ODD11 - 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 . . . . . . . . . . . . . . . . . . . . . . 39 Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . . 44 Section 11.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . 48 Section 11.4 Tangent Vectors and Normal Vectors Section 11.5 Arc Length and Curvature Review Exercises Problem Solving . . . . . . . . . . . . . . . 54 . . . . . . . . . . . . . . . . . . . . .60 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73 C H A P T E R 11 Vector-Valued Functions Section 11.1 Vector-Valued Functions Solutions to Odd-Numbered Exercises 1. r t 5t i 4t j 1 k t 5t 4t 1 t Domain: 0, 3. r t ln t i et j tk ln t et t Component functions: f t gt ht Domain: ,0 0, Component functions: f t gt ht 5. r t Ft Gt cos t i sin t j tk cos t i sin t j 2 cos t i tk Domain: 0, i sin t 0 j cos t sin t k 0 cos t 7. r t Ft Gt cos2 t i sin t cos t j sin2 t k Domain: , 9. r t 12 2t i 1 2i t 1j (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 tj 1 2i 1j tj sj 2i 2i j j 2 2t 1 j t ln 2 i 2t 1 2 t2i ti 2 11. r t ln t 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 39 40 Chapter 11 Vector-Valued Functions 13. rt rt sin 3t i sin 3t cos 3t j 2 tk 2 15. r t t2 1 t2 ut 3t 3t 3 1 t2 t2 2t 3 13 4t 8 5t 3 4 t3 t 2, a scalar. cos 3t 4t 3 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 19. r t x ti t, y t 2j t 2, z e0.75t k, e0.75t 2≤t≤2 Thus, z x 2. Matches (b) 20, 0, 0 Thus, y x 2. Matches (d) 21. (a) View from the negative x-axis: (c) View from the z-axis: 0, 0, 20 23. x y y y (b) View from above the first octant: 10, 20, 10 (d) View from the positive x-axis: 20, 0, 0 3t t x 3 1 1 25. x y t3, y x2 3 y 7 6 5 4 3 2 t2 27. x x2 cos , y y2 9 y 3 sin 1 Ellipse 2 1 x 12345 2 −5 −4 −3 −2 −1 x −3 −2 x 2 3 4 −2 −4 6 −2 −3 29. x x2 9 3 sec , y y2 4 12 9 6 3 2 tan 31. x y z 4t 2t t 1 2 3 33. x x2 4 z 2 cos t, y y2 4 t 1 2 sin t, z t 1 Hyperbola y Line passing through the points: 0, 6, 5 , 1, 2, 3 z x Circular helix z 7 −12 − 9 − 6 5 4 3 −3 −6 −9 −12 6 9 12 (0, 6, 5) (2, − 2, 1) 1 3 x (1, 2, 3) 3 45 6 y −3 3 3 y x 35. x x2 z 2 sin t, y y2 e t 2 cos t, z e t z 6 37. x y t x t, y x 2, z 2 2 4 16 3 t 2, z 23 3x 23 3t 6 4 2 z 4 (2, 4, 16 ) 3 1 1 1 2 3 0 0 0 0 1 1 1 2 3 2 2 4 16 3 2 x −2 −4 −6 5 y −3 3 x 3 y y z (− 2, 4, − 16 ) 3 Section 11.1 12 ti 2 32 tk 2 z −3 −2 2 3 x −2 −3 −4 −5 y x 2 1 −2 −3 Vector-Valued Functions 1 tj 2 1 cos t 2 3 k 2 41 39. r t Parabola tj 41. r t Helix sin t i 3 cos t 2 z 2 −1 1 −1 −2 2 3 4 y 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. 45. y 4 x t, then y ti 4 4 tj t. The radius of the helix is increased from 2 to 6. 47. y Let x rt x2 16 Let x rt z 8 7 6 5 4 3 2 1 4 3 2 1 x 2 2 Let x rt t, then y ti t t 2 2. 2 2j 49. x2 Let x y2 25 5 cos t, then y 5 cos t i 5 sin t. 51. y2 4 1 4 sec t, y 4 sec t i 2 tan t. 2 tan t j rt 5 sin t j 53. The parametric equations for the line are x 2 2t, y 3 5t, z 8t. (0, 8, 8) One possible answer is rt 2 2t i 3 5t j 8t k. x (2, 3, 0) 45 67 8 y 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 (Other answers possible) 42 Chapter 11 ti 2 4 t 2j, Vector-Valued Functions 0≤t≤2 y x2 59. z x2 y 2, x y 0 x t, z 2t2k (− 2, 4) 57. r1 t r2 t r3 t t i, 0 ≤ t ≤ 2 t j, 0 ≤ t ≤ 4 Let x t, then y Therefore, x rt ( 2, − t and z 2t 2. x2 y2 2t 2. (Other answers possible) t, y ti 2, 4 ) 5 tj z 2, −3 2 3 x 1 2 3 y 61. x2 x z t x y z rt 63. x2 Let x y2 4, z x2 2 cos t 4 z 2 sin t, y x2 0 0 2 0 2 sin t i y2 1 z2 1 4 sin2 t 6 1 3 4 2 2 2 2 cos tj 4, x z 2 2 0 4 4 sin2 t k 2 2 1 x sin t 2 3 4 2 2 2 0 2 0 −3 3 x 3 y z sin t, then z 2 1 2 sin t and 2 sin2 t x2 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 65. x2 z2 4, y 2 z2 4 y2 0 or y ± x. 3 z Subtracting, we have x2 (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) Section 11.1 t2 t2 4 j 2t Vector-Valued Functions 1 k t 1 k 2 43 67. y2 z2 2t cos t 2 z 2t sin t 2 4t2 4x2 69. lim t i t→2 2i 2j 16 12 8 4 5 4 since lim 8 t2 t→2 t 2 4 2t lim t→2 2t 2t 2 2. (L’Hôpital’s Rule) 7 x 6 12 16 y 71. lim t 2 i t→0 3t j 1 cos t k t 0 73. lim t→0 1 i t cos t j sin t k 1 does not exist. t since lim t→0 does not exist since lim 1 cos t t 1 j t , 0 , 0, lim t→0 sin t 1 t→0 0. (L’Hôpital’s Rule) 75. r t ti 77. r t ti arcsin t j 1, 1 t 1k Continuous on 79. r t e t, t 2, tan t Continuous on 81. See the definition on page 786. 