# A 1604x 3y 2z 3 2 69 fx ex fy ey b 14404x

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Unformatted text preview: (a) The line is parallel to the vector −2i + 3j; the slope is −3/2. (b) y = 0 in the xz -plane so 1 − 2t = 0, t = 1/2 thus x = 2 + 1/2 = 5/2 and z = 3(1/2) = 3/2; the coordinates are (5/2, 0, 3/2). 20. (a) x = 3 + 2t = 0, t = −3/2 so y = 5(−3/2) = −15/2 (b) x = t, y = 1 + 2t, z = −3t so 3(t) − (1 + 2t) − (−3t) = 2, t = 3/4; the point of intersection is (3/4, 5/2, −9/4). 21. (a) y (b) (0, 1) y (1, 1) x x (1, 0) (1, -1) 490 491 Chapter 14 z 22. (a) z (b) (0, 0, 1) (1, 1, 1) y (1, 1, 0) y (1, 1, 0) x x 23. r = (1 − t)(3i + 4j), 0 ≤ t ≤ 1 24. r = (1 − t)4k + t(2i + 3j), 0 ≤ t ≤ 1 25. x = 2 26. y = 2x + 10 y y 10 x x 2 -5 27. (x − 1)2 + (y − 3)2 = 1 28. x2 /4 + y 2 /25 = 1 y y 5 3 x 2 x 1 29. x2 − y 2 = 1, x ≥ 1 30. y = 2x2 + 4, x ≥ 0 y y 2 x 1 4 x 1 Exercise Set 14.1 492 z 31. z 32. (0, 2, π / 2) (0, 4, π /2) y (2, 0, 0) y (9, 0, 0) x x z 33. z 34. 2 y c o y x x 36. x = t, y = −t, z = 35. x = t, y = t, z = 2t2 z z √√ 2 1 − t2 y+x=0 y z= √ 2 – x 2 – y2 x x y 37. r = ti + t2 j ± 1 3 81 − 9t2 − t4 k 38. r = ti + tj + (1 − 2t)ks z z y=x x y x x+y+z=1 y 493 Chapter 14 39. x2 + y 2 = (t sin t)2 + (t cos t)2 = t2 (sin2 t + cos2 t) = t2 = z 40. x − y + z + 1 = t − (1 + t)/t + (1 − t2 )/t + 1 = [t2 − (1 + t) + (1 − t2 ) + t]/t = 0 √ √ = 41. x = sin t, y = 2 cos t, z = 3 sin t so x2 + y 2 + z 2 = sin2 t + 4 cos2 t + 3 sin2 t√ 4 and z = 3x; it is the curve of intersection of the sphere x2 + y 2 + z 2 = 4 and the plane z = 3x, which is a circle with center at (0, 0, 0) and radius 2. 42. x = 3 cos t, y = 3 sin t, z = 3 sin t so x2 + y 2 = 9 cos2 t + 9 sin2 t = 9 and z = y ; it is the curve of intersection of the circular cylinder x2 + y 2 = 9 and the plane z = y , which is an ellipse with √ major axis of length 6 2 and minor axis of length 6. 43. The helix makes one turn as t varies from 0 to 2π so z = c(2π ) = 3, c = 3/(2π ). 44. 0.2t = 10, t = 50; the helix has made one revolution when t = 2π so when t = 50 it has made 50/(2π ) = 25/π ≈ 7.96 revolutions. 45. x2 + y 2 = t2 cos2 t + t2 sin2 t = t2 , x2 + y 2 = t = z ; a conical helix. 46. The curve wraps around an elliptic cylinder with axis along the z -axis; an elliptical helix. 