Business Calc Homework w answers_Part_82

Business Calc Homework w answers_Part_82 - 406 Section 10.2...

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Unformatted text preview: 406 Section 10.2 (d) w v 17. The vector v is horizontal and 1 in. long. The vectors u and 11 w are in. long. w is vertical and u makes a 45 angle 16 with the horizontal. All vectors must be drawn to scale. (a) v u u+v u u+v+w=0 (b) 19. u+v+w v w 32 42 5; 1 3, 4 5 3 4 , 5 5 20. 42 ( 3)2 5; 1 4, 5 1 17 3 4 , 5 3 5 15 8 , 17 17 u 21. v u uv ( 15)2 82 17; 15, 8 (c) 22. ( 5)2 1 29 ( 2)2 2 29; 5 29 5, 1 2 t , 2 29 (d) u w uw 23. x ,y (y )2 2 1 1 t 17 . 2 ; for t 1, x 1 ,y 2 2, and (x )2 Tangent: Normal: 18. The angles between the vectors is 120 and vector u is horizontal. They are all 1 in. long. Draw to scale. (a) u 1 ,2 17 2 2 1 2, 2 17 1 17 4 , , 4 17 , 1 17 . 17 1 ,y 2 24. x v uv 1 t 1 ,y (y )2 2 1; for t 5 2 3, x 1, and (x )2 (b) Tangent: Normal v . 1 5 1 2 2 5 u 1 ,1 5 2 2 5 , 2 5 , 1 5 1, , . uv+w 25. x y w 4 sin t, y 5 , and 2 5 cos t; for t (y )2 2 3, 5 2 73 . 2 3 ,x 2 3, (x )2 2 73 Tangent (c) 12 21 9 , 5 73 2u 0.811, 0.585 , Normal v 2 5 ,2 73 2 3 5 73 , 12 219 2u v 0.585, 0.811 . Section 10.2 3 2 407 26. x y 3 sin t, y 3 2 3 cos t; for t (x )2 (y )2 1 2 3 2 2 1 2 4 ,x , , and 1 3 1 3 3 3. , , 1 2 1 2 32. Since u and v are nonzero, we know that u 0 and v 0. Therefore, the dot product u v u v cos is 0 if and only if cos 0, which occurs if and only if u and v are . orthogonal 2 Tangent: Normal: , 2 3 , . 33. (r1, r2) v u (u1, u2) u+v u v 3 2 , 27. AB E BA E AB E AC E 3, 1 , BC E 3, BA E CA E 1 , CB E 1, 3 , and AC E 2, 2. (v1, v2) 1, 3 , and CA E CB E 2, 2 . 10, and 10, BC E 2 2. cos cos cos 1 Angle at A AB AC E E AB AC E E 3(2) 1( 2) ( 1 5 10)(2 2) r1 r2 v1 v2 u1 so r1 u2 so r2 u1 u2 v1 v2 1 34. (a) To find u v, place both vectors with their initial points at the origin. The vector drawn from the terminal point of v to the terminal point of u is u v. Or, add u and v according to the parallelogram law. (b) 1 63.435 , Angle at B BC E E BA cos 1 BC BA E E (u1, u2) cos cos Angle at C 1 ( 1)( 3) ( 10)( ( 3)( 1) 10) v uv u uv v (r1, r2) 1 3 5 53.130 , and CB CA E E cos 1 CB CA E E v (v1, v2) u cos cos 28. AC E 1 1( 2) ( 3(2) 10)(2 2) 1 5 1 63.435 . 2. 0, so the angle measures 90 . w2) r1 r2 ( v1) ( v2) u1 r1 u2 r2 u1 u2 v1 v2 2, 4 and BD E 2(4) 4, 35. (a) Let P 1 OP E 2 (a (a, b) and Q 1 OQ E 2 c) (b d) , 2 2 2 OP E 3 1 OP E 3 (c, d). Then 1 a, b 2 1 c, d 2 AC BD E E 4( 2) 29. (a) u (v w) u1(v1 w1) u2(v2 (u1v1 u1w1) (u2v2 u2w2) (u1v1 u2v2) (u1w1 u2w2) u v u w (b) (u OM E v) w (u1 v1)w1 (u2 v2)w2 (u1w1 v1w1) (u2w2 v2w2) (u1w1 u2w2) (v1w1 v2w2) u w v w u12 u2 2 v) v1) u22 (v12 (u2 v22 v22) v2)(u2 v2) ( u12 u22)2 u2 (b) OM E (c) OM E 1 OQ E 3 2 OQ E 3 30. u u 31. (u u1 u v) (u (u1 2 v1)(u1 v1 2 (u12 2 u22) v2 408 Section 10.2 40. The indicated diagonal is (u between the diagonal and u is (u v) u u2 v u . u v u u v u But the cosine of the angle between the diagonal and v is (1 1 1 d 35. continued (d) M is a fraction of the way from P to Q. Let d be this fraction. Then OM E d OQ E (1 d)OP. E d)PQ, E MQ. E v). The cosine of the angle Proof: PM E so PQ E dPQ and MQ E E 1 PM and PQ E E d 1 d 1 1 d (u v) v u v v2 . u v v u v v If u and v are the same length then these two quantities are equal, and the diagonal makes the same angle with both sides. 