Business Calc Homework w answers_Part_73

Business Calc Homework w answers_Part_73 - 361 Chapter 8...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 361 Chapter 8 Review 56. 2 B x C x2 Bx 2(x 2) Cx(x 8 A x 3(x 2) x Ax 3 8 B)x 3 (A D x3 2) D(x D)x B 0, 2B C 0, 2C x3 x3 1 x1 1 x x3 x A x1 x x(x 1)(x 1) D 8 1, B 1, C 8 dx x3(x 2) 2, D dx x dx x2 ln x 57. t 3 3t 2 4t t 3t 3 A t 4 4t 4 Bt t2 A( t 2 (A 4 2 dx x2 2 ln x x 4 dx x3 2 x2 A C B A Bx(x C)x 2 B C 0, B (Bt 3, C (B C C)t 1, C 1) C)x Cx(x 1 1)(x 1) 1) A 1, and A 1. 0 dx dx x x B)t 2 4 and A 1, B 1 dx x x3 x3 Ct ln x 3 x3 x3 4x 2 4x 2 dx x 1 ln x 1 C A Equating coefficients of like terms gives B 1)(x x(x Equating coefficients of like terms gives C 1 1) (A A A( x x 1 1) t 2 1) 1 1 Solving the system simultaneously yields t(t 2 t 2 x1 x(x 2 1) B C x1 x1 x 0, and 2D Solving the system simultaneously yields A 1 x1 x+0 x1 0x2 0x2 2D Equating coefficients of like terms gives A x x3 x3 2) (2C C)x 2 (2B 59. x 3 60. x 2 4x 4. x 0x 3x 3x Solving the system simultaneously yields x3 A 4, B 3t 2 1, C 4t t3 4 t 4 x 4 dt t 4 dt t dt t4 dt t2 1 t dt 4 dt t2 1 t2 1 1 ln (t 2 1) 4 tan 1 t 2 4 ln t 58. t 4 (t 2 4t 2 (t 2 3 1 1)(t 2 3) (At C)t 3 At t2 B)(t 2 (A 1 3)(t 2 B 1 (B C 0, B D D 3 D)t 2 (3A C 0, B (t 2 dt 3)(t 2 1 ,C 2 1) 0, D 1 tan 1 t 2 C)t 3B 0, and 3B D D 1/2 dt t2 3 1 1t tan 2 3 1 C 3 Evaluate the integrals using Formula 16, with x in the first integral and a (A B)x 3A (x 3x 1)(x 3 in the second. t, a 1 3) 1) B Equating coefficients of like terms gives B x 1 2 1/2 dt t2 1 B(x x3 1) Solving the system simultaneously yields A 3) x 3, 3A B 0 3 ,B 2 Solving the system simultaneously yields A D)(t 2 0, 3A A( x 3x C Equating coefficients of like terms gives A 4x 2 3x x 4x 3 x 2 4x 3 3x A B 1)(x 3) x1 x3 3x A (Ct 3) (x 1) Ct t2 2 2 4x 2 dx 4x 3 x dx x2 2 x 3/2 dx 1 3 ln x 2 1 x 9 . 2 9/2 dx 3 9 ln x 2 3 C 362 61. Chapter 8 Review dy y(500 y) 1 y(500 y) 1 0.002 dx A y A(500 63. y B 500 y y) 1 1 tan , dy 3 9y 2 A)y By 9y 2 1/500 dy 500 y 1 ln 500 500 1 y ln 500 500 y C1 y 500 y C1 y 500 y 20 480 0.002x 0, y y 500 y 20. 1x e 24 24y 500e x (e x 24)y C2 64. t yex 500e x 500 e x e x 24 y y 1 dy y2 65. x 500 24 e x y2 x 1 dx 1 x tan 1 y C1 tan 1 y ln x Substitute x tan 1 tan 1 y tan ln x 4 y 9t 2 9t 2 dt 1 C 0, y C2 1 1 4 1 and x . 1 . 3 2 2 1 cos 3 9 sec2 9 2 3 sec 5 d tan , 0 1 4 2 9 tan2 9 3 sec tan d 3 tan 9 sec ln tan tan C cos2 d ln sec C ln x 3y d, cos 3 sec , dx 5 25x 1 1 1 cos 3 sin2 1 5 dx ln x 9y 2 1 1 sin , dt 3 25x 2 dx 1 dy C 1 cos2 d 3 1 cos 2 d 6 sin 2 C 6 12 sin cos C 6 6 sin 1 3t 3t 1 9t 2 C 6 6 1 1 sin 1 3t t 1 9t 2 C 6 2 1 Use Figure 8.18(b) with a and x t. 3 1 24 ke0 or k tan C1 1 ke x Substitute x d Use Figure 8.18(a) from the text with a 1 x sec Integrate by using Formula 88 with a C ln 62. ln y 2 d ln sec 1 ln y 500 2 sec 1 . 