2 n n, n Discontinuous at t Continuous on 2 3j 2k 2i 5j 2 83. r t (a) s t (b) s t (c) s t 85. Let r t t2i rt rt rt t tk t2i t2 t2i t 2i t 3j t 2j t 3j tk x2 t i y2 t j z2 t k. Then: x2 t z1 t j t→c t→c 3k tk x1 t ut y1 t j lim t→c t→c z1 t k and u t y1 t z2 t t→c lim r t t→c 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 x t i y t j z t k. Since r is continuous at 87. Let r t t c, then lim r t rc. t→c 89. True 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 rc Therefore, r is continuous at c. 44 Chapter 11 Vector-Valued Functions Section 11.2 1. r t xt x y2 r2 rt r2 4i 2t i 4i j 2j j t 2i t j, t0 t Differentiation and Integration of Vector-Valued Functions 2 4 2 y 3. r t (4, 2) cos t i sin t j, t0 sin t y 2 r′ (0, 1) t 2, y t r′ xt x cos t, y t y2 r 2 1 j sin t i i r 2 −2 −4 4 6 8 x2 r x 1 rt r 2 cos t j r t0 is tangent to the curve. r t0 is tangent to the curve. 3 2 5. r t (a) 8 16 6 16 4 16 2 16 ti y t 2j 7. r t x2 2 cos t i y2 rt 4, z 2 sin t j t 2 sin t i 2j 2i k 3 k 2 t k, t0 2 cos t j k r2 1 r4 1 r2 1 r r r4 1 3 2 3 2 z 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 (0, − 2, 32π ) r′ 2π rπ −2 2 x 1 2 y (c) r1 4 14 3 16 j . i 3 j 4 This vector approximates r 9. r t rt 13. r t rt 6t i 6i e ti ei 12 tj 2 3t2i 6t i tj j t 7t 2j 14t j 4j t 3k 3t 2k 11. r t rt 15. r t rt a cos3 t i a sin3 t j k 3a sin2 t cos t j 3a cos2 t sin t i t sin t, t cos t, t sin t t cos t, cos t t sin t, 1 17. r t (a) r t t3i (b) r t rt 3t2 6t t 18t3 t rt Section 11.2 Differentiation and Integration of Vector-Valued Functions 12 ti 2 ti i rt 13 tk 6 j tk t1 10 12 tt 2 t t3 2 12 tk 2 45 19. r t 4 cos t i 4 sin tj 4 cos t j 4 sin t j 4 sin t 0 4 cos t 4 cos t 4 sin t 21. r t tj (a) r t rt (b) r t 4 sin t i 4 cos t i rt (a) r t rt (b) r t 23. r t (a) r t cos t t sin t, sin t sin t sin t t cos t, t t cos t, cos t cos t t sin t, 1 t cos t, t sin t, 1 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 1 4 2t k t cos t t z 25. 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 r ′′ r ′′ 2 i 2 2 2 1 4 2 2 1 k 2 2 2 2 x r′ r ′ y 2 1 2 k 2k 2 2 1 4 4 2 1 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 t 3j 3t 2j 4 4 2 2i 4k 27. r t rt r0 t 2i 2t i 0 29. r r r n 2 2 cos3 i 6 cos2 0 3 sin3 j sin i 9 sin2 cos j Smooth on , 0 , 0, Smooth on n n1 , 2 2 1 j t 2t k 0: , n any integer. 31. r r r 1 2 sin 2 cos i i 1 1 2 cos 2 sin j j 33. r t rt i t 1i 1 j t2 t 2k 0 ,0 , 0, 0 for any value of , Smooth on r is smooth for all t 46 Chapter 11 ti i 3t j 3j Vector-Valued Functions tan t k sec2 t k 2 0 n 2n 2 2 4t i n, 2 t 3k (b) r t t5 9t 2 t4 8t 3i 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 1 . n 35. r t rt r is smooth for all t Smooth on intervals of form 37. r t ti 3t j i ut ut ut ut 3j t 2 k, u t 2t k 3t 3 8t 2t 4i t 2j (a) r t (c) r t Dt r t (e) r t Dt r t 4t 2 10 2t 2 10 t 2 π 39. 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 7 sin t cos t 7 sin t cos t 16 cos2 t 9 cos2 t −1 9 sin t cos t rt rt arccos rt rt 7 0 16 sin2 t 9 sin2 t 3.927 2.356 5 4 3 4 7 sin t cos t 16 cos2 t 9 cos2 t and t and t 0.785 5.498 4 . 16 sin2 t 1.855 maximum at t 1.287 minimum at t 1.571 for t 7 . 4 2 n ,n 2 t t t 2t t 2t tj rt 2i t 0, 1, 2, 3, . . . 41. r t lim lim lim rt 3t 3 ti t→0 1 t 3i 2 t t j t 2 j 3t 2i 1 t2 j t→0 t→0 lim 3i t→0 2t j 43. 2t i j k dt t 2i tj tk C 45. 1 i t j t 3 2k dt ln t i tj 252 tk 5 C 47. 2t 1i 4t 3j 3 t k dt t2 ti t 4j 2t 3 2k C 49. sec2 t i 1 1 t2 j dt tan t i arctan t j C Section 11.2 Differentiation and Integration of Vector-Valued Functions 47 1 1 51. 0 8t i tj k dt 4t 2i 0 t2 j 2 1 1 tk 0 0 4i 1 j 2 k 2 2 2 2 53. 