EVEN11 - C H A P T E R 11 Vector-Valued Functions Section...

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Unformatted text preview: C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . 268 Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . 273 Section 11.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . 278 Section 11.4 Tangent Vectors and Normal Vectors . . . . . . . . . . . . . . . 283 Section 11.5 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . 289 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304 C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions Solutions to Even-Numbered Exercises 2. r t 4 t2 i t 2j 6t k 4 t2 6t Domain: ln t i ln t ln t Domain: 0, i 8. r t Ft Gt t3 3 4. r t t 2 sin t i 4 cos tj tk sin t 4 cos t t Component functions: f t gt ht Domain: 6. r t Ft 2, 2 Gt Component functions: f t gt ht , 5t j 1i 1i 3t 2k 5t tj 4t j i 4t j 3t 2 3t 2k 3t 2 k 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: 10. r t cos t i i , 1, 2 sin t j (a) r 0 (b) r (c) r (d) r 4 2 i 2 cos t r 2j i cos 2 sin ti j 2 sin cos i tj 2 sin j cos i 2 sin j 6 ti k 6 e t4 6 6 6 6 12. r t (a) r 0 (b) r 4 (c) r c (d) r 9 t3 2j k 2i 2 t 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 3i 9 27j t4 e e 94 k 27 j 94 k 14. rt rt ti t t 3t j 2 4t k 3t 2 4t t1 2 9t2 16t2 25t 268 S ection 11.1 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 269 The dot product is a scalar-valued function. 1≤t≤1 2t k, 0.1 ≤ t ≤ 5 3 2t 3 ln x. Matches (a) 18. r t x cos cos ti t,y y2 sin sin tj t,z t 2k, t2 20. r t x ti t, y ln t j ln t, z 2 3x Thus, x2 1. Matches (c) Thus, z and y 22. r t x ti t, y tj t, z 2k 2⇒x y (b) 10, 0, 0 z y (a) 0, 0, 20 (c) 5, 5, 5 z 1 2 3 3 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 24. x y 1 1 t, y x t 26. x t2 y 5 4 t, y t2 t 28. x y x2 y2 2 cos t 2 sin t 4 y Domain: t ≥ 0 y 6 5 4 3 2 x 1 2 3 4 −1 3 2 1 x −1 1 2 3 4 5 −1 −1 1 1 x − 4 −3 −2 −1 −2 30. x x 2 2 cos3 t, y 23 2 sin3 t cos2 t 1 sin2 t 32. x y y t 2t 3t 5 34. x x2 9 z 3 cos t, y y2 16 t 2 1 4 sin t, z t 2 y 2 23 Line passing through the points: 3 x2 3 y 3 2 y2 3 22 0, 5 ( 2 , 0, 15 ) 2 5, 0 , z 6 4 5 15 , 0, 2 2 Elliptic helix z 4 (0, −5, 0) −3 −2 −2 −3 x 6 x 2 3 −6 2 −2 2 4 −4 −6 2 4 −4 −6 y 4 x 4 y 270 Chapter 11 Vector-Valued Functions 3 t 2 36. x x t x y z t2, y y2 ,z 4 2 4 4 3 2t, z 3 y 4 1 1 2 3 2 38. x y z cos t sin t t y2 t 1 t sin t t cos t 0 0 0 0 z 1 1 2 3 2 2 4 4 3 x2 z t2 1 z 2 or x2 y2 z2 1 Helix along a hyperboloid of one sheet z 4 3 2 −4 −3 −2 −1 1 1 −1 −2 5 x −3 2 3 4 y x 3 2 2 3 4 y 40. r t ti 32 tj 2 12 tk 2 z 42. r t Ellipse 2 sin t i 2 cos t j z 2 sin t k Parabola 2 1 1 −1 1 2 x 3 −2 −3 1 2 3 y x −2 −2 −3 −1 1 2 1 y −1 −1 44. r t ti z t2 j 13 th 2 (a) u t rt 2j is a translation 2 units to the left along the y-axis. z (b) u t 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 2 5 4 3 2 1 1 2 3 4 5 x x 5 x 5 4 4 3 12 34 5 −2 y 3 2 1 5 4 3 2 12 3 y −2 2 1 12 3 y 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 (e) u t r t reverses the orientation. z 5 4 3 2 1 1 2 3 4 5 x 12 3 y 5 S ection 11.1 46. 2x Let x rt 3y 5 0 1 2t 3 5j 5. 48. y 4 x2 t, then y ti 4 4 Vector-Valued Functions 271 t, then y ti Let x rt t 2. t2 j 1 2t 3 50. x Let x 2 2 y2 2 2 4 2 sin t. 2 sin t j 52. x2 16 Let x y2 9 1 4 cos t, y 4 cos t i 3 sin t. 3 sin t j 2 cos t, y 2 cos t i rt rt 54. One possible answer is rt 1.5 cos t i 1.5i 1.5 sin t j 2 k. r1 0 0, r1 10 4 10i r2 0 10i, r2 r3 0 4 5 2i 5 2i 5 2j 0 1 t k, 0 ≤ t ≤ 2 Note that r 2 56. r1 t r2 t r3 t t i, 0 ≤ t ≤ 10 sin t j , ti 10 cos t i 5 21 0≤t≤ t j, 5 21 0≤t≤1 5 2 j, r3 1 (Other answers possible) 0≤t≤1 y 1 t j, 58. r1 t r2 t ti 1 t j, ti x x 0≤t≤1 y (Other answers possible) 60. z x2 y 2, z y2 4 4 or 2 sin t, z 2sin t j 4k 2 x 2 y 6 z Therefore, x2 x rt 2 cos t, y 2 cos t i 4. 62. 