Name
MATH 152H
Exam 3
Sections 201/202 (circle one)
1-13
/52
Spring 2016
14
/20
P. Yasskin
15
/16
16
/18
Total
/106
Solutions
Multiple Choice: (4 points each. No part credit.)
3 1/n
1. The series S
3 1/ n
1
is
n 1
a. absolutely convergent.
correct choice
b. conditionally convergent.
c. divergent by the n th Term Divergence Test
d. divergent by the Alternating Series Test.
1
n
e. divergent because it is the difference between two p-series with p
1
p
n
1 and
1.
1
Solution: The series is telescoping and
k
3 1/n
Sk
3 1/ n
1
31
3 1/2
3 1/ k
1
3 1/2
3 1/3
3 1/k
3 1/ k
1
31
3 1/ k
1
n 1
So S
lim S k
k
lim 3 1
k
31
30
3
1
2. So it is convergent.
The related absolute series is the same series since 3 1/n
2. The series
n 1
1 n
n
3 1/ n
1
0.
is
a. convergent and absolutely convergent.
b. absolutely convergent but not convergent.
c. convergent but not absolutely convergent.
correct choice
d. divergent.
Solution:
n 1
1 n
n
is convergent because it is an alternating decreasing series and nlim 1
n
The related absolute series is
n 1
1
1. So
2
convergent.
p
n 1
1 n
n
1
n
0.
which is divergent because it is a p-series with
is convergent but not absolutely convergent i.e. conditionally
1
sin x 2 , compute f
3. If f x
a.
6 6!
b.
6 3!
6
0 .
c. 6 3!
d.
5!
correct choice
e. 5!
Solution: sin t
1 t3
6
t
The coefficient of x 6 is
4. Compute
n 0
1
e2
1
b.
e2
c. 1
2
d. e 2
a.
2
n!
6
0
6!
f
Equating we have
1 t5
1 x 10
sin x 2 x 2 1 x 6
5!
6
5!
1 . The coefficient of x 6 in the general Maclaurin series is
6
1 . So f 6 0
6!
5!
6
6
f
6
0
.
6!
n
HINT: Think about a known Maclaurin series.
correct choice
e. The series diverges.
Solution:
n 0
xn
n!
ex
n 0
cos x 3
5. Compute
a.
b.
So
lim
1
2
n!
n
e
1
e2
2
1 x6
2
x 12
x 0
1
24
1
4
c. 1
4
d. 1
6
e. 1
24
correct choice
cos x
3
Solution: lim
1
x 12
x 0
lim
x 0
1 x6
2
1
lim
x
24x 12
2
x3 4
4!
1
1 x6
2
x 12
x 0
3 4
x3
2
1
24
2
6. The series S
n 0
a. S
0
b. 0
S
c. S
4
d. S
4
3n
1 4n
satisfies
correct choice
4
e. The series diverges.
3n
4n
Solution:
n 0
Since |r|
is geometric.
1,
n 0
Also
n 0
3n
1 4n
3n
4n
1
1
So
4
3
4
3.
4
1 and r
a
n 0
1
3n
4n
n 0
3n
4n
0 because all terms are positive.
7. Find the equation of the sphere whose diameter has endpoints at
a.
b.
x
x
6
2
3
2
c.
x
3
2
d.
x
3
2
6
2
e.
x
2
2
1
2
y
1
2
y
1
2
2
2
y
y
y
4
z
14
2
7
24
z
7
2
6
z
7
2
z
14
4, 3, 8 .
6
2
z
and
2, 1, 6
correct choice
24
2
24
Solution: The midpoint is 1 2 4, 1 3, 6 8
3, 1, 7 .
