MATH 253
FINAL EXAM
Sections 501-503
Part I:
1. lim
nv.
a.
b.
c.
d.
Spring 1998
Solutions
Multiple Choice
P. Yasskin
(5 points each)
No Partial Credit
n2
n 2 Ÿ" 1 n n
0
1
correctchoice
2
4
e. divergent
!
Ÿ" 1 1
1
Ÿ" 1 n
1
n
n
n
n 1
is
2
n1
lim
nv.
2
nv.
.
2. The series
n2
n Ÿ" 1 n n
lim
a. absolutely convergent
b. conditionally convergent
correctchoice
c. divergent
d. none of these
!
.
n1
Ÿ" 1 n
2
n
n
1
is an alternating, decreasing series and
lim
nv.
n
n2
1
0.
So it is convergent by the Alternating Series Test.
!
.
Further, the related absolute series is
Integral Test, since
;
Thus the original series
1
!
2
n
n2 1
n1
n1
n
2
!
n . 1
n1
.
3.
.
dn
Ÿ" 1 1 lnŸn 2
2
which is divergent by the
1
1 .
1
..
n
n
n 1
2
n
n2
is conditionally convergent.
a. 0
b. 1
2
c. 1
d. 2
e. divergent
!
.
n1
n2
n
2
1
correctchoice
Since
lim
n1
n2
n
2
1
1 p 0,
the series
is divergent by the nth Term Divergence Test.
1
!
.
4. The series
n
n 1.5 1
n1
is
!
.
a. convergent by the Comparison Test with
!
.
b. conv. by the Limit Comp. Test with
n1
c. divergent by the Comparison Test with
!
.
d. div. by the Limit Comp. Test with
n1
n1
1
n .5
!
.
n1
1
n .5
1 .
n .5
but not by the Comp. Test.
1 .
n .5
but not by the Comp. Test.
correctchoice
e. none of these
n
is approximately like
so we
b n 1.5
1.5
1
n
n
.
1
want to compare to
which is a divergent p-series since
p .5 1.
n .5
n1
n 1 ,
n
Since
the Comparison Test does not apply. So we try the
1.5
n 1.5
n .5
n 1
a n lim
n
n .5 lim n 1.5 1,
the
Limit Comparison Test: Since
lim
nv. b n
nv. n 1.5 1
nv. n 1.5 1
1
.
n
series
also diverges.
n 1.5 1
For large n, the term
an
!
!
!
.
5.
n1
n1
2
4n 2 " 1
2
4n 2 " 1
(Note:
1
1
"
)
2n " 1
2n 1
a. 0
b. 1
2
c. 1
correctchoice
d. 2
e. divergent
We first compute the partial sum:
!
k
2
4n 2 " 1
1 " 1
1
3
!
k
1
1
"
2n " 1
2n 1
n1
n1
1 " 1 C
1
1 " 1
" 1
3 .5
2k " 3
2k " 1
2k " 1
2k 1
2
1
lim S k lim 1 "
1
So the infinite sum is
kv.
kv.
2k 1
4n 2 " 1
Sk
!
1"
1
2k 1
n1
2
for
What is the minimum
x0
fŸx e "x .
to within
degree of the Taylor polynomial you should use to approximate
e "0.1
"
8
o10 ?
Give the degree
n
of the highest power of
x
that you need to
keep.
a. 1
b. 3
c. 5
correctchoice
d. 7
e. 9
6. Consider the Taylor series about
!
.
Since
e
x
n0
e "x
we have
e "0.1
!
.
Ÿ" 1 n0
n
2
1x x
2!
Ÿ" 1 n0
!
.
xn
n!
n
Ÿ0.1 n!
n
xn
n!
!
.
Ÿ" 1 x3
3!
x4
4!
x5
5!
2
3
1"x x " x
2!
3!
10 "n
n!
n
n0
C
x4 " x5
4!
5!
1 " 10 "1
10 "2 " 10 "3
2
6
and
C
10 "4 " 10 "5
24
120
10 "6
720
Since this is an alternating decreasing series, the error is bounded by the next term.
10 "5 10 "8 10 "6 ,
Since
we must keep up to the
n5
term.
120
720
7. Find the volume of the solid under the plane
vertices
Ÿ1, 1 ,
Ÿ2, 1 a. 1
b. 2
correctchoice
c. 3
d. 4
e. 9
2
and
z
and above the triangle with
x
Ÿ1, 4 .
