Business Calc Homework w answers_Part_87

Business Calc Homework w answers_Part_87 - Section 10.6 22....

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Unformatted text preview: Section 10.6 22. 25. 431 [ 4.7, 4.7] by [ 3.1, 3.1] [ 6, 6] by [ 4, 4] Use the symmetries of the graphs: the shared area is /2 The area in question is half the circle minus two lobelike regions: 1 (2)2 2 /2 4 0 1 [2(1 2/2 cos )]2 d 2 cos 1 2 2 0 /2 8 0 (1 cos 2 )d /2 1 [2(1 2 sin )]2 d 4 sin2 ) d 1 2 1 sin 2 4 /2 8 23. For a 2 sin 1: 1 sin 2 4 2 6 16 2 26. 0 (4 4 8 sin 2 cos 0 8 0 [ 3, 3] by [ 2, 2] [ 3, 3] by [ 2, 2] (a) To find the integration limits, solve The curves intersect at the origin and when 3a cos 2 cos 3 2 cos 2 . 3 1 0 a(1 1 . cos ) Because of the curve's symmetry, the area inside the outer loop is 2 /3 Use the symmetries of the curves: the area in question is /3 2 0 1 (2 cos 2 1)2 d 4 cos 4 sin 0 2 /3 2 0 1 [(3a cos )2 2 /3 0 a (1 1 2 cos ) ] d cos2 ) d 1) d /3 2 (4 cos2 sin 2 1) d 2 /3 0 a2 a 2 0 (9 cos2 (8 cos 2 2 cos 2 3 2 3 /3 2 cos 2 sin 2 a2 4 24. 2 sin 2 a2 0 (b) Again, use the curve's symmetry. The inner loop's area is 2 1 2 /3 2 (2 cos sin 2 1)2 d 4 sin 2 /3 2 [ 3, 3] by [ 2, 2] 3 3 . 2 Subtract this from the answer in (a) to get 3 3 . The curves intersect when 6 cos 2 6 /6 3 or 5 . 6 /6 Use the symmetries of the curves. The area in question is 4 0 1 (6 cos 2 2 3) d 6 sin 2 0 3 3 . 432 27. Section 10.6 30. The integral given is incorrect because r out the circle twice as cos sweeps goes from 0 to 2 . Or, you can't use equation (2) from the text on the interval [0, 2 ] [ 9, 9] by [ 6, 6] because r cos is negative for To find the integration limits, solve 3 csc , 6 3 5 . The correct area is , which can be found 2 2 4 3 by computing the areas of the cardioid and the circle 2 4 5 . The area in question is 6 6 5 /6 1 2 (6 32 csc2 ) d /6 2 5 /6 1 36 9 cot 2 /6 separately and subtracting. dr d 31. 2 , so 5 Length 0 5 ( 2)2 2 (2 )2 d 4d 5 0 12 28. 9 3 0 1 2 ( 3 1 (27 3 [ 3, 3] by [-2, 2] 4)3/2 8) 19 3 To find the intersection points, solve 6 cos 2 48 cos4 cos 2 32. dr d e 2 , so e 0 9 sec2 4 2 Length 9 0 0 e 2 2 d 2 24 cos2 3 4 e d e 0 e 1 6 /6 . 33. dr d By the symmetry of the curves, the area in question is 2 0 sin , so 2 1 6 cos 2 2 3 sin 2 9 tan 4 9 sec2 4 /6 0 d 3 4 3 Length . 0 2 (1 2 0 2 cos )2 2 cos 4 cos2 2 2 ( sin )2 d d , 29. (a) Find the area of the right half in two parts, then double the result: Right half area /4 0 2 0 2 2 2d 1 tan2 2 /2 d /4 1 1 csc2 2 2 /2 d 2 cos 0 d d 8 1 tan 2 1 1 2 4 /4 0 1 4 cot /4 4 0 cos 2 1 (0 4 1) 3 2 3 4 4 8 . 34. dr d 8 sin 0 Total area is twice that, or (b) Yes. x y . a sin 2 0 cos 2 , so 2 tan sin x 1, cos y 1, x sin2 cos sin Length a 0 a2 sin4 sin2 cos 2 2 a2 sin2 d 2a 0 2 cos2 2 d tan lim /2 lim /2 y 2a /2 lim x /2 lim y Section 10.6 dr d 6 sin , so cos )2 /2 0 433 35. 41. 