# ODDREV11 - 68 Chapter 11 xti dr dt Vector-Valued Functions...

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 68 Chapter 11 xti dr dt Vector-Valued Functions ytj xt 2 85. Let r r z t k. Then r yt 2 r 2 xt 1 xt 2 r 2 2 yt yt 2 2 zt zt 2 2 and r x ti y tj 2y t y t z t k. Then, 2z t z t zt 12 2x t x t xtx t 87. Let r xi yj rr x2 yty t ztz t r. r. rr (using Exercise 77) z k where x, y, and z are functions of t, and r r dr dt r2 y2 z2 x i rr yj rr r2 zk r3 r r xx r 2r yy r r3 dr dt r zz xi yj zk 1 x y2 r3 1 yz r3 i x z2 xyy j xzz i x 2y k z 2y xx y 1 r r3 zz y j x 2z y 2z xx z yy z k yz x xz y xz xy z xy r r 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 r L GM r r e. y Planet Sun r Since r L and r are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate the coordinate system so that e lies along the positive x-axis and is the angle between e and r. Let e e . Then r e r e cos re cos . Also, L 2 θ e x L r L r r L r r L GM e r r GM r e r r r GM re cos r Thus, L 1 2 GM e cos r and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse. 91. A 1 2 rd 2 Thus, dA dt dA d d dt 1 2d r 2 dt 1 L 2 and r sweeps out area at a constant rate. Review Exercises for Chapter 11 1. r t ti csc t k n , n an integer n , n an integer 3. r t ln t i tj tk (a) Domain: t (a) Domain: 0, (b) Continuous for all t > 0 (b) Continuous except at t Review Exercises for Chapter 11 5. (a) r 0 (b) r (c) r c 2 1 i 3i 2c 2c (d) r 1 t r1 4j 1 1i 8 3k 1 3 69 1i c 21 2 ti c 1 2j t t t 1 2j 1 3 c 1 3k 1 3 1 3k c 1 3k 1 1 3 1i 2j 9. r t t 2j t3 i 1 t t2 ⇒ z z 1 t 3k 3 tk 3i j 3 t2 tj t 2k 7. r t xt x2 cos t i 2 sin2 t j 2 sin2 t 11. r t x t x y z i 1, y 0 1 0 1 sin t j sin t, z k 1 3 2 1 0 1 1 1 1 cos t, y t y 2 y 1 21 x y z y2 x2 2 1 1 1 z 1≤x≤1 y 2 1 1 2 1 y 3 2 1 −2 2 3 1 1 2 −1 x 1 x y x 13. r t ti z 3 2 1 1 ln t j 12 2t k 15. One possible answer is: r1 t r2 t r3 t 4t i 4i 4 3t j, 3 t i, t j, 0≤t≤1 0≤t≤3 0≤t≤4 1 2 3 2 y x 17. The vector joining the points is 7, 4, rt 2 7t, 3 4t, 8 10 . One path is 19. z x x2 t, y rt z 5 y 2, x t, z ti tj y 0, t 2t 2 2t 2k x 10t . −3 2 3 1 2 3 y x 21. lim t 2 i t→2 4 t2j k 4i k 70 23. r t Chapter 11 3t i 3i ut ut t j Vector-Valued Functions 1 j, u t ti t 2j 23 tk 3 (b) r t t2 t 3t 2 4t 1 t3 2t 2 (d) u t Dt u t (f ) r t 1 Dt r t ut ut 0 2r t 2r t 24 t 3 83 t 3 5t i 5i t3 i t2 2t 2t 4j 8t 3j 2t 2j 2j 3t 3 9t 2 23 tk 3 2t 2k t2 2t tk 1k (a) r t (c) r t Dt r t 3t 2 (e) r t Dt r t 10t 2 2t 1 1 2t 10t 10t 2 2t 2 i 25. x t and y t are increasing functions at t decreasing function at t t 0. t0, and z t is a 27. cos t i t cos t j dt sin t i t sin t cos t j C 29. cos t i sin t j t k dt 1 t 2 dt 1 t1 2 t2 ln t 1 t2 C 2 31. r t r0 rt j t2 2t i k 1i etj C i et e t k dt 3j 2j t 2i etj i e tk 2j C 4k 33. 