12 - CHAPTER 12 Vector-Valued Functions Section 12.1...

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Unformatted text preview: CHAPTER 12 Vector-Valued Functions Section 12.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . 77 Section 12.2 Differentiation and Integration of Vector-Valued Functions . . . . . . . . . . . . . 87 Section 12.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . 97 Section 12.4 Tangent Vectors and Normal Vectors . . . . . . . . . . . . 108 Section 12.5 Arc Length and Curvature . . . . . . . . . . . . . . . . . 125 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 144 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 154 CHAPTER 12 Vector-Valued Functions Section 12.1 1. r t 5t i 4t j Vector-Valued Functions 1 k t 5t 4t 1 t Domain: 2, 2 2. r t 4 t2 i t 2j 6t k 4 t2 6t t2 Component functions: f t gt ht Domain: 3. r t ln t i ,0 et j 0, tk Component functions: f t gt ht 4. r t ln t et t sin t i 4 cos tj tk sin t 4 cos t t Component functions: f t gt ht Domain: 0, 5. r t Ft Gt cos t i Component functions: f t gt ht Domain: , sin t j tk cos t i sin t j 2 cos t i tk Domain: 0, 6. r t Ft Gt ln t i ln t ln t Domain: 0, i sin t 0 j cos t sin t k 0 cos t 5t j 1i 1i 3t 2k 5t tj 4t j i 4t j 3t 2 3t 2k 3t 2 k 7. r t Ft Gt , cos2 t i sin t cos t j sin2 t k Domain: i 8. r t Ft Gt t3 3 j t 1 t 1 t k t 2 tt 2 t t 1 i t3 t 2 t3 t j t3 t 1 t3 t k t 1, Domain: , 1, 77 78 Chapter 12 12 2t i 1 2i Vector-Valued Functions t 1j 9. r t (a) r 1 (b) r 0 (c) r s (d) r 2 j 1 t 1 2 s r2 1 2i 1 2 s 2 1 t 2i 1j 2 1 2 1 2 s t 1 2i 1j 1 tj sj 2i 2i j j 2 2t 10. r t cos t i i 2 i 2 cos t r 2j i 2 sin t j 2t 1 2 ti tj 2 ti 2 (a) r 0 (b) r (c) r (d) r 4 2 sin ti j 2 sin cos i tj 2 sin j cos i 2 sin j 6 6 cos 6 6 6 6 11. r t ln t i 1 j t ln 2 i 3t k 1 j 2 6k ln 1 t 4 3 does not exist. j 3t 1 1 1 1 t t 4k j 31 1j tk 3 tk 0i j 3k (a) r 2 (b) r (c) r t (d) r 1 3 is not defined. 4 t ln t r1 4i ln 1 ln 1 ti ti 12. r t (a) r 0 (b) r 4 (c) r c (d) r 9 ti k 2i 2 t t3 2j e t4 k 8j c r9 e 1k 2i c 9 9 2 32 j 9 3i e c 24 k e t 32 9 t4 ti t t 9 32 j k e rt rt 3i 9 27j t4 e e 94 k 27 j 14. 94 k 4t k 3t 2 13. rt rt sin 3t i sin 3t cos 3t j 2 tk 2 ti t t 3t j 2 cos 3t t2 1 t2 4t 2 9t 2 16t 2 t1 25t 15. r t ut 3t 3t 3 1 t2 t2 2t 3 13 4t 8 5t 3 4 t3 t 2, a scalar. 4t 3 The dot product is a scalar-valued function. S ection 12.1 16. r t ut 3 cos t 4 sin t 2 sin t 6 cos t t 2 t2 t3 2t 2, a scalar. Vector-Valued Functions 79 The dot product is a scalar-valued function. 17. r t x ti t, y 2t j 2t, z t 2k, t2 2≤t≤2 18. r t x cos cos ti t,y y2 sin sin tj t,z t 2k , t2 1≤t≤1 Thus, z x 2. Matches (b) Thus, x2 1. Matches (c) 2t k, 0.1 ≤ t ≤ 5 3 2t 3 ln x. Matches (a) 19. r t x ti t, y t 2j t 2, z e0.75t k, e0.75t 2≤t≤2 20. r t x ti t, y ln t j ln t, z 2 3x Thus, y x 2. Matches (d) Thus, z 21. (a) View from the negative x-axis: (c) View from the z-axis: 0, 0, 20 22. r t x ti t, y tj t, z 2k 2⇒x y (b) 10, 0, 0 z y 1 2 3 3 and y 20, 0, 0 (b) View from above the first octant: 10, 20, 10 (d) View from the positive x-axis: 20, 0, 0 (a) 0, 0, 20 (c) 5, 5, 5 z 1 −3 2 2 −2 1 −1 2 3 y −3 −2 −1 2 1 3 3 y −2 −3 x 1 2 3 x 23. x y y y 3t t x 3 1 1 24. x y 1 1 t, y x t 25. x y t 3, y x2 3 y 7 6 5 4 3 2 t2 Domain: t ≥ 0 y 6 5 4 3 x 2 x 1 2 3 4 −5 −4 −3 −2 −1 −2 −3 2 x 12345 4 −2 −4 6 − 4 −3 −2 −1 −2 26. x t2 t, y t2 t 5 4 3 2 1 −1 y 27. x x2 cos , y y2 9 3 sin 2 1 −3 −2 y 1 Ellipse x 2 3 x −1 1 2 3 4 5 80 28. Chapter 12 x y x2 y2 2 cos t 2 sin t 4 Vector-Valued Functions y 29. x x2 9 x 3 sec , y y2 4 2 tan 12 9 6 3 −12 − 9 − 6 y 1 Hyperbola 1 x −1 −1 1 −3 −6 −9 −12 6 9 12 30. x x 2 2 cos3 t, y 23 2 sin3 t cos2 t 1 sin2 t 31. x y z 4t 2t t 1 2 3 32. x y y t 2t 3t 5 y 2 23 Line passing through the points: 3 Line passing through the points: 0, 5, 0 , z 6 4 −6 x2 3 y 3 2 y2 3 22 0, 6, 5 , 1, 2, 3 z 5 4 3 5 15 , 0, 2 2 (0, 6, 5) 5 ( 2 , 0, 15 ) 2 (2, − 2, 1) −3 −2 −2 −3 x x 2 3 3 1 (1, 2, 3) 3 (0, −5, 0) −6 45 6 y 2 −2 2 4 2 4 −4 −6 −4 x 6 y 33. x x2 4 z 2 cos t, y y2 4 t 1 2 sin t, z t 34. x x2 9 z 3 cos t, y y2 16 t 2 1 4 sin t, z t 2 35. x x2 z 2 sin t, y y2 e t z 6 2 cos t, z e t 4 Circular helix z 7 Elliptic helix z 4 −3 3 x 3 y −3 3 3 y 4 x 4 y x 36. x x t x y z t 2, y y2 ,z 4 2 4 4 3 2t, z 3 y 4 1 1 2 3 2 3 t 2 3 2 −4 −3 −2 −1 1 z 1 −1 −2 2 3 4 y 0 0 0 0 1 1 2 3 2 2 5 −3 4 4 3 x S ection 12.1 37. x y t x y z t, y x 2, z 2 2 4 16 3 Vector-Valued Functions 81 t 2, z 23 3x 23 3t 6 4 2 z 38. x cos t sin t t y2 t 1 t sin t t cos t (2, 4, 16 ) 3 y z 1 1 1 2 3 0 0 0 0 1 1 1 2 3 2 x 2 −2 −4 −6 5 y 2 4 16 3 x2 z t2 1 z 2 or x2 y2 z 4 z2 1 (− 2, 4, − 16 ) 3 Helix along a hyperboloid of one sheet 3 x 2 2 3 4 y 39. r t Parabola 12 ti 2 tj 32 tk 2 x z −3 −2 2 3 −2 −3 −4 −5 1 −2 −3 40. r t ti −1 32 tj 2 12 tk 2 1 z Parabola y 2 x 3 −1 1 −2 −3 1 2 −2 −3 3 y 41. r t Helix sin t i 3 cos t 2 z 2 −1 1 1 tj 2 1 cos t 2 3 k 2 42. r t Ellipse 2 sin t i 2 cos t j z 2 sin t k 2 −2 −1 1 2 3 4 y x −2 2 1 y 1 −1 2 x −1 43. 2π z (a) 2π z (b) 8π z (c) 2π z π −2 −2 2 x 2 y x −2 1 π −2 −3 −2 2 2 y x 4π −2 −2 2 2 y x π −2 1 2 y The helix is translated 2 units back on the x-axis. (d) 2 −2 2 −2 The height of the helix increases at a faster rate. (e) z The orientation of the helix is reversed. z −6 y π −6 π x 2π 6 x 6 y The axis of the helix is the x-axis. The radius of the helix is increased from 2 to 6. 82 Chapter 12 Vector-Valued Functions 13 th 2 13 t k has the roles 2 of x and y interchanged. The graph is a reflection in the plane x y. t2i tj z 5 4 3 1 12 3 2 y −2 2 3 4 5 x 1 12 3 y 44. r t ti z t2 j (a) u t rt 2j is a translation 2 units to the left along the y-axis. z (b) u t 5 4 3 2 1 1 2 3 4 5 x x 5 4 12 34 5 −2 y 3 2 5 4 3 2 rt 4k is an upward (c) u t shift 4 units. z 5 4 3 2 1 1 2 3 4 5 x 12 34 5 y (d) u t 13 t k shrinks the 8 z-value by a factor of 4. The curve rises more slowly. ti t2 j z 5 4 3 2 1 1 2 3 4 5 x 5 y r t reverses the (e) u t orientation. z 5 4 3 2 1 1 2 3 4 5 x 12 3 y 5 45. y 4 x t, then y ti 4 4 tj t. 46. 2x Let x 3y 5 0 1 2t 3 5j 5. 47. y Let x x 2 2 Let x rt t, then y ti t, then y ti t t 2 2. 2 2j rt rt 1 2t 3 48. y 4 x2 t, then y ti 4 4 t 2. t2 j 49. x2 Let x y2 25 5 cos t, then y 5 cos t i 5 sin t. 50. x Let x 2 2 y2 2 2 4 2 sin t. 2 sin t j Let x rt x2 16 Let x rt y2 4 2 cos t, y 2 cos t i rt 5 sin t j x2 16 Let x rt y2 9 rt 51. 1 4 sec t, y 4 sec t i 2 tan t. 2 tan t j 52. 1 4 cos t, y 4 cos t i 3 sin t. 3 sin t j 53. The parametric equations for the line are x rt 8 7 6 5 4 3 2 1 4 x 3 2 1 54. One possible answer is rt 1.5 cos t i 1.5i 1.5 sin t j 2 k. 1 t k, 0 ≤ t ≤ 2 2 2 z 2t, y 2t i 3 3 5t, z 5t j 8t. 8t k. One possible answer is Note that r 2 (0, 8, 8) (2, 3, 0) 45 67 8 y S ection 12.1 55. r1 t r2 t r3 t t i, 4 6 4t i t j, 6tj, 0≤t≤4 0≤t≤1 0≤t≤6 r1 0 r2 0 r3 0 0, r1 4 4i, r2 1 6j, r3 6 4i 6j 0 Vector-Valued Functions 83 (Other answers possible) 56. r1 t r2 t r3 t t i, 0 ≤ t ≤ 10 sin t j , ti r1 0 0, r1 10 4 10i r2 0 10i, r2 r3 0 4 5 2i 5 2i 5 2j 0 10 cos t i 5 21 0≤t≤ t j, 5 21 0≤t≤1 5 2 j, r3 1 (Other answers possible) 57. r1 t r2 t r3 t ti 2 4 t 2j , ti 0≤t≤2 y 4j, 0 ≤ t ≤ 2 x2 58. r1 t r2 t ti 1 t j, ti 0≤t≤1 y 1 t j, x x 0≤t≤1 y t j, 0 ≤ t ≤ 4 (Other answers possible) (Other answers possible) 59. z x2 y 2, x y 0 x t, z 2t2k 6 60. z t and z 2t 2. x2 y2 2t 2. x2 y 2, z x2 y 2 4 4 or 2 sin t, z 2sin t j 4k 4. Let x t, then y Therefore, x rt ( 2, − Therefore, x rt 2 cos t, y 2 cos t i z t, y ti 2, 4 ) 5 tj z (− 2, 2, 4) −3 2 3 x 1 2 3 y x 2 2 y 61. x2 x z t x y z rt y2 4, z x2 2 cos t 4 z 2 sin t, y x2 0 0 2 0 2 sin t i 4 sin2 t 6 1 3 1 4 2 2 2 2 cos tj 2 2 0 4 4 sin2 t k 3 4 2 2 2 0 2 0 −3 3 x 3 y 84 Chapter 12 4y2 z2 Vector-Valued Functions 16, x t2 and y z2 1 2 0 0 2 0 t2j 4 z 62. 4x2 If z t, then x 16 4t4 t2. 2 2 4 x t x y z rt 1.3 1.69 0.85 1.3 t2i 1 2 z2 1 1.2 1.44 1.25 1.2 16 1 1 1.66 1 4t4 1 1 1.66 1 tk 1.2 1.44 1.25 1.2 2 y 63. x2 Let x y2 4, x z 2 2 1 x sin t 2 z sin t, then z 2 1 2 sin t and x2 2 sin2 t y2 y2 4 z2 4. −3 3 1 y2 x z t x y z sin t y2 y ± ± 2 cos2 t, 1 1 2 cos t 2 cos t x 3 3 y sin t, y sin t −3 2 0 0 2 6 1 2 6 ± 2 3 2 ± 0 1 2 1 6 3 2 6 ± 2 1 2 2 2 0 0 rt rt 1 1 sin t i sin t i 2 cos tj 2 cos tj 1 1 sin t k and sin t k 64. x2 Let x y2 2 z2 10, x sin t, then y y 2 4 sin t and z 21 sin2 t 2 cos t. z t x y z 2 1 3 0 6 3 2 5 2 6 2 sin t i 2 0 2 2 2 6 5 2 3 2 6 2 2 3 1 0 2 2 x 4 4 4 y 2 rt 2 sin t j 2 cos t k 65. x2 z2 4, y 2 z2 2 4 y 2 z Subtracting, we have x 0 or y ± x. 3 (0, 0, 2) Therefore, in the first octant, if we let x rt ti tj 4 t2 k t, then x t, y t, z 4 t 2. 4 x 3 2 4 y (2, 2, 0) S ection 12.1 66. x2 Let x y z 1 t 8 y2 z2 t, then 4 t t4 and 16t 2 x2 16 8 43 y2 z2 t2 16 t2 z2 16. 4 x Vector-Valued Functions 85 16, xy 4 (first octant) 4 z 4 y 4 3≤t≤ t x y z rt 8 43 1.0 3.9 0 1.5 1.5 2.7 2.6 2 2 2 2.8 16t 2 2.5 2.5 1.6 2.7 16 k 3.0 3.0 1.3 2.3 3.5 3.5 1.1 1.6 8 43 3.9 1.0 0 ti 4 j t 1 t 2 z 16 12 8 4 t4 67. y2 z2 2t cos t 2t sin t 2 4t2 4x2 68. x2 y2 160 120 80 40 e t cos t z 2 e t sin t 2 e 2t z2 7 x 6 5 4 8 40 120 80 12 16 x y 40 80 120 y 69. lim t i t→2 t2 t2 4 j 2t 1 k t 2i 2j 1 k 2 70. lim et i t→0 sin t j t e tk i j k since lim t→2 since t2 t2 4 2t lim t→2 2t 2t 2 2. (L’Hôpital’s Rule) lim t→0 sin t t lim t→0 cos t 1 1 (L’Hôpital’s Rule) 71. lim t 2 i t→0 3t j 1 cos t k t 0 72. lim t→1 ti ln t j t2 1 2t 2 k i 1 j 2 2k since lim t→0 since 1 cos t t lim t→0 sin t 1 0. (L’Hôpital’s Rule) lim ln t 1 1 j t t→1 t 2 lim t→1 1t 2t t 1 . (L’Hôpital’s Rule) 2 73. lim t→0 1 i t cos t j sin t k 1 does not exist. t 74. lim e t i t→ t2 1 k 0 does not exist since lim t→0 since t→ lim e t 0, lim t→ 1 t 0, and lim t→ t t2 1 0. 86 Chapter 12 1 j t Vector-Valued Functions 75. r t ti 76. r t , 0 , 0, ti t 1j 77. r t ti arcsin t j 1, 1 t 1k Continuous on Continuous on 1, Continuous on 78. r t 2e t, e t, ln t 1 . 79. r t e t, t 2, tan t 2 2 n n, n Continuous on t 1 > 0 or t > 1: 1, Discontinuous at t Continuous on 2 80. r t 8, t, 3 t 81. See the definition on page 832. Continuous on 0, 82. No. The graph is the same because r t ut 2. For example, if r 0 is on the graph of r, then u 2 is the same point. 83. r t (a) s t (b) s t (c) s t t2i rt rt rt t 3j 3k 2i 5j tk t2i t2 t2i t 2i t 3j t 2j t 3j tk 3k tk 84. A vector-valued function r is continuous at t lim r t t→a a if the limit of r t exists as t → a and ra. i i j t≥0 is not continuous at t j t<0 0. The function r t 85. Let r t t→c x1 t ut y1 t j lim t→c t→c z1 t k and u t y1 t z2 t t→c x2 t i y2 t j z2 t k. Then: x2 t z1 t j t→c t→c lim r t y2 t z1 t i t→c x1 t z2 t t→c x1 t y2 t t→c x2 t y1 t k t→c lim y1 t lim z2 t lim y2 t lim z1 t i lim x1 t lim y2 t t→c t→c lim x1 t lim z2 t lim x2 t lim y1 t k t→c t→c lim x2 t lim z1 t j lim x1 t i t→c lim y1 t j t→c lim z1 t k t→c lim x2 t i t→c lim y2 t j t→c lim z2 t k t→c lim r t t→c lim u t t→c 86. Let r t t→c x1 t i ut y1 t j t→c t→c z1 t k and u t x2 t i y2 t j z2 t k. Then: lim z1 t lim z2 t t→c t→c lim r t lim x1 t x2 t lim x1 t lim x2 t t→c y1 t y2 t t→c z 1 t z2 t t→c lim y1 t lim y2 t lim z1 t k t→c lim x1 t i t→c lim y1 t j t→c lim x2 t i t→c lim y2 t j t→c lim z2 t k t→c lim r t t→c lim u t t→c 87. Let r t x t i y t j z t k. Since r is continuous at t c, then lim r t rc. t→c 88. Let ft 1, 1, if t ≥ 0 if t < 0 0, rc xci ycj zck ⇒ xc, yc, zc 2 2 2 2 are defined at c. r lim r t→c xt xc 2 2 yt yc zt zc and r t f t i. Then r is not continuous at c whereas, r 1 is continuous for all t. rc Therefore, r is continuous at c. S ection 12.2 89. True 4, 16 u 2 , they do not 91. False. Although r 4 collide. Their paths cross this point at different times. Differentiation and Integration of Vector-Valued Functions 90. False. The graph of x 92. True. y 2 z2 t 2 sin2 t y z t 3 represents a line. t2 x 87 t 2 cos 2 t Section 12.2 1. r t xt x y 2 Differentiation and Integration of Vector-Valued Functions 2 4 2 y t 2i t j, t0 t 2. r t (4, 2) ti t 3 j, t0 t3 1 4 3 2 y t 2, y t r′ xt y x t, y t x3 r1 rt i i i r' (1, 1) x 1 2 3 4 r r 1 r2 rt r2 4i 2t i 4i 2j −2 2 4 6 8 j 3t 2j 3j −4 −3 −2 −2 −3 −4 j −4 j r1 r t0 is tangent to the curve. 