Post on 11-Jan-2016
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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
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
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