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# Set8 - Math 21B-B Homework Set 8 Section 7.1 3 y2 2y 1 dy =...

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Math 21B-B - Homework Set 8 Section 7.1: 3. Z 2 y y 2 - 25 dy = Z 1 u du ( u = y 2 - 25 , du = 2 y dy ) = ln | u | + C = ln | y 2 - 25 | + C 6. Z sec y tan y 2 + sec y dy = Z 1 u du ( u = 2 + sec y, du = sec y tan y dy ) = ln | u | + C = ln | 2 + sec y | + C 8. Z sec x p ln(sec x + tan x ) dx = Z 1 u du ( u = ln(sec x + tan x ) , du = sec x (tan x + sec x ) sec x + tan x dx = sec x dx ) = 2 u + C = 2 p ln(sec x + tan x ) + C 15. Z e r r dr = Z 2 e u du ( u = r, du = 1 2 r dr ) = 2 e u + C = 2 e r + C 20. Z e - 1 /x 2 x 3 dx = Z 1 2 e u du ( u = - 1 x 2 , du = 2 x 3 dx ) = 1 2 e u + C = 1 2 e - 1 /x 2 + C 1

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51. dy dx = 1 + 1 x , y (1) = 3 y = Z 1 + 1 x y = Z dy dx dx = x + ln | x | + C We will use the initial condition y (1) = 3 to find the value of C . y (1) = 3 1 + ln 1 + C = 3 C = 2 Thus we get y = x + ln | x | + 2. 52. d 2 y dx 2 = sec 2 x, y (0) = 0 y 0 (0) = 1 dy dx = Z sec 2 x dx dy dx = Z d 2 y dx 2 dx = tan x + C 1 We will use the initial condition y 0 (0) = 1 to find C 1 . y 0 (0) = 1 tan(0) + C 1 = 1 C 1 = 1 Thus we get dy dx = tan x + 1. We now integrate dy dx to find y . y = Z tan x + 1 dx y = Z dy dx dx = Z sin x cos x + 1 dx = Z - 1 u + 1 du ( u = cos x, du = - sin x dx ) = - ln | u | + u + C 2 = - ln | cos x | + cos x + C 2 = ln | sec x | + cos x + C 2 We will use the initial condition y (0) = 0 to find C 2 . y (0) = 0 ln | sec(0) | + cos(0) + C 2 = 0 ln(1) + 1 + C 2 = 0 1 + C 2 = 0 C 2 = - 1 2
Thus we get y = ln | sec x | + cos x - 1. 57. a. L ( x ) = f (0) + f 0 (0) · x = ln(1 + 0) + 1 0 + 1 · x = x b. Let f ( x ) = ln(1 + x ) and consider f 0 ( x ) and f 00 ( x ). f 0 ( x ) = 1 1+ x f 00 ( x ) = - 1 (1+ x ) 2 On the interval [0 , 0 . 1] we see that f ( x ) is increasing (by f 0 ( x )) and concave down (by f 00 ( x )). Using what we know about the graphs of f ( x ) and L ( x ), we can see that the maximum approximation error on [0 , 0 . 1] occurs at x = 0 . 1. Therefore, the maximum error = L (0 . 1) - f (0 . 1) 0 . 00469. 58. a. L ( x ) = f (0) + f 0 (0) · x = e 0 + e 0 · x = 1 + x Section 7.2: 2. a. dp dh = kp ( k constant) p = p 0 when h = 0 Using the Law of Exponential Change (p509) we know that p = p 0 e kh . In the problem we are given that p (0) = 1013 and p (20) = 90. 3

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We will use these initial conditions to find the values of p 0 and k .
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Set8 - Math 21B-B Homework Set 8 Section 7.1 3 y2 2y 1 dy =...

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