MATH 172
EXAM 3
Section 502
Solutions
Fall 1999
P. Yasskin
Multiple Choice: (8 points each)
1. Which of the following is the direction field of the differential equation
a.
b.
d.
e.
dy
xy ?
dx
c.
The slope is m xy which is 0 on both axis, positive in quadrants I and III, and negative
in quadrants II and IV. This only happens in plot (b).
2. The limit lim
nv.
4Ÿ2 n
n
Ÿ5 a. converges to 0.
b. converges to
correctchoice
4
.
1" 2
5
c. converges to 4.
8
5 .
d. converges to
1" 2
5
e. diverges.
4Ÿ2 n
2 n 0
n 4 nlim
v. 5
Ÿ5 This is NOT a geometric series. It is not being added up.
lim
nv.
1
3n 2
3. lim
1 " 12
nv.
n
e "6
e6
1
e
e. e "3
a.
b.
c.
d.
correctchoice
3n 2
lim 1 " 12
nv.
n
e L where
3n 2
ln 1 " 12
L nlim
v.
n
1
3n 2 ln 1 " 12
nlim
v.
n
2
n3
1 " 12
n
3 nlim
v.
3 nlim
v.
"2
n3
3n 2
So nlim
1 " 12
v.
n
!
.
4. The series
"1 "3
1 " 12
n
e "3
Ÿ" 1 n1 n
3
n1
ln 1 " 12
n
3
nlim
v.
1
n2
2
n
a. diverges by the n th -Term Test.
b. converges to 2
c. converges to 2
correctchoice
5
d. converges to " 3
5
e. diverges by the Alternating Series Test
Ÿ" 1 Geometric series with a !
.
n1
Ÿ" 1 n1 n
3n
2
2
3
1 " "2
3
11 1
2
31
2 and r "2 .
3
3
2 2
32
5
2
!
.
5.
n1
n1 " n2
n
n1
a. 0
b. 1
correctchoice
c. 2
d. 3
e. .
!
k
n1 " n2
n
n1
n1
k " k1
2 " 3 3 " 4 C
1
2
2
3
k"1
k
2" k2
k1
S lim S k lim 2 " k 2 2 " 1 1
kv.
kv.
k1
Sk !
.
6. The series
n1
k1 " k2
k
k1
n!
2n
a. converges by the Integral Test.
b. diverges by the Integral Test.
c. converges by the Ratio Test.
d. diverges by the Ratio Test.
correctchoice
!
.
e. converges by Comparison with
n1
1 .
2n
Ÿn 1 !
a n1 a n n!n
2
2 n1
|a n1 |
Ÿn 1 ! 2 n
Ÿn 1 nlim
nlim
.1
L nlim
v. |a n |
v. 2 n1
v.
n!
2
Ratio Test:
Divergent
3
!
.
7. The series
n2
2
n n
3
a. diverges by the n th -Term Test.
b. converges by the Ratio Test.
c. diverges by the Ratio Test.
!
!
.
d. converges by Comparison to
.
e. diverges by Comparison to
n2
2
n3
correctchoice
2
n
n2
The n th -Term Test and the Ratio Test fail.
!
.
2 which is a p-series with
3
n
n2
23 and
p 3 1 and so is convergent. Further, n 3 n n 3 . So 3 2
n
n n
.
.
.
2
2 . Therefore
2
is also convergent.
3
3
3
n
n
n
n
n
n2
n2
n2
For large n, we have n 3 !
!
!
.
8. The series
n . So we compare to
n2
!
1
n1
a. converges by the Integral Test.
b. diverges by the Integral Test.
correctchoice
c. converges by the Ratio Test.
d. diverges by the Ratio Test.
e. diverges by the n th -Term Test.
The n th -Term Test and the Ratio Test fail.
Integral Test:
.
.
"1/2
1
dn dn 2Ÿn 1 1/2
Ÿn 1 2
2
n1
;
;
.
2
.
Series is divergent.
4
!
.
9. The series
n1
Ÿ" 1 n
is
n1
a. Absolutely Convergent
b. Conditionally Convergent
correctchoice
c. Convergent for Even n, Divergent for Odd n
d. Convergent for Odd n, Divergent for Even n
e. Divergent
!
.
The related absolute series is
1
n1
n1
problem.
!
.
The original series
Ÿ" 1 n
is alternating from the Ÿ"1 n , decreasing since
n1
gets smaller as n gets bigger and nlim
v.
n1
!
.
Series Test. Therefore,
Ÿ" 1 !
.
n1
Ÿ" 1 n 3/2
n1
1
n1
0. So it is convergent by the Alternating
is Conditionally Convergent.
n
is
a. Absolutely Convergent
b.
c.
d.
e.
1
n
n1
n1
10. The series
which is convergent from the previous
correctchoice
Conditionally Convergent
Convergent for Even n, Divergent for Odd n
Convergent for Odd n, Divergent for Even n
Divergent
!
.
The related absolute series is
n1
convergent.
Therefore the original series
!
.
n1
1 which is a p-series with p 3 1 and hence
2
n 3/2
Ÿ" 1 n 3/2
n
is Absolutely Convergent.
5
11. (15 points) Solve the differential equation 1
x
3
dy
3xy e "x with yŸ0 2
dx
We write the equation in standard linear form:
3
dy
3x 2 y xe "x
dx
We identify P 3x 2 and compute the integrating factor:
; P dx
; 3x 2 dx
e
ex
Ie
3
We multiply thru by e x and identify the left as the derivative of a product:
d e x 3 y e x 3 dy e x 3 3x 2 y x
dx
dx
We integrate:
3
e x y x dx 1 x 2 C
2
We use the initial condition y 2 when x 0:
®
C2
e02 1 02 C
2
Therefore
3
3
y 1 x 2 e "x 2e "x
2
3
;
12. (15 points) A tank contains 5000 gal of water. Initially, there are 10 lb of salt dissolved
in the water. Salt water containing 0.03 lb of salt per gal is added to the tank at the rate
of 2 gal per hour. The solution is kept mixed and is drained at the rate of 2 gal per
hour. Let SŸt be the amount of salt in the tank at time t.
a. State the differential equation and the initial condition satisfied by SŸt .
dS .03 lb
dt
gal
2
S lb
gal
"
hour
5000 gal
in
2
gal
.06 " 1 S
2500
hour
out
SŸ0 10
b. Solve this initial value problem.
;
dS 1 S .06
1 dt e t/2500
®
I exp
2500
dt
2500
d e t/2500 S .06e t/2500
®
dt
®
e t/2500 S .06e t/2500 dt .06 2500e t/2500 C 150e t/2500 C
;
S 150 Ce "t/2500
Initial condition:
10 150 C
Solution:
S 150 " 140e "t/2500
®
®
C "140
c. At large times, what is the asymptotic amount of salt in the tank.
lim SŸt lim 150 " 140e "t/2500 150 lb
tv.
tv.
6