Name
ID
MATH 253
Section
FINAL EXAM
Sections 501-503
Fall 1998
Solutions
P. Yasskin
Multiple Choice: (10 points each)
1
1"x
HINTS:
ex
1. Compute
a. 0
b.
c.
d.
e.
2. Find
b.
c.
d.
e.
1 x x2
2
1x x
2
x3
3
5
7
x" x x " x C
3!
5!
7!
2
4
6
cos x 1 " x x " x C
2!
4!
6!
C
x3
3!
sin x
C
3n 2
lim
nv. 1 n 3
correctchoice
1
2
3
Divergent
3n 2
lim
nv. 1 n 3
a.
r
2
5
"2
5
3
5
5
3
"2
3
5
1"r
1
2
3
n
3n
lim
nv. 1 n 3
1
n3
such that
lim
nv.
3
n
1
n3
5 5r 5r 2
0
5r 4
1
5r 3
C 3.
correctchoice
3
5
3
1"r
r
1" 5
3
"2
3
1
!
99
3. Compute
1
1
"
k1
k
correctchoice
k1
a. .9
b.
c.
d.
e.
!
99
k1
.99
1
1.1
Divergent
1
k
1
"
k1
1 " 1 .9
10
!
.
4. The series
1
"
1
1
n2
n1
1
"
2
1
2
C
1
"
99
1
100
is
n
1
2
!
!
.
a. divergent by comparison to
n1
b. convergent by comparison to
1 .
n
.
n1
1 .
n2
correctchoice
c. divergent by the ratio test.
d. convergent by the ratio test.
e. divergent by the n th -term test.
!
.
In the series
n1
1
n2
n
!
.
apply the Comparison Test by comparing with
n1
p-series since
!
.
So
n1
p
1
n2
n
2
1.
n2.
the largest term in the denominator is
Since
n2
n
n2
1
n2
So, we
which is a convergent
we have
1
n2
n
1 .
n2
is also convergent.
2
!
.
5. The series
Ÿ" 1 n
n1
3n 2
1 n3
is
a. absolutely convergent.
b. conditionally convergent.
correctchoice
c. divergent to ..
d. divergent to "..
e. oscillatory divergent.
!
.
The series
Ÿ" 1 n
n1
series and
3n 2
1 n3
is convergent because it is an alternating, decreasing
The related absolute series is
0.
;
is divergent by the Integral Test since
!
.
Ÿ" 1 n1
n
3n 2
1 n3
6. Compute
lim
xv0
!
.
3n 2
lim
nv. 1 n 3
.
1
3x 2 dx
1 x3
n1
lnŸ1 x 3 3n 2
1 n3
.
1
..
which
So
is conditionally convergent.
cosŸ2x " 1 2x 2
x4
a. 0
1
24
c. 1
12
d. 2
3
e. .
b.
correctchoice
cosŸ2x " 1 2x 2
lim
xv0
x4
Ÿ2x 1"
2!
lim
xv0
Ÿ2x 2
lim
xv0
4
4!
x4
"C
Ÿ2x 4
4!
x4
16
24
"C
" 1 2x 2
2
3
3
!
.
7. Given that
a.
b.
c.
d.
e.
1
1"x
1
2
Ÿ1 " x x
2
Ÿ1 " x x
1"x
n
1"x
xn
n0
(for |x|
then (for |x|
1),
1) we have
n xn
n0
correctchoice
!
.
Start with
!
.
1
1"x
x
n
n0
Multiply by
1 .
1"x
!
.
to get
x
nx n
n0
Ÿ1
!
.
d
dx
Apply
to get
nx n"1
n0
1
.
2
Ÿ1 " x x
.
" x 2
0 t x t 3,
8. Find the x-coordinate of the center of mass of the rectangle
> xy.
0tyt2
if the density is
a. .5
b. 1
c. 1.5
correctchoice
d. 2
e. 2.5
M
;;
x-mom
x
> dA
;;
x-mom
M
;;
2
0
x> dA
18
9
3
xy dx dy
0
;;
2
0
3
0
x2
2
x 2 y dx dy
3
0
y2
2
x3
3
2
0
3
0
9
2
y2
2
4
2
27
3
2
0
9
4
2
18
2
4
above the paraboloid
x2 y2 1
if the density is
> x2 y2.
