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# ODDREV10 - 256 91 x2 Chapter 10 y2 z2 z2 16 16 16 4 Vectors...

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Unformatted text preview: 256 91. x2 Chapter 10 y2 z2 z2 16, 16 16 4 Vectors and the Geometry of Space 93. x2 y2 z2 z2 2z 2z 0 0, r 2 0, z 1 2 (a) r 2 (b) 2 (a) r 2 (b) 2 1 0, 2 cos 2 cos 2 cos 95. x2 y2 4y 4r sin , r 4 sin sin , 4 sin 4 sin csc 0, 97. x2 y2 9 r 2 sin2 9 cos2 sin2 2 (a) r 2 (b) 2 (a) r 2 cos2 r2 (b) 2 9, sin 2 4 sin sin sin 4 sin , sin sin2 sin2 cos2 cos2 sin2 sin2 , sin2 9, 2 9 2 9 csc2 cos2 sin2 103. 0 ≤ 0≤ 0≤ ≤2 ≤ 6 99. 0 ≤ ≤ 2 101. 0 ≤ ≤2 0≤r≤2 0≤z≤4 z 0≤r≤a r≤z≤a z a ≤ a sec z 5 a 3 2 1 x 2 3 x 2 3 y y a a y −a −a 30° x 105. Rectangular 0 ≤ x ≤ 10 0 ≤ y ≤ 10 0 ≤ z ≤ 10 10 z 107. Spherical 4≤ ≤6 −8 10 8 z 8 y x −8 y 10 x 109. z z sin , r y r y 1 1 y y and the cylinder r 1. The curve of intersection is the ellipse formed by the intersection of the plane z Review Exercises for Chapter 10 1. P (a) u v (b) v (c) 2u v 1, 2 , Q \ 4, 1 , R 3, 4, 2 1 4i 25 3i 5, 4 j, 2j 3. v v cos i v sin j 8 cos 120 i 4i 4 3j 8 sin 120 j PQ \ PR 42 22 6, 2 4, 2 10, 0 10i Review Exercises for Chapter 10 5. 120 cos 100 arccos tan y 5 6 2 tan 2 11 5 5, 4, 0 10 11 3.015 ft y 120 lb 100 lb 2 ft 257 2 ⇒y y θ 2 tan arccos 5 6 4, x 5: 7. z 0, y 9. Looking down from the positive x-axis towards the yz-plane, the point is either in the first quadrant y > 0, z > 0 or in the third quadrant y < 0, z < 0 . The x-coordinate can be any number. 15 2 z2 2 11. x 3 2 y 2 2 z 6 2 13. x2 x 4x 2 2 4 y 3 y2 2 6y z2 9 9 4 4 9 15. v 4 z 2, 4 1, 7 3 2, 5, 10 Center: 2, 3, 0 Radius: 3 z 4 3 2 (2, − 1, 3) 3 2 1 12 3 5 5 x 4 3 −2 y 6 x 5 4 3 45 6 y (4, 4, − 7) 17. v w Since 21. P (a) u v (b) u (c) v 5 1 3, 6 3, 3 4, 9 4, 6 1 1 4, 2, 10 2, 1, 5 19. Unit vector: u u 2, 3, 5 38 2 , 38 3 , 38 5 38 2w v, the points lie in a straight line. 4, 4, 0 , R 1, 4, 0 3, 0, 6 1 9 3 i 4 2 i 3 36 3 45 3 j 4 2 j 3 3 52 2 i 40 i 3i 2, 0, 6 4j, 6k 06 3 23. u 7, 2, 3 , v v 1, 4, 5 5, 0, 0 , Q \ PQ \ Since u 0, the vectors are orthogonal. PR v v 25. u v u u v cos 5 cos 2 cos v 5 2 sin sin i 3j j 27. u 10, 5, 15 , v 2, 1, 3 u 5v ⇒ u is parallel to v and in the opposite direction. 52 1 2 uv uv arccos 2 5 22 1 52 6 4 15 3 2 4 6 258 Chapter 10 Vectors and the Geometry of Space ± 6, 29. There are many correct answers. For example: v 5, 0 . In Exercises 31–39, u 31. u u 33 14 2 14 < 2 3, 2 2, 1 , v 11 u 2 > < 2, 4, 3,w > < 1, 2, 2 . 