The base is shown. The top is
4
z
x.
The diagonal side of the base has slope
So its equation is
3
V
2
;;
;
2
1
2
1
1
0
1
"3x7
x dy dx
y"1
1
"3x 2
6x dx
;
" 3Ÿ x " 2 or
"3x7
2
xy
1
"x 3
y1
dx
2
3x 2
x1
m
;
2
1
"3.
y
¡xŸ"3x ¡"8 "3x 7.
7 ¢ " ¡x ¢ dx
12 ¢ " ¡"1 3 ¢
2
3
2
8. A 5 lb mass moves up the helix
the work done against the force of
a. "4=
b. "5=
c. "20=
correctchoice
d. "80=
e. "100=
;
;
F
v Ÿ"3 sin t, 3 cos t, 4 W
=
F v dt
0
=
0
rŸt Ÿ3 cos t, 3 sin t, 4t "5k.
F
gravity
"20 dt
"20=
;
F
ds
C
a.
b.
c.
d.
e.
;
;
C
counterclockwise around the circle
F
for the vector field
y2 4
2=
4=
8=
16=
32=
correctchoice
P dx Q dy
F
ds
where
Ÿ"yŸx
P
2
y 2 , xŸx 2
"yŸx 2
y2 y 2 .
"yx 2 " y 3
and
C
Q xŸx 2 y 2 x 3 xy 2 .
Let D denote the interior of the circle. Then, by Green’s Theorem,
Q
F
ds
" P dx dy Ÿ3x 2 y 2 " Ÿ"x 2 " 3y 2 dx dy y
x
;
;;
;;
D
C
;
Find
Ÿ0, 0, "5 9. Compute the line integral
x2
0 t t t =.
for
;;
D
Switch to polar coordinates.
ds
F
;;
2=
2
0
0
4r 2 r dr d2
x
2
2=¡r 4 ¢ 20
2
32=
y
r
2
dx dy
r dr d2
2
Ÿ4x 4y 2 dx dy
D
So:
C
4
2
Find the total mass of a plate bounded by the
10.
right half of the cardioid
r
1 sin 2
>
the y-axis if the mass density is
and
3x.
0
a. 4
1
correctchoice
b. =
c. 2
d. =
2
e. 1
2
M
;;
;
> dA
=/2
"=/2
; ;
=/2
1sin 2
"=/2
0
3Ÿr cos 2 r dr d2
cos 2 Ÿ1 sin 2 3 d2
Ÿ1 ;
=/2
=/2
sin 2 4
4
b.
c.
d.
e.
= Ÿ37 3/2 " 1
16
= Ÿ37 3/2 " 1
24
= Ÿ37 3/2
16
= Ÿ37 3/2
4
9=
4
correct
24
4
"=/2
11. Find the area of the piece of the paraboloid
a.
cos 2 ¡r 3 ¢ 1r0sin 2 d2
"=/2
z
0
4
"
9 " x2 " y2
4
in the first octant.
The paraboloid may be parametrized as
R
Ÿr, 2 Ÿr cos 2, r sin 2, 9
" r2 The tangent and normal vectors are
R
r
cos 2, sin 2, "2r
R
2
"r sin 2, r cos 2, 0
N
R
r
•
R
2
Ÿ2r
2
cos 2, 2r 2 sin 2, r 8
The length of the normal is
6
N
4
So the area is A
2
0
2
4r 4 cos 2 2 4r 4 sin 2 2 r 2
r 2
=
2
2
3
1 Ÿ4r 2
8
;;
N
dr d2
1 3/2
3
0
4r 4
; ;
=
24
=/2
3
0
0
Ÿ37 r2
r 4r 2
3/2
r 4r 2
1 dr d2
"1
5
1
Part II:
Work Out Problems
Partial credit will be given.
(10 points) The spiral at the right is made from an
12.
0.5
1
1.5
infinite number of semicircles whose centers are
all on the x-axis. The radius of each semicircle
is half of the radius of the previous semicircle.
a. Consider the infinite sequence of points where the spiral crosses the x-axis.
What is the x-coordinate of the limit of this sequence?
2"1
1 " 1
2
4
1 "C
8
!
.
n0
n
2 "1
2
2
1
1
2
4
3
b. What is the total length of the spiral (with an infinite number of semicircles)?
Or, is the length infinite?
Each semicircle has length
r n 1, 1 , 1 , 1 , C.