9. Find the mass of the solid inside the cylinder
z
a.
b.
c.
d.
e.
x2 y2
=
2
2=
3
=
3=
2
2=
and below the plane
z
correctchoice
In cylindrical coordinates, the cylinder is
the density is
> r2.
M
;;;
10. If
a.
b.
c.
d.
e.
11. If
a.
b.
c.
d.
> dV
; ;;
2=
1
0
0 r
1
0
4
6
2= 2r " r
4
6
2
r 2 r dz dr d2
2
2= 1 " 1
2
6
then
F Ÿxz 2 , "yz 2 , z 3 2 , 2 , 3z 2
Ÿz z
2,
2 , 3z 2
z
"
z
Ÿ
2
correctchoice
3z
2
5z
2Ÿy " x z
F
2
Ÿxz 2 Ÿ"yz 2 Ÿz 3 x
y
z
2=
then
•F
F Ÿxz 2 , "yz 2 , z 3 2Ÿy " x z
2Ÿx y z
Ÿ2yz, "2xz, 0 correctchoice
Ÿ2yz, 2xz, 0 ;
1
the paraboloid is
1
r3z
0
2
r2
dr
2=
2=
3
F
r
;
1
0
z
r2
and
r 3 Ÿ2 " r 2 dr
z2 " z2
3z 2
3z 2
e. 0
•F
î
F
k
x
y
z
îŸ0
" "2yz " F Ÿ0 " 2xz kŸ0 " 0 Ÿ2yz, 2xz, 0 xz 2 "yz 2 z 3
5
;
12. Compute
from
t
rŸt =
y 2 e xy dx Ÿ1 xy e xy dy
to
t
3=.
F
HINT:
Find a scalar potential for
a. "2=
b. 0
correctchoice
c. =
d. 3=
e. 4=
f
x
f
y
So
;
rŸt y 2 e xy
Ÿ1 f
®
f
xy e xy
rŸt Ÿt cos t, t sin t along the spiral
ye xy
e , Ÿ1 xy e xy .
f
y
®
0
2 xy
®
gŸ y dg
dy
®
Ÿy
g
C
e xy
xye xy
Take
dg
dy
Ÿ1 xy e xy
C
0.
" fŸ"=, 0 0e 0 " 0e 0
dg
dy
ye xy
y 2 e xy dx Ÿ1 xy e xy dy
fŸ"3=, 0 f rŸ3= " f rŸ= 0
Work-Out Problems
13. (25 points) You are given:
a. (10 pt) If
fŸx e
"x 2
!
.
"x 2
e
Ÿ" 1 n
n0
,
;
1 " x2
x4 " x6
2
6
C.
f Ÿ6 Ÿ0 .
find
f Ÿ6 Ÿ0 6
x
6!
The term with an
x6
is
f Ÿ6 Ÿ0 " 6! "5! "120.
6
x 2n
n!
6
"x .
6
So:
b. (10 pt) Use the quadratic Taylor polynomial approximation about
"x 2
e
to estimate
0.1
e
"x 2
0.1
e
0
"x 2
dx X
;
0.1
0
;
e
"x 2
1 " x 2 dx
and
0
for
(Keep 8 digits.)
dx.
;
0.1
3
x" x
3
0.1
e
"x 2
;
c. (5 pt) Your result in (b) is equal to
Since
0
The quadratic Taylor polynomial approximation is
;
x
0
0.1
e
e
.1 " .001
3
"x 2
dx
"x 2
X 1 " x2.
So:
.09966667
to within o how much? Why?
0
dx
are alternating decreasing series, the
0
error is at most the next term:
5
0.1 4
x dx x 5 0.1 Ÿ.1 2
10 0
10
0
10 "6
.000001
6
!
.
14. (15 points) Find the interval of convergence for the series
" 5 n
.