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 55. r t r0 rt 2i 4e2t i 3j 3 3et j dt C et 2e2t i 3et j 3j C 57. r t r0 rt rt C1 32j dt 600 3 i 32t j 600j C1 2i ⇒ C 1j 2e2ti 600 3 i 600 3 i 600 3 t i 600 600 600t 32t j 32t j dt 16t 2 j C r0 rt C 0 600t 16t 2 j 600 3 t i 59. r t r0 rt te 1 i 2 1 t 2i e tj j C t2 k dt 1 i 2 e t 1 e 2 t 2i e tj i tk 2j 2 e 2 k C j 2j k⇒C t 1k 1 e 2 t2 i i e t 2j t 1k 61. See “Definition of the Derivative of a Vector-Valued Function” and Figure 11.8 on page 794. 63. At t t0, the graph of u t is increasing in the x, y, and z directions simultaneously. cy t j cz t k and 65. Let r t xti ytj cx t i cx ti z t k. Then cr t cy t j y tj cz t k z tk cx t i Dt c r t cr t . f txti f ty t f tytj f tyt j ytj f t z t k. f tz t ztk f tzt k 67. Let r t xti ytj z t k, then f t r t f txt i y tj f trt z t k. Then r f t y f t f tj y ft j Dt f t r t f tx t f t x ti f tr t z tk f t xti 69. Let r t xti ytj xft i yft j z f t k and (Chain Rule) Dt r f t x f t f ti f t x ft i z f t f tk z ft k f tr f t . 48 Chapter 11 x1 t i rt Dt r t ut ut Vector-Valued Functions y1 t j vt vt z 1 t k, u t x1 t y2 t z3 t x1 t y2 t z 3 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 ut k. vt 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 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 ut x3 t i z2 t x3 t y3 t j z3 t k. Then: y2 t x3 t 71. Let r 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 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 y2 t x3 t y3 t z2 t y3 t z2 t y3 t z2 t rt ut y1 t y1 t y1 t vt z1 t x2 t y3 t z1 t x2 t y3 t 73. False. Let r t rt d rt dt rt rt cos t i 2 0 sin t i 1 sin tj cos t j Section 11.3 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 3. r t vt (3, 0) v x 4 6 t2 i rt rt t2, y tj 2t i 2i t, x 2. 4i 2i j y2 −2 −4 y j 4 at x (4, 2) 2 v a x 1, y x 3 1 −2 −4 2 4 6 8 At 3, 0 , t v1 At 4, 2 , t j, a 1 0 v2 a2 5. r t vt at x At v a 2 cos t i rt rt 2 cos t, y 2, 2 ,t 2 sin t j 2 sin t i 2 cos t i 2 sin t, x2 4 2i 2i . 2j 2j 2 cos t j 2 sin t j y2 4 7. r t vt at x At v a y 4 t rt rt t sin t, 1 1 cos t cos t, sin t sin t, cos t 1 . 2i 1 j cos t (cycloid) sin t, y ,2 , t 2, 0 0, 4 4 y 3 2 (π , 2) a v x v a −3 ( 2, 2) x 3 π 2π −3 Section 11.3 t2 k 2 tk t2 Velocity and Acceleration 49 9. r t vt st at ti i vt 0 2t 2j 5j 3k 1 4 3t k 11. r t vt ti i 1 2j t 2j 2t j 4t 2 k 9 14 st at 1 5t 2 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 15. r t vt st 4t, 3 cos t, 3 sin t 4, 16 0, 3 sin t, 3 cos t 9 sin2 t 9 4i cos2 t 3 sin t j 5 3 cos t j 3 sin t k 3 cos tk 1 9 32 t2 at 3 cos t, 3 sin t 9 k 17. (a) rt rt r1 x 1 t, 1, 1, t, y t 2, 2t, 2, t3 ,t 40 3t 2 4 1 (b) r 1 0.1 1 0.1, 1 2 0.1 , 1 4 3 0.1 4 1.100, 1.200, 0.325 3 4 1 2t, z 1 4 0 tj tk C ti k C vt k, rt 2j 2k t3 6 r1 rt r2 14 j 3 t3 6 17 j 3 j k 3 t 4 21. a t vt v1 1 j 2 t2 2 t2 2 tj tj t k, v 1 t k dt 1 k 2 9 j 2 9 j 2 9 tj 2 1 k 3 9 t 2 2 k 3 C 14 j 3 t3 6 C t2 2 t2 2 5j, r 1 t2 2 j t2 2 k 0 C 9 j 2 1 k 2 19. a t vt v0 rt r0 r2 i i C ti C 2i j j k, v 0 k dt ti t k dt t2 i 2 k 2i 0, r 0 ti tj 0, v t tj 0, r t j t k, v t t2 i 2 j j 5j ⇒ C 1 k 2 1 k dt 2 1 tk 2 C 14 j 3 1 k 3 1 k 3 0⇒C t3 6 1 t 2 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 50 Chapter 11 Vector-Valued Functions 12 gt j 2 3. 300 2 v0 2 27. r t v0 t 2 t v02 v0 cos ti h v0 sin v0 t 2 16t 2 t v0 ti 2 3 v0 t 2 16t 2 j 300 when 3 300 2 v0 300 2 , v0 v0 2 300 32 , v0 9600 16 0, 300 40 6 3002 32 v02 97.98 ft sec 0 40 6, v0 The maximum height is reached when the derivative of the vertical component is zero. yt yt t 3 tv0 2 16t 2 32t 53 4 53 4 3 40 3 53 4 16 53 4 2 3 0 40 6 t 2 16t 2 3 40 3 t 16t 2 40 3 40 3 32 Maximum height: y 78 feet 29. x t yt t v0 cos t v0 sin x v0 cos or t x v0 cos 16t 2 h 16 x2 v0 cos2 2 y v0 sin h tan x 16 sec2 v02 x2 h 31. r t (a) y ti 0.004t2 0.004x2 0.3667t 6 6 j, or (b) 18 0.3667x 0 0 120 (c) y 0.008x 0.3667 14.4 feet. 0⇒x 45.8375 and (d) From Exercise 29, tan 16 sec2 v02 ⇒ v0 0.3667 ⇒ 20.14 16 sec2 0.004 4000 cos2 y 45.8375 0.004 ⇒ v02 67.4 ft sec. 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 10 shows that 0 20 . 