1 j, t0 t 1 t r t0 is tangent to the curve. 3. r t xt x 1 y2 t 2i 2 2 y 4. (a) r t r 1 1 t3 t x 2i i i ti 1 j 3t 2j 3j 3 t 3 j, t0 1 3 y t 2, y t 1 (4, ( 1 2 r′ x x y (b) r 1 rt r1 r′ 2 1 (2, 1) r x 1 2 3 2 r2 rt r2 4i 2t i 4i 1 j 2 1 j t2 1 j 4 −1 −1 (1, 0) r 1 is tangent to the curve. r t0 is tangent to the curve. 5. r t xt x2 cos t i sin t j, t0 sin t 2 6. (a) r t x y e ti e t, y e 2t j, t0 e2t 0 x 2, x > 0 cos t, y t y2 r 2 rt r 2 1 j x2 ⇒ y 2 r′ sin t i i cos t j 1 (1, 1) r (0, 0) x 1 2 r t0 is tangent to the curve. y (b) r 0 rt i et i i j 2e 2 tj 2j r′ (0, 1) r0 x r 1 r 0 is tangent to the curve. 88 Chapter 12 ti y 8 16 Vector-Valued Functions 8. r t (a) 5 7. r t (a) 6 16 4 16 2 16 t 2j ti 4 y t2 j r (1) 1 1 3 r2 r4 r2 1 2 1 −3 −1 −1 r (1.25) r (1.25) − r (1) x 3 r4 1 x 2 16 4 16 6 16 8 16 (b) 1 r 4 r r 1 2 r 1 2 1 4 rt r r1 2 12 1 4 1 i 4 1 i 2 1 i 4 i i 1 j 16 1 j 4 3 j 16 2t j 1 j 2 1 4i 14 1 4 (b) r1 r 1.25 r 1.25 r1 rt i 3j 2.4375 j 0.5625 j 1.25 i 0.25 i i r1 2t j i (c) 2j 0.5625j 0.25 i 2.25 j (c) r 1.25 1.25 r1 1 0.25i This vector approximates r 1 . 3 16 j . 3 2 3 k, t0 2 3 j 4 r1 4 14 i This vector approximates r 9. (a) and (b) r t x2 y2 rt r r 3 2 3 2 z 2 cos t i t 2 sin t i 2j 2i k 3 k 2 2 sin t j t k, t0 10. r t y ti x 2, z rt r2 r2 z t2j 3 2 i 2i i 2 4, z 2 cos t j k 2t j 4j 4j 3 k 2 (0, − 2, 32π ) r′ 2π −2 2 −2 2 x 4 −4 rπ −2 2 x 1 2 y r −4 −6 y r' −6 11. r t rt 6t i 6i 7t 2j 14t j t 3k 3t 2k 12. r t rt 1 i t 1 i t2 4 ti 2 t 2 t i i 16t j 16 j t2 k 2 tk 13. r t rt a cos3 t i a sin3 t j k 3a sin2 t cos t j 14. r t rt t2 t j 2t t 5t3 2 j 2 ln t2 k t2 2t 2 k t j 2 k t 3a cos2 t sin t i S ection 12.2 15. r t rt e ti e ti 4j Differentiation and Integration of Vector-Valued Functions 16. r t rt sin t t cos t, cos t t sin t, t 2 89 t sin t, t cos t, 2t 17. r t rt t sin t, t cos t, t sin t t cos t, cos t t sin t, 1 18. r t rt arcsin t, arccos t, 0 1 1 t2 , 1 1 t2 ,0 19. r t t 3i 12 tj 2 3t 2i 6t i rt tj j 3t 2 6t 4 sin tj 4 cos t j 4 sin t j 4 sin t 0 4 cos t 4 cos t 4 sin t t 18t 3 t 20. r t (a) r t rt (b) r t t2 ti 2t 2i rt t2 1i 2j 2t tj 2t 1j (a) r t rt (b) r t 21. r t 12 2t 12 8t 4 cos t i 22. r t 8 cos t i 3 sin tj 3 cos t j 3 sin t j 8 cos t 3 cos t 3 sin t (a) r t rt (b) r t 4 sin t i 4 cos t i rt (a) r t rt (b) r t 8 sin t i 8 cos t i rt 8 sin t 55 sin t cos t 23. r t 12 ti 2 ti i tj j tk 13 tk 6 12 tk 2 24. r t ti i 2t 2j 3j 3k 3t 5k (a) r t rt (b) r t (a) r t rt (b) r t 0 rt 0 rt t1 10 12 tt 2 t cos t, t t cos t, cos t t t3 2 25. r t (a) r t cos t t sin t, sin t sin t sin t cos t t sin t, 1 t cos t, t sin t, 1 rt (b) r t 26. r t (a) r t rt (b) r t cos t rt t sin t, sin t t cos t cos t t cos t, 0 t sin t t sin t sin t t cos t t e t, t2, tan t e t, 2t, sec2 t e t, 2, 2 sec2 t tan t rt e 2t 4t 2 sec4 t tan t 90 Chapter 12 Vector-Valued Functions 1 4 r ′′ r ′′ 27. rt rt r r r r 1 4 14 14 rt r r r r 1 4 14 14 1 4 1 4 cos ti sin sin ti 2 j 2 tj cos t 2k, t0 tj 2t k z 2 i 2 2 2 1 4 2 2 1 k 2 2 2 2 2 1 2 k 2k 2 2 1 4 4 2 1 x r′ r ′ y 2 1 ti i 2 2 2i 2 2j sin j tj 2k 2 2 22 cos 2 2 2 2 2 22 2 2 2 2j 2 4 4 1 2 4 4 2 2i 4k 28. rt rt r 1 r 4 r14 r 14 rt r r 1 4 1 4 1 4 ti i i t 2j 2 tj 1 j 2 12 1 20 e0.75t k, t0 0.75e0.75t k 0.75e0.1875 k 1 2 9e3 8 2 1 4 ( 1) 1 r' ( 4 ) r' 4 z 2 ( 1) 1 r'' ( 4 ) r'' 4 −2 −1 i 2 16 1 j 2 33 e 4 5 4 −2 16k −1 1 1 33 e 4 2j 93 e 16 8 20 4 9e3 8 2 x −1 2 y −2 4i 3e3 16k 2i 2i 9 0.75t e k 16 93 e 16 22 1 1024 81e3 8 16k 93 e 16 2 16 4 9e3 16k 81 3 e 256 8 1024 16 81 e3 8 r 14 r14 32j 29. r t rt r0 t 2i 2t i 0 t 3j 3t 2j 30. r t rt , 0 , 0, 1 t t 1 i 1 1 3t j i 3j 1 2 Smooth on Not continuous when t Smooth on , 1 , 1, S ection 12.2 31. r r r n 2 0 , n any integer. 2 cos3 i 6 cos2 3 sin3 j sin i 9 sin2 cos j Differentiation and Integration of Vector-Valued Functions 32. r r r 2n 1 1 sin cos i i 1 cos j 91 sin j 0, n any integer 1 , 2n 1 Smooth on 2n n n1 , Smooth on 2 2 33. r r r 1 2 sin 2 cos i i 1 2 cos j j 34. r t rt rt 2t 8 16 t3 t3 i 2t2 8 t3 32t t3 j 2t 4 j 82 2 sin 0 for any value of , 4t3 i 82 Smooth on 0 for any value of t. 2. 2, r is not continuous when t Smooth on 1 j t 2t k 0: , 2, 35. r t rt i t 1i 1 j t2 t 2k 0 , 0 , 0, 36. r t rt et i et i e tj e tj 3t k 3k 0 , r is smooth for all t : r is smooth for all t 37. r t rt ti i 3t j 3j tan t k sec2 t k 2 0 n 2n 2 2 4t i n, 2 t 3k 1 38. r t rt . n ti 1 i 2t t2 2t j 1j 1 k 4 1 tk 4 0 r is smooth for all t r is smooth for all t > 0: 0, Smooth on intervals of form 39. r t ti 3t j i ut ut ut ut 3j t 2 k, u t 2t k 3t 3 8t 2t 4i 8t 3i 9t 2 t4 t 2j (a) r t (c) r t (b) r t t5 5t 4 4t 3 j 12t 2 t3 4t 3 j 12t 2 k 3t 2 24t k (d) 3r t Dt 3r t (f) r t Dt r t 2k ut ut 10t 2 t4 ti i 9t 9 t 10 t2 j 2t j t2 3t 2 6t t3 k 3t 2 k 4t 2 Dt r t (e) r t Dt r t 10 2t 2 10 t 2 40. r t ut ti 1 i t 2 sin tj 2 sin tj i ut ut 2 cos tk 2 cos t k 2 sin t k 4 cos2 t 5 (b) r t (d) 3r t Dt 3r t 2 sin tj ut ut 3t 3 2 cos tk 1 i t 4 sin t j 1 i t2 4 cos tk 4 sin tk (a) r t (c) r t 2 cos t j 1 4 sin2 t 0, t 0 Dt r t 4 cos tj —CONTINUED— 92 Chapter 12 Vector-Valued Functions 40. —CONTINUED— (e) r t ut i t 1t j k 2 sin t 2 cos t 2 sin t 2 cos t tj 1 t 2 sin t t t 1 t 1 k t 1 t2 1 t2 1j k (f) r t Dt r t t2 4 12 t 2 4 12 1 2 cos t t Dt r t ut 2t t t2 4 2 sin t 2 cos t 2 sin t 1 2 cos t t 41. r t (a) r t ti 2t 2 j ut ut t7 t 3 k, u t t 4k Dt r t 7t6 Alternate Solution: Dt r t ut rt ti 4t 6 (b) r t Dt r t ut ut i t 0 j 2t 2 0 12t 5i ut 2t j 3t 6 k t3 t4 5t 4j 2 rt tk 7t 6 2t 6i 3 ut 4t 3k i 4tj 3t 2k t 4k t 5j Alternate Solution: Dt r t ut rt i t 0 42. r t (a) r t cos t i ut ut sin t j sin t ut j 2t 2 0 t k, u t t2 cos t 2t k t3 4t 3 j rt i 1 0 tk ut j 4t 0 k 3t 2 t4 12t 5i 5t 4j Dt r t Alternate Solution: Dt r t ut rt ut rt sin t j t k t t sin t tk 2t ut k cos t ti cos t t cos t j t sin t j cos t k sin t k sin t i cos t j k j tk cos t i t (b) r t Dt r t ut ut cos t i j cos t sin t 0 1 t cos t t sin t 1i Alternate Solution: Dt r t ut rt ut rt k t 1 ut i j sin t cos t 0 1 k 1 t sin t t cos t 1i t sin t cos t j sin tk i j cos t sin t 0 0 S ection 12.2 43. rt rt rt rt cos 3 sin t i 3 cos t i 4 cos t j 4 sin tj 16 cos t sin t 9 sin2 t Differentiation and Integration of Vector-Valued Functions π 93 9 sin t cos t rt rt arccos rt rt 7 sin t cos t 7 sin t cos t 16 cos2 t 9 cos2 t −1 0 7 16 sin2 t 9 sin2 t 3.927 2.356 n ,n 2 7 sin t cos t 16 cos2 t 9 cos2 t 5 4 3 4 and t and t 0.785 5.498 16 sin2 t 4 . 1.855 maximum at t 1.287 minimum at t 1.571 for t 7 . 4 2 44. rt rt rt rt rt cos 0, 1, 2, 3, . . . t2i 2t i 2t 3 t4 t4 arccos tj j t t 2, r t 2t 3 t2 t4 t 4t 2 2t3 1 1 0.707 2 . 2 4t 2 1 0 1.0 − 0.5 t t2 4t 2 0.340 19.47 maximum at t 2 for any t. 45. r t lim lim rt 3 ti t t 2t rt t t t→0 lim t 3t 2 t 2i 1 t t t 2 j 3t 2i 1 t2 j t→0 j t→0 lim 3i t→0 2t tj 3i 2t j 46. r t lim rt t t t ti rt 3 t t 3 j t t→0 j 3 2t t 3 t tk ti 2tk lim t→0 lim t t→0 t t t t i t t t j 2k lim t→0 t t 1 t t i t i t 3 t tt 3t j tt t j 2k 2k lim t→0 t 3 j t2 t 2k 1 i 2t 94 Chapter 12 rt t 2t t Vector-Valued Functions t t t 2, 0, 2 t t t 2, 0, 2 t t t, 0, 2 t t2, 0, 2t rt rt 0, sin t 0, sin t t t t ,4 t t cos t t cos t, 4 t 1 sin t cos t t cos t sin t, 4 t t 0, sin t, 4t rt 47. r t lim lim lim t→0 48. r t lim lim t→0 t→0 t→0 t→0 lim t→0 lim 2t t→0 lim 0, t→0 sin t cos 2t, 0, 2 sin t t ,4 0, 0 0, cos t, 4 832 tk 3 49. 2t i j k dt t 2i tj tk C 50. 4t3i 6tj 4 t k dt t 4i 3t 2 j C 51. 1 i t j t 3 2k dt ln t i tj 252 tk 5 C 52. ln t i 1 j t k dt t ln t ti ln t j tk C (Integration by parts) 53. 2t 1i 4t 3j 3 t k dt t2 ti t 4j 2t 3 2k C 54. et i sin t j cos t k dt et i cos t j sin t k C 55. sec2 t i 1 1 t2 j dt tan t i arctan t j C 56. e t sin t i e t cos t j dt et 2 1 sin t cos t i et 2 cos t sin t j C 1 57. 0 8t i tj k dt 4t 2i 0 t2 j 2 1 1 1 1 tk 0 0 4i 1 j 2 1 k 1 58. 1 ti t 3j 3 t k dt t2 i 2 t4 j 4 1 1 343 tk 4 2 0 1 2 2 2 59. 0 a cos t i a sin t j k dt a sin t i 0 a cos t j 0 tk 0 ai aj 2 k 4 4 60. 0 sec t tan t i tan t j 2 sin t cos t k dt sec t i 2 ln sec t j 1i ln 2 j sin2tk 1 k 2 t 2j ti 0 2 61. 0 ti e tj te tk dt t2 i 2 2i 2 2 2 e tj 0 0 t e2 1 et k 0 62. t i 3 t2 3 t4 t1 0 t1 t 2dt 3 t 2 for t ≥ 0 e2 1j 1k 0 t 2j dt 1 1 3 1 10 3 3 2 t2 32 0 1 S ection 12.2 Differentiation and Integration of Vector-Valued Functions 95 63. r t r0 rt 2i 4e2t i 3j 3et j dt C 3 et 2e2t i 3et j 3j C 64. r t r0 rt C i 3t 2 j i 2 6 t k dt 2j t3 j 4t3 2k t3j 4t3 2k C 2i ⇒ C 1j 2e2ti 65. r t r0 rt rt C1 32j dt 600 3 i 32t j 600j C1 66. r t rt r0 rt 4 cos tj 4 sin tj 3k 3k 3 sin t k 3 cos t k C1 ⇒ C1 3 sin t k C2 0 4 j ⇒ C2 3 sin t k C1 0 600 3 i 600 3 i 600 3 t i 600 600 600t 32t j 32t j dt 16t 2 j C 4 cos tj 4j C2 r0 rt 4 cos tj r0 rt C 0 600t 16t 2 j 1 e 2 600 3 t i 67. r t r0 rt te 1 i 2 1 t2 i j e tj C t2 k dt 1 i 2 e t t2 i e tj i tk 2j 2 e 2 k C j 2j k⇒C t 1k 1 e 2 1 1 t2 j t2 i i e t 2j t 1k 68. r t r1 rt i C 1 j t2 1 k dt t arctan t i 2 1 j t i ln t k 1 j t j ln t k C 4 2 i 2i ⇒ C 1 4 4 arctan t i 69. See “Definition of the Derivative of a Vector-Valued Function” and Figure 12.8 on page 840. 70. To find the integral of a vector-valued function, you integrate each component function separately. The constant of integration C is a constant vector. 72. The graph of u t does not change position relative to the xy-plane. cz t k and 71. At t t0, the graph of u t is increasing in the x, y, and z directions simultaneously. 