47. (a) (b) (c) (d) III, since the curve is a subset of the plane y = −x IV, since only x is periodic in t, and y, z increase without bound II, since all three components are periodic in t I, since the projection onto the yz -plane is a circle and the curve increases without bound in the x-direction 49. (a) Let x = 3 cos t and y = 3 sin t, then z = 9 cos2 t. z (b) x 50. The plane is parallel to a line on the surface of the cone and does not go through the vertex so the curve of intersection is a parabola. Eliminate z to get y + 2 = x2 + y 2 , (y + 2)2 = x2 + y 2 , y = x2 /4 − 1; let x = t, then y = t2 /4 − 1 and z = t2 /4 + 1. y z x y Exercise Set 14.2 494 EXERCISE SET 14.2 y 1. y 2. 2 r(2p) – r(3p/2) r''(p) x -2 x 2 -2 r'(p/4) 3. r (t) = 5i + (1 − 2t)j 6. r (t) = 7. 5. r (t) = − 4. r (t) = sin tj 1 i + sec2 tj + 2e2t k t2 1 1 i + (cos t − t sin t)j − √ k 2 1+t 2t r (t) = 1, 2t , 8. r (t) = 3t2 i + 2tj, r (2) = 1, 4 , r (1) = 3i + 2j r(2) = 2, 4 r(1) = i + j y y 3 〈1, 4 〉 2 4 1 x x 1 2 9. r (t) = sec t tan ti + sec2 tj, r (0) = j r(0) = i y 1 10. 2 3 4 r (t) = 2 cos ti − 3 sin tj, √ 3 π = 3i − j r 6 2 √ 33 π =i+ j r 6 2 y 3 x 2 1.5 1 x -1 -2 -1 1 -1 -2 -3 2 495 Chapter 14 r (t) = 2 cos ti − 2 sin tk, 11. r (t) = − sin ti + cos tj + k, 12. r (π/2) = −2k, 1 1 r (π/4) = − √ i + √ j + k, 2 2 r(π/2) = 2i + j 1 π 1 r(π/4) = √ i + √ j + k 4 2 2 z z ( ) = − √12 i + √12 j + k r′ 3 y (2, 1, 0) 11 , ,3 √2 √2 ( x r ′ 6 = -2k () y ) x 13. 9i + 6j 14. √ √ 2/2, 2/2 15. 1/3, 0 16. j 17. 2i − 3j + 4k 18. 3, 1/2, sin 2 19. (a) continuous, lim r(t) = 0 = r(0) (b) not continuous, lim r(t) does not exist 20. (a) not continuous, lim r(t) does not exist. (b) continuous, lim r(t) = 5i − j + k = r(0) t→0 t→0 t→0 21. (a) t→0 lim (r(t) − r (t)) = i − j + k t→0 (b) lim (r(t) × r (t)) = lim (− cos ti − sin tj + k ) = −i + k t→0 (c) t→0 lim (r(t) · r (t)) = 0 t→0 22. r(t) · (r (t) × r (t)) = t 1 0 t2 2t 2 t3 3t2 6t = 2t3 , so lim r(t) · (r (t) × r (t)) = 2 t→1 1 23. r (t) = 2ti − j, r (1) = 2i − j, r(1) = i + 2j; x = 1 + 2t, y = 2 − t, z = 0 t 24. r (t) = 2e2t i + 6 sin 3tj, r (0) = 2i, r(0) = i − 2j; x = 1 + 2t, y = −2, z = 0 √ 25. r (t) = −2π sin πti + 2π cos πtj + 3k, r (1/3) = − 3 π i + π j + 3k, √ √ √ r(1/3) = i + 3 j + k; x = 1 − 3 πt, y = 3 + πt, z = 1 + 3t 1 1 i − e−t j + 3t2 k, r (2) = i − e−2 j + 12k, t 2 1 r(2) = ln 2i + e−2 j + 8k; x = ln 2 + t, y = e−2 − e−2 t, z = 8 + 12t 2 26. r (t) = Exercise Set 14.2 496 3 3 j, t = 0 at P0 so r (0) = 2i + j, 27. r (t) = 2i + √ 4 2 3t + 4 3 r(0) = −i + 2j; r = (−i + 2j) + t 2i + j 4 √ 28. r (t) = −4 sin ti − 3j, t = π/3 at P0 so r (π/3) = −2 3i − 3j, √ r(π/3) = 2i − π j; r = (2i − π j) + t(−2 3i − 3j) 29. r (t) = 2ti + 1 j − 2tk, t = −2 at P0 so r (...
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