Name
ID
1-10
/50
Fall 2015
11
/30
Version A Solutions P. Yasskin
12
/5
13
/20
Total
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MATH 251
Final Exam
Sections 512
Multiple Choice: (5 points each. No part credit.)
1. A triangle has vertices A
2, 2, 1 , B
3, 4, 2 and C
2, 5, 4 . Find the angle at A.
Correct Choice
a. 30°
b. 45°
c. 60°
d. 120°
e. 150°
Solution:
AB
1, 2, 1
AB AC
AB AC
cos
AC
0, 3, 3
3
2
9
63 2
30°
2. Find the tangent plane to the graph of z
x2y
y 3 x 3 at x, y
2, 1 .
Where does it cross the z-axis?
a.
72
b.
48
c.
12
Correct Choice
d. 12
e. 48
Solution:
z
f tan x, y
z-intercept
f x, y
x2y
f x x, y
2xy
f y x, y
2
f 2, 1
f tan 0, 0
x
y3x3
3y 3 x 2 f x 2, 1
2 3
3y x
f x 2, 1 x
12
f 2, 1
f y 2, 1
2
16 2
12
16
28
f y 2, 1 y
1
28 1
48
12
16 x
2
28 y
1
1
3. Find the tangent plane to the graph of hyperbolic paraboloid
9x 3 2 4y 1
at the point 2, 2, 3 . Where does it cross the z-axis?
a.
19
b.
19
2
2
4z
2
2
1
c. 0
d. 19
Correct Choice
2
e. 19
Solution:
F
P
18 x
N X
2, 2, 3
3 , 8y
N P
F
1 , 8z
18x
z-intercept at x
y
9x
8y
0:
2
3
2
N
8z
18 2
8z
4y
76
F
1
2
4z
2
2
18, 8, 8
P
8 2 8 3
76
19
z
8
2
/4
4. Sketch the region of integration for the integral
0
1
tan y
76
1
dx dy in problem (12),
arctan x
then select its value here:
a. 1
2
b.
c.
d.
4
3
2
Correct Choice
e. 1
Solution: Reverse the order of integration:
/4
0
1
tan y
1
0
1
0
1
dx dy
arctan x
y
arctan x
arctan x
dx
arctan x
1
arctan x
0
0
1
dy dx
arctan x
y
0.5
arctan x
dx
y 0
x
1
0
1
0.0
0.0
0.5
1.0
x
2
t, t 2 , t 3 . At t 2 hours,
he releases a trash pod which travels along the tangent line to the path of the Eagle with constant
velocity equal to the velocity of the Eagle at time of release. Where is the trash pod 1 hours after
release?
5. Ham Duet is flying the Millenium Eagle through the galaxy on the path r t
Correct Choice
a.
3, 8, 20
b.
3, 9, 27
c.
2, 4, 8
d.
1, 2, 12
e.
5, 10, 43
Solution:
P
r2
2, 4, 8
The path of the trash pod is X
1, 2t, 3t 2
v
P
tv
v2
2, 4, 8
1 hour after release, the trash pod is at X 1
1, 4, 12
t 1, 4, 12
2
1, 4
2
4, 8
t, 4
12
4t, 8
12t
3, 8, 20
6. Consider the surface of the cone given in cylindrical coordinates by z
r above the xy-plane. It
4
may be parametrized by
Its temperature is T
r .
R r,
r cos , r sin , 4
z. Find its average temperature.
32 2
3
b. 16 2
c. 2
3
d. 4
Correct Choice
3
e. 8
3
Solution:
We first find the normal and its length:
a.
î
N
e
e
î
k
er
cos
sin
1
e
r sin
r cos
0
r sin 2
k r cos 2
r cos
r sin
N
r 2 cos 2
r 2 sin 2
2 r dr d
2
r2
r cos , r sin , r
2r
The surface area is
2
4
S
0
0
The integral of the temperature, T
z
2
4
T dS
2
4
16 2
0
r, is
4
4
4
0
r2
2
2
r
2 r dr d
2
2
0
4r
r 2 dr
0
2 2r 2
r3
3
4
2
2 32
0
64 2
3
64
3
So the average temperature is
T ave
1
A
T dS
1
16 2
64 2
3
4
3
3
7. Death star is basically a spherical shell with a hole cut out of one end, which we will take as centered
at the south pole. In spherical coordinates, it fills the region between 1
the volume.
a. V
21
b. V
42
c. V
63
d. V
105
e. V
126
Solution:
Correct Choice
The volume is
2
2 /3
4
0
0
1
V
1
Compute
2
sin d d d
cos 2
3
2
8.
2 . Find
3
4 and 0
cos 0
4
2
3
1
3
cos
2
2 /3
0
1
2
3
3
4
1
1 21
63
xe y
f ds for f
y
R
counterclockwise around
4
2
the polar curve
r
3
cos 4
-4
shown at the right.