73. Let r t xti ytj cx t i cx ti 74. Let r t x1 t i rt ±ut Dt r t ± u t y1 t j z t k. Then cr t cy t j y tj cz t k z tk cr t . x2 t i y2 t j cx t i cy t j Dt c r t z1 t k and u t z 2 t k. z1 t ± z 2 t k z1 t ± z2 t k y2 t j z2 t k x1 t ± x2 t i x1 t ± x2 t i x1 t i y1 t j r t ±u t y1 t ± y2 t j y1 t ± y2 t j z1 t k ± x2 t i 96 Chapter 12 xti Vector-Valued Functions ytj z t k, then f t r t f txt i y tj f trt z1 t k and u t z1 t y2 t i y1 t z2 t x1 t y2 t z1 t y2 t i z1 t y2 t i rt ut xft i yft j z f t k and (Chain Rule) 75. Let r t f txti f ty t f tytj f tyt j ytj f t z t k. f tz t ztk f tzt k Dt f t r t f tx t f t x ti f tr t z tk f t xti 76. Let r t rt Dt r t x1 t i ut ut y1 t j x2 t i y2 t j z 2 t k. z1 t x2 t j z1 t y2 t i y1 t x2 t k z1 t x2 t j z1 t x2 t j x1 t y2 t x1 t y2 t y1 t x2 t k y1 t x2 t k x1 t y2 t x1 t z2 t y1 t x2 t k x1 t z2 t z1 t x2 t z1 t x2 t j y1 t z2 t y1 t z2 t x1 t y2 t y1 t z2 t x1 t z2 t z1 t y2 t y1 t x2 t x1 t z2 t y1 t z2 t rt 77. Let r t xti ytj ut x1 t z2 t z t k. Then r f t y f t f tj y ft j Dt r f t x f t f ti f t x ft i z f t f tk z ft k x ti f tr f t . z t k. xty t xtz t ytx t k x tz t ztx t z tx t j 78. Let r t rt Dt r t xti rt rt ytj z t k. Then r t ytz t ytz t xty t ytz t zty t i y tz t x ty t zty t i y tj xtz t zty t ytx t xtz t ztx t j z ty t i y tx t k ztx t j xty t ytx t k rt rt 79. Let r t rt Dt r t x1 t i ut ut y1 t j vt vt z 1 t k, u t x1 t y2 t z3 t x1 t y2 t z 3 t x2 t i y2 t j z2 t k, and v t y1 t x2 t z3 t x1 t y2 t z 3 t x3 t i z2 t x3 t y3 t j z3 t k. Then: y2 t x3 t z2 t y3 t x1 t y2 t z 3 t z1 t x2 t y3 t x1 t y3 t z 2 t y1 t x2 t z 3 t z1 t x2 t y3 t z1 t y2 t x3 t z2 t x3 t z2 t x3 t z2 t x3 t ut vt z1 t x2 t y3 t y2 t x3 t y2 t x3 t y2 t x3 t y1 t x2 t z3 t z1 t x2 t y3 t x1 t y3 t z2 t y1 t z2 t x3 t z1 t x2 t y3 t x1 t y2 t z3 t x1 t y2 t z3 t x1 t y2 t z3 t rt 80. Let r t xti ytj x2 t Dt 2x t x t 2xtx t x2 t ut vt x1 t y3 t z2 t y1 t z2 t x3 t z1 t y2 t x3 t y3 t z2 t y3 t z2 t y3 t z2 t rt ut y1 t y1 t x2 t z 3 t y1 t z2 t x3 t z1 t y2 t x3 t x2 t z3 t x2 t z3 t x2 t z3 t rt y1 t y1 t vt z1 t x2 t y3 t z1 t x2 t y3 t z t k. If r t y2 t y2 t z2 t z2 t r t is constant, then: C Dt C 0 0 0 2y t y t yty t 2z t z t ztz t rt 2rt Therefore, r t rt 0. S ection 12.3 81. r t (a) 5 Velocity and Acceleration 97 t sin t i 1 cos t j 82. r t 2 cos t i 3 sin t j y (a) Ellipse 2 1 0 0 40 −3 −1 x −1 −2 1 3 The curve is a cycloid. (b) rt rt rt 1 sin t i 1 cos t i cos t j 2 cos t cos2 t 0 sin2 t 2 2 cos t sin t j (b) rt rt rt 2 sin t i 2 cos t i 4 sin2 t 3 cos t j 3 sin t j 9 cos2 t 2 0 Minimum of r t is 0, t Maximum of r t is 2, t rt sin2 t cos2 t Minimum of r t is 2, t Maximum of r t is 3, t 1 rt 4 cos 2 t 9 sin2 t 0 2 Minimum and maximum of r t is 1. Minimum of r t is 2, t Maximum of r t is 3, t 83. True 85. False. Let r t rt d rt dt rt rt e t sin t i e t cos t e t sin t 2e t cos t i rt rt 1 e t cos t j e t sin t i e t cos t 2e t sin t j 2e 2 t sin t cos t 0 e t cos t e t sin t e t sin t j e t cos t i e t cos t 0 sin t i cos t j cos t i 2 sin tj k. 84. False. The definite integral is a vector, not a real number. 86. False Dt r t ut rt ut rt ut (See Theorem 11.2, part 4) 87. r t rt rt e t sin t e t sin t e t cos t j 2e 2 t sin t cos t Hence, r(t is always perpendicular to r (t . Section 12.3 1. r t vt at x 3t i rt rt 3t, y t 1. 3i t 3i 0 Velocity and Acceleration 1j j 2 y 2. r t vt (3, 0) 4 6 rt rt 6 ti i 0 tj j 4 y v x 6 at x v 1, y x 3 (3, 3) 1 −2 −4 t, y t, y 6 x 2 At 3, 0 , t v1 x 2 4 j, a 1 0 98 Chapter 12 t2 i rt rt t, y 2 Vector-Valued Functions y 3. r t vt at x tj 2t i 2i t, x 2. 4i 2i 2 sin t j 2 sin t i 2 cos t i 2 sin t, x 4 2i 2i . 2 4. r t vt (4, 2) v a x 2 4 6 8 t2i rt rt t2, y t3j 2t i 2i t3, x 3t2j 6tj y2 3 8 7 6 5 4 3 2 1 −1 y j 4 2 a at x v y2 −2 At 4, 2 , t v2 a2 5. r t vt at x At v a At 1, 1 , t v1 a1 6. r t 2i 2i 1. 3j 6j 2 sin tj 2 cos tj 2 sin tj 2 sin t, 0. x2 9 (1, 1) x 234567 8 j −4 2 cos t i rt rt 2 cos t, y 2, 2 ,t 3 cos ti 3 sin ti 3 cos ti 3 cos t, y 2 cos t j 2 sin t j y2 4 y 3 vt at x y2 4 1 Ellipse y 3 At 3, 0 , t v 4 4 2j a ( 2, 2) x 3 v0 a0 2j 3i −2 −1 1 v a x 2 1 2j −3 −1 (3, 0) −3 −3 7. r t vt at x At v a t t rt rt sin t, 1 1 cos t cos t, sin t 8. r t vt at x e e t, et rt rt t y e t, et e 1 ,y et 0. i i j j t, 2 sin t, cos t 1 . 2i 1 j 4 et 1 v (1, 1) a sin t, y ,2 , t 2, 0 0, cos t (cycloid) y et, y 1 x 1 2 x At 1, 1 , t (π , 2) a π 2π x 2 v v0 a0 1, 1 1, 1 9. r t vt st at ti i vt 0 2t 2j 5j 3k 1 4 3t k 10. r t vt 4t i 4i vt 0 4t j 4j 2t k 2k 16 16 4 6 11. r t vt st at ti i 1 2j t 2j 2t j 4t 2 k t2 k 2 tk t2 1 5t 2 9 14 st at 12. r t vt st at 3ti 3i tj j 9 12 tk 4 1 tk 2 1 12 t 4 10 12 t 4 13. r t vt st at ti i j tj 9 1 9 t2 9 t t2 k t2 t2 k 18 9 t2 t2 1 9 3 2k t2 1 k 2 9 S ection 12.3 14. r t vt st at 2i t2 i 2t i 4t 2 tj j 1 3 k 2t 2t 3 2k 3 tk 9t 4t 2 9t 1 15. r t vt st at Velocity and Acceleration 99 4t, 3 cos t, 3 sin t 4, 16 0, 3 sin t, 3 cos t 9 sin2 t 3 cos t, 4i 3 sin t j 5 3 cos t j 3 sin t k 3 cos tk 9 cos2 t 3 sin t 16. r t vt st at e t cos t, e t sin t, e t e t cos t e2 t cos t 2e t sin t i e t sin t i sin t 2 e t sin t e2 t cos t etk e t cos t j sin t 2 etk e2 t et 3 2e t cos t j t3 ,t 40 3t 2 4 17. (a) rt rt r1 x 1 t, 1, 1, t, y 0.1 t 2, 2t, 2, 1 18. (a) rt rt r3 t, 1, 1, 3 t, y 0.1 25 t 25 3 , 4 z 3 t 2, t2 3 4 4 , 25 t 25 t 2 , t0 t2 3 3 4 1 2t, z 0.1, 1 1 4 3 t 4 1 4 3 0.1 4 x (b) r 3 3 t 4 3 0.1 , 4 4 3 0.1 4 (b) r 1 1 2 0.1 , 0.1, 4 1.100, 1.200, 0.325 3.100, 3.925, 3.925 19. a t vt v0 rt r0 r2 21. a t vt v1 vt rt r1 rt r2 i i C ti C 2i tj j j k, v 0 k dt ti t k dt t2 i 2 k 2i 0, r 0 ti tj t2 2 j 2j tj 0 tk C ti k C j k 20. a t vt v0 rt r0 r2 2i 2i C 3k 3k dt 4j ⇒ v t 2ti 4j 2ti 2ti 3tk 4j t2i 4tj C 3tk 4tj 32 tk 2 32 tk 2 C 0, v t tj 0, r t j t k, v t i k, 2k 0 t2 k 2 C 9 j 2 j 3tk dt t2i C 4i 0 ⇒ rt 8j 6k t k, v 1 tj t k dt 1 k 2 9 j 2 t2 2 9 j 2 1 k 3 9 t 2 2 k 3 C C 5j, r 1 t2 j 2 1 j 2 t2 2 5j ⇒ C t2 2 t2 2 1 k 2 1 k 2 1 k dt 2 t3 6 14 j 3 1 k 3 9 tj 2 1 k 3 t3 6 1 tk 2 C 14 j 3 t3 6 17 j 3 0⇒C t3 6 1 t 2 14 j 3 100 22. a t vt v0 vt rt r0 rt r2 Chapter 12 cos t i cos t i j C sin t i sin t i i C j Vector-Valued Functions sin t j, v 0 sin t j dt k⇒C cos t j cos t j 0 tk 2k 24. (a) The speed is increasing. (b) The speed is decreasing. k k dt cos t i sin t j tk C k j k, r 0 sin t i i cos t j C i⇒C sin t j cos t i cos 2 i sin 2 j 23. The velocity of an object involves both magnitude and direction of motion, whereas speed involves only magnitude. 25. r t 88 cos 30 t i 44 3 t i 10 10 44t 88 sin 30 t 16t 2 j 16t 2 j 50 0 0 300 26. r t 900 cos 45 t i 450 2t i 3 3 900 sin 45 t 16t 2 j 16t 2 j 450 2t 450 2 The maximum height occurs when y t The maximum height reached by the projectile is y 3 450 2 225 2 16 16 225 2 16 3 2 32t 50,649 8 0, which implies that t 225 2 16. 6331.125 feet. 0 which implies that The range is determined by setting y t t Range: x 450 2 32 450 2 450 2 32 h v0 sin v0 t 2 16t 2 t 405,192 450 2t 16t 2 39.779 seconds. 405,192 25,315.500 feet v0 ti 2 v0 t 2 27. r t v0 t 2 t v02 v0 cos ti 12 gt j 2 3. 300 2 v0 2 3 16t 2 j 300 when 3 300 2 v0 300 2 , v0 v0 2 300 32 , v0 tv0 2 9600 16 0, 300 40 6 3002 32 v02 97.98 ft sec 0 40 6, v0 40 6 t 2 The maximum height is reached when the derivative of the vertical component is zero. yt yt t 3 16t 2 32t 53 4 53 4 3 40 3 53 4 16 53 4 2 3 0 16t 2 3 40 3 t 16t 2 40 3 40 3 32 Maximum height: y 78 feet S ection 12.3 220 ft/sec 3 220 cos 15 t i 3 5 220 sin 15 t 3 16t 2 j Velocity and Acceleration 101 28. 50 mph rt The ball is 90 feet from where it is thrown when x 220 cos 15 t 3 90 ⇒ t 27 22 cos 15 1.2706 seconds. The height of the ball at this time is y 5 220 sin 15 3 27 22 cos 15 x v0 cos h 16 x2 v0 cos2 2 16 27 22 cos 15 2 3.286 feet. 29. x t yt y t v0 cos t v0 sin x v0 cos x or t 16t 2 v0 sin h tan x 16 sec2 v02 x2 h 30. y 0.005x 2 is the coefficient of x. Therefore, tan 1, 4 rad 45 . Also From Exercise 29 we know that tan 16 sec2 v02 16 2 v02 rt negative of coefficient of x 2 0.005 or v0 40 2 t i 60, 32 4 40 2 40 2 32t j 24 2 j 8 2 5i 2j . direction 80 ft/sec 16t 2 j. Position function. 40 2t When 40 2 t t vt v 32 4 Speed 60 40 2 40 2 i 40 2 i 32 4 v 8 2 25 4 8 58 ft sec 31. r t (a) y (b) 18 ti 0.004t2 0.004x2 0.3667t 6 6j (c) y 0.008x 0.3667 14.4 feet. 0⇒x 45.8375 and 0.3667x y 45.8375 (d) From Exercise 29, tan 0 0 120 0.3667 ⇒ sec2 v0 2 20.14 16 sec2 0.004 4000 cos2 16 0.004 ⇒ v02 67.4 ft sec. ⇒ v0 102 Chapter 12 Vector-Valued Functions 60 32. r t 140 cos 22 t i 375, t 2.5 140 sin 22 t 20.47 feet. 16t2 j When x 2.889 and y Thus, the ball clears the 10-foot fence. 0 0 450 33. 100 mph (a) r t (b) 100 100 miles hr 0 5280 ti feet mile 3600 sec hour 440 sin 3 t 16t 2 j 440 ft sec 3 440 cos 3 3 0 Graphing these curves together with y θ 0 = 20 θ 0 = 25 10 shows that 0 20 . 0 0 θ 0 = 10 θ 0 = 15 500 (c) We want xt 440 cos 3 t ≥ 400 and yt 3 440 sin 3 30 11 cos 2 t 16t 2 ≥ 10. From x t , the minimum angle occurs when t 3 440 sin 3 30 11 cos 400 tan 14,400 1 121 14,400 tan2 tan2 16 30 11 cos . Substituting this for t in y t yields: 10 7 7 0 0 48,400 ± 48,4002 4 14,400 15,247 2 14,400 1,464,332,800 28,800 19.38 14,400 sec2 121 400 tan 48,400 tan 15,247 tan tan 1 48,400 34. h 7 feet, rt 35 , 30 yards v0 cos 35 t i 7 90 feet v0 sin 35 t v0 sin 35 t 16t 2 j 16t 2 t 4 90 v0 cos 35 4 3 v02 v0 129,600 v02 cos2 35 129,600 cos2 35 90 tan 35 54.088 feet per second 3 (a) v0 cos 35 t 90 when 7 7 v0 sin 35 90 v0 cos 35 16 90 v0 cos 35 2 90 tan 35 —CONTINUED— S ection 12.3 34. —CONTINUED— (b) The maximum height occurs when yt t v0 sin 35 v0 sin 35 32 32t 0. (c) x t t Velocity and Acceleration 103 90 ⇒ v0 cos 35 t 90 54.088 cos 35 90 2.0 seconds 0.969 second 22.0 feet. At this time, the height is y 0.969 35. r t v cos ti v sin t 16t 2 j (a) We want to find the minimum initial speed v as a function of the angle . Since the bale must be thrown to the position 16, 8 , we have 16 8 t v cos v sin t t 16t 2. 16 v cos 8 1 from the first equation. Substituting into the second equation and solving for v, we obtain: v sin 2 2 sin cos sin cos 2 sin cos 16 v cos 512 1 1 cos2 512 cos2 cos2 . 2 sin cos 512 cos2 16 1 v2 cos2 16 v cos 2 512 1 v2 cos2 1 v2 v2 512 2 sin cos We minimize f f f 512 512 2 sin cos 2 cos2 2 sin2 2 sin cos sin 2 tan 2 2 sin cos cos2 2 0 2 1.01722 58.28 0 ⇒ 2 cos 2 Substituting into the equation for v, v (b) If 16 8 45 , v cos v sin t t v 2 t 2 v 2 t 2 16t2 512 22 28.78 feet per second. 16t2 From part (a), v 2 2 22 22 2 512 12 1024 ⇒ v 32 ft sec. 104 Chapter 12 Vector-Valued Functions 0, v0 16t2 j 30,000 36. Place the origin directly below the plane. Then rt v0 cos 792 t i vt 792i ti 30,000 16t2 j 0 ⇒ t2 v0 sin t 792 and α 30,000 32t j. α At time of impact, 30,000 r 43.3 v 43.3 v 43.3 tan 34,294.6i 792i 16t2 1875 ⇒ t 43.3 seconds. (0, 0) 34,295 1385.6j 1088 mph 0.8748 ⇒ 0.7187 41.18 1596 ft sec 30,000 34,294.6 ti v0 sin 0 when t 37. r t v0 sin v0 cos t t 16t 2 j 0 and t v0 sin . 16 16t2 The range is x v0 cos t v0 cos v0 sin 16 v02 sin 2 . 32 1 ⇒ 15 Hence, x 12002 sin 2 32 3000 ⇒ sin 2 1.91 . 38. From Exercise 37, the range is x Hence, x v02 sin 2 . 32 150 v02 sin 24 32 66 ft sec 0 66 sin 10 t 16t 2 j 16t 2 j ⇒ v02 4800 ⇒ v0 sin 24 108.6 ft sec. 39. (a) rt rt 10 , v0 65t i (b) rt rt 10 , v0 146 ft sec 0 146 sin 10 t 16t 2 j 16t 2 j 66 cos 10 t i 11.46t 146 cos 10 t i 143.78t i 25.35t Maximum height: 2.052 feet Range: 46.557 feet 5 Maximum height: 10.043 feet Range: 227.828 feet 15 0 0 50 0 0 300 (c) rt rt 45 , v0 66 ft sec 0 66 sin 45 t 16t j 2 (d) 16t 2 j rt rt 45 , v0 146 ft sec 0 146 sin 45 t 16t 2 j 16t 2 j 66 cos 45 t i 46.67t i 146 cos 45 t i 103.24t i 46.67t 103.24t Maximum height: 34.031 feet Range: 136.125 feet 40 Maximum height: 166.531 feet Range: 666.125 feet 200 0 0 200 0 0 800 —CONTINUED— S ection 12.3 39. —CONTINUED— (e) rt rt 60 , v0 66 ft sec 0 66 sin 60 t 16t 2 j 16t 2 j (f ) rt rt 60 , v0 Velocity and Acceleration 105 146 ft sec 0 146 sin 60 t 16t 2 j 16t 2 j 66 cos 60 t i 33t i 57.16t 146 cos 60 t i 73t i 126.44t Maximum height: 51.047 feet Range: 117.888 feet 60 Maximum height: 249.797 feet Range: 576.881 feet 300 0 0 140 0 0 600 40. (a) r t t v0 cos 16t x i t v0 sin 0 when t v0 sin 32 16t 2 j v0 sin . 