4x2 If z 4y2 z2 16, x t2 and y z2 1 2 0 0 2 0 t2j 4 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 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 272 64. x2 Chapter 11 y2 2 z2 Vector-Valued Functions 10, x y 2 4 sin t and z 21 sin2 t 2 cos t. z Let x sin t, then y 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 66. x2 Let x y z 1 t y2 2 sin t j 2 cos t k z 4 z2 t, then 4 t t4 8 16, xy 4 (first octant) and 16t 2 x2 16 y2 z2 t2 16 t2 z2 16. 4 x 4 y 4 3≤t≤ 8 43 t x y z rt 8 43 1.0 3.9 0 1.5 1.5 2.7 2.6 1 t 2 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 t4 68. x2 y2 160 120 80 40 40 120 80 e t cos t z e t sin t 2 e 2t z2 70. lim et i t→0 sin t j t e tk i j k since lim t→0 40 sin t t lim t→0 cos t 1 1 (L’Hôpital’s Rule) x 80 120 y 72. lim t→1 ti ln t j t2 1 2t 2 k i 1 j 2 2k 74. lim e t i t→ 1 j t t t2 1 k 0 since lim ln t 1 t→1 t 2 since lim t→1 1t 2t 1 . (L’Hôpital’s Rule) 2 t→ lim e t 0, lim t→ 1 t 0, and lim t→ t t2 1 0. S ection 11.2 76. r t ti t 1j Differentiation and Integration of Vector-Valued Functions 78. r t 2e t, e t, ln t 1 . 273 Continuous on 1, 80. r t 8, t, 3 Continuous on t 1 > 0 or t > 1: 1, t Continuous on 0, ut 2. 82. No. The graph is the same because r t For example, if r 0 is on the graph of r, then u 2 is the same point. a if 84. A vector-valued function r is continuous at t the limit of r t exists as t → a and lim r t t→a ra. i i j t≥0 is not continuous at t j t<0 0. The function r t 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 z1 t z 2 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 88. Let ft 1, 1, if t ≥ 0 if t < 0 0, 90. False. The graph of x y z t 3 represents a line. and r t f t i. Then r is not continuous at c whereas, r 1 is continuous for all t. Section 11.2 2. r t xt y x3 r1 rt r1 i i i j 3t 2j 3j ti t 3 j, t0 t3 Differentiation and Integration of Vector-Valued Functions 1 4. r t xt x 1 y2 r2 rt r2 4i 2t i 4i 1 j 2 1 j t2 1 j 4 t 2i 1 j, t0 t 1 t 2 t, y t t 2, y t r t0 is tangent to the curve. y 4 3 2 r' (1, 1) x 1 2 3 4 r −4 −3 −2 r t0 is tangent to the curve. y 1 −2 −3 −4 2 r 1 r' x 8 −1 274 Chapter 11 Vector-Valued Functions 6. r t (a) ti 4 y 5 t2 j 8. r t y ti x 2, z rt t2j 3 2 i 2i i z 3 k, t0 2 2 r (1) 3 2 1 −3 −1 −1 2t j 4j 4j 3 k 2 r (1.25) r (1.25) − r (1) x 3 r2 r2 (b) r1 r 1.25 r 1.25 r1 rt i 3j 2.4375 j −2 2 1.25 i 0.25 i i r1 2t j i −4 −2 0.5625 j 2 x 4 (c) r −4 −6 y 2j 0.5625j 0.25 i 2.25 j r' −6 r 1.25 1.25 r1 1 0.25i This vector approximates r 1 . 10. r t rt 1 i t 1 i t2 16t j 16 j t2 k 2 tk 12. r t rt 4 ti 2 t i t2 t j 2t t ln t2 k t2 2t j 2 k t 14. r t rt sin t t cos t, cos t t sin t, t 2 16. r t rt arcsin t, arccos t, 0 1 1 t2 , 1 1 t2 ,0 t sin t, t cos t, 2t 18. r t (a) r t rt (b) r t t2 ti 2t 2i rt t2 1i 2j 2t tj 2t 1j 20. 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 8 sin t i 8 cos t i rt 12 2t 12 8t (b) r t 8 sin t 55 sin t cos t 22. r t ti i 2t 2j 3j 3k 3t 5k 24. r t (a) r t rt e t, t2, tan t e t, 2t, sec2 t e t, 2, 2 sec2 t tan t rt e 2t (a) r t rt (b) r t 0 rt 0 (b) r t 4t 2 sec4 t tan t S ection 11.2 1 4 Differentiation and Integration of Vector-Valued Functions 275 26. 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) 1 r' ( 4 ) r' 4 −2 −1 z 2 ( 1) 1 r'' ( 4 ) r'' 4 −2 −1 i 2 16 1 j 2 33 e 4 5 4 16k 2 x 1 −1 1 2 y 33 e 4 2j 93 e 16 8 20 4 9e 38 −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 1 t t 32j 28. rt rt 1 i 1 1 3t j i 3j 1 30. r r 1 sin cos i i 1 cos j sin j 2 r 2n 1 0, n any integer 1 , 2n 1 Not continuous when t Smooth on , 1 , 1, 2t2 8 t3 32t t3 Smooth on 2n 32. r t rt rt 2t 8 16 t3 i t3 4t3 i 82 j 2t 4 j 82 34. r t rt et i et i e tj e tj 3t k 3k 0 , 36. r t rt ti 1 i 2t t2 2t j 1j 1 k 4 1 tk 4 0 r is smooth for all t: 0 for any value of t. 2. 2, . r is smooth for all t > 0: 0, r is not continuous when t Smooth on 38. r t ut ti 1 i t i ut ut , 2 sin tj 2 sin tj 2, 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 Dt r t 2 cos t j 1 4 sin2 t 0, t 0 4 cos tj —CONTINUED— 276 Chapter 11 Vector-Valued Functions 38. —CONTINUED— (e) r t ut i t 1t 2 cos t Dt r t ut j k 2 sin t 2 cos t 2 sin t 2 cos t 1 t tj 1 t 2 sin t t t 1 t 1 k t 1 t2 1 t2 1j k 2 sin t 2 cos t 2 sin t 1 2 cos t t (f) r t Dt r t t2 4 12 t 2 t2i 2t i 2t 3 t4 tj j t t 2, r t 2t 3 t4 arccos 0.340 t t 2 4t 2 2t3 t4 4 12 2t t t2 4 1.0 40. rt rt rt rt rt cos 0 4t 2 1 − 0.5 1 1 0.707 2 . 