Transcript Exam 2 Fall 2012 - New Jersey Institute of Technology

Exam 2 Fall 2012
d
(cot x) = - csc2 x
dx
d
(csc x) = - csc x cot x
dx
p
2
2
dy
= -csc2 x + 2 csc x cot x
dx
dy p
p cot p = -1 -0 = -1
p
2
+
2
csc
= -csc
2
dx 2
2
2
y - y0 = m ( x - x0 )
dy = 0 =
dx
cos x
1
2
=0
sin x
sin x
= csc x ( 2 cot x - csc x )
2 cos x - 1 = 0
p
x=
3
d
dx
y2
2
x
= e + 2x
dy
x2
2y
= e 2x + 2
dx
e
dy
=
dx
dy
d
x2
2y
= e 2x + 2
dx
dx
dy
2 -2
dx
2
2
2
2
dy
dy + 2y d y = e x2 2x 2x + e x2 2
dx
dx2
dx
d2y
dx2
x2
2x + 2
2y
cos(x3) 3x2 +
1
1 - x2
+ 5-7x ln 5 (- 7)
esin x cos x [1 + ln(x3)] + esin x
1 3x2
x3
c.
d
dx
-1x
tan
ln y = ln
x
= ( tan-1x)( ln x )
1
1 dy
1 tan-1 x
=
ln
x
+
1 + x2
y dx
x
x2
y
x + sin x
(4x + 1)2/3
d
d
d
d - ln
ln
+ ln
ln
=
dx
dx
dx
dx
2 ln x 1/2 ln ( x + sin x) 2/3 ln ( 4x + 1)
1
1 dy
2
2 1
1
(
1
+
cos
x
)
4
+
=
y dx
x
3 4x + 1
2 x + sin x
(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.
A = 6 x2
A = 6 x2 = 6
x = *11 2
0.02
dx = *0.02
dA = 6 * 2 x dx
dA = 6 *2 x dx
relative change =
100% = 4%
= 0.24
5. (10 points) The position of a truck along a straight road from
3 3
2t -2tt2-32 where
s( ttst))=)===
15
15
time of 0 hours to time of 15 hours is given s(
by
15t
15
t --t t t3 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?
endpoints t = 00
*0 s(0) = 0
ds
30 t - 3-t23t2 = 0
= 15*(2t)
dx
30t
3t2 - t )
3
t (-10
d
d2s
= dx
dx2
15 s(15) = 0
t = *15
10
t = *10
10 s(10) = 500
= 30 - 6 t
= -30 at time t = 10
2
2
3
3
df -1
6. (5 points) let f(x) = 9x + ln x,
at
x x > 0 Find the value of
dx
the point x = 3 = f(1)
for the function f point of interest is ( x0,y0 ) = ( 11
1, 3 )
for the inverse function f-1(x) "symmetrical" point is ( x0,y0 ) = ( 3, 1)
d
(
dx
) =3
5
1
1
=
+
2
2 x
x
x=
d -1
f (x)
dx
=
x=
1
= 2
5
f'( f-1(x))
7. (10 points) The width of a rectangle is half of its length. At
what rate is the area of the rectangle increasing 1when its width is
*10
10cm and the width is increasing at the rate of 1/2 cm/s.
2
A = 2x*x = 2x2
x
x 2x x
dA
d
dt = dt
A = A(t)
= 4x
dx
dt
= 20
x = x(t)
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
q = 00
observer's line of site changing?
x = v*t
vt
-3
0-2
x
1 -1
d
dt
3 miles
q
2 0
31
42
d
x
tan q
= dt
3
2 q
dq
v cos
2
= sec q
= 66.7
dt
3
53
6
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.
9p 2
2
sin
sin x - + h 4 2
2
lim
9p
9p
x
x4
4
p
2
sin
4 = 2
p 9p
sinsin 9p
+ 2p = 2
4 44
2
f(a + h ) - f(a)
f'(a) = lim
h
h 0
9p
hh= x 4
9p
x→
4
x=h+
9p
4
→ 00
= hh →
sin(x)' = cos (x)
=
x=
9p
4
2
2
9 b.
ln ( 1 + x ) - ln (1)
'
= ln ( x )
x
0
lim
x
x=1
d
1
ln x =
dx
x
=1
x=1
f(a + h ) - f(a)
f'(a) = lim
h
h 0
9 c.
e
ln lim
x
0
(1 + x )1/x
lim ln (1 + x )1/x
= e x
0
ln (1 + x )
x
0
1 lim
lim (1 + x )1/x = e x
ln
x 0
x =x
e
sin x lim
x
9p
4
9p
x4
sin
x-
h
2
2
9p
x4
9p
4
sin x -
sin x -
h=x-
9p
4
p
+ 2p = 2
4
2
9p
4
2
2
And the answer is
f'( f-1(x))
disappear
(
)
A
Junk
wipe fast from bottom
wipe fast from left
erase center
a
4
f(x)
3
g(x)
d2y
dx2
dy
dx
wipe fast from left
x
1
3
A B C D E
erase
ax
ax
h
2
box filled
no fill
h
2
text moving down
1
3
(a + b)
aa
bb
F( x ) = f( x ) - f(a) -
g( x ) - g(a)
0
0
1
1
lim
x 0 sin (x) x
1
=t
x
lim 1 sin ( t )
t
0t
Junk
-3
-2
-1
0
1
2
3
aa
bb
f(a + h ) - f(a)
lim
h
h 0-
lim
x
0
lim
lim p
x
2
sec(x)
1 + tan x
x
∞
x
p
2