Business Calc Homework w answers_Part_71

Business Calc Homework w answers_Part_71 - 351 Section 8.4...

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Unformatted text preview: 351 Section 8.4 1 34. Volume 2x 2 1)(2 (x 0 1 4 0 x 1)(2 (x x) B 2x A x) x x 1 A(2 x) (A 5 sin , dt 25 x dx 1)(2 (x 36. t dx x) 25 25 sin t 2 dt 25 B(x B)x t 2 1) (2A B 1 and 2A B B) 1 x dx 1)(2 (x 0 4 3 1 0 1 0 ln x 35. y y2 9 25 sin 2 C cos C 25 t 25 t 25 2 25 5 t2 Use Figure 8.18(b) from the text with a 0 get 3 sec2 d, 9 tan2 t2 C C 5 and x t to x 4 ln 2 3 2 ln 2) 2 t2 25 7 sec , dx 2 37. x 2 9 sec2 3 sec y 2 4x 2 5 cos . 7 sec 2 49 sec 2 49 d dx 3 sec sec ln 4x 2 d ln sec tan 49 d ,0 2 49 tan 2 tan 1 2x ln 2 7 4x 2 7 1 ln 2x 2 4x 2 49 1 ln 7 2 1 ln 2x 2 4x 2 49 C C1 9 y2 y ln 3 ln 9 y2 y C Use Formula 88 for sec d with x Figure 8.18(a) from the text with a y2 3 sec . d 1 ln sec 2 y 3 3 7/2 sec tan 7 tan 1 sec 2 y2 9 49 C1 tan ln 9 25 sin 2 4 2 dy 9 cos 2 d 2 1 2 ln 2 0 3 tan , dy 9 2 dx 2x 1 4 ( ln 2 3 1 d 25 2 2 . 3 2 (5 cos ) d 25 t sin 1 2 5 25 t sin 1 2 5 1 4 3 1 ,B 3 2 25 cos 25 2 0 x) dx x1 2 2 5 cos 25 Solving the system simultaneously yields A 4 d, 25 cos 2 Equating coefficients of like terms gives A 5 cos C1 and a 1. Use 3 to get d C1 49 Use Figure 8.18(c) from the text with a x2 49 4 7 tan . 2 C1 7 to get 2 C1 352 Section 8.4 38. x sec 2 tan , dx x2 tan 2 1 x2 2 40. x 2 x sec 2 1 sec , dx 2 1 1 sec sec 2 x3 x2 tan 2 cos 2 1 2 1 csc cos 2 d 1 sin 2 2 C x2 1 sec Use Figure 8.18(a) from the text with a 39. x 1 sin , dx x2 1 1 . 41. z cos d, sin 2 2 2 3 cos sin 3 (1 d d cos 2 ) sin d 1 x x2 1 3 C 1 x2 1 (1 3 2 C x) 1 x2 3 Use Figure 18.8(b) from the text with a 1 x2 cos . C C 1 4 ln 4 sin ) d cot 4 cos 4 z 16 z z2 4 16 z z2 from the text with a 4 2 sin , dw w2 4 4 z2 16 4 4 z2 16 4 sin 2 z2 C . Use Figure 8.18(b) 16 4 and cos d, 2 z2 . 2 4 cos 2 8(2 cos ) d 4 sin 2 2 cos w2 2 csc 2 d 2 cot 2 4 w C w 2 C Use Figure 8.18(b) from the text with a get cot C 4 to get 2 cos 8 dw w2 C 1 and x 16 z 4 , cot z 42. w 16 cos 2 d Use Formula 89 with a csc 2 d 4 sin2 sin 4 ln d ,0 C (4 cos ) d 4 sin 4 ln csc 2 x2 1 1 x x x2 1 C x2 16 sin 2 (4 csc 1 cos 2 3 1 2 4 4 cos C dz 4 cos 2 sin 1 cos 3 3 cos z2 16 z cos x 16 4 cos cos to get z2 16 cos x2 1 C x 4 sin , dz cos 2 sin x 3 dx 1 to get 1 sec C x x2 d C cos sin x d d (sin ) 1 2 2 2d sec 2 (sin ) csc d ,0 2 sec tan d sec 3 tan 1 sec d tan 2 cos sin 2 tan 1 2 dx sec 2 d tan 2 sec dx x2 d, 4 w w 2 . 2 and x w to Section 8.4 x x x 45. For x dx 43. dy 2 9 9 9 sec 3 sec 2 tan 9 d ,0 9 tan 3 sin , dx 3 cos d ,0 9 x 2 9 3 d 9 0, x 2 9 cos /2 dx 0 /2 0 3 cos2 x2 3 x 3 ln Substitute x 5, y 5 ln 3 ln 3 9 C 1 cos 2 d 2 /2 sin 2 4 0 2 C or C 0 1 0 x x2 1 2 x sec 2 tan , dx 1 2.356 2 46. Volume 4 3 d 1 3 4 ln 3. dx 2 4 0 d ,0 4 The solution to the initial value problem is y x2 3 x 3 ln 9 1 44. (x dy x x2 1) 2 dy x dx dx (x 2 1)3/2 tan , dx 1 1 0 0, cos d, 2 2 x C 0, y 1 0 47. (a) 4 1. dx x(1000 1 x(1000 1 The solution to the initial value problem is x 1. 