0 0 500 —CONTINUED— Section 11.3 33. —CONTINUED— (c) We want xt 440 cos 3 t ≥ 400 and yt 3 440 sin 3 30 11 cos 2 Velocity and Acceleration 51 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 14,400 sec2 121 . 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 19.38 400 tan 48,400 tan 15,247 tan tan 1 48,400 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 16 v cos 8 1 512 1 v2 cos2 1 v2 v2 We minimize f f f 512 t t 16t 2. 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 1 512 cos cos2 512 2 sin cos 512 cos2 16 1 v2 cos2 16 v cos 2 2 sin cos2 512 . 2 sin cos cos2 2 sin cos cos2 2 0 2 1.01722 58.28 2 cos2 2 sin2 2 sin cos sin 2 tan 2 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 2 512 22 16t2 28.78 feet per second. 16t2 From part (a), v 2 2 22 22 2 512 12 1024 ⇒ v 32 ft sec. 52 Chapter 11 Vector-Valued Functions 37. r t v0 sin v0 cos t ti v0 sin 0 when 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 . 39. (a) rt rt 10 , v0 65t i 66 ft sec 0 66 sin 10 t 16t 2 j 16t j 2 (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 2 j 16t 2 j (d) 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 (e) rt rt 60 , v0 66 ft sec 0 66 sin 60 t 16t 2 j 16t 2 j (f ) rt rt 60 , v0 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: 51.074 feet Range: 117.888 feet 60 Maximum height: 249.797 feet Range: 576.881 feet 300 0 0 140 0 0 600 Section 11.3 41. r t v0 cos ti h v0 sin 1.5 t 4.9t 2 j 4.9t 2 j 100 1 2 Velocity and Acceleration 53 100 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 43. r t vt at vt at (a) v t b 1.5 bt b b 2 100 sin 30 5.102 sin t i cos t i sin t i 1 b 2 4.9 5.102 2 129.1 meters. b1 cos t j b1 b 2 b sin t j cos t j cos t i b sin tj sin t i cos t j 2b 2 cos t 0 when t 0, 2 , 4 , . . . . (b) v t is maximum when then v t 2b . t ,3 ,. . ., 45. rt vt vt b sin t i b cos t j b2 sin t cos t 0 b2 sin t cos t Therefore, r t and v t are orthogonal. 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. 49. a t 1 F m 32 m 2b 2b 1 2 32 2 10 4 10 rad sec vt b 8 10 ft sec 1 2 2g t 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 v0 sin t h. v02 sin2 g 2gh . 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 . v02 54 Chapter 11 Vector-Valued Functions 53. r t vt at xti x ti x ti Speed ytj y tj y tj vt z t k Position vector z t k Velocity vector z t k Acceleration vector xt 2 yt 2 zt 2 C, C is a constant. d xt dt 2x t x t 2x tx t 2 yt 2 zt 2 0 0 0 0 2y t y t y ty t 2z t z t z tz t vt at 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 33 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 sin2 t vt 6 6 cos t i (d) The speed is increasing when the angle between v and a is in the interval 9 0, −6 2 . The speed is decreasing when the angle is in the interval 2 , . Section 11.4 1. r t rt Tt T1 t2i 2ti rt rt 1 i 2 2tj Tangent Vectors and Normal Vectors 3. 4t2 t2 2 j 2 T 4 1 1 2 t2 ti j 1 rt rt rt Tt 4 cos t i 4 sin ti 16 rt rt 2 i 2 2 cos t i 2 sin t i 0, r 0 r0 r0 sin2 t 4 sin tj 4 cos tj 16 cos2 t sin ti 2 j 2 2 sin tj 2 cos t j 2j tk k 0 at 2, 0, 0 . k 2, c 2t, z 1 t cos tj 4 2j, r t 2ti 2j 2 t2 1 j 2 i 2 4 5. r t rt When t T0 ti i t 2j 2t j 0, r 0 r0 r0 tk k i k, t 2 i 2 1, b t, y k 0, c 0, z 1 t 0 at 0, 0, 0 . 7. r t rt When t T0 k, t 5 2j 5 0, b 2, y Direction numbers: a Parametric equations: x Direction numbers: a Parametric equations: x Section 11.4 9. r t rt When t T 2 cos t, 2 sin t, 4 2 sin t, 2 cos t, 0 ,r r r 2, 4 4 1 2 2, b 2t 2, 0 , 2, t at 2, 2, 4 . Tangent Vectors and Normal Vectors 55 4 4 4 4 2, 0 2, c 2, y 0 2t 2, z 4 Direction numbers: a Parametric equations: x 2 t, t 2, t 3 3 1, 2t, 2t 2 3, r 3 r3 r3 11. 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 t0 r1 j 1 1 t, y 6, c 3, y 1 i 3 at 3, 9, 18 . y Direction numbers: a Parametric equations: x 13. r t rt ti i T1 ln t j 1 j t rt rt 1 2t t k, k i 18 6t 9, z 18 t 18 15. r 4 j 1 k 2 2 i 3 1 t 2 1.05k cos r4 r4 u8 u8 16.29167 ⇒ 16.29513 1.2 2 j 3 1 k 3 u8 2, 16, 2 2, 16, 2 Hence the curves intersect. rt us 1, 2t, 1 ,r 4 2 23 1 2k 1 14 t, z 1.1 i 1 0.1 j 1, 8, 1 2 1 1 , 2, 4 12 Tangent line: x r t0 0.1 1 1 , 2, s 4 3 ,u 8 r 1.1 1.1, 0.1, 1.05 17. rt rt Tt Tt T2 N2 ti i 12 t j, t 2 tj 2 19. rt rt 6 cos ti 6 sin ti rt rt cos t i 2 i 2 6 sin tj 6 cos tj sin ti k, t 3 4 rt rt t2 2 53 2i i 1 t 1 3 2i tj t2 1 t2 1 3 2j Tt Tt N 3 4 cos tj 1 sin t j, T t 2 j 2 1 j 53 2 1 5 2i j 25 i 5 5 j 5 T2 T2 56 Chapter 11 Vector-Valued Functions 4t 2 i 8t i 8i vt vt O Tt Tt is undefined. 