33. projuw u u w 2 > u 2, 1 5 14 5 14 5 3, 14 15 10 ,, 14 14 15 5 ,, 14 7 i 2 1 j 4 2 k 3 2 35. n n n n v w 5 1 5 2i 2i j 37. V u 3, v w 2, 1, 0 4 4 2, 1 j 39. Area parallelogram u v 102 285 112 8 2 (See Exercises 36, 38) 41. F \ c cos 20 j 2k F \ sin 20 k PQ z PQ \ PQ 200 c F F 43. v j i 0 0 F j k 0 2 c cos 20 c sin 20 2c cos 20 2 ft 70° F 2c cos 20 i y PQ 100 cos 20 x 100 cos 20 j cos 20 100 1 tan2 20 sin 20 k 100 j tan 20 k 100 sec 20 106.4 lb 45. 3x 3y 7z 4, x y 2z 3 (a) x 1, y 2 t, z 3 (b) None Solving simultaneously, we have z 1. Substituting z 1 into the second equation we have y x 1. Substituting for x in this equation we obtain two points on the line of intersection, 0, 1, 1 , 1, 0, 1 . The direction vector of the line of intersection is v i j. (a) x (b) x t, y y 1, z 1 1 t, z 1 Review Exercises for Chapter 10 259 47. The two lines are parallel as they have the same direction numbers, 2, 1, 1. Therefore, a vector parallel to the plane is v 2i j k. A point on the first line is 1, 0, 1 and a point on the second line is 1, 1, 2 . The vector u 2i j 3k connecting these two points is also parallel to the plane. Therefore, a normal to the plane is v u i 2 2 2i j 1 1 4j k 1 3 2i 1 x 2j . 2y 2y 0 1 49. Q 2x 1, 0, 2 3y \ 6z 6 A point P on the plane is 3, 0, 0 . PQ n D 2, \ 2, 0, 2 3, 6 8 7 PQ n n Equation of the plane: x 51. Q 3, 2, 4 point 53. x 2y 3z 6 P 5, 0, 0 point on plane n \ Plane Intercepts: 6, 0, 0 , 0, 3, 0 , 0, 0, 2 z 2, 5, 1 normal to plane 2, \ PQ D 2, 4 3 PQ n n 10 30 30 3 6 x (0, 0, 2) 3 (0, 3, 0) (6, 0, 0) y 55. y 1 z 2 57. x2 16 y2 9 z2 1 2 −4 z Plane with rulings parallel to the x-axis z 2 Ellipsoid xy-trace: xz-trace: x2 16 x2 16 y2 9 y2 9 z2 z2 1 1 1 4 x 5 −2 y 2 6 x y yz-trace: 59. x2 16 y2 9 z2 1 2 −2 5 z Hyperboloid of two sheets y2 xy-trace: 4 xz-trace: None yz-trace: y2 9 z2 1 x2 16 1 x 5 y 260 Chapter 10 x2 Vectors and the Geometry of Space y2 2 z 61. (a) rz 2z x2 y2 2z 2 0 1 2 4 3 −2 2 x 1 2 3 y 2 (b) V 2 0 2 x3 2x 0 12 x 2 13 x dx 2 x4 8 2 0 1 dx 3 y 2 2 4 2 2 x2 1 x 12.6 cm3 x3 12 2 1 2 3 (c) V 2 2 12 12 x 2 13 x dx 2 2 12 y 1 dx 3 2x x2 31 64 x4 8 2 1 2 4 x 1 2 3 225 64 11.04 cm 3 63. 2 2, 2 2, 2 , rectangular (a) r (b) 22 22 2 22 22 2 4, 2 2 arctan 2 5, 1 3 ,z 4 3 , 4 2, arccos 4, 3 , 2 , cylindrical 4 arccos 1 , 5 2 5, 3 5 , arccos , spherical 4 5 2 2 2 25 65. 100, 6 , 50 , cylindrical 502 50 5 67. 