2 t t2 4t 2 19.47 maximum at t 2 for any t. 42. r t lim rt t t t ti rt 3 t t 3 j t t→0 j 2t t 3 tk ti 2tk lim t→0 lim t t→0 t t t t i t t t 3 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 44. 4t3i 6tj 4 t k dt t 4i 3t 2 j 832 tk 3 C 46. ln t i 1 j t k dt t ln t ti ln t j tk C (Integration by parts) 48. et i sin t j cos t k dt et i cos t j sin t k C S ection 11.2 Differentiation and Integration of Vector-Valued Functions 277 50. e t sin t i e t cos t j dt et 2 sin t cos t i et 2 cos t sin t j C 1 52. 1 ti t 3j 3 t k dt t2 i 2 1 1 t4 j 4 1 1 343 tk 4 1 0 1 2 54. 0 ti etj tetk dt t2 i 2 2i 2 2 2 etj 0 0 t e2 1 et k 0 56. r t r0 rt C i 3t 2 j i 2 6 t k dt 2j t3 j 4t3 2k t3j 4t3 2k C e2 1j 1k 58. r t rt r0 rt r0 rt 4 cos t i 4 sin t i 3k 3k 3 sin t k 3 cos t k C1 ⇒ C1 3 sin t k C2 4j 3 sin t k 1 k dt t 1 j t j 4i 4 j ⇒ C2 4i 1 j t2 C 4j C1 0 4 cos t i 4i C2 4 cos t 1 1 4 2 i j t2 60. r t r1 rt i arctan t i 2 1 j t i ln t k ln t k C 2i ⇒ C 1 4 4 arctan t i 62. To find the integral of a vector-valued function, you integrate each component function separately. The constant of integration C is a constant vector. 66. Let r t x1 t i rt ±ut Dt r t ± u t y1 t j z1 t k and u t x2 t i y2 t j 64. The graph of u t does not change position relative to the xy-plane. z 2 t k. z1 t ± z2 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 68. Let r t rt Dt r t x1 t i ut ut y1 t j z1 t k and u t z1 t y2 t i y1 t z2 t 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 x1 t y2 t z1 t y2 t i z1 t y2 t i rt ut y1 t z2 t rt ut x1 t z2 t 278 Chapter 11 xti rt Dt r t rt rt Vector-Valued Functions 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 x ti y tj z t k. xty t xtz t ytx t k x tz t ztx t z tx t j 70. Let r t 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 72. Let r t then: xti ytj x2 t z t k. If r t y2 t y2 t z2 t z2 t 2z t z t ztz t rt r t is constant, C Dt C 0 0 0 74. False Dt r t ut rt ut rt ut (See Theorem 11.2, part 4) Dt x2 t 2x t x t 2xtx t 2y t y t yty t 2rt Therefore, r t rt 0. Section 11.3 2. r t vt at x 6 6 rt rt t, y 0 t, y ti i tj Velocity and Acceleration y 4. r t vt t2i rt rt t2, y t3j 2t i 2i t3 1. 2i 2i 3j 6j x 3t2j 6tj y2 3 8 7 6 5 4 3 2 1 −1 y j 4 a at v (3, 3) v 6 x 2 x x At 1, 1 , t v1 a1 (1, 1) x 234567 8 2 4 6. r t vt at x 3 cos ti 3 sin ti 3 cos ti 3 cos t, y 0. 2j 3i 2 sin tj 2 cos tj 2 sin tj 2 sin t, x2 9 y2 4 1 Ellipse y 3 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 et 1 v (1, 1) a et, y 1 x 1 2 x At 3, 0 , t v0 a0 At 1, 1 , t v0 v a x 2 1 1, 1 1, 1 1 −2 −1 a0 −1 (3, 0) −3 10. r t vt st at 4t i 4i vt 0 4t j 4j 2t k 2k 16 16 4 6 Section 11.3 12 tk 4 1 tk 2 1 12 t 4 10 12 t 4 Velocity and Acceleration 279 12. r t vt st at 3ti 3i tj j 9 14. r t vt st at t2 i 2t i 4t 2i 2 tj j 1 2t 3 2k 3 tk 9t 4t 2 9t 1 3 k 2t 1 k 2 et cos t, et sin t, et et cos t e2t cos t et 3 et sin t i sin t 2 16. r t vt st 18. (a) et sin t e2t cos t et cos t j sin t 2 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 et k e2t at 2et sin t i 2et cos t j et k x (b) r 3 3 t 4 3 0.1 , 4 4 3 0.1 4 0.1, 4 3.100, 3.925, 3.925 20. a t vt v0 rt r0 r2 C 4i C 2i 2i 3k 3k dt 4j ⇒ v t 2ti 4j 2ti 2ti 3tk 4j t2i 4tj C 3tk 4tj 32 tk 2 32 tk 2 C 22. a t vt v0 vt rt j cos t i cos t i C sin t i sin t i cos t i r0 rt r2 24. (a) The speed is increasing. (b) The speed is decreasing. 26. r t 900 cos 45 t i 450 2t i 3 3 900 sin 45 t 16t 2 j 32t 50,649 8 0, which implies that t 225 2 16. 16t 2 j i C j sin t j, v 0 sin t j dt k⇒C cos t j cos t j tk 0 tk 2k k k dt C k j k, r 0 sin t i i cos t j C 3tk dt t2i 0 ⇒ rt 8j 6k sin t j i⇒C sin t j cos t i cos 2 i sin 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 6331.125 feet. 16t 2 0 which implies that The range is determined by setting y t t Range: x 450 2 32 450 2 450 2 32 405,192 450 2t 39.779 seconds. 