1) 1, 4 . d 8 x) x) B)x 1 2 1 dt 250 A B x 1000 x) A(1000 (A cos 2 d 2 /4 sin 2 4 0 1 4 1 4 8 C x2 cos 2 1 2 1 Substitute x x d /4 sec 2 1 sec 2 sec 4 0 C x2 (since 0 0 Volume y /4 dx (1 x 2)2 d sin 1 0 and when x dx (x 2 1)3/2 2 sec d sec 3 y dx x 2)2 sec 2 /4 sec 2 tan 2 1 2 tan 2 (1 . When x 2 . d /2 3 2 3 cos 0 C 3, 3 cos 3 3 tan 2 0 and when x 2 3 0 9 sin d ln sec 2 2 When x x2 9 3 sec tan 3 tan sec x dx 3 0 2 x2 9 Area dx y 0 on [0, 3] 3 3 sec , dx 2 0, y 353 8.076 x Bx 1000A Equating the coefficients and solving for A and B gives 1 A 1 ,B 1000 dx x(1000 x) 1 1000 (1/1000) dx (1/1000) dx x 1000 x 1 1 ln x ln (1000 x) 1000 1000 1 x ln C1 1000 1000 x t C2 x 250 1 x ln 1000 1000 x ln 4t C 1000 x x e 4t C Ae 4t 1000 x C1 354 Section 8.4 47. continued When t 49. (a) From the figure, tan 0, x 2 998 (b) 500 4t 499e cos x) sin x cos x)2 sin 2 x z 2(1 cos x)2 1 z 2(1 cos x)2 1 cos x 1 ds 1 sin 2 x 1 cos 2 x 1 x(1000 (1 (1 500 which occurs when z 2)2 z 2)2 z 2)2 (1 z 2)2 (1 z 2)2 (1 x) is greatest. 1 2z 2 (1 Only sin x (d) z 1 dx for 0 1 4z 2 z 2 )2 A 1 x where A x x2 1 dx 1 dz 1 1 2 tan 2 1 (1 2 z 2) dx 1 dx 2 x2 1 B x1 1 and B 2 dz 1 1 1 2 1 dx 1 1 1/2 1 1 x 0 x 1 1 ln x 1 1 2 ln 1 2 ln ln 3 1 2 x ln x 3 2 1 dx 1/2 1 0 1 1 x sec 2 dx 2 2 1 2 dz z2 z2 makes sense in this case. x 2 1 2 x x2 x2 0 1 z2 dz 1/2 Arc length 1 1 tan 2z 1 z 4 1 2z 2 (1 z 2)2 2z sin x 2x 2 dx 1 x2 x2 1 dx 1 x2 x2 x2 2 z2 z2 z2 1 1 x2 1 x2 x2 1 1 does not make sense in this case. 2x x2 x2 0 z2 z2 1 1 cos x 0 1 1.553 x4 2x2 1 (1 x 2)2 0 1) cos x 1.553 as shown in part (b). dy 48. dx 1/2 (z 2 or cos x) (c) From part (b), cos x This occurs when x x2 1 cos x) cos x 4t dx 1 (c) x(1000 x) dt 250 dx will be greatest when y dt x2 cos x)(1 z 2cos x 0 Half the population will have heard the rumor in about 1.553 days. t (1 cos x)(z 2 (1 cos 2 x 2 1 1 ln 4 499 t sin x cos x z 2(1 1 499 4t e z(1 1 cos x 1000 1 499e 1 (b) From part (a), z 1 499 1 4t e x 499 1000 4t x 4t x e e 499 499 4t 4t 1000e e 499 499 1000e 4t x 499 e 4t 1000 or x 1 499e 4t x1 sin x . cos x 1 2. A or A x 1000 x 2 x dx 2 z4 355 Chapter 8 Review 50. s Chapter 8 Review Exercises 2 dz 1 z2 dx 1 sin x (pp. 454–455) 2z 1 z2 1 z2 (z 1. lim 2 dz 2z 2 dz 1)2 1 2 z 2 3. lim x→0 1 1 C x cos x sin x sin x x sin x cos x lim cos x x→0 x sin x cos x lim x→0 x1/(1 cos x 2 2 dz 2z z lim x 1 e 1 e 5. The limit leads to the indeterminate form f (x) . ln x x ln x x lim x→ 1/x 1/x 1 0 ln f (x) lim x C 0 x1/x lim e0 lim e x→ 2 tan lim e x→1 ln f (x) 1 1 ln f (x) x→1 1 2 x) ln x 1x 1/x lim x→1 1 ln x lim 2 dz 1 z2 2z 1 1 z2 x→ 1 C 1 2 2 6. The limit leads to the indeterminate form 1 . C tan 2 1 cos t for t→0 and 3 5 2 t→0 5 sec 5t x x→1 1 1/(1 x) x→ 1 3 sec2 3t lim ln f (x) C x tan 2 2 dz (z 1)2 53. 2t 2t t→0 f (x) z2 dt sin t lim 4. The limit leads to the indeterminate form 1 . 1 z 1 tan 3t t→0 tan 5t dz z2 52. 1 The limit does not exist. C 1 C x 1 2 2 dz 1 z2 1 z2 1 1 z2 d 1 sin 2 1 2t) for t→0 tan dx 1 cos x ln (1 t2 t→0 2. lim 51. t 2 dz z2 1 z z2 z2 1 1 2z 1 f (x) 2 3x x 1 ln f (x) x ln 1 1 x dz z 1 lim ln z 3 x ln 1 1 t ln tan 2 C 1 C lim 1 x→ 7. lim r→ 8. lim → /2 lim 1 x x→ x→ 3x x cos r ln r 1 3/x 2 3/x 1 x2 lim eln f (x) x→ 0 since cos r 2 sec 3 x ln 1 3 x lim lim x→ 3x x 3 3 e3 1 and ln r→ as r→ . → /2 cos 2 lim → /2 1 sin 1 ...
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This document was uploaded on 10/31/2011 for the course MAC 2311 at University of Florida.

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