8t i 8t i 21. r t vt at Tt Tt Nt 4t i 4i O vt vt O Tt Tt is undefined. 4i 4 i 23. r t vt at Tt Tt Nt The path is a line and the speed is constant. The path is a line and the speed is variable. 25. r t at Tt T1 ti 1 j, v t t 2j t2 t4 j i 1 j, v 1 t2 i j, 27. r t vt at et cos t i et cos t et 2 et sin t j sin t i et cos t sin t j 2 j, a 1 t3 vt vt 1 i 2 Tt Tt 2 sin t i ,T v v et 2 cos t j 1 2 i j 2 2 i j. 1 2 i 2 i j i 1 j t2 1 t4 1 t 2i j At t Motion along r is counterclockwise. Therefore, 2t 3 4 t 1 N 32 Nt t4 2t 1 32 j aT aN 1 2 a a T N i j 2e 2e 2 2 2 i 2 j. 2t t4 1 t4 i j 1 t 2j 1 2 i 2 N1 aT aN a a 1 i 2 T N j 2 2 29. r t0 v t0 a t0 T t0 cos t0 2 2 t0 sin t0 i 2 sin t0 t0 cos t0 j t0 cos t0 i cos t0 cos t0 i t0 sin t0 j t0 cos t0 sin t0 j t0 sin t0 i sin t0 j v v Motion along r is counterclockwise. Therefore N t0 aT aN a a sin t0 i T N 2 2 cos t0 j. t0 3t 0 Section 11.4 31. 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 1 j, t0 t 2 2 Tangent Vectors and Normal Vectors a 0. 57 a sin t j a cos t j a 2 33. Speed: v t The speed is constant since aT cos t i sin t j cos t j sin t j sin t i cos t i 35. rt x rt Tt Nt r2 T2 N2 ti t, y i t2i t4 i 2 1 37. r t vt at Tt Nt ti i 0 v v T T 2t j 2j 3t k 3k 1 ⇒ xy t 1 j t2 j 1 1 i 14 2j 3k 14 i 14 2j 3k y is undefined. t 2j t4 1 1 j 2 j 4j 3 aT, aN are not defined. N 2 2i 1 2, 2 1 2 1 T 3 x 17 4i 17 17 i 17 39. r t vt v1 at Tt T1 ti i i 2j v v t 2j 2tj 2j k t2 k 2 tk k 41. rt vt v 2 at 4t i 4i 4i 3 cos tj 3 sin tj 3j 3 sin t k 3 cos t k 3 cos t j 3k v v 1 4i 5 T T k a a T N 0 3 1 4i 5 3j 3 sin t k 1 1 2j 5t 2 k i 2tj tk a 2 Tt 6 i 6 T T 30 30 a a T N 3 sin t j 3 cos t 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 T 2 Nt cos t j sin t k T z N1 aT aN N N 3 y 2 aT aN 2π 4π 3 x 58 Chapter 11 Vector-Valued Functions 43. T t rt rt Tt Tt Nt If a t aTT t aNN t , then aT is the tangential component of acceleration and aN is the normal component of acceleration. 45. If aN 0, then the motion is in a straight line. 47. r t t sin t, 1 cos t 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 2 1 cos t 1 :a 2T 1: aT 3 :a 2T 2 2 2 > 0 ⇒ the speed in increasing. 0 ⇒ the height is maximum. 2 2 2 < 0 ⇒ the speed is decreasing. Section 11.4 49. rt rt Tt Nt r T N 2 2 2 2j 2 cos t i 2 sin t i 2 17 17 cos t i 4 k 2i 1 k 2 17 17 4i k 2 sin t j 2 cos t j 2 sin t i sin t j t k, t0 2 1 k 2 2 cos t j 1 k 2 x Tangent Vectors and Normal Vectors z 59 2 3 2 −2 −1 1 2 1 B N −2 T ( 0, 2, π ) 2 y 2 17 17 j 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 51. From Theorem 11.3 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 53. r t (a) 0; (vertical component of velocity) 0 and aN 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 55. r t 0 and aN 1000 0 because the speed is constant. a cos t i a sin t j T 0 and a N a 2 From Exercise 31, we know a (a) Let a 0 . (b) Let a0 a 2. Then N a0 2 2 . Then N a 0 2 a2 2 4a 2 a or the centripetal acceleration is increased by a factor of 4 when the velocity is doubled. a 2 2 1 a 2 2 or the centripetal acceleration is halved when the radius is halved. 60 Chapter 11 Vector-Valued Functions 9.56 104 4385 57. v 9.56 104 4100 cos i dT dt 4.83 mi sec 59. v 4.67 mi sec 61. Let T t Tt M of 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 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 63. Using a v aTT a aNN, T vT v aT T v aN T aTT T N N T O, and T N 1, we have: aNN v aN T N v a v aN T v aN Thus, aN v v a . Section 11.5 1. r t dx dt s 0 4 Arc Length and Curvature 3. r t a cos3 t i a sin3 t j dy dt 3a sin2 t cos t sin t 2 ti 1, 4 3t j dy dt 1 3, dz dt 0 dx dt s 4 0 3a cos2 t sin t, 2 9 dt dt 3a cos2 t 2 3a sin2 t cos t 2 dt 10 0 12a 0 2 sin t cos t dt 2 4 10 t 0 y 4 10 3a 0 y 2 sin 2t dt 3a cos 2t 0 6a (4, 12) 12 a 8 −a x a −a 4 (0, 0) 4 8 12 x Section 11.5 12 gt j 2 1 32 t2 j 2 Arc Length and Curvature 61 5. (a) r t v0 cos ti h v0 sin 3 50 2t 32t j 25 2 16 t 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 7. r t dx dt s 0 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 x 2 −2 y 2 2 14 dt 0 14 t 0 2 14 9. 