# D x2 y 2 intersection and d is the distance between

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Unformatted text preview: −2) = −4i + j + 4k, (t + 1)2 r(−2) = 4i + j; r = (4i + j) + t(−4i + j + 4k) 30. r (t) = cos ti + sinh tj + 1 k, t = 0 at P0 so r (0) = i + k, r(0) = j; r = ti + j + tk 1 + t2 32. (sin t)i − (cos t)j + C 31. 3ti + 2t2 j + C 34. (t − 1)et , t(ln t − 1) + C 41. 1 13 t i + t4 j 3 4 38. 0 2 t(1 + t2 )1/2 dt = t2 + t4 dt = 0 40. 36. = 0, −2/3 2 39. 35. (t3 /3)i − t2 j + ln |t|k + C π /3 1 1 sin 3t, cos 3t 3 3 37. 33. (−t cos t + sin t)i + tj + C 0 2 2 − (3 − t)5/2 , (3 + t)5/2 , t 5 5 2 3/2 t i + 2t1/2 j 3 9 = 1 3 1 1 + t2 3 3/2 2 −e−t , et , t3 + C 1 = 0 1 1 i+ j 3 4 √ = (5 5 − 1)/3 0 √ √ = 72 6/5, 72 6/5, 6 −3 52 i + 4j 3 42. 12 1 (e − 1)i + (1 − e−1 )j + k 2 2 43. y(t) = y (t)dt = 1 t3 i + t2 j + C, y(0) = C = i + j, y(t) = ( 1 t3 + 1)i + (t2 + 1)j 3 3 44. y(t) = y (t)dt = (sin t)i − (cos t)j + C, y(0) = −j + C = i − j so C = i and y(t) = (1 + sin t)i − (cos t)j. 45. y (t) = y(t) = y(t) = y (t)dt = ti + et j + C1 , y (0) = j + C1 = j so C1 = 0 and y (t) = ti + et j. y (t)dt = 12 t i + et j + C2 , y(0) = j + C2 = 2i so C2 = 2i − j and 2 12 t + 2 i + (et − 1)j 2 497 Chapter 14 46. y (t) = y(t) = 47. y (t)dt = 4t3 i − t2 j + C1 , y (0) = C1 = 0, y (t) = 4t3 i − t2 j 1 1 y (t)dt = t4 i − t3 j + C2 , y(0) = C2 = 2i − 4j, y(t) = (t4 + 2)i − ( t3 + 4)j 3 3 0 0 6 48. 1.5 0 1.5 5 0 49. r (t) = −4 sin ti + 3 cos tj, r(t) = 16 cos2 t + 9 sin2 t, r (t) = 16 sin2 t + 9 cos2 t, −7 sin t cos t r r = 144 + 49 sin2 t cos2 t, θ = cos−1 144 + 49 sin2 t cos2 t r and r are parallel when θ = 0, so t = 0, π/2, π, 3π/2, 2π ; the angle between r and r is greatest at θmax = 0.28 (t ≈ 2.23, 5.53) and θmin = −0.28, (t ≈ 0.77, 3.95), so r and r are never perpendicular. 3 0 0 o √ √ 2 + 3t2 √ 50. r (t) = 2ti + 3t2 j, r(t) = t2 1 + t2 , r (t) = t 4 + 9t2 , cos θ = √ 1 + t2 4 + 9t2 √ cos θ = 1 when t = 0, ±23/4 / 3 and r and r are parallel; cos θ > 0, so they are never perpendicular. 0.3 0 1 0 51. (a) 2t − t2 − 3t = −2, t2 + t − 2 = 0, (t + 2)(t − 1) = 0 so t = −2, 1. The points of intersection are (−2, 4, 6) and (1, 1, −3). (b) r = i + 2tj − 3k; r (−2) = i − 4j − 3k, r (1) = i + 2j − 3k, and n = 2i − j + k is normal to the plane. Let θ be the acute angle, then √ for t = −2: cos θ = |n · r |/( n r ) = 3/ 156, θ ≈ 76◦ ; √ for t = 1: cos θ = |n · r |/( n r ) = 3/ 84, θ ≈ 71◦ . Exercise Set 14.2 498 52. r = −2e−2t i − sin tj + 3 cos tk, t = 0 at the point (1, 1, 0) so r (0) = −2i + 3k and hence the tangent line is x = 1 − 2t, y = 1, z = 3t. But x = 0 in the yz -plane so 1 − 2t = 0, t = 1/2. The point of intersection is (0, 1, 3/2). 