Solucionario larson (varias variables)

  • View
    76

  • Download
    26

Embed Size (px)

Text of Solucionario larson (varias variables)

  1. 1. www.elsolucionario.net www.elsolucionario.net
  2. 2. C H A P T E R 1 0 Vectors and the Geometry of Space Section 10.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . 474 Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 479 Section 10.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 483 Section 10.4 The Cross Product of Two Vectors in Space . . . . . . . . 487 Section 10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 491 Section 10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 496 Section 10.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 499 Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503 Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507 www.elsolucionario.net www.elsolucionario.net
  3. 3. C H A P T E R 1 0 Vectors and the Geometry of Space Section 10.1 Vectors in the Plane Solutions to Even-Numbered Exercises 474 2. (a) (b) x 12 6 5 4 3 2 1 3 2 31 v (0, 6) y v 3 3, 2 4 0, 6 4. (a) (b) x 3 12 3 2 1 ( 3, 2) v y v 1 2, 3 1 3, 2 6. u v v 7 2, 7 1 5, 8 u 1 4, 8 0 5, 8 8. u v v 25 0, 10 13 15, 3 u 11 4, 4 1 15, 3 10. (b) (a) and (c). v (1, 12) (3, 6) (2, 6) 1 4 2 4 6 8 10 12 6 1 2 3 4 5 6 7 y x v 3 2, 6 6 1, 12 12. (b) (a) and (c). x v 6 4 2 2 4 2 2 ( 5, 3) (0, 4) ( 5, 1) y v 5 0, 1 4 5, 3 14. (b) (a) and (c). x v 86422468 3 2 1 2 3 ( 10, 0) (7, 1)( , 1)3 y v 3 7, 1 1 10, 0 16. (b) (a) and (c). x v 1.000.750.25 1.250.50 1.25 1.00 0.75 0.50 0.25 (0.72, 0.65) (0.84, 1.25) (0.12, 0.60) y v 0.84 0.12, 1.25 0.60 0.72, 0.65 18. (a) CONTINUED x v 4v 12 20 844812 ( 4, 20) ( 1, 5) y 4v 4, 20 (b) x v v 43 2 2 2 1 5 11234 1 2 3 ( 1, 5) , ( ( y 1 2 v 1 2 , 5 2 www.elsolucionario.net www.elsolucionario.net
  4. 4. Section 10.1 Vectors in the Plane 475 18. CONTINUED (c) x v 0v 3 6 21123 ( 1, 5) y 0v 0, 0 (d) x v 15105510 10 15 20 25 30 15 ( 1, 5) (6, 30) 6v y 6v 6, 30 20. Twice as long as given vector x u 2u y u. 22. x u u v+ 2 2v y 24. (a) (b) (c) 2u 5v 23, 8 58, 25 34, 109 v u 8, 25 3, 8 11, 33 2 3u 2 33, 8 2, 16 3 26. x 2 1 1 v w u 321 y 3i j 3, 1 v 2i j i 2j 28. 4 2 2 6 8 10 12 4 6 8 10 5u 3w v y x v 5u 3w 52, 1 31, 2 7, 11 30. Q 7, 7 u2 7 u1 7 u2 2 9 u1 3 4 32. v 144 25 13 34. v 100 9 109 36. v 1 1 2 38. unit vectorv u u 5, 15 510 1 10 , 3 10 u 52 152 250 510 40. unit vectorv u u 6.2, 3.4 52 1.24 2 , 0.68 2 u 6.22 3.42 50 52 www.elsolucionario.net www.elsolucionario.net
  5. 5. 476 Chapter 10 Vectors and the Geometry of Space 42. (a) (b) (c) (d) (e) (f) u v u v 1 u v u v 1 13 3, 2 v v 1 v v 1 32 3, 3 u u 1 u u 0, 1 u v 9 4 13 u v 3, 2 v 9 9 32 u 0 1 1 u 0, 1, v 3, 3 44. (a) (b) (c) (d) (e) (f) u v u v 1 u v u v 1 52 7, 1 v v 1 v v 1 52 5, 5 u u 1 u u 1 25 2, 4 u v 49 1 52 u v 7, 1 v 25 25 52 u 4 16 25 u 2, 4, v 5, 5 46. u v u v u v 2 u v 2, 0 v 5 2.236 v 1, 2 u 13 3.606 u 3, 2 48. v 22, 22 4 u u 22 1, 1 u u 1 2 1, 1 50. v 0, 3 3 u u 0, 3 u u 1 3 0, 3 52. 5 2 i 53 2 j v 5cos 120i sin 120j 54. 0.9981i 0.0610j 0.9981, 0.0610 v cos 3.5i sin 3.5j 56. u v 5i 3j v i 3j u 4i 58. u v 10cos0.5i v 5cos0.5i 5sin0.5j 5cos0.5i 5sin0.5j u 5cos0.5i 5sin0.5j 60. See page 718: u ku (ku1, ku2) (u1, u2) u1 ku1 u2 ku2 u v u + v (u1 + v1, u2 + v2) (v1, v2) (u1, u2) u1 u2 v1 v2 www.elsolucionario.net www.elsolucionario.net