2 4 8
So the total length is
L
Ln
!
=r n
.
n0
where the radii are
!
.
=r n
n0
= 1
2
n
=
1" 1
2
2=.
6
2
!
.
13. (15 points) Find the interval of convergence for the series
" 3 n
2 n n ln n
Ÿx
n2
a. (2 pts) The center of convergence is
c
.
3
b. (7 pts) Find the radius of convergence. (Name the test you use.)
We apply the ratio test:
>
lim aan1
nv.
n
lim
nv.
an
" 3 n
2 n n lnn
Ÿx
a n1
" 3 n1
Ÿn 1 lnŸn 1 Ÿx
2 n1
2 n n ln n
" 3 n1
n
Ÿn 1 lnŸn 1 Ÿx " 3 Ÿx
2 n1
|x " 3|
lim n lim lnn
2 nv. n 1 nv. lnŸn 1 |x " 3|
|x " 3|
Ÿ1 Ÿ1 2
2
The series converges if
>
|x " 3|
2
1
n
1
|x " 3|
lim 1 lim
2 nv. 1 1 nv.
n
1
|x " 3|
or
n1
2
R
2
c. (2 pts) Check the left endpoint. (Name the test you use.)
x
3"2
!
.
The series becomes
1
n2
Ÿ" 2 !
.
n
2 n n ln n
n2
1
n lnn
So the series converges by the Alternating Series Test.
This is an alternating, decreasing series and
lim
nv.
Ÿ" 1 n
n ln n
.
0.
convergent
Circle:
divergent
d. (2 pts) Check the right endpoint. (Name the test you use.)
!
.
x
;
.
32
5
;
The series becomes
n2
Ÿ2 n
!
.
n
2 n ln n
1 dn . 1 du ¡ln u ¢ . ..
ln 2
u
2 n lnn
ln 2
So the series diverges by the Integral Test.
n2
1 .
n lnn
u
lnn
du
1 dn
n
convergent
Circle:
divergent
e. (2 pts) The interval of convergence is
¡1, 5 or
1tx
5
.
7
z 2 t 4 for z u 0.
Let H be the hemisphere surface x 2 y 2 z 2 4 for z u 0.
Let D be the disk x 2 y 2 t 4 with z 0.
Notice that H and D form the boundaryof V with outward normal provided
H is oriented upward and D is oriented downward. Then Gauss’ Theorem states
14. (10 points) Let V be the solid hemisphere x 2
;;
Compute
;;;
F dV
;;
V
F
dS for F
x3
y2
F
dS H
y2
z2, y3
F
dS
D
H
;;
x2
;;
z2, z3
x2
y2
using
one of the following methods: (Circle the method you choose.)
Method I: Parametrize H and compute
S
Method II: Parametrize D, compute
S
;;
;;
F
dS explicitly.
H
F dS
and
;;;
D
F
dS.
F dV and solve for
V
H
Method I:
(You don’t want to do it this way!) Parametrize H:
RŸI, 2 Ÿ2 sin I cos 2, 2 sin I sin 2, 2 cos I 0tIt =
0 t 2 t 2=
2
R
I Ÿ2 cos I cos 2, 2 cos I sin 2, "2 sin I R 2 "2 sin I sin 2, 2 sin I cos 2, 0
N
F
4 sin 2 I cos 2, 4 sin 2 I sin 2, 4 sin I cos I
y2 z2, y3 x2 z2, z3 x2 y2
3
2
2
3
3
2
3
2
2
2
Ÿ8 sin I cos 2 4 sin I sin 2 4 cos I, 8 sin I sin 2 4 sin I cos 2 4 cos I,
8 cos 3 I 4 sin 2 I cos 2 2 4 sin 2 I sin 2 2 3
2
2
F
N 8 sin I cos 3 2 4 sin I sin 2 4 cos 2 I
4 sin 2 I cos 2 8 sin 3 I sin 3 2 4 sin 2 I cos 2 2
2
3
8 cos I 4 sin I Ÿ4 sin I cos I 5
4
2
2
5
4
4
4
2
2
16¡2 sin I cos 2 sin I sin 2 cos 2 cos I sin I cos 2 2 sin I sin 2 sin I cos 2 sin 2