41. The slopes are the same. 42. v 0 since any other vector has positive magnitude. 25 115 north of east. 338.095, 725.046 Therefore, PM E But PM E 1 OM E d MQ. E OQ E 1 1 d OM E 1 OP E d OP and MQ E E 1 1 d OM, so E OQ E OM. E 43. 25 west of north is 90 800 cos 115 , sin 115 44. 10 east of south is 270 600 cos 280 , sin 280 Therefore, 1 OM E d 1 1 1 d(1 d) d 10 280 "north" of east. 104.189, 590.885 OM E 1 OP E d 1 OP E d OM E OM E 36. CA E v ( u u 1 OQ. E 1 d 1 OQ E 1 d 45. Initial velocity is 70 north of east: 325 cos 70 , sin 70 111.157, 305.400 . (1 d) OP E u v2 dOQ. E v. Since u2 0. Wind velocity is 130 north of east: 40 cos 130 , sin 130 25.712, 30.642 . 85.445, 336.042 . 346.735 mph. 75.734 north of east, or v and CB E v) u , these vectors are orthogonal, as v) (u Add the two vectors to get The speed is the magnitude, 336.042 The direction is tan 1 85.445 37. Two adjacent sides of the rhombus can be given by two vectors of the same length, u and v. Then the diagonals of the rhombus are (u v) and (u v). These two vectors are orthogonal since u v so (u v) (u v) u2 v 2 0. 38. Two adjacent sides of a rectangle can be given by two vectors u and v. The diagonals are then (u v) and (u v). These two vectors will be orthogonal if and only if u and v are the same length, since (u v) (u v) u2 v 2. 39. Let two adjacent sides of the parallelogram be given by two vectors u and v. The diagonals are then (u v) and (u v). So the lengths of the diagonals satisfy u v 2 (u v) (u v) u 2 2u v v2 2 and u v (u v) (u v) u 2 2u v v 2. The two lengths will be the same if and only if u v 0, which means that u and v are perpendicular and the parallelogram is a rectangle. 14.266 east of north. 46. w cos (33 Then w 15 ) 2.5 lb, so w 2.5 lb . cos 18 2.5 lb cos 33 , sin 33 cos 18 2.205, 1.432 . 47. Juana's pull 23 cos 18 , sin 18 21.874, 7.107 ; Diego's pull 18 cos ( 15 ), sin ( 15 ) 17.387, 4.659 . Add to get the combined pull of the children: 39.261, 2.449 . The puppy pulls with an opposite force of the same magnitude: 39.2612 2.4492 39.337 lb. 48. (a) 7 cos 45 , sin 45 has its terminal point at (4.950, 4.950). (b) 7 cos 45 , sin 45 terminal point at 49. AB E 50. AB E 2, 3 3 2 5 1, 0, 4 ( 4), 4 0 2 1 3 CD E 8 cos 210 , sin 210 has its ( 1.978, 0.950). 3, 4 1 4, 5 1 CD E Section 10.3 51. u u1, u2 , v (i) u (ii) v v1, v2 , w w1, w2 dy dx dy dx dy/dt dx/dt dy/dt dx/dt 5 cos t , which for t 4 sin t 5 cos t , which for t 4 sin t 409 3. 4. (iii) (iv) (v) (vi) (vii) (viii) (ix) u1 v1, u2 v2 v1 u1, v2 u2 v u (u v) w u1 v1, u2 v2 w1, w2 (u1 v1) w1, (u2 v2) w2 u1 (v1 w1), u2 (v2 w2) u (v w) u 0 u1, u2 0, 0 u1 0, u2 0 u1, u2 u u ( u) u1 , u 2 u1, u2 u1 u1, u2 u2 0, 0 0 0u 0 u1, u2 0u1, 0u2 0, 0 0 1u 1 u1, u2 1u1, 1u2 u1, u2 u a(bu) a(b u1, u2 ) a bu1, bu2 abu1, abu2 ab u1, u2 (ab)u a(u v) a u1 v1, u2 v2 au1 av1, au2 av2 au1, au2 av1, av2 a u1, u2 a v1, v2 au av (a b)u (a b) u1, u2 (a b)u1, (a b)u2 au1 bu1, au2 bu2 au1, au2 bu1, bu2 au bu 1 . Then their 3, a 1 1 , . 2 2 2 equals zero. is undefined: the tangent line is vertical. 