  6. 6. 62. See Theorem 10.1, page 719. For Exercises 6468, au bw ai 2j bi j a bi 2a bj. 64. Therefore, Solving simultaneously, we have a 1, b 1. 2a b 3.a b 0,v 3j. 66. Therefore, Solving simultaneously, we have a 2, b 1. 2a b 3.a b 3,v 3i 3j. 68. Therefore, Solving simultaneously, we have a 2, b 3. 2a b 7.a b 1,v i 7j. 70. at (a) Let then (b) Let then w w 1 145 12, 1. w 12, 1,m 1 12 . w w 1 145 1, 12. w 1, 12,m 12. x 2.y x3, y 3x2 12 72. (a) Let then (b) Let then w w 1 5 2, 1. w 2, 1,m 1 2. w w 1 5 1, 2. w 1, 2,m 2. fx sec2 x 2 at x 4 . f x tan x 74. v u v u 3 23i 33 2j u v 3i 33j u 23i 2j 76. magnitude direction 8.26 63.5 78. R F1F2F3 163.0 R F1 F2 F3 4.09 F3 3, F3 200 F2 4, F2 140 F1 2, F1 10 80. tan 250 1002 2503 1002 10.7 F1 F2 2503 10022 250 10022 584.6 lb 2503 1002i 250 1002j F1 F2 500 cos 30i 500 sin 30j 200 cos45i 200 sin45j Section 10.1 Vectors in the Plane 477 82. R arctan200 3152 2003 352 0.6908 39.6 R 2003 3522 200 31522 385.2483 newtons 2003 1402 1752i 200 1402 1752j 350cos135i sin135j 280cos45i sin45jF1 F2 F3 400cos30i sin30j www.elsolucionario.net www.elsolucionario.net
  7. 7. 478 Chapter 10 Vectors and the Geometry of Space 84. (a) (b) 0 0 40 2 500 400 cos 400 400 cos 100 cos2 100 sin2 F1 F2 20 10 cos , 10 sin F1 20, 0, F2 10cos , sin (c) The range is The maximum is 30, which occur at and The minimum is 10 at (d) The minimum of the resultant is 10. . 2. 0 10 F1 F2 30. 86. P2 1, 2 22, 1 5, 4 P1 1, 2 2, 1 3, 3 1 3 u 2, 1 u 7 1, 5 2 6, 3 88. Vertical components: Horizontal components: Solving this system, you obtain and v 3611.2.u 2169.4 u cos 1 v cos 2 0 u sin 1 v sin 2 5000 v vcos 2 i sin 2 j u ucos 1 i sin 1 j 2 arctan 24 10 1.9656 or 112.6 1 arctan24 20 0.8761 or 50.2 A B C v u y x 1 2 90. To lift the weight vertically, the sum of the vertical components of u and v must be 100 and the sum of the horizontal components must be 0. Thus, or And or u1 2 v cos 110 0 u cos 60 v cos 110 0 u3 2 v sin 110 100. u sin 60 v sin 110 100, v v cos 110i sin 110j u u cos 60i sin 60j 100 lb 20 30 u v Multiplying the last equation by and adding to the first equation gives Then, gives (a) The tension in each rope: (b) Vertical components: v sin 110 61.33 lb. u sin 60 38.67 lb. u 44.65 lb, v 65.27 lb. u 44.65 lb. u1 2 65.27 cos 110 0 usin 110 3 cos 110 100 v 65.27 lb. 3 www.elsolucionario.net www.elsolucionario.net
