Student (Print)
Section
Last, First Middle
Student (Sign)
Student ID
Instructor
MATH 152
Exam 3
Fall 2000
Test Form A
Solutions
1-9
/45
10
/10
11
/15
Part II is work out. Show all your work. Partial credit will be given.
12
/10
You may not use a calculator.
13
/10
14
/10
Part I is multiple choice. There is no partial credit.
TOTAL
1
Part I: Multiple Choice (5 points each)
There is no partial credit. You may not use a calculator.
!
.
1. Compute
a.
b.
c.
d.
25
3
125
3
125
2
75
2
3 n 5 2"n
n0
correctchoice
e. .
a 3 0 5 2 25, r 3 1
5
Geometric:
!
.
2. The series
n1
Ÿ" 1 S
25
1" 3
5
125
2
n
n
is
a. absolutely convergent
b. convergent but not absolutely convergent
correctchoice
c. divergent
d. none of the above
The series is convergent by the Alternating Series Test since it is (1) alternating due to the Ÿ"1 n , (2)
1 0.
decreasing in absolute value since 1n gets smaller, and nlim
v. n
!
.
However it is not absolutely convergent because the related absolute series is
n1
1 which is the
n
harmonic series which is divergent.
!
.
3. Determine the interval of convergence of the power series
n0
3 n
.
n1
Ÿx
a. Ÿ2, 4 ¢
b. ¡2, 4 c. Ÿ"4, "2 ¢
d. ¡"4, "2 correctchoice
e. Ÿ"., . Ratio Test:
a n1
L nlim
v. a n
|x 3|n1 n 1
nlim
|x 3| 1
v. n 2
|x 3|n
or "4 x "2
At x "4 the series is alternating and decreasing and so convergent.
At x "2 the series is harmonic and so divergent.
So the interval of convergence is "4 t x "2.
2
4. Find the Maclaurin series for fŸx sin 2x.
3 3
5 5
7 7
a. 2x " 2 x 2 x " 2 x C
3!
5!
3
5
2x
2x
2x "
"
3!
5!
3
5
7
x" x x " x
3!
5!
7!
3
5
7
x
x
x
x
3!
5!
7!
3
5
2x 2x 2x 3!
5!
b.
c.
d.
e.
7
correctchoice
7!
2x C
7!
C
C
2x 7 C
7!
3
5
7
The Maclaurin series for sin x is sin x x " x x " x C
3!
5!
7!
3 3
5 5
7 7
2
x
x
x
2
2
C
Substitute x v 2x: sin 2x 2x "
"
3!
5!
7!
5. Find the 3 rd degree Taylor polynomial for fŸx e "2x about x 1.
a. e "2 2e "2 Ÿx " 1 2e "2 Ÿx " 1 2 4 e "2 Ÿx " 1 3
3
b. e "2 " 2e "2 Ÿx " 1 2e "2 Ÿx " 1 2 " 4 e "2 Ÿx " 1 3
c. e "2 2e "2 Ÿx " 1 4e "2 Ÿx " 1 2
d. e "2 " 2e "2 Ÿx " 1 4e "2 Ÿx " 1 2
e. e "2 " 2e "2 Ÿx " 1 " 4e "2 Ÿx " 1 2
correctchoice
3
8e "2 Ÿx " 1 3
" 8e "2 Ÿx " 1 3
" 8e "2 Ÿx " 1 3
The function, 3 derivatives and their values at x 1 are
fŸx e "2x
fŸ1 e "2
f U Ÿx "2e "2x
f U Ÿ1 "2e "2
f UU Ÿx 4e "2x
f UU Ÿ1 4e "2
f UUU Ÿx "8e "2x
f UUU Ÿ1 "8e "2
So the Taylor polynomial is
f UUU Ÿ1 f UU Ÿ1 2
3
T 3 fŸ1 f U Ÿ1 Ÿx " 1 Ÿx " 1 Ÿx " 1 2
6
"2
"2
e "2 " 2e "2 Ÿx " 1 4e Ÿx " 1 2 " 8e Ÿx " 1 3
2
6
!
