 cos θ)

2v0 sin θ v 2 sin 2θ = 0 . g g

25. Letting r(t) = x(t)i + y(t)j + z(t)k, the equation dr/dt = v is equivalent to dx/dt = 6t2 x, dy/dt = −4ty 2 , dz/dt = 2t(z + 1). Separating the variables and integrating, we obtain x/x = 6t2 dt, dy/y 2 = −4tdt, dz/(z + 1) = 2tdt, and ln x = 2t3 + c1 , −1/y = 2t2 + c2 , ln(z + 1) + t2 + c3 . Thus, 3

r(t) = k1 e2t i +

2 1 j + (k3 et − 1)k. 2t2 + k2

26. We require the fact that dr/dt = v. Then d dp dr dL = (r × p = r + × p = τ + v × p = τ + v × mv = τ + m(v × v) = τ + 0 = τ. dt dt dt dt 27. (a) Since F is directed along r we have F = cr for some constant c. Then τ = r × F = r × (cr) = c(r × r) = 0. (b) If τ = 0 then dL/dt = 0 and L is constant.

12.4. CURVATURE AND ACCELERATION

69

28. (a) Since the cannon is pointing directly to the left, tha parmetric equations describing the path of the cannon ball are given by 1 x(t) = v0 t, y(t) = − gt2 + s0 2 r

2s0 The cannon ball will touch the groun when y = 0, which occurs at t = . At that g r r  2s0 2s0 time, x is given by x = = −v0 . Notice that this x value will be farther g g to the left with increasing values of v0 . Therefore, the cannon ball travels farther with more gunpowder. r 2s0 (b) As shown in part (a), the cannon ball will touch the groun when t = . This value g of t is independent of v0 . This occurs because v0 has no vertical component. (c) If the cannon ball is dropped, we have v0 = 0. Therefore, the parametric equations describing the cannon ball motion are given by 1 x(t) = 0, y(t) = − gt2 + s0 . 2 r 2s0 . Therefore the cannon ball touches the ground at the As before, y = 0 when t = g same time regardless of whether it is fired or dropped.

12.4

Curvature and Acceleration

1. r0 (t) = −t sin ti + t cos tj + 2tk; sin t cos t 2 T=− √ i+ √ j+ √ k 5 5 5

|r0 (t)| =

p √ t2 sin2 t + t2 cos2 t + 4t2 = 5t;

√ 2. r0 (t) = et (− sin t + cos t)i + et (cos t + sin t)i + 2et k, √ |r0 (t)| = [et (sin2 t−2 sin t cos t+cos2 t)+e2t (cos2√t+2 sin t cos t+sin2 t)+2e2t ]1/2 = 4e2t = 2et ; 1 1 2 T(t) = (− sin t + cos t)i + (cos t + sin t)j + k 2 2 2 p 3. √ We assume a > 0. r0 (t) = −a sin ti + a cos tj + ck; |r0 (t)| = a2 sin2 t + a2 cos2 t + c2 = a2 + c2 ; a sin t a cos t c dT a cos t a sin t T(t) − √ i+ √ j+ √ k; = −√ i− √ j, 2 + c2 2 + c2 2 + c2 2 + c2 dt a a a a a2 + c2 s 2 2 2 2 dT = a cos t + a sin t = √ a ; N = − cos ti − sin tj; dt 2 a2 + a2 + c2 a2 + c2 c i j k a cos t c a sin t = √c sin t i − √c cos t + √ a √ √ B = T × N = − √ 2 k; 2 2 2 2 2 a + c a + c a + c a2 + c2 a2 + c2 a2 + c2 − cos t − sin t 0 √ 2 2 |dT/dt| a/ a + c a κ= = √ = 2 2 2 r0 (t) a + c2 a +c