500 1 y ln 500 500 y d, 2 sec 1/500 dy y dy y(500 y) 2 sec2 500A 1 ,B 500 where A tan 3 dy 1 (B 1 1 sec2 3 5x 3 ln (5x tan 25x 2 3 25x 2 C1 9 C1 9) C 1 4 Integrate by using Formula 88 with a Use Figure 8.18(c) with a 3 . 5 1 and x . Chapter 8 Review 68. For x 66. x sin , dx cos d, 2 0, y Area 2 0 on [0, ). b x xe dx lim x2 1 (1 cos2 4x 2 dx x 2)3/2 4 sin2 cos u d 3 cos 4(1 cos2 ) d cos2 2 (4 sec du dv dx x xe 4 tan 4 x C 0 1 1. 69. (a) 1 b→0 b 2 u (ln x) dv 2 ln x dx x du (ln x)2 dx 2 v dx x (ln x) 1 a 2 ln x dx 2 ln x du dv 2 dx x v dx x 2x ln x 2 dx (ln x)2 dx x (ln x)2 2x ln x lim b→0 x(ln x)2 b→0 2 2 2 2 2x ln x 2x 2x C C 1 2x ln x 2x b lim [2 lim b→0 b(ln b)2 (ln b)2 1/b 2b ln b b→0 2 (ln b)(1/b) 1/b2 lim 2 (ln b) 1/b lim 2 /b 1/b2 b→0 b→0 /b 1/b2 2 lim b→0 2 lim ( b→0 2 2b] ln b 1/b 2 lim lim b→0 a x 1 kt 1/a x 2 ln x dx Area kt C1 1 b) lim 2 b b→0 2 1 a 1 C 0, t C 1 Evaluate 2 ln x dx by using integration by parts. u C1 b k dt x)2 dx (a x)2 1 C2 ax 1 kt ax Substitute x x 2 (ln x) dx kt xe x)2 k(a dx (ln x)2 dx by using integration by parts. Evaluate dx dt (a (ln x) dx 0 k dt b eb 1 lim b b→ e lim 2 lim 1] dx 0 b→ ( ln x)2 dx Volume e b b b→ 0 on (0, 1]. 1 x lim [ be 1 4 sin x2 Use Figure 8.18(b) with a 0, y e x b→ C x e x xe lim 1 a kt a x 1 kt dx dx e xe 4) d 4x 1 x x e v dx x 0 dx by using integration by parts. x Area 67. For x x Evaluate xe xe b→ 0 1/a 0 x e x C 363 364 Section 9.1 Chapter 9 Infinite Series 69. continued dx (b) k dt (a x)(b x) dx k dt (a x)(b x) 1 A (a x)(b x) ax 1 A( b x) (A B(a B)x kt C1 s Section 9.1 Power Series (pp. 457–468) B b x Exploration 1 x) bA 1. 1 aB 2. x 3. 1 Equating coefficients of like terms gives A B 0 and bA aB 4. 1 5. A a b dx x)(b (a ,B a 1/(a b) dx ax x) a a a b x x b C2 a b a b ab (x 3 x 4x 8x 1/(a b) dx bx ln b x ab a b x x C2 1n (x 3 C2 abe (a b)kt ab(e (a ae (a b)kt 1) 1 (x 3 1)2 … 3 … 1)3 …. 1)n x 1 x 1 1, 2. The interval of conver- Exploration 2 1. 1 2. tan x 2 1 x A Power Series for tan 4 x x x 1 01 x 0 … 6 t2 t3 3 x3 3 t 0. x …. n 2n ( 1) x dt t2 (1 1 t4 t5 5 x5 5 … t6 t7 7 x7 7 … … ( 1)nt 2n …) dt t 2n 1 … 2n 1 2n 1 x …. ( 1)n 2n 1 ( 1)n x 0 3. The graphs of the first four partial sums appear to be converging on the interval ( 1, 1). b)kt axe (a ab(e (a b) b)kt (2x) gence is (0, 2). C a (a b)kt e b bx 1 (x 3 1) …. n 2 which is equivalent to 0 C1 D x x ( 1) x … 3 …. nn1 (x 1) (x 1) ( 1)n(x 1)n …. 