16 v02 sin 2 32 (b) y t dy dt tv0 sin v0 sin 16t 2 32t 0 when t v0 sin . 32 t v0 sin Range: v0 cos Maximum height: y v0 sin 32 v02 sin2 32 16 v02 sin2 322 2 . v02 sin2 64 The range will be maximum when dx dt or 2 2 , 4 ti h rad. v02 2 cos 2 32 0 Minimum height when sin 1, or 41. r t v0 cos v0 sin 1.5 t 4.9t 2 j 4.9t 2 j 100 1 2 100 cos 30 t i 100 sin 30 t 4.9t2 The projectile hits the ground when t 1.5 0⇒t 10.234 seconds. The range is therefore 100 cos 30 10.234 The maximum height occurs when dy dt 100 sin 30 9.8t ⇒ t 5.102 sec 0. 886.3 meters. The maximum height is y 42. r t 1.5 v0 cos 100 sin 30 5.102 ti h v0 sin v0 sin 8 t 50 ⇒ t 4.9 t 4.9 5.102 4.9t 2 j 4.9t 2 j 50 . For this value of t, y v0 cos 8 2 2 129.1 meters. v0 cos 8 t i x 50 when v0 cos 8 t v0 sin 8 50 v0 cos 8 0: 50 v0 cos 8 0 4.9 2500 ⇒ v02 v02 cos2 8 ⇒ v0 4.9 50 tan 8 cos2 8 42.2 m sec 1777.698 50 tan 8 106 43. r t vt at Chapter 12 bt b b 2 Vector-Valued Functions b1 cos t j b1 b 2 sin t i cos t i sin t i 1 b b sin t j 2 cos t i b sin tj cos t j sin t i cos t j vt at (a) v t b 2b 2 cos t 0 when t 0, 2 , 4 , . . . . (b) v t is maximum when then v t 2b . t ,3 ,. . ., 44. r t vt Speed bt b 1 vt sin t i cos t i 2b b1 cos t j sin t j 1 cos t and has a maximum value of 2b when t 80.67 rad sec since since b 1 ,3 ,... . 55 mph 80.67 ft sec Therefore, the maximum speed of a point on the tire is twice the speed of the car: 2 80.67 ft sec 45. rt vt vt 110 mph b cos t j b2 sin t cos t 0 (b) − 10 10 b sin t i 46. (a) Speed v b2 b2 2 2 sin2 t sin2 t b2 2 cos2 t b b2 sin t cos t cos2 t Therefore, r t and v t are orthogonal. 10 − 10 The graphing utility draws the circle faster for greater values of . 47. a t b 2 cos t i b 2 sin t j b 2 cos t i sin t j 2r t a t is a negative multiple of a unit vector from 0, 0 to cos t, sin t and thus a t is directed toward the origin. 48. a t 49. a t 1 F m 32 m 2 b 2 2 cos t i 2 sin t j b 2 b, b 50. v t vt b 2 30 mph 44 ft sec 605 b 1 2 32 44 rad sec 300 b 2 10 n 3000 θ at F mb 4 10 rad sec vt b 8 10 ft sec 2 3000 44 300 32 300 2 605 lb Let n be normal to the road. n cos n sin 3000 605 Dividing the second equation by the first: tan 605 3000 arctan 605 3000 11.4 . S ection 12.3 h v0 sin t 1 gt 2 0 then 0 51. To find the range, set y t 2 By the Quadratic Formula, (discount the negative value) t v0 sin v0 sin 2 2 1 2g 4 1 2g h v0 sin 1 2 2g t Velocity and Acceleration h. 107 v0 sin t v02 sin2 g 2gh seconds At this time, xt v0 cos v0 sin v02 sin2 g 2gh v0 cos g v02 cos g v0 sin sin v02 sin2 sin2 2gh v02 2gh feet v02 52. h t 6 feet, v0 45 feet per second, 45 sin 42.5 32 69.02 feet. 2 2 42.5 . From Exercise 47, 2 32 6 2.08 seconds. 45 sin 42.5 At this time, x t 53. r t vt at xti x ti x ti Speed ytj z t k Position vector z t k Velocity vector z t k Acceleration vector xt 2 54. r t yt rt vt st 0 0 0 0 xti mxt xti x ti xt ytj b, m and b are constants. mxt mx t j 2 y tj y tj vt bj yt 2 zt 2 C, C is a constant. d xt dt 2x t x t 2x tx t 2 mx t C 1 m2 2 C, C is a constant. yt 2 zt 2 Thus, x t xt at 0 2y t y t y ty t 2z t z t z tz t vt at x ti mx t j 0. Orthogonal 55. r t (a) v t 6 cos t i rt vt 3 sin t j 6 sin t i 36 sin2 t 3 4 sin2 t 3 3 sin t at (c) −9 3 cos t j 9 cos2 t cos2 t 1 3 sin t j (b) t Speed 0 3 3 2 4 10 2 6 3 2 2 3 13 3 2 vt 6 6 cos t i (d) The speed is increasing when the angle between v and a is in the interval 9 0, 2 . −6 The speed is decreasing when the angle is in the interval 2 , . 108 Chapter 12 xti r1 2t Vector-Valued Functions ytj ztk 57. False. The acceleration is the derivative of the velocity. 56. (a) r1 t r2 t Velocity: r2 t 2r1 2t 4r1 2t r1 t , then: r1 t 2 Acceleration: r2 t (b) In general, if r3 t Velocity: r3 t Acceleration: r3 t 58. True r1 t 59. a t vt v0 vt rt r0 rt j sin t i a t dt i cos t i v t dt j sin t i i cos t j cos t i sin t j 0 C1 C1 ⇒ C1 sin t j sin t i cos t j 0 C2 C2 ⇒ C2 cos t j The path is a circle. Section 12.4 1. y Tangent Vectors and Normal Vectors 2. y 3. T T N N T x y 4. y N T T N N T x x N x 5. r t rt Tt T1 t2i 2ti rt rt 2tj, t 2j, r t 1 4t2 t2 2 j 2 4 1 1 2 t2 ti j 1 6. rt rt rt Tt T1 t3i 3t2i 9t rt rt 1 9 4 2t2j 4tj 16t2 1 9t4 16 3i 16t2 4j 3t2i 3 i 5 4tj 4 j 5 2ti 2j 2 t2 1 j 2 i 2 1 i 2 7. rt rt rt Tt T 4 cos t i 4 sin ti 16 sin2 t rt rt 2 i 2 4 sin tj, t 4 cos tj 16 cos2 t sin ti 2 j 2 4 8. rt rt 6 cos ti 6 sin ti 36 rt rt sin2 t 2 sin tj 2 cos tj 4 cos2 t 6 sin ti 36 sin2 t j 14 2 cos tj 4 cos2 t 1 28 3 3i j 4 rt Tt T cos tj 4 3 3 3i 36 3 4 S ection 12.4 9. rt rt re ln t i 1 i t 1 i e re re 0.1809i 2t j, t 2j 2j 1 i e 2j e 10. rt rt r0 T0 i 1 2e j 4e2 Tangent Vectors and Normal Vectors e t cos t i e t cos t i j i 2 j 2 i 2 2 j 2 e t j, t e t sin t i 0 et j 109 r0 r0 Te 1 4 e2 0.9835j 11. r t rt When t T0 ti i t 2j 2t j 0, r 0 r0 r0 t k, P 0, 0, 0 k i k, t 2 i 2 1, b t, y k 0, c 0, z 1 t 0 at 0, 0, 0 . 12. r t rt When t T1 t2i 2ti tj j 4 4 k, P 1, 1, 3 3 1, r t r1 r1 2i r1 j 5 2, b 2t 2i j t j 0 t 1 at 1, 1, 4 3 . Direction numbers: a Parametric equations: x 5 2i 5 1, c 1, y Direction numbers: a Parametric equations: x 1, z 4 3 13. r t rt When t T0 2 cos t i 2 sin t i 0, r 0 r0 r0 2 sin tj 2 cos t j 2j t k, P 2, 0, 0 k 0 at 2, 0, 0 . k 2, c 2t, z 1 t 14. r t rt When t T1 t, t, 1, 1, 4 4 t 2 , P 1, 1, 3 t t2 1, 1, 1 ,t 3 1 3 1 3 t 3 1, 1 at 1, 1, 3. k, t 5 2j 5 0, b 2, y 1, r 1 r1 r1 Direction numbers: a Parametric equations: x 21 1, 1, 7 1, b t 1, c 1, y 1 t 3 Direction numbers: a Parametric equations: x z 15. r t rt When t T 2 cos t, 2 sin t, 4 , P 2 sin t, 2 cos t, 0 ,r r r 2, 4 4 1 2 2, 2, 4 4 4 4 2, 0 , 2, t 4 at 2, 2, 4 . 2, 0 2, c 2, y 0 2t 2, z 4 Direction numbers: a Parametric equations: x 2, b 2t 110 16. r t rt Chapter 12 Vector-Valued Functions 3, 1 2 sin t, 2 cos t, 4 sin2 t , P 1, 2 cos t, ,r r r 2 sin t, 8 sin t cos t 3, 6 6 1 4 3, b 3t When t T 6 6 6 1, 2 3 , 3, 1, 2 3 1, c 1, y t 6 at 1, 3, 1 . Direction numbers: a Parametric equations: x 2 t, t 2, t 3 3 1, 2t, 2t 2 3, r 3 r3 r3 23 t 3, z 2 3t 1 17. r t rt When t T3 z 18 15 12 9 6 3 3 6 9 −3 x 12 15 18 1, 6, 18 , t 1 1 , 6, 18 19 1, b t 6, c 3, y 1 k 2 3 at 3, 9, 18 . y Direction numbers: a Parametric equations: x 18 6t 9, z 18 t 18 18. r t rt When t T 3 cos t i 3 sin t i ,r r r 4 sin t j 4 cos t j 3i 2 2 z 5 1 k 2 1 k, 2 2 37 3i 0, c 4, z t 2 1 k 2 1 t 4 e ti e ti i T0 at 0, 4, 1 37 4 . x 54 4 3 2 1 1 4 5 y 2 2 2 6i k Direction numbers: a Parametric equations: x 6, b 6t, y 19. r t rt T1 ti i ln t j 1 j t rt rt 1 t k, t0 1 i j 1 k 2 2 i 3 1 t 2 1.05k 2 j 3 1 k 3 20. r t rt r0 2 cos t j 2 sin t j 2 sin t k, t0 2 cos t k i i 5 1 s, y s 1 2k , r 0 2k 0 1 k; r 1 2t i 1 t, y j 2j, r 0 r0 r0 5 1 2k 1 14 t, z 1.1 i 1 0.1 j Tangent line: x r t0 0.1 Parametric equations: x s r t0 0.1 r0 0.1 2, z s 2s r 1.1 0.1, 2, 2 0.1 0.9, 2, 0.2 1.1, 0.1, 1.05 S ection 12.4 21. r 4 u8 2, 16, 2 2, 16, 2 Tangent Vectors and Normal Vectors 111 Hence the curves intersect. rt us cos r4 r4 0, 1, 0 0, 1, 0 1, 2t, 1 ,r 4 2 23 1, 8, 1 2 1 1 , 2, 4 12 1.2 1 1 , 2, s 4 3 u8 u8 ,u 8 16.29167 ⇒ 16.29513 22. r 0 u0 Hence the curves intersect. rt us u0 cos r0 r0 1, sin t, cos t , r 0 sin s cos s 1, 0, 1 u0 u0 12 t j, t 2 tj rt rt t2 2 53 2 1, 0, 1 sin s cos s cos s, 1 cos 2s 2 1 2 cos s, 0⇒ 2 6 j, t t 6 j t2 rt rt t2 t4 Tt t4 36 72t 36 1 1 i i 36 t 4 6 j t2 t4 12t3 36 2j 32 23. rt rt Tt Tt T2 N2 ti i 2 24. rt rt ti i 3 i 1 t 1 32 tj t2 1 t2 1 32 Tt j i 6 j t2 i i 1 j 53 2 1 5 2i j 25 i 5 5 j 5 32 j T2 T2 N2 T2 T2 1 3i 13 25. rt rt ln t i 1 i t rt rt t t2 2 53 2 t j 1 j, t 2 26. rt rt 3 cos t i 3 sin t i sin t i cos t i 2 i 2 2 i 2 3 sin t j, t 4 3 3 cos t j, r t cos t j sin t j 2 j 2 2 j 2 Tt 1 i t 1 t2 32 Tt j 1 1 1 t2 32 i tj 1 t2 T Tt 4 4 Tt T2 N2 1 i i j N 1 j 53 2 25 i 5 5 j 5 T2 T2 112 27. Chapter 12 rt rt ti i t2j 2t j Vector-Valued Functions ln t k, t 1 k t i 1 2t j 4t 2 i 9 63 2 1 Tt rt rt 1 4t 4 3 63 i 2 1 k t ti 4t 4 Tt T1 N1 4t 4 t2 1 32 1 t2 2t 3 4t 4 4t t2 1 3 63 2 i 2t 2 j k t2 1 8t 3 t2 t 1 32 j 4t 4 3k 3 14 k 14 32 k i 6 j 63 2 2j 3k 14 et j et j e2t k 2j 14 i 14 e t k, t e tk 2 14 j 14 28. rt rt rt Tt Tt T0 N0 2t i 2i 2 rt rt 2e t et e 1 j 2 2 j 2 0 29. rt rt 6 cos ti 6 sin ti rt rt cos t i T3 T3 4 4 6 sin tj 6 cos tj sin ti k, t 3 4 e 2i et t2 2t et e tk t e t2 et e t et j t e e et Tt Tt cos tj 1 2 j 2 sin t j, T t 2 i 2 i 2 e t 2j 2 et e t 2k N 3 4 1 k 2 2 k 2 31. r t vt 2 cos t j 4 cos2 t at Tt Tt Nt 4t i 4i 0 vt vt 0 Tt Tt is undefined. 4i 4 i 30. rt rt Tt cos ti sin t i rt rt 2 sin t j 2 cos t j sin t i sin2 t k, t 4 The unit normal vector is perpendicular to this vector and points toward the z-axis: Nt N 2 cos ti sin tj . sin2 t 4 cos2 t 25 5 , ,0 5 5 4t i 4i 0 vt vt 0 Tt Tt is undefined. 1 2i 5 j 2t j 2j The path is a line and the speed is constant. 4 32. r t vt at Tt Tt Nt 33. r t vt at Tt Tt Nt 4t 2 i 8t i 8i vt vt 0 Tt Tt is undefined. 8t i 8t i The path is a line and the speed is constant. The path is a line and the speed is variable. S ection 12.4 Tangent Vectors and Normal Vectors 1 j, v t t 2j t2 t4 j 1 2 i 2 t4 2t 1 3 2i 113 34. r t vt at Tt Tt Nt t2j 2t j 2j vt vt 0 Tt Tt k 35. r t at 2t j 2t j Tt T1 is undefined. Nt ti i 1 j, v 1 t2 i j, 2 j, a 1 t3 vt vt 1 i 2 Tt Tt i j 1 j t2 1 t4 1 t 2i j The path is a line and the speed is variable. t4 2t 1 t 2j j 2t 3 1 3 2j t4 1 t4 N1 aT aN 36. rt vt at Tt T1 t 2i 2ti 2tj, t 2j, v 1 2i 1 4t2 j 4 2 i 2 1 Nt Tt Tt 1 t2 N1 aT aN a a 2 i 2 T N t 2 1 2 i 2 i 1 i 2 a a t 1 6ti vt vt 2i 20 T N j 2 2 t3 i 3t 2 i 1 2i 2j 37. rt vt at 2t 2j, t 1 2i 6i 4j 4j 4t j, v 1 2i, a 1 vt vt 1 i 2 4j, a 1 2ti 2j 2 j 2 t 1 1 2 1 t2 1 ti j Tt T1 1 3t 2 i 4t j 9t 4 10t 2 1 4j i 5 1 1 3 2i 2j 5 5 4 9t 4 i 2j 3 2j 1 3 2i t 1 3 2j Tt T1 N1 aT aN 16t 3t 2 9t 10t 2 4 36t 4 10t 2 1 t2 1 i 2 j 2 2 2 tj 64 i 20 3 2 2i 5 a a T N j 32 j 20 3 2 1 6 5 1 12 5 8 4 14 5 5 85 5 114 38. Chapter 12 rt vt at Tt T0 Tt T1 N1 aT aN t3 3t 2 6ti vt vt 4i 16 9t 4 4t i 4i Vector-Valued Functions t2 1 j, t 0 4i 39. rt vt at Tt T0 4 16 i 32 18t 4 20t 2 16 j N0 aT aN i 2i 5 a a T N 1 1 5 1 2 5 8 4 75 5 65 5 et i et i et i vt vt 2j 5 j e 2e 4e 2t j, t 0 v0 a0 i i 2j 4j 2t j, v 0 2j 2t j, 2t j, 2j, a 0 3t 2 9t 4 i 4 i 2t j 20t 2 16 et i 2e 2t j 4e 4t e 2 t 4t 3t 2 20t 2 1 j 2 32 9t 4 32 32 j 16 3 2 j a a et i et i et i vt vt i j 3 e 2t e 2 t e e 2t 4t T N 0 2 e tj e tj 40. rt vt at Tt T0 Tt T0 N0 aT aN t k, t k, v 0 i 0 i j j k e t j, a 0 et i e t j k e 2t e 2 t 1 k 2 1 e 2t 2e 2 t e e 2t 4t 3 2i 1 j 132 et 1 e 4t e e 2t 1 4t 3 2k 3 i 33 2 2 i 2 a a T N 3 j 33 2 2 j 2 0 2 et sin t j sin t i et cos t sin t j 42. r t vt v0 j 2 2 i j. at a0 T0 j. N0 2 41. r t vt at At t et cos t i et cos t et 2 a cos t i a sin t i bj a a v0 v0 2 2 b sin t j b cos t j 2 sin t i ,T v v et 2 cos t j 1 2 i cos t i i j b 2 sin t j Motion along r is counterclockwise. Therefore, N aT aN 1 2 a a T N i j 2e 2e 2 2 i 2 Motion along r t is counterclockwise. Therefore, i. a a T N 0 a 2 aT aN S ection 12.4 43. r t0 v t0 a t0 T t0 cos t0 2t 0 2 Tangent Vectors and Normal Vectors 115 t0 sin t0 i 2t 0 sin t0 t0 cos t0 j cos t0 i sin t0 j t0 cos t0 sin t0 j cos t0 cos t0 i t0 sin t0 i sin t0 j v v Motion along r is counterclockwise. Therefore N t0 aT aN 44. r t0 v t0 a t0 T 2 sin t0 i a a t0 1 T N 2 2 cos t0 j. t0 1 3t 0 sin t0 i cos t0 i cos t0 j sin t0 j cos t0 j sin t0 i 1 v v cos t0 i sin t0 j 2 1 cos t0 sin t0 i 1 cos t0 j . 