70

CHAPTER 12. VECTOR-VALUED FUNCTIONS √

√ |r0 ; (1)| = 3; 1 T(t) = (1 + t2 + t4 )−1/2 (i + tj + t2 k), T(1) √ (i + j + k); 3 dT 1 t 2 4 −3/2 3 2 = − (1 + t + t ) (2t + 4t )i + [(1 + t + t)−1/2 − (1 + t2 + t)−3/2 (2t + 4t3 )]j dt 2 2 2 2 4 −1/2 t 2 4 −3/2 3 [2t(1 + t + t ) (1 + t + t ) (2t + 4t )]k; √ 2 r d 1 1 1 2 1 d 1 T(1) = − √ i + √ k, T(1) = + = √ ; N(1) = − √ (i − k)k, dt dt 3 3 3 3 3 2 i j k √ √ √ 1 B(1) = 1/ √3 1/ 3 1/√3 = √ (i − 2j + k); 6 −1/ 2 0 1/ 2 √ √ √ d 2/ 3 2 κ = T(1) = |r0 (1)| = √ = dt 3 3 √ 1 5. From Example 1 in the text, a normal to the osculating plane is B(π/4) = 26 (3i − 3j + 2 2k). √ √ The point on√the curve An equation √ √ when t = π/4 is ( 2, 2, 3π/4). √ √ of the√plane is 3(x − 2) − 3(y − 2) + 2 2(z − 3π/4(= 0, 3x − 3y + 2 2z = 3π/2, or 3 2x − 3 2y + 4z = 3π. 4. r0 (t) = i + tj + t2 k;

|r0 (t)| =

1 + t2 + t4 ,

6. From Problem 4, a normal to the osculating plane is B(1) = √16 (i − 2j + k). The point on the curve when t = 1 is (1, 1/2, 1/3). An equaiton of the plane is (x−1)−2(y −1/2)+(z −1/3) = 0 or x − 2y + z = 1/3. √ 7. v(t) = j + 2tk, |v(t)| = 1 + 4t2 ; a(t) = 2k; v · a = 4t, v × a = 2i, |v × a| = 2; 4t 2 aT = √ , aN = √ 2 1 + 4t 1 + 4t2 8. v(t) = −3 p psin ti + 2 cos tj + k, √ p |v(t)| = 9 sin2 t + 4 cos2 t = 1 = 5 sin2 t + 4 sin2 t + 4 cos2 t + 1 = 5 sin2 +1; a(t) = −3 cos ti − 2 sin tj; v · a = 9 sin t cos cos t, qt − 4 sin t cos t = 5 sin t√ √ 2 v × a = 2 sin ti − 3 cos tj + 6k, |v × a| = 4 sin +(cos2 t + 36 = 5 cos2 t + 8; s √ 5 sin t cos t cos2 t + 8 aT p , aN = sin2 t + 1 sin2 t + 1 √ 9. v(t) = 2ti + 2tj + 4tk, |v(t)| = 2 6t, t > 0; a(t) = 2i + 2j + 4k; v · a = 24t, v × a = 0; √ 24t aT = √ = 2 6, aN = 0, t > 0 2 6t √ 10. v(t) = 2ti − 3t2 j = 4t3 k, |v(t)| = t 4 + 9t2 + 16t4 , t >); a(t) = 2i −√ 6tj + 12t2 k; 3 5 4 3 2 2 v · a = 4t + 18t + 48t ; v × a =√−12t i − 16t j − 6t k, |v × a| = 2t 36t4 + 64t2 + 9; 4 + 18t2 + 48t4 2t 36t4 + 64t2 + 9 aT = √ , aN = √ t>0 4 + 9t2 + 16t4 4 + 9t2 + 16t4 √ 11. v(t) = 2i + 2tj, |v(t)| = 2 1 + t2 ; a(t) = 2j; v × a = 4k, |v × a| = 4; 2t 2 aT = √ , aN = √ 2 1+t 1 + t2

12.4. CURVATURE AND ACCELERATION

71

√ t 2t 1 − t2 1 1 + t2 i+ j, |v(t)| = ; a(t) = − i+ j; 12. v(t) = 2 2 2 2 2 1+t 1+t 1+t (1 + t ) (1 + t2 )2 2t t − t3 1 1 v·a=− + ; v×a= k, |v × a| = ; (1 + t2 )3 (1 + t2 )3 (1 + t2 )2 (1 + t2 )2 2 3 2 2 t/(1 + t ) t 1 a/(1 + t ) aT = − √ =− = , aN = √ 2 2 2 2 2 (1 + t )3/2 (1 + t2 )3/2 1 + t )/(1 + t 1 + t /(1 + t ) 13. v(t) = −5 sin ti + 5 cos tj, |v(t)| = 5; a(t) = −5 cos ti − 5 sin tj; v × a = 25k, |v × a| = 25; aT = 0, aN = 5