25, r2 3 , , spherical 44 25 sin 3 4 2 1002 6 arccos 50 5, ⇒r 25 2 2 50 50 5 arccos 1 5 4 63.4 z 25 cos 2 , 2 , 25 cos 3 4 25 2 2 6 , 63.4 , spherical 4 25 2 , cylindrical 2 69. x2 y2 2z r 2 sin2 cos2 2 (a) Cylindrical: r 2 cos2 (b) Spherical: 2 2z, r 2 cos 2 sin2 sin2 2z 2 cos , sin2 cos 2 2 cos 0, 2 sec 2 cos csc2 sin2 Problem Solving for Chapter 10 261 Problem Solving for Chapter 10 1. b b a b a a a a b c b b b b b c c c c \ 0 0 0 b c a 3. Label the figure as indicated. From the figure, you see that \ SP SR \ 1 a 2 1 a 2 \ 1 b 2 1 b 2 \ RQ and \ b c sin A a b sin C PQ . \ \ Since SP RQ and SR Q PQ , PSRQ is a parallelogram. Then, sin A a b a sin C . c The other case, sin A a sin B is similar. b z 6 5 4 3 c a P abc b 1 2 R abc a− 1b S 2 b 1 2 a+ 1b 2 5. (a) u D 0, 1, 1 direction vector of line determined by P1 and P2. \ P1Q u u 2, 0, 1, 1 2 x P2 P1 0, 1, 1 4 3 1 2 Q 23 4 y 2, 2 2 3 2 32 2 2, 0, 1 5. (b) The shortest distance to the line segment is P1Q 1 7. (a) V 0 2 dz 2 z2 2 1 0 1 2 1 1 2 9. (a) 2 sin Torus −3 2 z 1 Note: base altitude 2 (b) x2 a2 x2 ca At z y2 b2 2 1 2 c) z : (slice at z y2 cb 1 3 x 3 −2 y 2 (b) 2 cos Sphere −3 −2 1 2 3 x z c, figure is ellipse of area ca k cb dc abc. abc2 k 2 0 V 0 abc 1 2 abk k abk2 2 1 −2 2 3 y (c) V 1 base height 2 v w z u v zw u v w z. 11. From Exercise 64, Section 10.4, u 262 Chapter 10 u cos 0 i Vectors and the Geometry of Space sin 0 j j T 13. (a) u ui (d) 2.5 Downward force w T T cos 90 T 0 u u 1 If w sin cos 30 , u ⇒T and u 1 2 2 3 2 3 sin i T T T i sin 90 j 0 0 u 60 cos j ui j T sin i cos j (e) Both are increasing functions. (f) → lim 2 T and → lim 2 u . 1 2 T and 1 1.1547 lb 32T 0.5774 lb tan ≤ 90 and T sec . (b) From part (a), u Domain: 0 ≤ (c) T u 0 1 0 10 1.0154 0.1763 20 1.0642 0.3640 30 1.1547 0.5774 40 1.3054 0.8391 50 1.5557 1.1918 60 2 1.7321 15. Let sin For u v u , the angle between u and v. Then uv uv vu . uv cos , sin , 0 , u sin sin cos cos cos cos v sin 1 and k. cos , sin , 0 and v i cos cos j sin sin v u k 0 0 Thus, sin sin . 17. From Theorem 10.13 and Theorem 10.7 (6) we have \ D PQ n n w u u v v u u v v w u u v v w . 19. a1, b1, c1, and a2, b2, c2 are two sets of direction numbers for the same line. The line is parallel to both u and v a 2 i b2 j c2 k. Therefore, u and v are parallel, and there exists a scalar d such that u d v, a1i b1 j c1k d a2i b2 j c2k , a1 a2d, b1 b2d, c1 c2d. a1i b1j c1k ...
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