5. dy dx dy/dt dx/dt 6 5 cos t , which for t 4 sin t 6 equals 5 3 . 4 Also, at t ,x 2 5 2 3 and y 5 4 3 5 . The equation for the 2 tangent line is y y 6. dy dx 5 3 x 4 dy/dt dx/dt 6 (x 2 3), or 10. 5 cos t , which for t 4 sin t 6 equals 5 3 . 4 Also, at t ,x 2 5 2 3 and y 4 5 3 5 . The equation for the 2 normal line is y y 7. lim 8. y x 4 15 2 4 3 (x 2 3), or x 9 . 10 2 x2 x x2 (x lim x 2 2)(x 2) 2 x2 x lim 1 2 1 4 3 2x, and Length 0 1 3.400. (3 2x)2 dx, which 52. Write the two vectors as a 1, 1 and b 1, sum is a a 7 ,b 2 using NINT evaluates to b, a b , so solve a 7 7 , 2 2 b b 4 to get 9. x t cos t 2 1 . So 3, 4 2 1 1 sin t, and y (t cos t sin t)2 t sin t cos t, and cos t)2 dt 53. (a) Slope y 1 1, so y 2) or y y1 x m(x 1. x1) becomes Length 0 2 0 ( t sin t (x t2 1 dt, 2.958. (b) Slope y 1 1, so y y1 m(x x 2 or y x 3. x1) becomes which using NINT evaluates to 10. y 2 y 54. The slopes of the lines are 3 and 1, which means that 4 xe x e x C (use integration by parts), so 0 e0 C and C 3: xe x e x 3 vectors 4, 3 and 1, 1 are parallel to the respective lines. cos 1 Section 10.3 Exercises 1. [5 ( 1)]i 3)i [0 (0 [0 (1 4)j 6i 3i 3j 4j 3i 4i i 7i 2j 6j 3(4)j] i 16j 2j and 3j. j 5j 2. (0 3. E AB CD E (a) [3 (b) [3 4. (a) (5 (b) (5 ( 4)]j 4 1 3 1 cos 1 7 5 2 2 10 8.130 . s Section 10.3 Vector-valued Functions (pp. 529539) Quick Review 10.3 1. f (x) 4 x x2 1 3 1 3 ( 3)]i (2 0)j 4)i ( 3 0)j ( 4)]i ( 4)]i 3)i 3)i [2 [2 ( 3)]j ( 3)]j 4]j 4]j 15i [3(3)i 8i 2i 6j [( 2) [( 2) 3( 2)j 2( 2)j] , so for x 3 1, f (x) 1 3 3 and (x 1) or (c) 3(5)i (d) [2(5)i f (x) y 2. f (x) f (x) . Then y x x 4 3 x2 . 1, f (x) 3 3(x 3 and 1) or y 3x. , so for x 1 4 3 . Then y 410 5. (a) Section 10.3 8. (a) [ 6, 6] by [ 4, 4] [ 6, 6] by [ 3, 5] (b) v(t) d (2 cos t)i dt d (3 sin t)j dt (b) v(t) ( 2 sin t)i a(t) d ( 2 sin t)i dt (3 cos t)j d (3 cos t)j dt ( 2 cos t)i (c) v (3 sin t)j ( 2)2 1, 0 02 2, a(t) (c) v(1) d d 2 (2 ln (t 1))i (t )j dt dt 2 i (2t)j t 1 d 2 d 2 i (2t)j i dt t 1 dt (t 1)2 2j 2 2, 0 ; speed 1 2 1, 2 ; speed 1 5 12 1 5 22 , 2 5 5, direction (d) Velocity 6. (a) 2, 0 1, 0 direction (d) Velocity 9. v(t) (cos t)i 1, 2 1 5 2 5 (2t , 2 5 sin t)j, r(0) j and v(0) i. So the slope is zero (the velocity vector is horizontal). [ 4.5, 4.5] by [ 3, 3] (a) The horizontal line through (0, (b) The vertical line through (0, 10. v(t) r v 22 2, 4 4 1): y 1): x 0. 1. (b) v(t) d (cos 2t)i dt d (2 sin t)j dt ( 2 sin 2t)i a(t) d ( 2 sin 2t)i dt (2 cos t)j d (2 cos t)j dt ( 2 sin t)i ( 2 ( 3 2 3 x 2 3 2 2 x 3 (3 cos t)j, 3 2 3 2 3)i 2)i 1 j and j. So the slope is 3/ 2 2 3 . 2 ( 4 cos t)i (c) v(0) 0, 2 ; speed 1 0, 2 2 (2 sin t)j 02 0, 1 (a) y y 1 6 2 2 direction (d) Velocity 7. (a) 3 [x 2 7 ( 2 3)] or 2 0, 1 (b) y y [ 6, 6] by [ 4, 4] 2 1 2 [x 3 ( 2 3)] or 5 2 18 6 2 (b) v(t) a(t) d d (sec t)i (tan t)i (sec t tan t)i dt dt d d (sec t tan t)i (sec2 t)j dt dt 11. (sec2 t)j 1 (6 6t 3i /4 6t) dt i 1 3 2t 3/2 j 1 2 t dt j 3t 2 i 1 2 (sec t tan2 t sec3 t)i (2 sec2 t tan t)j 2 5 , 3 (4 2 2)j /4 (c) v 2 4 , ; speed 3 3 3 2 4 , direction 2 5 3 3 6 2 5 3 1 5 2 2 4 2 3 3 1 2 12. sin t dt i /4 (1 /4 cos t) dt j /4 , /4 5 2 5 5 cos t 2 j i /4 t sin t j /4 (d) Velocity , 2 ...
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