2 1 cos t0 2 Motion along r is clockwise. Therefore, N aT aN a a T N sin t0 2 1 cos t0 2 2 2 1 cos t0 2 1 cos t0 45. r t vt at Tt Nt aT aN a cos t i a sin t i a vt vt Tt Tt a a T N 0 a a 2 2 a sin t j a cos t j a 2 46. T t points in the direction that r is moving. N t points in the direction that r is turning, toward the concave side of the curve. y a cos t i sin t j cos t j sin t j sin t i cos t i T N a x 47. Speed: v t 48. If the angular velocity 0. 2 is halved, 2 The speed is constant since aT aN a a 4 2 . aN is changed by a factor of 1 . 4 116 Chapter 12 1 j, t0 t Vector-Valued Functions 49. rt x rt Tt Nt r2 T2 N2 ti t, y i t2i t4 i 2 1 50. r t x rt Tt y t 3i t 3, y 3t 2 i rt rt 3i t j, t0 t⇒x j 1 y3 or y x1 3 1 ⇒ xy t 1 j t2 j 1 3t 2 i 9t 4 3 10 i 10 3 10 j 10 j 1 10 j 10 y T1 N1 j 10 t 2j 4 t 1 1 j 2 j 4j 3 2 10 i 10 2 N 2i 1 1 (1, 1) N T 2, 1 2 T 1 2 3 x −1 −1 1 2 x 17 4i 17 17 i 17 51. rt x rt Tt Nt r T N 4 4 4 ti i 0 2 cos t i 2 cos t, y 2 sin t i 1 2 2 sin t j, t0 4 y2 4 52. rt x 3 rt 3 cos t i cos t, y 2 2 sin t j, t0 sin t ⇒ 2 cos t j j y 3 2 sin t ⇒ x 2 2 cos t j 2 cos t j x2 9 y2 4 1 Ellipse 3 sin t i 2j ⇒ T i 2 sin t i cos t i 2i sin t i y cos t j r N sin t j 2j 1 T N ( 2, 2 ) 1 x 2 2 2 2 2t j 2j N T −2 −1 −1 1 2 x i i j j −1 −1 1 −3 53. r t vt at Tt 3t k, t 3k 1 54. r t vt at 4ti 4i 0 v v T T 4t j 4j 2t k 2k v v T1 1 i 14 2j 3k 14 i 14 2j 3k Tt Nt 1 2i 3 2j k is undefined. Nt T T is undefined. aT, aN are not defined. aT, aN are not defined. S ection 12.4 t2 k, t 2 tk k Tangent Vectors and Normal Vectors 117 55. r t vt v1 at Tt T1 ti i i 2j v v t 2j 2tj 2j k 1 56. rt vt v0 at et sin t i et cos t i 2et 2i v v 1 i 3 1 2 2 i 2 a a T N j k et cos tj et sin t i et k et sin t et cos t j et k cos ti k 2et sin tj et k 1 1 2j 5t 2 k a0 i 2tj tk Tt T0 6 i 6 T T 30 30 a a T N 1 cos t 3 j sin t 2 j 2 3 2 k sin t i sin t cos t j k Nt 5t i 2j k 1 5t 2 3 2 5 1 5t 2 5i 56 6 30 6 2j k 5t i 2j k 5 1 5t 2 Nt N0 aT aN cos t i cos t sin t j N1 aT aN 57. rt vt v 2 at a 2 Tt T 4t i 4i 4i 3 cos tj 3 sin tj 3j 3 sin t k, t 3 cos t k 2 58. rt vt v2 at ti i i 6j v v 3t 2j 6tj 12j k t2 k 2 tk 2k x 4 z 4 2 N 2 4 6 8 T y 3 cos t j 3k 3 sin t k Tt T2 1 1 1 i 149 37t 2 12j 1 37t 2 i 2k 6t j tk v v 1 4i 5 T T k a a T N 1 4i 5 3j 3 sin t j 3 cos t k Nt T T 1 37 1 T N z 1 32 37t i 37 372 6j k 6j k 2 Nt 1 cos t j sin t k 37t 2 74 i 6j 37t i 6j k N 2 aT aN N2 y 0 3 4π 2π 3 1 37 149 1 5513 74i 3 k aT x a a T N 74 149 37 5513 37 149 aN 118 Chapter 12 rt rt Tt Tt Vector-Valued Functions 59. T t 60. The unit tangent vector points in the direction of motion. Nt If a t aTT t aNN t , then aT is the tangential component of acceleration and aN is the normal component of acceleration. 61. If aN 63. r t 0, then the motion is in a straight line. t sin t, 1 cos t 62. If aT 0, then the speed is constant. The graph is a cycloid. (a) r t vt at Tt 2 t sin t, 1 cos t sin t y cos t, sin t, 2 cos t 1 1 cos t, sin t 1 t= 2 t=1 3 t= 2 vt vt Tt Tt a T x 21 cos t 1 Nt 21 1 21 1 21 2 cos t sin t, 2 1 cos t 2 2 aT aN When t When t When t cos t 2 sin t 1 sin2 t 2 2 2 cos t 2 cos t sin t 2 21 1 21 sin t cos t 2 a N 1 :a 2T 1: aT 3 :a 2T vt 2 cos t 2 2 2 2 2 cos t 1 cos t cos t cos t 21 2 cos t 2 0, aN 2 2 , aN , aN cos t aT 2 2 2 (b) Speed: s ds dt When t When t When t 21 sin t cos t 2 2 2 21 1 :a 2T 1: aT 3 :a 2T > 0 ⇒ the speed in increasing. 0 ⇒ the height is maximum. 2 2 2 < 0 ⇒ the speed is decreasing. S ection 12.4 64. (a) r t vt at Tt aT aN When t (b) Since aT 2 Tangent Vectors and Normal Vectors 119 cos t sin t cos t t sin t, sin t sin t 3 2 2 t cos t t cos t, cos t 3 cos t 2 t sin t 2 t cos t, 2 t sin t t sin t, sin t t cos t vt vt a T a 2 cos t, sin t cos t aT2 2 2 cos t 4 3 22 t sin t 4 sin t 3 2 sin t 3 t cos t 2 1 3 t t 2 1, aT 2 , aN . When t 2, aT , aN 2 3 . 1 and t 2. > 0 for all values of t, the speed is increasing when t t k, t0 2 1 k 2 2 cos t j 1 k 2 x 65. rt rt Tt Nt r T N 2 2 2 2 cos t i 2 sin t i 2 17 17 cos t i 2j 4 k 2 sin t j 2 cos t j 2 sin t i sin t j z 2 3 2 −2 −1 1 2 1 B N −2 T ( 0, 2, π ) 2 y 2 17 17 j 2i 1 k 2 17 17 4i k i B 2 T 2 N 2 j k 17 17 0 17 i 17 4 17 k 17 17 i 17 4k 4 17 0 17 0 1 66. rt rt Tt Nt r1 T1 N1 i ti i 1 1 j 1 i 6 t2j 2t j 1 4t 2 4t 2 1 k 3 2j 3i t3 k, t0 3 t2k 1 1 2 z N B t4 i 2t j t2k x 1 2 T −1 2 1 ( 1, 1, 3 ) 1 y 1 t4 1 t2 t4 2t t3 i 1 t4 j t 2t 3 k k 3k i 2 2 j 6 3 i k 6 6 2 2 3 i 3 3 j 3 3 k 3 3 i 3 j k k 1 63 B1 T1 N1 6 6 20 2 120 Chapter 12 Vector-Valued Functions 67. rt rt rt r 4 Tt N 4 i sin t j cos t k, t0 4 cos t j 1 T 4 sin t j 2 j 2 sin t k, 2 j 2 cos t k, 2 k 2 2 k 2 i B T N 0 4 0 j 2 2 2 2 0 e t sin t j k 2 2 2 2 i 4 4 68. rt rt r0 rt 2 2e t i 2e t i 2i 4e 2 t 4e 2 t j e t cos t j e t cos t e t sin t k, t0 e t sin t j e t cos t k k e 2 t sin2 t e 2 t cos2 t 2e 2 t sin t cos t k ⇒ T0 e 2 t cos2 t 2e 2 t cos 2 t 1 2i 6 e 2 t sin2 t sin2 t 2e 2 t cos t sin t 6e 2t rt Tt Tt T0 6et rt rt 1 6 1 6 j 1 2i 6 sin t cos t sin t j cos t 2 2 j 1 6 2 2 j k 1 6 2 2 sin t cos t k cos t j sin t k 2 k 2 k ⇒ N0 i 2 6 0 B0 T0 N0 3 i 3 3 j 3 3 k 3 S ection 12.4 Tangent Vectors and Normal Vectors 121 69. rt rt rt r T 3 3 4 sin t i 4 cos ti 4 cos tj 4 sin t j 2t k, t0 2 k, 4 3 16 cos 2 t 2i 1 25 1 25 3 i 2 2 3j 2i 16 sin 2 t 2k 2 3j 2k 4 cos tj 20 25 5 i 5 15 j 5 5 k 5 5 i 5 3j k Tt N 4 sin t i 1 j 2 3 B 3 T 3 N 3 i 5 5 3 2 j 15 5 1 2 k 5 5 0 5 i 10 15 j 10 45 k 10 5 i 10 3j 4k 70. rt rt rt r T 4 4 2 cos 2t i 4 sin 2t i 17 4i 17 17 1 17 j k 4i 2 sin 2t j 4 cos 2t j t k, t0 k 4 k 8 sin 2tj Tt N 8 cos 2t i 4 i B 4 T 4 N 4 j k 17 17 0 17 i 17 4 17 k 17 4 17 0 17 0 1 71. From Theorem 12.3 we have: rt vt at Tt Nt aT aN a a v0t cos v0 cos i 32 j v0 cos i v02 cos2 v0 sin v02 cos2 T N v0 sin v0 sin 32t j 32t 2 2 i h v0 sin v0t sin 32t j 16t2 j 32t i v0 cos j v0 sin 32t (Motion is clockwise.) 32 v0 sin 32t v02 cos2 v0 sin 32t v0 2 2 cos2 32v0 cos v0 sin 32t 32. 32t 2 Maximum height when v0 sin At maximum height, aT 0; (vertical component of velocity) 0 and aN 122 72. Chapter 12 45 , v0 v0 cos v0 sin aT aN 150 Vector-Valued Functions 150 32t 2 2 150 75 2 2 2 32t 32t 32t 2 32t 75 2 32t 75 2 5625 5625 32 75 2 11250 11250 75 2 32 75 2 75 2 16 32t 256t2 256t2 32. 12 gt j 2 1200 2t 1200 2 1200 2t 2 At the maximum height, aT 73. (a) r t v0 cos ti h 0 and aN v0 sin 5 50t t (b) 16t 2 j −20 60 100 cos 30 t i 50 3 t i 5 100 sin 30 t 16t 2 j Maximum height Range 300 −10 44.0625 279.0325 (c) Speed vt vt 50 3 i 2500 3 4 64t 2 50 50 200t 32t j 32t 625 2 (d) t Speed 0.5 93.04 1.0 88.45 1.5 86.63 2.0 87.73 2.5 91.65 3.0 98.06 at (e) 0 50 32 j aN 3 aT −50 The speed is increasing when aT and aN have opposite signs. 74. (a) r t v0 cos ti h 4 v0 sin t 12 gt j 2 16t 2 j 0 1400 0 (b) 400 200 cos 45 t i 100 2 t i (c) vt vt at (d) t Speed 100 2 i 20,000 32j 0.5 189.03 5 200 sin 45 t 16t 2 j Maximum height Range 1258 ft 317.5 ft 100 2t 100 2 100 2 32t j 32t 2 1.0 178.81 1.5 169.49 2.0 161.23 2.5 154.18 3.0 148.54 75. r t (a) 10 cos 10 t, 10 sin 10 t, 4 rt rt 4t , 0 ≤ t ≤ 1 20 100 sin 10 t , 100 cos 10 t , 4 100 100 2 2 sin2 10 t 16 2 100 2 2 cos2 10 t 1 16 4 625 314 mi hr (b) aT aT 0 and aN 1000 0 because the speed is constant. S ection 12.4 76. 600 mph rt vt at Tt 880 ft sec 16t2 32tj 36,000 j 77. r t Tangent Vectors and Normal Vectors a cos t i a sin t j T 0 and a N a 123 880ti 880i 32j From Exercise 45, we know a (a) Let a 0 2 . 2 . Then N a 0 2 a2 2 4a 2 880i 32tj 16 4t2 3025 55i 4t2 2tj 3025 or the centripetal acceleration is increased by a factor of 4 when the velocity is doubled. (b) Let a0 a N a 2. Then a0 2 Motion along r is clockwise, therefore Nt aT aN a a 2ti 4t2 T N 55j 3025 64t 4t 2 a 2 2 1 a 2 2 3025 or the centripetal acceleration is halved when the radius is halved. 1760 4t2 3025 9.56 104 4100 78. rt vt vt at at (a) F r cos t i r sin t i r r r 2 r sin t j r cos t j v r 2 79. v 4.83 mi sec 1 2 r cos t i sin t j mat mr 2 m2 r r 2 mv 2 r (b) By Newton’s Law: mv2 r GMm 2 ,v r2 GM ,v r GM r 9.56 104 4385 80. v 9.56 104 4200 4.77 mi sec 81. v 4.67 mi sec 82. Let x distance from the satellite to the center of the earth x v 2x t 9.56 2 r 4000 . Then: 2x 24 3600 104 x 104 24 42 2 9.56 x 10 4 4 2x 2 24 2 3600 x3 v 9.56 3600 2 ⇒x 26,245 mi 2 26,245 24 3600 1.92 mi sec 6871 mph 83. False. You could be turning. 85. (a) r t x x2 cosh bt i cosh bt , y y2 sinh bt j, b > 0 sinh bt sinh2 bt 1, hyperbola 84. True. All the motion is in the tangential direction. (b) v t at b sinh bt i b2 cosh bt i b cosh bt j b 2 sinh bt j cosh2 bt b 2r t 124 Chapter 12 cos i dT dt Vector-Valued Functions sin j be the unit tangent vector. Then dT d d dt cos sin i cos j 2i d dt M d . dt 2 j and is rotated counterclockwise through an angle If d dt < 0, then the curve bends to the right and M has the opposite direction as T . Thus, N T T 86. Let T t Tt M of sin i cos j 2 from T. sin If d dt > 0, then the curve bends to the left and M has the same direction as T . Thus, M has the same direction as N T , T which is toward the concave side of the curve. y again points to the concave side of the curve. y T T φ φ M x M N x 87. rt yt rt vt vt Tt xti mxt xti x ti xt vt vt ytj b, m and b are constants. mxt mx t j 2 88. Using a we have: v aTT a aNN, T vT v aT T aTT T N N T O, and T N 1, bj aNN v aN T N mx t 1 2 xt 1 m2 v a v aN T v aN T v aN Thus, aN v v dv dt ct 3 2bt a . ±i mj , constant m2 Hence, T t 0. 89. a 2 a a aNN 2 90. F aTT N aNN aN2 N 2 ma at dx dt 2b 4b 2 4b 2 4b 2 1 bt 2 a 6ct dv dt Force aTT aT T aT2 aN 2 a 2 2 x v dv dt F2 2aTaNT aN2 aT2 a 2 3ct 2 Since aN > 0, we have aN aT 2. 24bct 12c 12c ± 36c 2t 2 2bt v a 12ac 12cv 6ct, which 3ct 2 F fv 4b 2 The sign of the radical is the sign of 2b cannot change. Section 12.5 Arc Length and Curvature 125 Section 12.5 1. r t dx dt s 0 4 Arc Length and Curvature 2. r t 3, dz dt 0 dx dt s 0 ti 1, 4 3t j dy dt 1 ti 1, 4 t 2k dy dt 1 0, dz dt 2t 9 dt dt 4t 2 dt 4 10 0 1 2t 1 4 4 10 1 8 65 4 y 4t 2 ln 8 ln 2t 65 1 4t 0 4 10 t 0 y 16.819 (4, 12) 12 16 12 8 (4, 16) 8 4 4 (0, 0) 4 8 12 x x (0, 0) 1 2 3 4 3. r t dx dt s 0 t 3i 3t 2, 2 t 2 j, dy dt 9t 4 2t 0≤t≤2 4. r t dx dt 1, 6 t dy dt 1 0 1i 2t t 2 j, 0≤t≤6 2 4t 2 dt 0 2 9t 2 4 t dt s 1 ln 4 1 ln 4 y 40 35 30 25 20 15 10 5 −1 4t 2 dt 4t 2 1 12 2t 1 t 4t 2 2 6 1 9t 2 27 1 40 3 27 2 4 32 0 1 0 43 2 145 3 145 1 80 10 27 y 5 8 (7, 36) (8, 4) 4 3 2 1 (1, 0) x 12345678 (0, 0) −2 −1 2 4 6 8 10 x 5. r t dx dt s 4 a cos3 t i a sin3 t j dy dt 3a sin2 t cos t 2 a y 3a cos2 t sin t, 2 3a cos2 t sin t 0 2 3a sin2 t cos t 2 dt −a −a x a 12a 0 2 sin t cos t dt 2 3a 0 2 sin 2t dt 3a cos 2t 0 6a 126 6. r t dx dt Chapter 12 a cos t i a sin t, 2 Vector-Valued Functions a sin t j dy dt a y a cos t a2 cos2 t dt 2 a x s 0 2 a2 sin2 t a dt 0 at 0 2a 7. (a) r t v0 cos ti h v0 sin 3 50 2t 32t j 25 2 16 t 12 gt j 2 1 32 t2 j 2 100 cos 45 ti 50 2t i (b) v t 50 2 50 2i 32t 3 100 sin 45 t 16t2 j 50 2 0⇒t Maximum height: 3 (c) 3 50 2t 16t2 50 2 0⇒t 25 2 16 4.4614 16 25 2 16 2 81.125 ft Range: 50 2 4.4614 4.4614 315.5 feet 50 2 32t dt 2 (d) s 0 50 2 2 362.9 feet 8. (a) r t yt yt v0 cos v0 sin v0 sin ti t v0 sin 12 gt 2 gt t 12 gt j 2 (b) y t v0 sin t 12 gt 2 0⇒t 2v0 sin g 2v0 sin g v02 2 sin g 1, or 4 . Range: x t v0 sin . g 2 . v0 cos 0 when t 1, or The range x t is a maximum for sin 2 Maximum height when sin (c) x t yt xt 2 v0 cos v0 sin yt 2 gt v02 cos2 v02 cos2 v02 v0 sin v02 sin2 g 2t2 12 gt 2 2v02g sin t g2t2 2v0 g sin t 2v0 g sin t 2v0 sin g s 0 v02 g 2t2 dt Since v0 s 96 ft sec, we have 6 sin 12 962 0 6144 sin t 1024t2 dt . 