MATH 151, FALL SEMESTER 2011
COMMON EXAMINATION 3 - VERSION B - SOLUTIONS
Name (print):
Instructor’s name:
Signature:
Section No:
Part 1 – Multiple Choice (13 questions, 4 points each, No Calculators)
Write your name, section number, and version letter (B) of the exam on the ScanTron form.
Mark your responses on the ScanTron form and on the exam itself
1.
The electric field between parallel plates produces the acceleration a t
5 cos t cm/sec 2 on a
charged particle, where t is in seconds. If the particle has initial position x 0
3 cm and initial
2 cm/sec, where is the particle at time t
sec?
velocity v 0
a.
b.
c.
d.
e.
x
x
x
x
x
2
2
2
2
2
3 cm
3 cm
13 cm
8 cm
8 cm
Correct Choice
SOLUTION: v t
5 sin t C
v0
C
xt
5 cos t 2t K
x0
5 K 3
x
5 cos
2
8 2
13
2.
Find all solutions of ln x
a.
b.
c.
d.
e.
ln x
2
2
K
vt
8
Compute arcsin cos 5
4
b.
c.
1
4
3
4
d.
e.
5
4
8
5, 3 only
5, 0 only
3 only
Correct Choice
5 only
0, 3 only
x
x
x
x
x
a.
2t
ln 15 .
SOLUTION: ln x x 2
ln 15
x 2 2x 15
x
5:
ln 5 is not defined, so not a solution.
x 3:
ln 3 ln 3 2
ln 15
3.
5 sin t 2
xt
5 cos t
1
4
0
x
5 x
3
0
.
Correct Choice
5
4
SOLUTION: cos 5
1 2 because 5 is in quadrant III.
4
4
1 .
arcsin cos 5
arcsin 1 2
arcsin 1 2
4
4
1
4.
Compute lim
ln 4
x
0
ln 4
5
ln 2
3
ln 4
ln 5
ln 2
ln 3
a.
b.
c.
d.
e.
2x
ln 5
Correct Choice
SOLUTION: lim
ln 4
x
5.
If f x
a.
b.
c.
d.
e.
ln 1
x 1
1 x2
1
2 1 x2
1 x
1 x2
x 1
1 x2
1
2 1 x2
P0 4
x2
t/
1/2
lim
ln 4
x
5
3x
2x
3x
4
ln lim
x
5
2x
3x
ln 2
3
P
1 ln 1
2
500 4
3/2
500 2
x2
arctan x
1
ln x
at x
2
df
dx
1 2x
2 1 x2
e. Its y-intercept is
1
1
3
Correct Choice
2e
1 2e
1
ln x
SOLUTION: f x
f x
y
12/8
2 ln x
fe
3
4000
arctan x , then f x
Find the line tangent to y
a.
b.
c.
d.
e.
500 4
Correct Choice
SOLUTION: f x
2
ln 5
3000
4000
Correct Choice
6
500e
500e 1/6
500e ln 4 2/3
SOLUTION: P
7.
2x
A shrimp farm starts off with 500 shrimp. The number of shrimp quadruples after 8 weeks (to
2000 shrimp). Assuming an exponential growth, how many shrimp are there after 12 weeks?
a.
b.
c.
d.
e.
6.
3x
3
f e x
1
x
e
2
2
x ln x 3
1 2e x
ln x
2
f e
e
1
ln e
fe
2
e ln e
2x 3
e
3
2
2
e
1
1
1
x2
x 1
1 x2
8.
a.
b.
c.
d.
e.
1 .
x 1
x x2 1
x 1
x x2 1
2x
x
2
x 1 ln x 2
x
x 2 1 ln x 2
x2
1
x
1
1 2x
ln x 2
e x ln
f x
x2 1
2x 2
1
x
x2
SOLUTION: f x
9.
x
x2
Compute the derivative of f x
1
d x ln x 2
dx
2
x
1
Correct Choice
1
e ln
x
x2 1
e x ln
x2 1
e x ln
ln x 2
x2 1
2x 2
1
x
2
1
x2
x
1
x3
Find the locations of the absolute maximum and absolute minimum of f x
interval 0, 5 .
