Pre-Calc Homework Solutions 117

# Pre-Calc Homework Solutions 117 - Section 3.9 41 The line...

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Unformatted text preview: Section 3.9 41. The line passes through (a, e a) for some value of a and has slope m e a. Since the line also passes through the origin, ea a 0 and we have 0 117 45. y ln y ln y ln y 5 (x (x 3)4(x 2 1) (2x 5)3 (x 3)4(x 2 1) 1/5 (2x 5)3 the slope is also given by m ea y ea , so a a ln 3)4(x 2 1) 1/5 (2x 5)3 1. Hence, the slope is e and the equation is 1 (x 3)4(x 2 1) ln 5 (2x 5)3 1 [4 ln (x 5 4 d ln (x 5 dx ex. xe x, we have y (x)(e x) (e x)(1) (x 1)e x, so 3) 3) ln (x 2 1) 3 ln (2x 5)] 42. For y d (ln y) dx 1 d ln (x 2 5 dx 1 dy y dx dy dx dy dx 4 5(x 3) the normal line through the point (a, ae a) has slope m y (a (a 1 and its equation is 1)e a 1) 3 d ln (2x 5 dx 5) 3 1 (2) 5 2x 5 6 5(2x 5) 1 (x 1)e a a) ae a. The desired normal line 4 1 5 x 3 1 1 (2x) 5 x2 1 3) 2x 5(x 2 1) y 4 5(x (x includes the point (0, 0), so we have: 1 0 (0 a) ae a (a 1)e a a 0 ae a (a 1)e a 1 0 a ea (a 1)e a 1 a 0 or ea 0 (a 1)e a 1 The equation e a 0 has no solution, so we (a 1)e a 3)4(x 2 1) 1/5 (2x 5)3 6 5(2x 5) x(x 2 1)1/2 (x 1)2/3 2x 5(x 2 1) 46. y ln y ln y d ln y dx 1 dy y dx dy dx dy dx x x2 1 (x 1)2/3 ln x(x 2 1)1/2 (x 1)2/3 1 ln (x 2 2 need to use a y 43. 0. The equation of the normal line is 0) 0e 0, or y x. ln x d ln x dx 1 x 1) 2 ln (x 3 1) 1) 1 (x (0 1)e 0 1 d ln (x 2 2 dx 1) 2 d ln (x 3 dx y ln y ln y d ln y dx 1 dy y dx dy dx dy dx (sin x)x ln (sin x)x x ln (sin x) d [x ln (sin x)] dx 1 (x) (cos x) sin x 1 1 (2x) 2 x2 1 1 x 2 2 1 (1) 3 x 1 2 1) 2 1 3(x 1) y x x2 1 3(x x x2 x x 1 1 (x 1)2/3 x ln (sin x)(1) 47. dA dt y[x cot x ln (sin x)] ln (sin x)] d 1 t/140 dt 2 d 20 2 t/140 dt 20 (sin x)x[x cot x x tan x ln (x tan x) (tan x)(ln x) 20 (2 20(2 (2 t/140 t/140 t/140 )(ln 2) d dt t 140 1 140 44. y ln y ln y d ln y dx 1 dy y dx dy dx dy dx )(ln 2) )(ln 2) 7 At t dA dt 2 days, we have (2 1/70 d [(tan x)(ln x)] dx 1 (tan x) (ln x)(sec 2 x) x tan x y (ln x)(sec 2 x) x tan x x tan x (ln x)(sec 2 x) x )(ln 2) 7 0.098 grams/day. This means that the rate of decay is the positive rate of approximately 0.098 grams/day. 48. (a) (b) d ln (kx) dx 1 d kx kx dx k kx 1 x d d ln (kx) (ln k dx dx 1 d 0 ln x x dx ln x) ...
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