v · a = 0,

p 0 14. v(t) = sinh ti + cosh tj, |v(t)| = sinh t2 + cosh2 t a(t) = cosh ti + sinh tj v · a = 2 sinh t cosh t; v × a = (sinh2 t − cosh2 t)k = −k, |v × a| = 1; 1 2 sinh t cosh t , aN = p aT = p 2 2 2 sinh + cosh sinh + cosh2 √ 15. v(t) = et (i + j + k), √|v(t)| = 3e−t ; a(t) = e−t (i + j + k); v · a = −3e−2t ; v × a = 0, |v × a| = 0; aT = − 3e−t , aN = 0 √ 16. v(t) = i + 2j + 4k, |v(t)| = 21; a(t) = 0; v · a = 0, v × a = 0, |v × a| = 0; aT = 0, aN = 0 p 17. v(t) = −a sin ti + b cos tj + ck, |v(t)| = a2p sin2 t + b2 cos2 +c2 ; a(t) = −a cos ti − b sin tj; v × a = bc sin tip− ac cos tj + abk, |v × a| = b2 c2 sin2 t + a2 c2 cos2 t + a2 b2 b2 c2 sin2 t + a2 c2 cos2 t + a2 b2 |v × a| = κ= |v|3 (a2 sin2 t + b2 cos2 t + c2 )3/2 p 18. (a) v(t) = −a sin ti + b cos tj, |v(t)| = a2 sin2 t + b2 cos2 t; a(t) = −a cos ti − b sin tj; ab v × a = abk; |v × a| = ab; κ = 2 2 (a sin t + b2 cos2 t)3/2 (b) When a = b, |v(t)| = a, |v × a| = a2 , and κ = a2 /a3 = 1/a. 19. The equation of a line is v(t) = b + tc, when b and c are constant vectors. v(t) = c, |v(t)| = |c|; a(t) = 0; v × a = 0; κ = |v × a|/|v|3 = 0 20. v(t) = a(1 − cos t)i + a sin tj; v(π) = 2ai, |v(π)| = 2a; a(t) = a sin ti + a cos tj, i j k |v × a| 2a2 1 a(π) = −aj; |v × a| = 2a 0 0 = −2a2 k; |v × a| = 2a2 ; κ = = = 3 3 |v| 8a 4a 0 −a 0 p 21. v(t) = f 0 (t)i + g 0 (t)j, |v(t)| = [f 0 (t)]2 + [g 0 (t)]2 ; a(t) = f 00 (t)i + g 00 (t)j; v × a = [f 0 (t)g 00 (t) − g 0 (t)f 00 (t)]k, |v × a| = |f 0 (t)g 00 (t) − g 0 (t)f 00 (t)|; |v × a| |f 0 (t)g 00 (t) − g 0 (t)f 00 (t)| κ= = |v|3 ([f ”(t)]2 + [g 0 (t)]2 )3/2 22. For y = F (x), r = xi + F (x)j. We identify f (x) = x and g(x) = F (x) in Problem 21. Then f 0 (x) = 1, f 00 (x) = 0, g 0 (x) = F 0 (x), g 00 (x) = F 00 (x), and κ = |F 00 (x)|/(1 + [F 0 (x)]2 )3/2 .

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CHAPTER 12. VECTOR-VALUED FUNCTIONS

23. F (x) = x2 , F 00 (x) = 2,

F (0) = 0,

F (1) = 1;

F 00 (0) = 2,

F 00 (1) = 2;

F 0 (x) = 2x,

F 0 (0) = 0, F 0 (1) = 2; 2 1 κ(0) = = 2; ρ(0) = ; 2 3/2 2 (1 + 0 )

2 2 = √ ≈ 0.18; 2 3/2 (1√+ 2 ) 5 5 √ 5 5 ≈ 5.59; Since 2 > 2/5 5, the curve is ”sharper” at (0, 0). ρ(1) = 2 κ(1) =

24. F (x) = x3 ,

F (−1) = −1,

F 0 (1/2) = 3/4;

F (1/2) = 1/8;