405,192 25,315.500 feet 280 Chapter 11 Vector-Valued Functions 28. 50 mph rt 220 ft/sec 3 220 cos 15 t i 3 5 220 sin 15 t 3 16t 2 j 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 16 27 22 cos 15 2 3.286 feet. 30. y x 0.005x 2 is the coefficient of x. Therefore, tan 1, 4 rad 45 . Also From Exercise 34 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 32. Wind: 8 mph rt 50 176 ft sec 15 176 i 15 2.5 140 sin 22 t 16t2 j 140 cos 22 t 0 0 450 When x 375, t 2.98 and y 16.7 feet. Thus, the ball clears the 10-foot fence. Section 11.3 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 7 v0 sin 35 90 v0 cos 35 16 90 v0 cos 35 2 Velocity and Acceleration 281 (a) v0 cos 35 t 90 when 7 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 (c) x t t 90 ⇒ v0 cos 35 t 90 54.088 cos 35 90 3 90 tan 35 (b) The maximum height occurs when yt t v0 sin 35 v0 sin 35 32 32t 0. 2.0 seconds 0.969 second 22.0 feet. 0, v0 16t2 j 30,000 At this time, the height is y 0.969 36. Place the origin directly below the plane. Then rt v0 cos 792 t i vt 792i ti 30,000 16t2 j 0⇒ 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 t2 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 38. From Exercise 37, the range is x Hence, x v02 sin 2 . 32 150 v02 sin 24 32 i ⇒ v02 4800 ⇒ v0 sin 24 108.6 ft sec. 40. (a) r t t v0 cos 16t x 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 rad. v02 2 cos 2 32 0 Minimum height when sin 1, or 282 42. r t Chapter 11 v0 cos ti Vector-Valued Functions h v0 sin v0 sin 8 t 50 ⇒ t 4.9 t 4.9t 2 j 4.9t 2 j 50 . For this value of t, y v0 cos 8 2 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 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 46. (a) Speed v 110 mph b2 b2 2 2 sin2 t sin2 t b2 2 cos2 t b (b) − 10 10 cos2 t 10 − 10 The graphing utility draws the circle faster for greater values of . 48. a t 50. v t vt b at F mb b 2 2 b 2 cos t i sin t j b 2 30 mph 44 ft sec 605 44 rad sec 300 3000 n θ 44 3000 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 . 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 S ection 11.4 54. r t yt rt vt st Thus, x t xt at 0 x ti mx t j 0. xti mxt xti x ti xt ytj b, m and b are constants. mxt mx t j 2 Tangent Vectors and Normal Vectors xti r1 2t 2r1 2t 4r1 2t r1 t , then: r1 t 2 283 56. r1 t r2 t bj ytj ztk Velocity: r2 t C, C is a constant. In general, if r3 t Velocity: r3 t Acceleration: r2 t 2 mx t C 1 m2 Acceleration: r3 t r1 t Section 11.4 2. rt rt rt Tt T1 t3i 3t2i 9t rt rt 1 9 16 4 Tangent Vectors and Normal Vectors 2t2j 4tj 16t 2 4. rt rt rt 6 cos ti 6 sin ti 36 sin t rt rt 2 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 1 9t4 3i 16t2 4j 3t2i 3 i 5 4tj 4 j 5 T Tt 3 3 3i 36 3 4 6. r t rt When t T1 t2i 2ti tj j 4 k 3 8. r t rt t, t, 1, 1, 4 4 t2 t t2 1, 1, 1 ,t 3 1 3 1 3 t 3 1, 1 at 1, 1, 3. 1, r t r1 r1 2i r1 j 5 2, b 2t 2i j t j 0 t 4 1 at 1, 1, 3 . When t T1 1, r 1 r1 r1 5 2i 5 1, c 1, y 21 1, 1, 7 1, b t 1, c 1, y 1 t 3 Direction numbers: a Parametric equations: x 1, z 4 3 Direction numbers: a Parametric equations: x z 10. r t rt When t T 2 sin t, 2 cos t, 4 sin2 t 2 cos t, ,r r r 2 sin t, 8 sin t cos t 3, 6 6 1 4 3, b 3t 1, 2 3 , 3, 1, 2 3 1, c 1, y 23 t 3, z 2 3t 1 t at 1, 3, 1 . 6 6 6 6 Direction numbers: a Parametric equations: x 284 12. r t rt Chapter 11 3 cos t i 3 sin t i ,r r r Vector-Valued Functions 4 sin t j 4 cos t j 3i 2 2 1 k 2 5 z 1 k 2 1 k, 2 2 37 3i 0, c 4, z t 2 1 k 2 1 t 4 at 0, 4, 1 37 4 . x 54 4 3 2 When t T 1 1 4 5 y 2 2 2 6i k Direction numbers: a Parametric equations: x 6, b 6t, y 14. r t rt r0 e ti e i T0 ti 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 2j, r 0 r0 r0 5 Parametric equations: x s r t0 0.1 r0 0.1 2, z s 2s 0.1, 2, 2 0.1 0.9, 2, 0.2 16. r 0 u0 0, 1, 0 0, 1, 0 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 0⇒ cos s, 1, 0, 1 sin s cos s cos s, 1 cos 2s 2 1 2 2 18. rt rt Tt ti i 6 j, t t 6 j t2 3 20. 