Presentation is loading. Please wait.

Presentation is loading. Please wait.

Exam 2 Fall 2012 dy dx = d dx (cot x) = - csc 2 x -csc 2 x + 2 csc x cot x d dx (csc x) = - csc x cot x dy dx 22 = -csc 2 22 + 2 csc cot 22 22.

Similar presentations


Presentation on theme: "Exam 2 Fall 2012 dy dx = d dx (cot x) = - csc 2 x -csc 2 x + 2 csc x cot x d dx (csc x) = - csc x cot x dy dx 22 = -csc 2 22 + 2 csc cot 22 22."— Presentation transcript:

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

2 d dx y 2 = e + 2x x2x2 2y dy dx = x2x2 e 2x+ 2 d dx 2y dy dx = x2x2 e 2x+ 2 2 dy dx dy dx + 2y d 2 y dx 2 = x2x2 e 2x + x2x2 e 2 x2x2 e + 2 2y dy dx = d 2 y dx 2 2 -2 dy dx 2 2

3 cos(x 3 ) 3x 2 1 - x 2 1 + + 5 -7x ln 5 (- 7)

4 e sin x cos x [1 + ln(x 3 )] + e sin x 3x 2 1x31x3

5 ln = ( tan -1 x)( ln x ) 1y1y x tan -1 x c. ln y = d dx dy dx = 1 1 + x 2 ln x + 1x1x tan -1 x

6 ln y = x2x2 x + sin x + ln - (4x + 1) 2 / 3 2 ln x 1 / 2 ln ( x + sin x) 2 / 3 ln ( 4x + 1) d dx dy dx 1y1y = 2x2x d dx + d dx 1 x + sin x 1212 ( 1 + cos x ) d dx - 2323 1 4x + 1 4

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

8 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 = 15t 2 - t 3 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 = 15t 2 - t 3 ds dx = 15*(2t) - 3t 2 30t - 3t 2 3 t ( 10 - t ) = 0 endpoints t = 0 s( t ) = 15 t 2 - t 3 0 *0*0 0 s(0) = 0 t = 15 s( t ) = 15 t 2 - t 3 15 * 15 15 2 3 s(15) = 0 t = 10 s( t ) = 15 t 2 - t 3 10 * 10 10 2 3 s(10) = 500 d dx 30 t - 3 t 2 = 30 - 6 t * 10 = -30 at time t = 10 d 2 s dx 2 =

9 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 for the inverse function f -1 (x) "symmetrical" point is ( x 0,y 0 ) = ( 3, 1) for the function f point of interest is ( x 0,y 0 ) = ( 1, 3 ) d dx ( ) = 3 x = = d dx f -1 (x) = 1 f ' ( f -1 (x)) x = 9x + ln x 1 2 x + 1x1x 1 1 1 5252 3 5252 2525 =

10 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. x x x 2x A = 2x*x A = A(t) x = x(t) = 2x 2 dA dt = d dt = 4 x dx dt 1212 * 10 = 20 2x 2

11 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?  3 miles -3 -2 -1 0 1 2 3 x = v*t x x3x3 = 0 1 2 3 4 5 6 v t tan  d dt d dt v3v3 = sec 2  d  dt cos 2   = 0 0 200 = 66.7

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 lim h 0 f(a + h ) - f(a) h lim x 9  4 sin x - 2222 9  4 x - h = x - 9  4 x = h + 9  4 x → 9  4 = h → 0 h 9   + h+ h sin - 2222 sin 44 = 2222 44 + 2  2222 9  4 sin 9  4 f ' (a) = sin(x) ' = cos (x) x = 9  4 2222 = 9 a.

14 9 b. lim x 0 ln ( 1 + x ) x lim h 0 f(a + h ) - f(a) h f ' (a) = h h h - ln (1) = ln ( x ) ' x = 1 d dx ln x = 1x1x x = 1 = 1

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

16 sin x - 2222 9  4 x - h h = x - 9  4 x - 9  4 9  4 sin x - 2222 sin 44 + 2  2222 lim x 9  4 sin x - 9  4 x -

17 wipe fast from left f ' ( f -1 (x)) ( ) axax And the answer is A erase center disappear erase wipe fast from bottom box filled axax no fill dy dx wipe fast from left h2h2 h2h2 A B C D E text moving down a 1313 4343 x 1313 Junk (a + b) aa bb f(x) g(x) d 2 y dx 2

18 1x1x = t 0 lim t 0 sin ( t ) 1 t lim x 0 1x1x 1 sin (x) - F( x ) = f( x ) - f(a) - g( x ) - g(a) 0

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


Download ppt "Exam 2 Fall 2012 dy dx = d dx (cot x) = - csc 2 x -csc 2 x + 2 csc x cot x d dx (csc x) = - csc x cot x dy dx 22 = -csc 2 22 + 2 csc cot 22 22."

Similar presentations


Ads by Google