F 00 (x) = 6x, F 00 (−1) = −6,

F 0 (x) = 3x2 ,

F 0 (−1) = 3,

F 00 (1/2) = 3;

κ(−1) =

3 √ ≈ 0.19; 5 10 √ 5 10 ρ(−1) = ≈ 5.27; 3 3 3 125 = ≈ 1.54; ρ( 21 ) = ≈ 0.65 κ( 12 ) = 125/64 192 [1 + (3/4)2 ]3/2 Since 1.54 > 0.19, the curve is ”sharper” at (1/2, 1/8).

|−6| (1+32 )3/2

=

6 √

10 10

=

|F 00 (x)| . |1 + (F 0 (x))2 |3/2 2 Now, F 0 (x)2x, F 00 (x) = 2, and (F 0 (x))2 = 4x2 so that κ = . (1 + 4x2 )3/2 As x → ±∞, the denominator grows without bound. Therefore, κ(x) → 0 as x → ±∞.

25. Letting F (x) = x2 , we can use Problem 22 to get κ(x) =

y

x 26. (a) √ 3t(2t2 + 1) t4 + 4t2 + 1 2t(t2 + 2) √ − ; (t4 + t2 + 1)5/2 (t4 + t2 + 1)3/2 t4 + 4t2 + 1 critical numbers occur at t = −.271469, t = 0, and t = .271469.

(b) κ0 (t) =

(c) Maximum of 1.017182 occurs at t = −.271469 and t = .271469. 27. Since (c, F (c)) is an inflection point and F 00 exists on an interval containg c, we must have F 00 (c) = 0. Therefore, using the formula from Problem 22, we see that the curvature is zero. 28. We use the fact that T · N = 0 and T · T = N · N = 1. Then |a(t)|2 = a · a = (an N + at T) · (an N + at T) = a2N N · N + 2an at N · T + a2T T · T = a2N + a2T .

CHAPTER 12 IN REVIEW

73

Chapter 12 in Review A. True/False 1. True; |v(t)| =

2

2. True; the curvature of a circle of radius a is κ = a1 . 3. True 4. False; consider r(t) = t2 i. In this case, v(t) = 2ti and a(t) = 2i. Since v · a = 4t, the velocity and acceleration vectors are not orthogonal for t 6= 0. 5. True 6. False; see Problem 20c in Section 14.2 7. True 8. True 9. False; consider r1 (t) = r2 (t) = i. 10. True,

d dr dr dr d |r(t)|2 = (r · r) = r · + · r = 2r · . dt dt dt dt dt

B. Fill in the Blanks 1. y = 4 2. 0 3. r0 (t) = h1, 2t, t2 i so r0 (1) = h1, 2, 1i 4. r00 (t) = h0, 2, 2ti so r00 (1) = h0, 2, 2i i j j √ 5. r0 (1) × r00 (1) = 1 2 1 = h2, −2, 2i so r0 (1) × r00 (1) = 12. 0 2 2 √ √ √ r0 (1) × r00 (1) 12 2 √ Since r0 (1)| = 6, we have κ(1) = = . = |r0 (1)|3 6 6 6   r0 (1) h1, 2, 1i 1 2 1 6. T(1) = 0 = √ = √ ,√ ,√ |r (1)| 6 6 6 6   0 2 r (t) h1, 2t, t i 1 2t t2 7. T(t) = 0 =√ = √ ,√ ,√ |r (t)| 1 + 4t2 + t4 1 + 4t2 + t4 1 + 4t2 + t4  1 + 4t2 + t4  2 4 −2(t + 2) −2(t − 1) 2t(2t2 + 1) . So T0 (t) = , 4 , 4 4 2 3/2 2 3/2 (t + 4t (t + 4t + 1) (t + 4t2 + 1)3/2  + 1) q −6 1 6 −1 1 1 1 0 √ √ √ . This gives T0 (1) = , 0, = , 0, and |T (1)| = 6 + 6 = 3/2 63/2 6 6 3 D E 6 −1 √ √1 , 0, 0 T (1) −1 1 6 6 Therefore N(1) = 0 = = h √ , 0, √ i. 1 √ |T (1)| ( 3) 2 2