2 3t i 2t 2j t 3k dt 3t 2 i 2 2t 3 j 3 t4 k 4 2 2 32 j 3 5k ⇒ C e t 4k 2 2 35. 0 et 2i 3t2j k dt 2et 2i t3j tk 0 2e 2i 8j 2k 37. r t vt cos3 t, sin3 t, 3t rt vt 3 3 cos2 t sin t, 3 sin2 t cos t, 3 9 cos4 t sin2 t cos2 t sin2 t 9 sin4 t cos2 t cos2 1 sin2 t 2 2 9 1 t sin2 t 3 cos2 t sin2 t at vt 6 cos t 3 cos2 t cos t, 6 sin t cos2 t cos t , 3 sin t 2 cos t 2 3 sin2 t sin t , 0 3 cos t 2 sin t 1 3 , t 2, t , t 0 2 3 , 2t, 1 2 sin t , 0 v02 sin 2 32 75 2 sin 60 32 2 39. rt rt r4 ln t 1 t 4 41. Range x 152 feet 1, 8, 1 2 direction numbers 0, 16, 2 , the parametric equations are Since r 4 1 x t, y 16 8t, z 2 2 t. r t0 0.1 r 4.1 v02 sin 2 9.8 0.1, 16.8, 2.05 43. Range x 80 ⇒ v0 80 9.8 sin 40 34.9 m sec Review Exercises for Chapter 11 45. rt vt vt at Tt 5t i 5i 5 0 i at Tt a N does not exist Nt a a T N i (The curve is a line.) i vt 47. rt vt ti i tj 1 j 2t 4t 1 2t 1 j 4t t 12 tj 4t 1 2 t 2 tj 4t 1 1 1 12 tk 2 tk 5t 2 k 2t j t k 1 5t2 5t i 2j k 5 1 5t2 5t 1 5t 2 5 5t 2 1 5t 2 5 51 2 ti 4t j 1 71 N t does not exist a T 0 1 4t t 4t 1 2t 4t ti i 1 2j i t 2j 2t j 49. rt vt vt at Tt Nt a a T N et i et i e2t et i e tj e tj e e tj 2t 51. rt vt v at Tt Nt a a T N et i e t j e2t e 2t e ti e2t etj e 2t e2t e 2t e2t e 2t 2 e2t e 2t 53. r t 2 cos t i 3 ,x 4 2 sin t j t k, x 2, y 2 cos t, y 2, z k 2, b 2, z t 3 . 4 2 sin t, z t When t rt 2 sin t i 2 cos t j 3 ,a 4 2t Direction numbers when t x 2t 2, y 2, c 3 4 1 55. v 9.56 104 4600 2t i 2i b 4.56 mi sec 57. r t rt s 3tj, 0 ≤ t ≤ 5 3j 5 2 −4 −2 y (0, 0) x 246 8 10 12 14 r t dt a 5 0 4 9 dt −4 −6 −8 − 10 − 12 − 14 − 16 13t 0 5 13 (10, − 15) 72 59. Chapter 11 rt rt rt 10 cos3 t i Vector-Valued Functions 10 sin3 tj 30 sin2 t cos tj sin4 t cos2 t − 10 10 y 30 cos2 t sin ti 30 cos4 t sin2 t 30 cos t sin t 2 2 −2 x 2 10 s 4 0 30 cos t sin t dt 120 sin2 t 2 2 60 0 − 10 61. r t rt s a 3ti 3i b 2tj 2j 4tk, 0 ≤ t ≤ 3 4k 3 3 12 10 8 6 4 2 2 x z (−9, 6, 12) r t dt 0 9 4 16 dt 0 29 dt 3 29 (0, 0, 0) 24 68 10 y 63. r t rt s 8 cos t, 8 sin t, t , 0 ≤ t ≤ < b 2 65 65 2 65. r t rt s 1 ti 2 1 i 2 sin t j cos t j r t dt cos t k, 0 ≤ t ≤ sin t k 8 sin t, 8 cos t, 1 , r t 2 r t dt a z 65 dt 0 0 π 2 (0, 8, π ) 2 4 6 8 y 0 1 4 dt 0 cos2 t 5 t 2 sin2 t dt 5 2 4 86 x (8, 0, 0) 5 2 0 67. r t Line k 0 3ti 2tj 69. rt rt rt r r 2ti 2i j i 2 0 r r 12 tj 2 tj 2k j t 1 r 3 t2k 2tk, r 5t2 4 k 2t 2 5t2 4j 20 4 2k, r r 25 5t2 20 K 32 4 32 71. y y y K At x 12 x 2 x 1 1 2 73. y y K ln x 1 ,y x 1 y y 1, K 1 x2 232 y y 4, K 232 1 1 x2 1 1 22 1 x2 1 x 2]3 2 32 At x 32 1 and r 173 2 1 23 2 2 and r 4 2 2. 