2
The radius is the distance from 2, 1, 6 to 3, 1, 7 :
r
d 2, 1, 6 , 3, 1, 7
The circle is
x
3
3
2
y
1
8. A triangle has vertices A
2
2
2
1
z
7
2
1, 2, 3 , B
1
2
7
6
2
1
4
1
6
6.
and C
2, 3, 3
1, 3, 2 . Find the angle at A.
a. 0°
b. 30°
c. 45°
d. 60°
correct choice
e. 90°
Solution: AB
B
A
2, 3, 3
AB AC
0
1
AB
cos
0
1
AB AC
AB AC
1
2 2
1, 2, 3
1
1
2
1
0
1, 1, 0
AC
2
AC
C
A
0
1, 3, 2
1
1
1, 2, 3
0, 1, 1
2
60°
3
1
at x 4 is
x
1f 4 x 4 2 1f 4 x 4 3
T3 x
f4 f 4 x 4
2
6
1
1 x 4
3 x 4 2
5 x 4 3
2
16
256
2048
15
7/2
and the third derivative is f x
x .
8
If the 2 nd degree Taylor polynomial T 2 x is used to approximate f 6
9. The 3 rd degree Taylor polynomial for f x
1
6
we get
1
1 6 4
3 6 4 2
27
. 4219
2
16
256
64
Which of the following is the best bound on the error in this approximation?
T2 6
a. |E 2 |
b. |E 2 |
c. |E 2 |
d. |E 2 |
e. |E 2 |
5
128
5
256
15
128
15
256
15
512
. 0391
. 0195
correct choice
. 1172
. 0586
. 0293
Solution Method 1: This is the beginning of an alternating series.
So the error in the approximation is less than the absolute value of the next term.
5 6 4 3
5
. 0195.
|E 2 | |a 3 |
2048
256
Solution Methiod 2: The Taylor Remainder Theorem says:
If you use T 2 x , to approximate the function, f x , then the remainder, E 2 x
f x T 2 x , is
Mx a 3
bounded by |E 2 x |
where a is the center and M |f 3 c | for all c between a
3!
and x.
With x 6, this says:
M6 4 3
E 2 6 is bounded by |E 2 6 |
where M |f 3 c | for all c between 4 and 6.
3!
15 x 7/2 on 4, 6 occurs at x 4. So we take M
15
15 .
The maximum of |f 3 x |
7
8
1024
8 4
3
6
4
15
5
Then |R 2 x |
. 0195.
1024
256
3 !
Both methods give the same answer.
10. Suppose the function y
f1
fx
is the solution of
dy
dx
x2
y 2 satisfying the initial condition
2. Find f 1 .
a. 1
b. 2
c. 3
d. 1
e. 2
correct choice
Solution: As a solution, y
fx
satisfies
df
dx
x2
f x 2 . So f 1
12
f1
2
1
22
3.
4
11. Find the solution, y
dy
dx
x2
y2
satisfying the initial condition
3. What is f 3 ?
f0
a.
b.
c.
d.
e.
f x , of the differential equation
3 2
332
3 36
6
18
correct choice
Solution: We separate: y 2 dy
From the initial condition, when x
Then y 3
x3
3C
x3
0, y
27 and y
x3
3. So
27
y3
3
x 3 C.
3
27
0 3 C. So C
3
3
f x . So f 3
54 1/3
x 2 dx and integrate:
1/3
12. Find the integrating factor for the linear differential equation x 4
dy
dx
4x 3 y
9.
332.
x2.
a. x 4
b. 4 ln x
c.
4 ln x
d. e
4/x
e. x
4
correct choice
Solution: In standard form the equation is
So P
4,
x
4 dx
x
P dx
dy
dx
4 ln x and I
4y
x
e
P dx
1 .
x2
e
4 ln x
x
13. Which of the following is the direction field of the differential equation
a.
b.
c.
4
dy
dx
x 2 y?
d.
correct choice
Solution: The slope, x 2 y, must be positive in the 1 st and 2 nd quadrants and negative in the 3 rd
and 4 th quadrants.
5
Work Out: (Points indicated. Part credit possible. Show all work.)
14. (20 points) For each series, determine if it is convergent or divergent.
Be sure to identify the Convergence Test(s) and check out their hypotheses.
a. (6 pts)
ne
n2
n 2
Solution: Use Integral Test:
Consider f x
xe
x2
Note: The terms are positive, f interpolates: f n
f is decreasing because f x
e
x2
We integrate using the substitution u
2
1 e u du
1 e x2
x e x dx
2
2
2
So
ne
n2
xe
x
2
x2
2
1e
2
2x
du
4
ne
e
n2
x2
1
and
2x 2
0 for x
2.