2
3
2
4
cos I sin I sin 2 2 cos I sin I sin I cos I¢
5
4
2
2
5
4
4
4
2
2
16¡2 sin I cos 2 sin I sin 2 cos 2 cos I sin I cos 2 2 sin I sin 2 sin I cos 2 sin 2
2
3
2
4
cos I sin I sin 2 2 cos I sin I sin I cos I¢
;;
x3
F
dS
H
;
;
;
16
;;
;;
since
2=
sin 2 d2
0
2=
=/2
0
0
2=
=/2
0
0
0
F
N dI d2
¡2 sin
;
5
2=
0
I cos 4 2 2 sin 5 I sin 4 2 2 cos 4 I sin I sin 3 I cos I¢ dI d2
cos 2 d2
;
0
;
;
2=
0
sin 2 2 cos 2 d2
Further, after much work:
=/2
2=
2=
sin 5 I dI 8
cos 4 2 d2 sin 4 2 d2
15
0
0
0
=/2
=/2
cos 4 I sin I dI 1
sin 3 I cos I dI 1
5
4
0
0
;
3=
4
and
;
0
;
2=
0
2=
0
cos 2 2 sin 2 d2
1 d2
0
2=
8
;;
So
2=
=/2
0
0
F
N dI d2
8
15
16 2
3=
4
Method II:
Parametrize D:
0trt2
RŸr, 2 Ÿr cos 2, r sin 2, 0 cos 2, sin 2, 0
Rr R
2
N
F
8
15
3=
4
2 1
5
Ÿ2= 1 Ÿ2= 4
232 =
5
0 t 2 t 2=
Ÿ"r sin 2, r cos 2, 0 This is up but we need down; so we reverse it to get N
0, 0, r
x3
F
N
"r 3
;;
2
F
dS
y2
z2, y3
;;
2=
2
0
0
x2
z2, z3
F
N dr d2
x2
;;
2=
2
0
0
y2
Ÿ0, 0, "r r 3 cos 3 2 r 2 sin 2 2, r 3 sin 3 2 r 2 cos 2 2, r 2
"r 3 dr d2
4
"2= r
4
2
0
"8=
D
Use spherical coordinates for V:
, ,
R
Ÿ> I 2 Ÿ> sin I cos 2, > sin I sin 2, > cos I 0 t > t 2,
dV > 2 sin I d> dI d2
F 3x 2 3y 2 3z 2 3> 2
;;;
F dV
;; ;
2=
=/2
2
0
0
0
3> 2 > 2 sin I d> dI d2
2=¡" cos I ¢ =I/20
V
2=¡1 ¢ 3 32
5
We combine these to get
dS ;; F
H
;;;
V
F dV "
;;
0tIt =
2
3> 5
5
0 t 2 t 2=
2
> 0
192 =
5
F
dS
192 = " "8=
5
232 =
5
D
Notice that Methods I and II give the same answer.
9
15. (10 points) Find the point Ÿx, y, z in the first octant on the surface
z
27
x
64
y
which is closest to the origin.
Minimize the square of the distance to the origin
f x2 y2
64 .
27
constraint that the point lies on the surface
z
y
x
Method I:
z2
subject to the
Eliminate a constraint:
2
Minimize
f
x2
y2
27
x
64
y
"27 , 2y 2 27 64
"64
Ÿ0, 0 2
x
y2
x
y
y2
x2
Multiply the first equation by
and the second equation by
:
54
128
y3
27 64
x 3 27 64
(1.)
(2.)
27
64
y
y
x
x
y
x
Equate these to obtain
and plug back into (1.) to obtain:
3
4
x 3 27 16 3 75
27
x
x
x
4
Cross multiply:
So
and
x 75 27 3 4 5 2
x3 5
y4 5
64
27
64
25
27
and
5 5.
z
y
x
3 5
4 5
5
f
2x 2
Method II:
Minimize
27
x
64
y
Lagrange multipliers:
subject to
f x2 y2 z2
g
z " 27 " 64
y
x
0
27 , 64 , 1
x2 y2
Lagrange equations:
f 5
g:
2x 2725
2y 6425
2z 5
x
y
Cross multiply in the first two equations and replace the 5 by 2z:
2y 3 128z
2x 3 54z
Divide by 2 and take the cube root:
x 33 z
y 43 z
z 4/3
Plug into the constraint and solve:
z 27 64 25
3 z
43 z
33 z
z 5 3/2 5 5
x 33 z 3 5
y 43 z 4 5
f
Ÿ2x, 2y, 2z g
52
10