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 11. r t dx dt s 1 t 2i 2t, 3 tj dy dt 2t 1, 2 ln t k dz dt 1 t2 t2 t2 t 1 2 a cos t, a2 cos2 t 1 t 1 t 1 dt 8.37 2 s 0 2 a2 sin2 t a2 0 z dt 3 b2 dt a2 b2 t 0 2 a2 b2 1 3 1 4t 4 4t 4 (a, 0, 2π b) dt 2π b πb x (a, 0, 0) y 13. r t ti 4 t2 j t 3k, 0≤t≤2 2, 0, 8 82 84 2 21 9.165 (a) r 0 distance 0, 4, 0 , r 2 22 42 —CONTINUED— 62 Chapter 11 Vector-Valued Functions 13. —CONTINUED— (b) r0 r 0.5 r1 r 1.5 r2 0, 4, 0 0.5, 3.75, .125 1, 3, 1 1.5, 1.75, 3.375 2, 0, 8 0.5 0.5 0.5728 2 2 distance .25 2 2 .125 2 2 .5 2 .75 2 .875 2 0.5 2 1.25 2 2.375 2 1.75 1.2562 4.625 2.7300 4.9702 9.529 (c) Increase the number of line segments. (d) Using a graphing utility, you obtain 9.57057. 15. r t (a) s 0 t 2 cos t, 2 sin t, t t xu 2 yu 2 zu 2 du (b) s 5 x t s 5 s 5 s 5 s 5 s 5 s 5 k 2 sin u 0 t 2 2 cos u t 2 1 2 du 2 cos ,y 2 sin ,z 5 du 0 5u 0 5t 2 cos 1 2 sin 1 1 1.081 1.683 rs 2 cos i 2 sin j (c) When s 5: x y z 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 2 si 2 2 j 2 rs Ts 0 (The curve is a line.) s 5 2 2 cos 5 s 5 2 1 5 2 4 5 1 5 s 5 1 17. rs rs Ts Ts 1 2 i 2 rs rs 1 and 2 sj 2 rs 1 2 1 2 1 19. r s Ts Ts K 2 cos rs 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 i 2 sin 5 0⇒K Section 11.5 1 j t 1 j t2 j Arc Length and Curvature 63 21. rt vt Tt Tt K 4t i 4i 2t j 2j 23. rt vt ti i i 2 j t3 2j t 2i t4 1 2i 5 0 Tt rt j v1 at 0 (The curve is a line.) a1 Tt Nt N1 K j 1 1 1 12 t4 1 i 2 a v N 2 i t 2j j 2 2 a sin ti cos ti a 1 a t cos t tj tj 25. rt rt Tt Tt K 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 27. rt rt Tt Tt K a cos t i a sin sin ti a cos tj sin tj 2 cos 2 t i Tt rt et cos t i e sin t 1 2 1 2 Tt rt sin t cos t 1 2 et t2 k 2 tk t cos Tt rt cos t 2 8 29. rt rt Tt Tt K et sin t j e cos t i cos t i sin t i 2 e 2 t t 31. r t e cos t cos t sin t t t sin t, sin t e sin t j sin t j cos t j t From Exercise 21, Section 11.4, we have: a N K 3t at v Nt 2 3t 4t 2 1 t 33. rt rt Tt Tt K ti i i t 2j 2t j 35. rt rt Tt Tt K 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 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 Tt rt 1 32 64 37. y Chapter 11 3x 2 Vector-Valued Functions 39. y 0, and the radius of curvature y y K 1 K 2x2 4x 4 4 1 173 4 2 3 0, K Since y is undefined. 4 232 4 173 2 0.057 17.523 (radius of curvature) 41. y y y At x a2 x a2 2x a2 0: 2 x2 x2 a2 x2 32 43. (a) Point on circle: Center: Equation: 0 1 a 1a 1 02 a 32 2 ,1 2 ,0 2 x 2 y2 1 y y K 1 K (b) The circles have different radii since the curvature is different and 1 a r 1 . K (radius of curvature) 45. y K x 1 ,y x 1 2 02 32 1 2 1 ,y x2 2 x3 47. y y y0 K ex, ex, 1, 1 x y y0 1 12 32 0 ex 1 1 23 2 1 ,r 22 1. x or y x 1 1 K 22 1. 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 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 1 2 y 5 2 2 1 4 (1, 2) −6 6 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 22 8 4 ±2 −4 x2 x2 x Since the circle is above the curve, x Center of circle: 6 2 and y 3 2 3. 2, 3 2 2 Equation of circle: x y 8 −6 0 (0, 1) 3 Section 11.5 49. π Arc Length and Curvature 2x 232 65 y 51. y x K 1 1 2 3, y 2 2x 1 1, y 1 2 2 4x 1 232 π x B A (a) K is maximum when x (b) lim K x→ 1 or at the vertex 1, 3 . 0 − 2π 53. y x2 3, y K 2 x 3 13 ,y 43 2332 2 x 9 43 55. y 6 y 4 32 x 3x 6x 1 1 3 2 3 1 2 9x 4 9x 1 1 y y x1 3 9x2 3 y K (a) K ⇒ (b) lim K x→ as x ⇒ 0. No maximum 0 232 1 6x 9x 1 1 432 0 at x 1. Curvature is 0 at 1, 3 . y y b 232 57. K 1 59. s a r t dt 61. The curve is a line. The curvature is zero when y 0. 