A ci 5 b o 1 0 1 2 o l i f z0 c 5 4 3 2 1 x 0 9 6 3 0

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Unformatted text preview: (d) (2 10, π, tan−1 3) (b) (c) (5, π/2, tan−1 (4/3)) √ 11. (a) 5 3/2, π/4, −5/2 (c) (0, 0, 3) (b) (0, 7π/6, −1) (d) (4, π/6, 0) 12. (a) (0, π/2, −5/2) √ √ (b) (3 2, 0, −3 2) √ (d) (5/2, 2π/3, −5 3/2) (c) (0, 3π/4, −3) z 15. z 16. z 17. y y (3, 0, 0) x y 2 z= x +y x 2 2 x +y =9 x y = x, x ≥ 0 2 Exercise Set 13.8 482 z 18. z 19. z 20. y (0, 4, 0) y y (2, 0, 0) x x=2 x z=x x 2 z 21. 2 x + ( y – 2) = 4 z 22. y x (1, 0, 0) x 2 2 y x2 – y2 = z 2 x +y +z =1 z 23. z 24. y y y = √3x x x (3, 0, 0) 2 2 2 x +y +z =9 z 25. z 26. (0, 0, 2) y y x z = √x2 + y2 x z=2 483 Chapter 13 z 27. z 28. (0, 0, 2) y y (1, 0, 0) x 2 2 x 2 x2 + y2 = 1 x + y + (z – 2) = 4 29. z z 30. y y (1, 0, 0) (1, 0, 0) x (x – 1)2 + y2 + z2 = 1 x (x – 1)2 + y2 = 1 31. (a) z = 3 (b) ρ cos φ = 3, ρ = 3 sec φ 32. (a) r sin θ = 2, r = 2 csc θ (b) ρ sin φ sin θ = 2, ρ = 2 csc φ csc θ 33. (a) z = 3r2 (b) ρ cos φ = 3ρ2 sin2 φ, ρ = 34. (a) z = √ 3r (b) ρ cos φ = 1 csc φ cot φ 3 √ 1 π 3ρ sin φ, tan φ = √ , φ = 6 3 35. (a) r = 2 (b) ρ sin φ = 2, ρ = 2 csc φ 36. (a) r2 − 6r sin θ = 0, r = 6 sin θ (b) ρ sin φ = 6 sin θ, ρ = 6 sin θ csc φ 37. (a) r2 + z 2 = 9 (b) ρ = 3 38. (a) z 2 = r2 cos2 θ − r2 sin2 θ = r2 (cos2 θ − sin2 θ), z 2 = r2 cos 2θ (b) Use the result in part (a) with r = ρ sin φ, z = ρ cos φ to get ρ2 cos2 φ = ρ2 sin2 φ cos 2θ, cot2 φ = cos 2θ 39. (a) 2r cos θ + 3r sin θ + 4z = 1 (b) 2ρ sin φ cos θ + 3ρ sin φ sin θ + 4ρ cos φ = 1 40. (a) r2 − z 2 = 1 (b) Use the result of part (a) with r = ρ sin φ, z = ρ cos φ to get ρ2 sin2 φ − ρ2 cos2 φ = 1, ρ2 cos 2φ = −1 Exercise Set 13.8 484 41. (a) r2 cos2 θ = 16 − z 2 (b) x2 = 16 − z 2 , x2 + y 2 + z 2 = 16 + y 2 , ρ2 = 16 + ρ2 sin2 φ sin2 θ, ρ2 1 − sin2 φ sin2 θ = 16 42. (a) r2 + z 2 = 2z (b) ρ2 = 2ρ cos φ, ρ = 2 cos φ 43. all points on or above the paraboloid z = x2 + y 2 , that are also on or below the plane z = 4 44. a right circular cylindrical solid of height 3 and radius 1 whose axis is the line x = 0, y = 1 45. all points on or between concentric spheres of radii 1 and 3 46. all points on or above the cone φ = π/6, that are also on or below the sphere ρ = 2 √ √ 47. θ = π/6, φ = π/6, spherical (4000, π/6, π/6), rectangular 1000 3, 1000, 2000 3 48. (a) y = r sin θ = a sin θ but az = a sin θ so y = az , which is a plane that contains the curve of intersection of z = sin θ and the circular cylinder r = a. From Exercise 60, Section 12.4, the curve of