2x dx
which converges
converges.
n 2
b. (6 pts)
n 2
n5
n6
3
1
Solution: Compare to
n 2
Note:
1
n
which is a divergent harmonic series (p-series with p
n5
n6
The Simple Comparison Test will not work because
3
1
n5
n6
1).
1
n
Use Limit Comparison Test:
5
lim a n
lim n 6 3 n
1 which is finite and non-zero.
n
n
bn
n 1 1
So
n 2
n5
n6
c. (8 pts)
n 2
3
1
diverges also.
sin n
n2 1
Solution: The series is not positive and is not alternating. So we cannot use a test for positive
series nor the alternating series test.
Use Related Absolute Series Test:
The related absolute series is
n 2
and so
n 2
1
n2
|sin n|
n2 1
which is
n 2
is convergent. Consequently,
n 2
test. Consequently,
n 2
sin n
n2 1
|sin n|
n2 1
1
n2
which is a p-series with p
2
is convergent by the comparison
converges because its related absolute series converges.
6
15. (16 points) Find the radius and interval of convergence of the series
n 1
1 n
x
n 3n
4 n.
Be sure to identify the Convergence Test(s) and check out their hypotheses.
|x
4| n
n 3n
|x 4| n 1
n 3n
n 1 |x
4| n
n 1 3
|x
Solution: Use Ratio Test:
lim aann1
n
lim
n
|a n |
|x 4| n 1
n 1 3n 1
|a n 1 |
4|
3
lim
n
n
4|
|x
n
1
3
4|
1 or |x 4| 3.
3
3 and the series is absolutely convergent on
|x
The series is absolutely convergent when
So the radius of convergence is R
1, 7 .
We check the endpoints:
x
1:
n 1
x
7:
n 1
1 n
1
n 3n
4
1 n
7
n 3n
4
1
n
n
which is a divergent harmonic series.
n 1
1
n
n
n
which is convergent by the Alternating Series Test.
n 1
So the interval of convergence is 1
x
7 or
1, 7 .
7
16. (18 points) A salt water fish tank contains 20 liters of water with 700 grams of salt. In order to
reduce the salt concentration, you pour in salt water with a concentration of 15 grams of salt per liter
at 2 liters per minute. You keep the tank well mixed and drain the mixture at 2 liters per minute. Let
S t be the amount of salt (in grams) in the tank at time t (in minutes).
a. (6 pts) Write the differential equation and initial condition for S t .
Solution:
S0
dS g
dt min
g
L
15
2
Sg
20 L
L
min
2
L
or dS
dt
min
30
1 S
10
700
b. (9 pts) Solve the initial value problem for S t .
Solution Method 1: The equation is separable. Separate and integrate.
dS
1 S
dt
10 ln 30
t C
1
10
30
S
10
C
1 S
t
1 S
Solve:
ln 30
30
e C/10 e t/10
10
10
10
10
1 S Ae t/10
30
where A
e C/10
S 300 10Ae t/10
10
Use the initial condition:
At t 0, we have S 700. So
700
300
10A
A
40
Substitute back:
S
300
400e
t/10
Solution Method 2: The equation is linear: Put it in standard form:
dS
dt
1 S
10
30
1 dt
e t/10
10
1 e t/10 S 30e t/10
Multiply the standard equation by I:
e t/10 dS
10
dt
d e t/10 S
Recognize the left side as the derivative of a product:
30e t/10
dt
Integrate and solve:
e t/10 S
30e t/10 dt 300e t/10 C
S 300 Ce t/10
Find the integrating factor:
Use the initial condition:
700
300
C
C
Substitute back:
I
exp
At t
0, we have S
700. So
400
S
300
400e
t/10
c. (3 pts) After how many minutes will the amount of salt in the tank drop to 400 grams?
Solution: Set S
400
t
300
400e
10 ln 1
4
400 and solve for t:
t/10
400e
t/10
100
e
t/10
1
4
t
10
ln 1
4
10 ln 4
8