.
6. Compute
a. "4
b. "2
n0
n 1
2
n"1
!
.
.
HINT: Differentiate
xn.
n0
c. 4
9
d. 2
e. 4
!
.
n0
xn correctchoice
1
1"x
Differentiate:
provided |x| 1
!
.
nx n"1 n0
Evaluate at x 1 :
2
!
.
n0
1
Ÿ1
n 1
2
provided |x| 1
" x 2
n"1
1
1"
1
2
2
4
3
7. Find a unit vector which is orthogonal to the plane containing the points P Ÿ1, 0, 0 , Q Ÿ0, 2, 0 ,
and R Ÿ0, 0, 3 .
6 , "3 , 2
a.
11 11 11
b. Ÿ6, "3, 2 c. Ÿ6, 3, 2 6 , "3 , 2
d.
7 7 7
6, 3, 2
e.
7 7 7
correctchoice
PQ Q " P Ÿ"1, 2, 0 , PR R " P Ÿ"1, 0, 3 i
N PQ • PR j k
"1 2 0
"1 0 3
Ÿ6, 3, 2 N 36 9 4 49 7
N
N
6, 3, 2
7 7 7
8. Find the equation of the circle for which one diameter has endpoints P Ÿ1, 1 and Q Ÿ7, 9 .
a. Ÿx 3 2 Ÿy 4 2 25
b. Ÿx 3 2 Ÿy 4 2 5
c. Ÿx " 4 2 Ÿy " 5 2 25
correctchoice
d. Ÿx " 3 Ÿy " 4 5
e. Ÿx 4 2 Ÿy 5 2 25
2
2
The center is at C PQ
Ÿ4, 5 and the radius is r CP |Ÿ3, 4 | 5.
2
9. Consider the triangle with vertices A Ÿ0, 0, 0 , B Ÿ1, 1, 0 and C Ÿ0, 2, 2 . Find the angle at A.
a. 0°
b. 30°
c. 45°
d. 60°
correctchoice
e. 90°
AB 11 2
AC 44 cos 2 AB AC
AB AC
2
1
2
22 2
8 2 2
AB AC 2
2 60°
4
Part II: Work Out
Show all your work. Partial credit will be given.
You may not use a calculator.
10. (10 points)
a. Find the power series representation for fŸx convergence.
1
1
1 x2
centered at x 0 and its radius of
!x
.
n
provided |x| 1
1"x
n0
Substitute x v "x 2
.
.
n 2n
1
2 n
"
provided |x| 1
x
Ÿ
Ÿ" 1 x
1 x2
n0
n0
!
!
;
So R 1
b. Find the power series representation for fŸx x tan "1 x centered at x 0 and its radius of
HINT: tan "1 x convergence.
tan "1 x ;
x
0
1
1t
fŸx x tan "1 x 2
dt ! "1
.
n0
Ÿ
;!
x .
Ÿ" 1 0 n0
2n2
n
x
2n 1
x
0
1
1 t2
t dt n 2n
dt
! "1
.
n0
Ÿ
n
x 2n1
2n 1
and R 1
and R 1
5
11. (15 points) Determine if each of the following series converges or diverges. Say why.
Be sure to name or quote the test(s) you use and check out all requirements of the test.
! n "ln1n
.
a.
n2
Ÿ
Ÿ
n
Circle one:
2
Converges
Diverges
Explain:
The related absolute series is
;
! n lnn1
.
n2
Ÿ
2
. Apply the Integral Test:
1
nŸlnn 2
is continuous,
decreasing and
.
1
"1 . 0 " "1 1
2
lnn 2
ln 2
ln 2
2 nŸlnn .
n
Ÿ" 1 So
also converges.
2
n2 nŸlnn !
!
.
b.
Ÿ" 3 n
Circle one:
n!