intersection of a plane and a circular cylinder is an ellipse. z (b) z = sin θ y x 49. (a) (10, π/2, 1) 50. (b) (0, 10, 1) √ (c) ( 101, π/2, tan−1 10) 20 0 30 0 51. Using spherical coordinates: for point A, θA = 360◦ − 60◦ = 300◦ , φA = 90◦ − 40◦ = 50◦ ; for point B , θB = 360◦ − 40◦ = 320◦ , φB = 90◦ − 20◦ = 70◦ . Unit vectors directed from the origin to the points A and B , respectively, are uA = sin 50◦ cos 300◦ i + sin 50◦ sin 300◦ j + cos 50◦ k, uB = sin 70◦ cos 320◦ i + sin 70◦ sin 320◦ j + cos 70◦ k The angle α between uA and uB is α = cos−1 (uA · uB ) ≈ 0.459486 so the shortest distance is 6370α ≈ 2,927 km. 485 Chapter 13 CHAPTER 13 SUPPLEMENTARY EXERCISES 2. (c) F = −i − j (d) 1, −2, 2 = 3, so r − 1, −2, 2 = 3, or (x − 1)2 + (y + 2)2 + (z − 2)2 = 9 √ 3. (b) x = cos 120◦ = −1/2, y = ± sin 120◦ = ± 3/2 (d) true: u × v = u 4. (d) v | sin(θ)| = 1 x + 2y − z = 0 5. (b) (y, x, z ), (x, z, y ), (z, y, x) (c) circle of radius 5 in plane z = 1 with center at (0, 0, 1) (rectangular coordinates) (d) the two half-lines z = ±x, x ≥ 0 in the xz -plane 6. (x + 3)2 + (y − 5)2 + (z + 4)2 = r2 , (b) r2 = 52 = 25 (a) r2 = 42 = 16 −→ −→ −→ (c) r2 = 32 = 9 −→ 7. (a) AB = −i + 2j + 2k, AC = i + j − k, AB × AC = −4i + j − 3k, area = (b) area = √ 1√ 3 1 −→ h AB = h = 26, h = 26/3 2 2 2 −→ √ 1 −→ AB × AC = 26/2 2 8. The sphere x2 + (y − 1)2 + (z + 3)2 = 16 has center Q(0, 1, −3) and radius 4, and −→ √ √ √ √ P Q = 12 + 42 = 17, so minimum distance is 17 − 4, maximum distance is 17 + 4. 9. (a) a · b = 0, 4c + 3 = 0, c = −3/4 √ (b) Use a · b = a b cos θ to get 4c + 3 = c2 + 1(5) cos(π/4), 4c + 3 = 5 c2 + 1/ 2 Square both sides and rearrange to get 7c2 + 48c − 7 = 0, (7c − 1)(c + 7) = 0 so c = −7 (invalid) or c = 1/7. (c) Proceed as in (b) with θ = π/6 to get 11c2 − 96c + 39 = 0 and use the quadratic formula to √ get c = 48 ± 25 3 /11. (d) a must be a scalar multiple of b, so ci + j = k (4i + 3j), k = 1/3, c = 4/3. −→ −→ −→ −→ 10. OS = OP + P S = 3i + 4j+ QR = 3i + 4j + (4i + j) = 7i + 5j 11. (a) the plane through the origin which is perpendicular to r0 (b) the plane through the tip of r0 which is perpendicular to r0 12. The normals to the planes are given by a1 , b1 , c1 and a2 , b2 , c2 , so the condition is a1 a2 + b1 b2 + c1 c2 = 0. −→ −→ −→ −→ −→ −→ −→ −→ −→ 13. Since AC · (AB × AD) =AC · (AB × CD) + AC · (AB × AC ) = 0 + 0 = 0, the volume of th...
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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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