63. Endpoints of the major axis: ± 2, 0 Endpoints of the minor axis: 0, ± 1 x2 2x 4y2 8yy y y K Therefore, since 65. f x (a) K (b) For x x2 For x 5 2 x4 x2 16x 6 4 0 x 4y 4y 1 16y2 1 1 4y3 x 4y 232 x 4y 4y 16 x2 x2 y 16y2 4y2 x2 16y3 16 12y2 4 32 1 4y3 16 16 3x2 0. 32 16y2 32 2 ≤ x ≤ 2, K is largest when x ± 2 and smallest when x 2 6x2 16x4 1 4x2 1 32 0, K y 1, K 1 . K 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 Using the graph of f, you see that the center of curvature is 0, 1 . Thus, 2 x2 y 1 2 2 2 f −3 3 5 . 4 To graph these circles, use y 1 ± 2 1 4 x2 and y 1 ± 2 5 4 −2 x2. —CONTINUED— 66 Chapter 11 Vector-Valued Functions 65. —CONTINUED— (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 . 5 −3 −2 3 0, ± .7647, ± 0.4082 x2. The circle will drop y k 2 16 into the parabola y 67. (a) Imagine dropping the circle x2 to the point where the tangents to the circle and parabola are equal. y x2 and x2 y 2y k 2 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. 1 16.25, and the center of the circle is 16.25 units from the vertex of the parabola. Since the radius of Finally, k x2 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 y y x2 has a curvature of K 1 2. 2 at 0, 0 . The radius of the largest sphere that will 69. Given y f x: K R 1 1 K 232 The center of the circle is on the normal line at a distance of R from x, y . Equation of normal line: y x x0 2 y0 2 1 x y 1 1 2 x0 y y y y 2 2 232 1 x y x0 2 x0 1 y x x Thus, x0, y0 For y When x ex, y 0: x0 y0 x y z, y z. 1 e x 23 ex, y x y ex, z yz z 2, 3 1 0 2 e2x e 2 x ex. x 1 12 3 x0 2 y 1 y 2 y y 2 22 Center of curvature: yz (See Exercise 47) x0 x0 y1 y x yz 1 x y y z y y0 y0 x yz z Section 11.5 71. r r r K 1 cos sin 2r r 2 2 Arc Length and Curvature 67 sin 73. r r r rr r2 r2 32 a sin a cos a sin 2r r 2 2 K sin 1 3 2 21 sin sin 77. r r 1 23 rr r2 a2 r2 32 2 cos2 31 81 75. r r r K sin sin 1 sin sin 2 2a 2 cos2 2a2 a3 4 sin 2 8 cos 2 a2 sin2 cos2 a2 a2 sin2 sin2 3 cos2 3 2 ,a > 0 a ea , a > 0 aea a2ea 2r r ea 2 2 At the pole: K rr r2 1 , K ⇒ 0. , K ⇒ 0. 81. x x dy dt dx dt gt ft f tg t ft ft f tg t ft y y g tf t 2 r2 32 2a2e2a a2e2a a2e2a e2a e2a 32 2 r0 2 8 1 4 1 a2 ⇒ (a) As (b) As a ⇒ 79. x y ft gt dy dx a a1 a sin K x x sin cos y y y a1 a sin a cos x cos x y y 2 y y 232 y dgt dt f t dx dt a2 1 cos cos a2 sin2 21 2 2 sin2 3 a cos a 1 cos 1 a 2 2 cos 3 2 2 g tf t 3 1 1 cos a 2 2 1 cos 1 2a 2 Minimum: 1 4a 2 cos 32 1 cos ≥ 0 f tg t K 1 232 1 g tf t ft3 g t 2 32 ft g tf t ft 3 2 1 csc 4a 2 f tg t ft Maximum: none gt 2 23 K→ as →0 ft f tg t ft2 ds dt 2 g tf t g t 232 1 100 ft 30 5280 ft 3600 sec 2 83. aN mK 5500 lb 32 ft sec2 3327.5 lb 68 Chapter 11 xti dr dt Vector-Valued Functions ytj xt 2 85. Let r r z t k. Then r yt 2 r 2 xt 1 xt 2 r 2 2 yt yt 2 2 zt zt 2 2 and r x ti y tj 2y t y t z t k. Then, 2z t z t zt 12 2x t x t xtx t 87. Let r xi yj rr x2 yty t ztz t r. r. rr (using Exercise 77) z k where x, y, and z are functions of t, and r r dr dt r2 y2 z2 x i rr yj rr r2 zk r3 r r xx r 2r yy r r3 dr dt r zz xi yj zk 1 x y2 r3 1 yz r3 i x z2 xyy j xzz i x 2y k z 2y xx y 1 r r3 zz y j x 2z y 2z xx z yy z k yz x xz y xz xy z xy r r 89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have r L GM r r e. y Planet Sun r Since r L and r are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate the coordinate system so that e lies along the positive x-axis and is the angle between e and r. Let e e . Then r e r e cos re cos . Also, L 2 θ e x L r L r r L r r L GM e r r GM r e r r r GM re cos r Thus, L 1 2 GM e cos r and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse. 91. A 1 2 rd 2 Thus, dA dt dA d d dt 1 2d r 2 dt 1 L 2 and r sweeps out area at a constant rate. Review Exercises for Chapter 11 1. r t ti csc t k n , n an integer n , n an integer 3. r t ln t i tj tk (a) Domain: t (a) Domain: 0, (b) Continuous for all t > 0 (b) Continuous except at t Review Exercises for Chapter 11 5. (a) r 0 (b) r (c) r c 2 1 i 3i 2c 2c (d) r 1 t r1 4j 1 1i 8 3k 1 3 69 1i c 21 2 ti c 1 2j t t t 1 2j 1 3 c 1 3k 1 3 1 3k c 1 3k 1 1 3 1i 2j 9. r t