17 17 17. 75. The curvature changes abruptly from zero to a nonzero constant at the points B and C. P roblem Solving for Chapter 11 73 Problem Solving for Chapter 11 t 1. x t 0 cos u2 du, y t 2 t sin 0 u2 du 2 3. Bomb: r1 t Projectile: r2 t 5000 400t, 3200 t, v0 sin 16t2 t 16t2 xt (a) s t2 cos ,y t 2 a t2 sin 2 a v0 cos At 1600 feet: Bomb: dt 0 xt 0 2 y t 2 dt t2 2 t2 2 1 a. ,y t a t2 2 3200 16t2 1600 ⇒ t 10 seconds. Projectile will travel 5 seconds: t cos 5 v0 sin 16 25 v0 sin t Horizontal position: At t At t 10, bomb is at 5000 400 10 1000. 1600 400. (b) x t t sin t cos2 t sin2 K At t (c) K a, K a t2 2 5, projectile is at 5v0 cos . 200. 400 ⇒ tan 200 447.2 ft sec 2⇒ 63.4 . length Thus, v0 cos Combining, v0 sin , 0 ≤ ≤2 v0 sin v0 cos 200 cos 5. x x 2 1 y cos , y 2 1 2 cos 2 cos 2 sin2 4 sin2 2 t t st x K 2 sin 2 d 4 cos cos cos 3 2 4 cos t 2 sin , y 11 cos sin sin 2 1 cos 8 sin3 1 4 sin 2 2 1 2 sin Thus, s2 1 K 2 t 4 sin and 2 16 cos2 t 2 16 sin2 t 2 16. 7. r2 t d rt dt rt 2 rt 2rt rt d rt dt rt rt rt ⇒ d rt dt rt rt rt 74 Chapter 11 Vector-Valued Functions dB ds N 9. r t rt rt T T N B At t T 4 cos ti 4 sin ti 4 cos ti 4 sin ti 5 4 cos ti 5 cos ti N ,T N B z 6π 4 sin tj 4 cos tj 4 sin tj 4 cos tj 5 4 sin tj 5 sin tj 3 sin ti 5 3tk, t 2 5 11. (a) B T dB ds N d T ds T T T 1 constant length ⇒ N N N T T T dB ds N T T T N T T N B 3k, r t 3 k 5 T dB ds T 0 3 cos tj 5 3 k 5 4 k 5 dB dB B and ds ds for some scalar . Hence, (b) B B T N T⇒ 2 2 2 2 3 i 5 4 i 5 j N. Using Exercise 10.3, number 64, T N N N N T T NT N N TN 4 k 5 T B T T N T T T T NT N T TN B N B N T N. Now, KN Finally, dT ds Ts Ts Ts dT . ds 3π 4 x 3 2 1 4 y Ns d B ds T B B KT T KN B. B T N T 13. r t (a) −3 t cos t, t sin t , 0 ≤ t ≤ 2 2 2 (b) Length 0 2 3 r t dt 2t 2 0 1 dt 6.766 (graphing utility) −2 22 (c) K K0 K1 K2 2 t 2 1 32 22 t (d) 5 2 2 2 1 32 1.04 0 0 5 0.51 0 (f ) As t → , the graph spirals outward and the curvature decreases. (e) lim K t→ ...
View Full Document

## This note was uploaded on 05/18/2011 for the course MAC 2311 taught by Professor All during the Fall '08 term at University of Florida.

Ask a homework question - tutors are online