Name
ID
1-10
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Fall 2015
11
/30
Version B Solutions P. Yasskin
12
/5
13
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Total
/105
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 its area.
a. 3
2
b. 3 3
2
Correct Choice
3
c.
d. 3 3
e. 3
Solution:
AB
î
AB
AC
1, 2, 1
AC
0, 3, 3
k
3î
1 2 1
3
3k
1 AB
2
A
1 32
2
AC
32
3 3
2
32
0 3 3
2. Find the tangent plane to the graph of z
x3y
y 3 x 2 at x, y
1, 2 .
Where does it cross the z-axis?
a.
38
b.
14
c.
10
Correct Choice
d. 10
e. 38
Solution:
z
f tan x, y
z-intercept
f x, y
x3y
y3x2
2
f x x, y
3x y
f y x, y
x3
f 1, 2
f tan 0, 0
2y x f x 1, 2
3y 2 x 2
f x 1, 2 x
10
f 1, 2
3
f y 1, 2
1
22 1
10
22
13
f y 1, 2 y
2
13 2
38
10
22 x
1
13 y
2
1
3. Find the tangent plane to the graph of hyperbolic paraboloid
4x 3 2 4y 1
at the point 4, 2, 1 . Where does it cross the z-axis?
2
9z
2
2
1
11
9
11
3
a.
b.
c. 0
d. 11
Correct Choice
3
e. 11
9
Solution:
F
P
8x
N X
4, 2, 1
3 , 8y
N P
z-intercept at x
F
1 , 18 z
8x
y
4x
8y
2
2
4y
N
18z
32
18z
66
0:
3
F
1
2
9z
2
2
8, 8, 18
P
16
18
66
z
18
66
11
3
4. Queen Lean is flying the Millenium Eagle through the galaxy. Her current galactic position is
P
4, 3, 1 lightyears and her current velocity is v
0. 1, 0. 2, 0. 3 lightyears/year. She is passing
through a deadly polaron field whose density is related to the dark energy intensity I and the dark
matter pressure P by
IP.
She measures the dark energy intensity and its gradient are currently
I
7 lumens and I
2, 3, 1 lumens /lightyear
She measures the dark matter pressure and its gradient are currently
P
2
0. 8 dynes/lightyear and P
0. 4, 0. 1, 0. 2 dynes/lightyear
3
At what rate does she see the polaron density changing?
a. d
7. 724
b.
1. 22
c.
d.
e.
dt
d
dt
d
dt
d
dt
d
dt
1. 06
1. 06
Correct Choice
7. 724
Solution:
dI
dt
dP
dt
d
dt
I v
P v
dI
I dt
2, 3, 1
0. 1, 0. 2, 0. 3
0. 4, 0. 1, 0. 2
dP
P dt
P dI
dt
.2 .6 .3
0. 1, 0. 2, 0. 3
I dP
dt
1. 1
. 04 . 02 . 06 . 12
0. 2 1. 1
7 . 12
1. 06
2
8 y2
2
5. Sketch the region of integration for the integral
0
ex
2
y2
dx dy in problem (12),
y
then select its value here:
a.
b.
c.
d.
e.
4
8
8
16
16
e8
1
e 16
e8
1
Correct Choice
1
e 16
e8
1
1
Solution: Switch to polar coordinates:
/4
2 2
0
/4
2
e r r dr d
0
0
1 e8
2
4
1 e r2
2
1
8
e8
y
2
2 2
1
0
1
0
0
1
2
x
6. The surface of the Death star is a sphere of radius 2 with a hole cut out of one end, which we will take
as centered at the south pole. It may be parametrized by
R ,
2 sin cos , 2 sin sin , 2 cos
2 . Find the surface area.
for 0
3
a. A
b. A
2
c. A
3
d. A
6
e. A
12
Solution:
Correct Choice
We first find the normal and its length:
î
N
e
e
k
e
2 cos cos
2 cos sin
2 sin
e
2 sin sin
2 sin cos
0
î 4 sin 2 cos
4 sin 2 sin
4 sin cos sin 2
k 4 sin cos cos 2
4 sin 2 cos , 4 sin 2 sin , 4 sin cos
2
4 sin 2 cos
N
16 sin 4
4 sin 2 sin
16 sin 2 cos 2
2
4 sin cos
2
4 sin
The surface area is
A
2
2 /3
0
0
1 dS
8
4 sin d d
cos 2
3
cos 0
8
2
cos
1
2
1
2 /3
0
12
3
7. Consider the solid below the cone given in cylindrical coordinates by z
Its temperature is T
r above the xy-plane.