74

CHAPTER 12. VECTOR-VALUED FUNCTIONS i 1 8. B(1) = T(1) × N(1) = √6 −1 √

j √2 6

2

0

k  1 −1 1 1 √ = √ ,√ ,√ 6 3 3 3 √1 2

9. A normal to the normal plane is T(1) =

D

√1 , √2 , √1 6 6 6

E

so we can use n = h1, 2, 1i as a vector

1 3 i,

normal to the plane. Since r(1) = h1, 1, the point (1, 1, 13 ) lies on the normal plane at t = 1. Thus an equation of the normal plane is (x − 1) + 2(y − 1) + (z − 13 ) = 0 or x + 2y + z = 1) 3 or 3x + 6y + 3z = 10 D E −1 √1 10. A normal to the osculating plane is B(1) = √13 , √ , . So we can use n = h1, −1, 1i as a 3 3 normal vector. Using the point (1, 1, 31 ), an equation of the osculating plane is (z − 1) − (y − 1) + (z − 13 ) = 0 or x − y + z = 31 or 3x − 3y + 3z = 1.

C. Exercises √ Rπ√ cos2 t + sin2 +1dt = 0 2dt = 2π √ √ √ √ Rt√ 2. r0 (t) = 5i + j + 7k; s(t) = 0 25 + 1 + 49du = 5 3t; s(3) = 15 3. Solving 5 3t = 80 3, √ we see that the distance traveled will be 80 3 when t = 16 or at the point (80, 17, 112).

1. r0 (t) = cos ti + sin tj + k;

3. r(3) = −27i + 8j + k;

s=

Rπp 0

r0 (t) = −6ti = √

is x = −27 − 18t, y = 8 + t, z = 1 + t.

2 + k; t+1

r0 (2) = −18i + j + k. The tangent line

5.

4. z

z

y x

y

x

6.

d d d [r1 (t) × r2 (t)] = r1 (t) × r2 (t) + r1 (t) × r2 (t) dt dt dt = (t2 i + 2tj + t3 k) × (−i + 2tj + 2tk) + (2ti + 2j + 2t2 k) × [−ti + t2 j + (t2 + 1)k] = (4t2 − 2t4 )i − 3t3 j + (2t3 + 2t)k + (2t2 + 2 − 3t4 )i − (5t3 + 2t)j + (2t3 + 2t)k = (2 + 6t2 − 5t4 )i − (8t3 + 2t)j + (4t3 + 4t)k d d [r1 (t) × r2 (t)] = [(2t3 + 2t − t5 )i − (2t4 + t2 )j + (t4 + 2t2 )k] dt dt = (2 + 6t2 − 5t4 )i − (8t3 + 2t)j + (4t3 + 4t)k

CHAPTER 12 IN REVIEW 7.

75

d d d [r1 (t) · r2 (t)] = r1 (t) · r2 (t) + r1 (t) · r2 (t) dt dt dt = (cos ti − sin tj + 4t3 k) · (2ti + sin tj + 2e2t k) (− sin ti − cos tj + 12t2 k) · (t2 i + sin tj + e2t k) = (2t cos t − sin t cos t + 8t3 e2t − t2 sin t − sin t cos t + 12t2 e2t = 2t cos t − t2 sin t − 2 sin t cos t + 8t3 e2t + 12t2 e2t d d [r1 (t) · r2 (t)] = [t2 cos t − sin2 t + 4t3 e2t ] = −t2 sin t + 2t cos t − 2 sin t cos t + 8t3 e2t + 12t2 e2t dt dt

8.

d d [r1 (t) · (r2 (t) × r3 (t))] = r1 (t) · [r2 (t) × r3 (t)] + r0 (t) · [r2 (t) × r3 (t)] dt dt = r1 (t) · [(r2 (t) × r03 (t)) + (r02 (t) × r3 (t))] + r01 (t) · (r2 (t) × r3 (t)) = r1 (t) · (r2 (t) × r03 (t)) + r1 (t) · r02 (t) × r3 (t)) = r01 (t) · (r2 (t) × r3 (t))

9. We are given F = ma = 2j; v(0) = i + j + k. and r(0) = i + j. Then Z Z 2 2 v(t) = a(t)dt = jdt = tj + c m m i = j + k = v(0) = c   2 v(t) = i + t+1 j+k m   1 2 t + t j + tk + b r(t) = ti + m i + j = r(0) = b   1 2 r(t) = (t + 1)i + t + t + 1 j + tk m The parametric equations are x = t, y =