53. r1 (1) = r2 (2) = i + j + 3k so the graphs intersect at P; r1 (t) = 2ti + j + 9t2 k and 1 r2 (t) = i + tj − k so r1 (1) = 2i + j + 9k and r2 (2) = i + j − k are tangent to the graphs at P, 2 √ r1 (1) · r2 (2) 6 thus cos θ = = − √ √ , θ = cos−1 (6/ 258) ≈ 68◦ . r1 (1) r2 (2) 86 3 54. r1 (0) = r2 (−1) = 2i + j + 3k so the graphs intersect at P; r1 (t) = −2e−t i − (sin t)j + 2tk and r2 (t) = −i + 2tj + 3t2 k so r1 (0) = −2i and r2 (−1) = −i − 2j + 3k are tangent to the graphs at P, thus cos θ = r1 (0) · r2 (−1) 1 = √ , θ ≈ 74◦ . r1 (0) r2 (−1) 14 55. (a) r1 = 2i + 6tj + 3t2 k, r2 = 4t3 k, r1 · r2 = t7 ; (b) r1 × r2 = 3t6 i − 2t5 j, d (r1 · r2 ) = 7t6 = r1 · r2 + r1 · r2 dt d (r1 × r2 ) = 18t5 i − 10t4 j = r1 × r2 + r1 × r2 dt 56. (a) r1 = − sin ti +cos tj + k, r2 = k, r1 · r2 = cos t + t2 ; d (r1 · r2 ) = − sin t +2t = r1 · r2 + r1 · r2 dt (b) r1 × r2 = t sin ti + t(1 − cos t)j − sin tk, d (r1 × r2 ) = (sin t + t cos t)i + (1 + t sin t − cos t)j − cos tk = r1 × r2 + r1 × r2 dt 57. d [r(t) × r (t)] = r(t) × r (t) + r (t) × r (t) = r(t) × r (t) + 0 = r(t) × r (t) dt 58. d du d [u · (v × w)] = u · [v × w] + · [v × w] = u · dt dt dt =u· v× dw +u· dt v× dw dv du + ×w + · [v × w] dt dt dt du dv ×w + · [v × w] dt dt 59. In Exercise 60, write each scalar triple product as a determinant. 60. Let c = c1 i + c2 j, r(t) = x(t)i + y (t)j, r1 (t) = x1 (t)i + y1 (t)j, r2 (t) = x2 (t)i + y2 (t)j and use properties of derivatives. 61. Let r1 (t) = x1 (t)i + y1 (t)j + z1 (t)k and r2 (t) = x2 (t)i + y2 (t)j + z2 (t)k, in both (8) and (9); show that the left and right members of the equalities are the same. 62. (a) k r(t) dt = =k k (x(t)i + y (t)j + z (t)k) dt x(t) dt i + k (b) Similar to Part (a) y (t) dt j + k (c) z (t) dt k = k r(t) dt Use Part (a) on Part (b) with k = −1 499 Chapter 14 EXERCISE SET 14.3 1. (a) The tangent vector reverses direction at the four cusps. (b) r (t) = −3 cos2 t sin ti + 3 sin2 t cos tj = 0 when t = 0, π/2, π, 3π/2, 2π . 2. r (t) = cos ti + 2 sin t cos tj = 0 when t = π/2, 3π/2. The tangent vector reverses direction at (1, 1) and (−1, 1). 3. r (t) = 3t2 i + (6t − 2)j + 2tk; smooth 4. r (t) = −2t sin(t2 )i + 2t cos(t2 )j − e−t k; smooth 5. r (t) = (1 − t)e−t i + (2t − 2)j − π sin(πt)k; not smooth, r (1) = 0 6. r (t) = π cos(π...
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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