Pre-Calc Homework Solutions 117

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

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: 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) ...
View Full Document

Ask a homework question - tutors are online