Explain:
n0
Apply Alternating Series Test:
It is alternating because of the
Converges
Diverges
n
. It is decreasing in absolute value because 3
gets
n!
n
3
3
3
3
C
3
smaller as n gets larger provided n 3. Finally, lim
nlim
0
because
nv. n!
v. 1 2 3 C n
.
n
Ÿ" 3 beyond n 3 we keep multiplying by fractions less than 1.
So
converges.
n!
n0
OR
Apply the Ratio Test:
n
n1
|a n | 3
|a n1 | 3
n!
Ÿn 1 !
.
n
Ÿ" 3 a n1 lim 3 n1 n! lim 3 0 1
So
converges.
L nlim
v. a n
nv. Ÿn 1 ! 3 n
nv. n 1
n!
n0
Ÿ" 3 n
!
!
!
.
c.
Ÿ" 1 n1 n"1
5
2 n
4
n
n1
Explain:
Circle one: Converges
Apply the Ratio Test:
n"1
5n
|a n | 52 n
|a n1 | 2 n1
n 4
Ÿn 1 4
a n1 lim
5n
n 2 4 n 5 lim
n2
5 1
L nlim
n
"
1
2
a
n
v.
nv. n 1 4 1 5
nv. n 1 2
n
4
4
Ÿ
Ÿ
!
.
So
n0
Ÿ" 3 n!
Diverges
n
diverges.
6
12. (10 points) Determine if each of the following series converges or diverges. Say why.
Be sure to name or quote the test(s) you use and check out all requirements of the test.
If it converges, find the sum. If it diverges, does it diverge to ., ". or neither?
!
.
a.
n1
1
1 3n
Circle one: Converges
Diverges
Explain:
!
.
1
n
diverges because it is a p-series with p 1 1.
3
n1
1
Apply the Limit Comparison Test: a n bn 1
3
3 n
1 n
a n lim 3 n
and 0 L .
1
L nlim
v. b n
nv. 1 3 n
3
!
.
So
n1
!
.
b.
n1
1
1 3n
also diverges. Since
1
0, it diverges to ..
1 3n
n"1 " n
n
n1
Circle one:
Converges
Diverges
Explain:
!
k
n"1 " n
n
n1
n1
S lim S k lim "k "1
kv.
kv. k 1
Sk 0" 1
2
1 " 2
2
3
C
k"1 " k
k
k1
"k
k1
7
13. (10 points) Find the volume of the parallelepiped with adjacent edges PQ, PR and PS, where
P Ÿ1, 1, 1 , Q Ÿ2, 3, 1 , R Ÿ4, 1, 5 and S Ÿ1, 3, 4 .
PQ Q " P Ÿ1, 2, 0 , PR R " P Ÿ3, 0, 4 and PS S " P Ÿ0, 2, 3 1 2 0
V PQ PR • PS
det
3 0 4
|"8 " 18| 26
0 2 3
!
.
14. (10 points) Determine the interval of convergence of the power series
n1
Ÿx
n
" 2 n . Be sure to
1/3 n
3
show how you checked the convergence at the endpoints.
Apply the Ratio Test:
|x " 2|n
|x " 2|n1
|a n | 1/3 n
|a n1 | 1/3 n1
n 3
Ÿn 1 3
1/3
" 2|
a n1 lim |x " 2|n1
n 1/3 3 n |x " 2| lim
n
|x
L nlim
n
1/3 n1 |x " 2|
nv.
nv. n 1
v. a n
3
3
Ÿn 1 3
|x " 2|
The series converges when L 1 or |x " 2| 3 or " 1 x 5
3
We check the endpoint at x "1.
.
.
n
n
Ÿ" 3 Ÿ" 1 The series is
which is convergent because it is an alternating, decreasing
n 1/3 3 n
n 1/3
n1
n1
series.
We check the endpoint at x 5.
.
.
n
Ÿ3 1
The series is
which is a divergent p-series since p 1 1.
1/3 n
1/3
3
n
n 3
n1
n1
!
!
!
!
Thus the interval of convergence is "1 t x 5 or
¡"1, 5 .
8