# Exam 2 Fall 2012 d dx (cot x) = - csc2 x d dx (csc x) = - csc x cot x

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

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

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

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

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

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

(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. x = 1 *1 *12 A = 6 x2 A = 6 x2 = 6 6 dx = 0.02 *0.02 dA = 6 * 2 x dx dA = 6 *2 x dx = 0.24 0.24 relative change = 100% = 4%

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 t = 0 *0 s(0) = 0 ds dx t = 15 *15 15 15 s(15) = 0 2 3 = 15*(2t) 30 t - 3 t2 - 3t2 = 0 3 t ( t ) 30t - 3t2 t = 10 *10 *10 10 10 s(10) = 500 2 3 d2s dx2 d dx = = t = -30 at time t = 10

6. (5 points) let f(x) = 9x + ln x, x > 0 Find the value of at
the point x = 3 = f(1) df -1 dx 9x + ln 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) d dx 1 2 x 1 x 5 2 5 2 ( ) = 3 + = x = d dx f-1(x) = 1 f'( f-1(x)) 2 5 x = =

7. (10 points) The width of a rectangle is half of its length
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 1/2 cm/s. 1 2 *10 A = 2x*x = 2x2 2x2 A = A(t) x = x(t) x dA dt d dt dx dt x 2x x 4 x = = = 20

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? 200 q = 0 x = v*t v t x 3 miles d dt d dt x 3 = tan q dq dt v 3 cos2 q = sec2 q = 66.7 q

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).

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

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

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

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

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

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

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

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