5
4
0
4
Abs Max at x
Abs Max at x
Abs Max at x
Abs Max at x
0
0
5
d.
Abs Min at x
Abs Min at x
Abs Min at x
Abs Min at x
e.
Abs Min at x
4 Abs Max at x
5
a.
b.
c.
a.
b.
c.
d.
e.
3 sin x
x3
1
5x 2
2x 2
x
2
1
8x on the
Correct Choice
2
3
SOLUTION: f x
3x 2 10x 8
3x 2 x 4
Critical point at x
2
(x
is not in the interval.) Endpoints at x 0, 5.
3
f0
0
f4
64 80 32
48
f5
125 125 40
40
10. For f x
ln x 2
4.
3x which of the following is TRUE?
The First Derivative Test says x 0 is a Local Maximum
The Second Derivative Test says x 0 is a Local Minimum
The Second Derivative Test says x 0 is a Local Maximum
The Second Derivative Test says x 0 is an Inflection Point
The Second Derivative Test Fails at x 0.
SOLUTION: f x
3 cos x 3x 2 3
The Second Derivative Test Fails.
f 0
0
f x
3 sin x
Correct Choice
6x
f 0
0
3
11.
The plot at the right shows the graph of the
f ' (x)
FIRST DERIVATIVE f x of a function f x .
Which of the following is FALSE?
a. f x has a local maximum at x
10
0 Correct Choice
-1
b. f x is increasing on 0, 1
c. f x is increasing on 1, 3
1
2
3
4
5
x
-10
d. f x has an inflection point at x
e. f x is concave up on
20
3
1, 1
SOLUTION:
(a) is false because f x has a local minimum at x 0 because to the left of 0, f is decreasing:
f x
0 and to the right of 0 f is increasing: f x
0.
(b) is true because f x
0 on 0, 1
(c) is true because f x
0 on 1, 3
(d) is true because to the left of 3 f is increasing: f x
0.and to the right of 1 f is decreasing:
f x
0. So the concavity changes.
(e) is true because f is increasing: f x
0
x
12. Compute lim e
x 0
1
x2
x
a.
b.
c.
d.
1
2
0
1
2
Correct Choice
e.
SOLUTION: Apply l’Hospital’s Rule twice:
x
x
x
lH
lH
1
lim e 12 x
lim e 1
lim e
x 0
x 0
x 0 2
2x
2
x
5
13. Compute
i2 .
2
i 2
a.
b.
c.
d.
e.
56
58
60
62
64
Correct Choice
5
SOLUTION:
2
i 2
4
i2
2
22
2
32
2
42
2
52
62
Part 2 – Work Out Problems (5 questions. Points indicated. No Calculators)
Solve each problem in the space provided. Show all your work neatly and concisely, and
indicate your final answer clearly. You will be graded, not merely on the final answer, but also on
the quality and correctness of the work leading up to it.
ax 2
14. (6 points) Find the values of a and b so that f x
b ln x will have an inflection point at
1, 5 .
SOLUTION: f 1
a 5
fx
5x 2 b ln x
b
f x
10x bx
f x
10
f 1
x2
Note:
To be an inflection point, not only must f 1
change. In fact when a 5 and b 10, we have f x
for 0
x
1 and positive for x
15. (8 points) Compute L
lim
x
4
x
1
3
x2
L
lim
x
where
1
P
3
x2
x
ln
lim
x
lim
x ln 1
x
4
x
e
0
b
10
x
1 4
x
3
x2
ln 1
3
x2
b
0, but the concavity must actually
2
10 102
10 x 2 1 which is negative
x
x
1.