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 rt rt t t4 2 1 1 36 72t 36 i i 36 t4 6 j t2 12t3 t4 36 2j i 6 j t2 2 cos t j 4 cos2 t The unit normal vector is perpendicular to this vector and points toward the z-axis: Nt 2 cos ti sin tj . sin2 t 4 cos2 t Tt N2 t4 32 32 j T2 T2 1 3i 13 S ection 11.4 22. r t vt at Tt Tt Nt 4t i 4i O vt vt O Tt Tt is undefined. 1 2i 5 j 2t j 2j 24. r t vt at Tt Tt Nt t2j 2t j 2j Tangent Vectors and Normal Vectors k 285 vt vt O Tt Tt 2t j 2t j is undefined. The path is a line and the speed is constant. 26. rt vt at Tt T1 t2i 2ti 2tj, t 2j, v 1 2i 1 4t2 j 4 2 i 2 1 Nt Tt Tt 1 t2 N1 aT aN 30. r t0 v t0 a t0 T 2 The path is a line and the speed is variable. 28. r t a cos t i a sin t i bj a a v0 v0 2 1 2i 2j b sin t j b cos t j vt v0 2i, a 1 vt vt 1 i 2 2ti 2j 2 j 2 t2 1 1 t 1 1 t2 1 ti j at a0 T0 cos t i b 2 sin t j 2i j t2 1 3 2i 3 2j Motion along r t is counterclockwise. Therefore, N0 aT aN a a i. T N 0 a 2 t2 1 i 2 j 2 2 2 sin t0 i cos t0 i 1 tj 2 i 2 a a t0 1 T N 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 32. 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 T N a x 286 Chapter 11 Vector-Valued Functions 34. If the angular velocity 2 is halved, 2 aN a a 4 2 . aN is changed by a factor of 1 . 4 36. rt x rt Tt Nt r T N 4 4 4 2 cos t i 2 cos t, y 2 sin t i 1 2 2 sin t j, t0 4 y2 4 38. r t vt at sin t i y 4ti 4i 0 v v T T 4t j 4j 2t k 2k 2 sin t ⇒ x 2 2 cos t j 2 cos t j 2 sin t i cos t i 2i cos t j Tt Nt 1 2i 3 2j k sin t j 2j 1 is undefined. T N ( 2, 2 ) x aT, aN are not defined. 2 2 2 2 i i j j −1 −1 1 40. rt vt v0 at a0 Tt T0 Nt N0 aT aN et sin t i et cos t i j k et cos tj et sin t i et k et sin t et cos t j et k 42. rt vt v2 ti i i 6j v v 3t 2j 6tj 12j k t2 k 2 tk 2k 2et cos ti 2i v v 1 i 3 1 2 2 i 2 a a T N k 2et sin tj et k at Tt 1 1 12j 37t 2 1 cos t 3 j sin t 2 j 2 3 2 k i 6t j tk sin t i sin t cos t j k T2 1 i 149 T T 1 37 1 2k 1 37t 2 37t i 37 372 6j k k 6j k cos t i cos t sin t j Nt 1 32 1 37t 2 74 i 6j 37t i 6j N2 1 37 149 1 5513 74i k aT aN z 4 2 a a T N 74 149 37 5513 37 149 N x 4 2 4 6 8 T y S ection 11.4 44. The unit tangent vector points in the direction of motion. 48. (a) r t vt at Tt aT aN When t (b) Since aT t2j 2t j 1 1 i j 1 i 6 1 63 1 4t 2 4t 2 1 k 3 2j 3i k 3k i B1 T1 N1 6 6 2 2 j 6 3 i k 6 6 2 2 3 i 3 3 j 3 3 k 3 3 i 3 k 2 Tangent Vectors and Normal Vectors 0, then the speed is constant. 287 46. If aT cos t sin t cos t t sin t, sin t sin t 3t 2t 2 t cos t cos t, cos t 3t cos t 2t sin t 2t cos t, 2t sin t sin t, sin t cos t vt vt a T a 2 cos t, sin t cos t aT2 2, 2 cos t 4 3t 2t 2 sin t 4 sin t 3t 2, 2 sin t 3t cos t 2 1 3. 1, aT 2 aN When t 2, aT aN 2 3. > 0 for all values of t, the speed is increasing when t t3 k, t0 3 t2k 1 and t 2. 50. rt rt Tt Nt r1 T1 N1 ti i 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 j k 20 2 52. (a) r t v0 cos ti h 5 50t v0 sin t 12 gt j 2 16t 2 j (b) 60 100 cos 30 t i 50 3 t i 5 100 sin 30 t 16t 2 j −20 −10 300 Maximum height Range (c) Speed vt vt 4 —CONTINUED— 50 3 i 2500 3 64t 2 50 50 200t 32t j 32t 2 44.0625 279.0325 (d) t 32 j 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 625a t 288 Chapter 11 Vector-Valued Functions 52. —CONTINUED— (e) Tt Nt aT aN aTT aNN 25 3 i 2 64t 2 25 16t j 200t 625 (f ) 0 50 25 16t i 25 3 2 64t 2 200t 625 a a T N 32j 880 ft sec 16t2 32tj 36,000 j 56. 16 16t 25 64t2 200t 625 400 3 64t2 200t 625 3 −50 The speed is increasing when aT and aN have opposite signs. 54. 600 mph rt vt at Tt rt vt vt at r cos t i r sin t i r r r 2 r sin t j r cos t j v r 2 880ti 880i 32j 1 2 r cos t i sin t j 880i 32tj 16 4t2 3025 55i 4t2 2tj 3025 at (a) F Motion along r is clockwise, therefore Nt aT aN a a 2ti 4t2 T N 55j 3025 64t 4t2 2 mat mr 2 m2 r r 2 mv 2 r (b) By Newton’s Law: mv2 r 3025 GMm 2 ,v r2 GM ,v r GM r 1760 4t 3025 58. v 9.56 104 4200 4.77 mi sec 60. 