0.9855 56.5 . Using a computer algebra system, s is a maximum for S ection 12.5 9. r t dx dt s 0 x Arc Length and Curvature 127 2t i 2, 2 3t j 3, 3 tk dz dt 2 z dy dt 22 (4, − 6, 2) 4 2 1 (0, 0, 0) 12 dt 2 2 −2 y 2 14 dt 0 14 t 0 2 14 10. r t dx dt s 0 i 0, 2 t 2j dy dt 4t 2 2t, t3k, dz dt 0, 2 3t 2 2 11. r t dx dt 4 9t 2t dt z 3t, 2 cos t, 2 sin t 3, dy dt 2 2 sin t, 32 dz dt 2 2 cos t 2 cos t 2 dt 2 9t 4 dt 0 2 32 0 s 0 2 2 sin t 1 4 27 1 40 3 27 9t 2 2 43 2 1 80 10 27 8 x 8 7 6 5 4 3 2 1 (1, 0, 0) 3 4 13 dt (1, 4, 8) 13t 0 0 z 13 2 −4 5 4 3 56 y 2 1 2 ( 32π , 0, 2) 2 3 5 x (0, 2, 0) 3 4 5 y 12. r t dx dt s 0 2 sin t, 5t, 2 cos t , 2 cos t, dy dt 5, 25 29 dz dt 0, 2 sin t 4 sin2 t dt 13. r t dx dt a cos t i a sin t, 2 a sin t j dy dt bt k dz dt b b2 dt 2 a cos t, a2 cos2 t 4 cos2 t 29 dt 0 z s 0 2 a2 sin2 t a2 0 z b2 dt a2 b2 t 0 2 a2 b2 12 10 8 6 4 (0, 0, 2) (a, 0, 2π b) − 10 −8 −6 468 10 12 14 y 2π b 4 πb x −4 −6 (0, 5 , − 2) x (a, 0, 0) y 14. r t dx dt s 0 cos t t cos t, 2 t sin t, sin t dy dt t sin t, 2 t cos t, t2 3 z dz dt 2 2t 2t 2 dt 5 8 2 x 2 (π2 , 1, π4 ) 2 (1, 0, 0) 1 2 3 2 3 y t cos t 2 t sin t t2 2 5t 2 dt 0 2 0 5 128 15. r t dx dt s Chapter 12 t 2i 2t, 3 Vector-Valued Functions ln t k 16. r t 1 t 2 tj dy dt 2t 1, 2 sin t i cos t, 2 cos t j dy dt 2 t3k sin t, dz dt 2 dz dt 1 t2 dx dt 1 t 2 3t 2 3t 2 2 dt dt 1 3 1 3 1 s 0 2 2 0 cos t sin t 11.15 4t 4 4t 4 t t2 1 1 t2 dt dt 8.37 9t 4 dt 17. r t ti 4 t2 j t 3k, 0≤t≤2 2, 0, 8 82 84 2 21 9.165 (a) r 0 distance (b) r0 r 0.5 r1 r 1.5 r2 0, 4, 0 , r 2 22 0, 4, 0 0.5, 3.75, 0.125 1, 3, 1 1.5, 1.75, 3.375 2, 0, 8 0.5 0.5 0.5728 2 2 42 distance 0.25 1.75 1.2562 2 2 0.125 4.625 2.7300 2 2 0.5 2 0.75 2 0.875 2 0.5 2 1.25 2 2.375 2 4.9702 9.529 (d) Using a graphing utility, you obtain 9.57057. (c) Increase the number of line segments. t i 4 t j 4 t k, 0 ≤ t ≤ 2 18. r t 6 cos 6i 2j 2 sin 6, 0, 0 (a) r 0 r2 (c) Increase the number of line segments. 0, 2, 2 (d) Using a graphing utility, you obtain 44 2 11 6.633 s 0 2 2k 62 6, 0, 0 distance (b) r0 r 0.5 r 1.0 r 1.5 r 2.0 distance 19. r t (a) s 0 t 22 22 r t dt 7.0105. 5.543, 0.765, 0.5 4.243, 1.414, 1.0 2.296, 1.848, 1.5 0, 2, 2 6.9698 2 cos t, 2 sin t, t t xu 2 yu 2 2 zu 2 2 du (b) x s 5 t 2 cos 2 cos s 5 s 5 ,y i 2 sin 2 sin s 5 s 5 ,z j s 5 s 5 k 2 sin u 0 t 2 cos u t 1 2 du 5 du 0 5u 0 5t rs —CONTINUED— S ection 12.5 19. —CONTINUED— (c) When s 5: x y z 2 cos 1 2 sin 1 1 1.081 1.683 Arc Length and Curvature 129 1.081, 1.683, 1.000 When s 4: x y z 2 cos 2 sin 4 5 4 5 4 5 0.433 1.953 1.789 0.433, 1.953, 1.789 (d) r s 2 sin 5 s 5 2 2 cos 5 32 t 2 s 5 2 1 5 2 4 5 1 5 1 20. r t (a) s 4 sin t t t cos t , 4 cos t 2 t sin t , zu 2 2 xu 0 t yu 2 2 du t t 2 4u sin u 0 4u cos u 3u du 0 16u 9u 2 du 0 5u du 52 t 2 (b) t x y z rs 2s 5 4 sin 4 cos 3 2 2s 5 4 sin 5: 4 sin 25 5 25 5 1.342 25 cos 5 25 sin 5 25 5 25 5 1.030 2s 5 2s 5 2 2s cos 5 2s sin 5 3s 5 2s 5 2s 5 2s 5 2s cos 5 2s i 5 4 cos 2s 5 2s sin 5 2s j 5 When s x y z 3s k 5 4: 4 sin 4 cos 12 5 2.4 8 5 8 5 8 cos 5 8 sin 5 8 5 8 5 2.291 6.029 (c) When s x y 4 cos 35 5 5.408 z 1.030, 5.408, 1.342 (d) r s 4 sin 5 2s 5 2 2.291, 6.029, 2.400 4 cos 5 2s 5 2 3 5 2 16 25 9 25 1 130 Chapter 12 Vector-Valued Functions 2 si 2 2 j 2 rs Ts 0 (The curve is a line.) 21. rs rs Ts Ts 1 2 i 2 rs rs 1 and 2 sj 2 rs 1 2 1 2 1 22. r s rs Ts Ts 3 i rs si and j rs 1 0⇒K Ts 0 (The curve is a line.) 0⇒K 23. r s Ts Ts K 2 cos rs s 5 i 2 sin 2 sin 5 s s 5 5 i j s 5 k s 5 j 1 k 5 2 cos 5 s 5 j 2 cos 5 Ts s 5 2 5 2s 5 4 sin 5 5 cos 2s 4 25 i 2 sin 5 24. rs Ts Ts K 4 sin rs 4 25 Ts 2s cos 5 2s i 5 2s i 5 5 2s 4 cos 5 4 25 2 10s 25s 2s i 5 2s j 5 5 sin 2s 4 cos 3 k 5 2s j 5 2s 5 2s sin 5 2s j 5 3s k 5 25. rt vt Tt Tt K 4t i 4i 2t j 2j j 26. rt vt Tt Tt K t 2j 2t j j 0 Tt rt k 27. rt vt v1 ti i i 2 j t3 2j t 2i t4 1 j t 1 j t2 j 1 2i 5 0 Tt rt 0 at a1 Tt Nt N1 K 0 (The curve is a line.) j 1 1 1 12 t4 1 i 2 a v N 2 i t 2j j 2 2 S ection 12.5 28. rt vt v1 at a1 Tt Nt N1 K a v ti i i 2j 2j i 1 1 1 1 5 N 2 Arc Length and Curvature 0 1 sin2 t 131 t2j 2t j 2j 29. rt rt Tt Tt ti i i 1 1 cos t j, t sin t j, r t sin t j sin2 t sin t cos t i sin2 t 3 2 j, r 0 i 1 1 cos t sin2 t 3 2j 2t j 4t 2 4t 2 2i 2 55 2t i j j T0 K T0 r0 Alternate Solution: x t K K0 1 xy x2 yx y2 32 t, y t 1 cos t cos t sin2 t 32 30. r t xt K 5 cos t i 5 cos t, y t xy x2 4 sin t j, t 4 sin t 32 3 31. rt rt Tt Tt 4 cos 2 t i 8 sin 2 t i sin 2 t i 4 sin 2 t j 8 cos 2 t j cos 2 t j 2 sin 2 t j 1 4 yx y2 2 cos 2 t i Tt rt 2 8 5 sin t 4 sin t 4 cos t 25 sin2 t 16 cos2 t 3 20 16 cos2 t 20 16 1 4 32 5 cos t 2 K 25 sin2 t K 3 25 3 4 32 160 91 8281 32. rt rt rt Tt Tt 2 cos t i 2 sin t i 4 sin2 t 2 sin t i 4 sin2 t sin t j cos t j cos2 t cos tj cos2 t 33. rt rt Tt Tt K a cos t i a sin t i sin t i cos t i Tt rt a 1 a a sin t j a cos t j cos t j sin t j 2 cos t i 4 sin t j 4 sin2 t cos2 t 3 2 Tt rt 4 sin2 t 2 4 sin2 t cos2 t 4 sin2 t cos2 t 2 cos2 t 32 K 132 34. Chapter 12 rt rt Tt Tt a cos t i a sin Vector-Valued Functions b sin ti tj tj 35. r t at sin t , a 1 cos t b cos From Exercise 44, Section 12.4, we have: a N K a at vt a 2a2 2 2 a sin t i a2 sin2 t b cos t j b2 cos2 t 2 2 1 Nt cos t ab2 cos t i a2b sin t j a2 sin2 t b2 cos2 t 3 2 Tt rt a2 sin2 t ab a2 sin2 t b2cos2 t 2 2 a sin t b2 cos2 t ab b2 cos2 t 32 K 2 2 1 1 cos t cos t 4a 1 2 cos t 36. r t cos t t sin t, sin t t cos t 37. rt rt Tt Tt K ti i i t 2j 2t j t2 k 2 tk From Exercise 43, Section 12.4, we have: a N K 3t at v Nt 2 3t 4t 2 1 t 2t j t k 1 5t 2 5t i 2 j k 1 5t 2 3 2 Tt rt 5 1 5t 2 1 5t 2 5 5t 2 1 32 38. rt rt Tt Tt K 2t 2 i 4t i 4t i 1 4i 1 Tt rt j j tj 12 tk 2 tk 39. rt rt Tt 4t i 4i 1 4i 5 1 5 3 cos t j 3 sin t j 3 sin t j 3 cos t j 35 5 3 sin t k 3 cos t k 3 cos t k 3 sin t k 3 25 tk 17t2 k 32 17t j 17t 2 Tt K 289t2 17 1 1 17t 2 3 2 1 17 17t2 32 17t2 12 Tt rt 40. rt rt Tt Tt K et cos t i et sin t 1 3 1 3 Tt rt et sin t j et cos t i cos t i sin t i 3 et k et cos t cos t sin t cos t et sin t j sin t j cos t j sin t cos t 2 et k sin t cos t 1 k sin t 2 3et 2 3et S ection 12.5 41. y 3x 2 0, and the radius of curvature 42. y mx b Arc Length and Curvature 133 Since y 0, K is undefined. 0, K Since y undefined. 4 ,x x 4 ,y 1 x2 8 0, and the radius of curvature is 43. y y y K 1 K 2x2 4x 4 3, x 1 44. y y 2x 2 1 2 4 1 173 4 2 4 232 4 173 2 0.057 y K 1 K 8 ,y 1 x3 1 53 2 8 3 4 y y 17.523 (radius of curvature) 232 1 8 4 32 8 53 2 (radius of curvature) 45. y y y At x a2 x a2 a2 a2 0: x2, x x2 x2 32 0 46. y y y 16 x2 9x 16y 9 0: 16y 16y y y 2 y y K 1 K 0 1 a 1 a 1a 02 32 At x 0 3 16 1 16 3 3 16 02 3 2 1 a K 1 K 3 16 (radius of curvature) (radius of curvature) 47. (a) Point on circle: Center: Equation: ,0 2 ,1 2 2 y2 1 2 (b) The circles have different radii since the curvature is different and x r 1 . K 134 Chapter 12 4x2 x 2 Vector-Valued Functions 1 ,y x 1 2 02 32 48. (a) y y y At x 3 24x 3 2 49. y K x 1 1 ,y x2 2 at 1, 2 2 x3 x2 72 1 x2 0: y y x2 33 0 72 27 8 3 32 Radius of curvature 1 2. Since the tangent line is horizontal at 1, 2 , the normal line is vertical. The center of the circle is 1 2 unit above the point 1, 2 at 1, 5 2 . Circle: x 4 1 2 y 5 2 2 1 4 K r Center: 3 0, 8 y 83 1 02 1 K 3 8 8 3 (1, 2) −6 6 −4 Equation: x2 3 8 2 9 64 (b) The circles have different radii since the curvature is different and r 1 . K ln x, 1 , x 1, 1 1 12 x y y1 1 32 50. y y y1 K 1 1 x2 1 23 ,r 2 1 K 51. y y y0 K 1 ex, ex, 1, 1 12 x y y0 32 0 ex 1 1 23 2 1 ,r 22 1 K 22 1. 23 2 22 1. The slope of the tangent line at 0, 1 is y 0 The slope of the normal line is Equation of normal line: y 1 x 1 1 1. x or y The slope of the tangent line at 1, 0 is y 1 The slope of the normal line is Equation of normal line: y 1. x x 1 The center of the circle is on the normal line 2 2 units away from the point 1, 0 . 1 1 x x 2 2 The center of the circle is on the normal line 2 2 units away from the point 0, 1 . 0 x 2 1 x2 y 2 2 22 8 4 ±2 0 x 4x 2x 3x y 1 2 2 22 8 8 0 0 3 or x 1 3 and y 2. x x2 x 2x2 2 x2 2x 2 3 1 x Since the circle is above the curve, x Center of circle: Equation of circle: x 22 y3 2, 3 2 2 and y 6 3. Since the circle is below the curve, x Center of circle: 3, Equation of circle: x 32 y2 2 8 −6 2 2 y (0, 1) 0 3 8 −2 −4 (1, 0) x 2 4 6 S ection 12.5 13 x, 3 x2, 1, 1 2 1 Arc Length and Curvature 135 52. y y y1 K x y1 y1 32 1 2x 2 1, r 2 1 K 2 1. x 4 3 y 3 2 1 1 The slope of the tangent line at 1, 3 is y 1 The slope of the normal line is Equation of normal line: y 1 3 2 1. x 1 or y 2 units away from the point 1, 1 . 3 1 3 The center of the circle is on the normal line 1 1 x x 2 2 y 1 1 2 2 1 0 or x 2 0 and y 2 54. 4 3 y −2 x x 2 2 1 ( 1, 3 ) x x 4 Center of circle: 0, 3 1 −1 2 Since the circle is above the curve, x Equation of circle: x2 53. π 4 3. y 42 3 y π x B A −4 −2 2 4 x B A −3 − 2π −4 55. y K x 1 1 2 3, y 2x 232 1, y 1 2 232 56. y K x3, y 1 3x2, y 6x 9x 4 32 6x 2 2x 1 2 4x 1 (a) K is maximum when x (b) lim K x→ 1 or at the vertex 1, 3 . (a) K is maximum at (b) lim K x→ 0 0 4 1 , 45 4 1 , 453 4 1 , 45 4 1 . 453 57. y K x2 3, y 2 x 3 2 9x 4 9x 1 3, 43 y 2 x 9 x1 3 43 58. y 6 K dK dx 1 ,y x 1 y y 1 ,y x2 232 2 . Assume x > 0. x3 2 x3 1 1 x4 32 1 2332 9x2 3 4 32 x4 2x 3 1 32 (a) K ⇒ (b) lim K x→ as x ⇒ 0. No maximum 0 6x2 1 x 4 x4 1 5 2 1 and x 1 by symmetry . (a) K has a maximum at x (b) lim K x→ 0 136 Chapter 12 1 ,y x 1 x2 1 x2 2x2 x2 1 1 52 Vector-Valued Functions 1 x2 x 32 59. y K dK dx ln x, y 60. y K 1 32 e x, y 1 y y y 232 ex 1 ex e2x 32 1 x2 dK dx 1 . 2 (a) 1 e x 1 2e 2 x 1 e 2x 5 2 2e 2 x 0 ⇒ e2x 1 ⇒x 2 11 ln 22 1 ln 2. 2 1 ln 2 2 (a) K has a maximum when x (b) lim K x→ 0 K has maximum curvature at x (b) lim K x→ 0 61. y K 1 1 x3, y 6x 9x 4 32 3x2, y 6x 62. y K x 1 1 3 3, y 232 3x 1 6x 9x 1 2, y 1 1 432 6x 1 0 at x 1. y y Curvature is 0 at x 0: 0, 1 . Curvature is 0 at 1, 3 . cos x 64. y 0 for x 2 K. K sin x, y 1 cos x, y 32 63. y K cos x, y 1 y y 232 sin x, y 1 sin x n. cos x sin2 x 32 sin x cos2 x 0 for x Curvature is 0 at 2 K ,0 . Curvature is 0 for x n : n ,0 65. The curve is a line. 66. K 1 y y 232 0, so K y. At the smooth relative extremum y Yes, for example, y x 4 has a curvature of 0 at its relative minimum 0, 0 . The curvature is positive for any other point of the curvature. 