1 (x 3 …. ( x)n … 4 2 x x(ae (a x 0, x ln kt b ax ln (a b)kt bx ax De(a b)kt bx Substitute t 2x … x3 This geometric series converges for b 1 ln 1 3 x 1 ln a x ab 1 2 1 Solving the system simultaneously yields 1 x Power Series for Other Functions x2 x b)kt b)kt 1) 1) b [ 5, 5] by [ 3, 3] 4. When x e bkt Multiply the rational expression by bkt . e ab(e akt e bkt) x akt bkt ae be 1 1 3 1, the series becomes 1 5 1 7 … ( 1)n 2n 1 …. This series does appear to converge. The terms are getting smaller, and because they alternate in sign they cause the partial sums to oscillate above and below a limit. The two calculator statements shown below will cause the successive partial sums to appear on the calculator each time the ENTER button is pushed. The partial sums will appear to be approaching a limit of /4 (which is tan 1(1)), although very slowly. Section 9.1 Exploration 3 A Series with a Curious Property 2 1. f (x) 1 2. f (0) x 2! x 1 3 0 x 3! … 0 … n x n! …. u2 1. u3 3. Since this function is its own derivative and takes on the value 1 at x 0, we suspect that it must be e x. 4. If y dy f (x), then dx y and y ( 1)1 1 ( 1)2 2 ( 1)3 3 ( 1)4 4 ( 1)30 30 2. u1 1 when x 0. u4 u30 3. (a) Since 5. The differential equation is separable. dy dx y dy dx y ln y x (c) an 1 Ke0 ⇒ K y e x. 1 30 18 6 1 54 18 3, the common ratio is 3. ) 4 8 (c) an 1 1 4 8 1 2 2 4 19 2 k Ke 1 3 2(3n (b) 8 y 1 2 39,366 4. (a) Since C 1 6 2 (b) 2(39) 1 , the common ratio is 2 1 . 2 1 64 1n1 2 8( 0.5)n 5. (a) We graph the points n, 6. The first three partial sums are shown in the graph below. It is risky to draw any conclusions about the interval of convergence from just three partial sums, but so far the convergence to the graph of y e x only looks good on ( 1, 1). Your answer might differ. 365 1 1 n n2 for n 1, 2, 3, … . (Note that there is a point at (1, 0) that does not show in the graph.) [0, 25] by [ 0.5, 0.5] [ 5, 5] by [ 3, 3] 7. The next three partial sums show that the convergence extends outside the interval ( 1, 1) in both directions, so ( 1, 1) was apparently an underestimate. Your answer in #6 might have been better, but unless you guessed “all real numbers,” you still underestimated! (See Example 3 in Section 9.3.) (b) lim an n→ lim n→ 1 n n2 0 6. (a) We graph the points n, 1 1n for n n 1, 2, 3, … . [0, 23.5] by [ 1, 4] (b) lim an [ 5, 5] by [ 3, 3] n→ Quick Review 9.1 1. u1 1 2 4 u2 2 u3 3 u4 u30 4 3 4 4 4 5 4 6 4 2 4 2 4 4 30 2 4 2 lim 1 n→ 1n n e 7. (a) We graph the points (n, ( 1)n) for n 1, 2, 3, … . 1 2 3 4 32 [0, 23.5] by [ 2, 2] 1 8 (b) lim an does not exist because the values of an n→ oscillate between 1 and 1. ...
View Full Document

Ask a homework question - tutors are online