3nn3
Ÿx
n1
Be sure to identify each of the following and give reasons:
(1 pt) Center of Convergence:
an
>
" 5 n
3nn3
Ÿx
a n1
lim
an
nv.
a n1
Ÿx " 5 3 n1 Ÿn lim
nv.
a
Ÿx " 5 3 n1 Ÿn n1
n1
1 3
3nn3
" 5 n
1 3 Ÿx
|x " 5|
3
The series converges if
5
or
1
lim
nv.
Ÿx
" 5 3
|x " 5|
!
.
The series
" 5 n
3nn3
Ÿ8
n1
p
3
x
53
!
.
n1
1
n3
n1
|x " 5|
3
3
Radius of Convergence:
(1 pt) Right Endpoint:
3
n
R
3
(5 pt)
8
converges because it is s p-series with
1.
At the Right Endpoint the Series
Converges
(circle one)
(3 pt)
Diverges
(1 pt) Left Endpoint:
!
.
The series
!
.
series
n1
and
lim 1
nv. n 3
n1
" 5 n
3nn3
Ÿ2
1
n3
x
5"3
!
.
n1
converges
Ÿ" 1 n
2
n
3
OR
converges because its related absolute
because it is an alternating decreasing series
0.
At the Left Endpoint the Series
Converges
(circle one)
(3 pt)
Diverges
(1 pt) Interval of Convergence:
2txt8
or
¡2, 8 ¢
7
15. (30 points) Stokes’ Theorem states that if S is a surface in 3-space and S is its
boundary curve traversed counterclockwise as seen from the tip of the normal to S
then
;;
• F dS
•F
Ÿ"yz, xz, z 2
and S is the part of the cone
with normal pointing in and up.
.
•F
a. (5 pt) Compute
F
2
(HINT: Use rectangular coordinates.)
î
F
k
x
y
z
xz
2
" yz
F
ds
S
S
Verify Stokes’ Theorem if
z x2 y2
below
z
>
z
îŸ0
" x " F Ÿ0 " "y k Ÿz " "z Ÿ"x, "y, 2z • F dS.
;; b. (10 pt) Compute
S
(HINT: Here is the parametrization of the cone and the steps you should use.
Remember to check the orientation of the surface.)
R
Ÿr, 2 Ÿr cos 2, r sin 2, r R
r
Ÿcos 2, sin 2, 1 R
2
Ÿ"r sin 2, r cos 2, 0 N
iŸ"r cos 2 •F
•F
" jŸr sin 2 kŸr R
Ÿr, 2 N
• F dS ;; S
Ÿ"r cos 2, "r sin 2, r Ÿ"r cos 2, "r sin 2, 2r r 2 cos 2 2 r 2 sin 2 2 2r 2
;;
2=
2
0
0
3r 2 dr d2
ds
> F
c. (15 pt) Compute
.
2=¡r 3 ¢ 20
3r 2
Recall
16=
F
2
Ÿ"yz, xz, z .
S
(HINT: Parametrize the boundary circle. Remember to check the orientation of the
curve.)
rŸ2 Ÿ2 cos 2, 2 sin 2, 2 r Ÿ2 F
Ÿ"4 sin 2, 4 cos 2, 4 ds
> F
S
;
2=
0
8 d2
vŸ2 Ÿ"2 sin 2, 2 cos 2, 0 F
v
8 sin 2 2 8 cos 2 2
8
16=
8
16. (20 points) Find the minimum value and its location(s) for the function
9x 2
on the ellipse
f
f
xy
g
9x 2
5
g:
y
x
4y 2
4y 2 " 72
518x
58y
®
®
Constraint:
9x 2 4y 2 72
y o 3 x o 3 Ÿ o 2 o3
2
2
f
Ÿy, x 5
y
18x
x
8y
5
18x 2
g
"6
at
xy
4y 2
Ÿ18x, 8y y
18x
®
72
Critical points:
Ÿ2, 3 ,
Ÿ2, "3 ,
Ÿ"2, 3 ,
fŸ2, 3 6,
fŸ2, "3 "6,
fŸ"2, 3 "6,
The minimum is
fŸx, y 72.
Ÿx, y Ÿ2, "3 x2
4
x
x
8y
®
9x 2
o2
Ÿ"2, "3 fŸ"2, "3 6
and
Ÿx, y Ÿ"2, 3 .
9