62 cos )2 62 sin2 d , which using (1 cos )4 [ 1.5, 1.5] by [ 1, 1] 3 (1 Length (1 NINT evaluates to 6.887. 2 (Note: the integrand can simplify to 3 sec dr 36. d 2 sin , so (1 cos )2 /2 .) 2r dr d dr d 2 sin 2 sin 2 cos 2 . Use the curve's symmetry and note that r is defined for 22 sin2 d , which using (1 cos )4 Length (1 22 cos )2 0 /4 4 : Surface area cos 2 sin d /4 NINT evalutes to 2.296. 3 2 0 /4 2 0 cos 2 sin2 2 d cos 2 (Note: the integrand can simplify to csc dr d 2 ). 2 4 42. For a 2 sin cos 0 (4 2 2) 3.681 37. cos2 3 /4 sin cos6 cos4 3 3 3 , so cos4 d /4 0 8 3 3 1: Length 0 /4 0 sin2 3 d 1 2 dr 38. d cos 2 1 sin 2 2 3 2 sin 4 3 . dr d [ 3, 3] by [ 2, 2] 2a sin , so surface area /2 , so cos2 2 1 sin 2 cos 2 1 2 2 0 2 (2a cos ) cos /2 0 4a2 cos2 /2 4a2 sin2 d Length 0 2 0 1 sin 2 d 2 16a2 16a2 43. dx 2 d 1 2 cos2 d 4a2 2 0 2 (1 sin 2 ) d sin 2 1 sin 2 4 2d 0 2 . sin 2 c os 2 2 dy 2 d ( f ( ) cos , so f ( ) sin )2 f ( ) cos )2 ( f ( ) sin )2 ( f ( ) sin )2 39. dr d 1 2 /4 ( sin 2 )(2) ( f ( ) sin ( f ( ) cos )2 sin 2 c os 2 cos 2 Surface area 2 0 /4 cos 2 cos cos cos2 2 d 2 0 ( cos 2 )2 sin2 2 d ( f ( ) cos )2 ( f ( ))2(cos2 sin2 ) sin2 ) r2 cos ) d a 2 /2 d 2 0 /4 ( f ( ))2(cos2 ( f ( ))2 ( f ( ))2 2 2 0 cos /4 dr 2 d a 2 2 2 dr d sin 2 2 /2 4.443. 44. (a) 1 2 1 2 0 1 ( 0 a(1 0 2 sin 0 a 40. e /2 , so surface area /2 2 (b) /2 2 ad 0 a 0 2 0 /2 2e sin 5d ( 2e ) 2 2 2 (e /2 2 ) d (c) /2 2 e sin 0 /2) a cos /2 d a /2 sin /2 2a 2 5 5 (e 1 e (sin 2 /2 /2 cos ) 0 45. If g( ) 2f ( ), then (g( )2 g ( )2) 2 ( f ( )2 g is 2 times the length of f. f ( )2), so the length of 1) 40.818 434 Chapter 10 Review 46. If g( ) 2f ( ), then (g( )2 g ( )2) 2 g( ) sin 4[2 f ( ) sin ( f ( )2 f ( )2)], so the area generated by g is 4 times that of f. 47. (a) Let r 1.75 0.06 . 2 50. 2 3 0 a3 sin 0 d 2 3 a 3 cos 0 4 3 a , and 3 a2 d 4 3 a 3 a2. So 4a . By symmetry, x 3 dr (b) Since d b , this is just Equation 4 for the 2 y length of the curve. 80 a2 4a 0, . 3 0, so the centroid is (c) Using NINT, 0 1.75 741.420 cm 0.06 2 2 0.06 2 d 2 s Chapter 10 Review Exercises (pp. 569572) 1. (a) 3 3, 4 4 2, 17, 32 172 3 1 6 2 evaluates to 7.414 m. b 2 r 2 1/2 5 1313 9 8, 12 20 (d) r 2 since b 2 1/2 2 b 2 r 2 r 1 r (b) 2. (a) (b) 3. (a) (b) (b) 322 2, 4 1 8 2 5 2 2(4) 10 10, 725 1, 6, 25 5 8 1 is a very small quantity squared. 2( 3), 2 2 (e) L La 741.420 cm (from part (c)), 80 0 1.75 0.03 2 0.06 2 2 80 4. (a) 5(2), 5( 5) d 236 741.416 cm 5. 10 2 25 2 29 3 2 1.75 0 6 radians below the negative x-axis: , 1 2 48. (a) Use the approximation, La, from 47(e). If the reel has made n complete turns, then the angle is 2 n. So from the integral, La n r0 b bL r02 [assuming counterclockwise]. 6. 7. 2 8. 5 dy dx 3 1 , 2 1 42 12 1 (3/5)2 dy/dt dx/dt 3 (4/5)2 2 bn 2 1 2 r0n. Solving for n gives 1 . 4, 1 3 4 , 5 5 8 17 , 2 17 (b) The take up reel slows down as time progresses. (c) Since L is proportional to time, the formula in part (a) shows that n will grow roughly as the square root of time. 2 49. 