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 104 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 62. 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 64. a 2 a a aNN 2 aTT aT2 T aT2 aTT N aNN aN2 N 2 bj 2aTaNT aN2 aT2 a 2 mx t 1 2 xt 1 m2 aN 2 a 2 ±i mj , constant m2 Since aN > 0, we have aN aT2. Hence, T t 0. Section 11.5 Arc Length and Curvature 289 Section 11.5 2. r t dx dt s 0 Arc Length and Curvature y ti 1, 4 t 2k dy dt 1 0, dz dt 16 (4, 16) 2t 12 8 4 4t 2 dt 4 1 2t 1 4 1 8 65 4 4. r t dx dt 2 4t 2 ln 8 x ln 2t 65 1 4t 0 (0, 0) 1 2 3 4 16.819 a cos t i a sin t, 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 6. (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 290 Chapter 11 Vector-Valued Functions 8. r t dx dt s 0 3t, 2 cos t, 2 sin t 3, dy dt 2 10. r t 2 cos t 2 cos t t cos t, 2 t sin t, sin t dy dt t sin t, 2 t cos t, t2 2t 2 2 sin t, 32 dz dt dx dt s 0 dz dt 2 sin t 2 cos t 2 dt 2 t cos t 2 t sin t t2 2 2t 2 dt 5 8 2 2 13 dt 0 z 5 4 3 2 1 2 13t 0 13 2 5t 2 dt 0 z 3 2 2 0 5 (π2 , 1, π4 ) 2 ( 32π , 0, 2) 2 3 5 x (0, 2, 0) 3 4 5 y (1, 0, 0) 1 2 3 x 2 3 y 12. r t dx dt sin t i cos t, 2 cos t j dy dt 2 t3k sin t, dz dt 2 14. r t 3t 2 3t 2 2 dt (b) 6 cos 6i 2j t i 4 2 sin 6, 0, 0 t j 4 t k, 0 ≤ t ≤ 2 (a) r 0 r2 2k 62 6, 0, 0 0, 2, 2 22 22 44 2 11 6.633 s 0 2 2 0 cos t sin t 11.15 distance r0 r 0.5 r 1.0 r 1.5 r 2.0 9t 4 dt 5.543, 0.765, 0.5 4.243, 1.414, 1.0 2.296, 1.848, 1.5 0, 2, 2 (c) Increase the number of line segments. (d) Using a graphing utility, you obtain 2 s 0 r t dt 7.0105. 16. r t (a) s 4 sin t t t cos t , 4 cos t 2 t sin t , zu 2 2 32 t 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 —CONTINUED— Section 11.5 16. —CONTINUED— (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 6.956 2s 5 2s 5 2 Arc Length and Curvature 291 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 14.169 z 6.956, 14.169, 1.342 (d) r s 4 sin 5 si and j rs 1 2s 5 2 2.291, 6.029, 2.400 4 cos 5 2s 5 2 3 5 2 16 25 9 25 1 18. r s rs Ts Ts 3 i rs 0⇒K 2s 5 Ts 0 2s cos 5 2s i 5 2s i 5 5 2s (The curve is a line.) 20. rs Ts Ts K 4 sin rs 4 25 Ts 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 4 sin 5 5 cos 2s 4 25 k 4 cos 5 4 25 2 10s 25s 22. rt vt Tt Tt K t 2j 2t j j 0 Tt rt 0 292 Chapter 11 t2j 2t j 2j Vector-Valued Functions 24. rt vt v1 at a1 Tt Nt N1 K ti i i 2j 2j i 26. rt rt rt Tt 2 cos t i 2 sin t i sin t j cos t j cos2 t cos tj cos2 t 4 sin2 t 2 sin t i 4 sin2 t 2t j 1 4t 2 1 1 4t 2 2i 2t i j j Tt 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 1 5 a v N 2 2 55 b sin ti tj tj 30. r t at sin t , a 1 cos t 28. rt rt Tt Tt a cos t i a sin b cos From Exercise 22, Section 11.4, 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 sin2 a t b2 cos2 t ab b2 cos2 t 32 K 2 2 1 1 cos t cos t 4a 1 2 cos t 32. rt rt Tt Tt K 4t i 4i 1 2i 3 0 Tt rt 4t j 4j 2j 2t k 2k k 34. rt rt Tt Tt 2t2 i 4ti 4ti 1 4i 1 Tt rt j j tj 12 tk 2 tk tk 17t2 k 32 0 K 17tj 17t2 289t2 17 1 1 17t2 3 2 1 17 17t2 32 17t2 12 36. 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 11.5 38. y mx b 0, K 4 ,x x 4 ,y 1 x2 8 8 4 8 53 2 0, and the radius of curvature is undefined. 3 4 Arc Length and Curvature 293 Since y 40. y y y K 1 K 2x 2 1 2 42. y y y At x 16 x2 9x 16y 9 0: 16y 16y y y 2 8 ,y 1 x3 1 53 2 8 y y 0 3 16 1 16 3 3 16 02 3 2 232 1 32 (radius of curvature) K 1 K 4x2 x2 x2 3 24x 3 2 3 16 (radius of curvature) 44. (a) y y y At x 46. y y y1 K ln x, 1 , x 1, x y y1 1 12 1 1 x2 1 1 ,r 23 2 1 K 23 2 72 1 x2 0: y y x2 33 0 72 27 1 1 K 8 3 83 02 3 8 32 1 32 22 1. x 1 The slope of the tangent line at 1, 0 is y 1 The slope of the normal line is 8 3 Equation of normal line: y 1. x 1 K r Center: 0, 3 8 y 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 0 x 4x 2x 3x y 1 2 2 22 8 8 0 0 3 or x 1 3 and y 2 2 2x2 3 8 2 2 3 1 x Equation: x2 9 64 2 x2 2x (b) The circles have different radii since the curvature is different and r 1 . K Since the circle is below the curve, x Center of circle: 3, Equation of circle: x y 2. 