1 2 t + t + 1, z = t. m

10. y

x v

a

v(t) = i − 3t2 j, v(1) √ a(t) = −6tj, a(1) = −6j √ = i − 3j; |v(1)| = |i − 3j| = 1 + 9 = 10

76

CHAPTER 12. VECTOR-VALUED FUNCTIONS

11. v(t) = 6i + j + 2tk; a(t) = 2k. To find when the particle passes through the plane, we solve −6t + t + t2 = −4 or t2 − 5t + 4 = 0. This gives t = 1 and t = 4. v(1) = 6i + j + 2k, a(1) = 2k; v(4) = 6i + j + 8k, a(4) = 2k 12. We are given r(0) = i + 2j + 3k. Z r(t) =

Z v(t)dt =

(−10ti + (3t2 − 4t)j + k)dt = −5t2 i + (t3 − 2t2 )j + tk + c i + 2j + 3k = r(0) = c

r(t) = (1 − 5t2 )i + (t3 − 2t2 + 2)j + (t + 3)k r(t) = −19i + 2j + 5k √ √ √ R √ 13. v(t) = a(t)dt = ( 2 sin ti + 2 cos tj)dt = − 2 cos √ ti + 2 sin √tj + c; −i + j + √ k = v(π/4) = −i + j + c, c = k; v(t) = − 2 cos ti + 2 sin tj + k; √ r(t) = − 2 sin ti− 2 cos tj+tk+b; i+2j+(π/4)k = r(π/4) = −i−j+(π/4)k+b, b = 2i+3j; √ √ r(t) = (2 − 2 2 sin t)i + (3 − 2 cos t)j + tk; r(3π/4) = i + 4j + (3π/4)k √ 3 3 14. v(t) = ti + t2 j − tk; |v| = t t2 + 2, t > 0; a(t) = i + 2tj − k; v · a = t + 2t √ + t = 2t√+ 2t ; 3 2 2 √ 2t + 2t 2 + 2t t 2 2t v × a = t2 bi + t2 k, |v × at2 2; aT = √ = √ , aN = √ = √ ; 2 2 2 2 t t +2 t +2 t t +2 t +2 √ √ t2 2 2 = κ= 3 2 3/2 2 t (t + 2) t(t + 2)3/2 R

0 15. r0 (t) = sinh = sinh 1i + cosh 1j + k; p ti + cosh tj + k, r (1)√ √ √ 2 2 0 |r (t)| = sinh t + cosh t + 1 = 2 cosh2 t = 2 cosh t; |r0 (1)| = 2 cosh 1; 1 1 1 1 T = √ tanh ti + √ j + √ sech tk, T(1) = √ (tanh 1i + j + sech 1k); 2 2 2 2 dT 1 d 1 1 1 2 = √ sech ti − √ sech t tanh tk; T(1) = √ sech2 1i − √ sech 1 tanh 1k, dt 2 2 2 dt 2 p d sech 1 1 2 2 T(1) == √ sech 1 + tanh +1 = √ sech 1; N(1) = sech 1i − tanh 1k; dt 2 2 1 1 1 B(1) = T(1) × N(1) = − √ tanh 1i + √ (tanh2 1 + sech2 1)j − √ sech 1k 2 2 2 1 = √ (− tanh 1i + j − sech 1k) 2 √ d (sech 1)/ 2 1 = sech2 1 κ = T(1) /|r0 (1)| = √ dt 2 2 cosh 1

16. The parametric equations describing the path of the ball are √ x(t) = 66 cos(30◦ )t = 33 3ty(t) = −16t2 + 66 sin(30◦ )t + 148 = −16t2 + 33t + 148 The ball touches the ground when y(t) = 0 or −16t2 + 33t + 148 = 0. This occurs when t ≈ 4.243. The ball therefore strikes the ground √ at x(4.243) = 242.52 ft. The velocity of the √ ball at time t is v(t) = h33 3, −32t + 33i. The impact velocity is given by v(4.243) = h33 3, −32(4.243) + 33i ≈ h57.158, −102.776i. The impact speed is then |v(4.243)| ≈ 117.6 ft/s.