67. Endpoints of the major axis: ± 2, 0 Endpoints of the minor axis: 0, ± 1 x2 2x 4y2 8yy y y K Therefore, since 4 0 x 4y 4y 1 16y2 1 1 4y3 x 4y 232 x 4y 4y x2 y 16y2 32 4y2 x2 16y3 16 12y2 4 32 1 4y3 16 16 3x2 0. 32 16y2 16 x2 2 ≤ x ≤ 2, K is largest when x ± 2 and smallest when x S ection 12.5 x x 2 Arc Length and Curvature 137 68. y1 ax b x , y2 y 4 2 We observe that 0, 0 is a solution point to both equations. Therefore, the point P is the origin. y1 y2 At P, y1 0 ab and y2 0 2 0 2 2 y2 ax b x x 2 , x, y1 y2 ab 2 x 2 2x , 2 y1 y2 x 2a −4 −2 P 2 y2 x 4 , 4 2 3 −4 y1 1 . 2 y2 0 or ab 1 2. Since the curves have a common tangent at P, y1 0 same curvature at P, K1 0 K2 0 . K1 0 K2 0 Therefore, 2a Thus, a y1 1 4 and 1 1 y1 0 y1 0 y2 0 y2 0 ± 2 or a 1 232 Therefore, y1 0 1 2. Since the curves have the 1 1 2a 12 232 232 12 1 22 32 ± 4 . In order that the curves intersect at only one point, the parabola must be concave downward. 1 b x 1 2a and 2. y2 x x 2 1 x2 4 x4 x2 16x6 0, K y 1, K 1 . K 69. f x (a) K 2 6x2 16x4 1 4x2 1 32 (b) For x x2 For x 5 2 2. f 0 1 2 2 0. At 0, 0 , the circle of curvature has radius 1 . Using the symmetry of the graph of f, you obtain 2 1 . 4 0. At 1, 0 , the circle of curvature has radius 2 5 5. f 1 1 Using the graph of f, you see that the center of curvature is 0, 2 . Thus, 2 f −3 3 x 2 y 1 2 2 5 . 4 To graph these circles, use y 1 ± 2 1 4 x2 and y 1 ± 2 5 4 −2 x2. . 5 (c) The curvature tends to be greatest near the extrema of f, and K decreases as x → ± However, f and K do not have the same critical numbers. Critical numbers of f: x Critical numbers of K: x 0, ± 2 2 ± 0.7071 −3 −2 3 0, ± 0.7647, ± 0.4082 138 Chapter 12 Vector-Valued Functions 70. y (a) 1 85 x ,0≤x≤5 4 5 z 4 2 (b) V 0 2x 58 5 4 x8 5 dx 4 125 5 3 5 9 2 4 y 2 4 x 114.6 cm3 (rotated about y-axis) (c) y 2 35 x ,y 5 6 x 25 25 6 25x2 5 K 1 1 6 25x2 5 4 65 x 25 (d) No, the curvature approaches as x → 0 . Hence, any spherical object will hit the sides of the goblet before touching the bottom 0, 0 . 6 46 x 25 32 5 32 25x 2 5 1 0 0 5 71. (a) Imagine dropping the circle x2 y k 2 16 into the parabola y to the point where the tangents to the circle and parabola are equal. y x2 and x2 y 2y k 2 x2. The circle will drop y 16 ⇒ x2 0 k . and y x2 k 2 16 15 10 Taking derivatives, 2x y Thus, x y Thus, x2 x2 k 2 ky x y 2x. Hence, ky x⇒y − 10 −5 x 5 10 k 2x ⇒ x 2x y k⇒ 1 2 x2 k ⇒ x2 k 1 . 2 x2 1 2 2 16 ⇒ x2 15.75. Finally, k x2 1 16.25, and the center of the circle is 16.25 units from the vertex of the parabola. Since the radius of 2 the circle is 4, the circle is 12.25 units from the vertex. (b) In 2-space, the parabola z y2 or z touch the vertex has radius 1 K c K 13 x 3 x2 2x 2x 1 x4 At x 32 x2 has a curvature of K 1 2. 2 at 0, 0 . The radius of the largest sphere that will 72. s y y y K When x 1: s K c 1 30 1 2 4 2c 30 2 2 4 2c ⇒ c 4 3 ,K 2 s [1 3 2 3 81 16 c K 32 0.201 4 30 2 K 56.27 mi hr. S ection 12.5 Arc Length and Curvature 1 . K 139 73. P x0, y0 point on curve y y f x . Let , be the center of curvature. The radius of curvature is 1 . f x0 x0 x0 2 f x . Slope of normal line at x 0, y0 is y0 1 x f x0 y0 y0 Equation of normal line: y , is on the normal line: f x0 2 Equation 1 2 x0, y0 lies on the circle: x 0 1 K 1 f x0 f x0 232 2 Equation 2 Substituting Equation 1 into Equation 2: f x0 y0 y0 2 y0 1 1 2 y0 f x0 2 2 1 K 2 y 1 f x0 2 f x0 2 3 P ( x0, y0) 1 K 2 f x0 f x0 2 22 (, ) x When f x0 > 0, Hence y0 1 y0 > 0, and if f x0 < 0, then y0 < 0. f x0 f x0 1 2 y0 Similarly, x0 f x0 f x0 2 y0 z f x0 z. 74. (a) y z , (c) y z , fx 1 e x, f x f0 f0 0 2 fx e x, 0, 1 (b) y z x2 ,y 2 1 f1 f1 1 x, y 2 1, 1, 2 1 2 2 2, 1 2x, y 2 2 2, 0, 0 2, 3 , 2, x2 1 ,y f0 f0 1 2 2 1, 5 2 1 2 1 2 r sin j 0, 1 2 cos i f sin j 0, 0 75. r x y x y x y K r cos i f f f f f f xy x 2 f cos sin sin cos cos sin yx y2 f f f 32 f cos sin sin cos f2 f 2 f f f sin cos f f f f 2f 232 cos sin 2 f f r2 r2 cos sin rr r 2f 2f 2r 232 2 sin cos f f cos sin 140 Chapter 12 1 cos sin 2r r 2 cos 31 81 (b) r r r K 1 0 2r r 2 2 2 2 2 Vector-Valued Functions (c) r r r rr r2 r2 32 76. (a) r r r K sin a sin a cos a sin 2r r 2a 2 2a2 a3 2 2 K sin 1 3 2 21 sin sin (d) r r r 1 23 rr r2 r2 32 1 sin sin sin sin 2 cos2 a2 sin2 a2 sin2 3 cos2 3 a2 cos2 2 ,a > 0 a a2 sin2 e e e 2r r 2 2 rr r2 r2 32 2 1 2 232 K rr r2 r2 32 2e2 2e2 3 2 1 2e 77. r r r K ea , a > 0 aea a2ea 2r r e a 2 2 78. At the pole, r K 2r r 2r r 2 3 2 2 0. rr r2 2 r r2 32 rr r2 1 r2 32 2a2e2a a2e2a 2 2a ae e2a e2a 32 1 a2 ⇒ (a) As , K ⇒ 0. , K ⇒ 0. 80. r r 2 r0 2 8 1 4 6 cos 3 18 sin 3 (b) As a ⇒ 79. r r 4 sin 2 8 cos 2 At the pole, 6 and K 2 r 6 2 18 1 . 9 ,r 6 18, At the pole: K 81. x f t,y dy dx dy dt dx dt gt gt ft f tg t ft ft g tf t 2 y y dgt dt f t dx dt y y f tg t ft g tf t 3 f tg t K 1 232 1 g tf t ft3 g t 2 32 ft f tg t ft ft 2 g tf t 3 gt 2 23 f tg t ft2 g tf t g t 232 ft S ection 12.5 82. x t yt K t3, x t 12 t,y t 2 3t 2 1 3t 2 2 3t 2 t 3 Arc Length and Curvature y y y a1 a sin a cos x 232 141 3t 2, x t t, y t t 6t t 1 232 6t 1 83. x x x K 3 a a1 a sin x x sin cos cos y 2 y y 9t 2 32 t 9t 2 1 32 a2 1 cos cos a2 sin2 a 2 1 cos 2 a 2 sin2 3 1 cos 1 a 2 2 cos 3 1 1 cos a 2 2 1 cos 2 2 K → 0 as t → ± 5 32 1 cos ≥ 0 −4 0 4 1 2a 2 Minimum: 1 4a 2 cos 1 csc 4a 2 Maximum: none 12 tk 2 tk 5t2 1 K→ as →0 84. (a) r t vt ds dt K aT aN 3t 2i 6t i vt 2 31 d 2s dt 2 K ds dt 3t 3 t3 j 3t 2 j (b) r t vt 6t ds dt d 2s dt2 at ti i vt t2j 2tj 31 d 2s t2 , 2 dt t2 2 6t 2 5t 5t2 2j rt k 1 ijk 1 2t t 021 5 1 32 2 31 t2 2 rt 91 t2 2 vt at j 2k 6 K aT aN rt rt rt3 d 2s dt2 K ds dt 2 5t2 1 5 1 32 5t 5t2 5t2 5t2 1 5 5t2 1 85. F maN mK ds dt ds dt 2 5500 lb 32 ft sec2 6400 lb 32 ft sec2 x 1 100 ft 1 250 ft 30 5280 ft 3600 sec 35 5280 ft 3600 sec 2 3327.5 lbs 2 2 86. F maN mK 94864 45 2108.1 lbs 87. y y y K cosh x ex 2 ex 2 1 e x ex 2 x e e sinh x cosh x cosh x cosh2 x 3 1 cosh2 x 1 y2 cosh x sinh x 232 2 142 Chapter 12 Vector-Valued Functions dT ds Tt vt rt rt ds Tt dt d 2s Tt dt 2 ds dt Tt rt rt ds Tt dt Tt ds dt 2 88. (a) K Ts d T dt ds dt d T dt , by the Chain Rule dt ds Tt rt (b) Tt rt rt rt rt rt ds dt Since T t rt rt Therefore, d 2s Tt dt 2 ds 0 and dt rt rt 2 2 Tt Tt r t , we have: Tt Tt rt 2 Tt Tt K. Tt Tt rt 2 1K r t from (a) rt rt rt3 rt rt 2 vt vt rt at 2 (c) K rt rt r t3 rt rt at Nt rt2 91. True 92. True aN K ds dt 2 89. False 90. False Curvature 1 radius xt 1 xt 2 r 2 2 93. Let r r dr dt xti ytj xt xtx t 2 z t k. Then r yt 2 r 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 yty t GmM r r3 GM r r3 ztz t r. 94. F ma ⇒ m a a Since r is a constant multiple of a, they are parallel. Since a d r dt Thus, r r r r r r 0 0 0. r is parallel to r, r r 0. Also, r is a constant vector which we will denote by L. S ection 12.5 95. Let r xi yj rr x2 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 1 x y2 r3 1 yz r3 i yz x 1 r GM 1 0 GM r r3 1 r r3 Thus, r GM L r r r r xz y x z2 xyy j xz xy z 1 r r3 r 1 r r3 r r r r r r 0 xzz i x 2y k xy z 2y xx y 1 r r3 zz y j x 2z r r xx r 2r yy r r3 r. rr (using Exercise 93) Arc Length and Curvature 143 dr dt r zz xi yj zk y 2z xx z yy z k r r 96. dr dt GM L r r 0 r GM r r3 L r r 1 r r3 r r r is a constant vector which we will denote by e. r y 97. From Exercise 94, 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 96, we have r L GM r r e. 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. 1 2 98. L Let: r r r r r cos i r sin i sin j d cos j dt i dr dt j r sin d dt r cos r r d dt r2 dr d d dt k 0 0 d . dt 99. A 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. Then: r r r cos r sin r2 d k and L dt 144 Chapter 12 Vector-Valued Functions 100. Let P denote the period. Then P A 0 dA dt dt 1 L P. 2 Also, the area of an ellipse is ab where 2a and 2b are the lengths of the major and minor axes. ab P P2 1 LP 2 2 ab L 4 4 4 22 a L 2 a2 c2 4 2 4 22 a2 a1 L2 e2 a ed L2 a 2 24 ed 2 L a3 423 a GM Ka 3 L 2 GM 3 a L2 Review Exercises for Chapter 12 1. r t ti csc t k n , n an integer n , n an integer 2. r t ti 1 t 4 j k (a) Domain: t (b) Continuous except at t (a) Domain: 0, 4 and 4, (b) Continuous except at t 4. r t 2t 1i t 2j , tk 4 3. r t ln t i tj tk (a) Domain: 0, (b) Continuous for all t > 0 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 (a) Domain: (b) Continuous for all t 1i c 21 2 ti c 1 2j t t t 1 2j 1 3 1 3 c 1 3k c 1 3k 1 1 3 1i 2j t 2j t3 1 3 1 t 3k 3 tk 3i j 1 3k 3 t2 6. (a) r 0 (b) r 2 3i 2 j k i 3 cos 3 cos t 1 sin s ti 3i 1 j sin tk s k tj tk 3i j k (c) r s (d) r t 3 cos s r sin t R eview Exercises for Chapter 12 t t 1 t 145 7. r t xt x2 cos t i 2 sin2 t j 2 sin2 t 8. r t xt y x ti j t 1 9. r t x y z i 1 t tj t 2k cos t, y t y 2 y 1 21 t, y t x 1 y 4 x2 t2 ⇒ z z 2 y2 1≤x≤1 y 3 2 1 1 1 x −2 −1 1 2 3 4 2 x 1 y −1 x 1 −2 10. r t x y t x y z 2t i 2t, y 1 2 x, tj t, z y2 t 2k t 2, 11. r t x t i 1, y 0 1 0 1 sin t j sin t, z k 1 3 2 1 0 1 1 1 1 12. r t x x 2 2 cos t i 2 cos t, y z 2 tj t, z 2 sin t k 2 sin t z 4 3 2 2 0 3 2 0 2 0 0 0 0 1 2 1 1 z 1 2 1 1 2 4 2 4 x y z 2 1 1 1 z 3 t x y z 0 2 0 0 z 2 0 2 2 5 2 3 1 −2 2 3 1 1 2 y 2 x 2π y 4 x x 3 y 13. r t ti ln t j 12 2t k 3 2 1 z 14. r t 1 2 ti tj 13 4t k 6 5 4 3 z 1 2 3 1 2 y −3 −2 1 −1 2 2 1 −1 3 1 2 −2 3 −3 x x y 15. One possible answer is: r1 t r2 t r3 t 4t i 4i 4 3t j, 3 t i, t j, 0t≤1 0≤t≤3 0≤t≤4 16. One possible answer is: r1 t r2 t r3 t 4t i, 4 cos t i 4 t j, 4 sin t j, 0≤t≤1 0≤t≤ 2 0≤t≤4 17. The vector joining the points is 7, 4, rt 2 7t, 3 4t, 8 10 . One path is 10t . 146 Chapter 12 Vector-Valued Functions 19. z x 3 4 18. The x- and y-components are 2 cos t and 2 sin t. At t 3 , 2 x2 t, y rt y 2, x t, z ti tj y 0, t 2t 2 2t 2k x 5 z the staircase has made of a revolution and is 2 meters high. Thus, one answer is rt 2 cos t i 2 sin t j 4 t k. 3 x t2 t2k t2k 4 x 5 y 3 z −3 2 3 1 2 3 y x 20. x2 x z2 t, y rt rt 4, x t, z ti ti tj tj y ± 0, t 4 4 4 21. lim t 2 i t→2 4 t2j k 4i k 22. lim t→0 sin 2t i t e tj et k lim t→0 2 cos 2t i 1 23 tk 3 j k 2i j k 23. r t 3t i 3i ut t j 1 j, u t ti t 2j (a) r t (c) r t Dt r t (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 0 2r t 2r t ut ut 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 3t 2 ut (e) r t Dt r t 10t 2 2t 1 1 2t 10t 10t 2 2t 2 i 24. r t sin t i cos tj t k, u t k sin t i cos t j 1 k t (b) r t (d) u t sin t i 2r t 2r t cos t j sin t i cos t j sin t j 1 t 2t k 1 t2 2k (a) r t (c) r t Dt r t cos t i ut ut 2 sin t j 0 Dt u t cos t i (e) r t Dt r t (f) r t Dt r t ut 1 t2 t 1 t2 t cos t i 1 cos t t2 1 sin t t t sin t t sin t j cos t i 1 cos t t 1 sin t t2 t cos t sin t j 1 cos t t ut 1 sin t t 25. x t and y t are increasing functions at t decreasing function at t t 0. t0, and z t is a 26. The graph of u is parallel to the yz-plane. R eview Exercises for Chapter 12 147 27. cos t i t cos t j dt sin t i t sin t cos t j C 28. ln t i t ln tj k dt t ln t ti t2 4 1 2 ln t j tk C 29. cos t i sin t j t k dt 1 t 2 dt 1 t1 2 t2 ln t 1 t2 C 30. tj t 2k i tj t k dt t2 t3 i t 2j t k dt t3 3 t4 i 4 t3 j 3 t2 k 2 C 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 32. r t sec t i ln sec t tan t j tan t i t 2k dt ln cos t j t3 k 3 C 5k ⇒ C e t 4k r0 rt C 3k tan t i ln cos t j t3 3 3k ln sec t 2 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 1 34. 