  8. 8. Section 10.2 Space Coordinates and Vectors in Space 479 92. Direction North of East: Speed: 336.35 mph N 84.46 E tan 35.36 364.64 5.54 u v 400 252i 252j 364.64i 35.36j v 50cos 135i sin 135j 252i 252j wind u 400iplane 94. v sin2 cos2 1 u cos2 sin2 1, 96. Let u and v be the vectors that determine the parallelogram, as indicated in the figure. The two diagonals are and Therefore, But, Therefore, and Solving we have u s r v x y 1 2 .x y 0.x y 1 x yu x yv. xu v yv u u r s s yv u.r xu v,v u.u v 98. The set is a circle of radius 5, centered at the origin. u x, y x2 y2 5 x2 y2 25 100. True 102. False a b 0 104. True Section 10.2 Space Coordinates and Vectors in Space 2. x y (3, 2, 5) 2 3 , 4, 2( ( 8 z 4. x y (4, 0, 5) (0, 4, 5) 8 z 6. B3, 1, 4 A2, 3, 1 8. 7, 2, 1 x 7, y 2, z 1: 10. x 0, y 3, z 2: 0, 3, 2 12. The x-coordinate is 0. 14. The point is 2 units in front of the xz-plane. 16. The point is on the plane z 3. 18. The point is behind the yz-plane. www.elsolucionario.net www.elsolucionario.net
  9. 9. 480 Chapter 10 Vectors and the Geometry of Space 20. The point is in front of the plane x 4. 22. The point is 4 units above the xy-plane, and above either quadrant II or IV. x, y, z 24. The point could be above the xy-plane, and thus above quadrants I or III, or below the xy-plane, and thus below quadrants II or IV. 26. 16 64 16 96 46 d 2 22 5 32 2 22 28. 4 49 9 62 d 4 22 5 22 6 32 30. Since the triangle is isosceles.AB AC, BC 16 16 0 42 AC 4 4 1 3 AB 4 4 1 3 A5, 3, 4, B7, 1, 3, C3, 5, 3 32. Neither BC 0 4 9 13 AC 25 0 9 34 AB 25 4 0 29 A5, 0, 0, B0, 2, 0, C0, 0, 3 34. The y-coordinate is changed by 3 units: 5, 6, 4, 7, 4, 3, 3, 8, 3 36. 4 8 2 , 0 8 2 , 6 20 2 6, 4, 7 38. Center: Radius: 5 x2 y2 z2 8x 2y 2z 7 0 x 42 y 12 z 12 25 4, 1, 1 40. Center: (tangent to yz-plane) x 32 y 22 z 42 9 r 3 3, 2, 4 42. Center: Radius: 109 2 9 2 , 1, 5 x 9 2 2 y 12 z 52 109 4 x2 9x 81 4 y2 2y 1 z2 10z 25 19 81 4 1 25 x2 y2 z2 9x 2y 10z 19 0 44. Center: Radius: 3 1 2 , 4, 1 x 1 2 2 y 42 z 12 9 x2 x 1 4 y2 8y 16 z2 2z 1 33 4 1 4 16 1 x2 y2 z2 x 8y 2z 33 4 0 4x2 4y2 4z2 4x 32y 8z 33 0 www.elsolucionario.net www.elsolucionario.net
  10. 10. Section 10.2 Space Coordinates and Vectors in Space 481 46. Interior of sphere of radius 4 centered at 2, 3, 4. x 22 y 32 z 42 < 16 x2 4x 4 y2 6y 9 z2 8z 16 < 4 9 16 13 x2 y2 z2 < 4x 6y 8z 13 48. (a) (b) x y 4, 5, 2 8 z 4i 5j 2k 4, 5, 2 v 4 0i 0 5j 3 1k 50. (a) (b) x y 0, 0, 44 z 4k 0, 0, 4 v 2 2i 3 3j 4 0k 52. Unit vector: 5, 12, 5 194 5 194 , 12 194 , 5 194 5, 12, 5 25 144 25 194 1 4, 7 5, 3 2 5, 12, 5 54. Unit vector: 1 73 , 6 73 , 6 73 1, 6, 6 1 36 36 73 2 1, 4 2, 2 4 1, 6, 6 56. (b) (a) and (c). 6i 4j 9k 6, 4, 9 v 4 2i 3 1j 7 2k x y 12 ( 4, 3, 7) ( 6, 4, 9) (2, 1, 2) z 58. Q 1, 8 3, 3 q1, q2, q3 0, 2, 5 2 1, 2 3, 1 2 60. (a) (c) x y 2 2 1 1, 1, z 1 2 v 1, 1, 1 2 x y 4 2, 2, 1 z v 2, 2, 1 (b) (d) x y 8 2 5 5, 5, z 5 2 v 5, 5, 5 2 x y 8 4, 4, 2 z 2v 4, 4, 2 www.elsolucionario.net www.elsolucionario.net
  11. 11. 482 Chapter 10 Vectors and the Geometry of Space 62. 7, 0, 4z u v 2w 1, 2, 3 2, 2, 1 8, 0, 8 64. z 5u 3v 1 2 w 5, 10, 15 6, 6, 3 2, 0, 2 3, 4, 20 66. z 0, 2, 3 9 3z3 0 z3 3 6 3z2 0 z2 2 0 3z1 0 z1 0 0, 6, 9 3z1, 3z2, 3z3 0, 0, 0 2u v w 3z 21, 2, 3 2, 2, 1 4, 0, 4 3z1, z2, z3 0, 0, 0 68. (b) and (d) are parallel since and 3 4 i j 9 8 k 3 2 1 2 i 2 3 j 3 4 k.i 4 3 j 3 2 k 21 2 i 2 3 j 3 4 k 70. (b) is parallel since zz 14, 16, 6. z 7, 8, 3 72. Therefore, and are parallel. The points are collinear. PR P