t 2j t3 i 1 t t2 ⇒ z z 1 t 3k 3 tk 3i j 3 t2 tj t 2k 7. r t xt x2 cos t i 2 sin2 t j 2 sin2 t 11. r t x t x y z i 1, y 0 1 0 1 sin t j sin t, z k 1 3 2 1 0 1 1 1 1 cos t, y t y 2 y 1 21 x y z y2 x2 2 1 1 1 z 1≤x≤1 y 2 1 1 2 1 y 3 2 1 −2 2 3 1 1 2 −1 x 1 x y x 13. r t ti z 3 2 1 1 ln t j 12 2t k 15. One possible answer is: r1 t r2 t r3 t 4t i 4i 4 3t j, 3 t i, t j, 0≤t≤1 0≤t≤3 0≤t≤4 1 2 3 2 y x 17. The vector joining the points is 7, 4, rt 2 7t, 3 4t, 8 10 . One path is 19. z x x2 t, y rt z 5 y 2, x t, z ti tj y 0, t 2t 2 2t 2k x 10t . −3 2 3 1 2 3 y x 21. lim t 2 i t→2 4 t2j k 4i k 70 23. r t Chapter 11 3t i 3i ut ut t j Vector-Valued Functions 1 j, u t ti t 2j 23 tk 3 (b) r t t2 t 3t 2 4t 1 t3 2t 2 (d) u t Dt u t (f ) r t 1 Dt r t ut ut 0 2r t 2r t 24 t 3 83 t 3 5t i 5i t3 i t2 2t 2t 4j 8t 3j 2t 2j 2j 3t 3 9t 2 23 tk 3 2t 2k t2 2t tk 1k (a) r t (c) r t Dt r t 3t 2 (e) r t Dt r t 10t 2 2t 1 1 2t 10t 10t 2 2t 2 i 25. x t and y t are increasing functions at t decreasing function at t t 0. t0, and z t is a 27. cos t i t cos t j dt sin t i t sin t cos t j C 29. cos t i sin t j t k dt 1 t 2 dt 1 t1 2 t2 ln t 1 t2 C 2 31. r t r0 rt j t2 2t i k 1i etj C i et e t k dt 3j 2j t 2i etj i e tk 2j C 4k 33. 2 3t i 2t 2j t 3k dt 3t 2 i 2 2t 3 j 3 t4 k 4 2 2 32 j 3 5k ⇒ C e t 4k 2 2 35. 0 et 2i 3t2j k dt 2et 2i t3j tk 0 2e 2i 8j 2k 37. r t vt cos3 t, sin3 t, 3t rt vt 3 3 cos2 t sin t, 3 sin2 t cos t, 3 9 cos4 t sin2 t cos2 t sin2 t 9 sin4 t cos2 t cos2 1 sin2 t 2 2 9 1 t sin2 t 3 cos2 t sin2 t at vt 6 cos t 3 cos2 t cos t, 6 sin t cos2 t cos t , 3 sin t 2 cos t 2 3 sin2 t sin t , 0 3 cos t 2 sin t 1 3 , t 2, t , t 0 2 3 , 2t, 1 2 sin t , 0 v02 sin 2 32 75 2 sin 60 32 2 39. rt rt r4 ln t 1 t 4 41. Range x 152 feet 1, 8, 1 2 direction numbers 0, 16, 2 , the parametric equations are Since r 4 1 x t, y 16 8t, z 2 2 t. r t0 0.1 r 4.1 v02 sin 2 9.8 0.1, 16.8, 2.05 43. Range x 80 ⇒ v0 80 9.8 sin 40 34.9 m sec Review Exercises for Chapter 11 45. rt vt vt at Tt 5t i 5i 5 0 i at Tt a N does not exist Nt a a T N i (The curve is a line.) i vt 47. rt vt ti i tj 1 j 2t 4t 1 2t 1 j 4t t 12 tj 4t 1 2 t 2 tj 4t 1 1 1 12 tk 2 tk 5t 2 k 2t j t k 1 5t2 5t i 2j k 5 1 5t2 5t 1 5t 2 5 5t 2 1 5t 2 5 51 2 ti 4t j 1 71 N t does not exist a T 0 1 4t t 4t 1 2t 4t ti i 1 2j i t 2j 2t j 49. rt vt vt at Tt Nt a a T N et i et i e2t et i e tj e tj e e tj 2t 51. rt vt v at Tt Nt a a T N et i e t j e2t e 2t e ti e2t etj e 2t e2t e 2t e2t e 2t 2 e2t e 2t 53. r t 2 cos t i 3 ,x 4 2 sin t j t k, x 2, y 2 cos t, y 2, z k 2, b 2, z t 3 . 4 2 sin t, z t When t rt 2 sin t i 2 cos t j 3 ,a 4 2t Direction numbers when t x 2t 2, y 2, c 3 4 1 55. v 9.56 104 4600 2t i 2i b 4.56 mi sec 57. r t rt s 3tj, 0 ≤ t ≤ 5 3j 5 2 −4 −2 y (0, 0) x 246 8 10 12 14 r t dt a 5 0 4 9 dt −4 −6 −8 − 10 − 12 − 14 − 16 13t 0 5 13 (10, − 15) 72 59. Chapter 11 rt rt rt 10 cos3 t i Vector-Valued Functions 10 sin3 tj 30 sin2 t cos tj sin4 t cos2 t − 10 10 y 30 cos2 t sin ti 30 cos4 t sin2 t 30 cos t sin t 2 2 −2 x 2 10 s 4 0 30 cos t sin t dt 120 sin2 t 2 2 60 0 − 10 61. r t rt s a 3ti 3i b 2tj 2j 4tk, 0 ≤ t ≤ 3 4k 3 3 12 10 8 6 4 2 2 x z (−9, 6, 12) r t dt 0 9 4 16 dt 0 29 dt 3 29 (0, 0, 0) 24 68 10 y 63. r t rt s 8 cos t, 8 sin t, t , 0 ≤ t ≤ < b 2 65 65 2 65. r t rt s 1 ti 2 1 i 2 sin t j cos t j r t dt cos t k, 0 ≤ t ≤ sin t k 8 sin t, 8 cos t, 1 , r t 2 r t dt a z 65 dt 0 0 π 2 (0, 8, π ) 2 4 6 8 y 0 1 4 dt 0 cos2 t 5 t 2 sin2 t dt 5 2 4 86 x (8, 0, 0) 5 2 0 67. r t Line k 0 3ti 2tj 69. rt rt rt r r 2ti 2i j i 2 0 r r 12 tj 2 tj 2k j t 1 r 3 t2k 2tk, r 5t2 4 k 2t 2 5t2 4j 20 4 2k, r r 25 5t2 20 K 32 4 32 71. y y y K At x 12 x 2 x 1 1 2 73. y y K ln x 1 ,y x 1 y y 1, K 1 x2 232 y y 4, K 232 1 1 x2 1 1 22 1 x2 1 x 2]3 2 32 At x 32 1 and r 173 2 1 23 2 2 and r 4 2 2. 17 17 17. 75. The curvature changes abruptly from zero to a nonzero constant at the points B and C. ...
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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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