2 3 2 y 8 2 (1, 0) x 2 −2 −4 4 6 294 Chapter 11 13 x, 3 x2, 1, K 1 Vector-Valued Functions 48. y y y1 x y1 y1 2 1 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 1. x 1 or y The center of the circle is on the normal line 1 1 x x 2 2 2 units away from the point 1, 1 . 3 1 3 y 1 1 2 2 2 1 0 or x 2 0 and y 2 52. y x3, y K 1 3x2, y 6x 9x 4 32 −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 50. 4 3 y 4 3. y 42 3 6x B A −4 −2 2 4 x (a) K is maximum at (b) lim K x→ 4 1 , 45 4 1 , 453 4 1 , 45 4 1 . 453 −3 −4 0 54. y ln x, y K dK dx 1 ,y x 1 x2 1 x2 2x2 x2 1 1 52 32 1 x2 x x2 1 32 56. y y y K 1 . 2 cos x sin x cos x 1 y y 232 1 1 cos x sin2 x 32 0 for x 2 K. (a) K has a maximum when x (b) lim K x→ Curvature is 0 at 2 K ,0 . 0 58. y y y K cosh x ex 2 ex 2 1 e x ex 2 x e x 60. See page 828. e sinh x cosh x cosh x cosh2 x 3 1 cosh2 x 1 y2 cosh x sinh x 232 2 S ection 11.5 y y Arc Length and Curvature 295 62. K 1 232 At the smooth relative extremum y 0, so K y . Yes, for example, y 0, 0 . The curvature is positive for any other point of the curvature. x x 2 x 4 has a curvature of 0 at its relative minimum 64. 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 66. y (a) 1 85 x ,0≤x≤5 4 5 z 4 2 (b) V 0 2x 5 18 x 4 5 dx (shells) 5 0 2 2 4 x 2 4 y x13 5 dx 0 5 x18 5 2 18 5 5 18 5 36 143.25 cm3 (rotated about y-axis) (c) y 2 35 x ,y 5 K 1 1 6 25 6 x 25 25x2 5 6 25x2 5 4 65 32 x 25x 2 25 6 5 (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 . 46 x 25 32 5 1 0 0 5 296 Chapter 11 c K 13 x 3 x2 2x 1 2x x4 1: s 32 Vector-Valued Functions 68. s y y y K When x K c 1 30 1 2 4 2c 30 2 2 4 2c ⇒ c 4 At x 3 ,K 2 s [1 3 2 3 81 16 c K r sin j 32 0.201 4 30 2 K f cos i 56.27 mi hr. 70. r x y x y x y K r cos i f f f f f f xy x 2 f sin j cos sin sin cos cos sin yx y2 f f f 32 f cos sin sin cos f 2 f f f f2 sin cos f f f f 2f 232 cos sin 2 f f r2 r2 74. r cos sin rr r e e e 2r r 2f 2f 2r 232 2 sin cos f f cos sin 72. r r r K 1 0 2r r 2 2 r r rr r2 r2 32 2 1 2 232 2 2 K rr r2 r2 32 2e2 2e2 3 2 1 2e 76. At the pole, r K 2r r 2r r 2 3 2 2 0. rr r2 2 r r2 32 78. r r 6 cos 3 18 sin 3 At the pole, 6 and K 2 r 6 2 18 1 . 9 ,r 6 18, S ection 11.5 80. x t yt K t3, x t 12 t,y t 2 3t 2, x t t, y t 6t 1 −4 0 4 5 Arc Length and Curvature 297 3t 2 1 t 6t 3t 2 2 t232 3t 2 3 3 9t 2 t 132 t 9t 2 1 32 K → 0 as t → ± 82. (a) r t vt ds dt K aT aN 3t 2i 6t i vt 2 31 d 2s dt 2 ds K dt t2 6t 2 2 3t 3 t3 j 3t 2 j (b) r t vt 6t ds dt d 2s dt2 at ti i vt t2j 2tj 12 tk 2 tk 5t2 1 31 d 2s t, 2 dt 2 5t 5t2 2j rt k 1 2 31 t2 2 91 t2 2 6 rt vt at ijk 1 2t t 021 5 1 32 j 2k K aT aN 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 rt rt rt3 d 2s dt2 K ds dt 2 5t2 1 5 1 32 5t 5t2 5t2 5t2 1 5 5t2 1 84. (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 298 Chapter 11 Vector-Valued Functions GmM r r3 GM r r3 r is parallel to r, r r 0. Also, 86. 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 a constant vector which we will denote by L. 88. dr dt GM L r r 1 r GM 1 0 GM r r3 1 r r3 r r 0 r GM r r3 r r L r 1 r r3 r r 1 r r3 r r r r r r 1 r r3 r r 0 Thus, r GM L r is a constant vector which we will denote by e. r 90. L Let: r r r r r cos i r sin i i sin j cos j d dt dr dt j r sin r cos d dt r r2 dr d d dt k 0 0 92. 