Exam 2 Fall 2012

dydx =

ddx

(cot x) = - csc2 x

-csc2 x + 2 csc x cot x

ddx

(csc x) = - csc x cot x

dydx

2

= -csc2 2

+ 2 csc cot 2

2

= -1 -0

y - y0 = m ( x - x0 )

= -1-1

2

2

dydx

= 0

-csc2 x + 2 csc x cot x

= = csc x ( )2 cot x - csc xcsc x ( 2 cot x - csc x )

2 - = 0 cos xsin x

1sin x

2 cos x - 1 = 0 x =3

ddx

y2 = e + 2xx2

2ydydx

= x2e 2x + 2

ddx

2ydydx

= x2e 2x + 2

2dydx

dydx + 2y

d2ydx2

= x2e 2x 2x + x2

e 2

x2e 2x + 2

2y

dydx =

d2ydx2

2

-2dydx

2 2

cos(x3)3x2

1 - x2

1+ + 5-7x ln 5 (- 7)

esin x cos x [1 + ln(x3)] + esin x 3x21x3

ln = ( tan-1x)( ln x )

1y

x tan-1xc. lny =ddx

dydx =

11 + x2

ln x +1x

tan-1 x

ln ln

y

=

x2 x + sin x

+ ln - ln

(4x + 1)2/3

2 ln x 1/2 ln ( x + sin x) 2/3 ln ( 4x + 1)

ddx

dydx

1y

=2x

ddx

+

ddx

1x + sin x

12

( 1 + cos x )

ddx

- 23

14x + 1

4

A = 6 x2

dA = 6 * 2 x dx

x = 1dx = 0.02

dA = 6 *2 x dx

*1*0.02

= 0.24

A = 6 x2*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.

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 = 15t2 - t3

dsdx = 15*(2t) - 3t2

30t - 3t2 3 t ( 10 - t )

= 0

endpoints t = 0

s( t ) = 15 t 2 - t 3

0*00 s(0) = 0

t = 15

s( t ) = 15 t 2 - t 3

15*1515 2 3s(15) = 0

t = 10

s( t ) = 15 t 2 - t 3

10*1010 2 3s(10) = 500

ddx

30 t - 3 t2

= 30 - 6 t

*10

= -30 at time t = 10d2sdx2

=

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 ( x0,y0 ) = ( 3, 1)

for the function f point of interest is ( x0,y0 ) = ( 1, 3 )

ddx

( ) = 3

x =

=

ddx

f-1(x) =1

f'( f-1(x))x =

9x + ln x

12 x

+ 1x

111

52

3

52

25

=

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 x2x

A = 2x*x A = A(t) x = x(t)= 2x2

dAdt =

ddt

= 4 x dxdt

12

*10

= 20

2x2

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

x3

=

0 1 2 3 4 5 6

v t

tan ddt

ddt

v3

= sec2 ddt

cos2

= 00

200

= 66.7

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).

lim

h 0

f(a + h ) - f(a) hlim

x 9 4

sin x - 22

9 4

x -

h = x - 9 4

x = h + 9 4

x → 9 4

= h → 0h → 0

h

9 + hsin - 2

2

sin 4 = 2

2

sin 4 + 2 2

29 4

sin 9 4

f'(a) =

sin(x)' = cos (x)

x =9 4

22

=

9 a.

9 b.

lim x 0

ln ( 1 + x ) x lim

h 0

f(a + h ) - f(a) h

f'(a) = hhh- ln (1)

= ln ( x )'

x = 1

ddx

ln x =1x

x = 1

= 1

9 c.

lim (1 + x )1/x x 0

lne x = x

lne

lim ln (1 + x )1/x x 0= e

=lim

x 0e

ln (1 + x ) x1

lim (1 + x )1/x x 0

sin x - 22

9 4

x -

h

h = x - 9 4

x - 9 4

9 4

sin x - 22

sin 4 + 2 2

2

lim x 9

4

sin x -

9 4

x -

wipe fast from left

f'( f-1(x))

( )

ax

And the answer is

A

erase center

disappear

erase

wipe fast from bottom

box filled

ax no fill

dydx

wipe fast from left

h2

h2

A B C D E

text moving down

a

13

43

x13

Junk

(a + b) aabb

f(x)g(x)

d2ydx2

1x

= t0

lim

t 0

sin ( t )1 t

lim x 0

1x

1sin (x)

-

F( x ) = f( x ) - f(a) - g( x ) - g(a)

0

Junk

-3 -2 -1 0 1 2 3

aabb

lim

h 0-

f(a + h ) - f(a) h

sec(x)1 + tan x

lim x

2x

2

lim x 0

lim x ∞