SOLUTION:
4
x
10
lim
x
x
lim
e
x
4
x
1
x
3
x ln 1 4
x x2
eP
3
x2
We can apply l’Hospital’s Rule because the numerator and denominator go to zero.
We then multiply top and bottom by x 2 .
4
6
x2
x3
3
1 4x
4 6x
2
l’H
x
P
lim
lim
4
So L e P e 4
x
x
1
4
3
1 x
x2
x2
ALTERNATIVE SOLUTION FOR P :
P
lim
x ln 1
x
4
x
3
x2
lim
t 0
ln 1
4t
t
3t 2
l’H
lim
t 0
1
4 6t
4t 3t 2
1
4
5
16. (10 points) A paint can needs to hold a liter of paint (1000 cm 3 ). The shape will be a cylinder of
radius r and height h. The sides and bottom will be made from aluminum which costs $0. 20 per
cm 2 . The top will be made from plastic which costs $0. 05 per cm 2 . What are the DIMENSIONS
and COST of the cheapest such can? (Keep answers in terms of .)
SOLUTION: V
C . 20 2 rh
r2
400 . 5 r
C
r2
800 1/3
r
C
400
r
r2h
. 05 r 2
0
. 25 r 2
OR C . 4 rh . 25 r 2
1000
r2
400 . 25 r 2
. 4 rh . 25 r 2
r
400
400
800
.5 r
r3
.5
r2
1000
1000 800 1/3
5 800
2/3
800
4
800
1000
h
400
800
2
5
h
800
. 25
1/3
800
1/3
2/3
2/3
800
2
5
4
800
1/3
2/3
800
3
4
4
1/3
800
4
2/3
3
4
800
800
2/3
2/3
To see that C is the absolute minimum, we note there is only one critical point and
400 . 25 r 2 approaches as r approaches 0 or .
C
r
1
17. (8 points) Derive the formula d arcsin x
.
1 x2
Note: The inverse function of sin x is denoted by either arcsin x or by sin
dx
1
x .
SOLUTION 1: If y arcsin x then x sin y . Use implicit differentaition to differentiate both
sides with respect to x:
dy
dy
1
So
But since y arcsin x , this says:
1 cos y
dx
dx
cos y
d arcsin x
1
1
dx
cos y
cos arcsin x
To get this in the correct form, if sin y
x then cos y
1 sin 2 y
1 x2 .
Notice we use the positive square root because y arcsin x is between
and
so that
2
2
1
1
cos y is positive. So d arcsin x
dx
cos y
1 x2
SOLUTION 2: If b
function says:
dg
1
dx x a
df
dx x b
6
arcsin a then a
So
d arcsin x
dx
sin b . The formula for the derivative of the inverse
x a
1
d sin x
dx
1
cos x | x b
x b
1
cos arcsin a
18. (16 points) For the function f x
a.
xe x find
all intervals where f is increasing or decreasing.
e x xe x
1 x ex.
SOLUTION: f x
For x
1, f x
0 and f is decreasing.
1, f x
0 and f is increasing.
For x
b.
critical point: x
all intervals where f is concave up or concave down.
SOLUTION: f x
e x e x xe x
2 x ex.
For x
2, f x
0 and f is concave down.
For x
2, f x
0 and f is concave up.
c.
1
inflection point: x
2
all local minima, local maxima, absolute minima and absolute maxima (give location
and value).
SOLUTION: Since f is decreasing on the left of 1 and f is increasing on the right of
1, there is a local and absolute minimum at x
1 and f 1
e x.
Since there are no other critical points and no endpoints, there are no local or absolute
maxima.
d.
all horizontal asymptotes (be sure to compute the limits).
SOLUTION: xlim xe x
x
e
no horizontal asymptote as x
.
0 "
indeterminate form. Apply l’Hospital’s Rule:
lH
lim xe x xlim x x
lim 1 x
lim e x 0
x
x
x
e
e
So y 0 is the horizontal asymptote as x
.
x
lim xe
“
7