-2
2
4
x
-2
Hint: Use a Theorem.
-4
a. 4e 3
b. 3e 4
c. 4e 4
3e 3
d. 3e 3
4e 4
e. 0
Correct Choice
Solution: By the FTCC,
f ds
fB
fA
0 because B
A no matter what point you start at.
R
4
3
y
9.
Compute
F ds for F
4x 3
2y, x
2
3y 4
1
P
counterclockwise around the complete
-3
boundary of the plus sign shown at the right.
-2
-1
1
-1
Hint: Use a Theorem.
2
3
x
-2
-3
a. 120
b. 60
Correct Choice
c. 20
d.
20
e.
60
Solution:
P 4x 3 2y
By Green’s Theorem,
Q
F ds
xQ
R
10. Compute
3y 4
x
xQ
yP
yP
dx dy
1
2
3 area
3 dx dy
R
3
3 20
60
R
F dS over the complete surface of the hemisphere 0
z
9
x2
y 2 oriented
V
2 x 3 z, 2 y 3 z, 1 z 4
3
2
3
Hint: Use a Theorem.
outward,for F
a. 27
b. 54
c. 81
d. 162
e. 243
Solution:
Correct Choice
By Gauss’ Theorem
F dV. The divergence is
F dS
H
F 2x 2 z 2y 2 z
So the integral is:
2z 3
2z x 2
y2
z2
2
2
H
3
cos and the volume element is dV
/2
3
F dV
2
0
H
2
0
sin 2
2
3
2
cos
2
sin d d d .
sin d d d
0
/2
6
2
0
6
3
35
243
0
5
Work Out: (Points indicated. Part credit possible. Show all work.)
11.
(30 points) Verify Stokes’ Theorem
F dS
F ds
C
C
yz 2 , xz 2 , z 3
for the vector field F
x2
the cone z
y2
for z
and the surface of
2, oriented down and out.
Be careful with orientations. Use the following steps:
Left Hand Side:
The cone, C, may be parametrized as R r,
r cos , r sin , r
a. Compute the tangent vectors:
er
cos ,
e
r sin ,
sin ,
1
r cos ,
0
b. Compute the normal vector:
N
î r cos
r sin 2
k r cos 2
r sin
r cos , r sin , r
This is oriented up and in. Need down and out. Reverse: N
2
c. Compute the curl of the vector field F
î
F
r cos , r sin , r
3
yz , xz , z :
k
x
yz 2
2
y
z
î0
2xz
0
2yz
k z2
z2
2xz, 2yz, 2z 2
xz 2 z 3
d. Evaluate the curl of F on the cone:
F
2r 2 cos , 2r 2 sin , 2r 2
R r,
e. Compute the dot product of the curl of F and the normal N.
F N
2r 3 cos 2
2r 3 sin 2
2r 3
4r 3
f. Compute the left hand side:
2
F dS
C
2
F N dr d
C
0
0
4r 3 dr d
2
r4
2
0
32
6
Right Hand Side:
g. Parametrize the circle,
r
C:
2 cos , 2 sin , 2
h. Find the tangent vector on the curve:
2 sin , 2 cos , 0
v
This is counterclockwise. Need clockwise. Reverse v
2
i. Evaluate F
F
2
yz , xz , z
3
2 sin , 2 cos , 0
on the circle:
8 sin , 8 cos , 8
r
j. Compute the dot product of F and the tangent vector v:
F v
16 sin 2
16 cos 2
16
k. Compute the right hand side:
2
F ds
C
2
F vd
0
16 d
32
which agrees with (f).
0
7
12.
y
(5 points) Sketch the region of integration
for the integral
/4
0
1
tan y
0.5
1
dx dy.
arctan x
Shade in the region.
0.0
0.0
You computed its value in problem (4).
0.5
1.0
x
13. (20 points) A cardboard box needs to hold 18 cm 3 . The cardboard for the vertical sides costs 12¢ per
cm 2 while the thicker bottom costs 16¢ per cm 2 . There is no top. What are the length, width, height
and cost of the box which costs the least?
Solution:
C 16LW
The volume constraint is V LWH 18. The cost is
12 2LH 2WH
16LW 24LH 24WH which needs to be minimized.
Lagrange Multipliers:
C
16W 24H, 16L
24H, 24L
24W
V
16W 24H
16L 24H
WH
LH
16LW 24LH
16LW 24WH
LWH
LWH
24L
LW
24LH
LWH
24W
LWH
C
WW 2 W
3
16LW
24LH
18
W3
24WH
24WH
27
16 3 3
W
3
WH, LH, LW
C
V
16LW 24LH 16LW 24WH
24LH 24WH
L W
16LW 24WH 24LH 24WH
16LW 24LH
2W 3H
L
24 3 2
3
H
24 3 2
2
432¢
8