Exam 2 Fall 2012

1. d dx d dx p 3 p 2 dy dx dy dx dy dx (cot x) = - csc2 x cos x sin x 1 sin x p 2 p 2 p 2 p 2 (csc x) = - csc x cot x 2 - = 0 Exam 2 Fall 2012 2 = -csc2 x + 2 csc x cot x -csc2 x + 2 csc x cot x = -1 -1 + 2 csc cot = -1 -0 -csc2 = y - y0 = m ( x - x0 ) = csc x ( ) = 0 = 2 cot x - csc x csc x ( 2 cot x - csc x ) x = 2 cos x - 1 = 0

2. d dx dy dx d2y dx2 dy dx dy dx dy dx dy dx d dx d2y dx2 dy dx = 2y 2y 2x + 2 2y x2 x2 x2 x2 x2 e e e e e x2 y2 = e + 2x = 2x + 2 2 -2 = 2x + 2 2 2 + = 2x 2 2x 2 + 2y

3. 1 + 1 - x2 + 5-7x ln 5 (- 7) cos(x3) 3x2

4. 1 x3 3x2 esin x + [1 + ln(x3)] esin x cos x

5. d dx dy dx 1 1 + x2 1 y 1 x tan-1x c. x y = ln = ( tan-1x)( ln x ) ln = tan-1 x ln x +

6. d dx d dx d dx d dx dy dx 2 3 1 2 2 x 1 y x2 x + sin x y (4x + 1)2/3 ln + - ln ln ln = 1/2 ln ( x + sin x) 2 ln x 2/3 ln ( 4x + 1) 1 x + sin x 1 4x + 1 ( 1 + cos x ) 4 - + =

7. (8 points) Use differentials to approximate the change in the surface area of a cube when the length of each side increases from 1 ft to 1.02 ft. Then compute the (estimated) relative change in surface area. *1 *12 x = 1 A = 6 x2 A = 6 x2 = 6 6 *0.02 dx = 0.02 dA = 6 * 2 x dx dA = 6 *2 x dx = 0.24 0.24 = 4% 100% relative change =

8. d2s dx2 ds dx d dx 5. (10 points) The position of a truck along a straight road from time of 0 hours to time of 15 hours is given by s = 15t2 - t3 where s is given in miles. At what time is the truck furthest from its original point? How far is the truck from its original point at that instant? What is the truck's acceleration at that instant? s( t ) = 15 t 2 - t 3 s( t ) = 15 t 2 - t 3 s = 15t2 - t3 s( t ) = 15 t 2 - t 3 endpoints 0 t = 0 *0 s(0) = 0 0 2 s(15) = 0 3 15 15 t = 15 *15 30 t - 3 t2 - = 0 3t2 15*(2t) = 3 2 *10 s(10) = 500 30t - 3t2 10 t = 10 10 *10 3 t ( 10 - t ) = = -30 at time t = 10 = 30 - 6 t

9. 6. (5 points) let f(x) = 9x + ln x, x > 0 Find the value of at the point x = 3 = f(1) 5 2 df -1 dx d dx d dx 1 = f'( f-1(x)) f-1(x) 5 2 2 5 1 x 1 2 x 9x + ln x x = x = 3 for the function f point of interest is ( x0,y0 ) = ( 1, 3 ) 1 1 1 for the inverse function f-1(x) "symmetrical" point is ( x0,y0 ) = ( 3, 1) ( ) = 3 = + =

10. 1 2 dA dt dx dt d dt 7. (10 points) The width of a rectangle is half of its length. At what rate is the area of the rectangle increasing when its width is 10 cm and the width is increasing at the rate of 1/2cm/s. *10 A = 2x*x 2x2 = 2x2 A = A(t) x = x(t) x x 2x x 4 x = 20 = =

11. -3 -2 -1 0 1 2 3 0 1 2 3 4 5 6 dq dt x 3 d dt d dt v 3 3 miles 200 8. (10 points) As the airplane flies at the speed of 200 miles per hour at constant altitude of 3 miles directly over the head of an observer, how fast is the angle between vertical and the observer's line of site changing? q = 0 0 x = v*t v t x tan q = cos2q sec2q = = 66.7 q

12. 9. (15 points) In parts (a) and (b) the limits can be found by noticing that they can be thought of as derivatives. Evaluate these limits [ you may not use L'Hopital's Rule ] and use the result(s) to evaluate the limit in part (c).

13. sin x - 9p 4 9p 4 lim lim x - h = x - x = h+ x → f(a+ h) - f(a) h 2 2 2 2 2 2 2 2 2 2 sin sin h 0 = x 9p 4 9p 4 9p 4 9p 4 9p 4 9p 4 p 4 p 4 9p 4 sin - +h x = + 2p = sin 9 a. f'(a) = h h → 0 = h → 0 sin(x)' = cos (x) =

14. ln ( 1 + x ) x lim lim d dx ' = ln ( x ) ln x f(a+ h) - f(a) h 1 x h 0 x 0 x = 1 x = 1 9 b. - ln (1) f'(a) = h h h = 1 =

15. lim (1 + x )1/x lim lim (1 + x )1/x lim ln (1 + x )1/x ln (1 + x ) x x 0 x 0 x 0 x 0 9 c. ln = e e 1 e = ln = x x e

16. sin x - 9p 4 9p 4 9p 4 lim x - x - h = x - 2 2 2 2 2 2 sin x 9p 4 9p 4 9p 4 p 4 sin x - + 2p = x - sin x - h

17. f'( f-1(x)) And the answer is 1 3 1 3 h 2 h 2 dy dx d2y dx2 f(x) g(x) 4 3 aa bb x A Junk disappear wipe fast from bottom ( ) wipe fast from left wipe fast from left B C A D E erase center a erase ax box filled no fill ax (a + b) text moving down

18. 1 sin (x) sin ( t ) lim lim 1 t 1 x t 0 x 0 F( x ) = f( x ) - f(a) - g( x ) - g(a) 0 1 x = t 0 -

19. sec(x) 1 + tan x -3 -2 -1 0 1 2 3 lim lim lim x p 2 p 2 f(a+ h) - f(a) h h 0- x 0 x ∞ aa bb Junk lim x