3 2 0 3, 4 9. (a) (1/2)sec t tan t (1/2)sec2 t 3 2 sin t dy dx 3 2 3 2 For t a (1 2 3 a 3 2 ;x ,y 1, and 3 2 . So the 3 cos ) cos (cos 3 cos2 15 sin 8 1 cos3 4 3 d 3 cos3 cos 2 tangent line is y cos4 ) d cos2 sin y (b) d 2y dx 2 3 2 1 x or 0 x 1 . 4 dy /dt dx/dt cos t (1/2)sec2 t 1 equals . 3 4 3/t 2 2/t 3 5 ,y 4 2 3 a 3 sin 3 15 8 dy dx 2 cos3 t, sin 0 which for t 10. (a) dy dx dy/dt dx/dt 5 3 a , and 2 2 0 a2(1 cos )2 d cos2 ) d 1 sin 2 4 2 0 a2 a2 2 (1 0 2 cos 2 sin 3 2 3 a2. (b) 5a ,0 . 6 3 t 2 1 dy For t 2: x , and 2 dx 1 So the tangent line is y 3 x 2 13 y 3x . 4 d 2y dx 2 dy dx dy /dt dx/dt 3/2 2/t 3 3. 5 or 4 So x 5 3 a 2 3 a2 5a . 6 3t 3 , which for t 4 2 equals 6. By symmetry, y 0, so the centroid is Chapter 10 Review 1 tan t sec t equals zero for t k , where k is any 2 dx 1 integer. sec2 t never equals zero. dt 2 dy dt 435 11. 17. (a) (a) Horizontal tangents at 1 tan 2 1 1 tan 0, sec 0 2 2 0, 1 and 2 [ 1.5, 1.5] by [ 1, 1] , 1 sec 2 0, 1 . 2 dx never equals dt (b) 2 18. (a) (b) There are no vertical tangents, since zero. 12. dy dt 2 cos t equals zero for dx dt k , where k is any odd 2 [ 3, 3] by [ 2, 2] (b) 19. (a) integer. integer. 2 sin t equals zero for t k , where k is any (a) Horizontal tangents at and 2 cos 3 3 , 2 sin 2 2 2 cos 2 , 2 sin 2). 2 (0, 2) [ 1.5, 1.5] by [ 1, 1] (0, (b) ( 2, 0) and k , 2 (b) Vertical tangents at ( 2 cos 0, 2 sin 0) ( 2 cos , 2 sin ) (2, 0). 13. dy dt 2 20. (a) 2 sin t cos t sin 2t equals zero for t dx dt where k is any integer. sin t equals zero for t k , [ 1.5, 1.5] by [ 1, 1] where k is any integer. Where they are both zero, use L'Hpital's rule: dy/dt tk dx/dt (b) 21. dy dx f ( ) sin f ( ) cos f ( ) cos f ( ) sin cos 2 cos cos 2 sin lim lim tk sin 2t sin t lim tk 2 cos 2t cos t 2. (0, 0). (a) Horizontal tangent at cos 2 , cos 2 2 sin 2 sin 2 sin 2 cos 2 (b) There are no vertical tangents. 14. dy dt 0, dy dx dy dx 4 , 0, 3 5 , 0, 4 4 2/ 2 and 0, dy dx 7 4 are polar solutions. 2/ 2 9 cos t equals zero for t k , where k is any odd 2 /4 1, 1, dx integer. dt 4 sin t equals zero for t k , where k is 2/ 2 2/ 2 2/ 2 3 /4 1, 1. 5 /4 any integer. (a) Horizontal tangents at 4 cos 4 cos 3 3 , 9 sin 2 2 2 dy dx 2/ 2 2 2 2 2 7 /4 , 9 sin 2 (0, 9) and 22. (4, 0) and The Cartesian equations are y dy dx f ( ) sin f ( ) cos 2 sin 2 sin 2 sin 2 cos 4 sin2 cos 4 cos2 sin 4 sin2 6 cos f ( ) cos f ( ) sin (1 (1 cos 2 )cos cos 2 ) sin x. (0, 9). (b) Vertical tangents at (4 cos 0, 9 sin 0) (4 cos , 9 sin ) ( 4, 0). 15. cos 2 cos3 cos sin 2 cos2 sin sin 2 cos2 sin [ 7.5, 7.5] by [ 5, 5] 4 sin2 2 cos2 . 3 sin 2 16. 0, dy dx [ 7.5, 7.5] by [ 5, 5] 2 and 0, dy dx 3 2 are polar solutions. 4 is undefined, so the tangent lines 0 /2 3 /2 are vertical with equation x 0. ...
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