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 1 LP 2 2 ab L 4 4 4 2a 2 Then: r r r cos r sin d dt r2 d k and L dt r d . dt P2 L L 2 2 a2 ed a c2 4 4 2a 2 a2 1 L2 e2 2a 4 2 2ed L 2 a3 423 a GM Ka 3 L 2 GM 3 a L2 Review Exercises for Chapter 11 2. r t ti 1 t 4 j k 4. r t 2t 1i t 2j , tk (a) Domain: 0, 4 and 4, (b) Continuous except at t 4 (a) Domain: (b) Continuous for all t Review Exercises for Chapter 11 6. (a) r 0 (b) r 2 3i 2 j k i 3 cos 3 cos t t t 1 t 1 y 299 (c) r s (d) r t 3 cos s r 1 sin s ti 3i 1 j sin s k tj tk tk 3i j k sin t 8. r t xt y ti t, y t j 4 10. r t t 1 −2 −1 −2 3 2 1 x 1 2 3 4 2t i 1 2 x, tj t, z y2 1 2 1 1 t 2k t 2, 5 z x y t x y z 2t, y z 0 0 0 0 x x 1 2 1 1 2 4 2 4 4 x 3 y 12. r t x x2 t x y z 2 cos t i 2 cos t, y z2 0 2 0 0 4 tj t, z 2 sin t k 2 sin t 2 x 3 z 14. r t 1 2 ti z 6 5 4 tj 13 4t k 2 0 2 2 0 2 3 2 0 3 2 2 2π y −3 −2 1 −1 2 3 2 1 −1 3 1 2 −2 3 −3 x y 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 18. The x- and y-components are 2 cos t and 2 sin t. At t 3 , 2 0≤t≤4 the staircase has made 3 of a revolution and is 2 meters 4 high. Thus, one answer is rt 2 cos t i 2 sin t j 4 t k. 3 2 cos 2t i 1 j k 20. x2 x z2 t, y rt rt 4, x t, z ti ti tj tj y ± 0, t 4 4 4 t2 x z 22. lim t→0 sin 2t i t e tj et k lim t→0 j k 2i t2k t2k 4 x 5 y 3 300 Chapter 11 Vector-Valued Functions 1 k t (b) r t (d) u t 0 Dt u t 2r t cos t i sin t j sin t i 2r t cos t j sin t i cos t j 1 t 2t k 1 t2 2k 24. r t sin t i cos tj t k, u t k sin t i cos t j (a) r t (c) r t Dt r t cos t i ut ut 2 sin t j (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 26. The graph of u is parallel to the yz-plane. t2 4 28. ln t i t ln tj k dt t ln t ti 1 2 ln t j tk 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 32. r t r0 rt C sec t i 3k tan t j t 2k dt ln sec t tan t i ln cos t j t3 k 3 C ln sec t tan t i ln cos t j t3 3 3k 1 34. 0 tj t sin t k dt 2 32 tj 3 1 sin t t cos t k 0 2 j 3 sin 1 cos 1 k 1 36. 1 t3i arcsin tj t2k dt t4 i 4 2 k 3 t arcsin t 1 t2 j t3 k 3 1 1 38. rt rt vt rt at t, vt 1 0, tan t, et 1, sec t 2 sec t 2 4 40. sec2 t, et e 2t t 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 2 direction numbers 3, tan t, e 3, 0, 0 , the parametric equations are x 2t. 0.1 r 0.1 3, 0.1, 0.2 Review Exercises for Chapter 11 v02 sin cos 16 6 2 13 4 2 13 3v02 104 2 13 6 301 42. Range 4 v02 16 416 3 v02 ⇒ v0 11.776 ft sec 4 44. r t (a) r t 20 v0 cos ti v0 sin t 1 2 9.8 t2 j 4.9t 2 j (b) r t 20 20 cos 30 t i 20 sin 30 t 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 2 j 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) 2 t 1 14 12 3j 2j 46. rt vt v at Tt 1 4i 5 0 1 4i 5 4t i 3j 2 3t j 48. rt vt vt 2t 2i 2 t t 4 t t i t t t t 1 1 1 1 1i 2 t 1 j 1 3j at Tt Nt a a T N N t does not exist a a T 0 1 2i t 14 t 4 j 1 N does not exist 1 2j 1 1 4 3 t 1 t 4 t 1 2 4 1 1 1 4t 3 1 1 4 4 302 50. Chapter 11 rt vt vt at Tt Nt at at Tt Nt t 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 2 1 2 1 1 k t t 2 52. rt vt vt at Tt Nt a a T N i t 1i j 2t 4 t2 tj 1 k t2 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 When t rt 2, y 2t j 2t 2k 2, a 4, z 8t 1, b 16 3 Direction numbers when t x t 2, y 4t 4, c 8 2 k t3 t 2i i t2j k 2t 4 1 2t 2k 2 2t 4 1 j 1 1 2 t 3 2t 4 4 t 2 2t 4 56. Factor of 4 t2i 2t i b 58. r t rt s 2tk, 0 ≤ t ≤ 3 2k 3 6 5 4 z r t dt a 0 4t2 1 t t t2 4 dt 3 3 2 ln ln t2 10 1 0 −2 (0, 0, 0) 1 2 1 2 y (9, 0, 6) 3 4 5 6 7 9 x 3 3 10 11.3053 Review Exercises for Chapter 11 60. rt rt rt s 0 y z 8 6 4 2 − 8 − 6 − 4 −2 −4 −6 −8 2 x x 2468 2 1 1 1 4 3 303 10 cos t i 10 sin t i 10 2 10 sin tj 10 cos tj 62. r t rt s ti i b t2j 2tj 2tk, 0 ≤ t ≤ 2 2k, r t 2 5 5 4t2 dt 105 4t2 r t dt a 0 10 dt 20 21 5 ln 5 4 5 ln 4 45 6.2638 (2, 4, 4) 2 3 4 y 64. r t xt s 2 sin t t cos t , 2 cos t t sin t , t , 0 ≤ t ≤ 4t2 1 dt 4.6468 1 2 66. rt rt rt et sin t i et cos t et cos t 2et et cos t k, 0 ≤ t ≤ et sin t i et sin t 2 2t sin t, 2t cos t, 1 , r t b 2 et sin t et sin t et cos t k et cos t 2 r t dt a 0 4t2 17 4 17 1 ln 4 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 1 t 9 1 t 9t 1 t 2 i 1 t 1 t 2 j 3 32 k 0 0 3 2t3 2 1 9t 3 2 t3 5 sin tk 5 cos tk, r t 29 3 2 r r 3 t 2 3 2k; r r 3 2t 3 2 0 K rt rt rt3 2ti 2i 5 cos t j 5 sin tj 21 9t 32 70. rt rt rt r r r r K 5 cos tj i 2 0 725 r r 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 ...
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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.

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