0 tj t sin t k dt 2 32 tj 3 2 j 3 sin 1 1 sin t t cos t k 0 cos 1 k 2 2 35. 0 et 2i 3t2j k dt 2et 2i t3j tk 0 2e 2i 8j 2k 1 36. 1 t3i arcsin tj t2k dt t4 i 4 t arcsin t 1 t2 j t3 k 3 1 1 2 k 3 37. r t vt cos3 t, sin3 t, 3t rt vt 3 cos2 t sin t, 3 sin2 t cos t, 3 9 cos4 t sin2 t 9 sin4 t cos2 t sin2 t 1 9 3 cos2 t sin2 t cos2 t 3 at vt cos2 t sin2 t 1 sin2 t 6 cos t 3 cos2 t cos t, 6 sin t cos2 t cos2 t , 3 sin t 2 cos2 t sin2 t , 0 3 sin2 t sin t , 0 3 cos t 2 sin2 t 38. rt rt vt rt at t, vt 1 0, tan t, et 1, sec4 t 2 sec2 t sec t, e e2t tan t, et 2 t 39. rt rt r4 ln t 1 t 3 1 3 , t 2, t , t 0 2 , 2t, 1 2 4 1, 8, 1 2 direction numbers Since r 4 0, 16, 2 , the parametric equations are 1 x t, y 16 8t, z 2 2 t. r t0 0.1 r 4.1 0.1, 16.8, 2.05 148 Chapter 12 Vector-Valued Functions 12 gt 2 40. rt rt r0 Since r 0 y t, z r t0 3 cosh t, sinh t, 3 sinh t, cosh t, 0, 1, 2t , t0 2 0 41. r t v0 t cos , v0 t sin 75 3 75 t, t 2 2 16t 2 75 32 2 direction numbers 3, 75 t 2 Range 3, 0, 0 , the parametric equations are x 2t. 0.1 r 0.1 3, 0.1, 0.2 16t 2 0⇒t 75 3 75 2 32 v0 cos 5625 64 v0 sin 1 2g 3 152.2 feet v02 sin 2 g or, Range 75 2 sin 60 32 6 4 ⇒ v0 16 6 152.2 feet 42. y x v0 16t2 v0t ⇒ 4 86 3 6 0⇒t v0 6 4 6.532 ft sec 43. Range x v02 sin 2 9.8 ti 80 ⇒ v0 80 9.8 sin 40 9.8 t2 j 4.9t 2 j 34.9 m sec (see Exercise 41) 44. r t (a) r t 20 v0 cos v0 sin t 1 2 20 cos 30 t i 20 sin 30 t (b) r t 20 20 cos 45 t i 20 sin 45 t 4.9t 2 j 0 0 45 0 0 45 Maximum height (c) r t 20 5.1 m; Range 35.3 m 4.9t j 2 Maximum height 10.2 m; Range 40.8 m 20 cos 60 t i 20 sin 60 t 0 0 45 Maximum height 15.3 m; Range 35.3 m (Note that 45 gives the longest range) R eview Exercises for Chapter 12 45. rt vt vt at Tt 5t i 5i 5 0 i 46. rt vt v at Tt 1 4i 5 0 1 4i 5 3j 4t i 3j 2 3t j 149 N t does not exist. a a T 0 a a T N t does not exist. 0 N does not exist. (The curve is a line.) N does not exist. 47. rt vt vt at Tt Nt a a T N ti i tj 1 j 2t 4t 1 2t 1 j 4t t 48. rt vt vt at 2t 2i 2 t t 4 t t i t t t t 1 1 1 1 1i 2 t 1 14 12 3 2 t 2j 1 j 1 j j 1 i i 12 tj 4t 1 2 t 2 tj 4t 1 1 1 2 ti 4t j 1 Tt Nt a a T N 1 2i t 14 t 4 1 2j 1 1 4 3 1 4t t 4t 1 2t 4t t 1 t 4 t 1 2 4 1 1 1 4t 3 1 1 4 4 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 et i e t j e2t e 2t e ti e2t etj e 2t e2t e 2t e2t e 2t 2 e2t e 2t 150 50. Chapter 12 rt vt vt at Tt Nt at at Tt Nt t cos t i rt speed rt vt vt Vector-Valued Functions t sin t j t sin t cos t i cos t 2 sin t i cos t i t2 t sin t 1 2 t cos t sin t j sin t 2 t sin t t cos t t sin t sin t i t2 t cos t t sin t t cos t 1 cos t j t2 1 2 cos t j sin t j t cos t t t2 1 t2 2 t2 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 a a T N Nt 5 51 vt at Tt 2t 4 t2 2 k t3 t 2i i t2j k 2t 4 1 2t 2k 2 2t 4 1 j 1 1 1 k t 51. rt vt v at Tt Nt a a T N ti i 1 2j i t 2j 2t j 52. rt vt i t 1i j tj 1 k t2 1 2 t 3 2t 4 4 t 2 2t 4 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 54. r t ti t2j 2, x i 23 3 t k, x t, y 4, z t 2, z 16 3. 23 3t 55. v When t rt 2, y 2t j 9.56 104 4.56 mi sec 4600 (see Exercise 56, Section 11.4) 2t 2k 2, a 4, z 8t 1, b 16 3 Direction numbers when t x t 2, y 4t 4, c 8 R eview Exercises for Chapter 12 56. Factor of 4 57. r t rt s a 5 151 2t i 2i b 3tj, 0 ≤ t ≤ 5 3j 5 58. r t rt t2i 2t i b 2tk, 0 ≤ t ≤ 3 2k 3 r t dt 0 4 9 dt s a r t dt 0 4t2 1 t t t2 4 dt 3 13t 0 y 2 −4 −2 −4 −6 −8 − 10 − 12 − 14 − 16 5 13 ln ln t2 10 z 1 0 3 3 10 11.3053 (0, 0) x 246 8 10 12 14 6 5 4 3 2 (10, − 15) (9, 0, 6) −2 1 2 3 4 5 6 7 9 x (0, 0, 0) 1 2 y 59. rt rt rt 10 cos3 t i 10 sin3 tj 30 sin2 t cos tj sin4 t cos2 t 60. rt rt rt s 10 cos t i 10 sin t i 10 2 10 sin tj 10 cos tj 30 cos2 t sin ti 30 cos4 t sin2 t 30 cos t sin t 2 10 dt 0 y 20 s 4 0 y 30 cos t sin t dt sin2 t 120 2 2 60 0 8 6 4 2 − 8 − 6 − 4 −2 10 x 2468 2 − 10 −2 x 2 10 −4 −6 −8 − 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 152 Chapter 12 Vector-Valued Functions 62. r t rt s ti i b t2j 2tj 2tk, 0 ≤ t ≤ 2 2k, r t 2 63. r t 5 4t 2 8 cos t, 8 sin t, t , 0 ≤ t ≤ 8 sin t, 8 cos t, 1 , r t b 2 2 65 65 2 rt s a r t dt a 0 5 5 ln 4 4t2 dt 105 45 6.2638 r t dt 0 z 65 dt 21 z 4 3 2 1 1 2 x 1 5 ln 5 4 π 2 (0, 8, π ) 2 4 6 8 y 4 86 x (8, 0, 0) (2, 4, 4) 2 3 4 y 64. r t rt s 2 sin t t cos t , 2 cos t t sin t , t , 0 ≤ t ≤ 4t2 1 dt 1 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 2t sin t, 2t cos t, 1 , r t b 2 r t dt a 0 2 4t2 1 1 ln 4 3.055 0 2 4 1 0 1 4 dt 0 cos2 t 5 t 2 sin2 t dt 5 2 5 2 et cos t k, 0 ≤ t ≤ et sin t i et sin t 2 0 66. rt rt rt et sin t i et cos t et 2et cos t 67. r t et cos t k et cos t 2 3ti 2tj et sin t et sin t Line K 0 s 0 r t dt 2 0 et dt 2et 0 2e 1 68. rt rt rt 2 ti 1 t i 3tj 3j, r t 3 2i 69. 1 t 9 1 t 9t 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 1 t 2 i 1 t 1 t 2 j 3 32 k 0 0 3 2t3 2 1 9t 3 2 t3 3 2 k 2t 2 5t2 4j 20 4 2k, r r 25 5t2 20 r r 3 t 2 3 2k; r r 3 2t 3 2 K 32 0 4 32 K rt rt rt3 21 9t 32 R eview Exercises for Chapter 12 70. rt rt rt r r r r K r r 2ti 2i 5 cos t j 5 sin tj 5 sin tk 5 cos tk, r t 29 153 5 cos tj i 2 0 725 r 3 5 sin tk k 5 cos t 5 sin t 25i 10 sin tj 10 cos tk j 5 sin t 5 cos t 725 29 3 2 25 29 29 29 5 29 71. y y y K At x 12 x 2 x 1 1 2 72. y y e x2 1 e 2 x 2, y 1 e 4 x2 y y 4, K 232 1 2 1 x2 32 K 2 1 y y 232 1 173 1x e 4 1 1 e 4 2 2 2 x 32 and r 173 17 17. At x 0, K 14 5 43 2 55 53 2 25 ,r 25 55 . 2 73. y y K At x ln x 1 ,y x 1 y y 1, K 1 x2 232 74. y y y 1 1 22 1 1 x 2]3 x2 2 tan x sec2 x 2 sec2 x tan x 1 4 y y ,K 232 K 2 2. At x 2 sec2 x tan x 1 sec4 x]3 2 4 55 45 and r 25 55 . 4 1 23 2 2 and r 4 4 53 2 75. The curvature changes abruptly from zero to a nonzero constant at the points B and C. 76. y y y K At x ax 5 5ax 4 20ax3 1 1: bx 3 3bx 2 6bx 20ax 4 6bx 5ax 4 3bx 2 c k y y1 0 ⇒ 20a 0 ⇒ 5a 1⇒a b −1 x 1 cx c 1 y (1, 1) 232 (− 1, − 1) −1 6b 3b c c 0 0 1 3 8, Solving these 3 equations for a, b, c, you obtain a y 35 x 8 53 x 4 15 x 8 b 5 4, c 15 8. By symmetry, the same holds at x 1. 154 Chapter 12 Vector-Valued Functions Problem Solving for Chapter 12 t 1. x t 0 cos u2 du, y t 2 t sin 0 u2 du 2 2. 2 x 3 13 x2 3 y2 1 3y 3 a2 0 3 xt (a) s t2 cos ,y t 2 a t2 sin 2 a 2 y 3 xt 0 2 y t 2 dt 0 y dt a rt t2 t cos 2 rt r ti t Tt Tt cos3 ti y1 x 13 3 Slope at P x, y . sin3 tj 3 sin2 t cos tj (b) x t t2 t sin ,y t 2 t cos2 t2 2 1 a, K a a. length t sin2 3 cos2 t sin ti 3 cos t sin t rt rt sin ti cos ti cos tj K At t (c) K t2 2 sin tj Q 0, 0, 0 origin P \ cos3 t, sin3 t, 0 on curve. T i cos3 t cos t cos3 t sin t \ PQ j sin3 t sin t k 0 0 sin3 t cos t k D K PQ T T Tt rt cos t sin t 1 3 cos t sin t 1 Thus, the radius of curvature, , is three times the K distance from the origin to the tangent line. 3. Bomb: r1 t Projectile: r2 t 5000 400t, 3200 t, v0 sin 16t2 t 16t 2 4. Bomb: r1 t Projectile: r2 t 5000 400t, 3200 t, v0 sin 16t2 t 16t2 v0 cos v0 cos At 1600 feet: Bomb: 3200 16t 2 1600 ⇒ t 10 seconds. At 1600 feet: Bomb: 3200 16t 2 1600 ⇒ t 10 Projectile will travel 5 seconds: 5 v0 sin 16 25 v0 sin Horizontal position: At t At t 10, bomb is at 5000 400 10 1000. 1600 400. Projectile will travel 5 seconds: 5 v0 sin 16 25 v0 sin Horizontal position: At t At t Thus, 2⇒ 5v0 cos 63.4 . v0 cos Combining, v0 v0 sin v0 cos 9000 1800. 400 ⇒ tan 1800 1843.9 ft sec 2 ⇒ 9 12.5 . 10, bomb is at 5000 400 10 5. 9000. 1600 400. 5, projectile is at 5v0 cos . 200. 400 ⇒ tan 200 447.2 ft sec 5, projectile is at v0 cos Thus, v0 cos Combining, v0 v0 sin v0 cos 200 cos 1800 cos P roblem Solving for Chapter 12 5. x x 2 155 1 y cos , y 2 sin , 0 ≤ 1 2 cos 2 cos t 2 ≤2 6. r r 1 sin t cos sin2 4 sin2 t st 2 t 1 2 sin cos d 2 sin2 d t 2 4 cos t 2 2 cos d t st x K 1 2 sin 2 d 4 cos cos 2 t 4 cos 2 K cos 8 sin3 1 4 sin 2 3 4 1 2 3 2r 2 4 cos r2 32 2 2 sin , y cos cos r 2 sin2 2 rr r2 1 sin sin 3 cos cos 2 1 cos 2 2 sin 2 8 sin3 3 cos 8 sin3 sin2 sin3 2 3 4 sin 2 Thus, s 2 1 K 2 t 4 sin and 2 t 16 cos 2 2 2 2 t 16 sin 2 2 16. 1 K s2 7. r t 2 4 sin 3 2 2 9 16 cos2 2 16 sin2 2 16 rt 2 rt 2rt d rt dt rt rt rt rt ⇒ d rt dt rt rt rt d rt dt 8. (a) r r dr dt a xi yj position vector r sin j r sin d i dt dr sin dt dr cos dt dr sin dt d dt d dt dr sin dt dr cos dt r cos d dt d dt d j dt r cos r sin d dt d dt 2 r cos i dr cos dt d 2r dt2 d 2r cos dt2 d 2r sin dt2 r sin 2 d2 i dt2 d2 dt2 r cos ar a ur a dt2 cos i d 2r cos2 d 2r 2 sin dt2 sin j 2 dr sin dt 2 2 cos d dt d dt r cos2 r sin2 d dt 2 r cos sin 2 d2 dt2 d2 dt2 dr sin cos dt d dt r cos sin d 2r dt2 a a a a dt 2 r d dt u ur ur r a a d dt 2 sin i uu ur 2 cos j 2 dr d dr dt r d2 dt2 d 2r dr d dt dt r d2 u dt2 —CONTINUED— 156 Chapter 12 Vector-Valued Functions 8. —CONTINUED— (b) r r d dt t 42,000 cos i 12 42,000, d2 dr dt 0 0, d 2r dt2 t 42,000 sin j 12 0 9. r t rt rt T T 2 4 cos ti 4 sin ti 4 cos ti 4 sin ti 5 4 cos ti 5 cos ti T 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 3k, r t 3 k 5 , 12 d t 2 Therefore, a 42000 12 2 ur 875 3 2u . r N B At t Radial component: 875 3 3 cos tj 5 3 k 5 4 k 5 Angular component: 0 2 2 2 2 3 i 5 4 i 5 j 4 k 5 T B N B N T 3π 4 x 3 2 1 4 y 10. r t rt T T N B At t T cos ti sin ti sin ti cos ti cos ti N ,T N B 4 4 4 sin tj k, t z 4 2 cos tj, r t cos tj sin tj sin tj k 2 i 2 2 i 2 k 1 1 −1 −2 −1 −2 2 −1 2 x −2 y 4 2 j 2 2 j 2 P roblem Solving for Chapter 12 dB ds N 157 11. (a) B T dB ds N d T ds T T T 1 constant length ⇒ N N N T T T N T T T N B (b) B B T N N. Using Exercise 11.4, number 64, T N N N N T T NT N N TN T dB ds B T T T N 0 T T N T T T N. T NT N T TN dB dB B and Hence, ds ds for some scalar . dB T⇒ ds Now, KN Finally, Ns d B ds dT ds Ts Ts Ts dT . ds T B B KT T KN B. B T N T 12. y y y 15 x 32 53 x 64 2 2 15 1 x 128 2 K 15 1 2 x 128 25 3 x 1 4096 32 At the point 4, 1 , K 120 ⇒r 89 3 2 1 K 89 3 2 120 7. 13. r t (a) −3 t cos t, t sin t , 0 ≤ t ≤ 2 2 2 (b) Length 0 2 3 r t dt 22 t 1 dt 6.766 (graphing utility) 0 −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→ 158 Chapter 12 Vector-Valued Functions 14. (a) Eliminate the parameter to see that the Ferris wheel has a radius of 15 meters and is centered at 16 j. j, which is the low point on the Ferris wheel. At t 0, the friend is located at r1 0 (b) If a revolution takes t seconds, then t 10 and so t t t 10 2 20 seconds. The Ferris wheel makes three revolutions per minute. 8.032 11.472 14 m sec. The angle of 20 8.03i 11.47j. The speed is (c) The initial velocity is r 2 t 0 0.96 radians or 55 . inclination is arctan 11.47 8 03 (d) Although you may start with other values, t0 0 is a fine choice. The graph at the right shows two points of intersection. At t 3.15 sec the friend is near the vertex of the parabola, which the object reaches when t t0 2 11.47 4.9 1.17 sec. 0 30 0 Thus, after the friend reaches the low point on the Ferris wheel, wait t0 order to allow it to be within reach. 2 sec before throwing the object in (e) The approximate time is 3.15 seconds after starting to rise from the low point on the Ferris wheel. The friend has a constant speed of r 1 t 15 m sec. The speed of the object at that time is r 2 3.15 8.032 11.47 9.8 3.15 2 2 8.03 m sec. ...
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