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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 (...
View Full Document

Ask a homework question - tutors are online