4
z. Find its average temperature.
a.
b. 1
Correct Choice
c. 1
2
d. 32
3
e. 64
3
Solution:
The volume is
4
2
4 r
4
1 r dz d dr
V
0
0
4
2
0
0
4
r3
3 0
z, is
2
z r dz d dr
2
The integral of the temperature, T
4 r
T dV
0
16r
0
8r 2
r 3 dr
4 2 81
3
So the average temperature is
Compute
f ds for f
4
2
4r
r 2 dr
0
64
3
4 r
4
dr
r4
3
r 2 dr
0
0
8r
3
r4
4
T dV
1
4
8 42
0
3
44
4
84
3
64
3
1
1
V
T ave
8.
2
rz
2
8r 2
0
3
4 r
0 dr
64
3
32
4
0
4
rz
0
2r 2
2
2
xe y
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. 0
Correct Choice
b. 3e 4
c. 4e 3
d. 3e 3
4e 4
e. 4e 4
3e 3
Solution:
By the FTCC,
f ds
fB
fA
0 because B
A no matter what point you start at.
R
4
y
9.
Compute
8x 3
F ds for F
2
5y 4
5y, x
3
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.
20
b.
40
c.
80
Correct Choice
d. 20
e. 80
Solution:
P 8x 3 5y
By Green’s Theorem,
Q
F ds
xQ
R
10. Compute
x
5y 4
yP
xQ
yP
dx dy
1
5
4 area
4 dx dy
R
4
4 20
80
R
F dS over the complete surface of the hemisphere 0
z
4
x2
y 2 oriented
V
x 3 z, y 3 z, 3 z 4
4
Hint: Use a Theorem.
outward,for F
a.
b.
128
64
c.
32
d. 32
Correct Choice
e. 64
Solution:
By Gauss’ Theorem
F 3x 2 z 3y 2 z
So the integral is:
3z 3
3z x 2
F dV. The divergence is
F dS
H
2
y
z2
3
2
H
3
cos and the volume element is dV
/2
2
F dV
3
0
H
2
0
sin 2
2
3
2
cos
2
sin d d d .
sin d d d
0
/2
6
3
0
6
2
25
32
0
5
Work Out: (Points indicated. Part credit possible. Show all work.)
11.
(30 points) Verify Stokes’ Theorem
F dS
yz 2 , xz 2 , z 3
for the vector field F
the paraboloid z
x
2
y
2
for z
F ds
P
P
and the surface of
4, oriented down and out.
Be careful with orientations. Use the following steps:
Left Hand Side:
r cos , r sin , r 2
The paraboloid, P, may be parametrized as R r,
a. Compute the tangent vectors:
cos ,
er
e
r sin ,
sin ,
2r
r cos ,
0
b. Compute the normal vector:
N
î 2r 2 cos
2r 2 sin
r sin 2
k r cos 2
2r 2 cos , 2r 2 sin , r
This is oriented up and in. Need down and out. Reverse: N
yz 2 , xz 2 , z 3 :
c. Compute the curl of the vector field F
î
F
k
x
yz 2
2r 2 cos , 2r 2 sin , r
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 paraboloid:
F
2r 3 cos , 2r 3 sin , 2r 4
R r,
e. Compute the dot product of the curl of F and the normal N.
F N
4r 5 cos 2
4r 5 sin 2
2r 5
6r 5
f. Compute the left hand side:
2
F dS
P
2
F N dr d
P
0
0
6r 5 dr d
2
r6
2
0
128
6
Right Hand Side:
g. Parametrize the circle,
r
C:
2 cos , 2 sin , 4
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:
32 sin , 32 cos , 64
r
j. Compute the dot product of F and the tangent vector v:
F v
64 sin 2
64 cos 2
64
k. Compute the right hand side:
2
F ds
P
2
F vd
0
64 d
128
which agrees with (f).
0
7
12.
(5 points) Sketch the region of integration
y
2
for the integral
8 y2
2
0
1
ex
2
y2
dx dy.
y
0
You computed its value in problem (5).
0.0 0.5 1.0 1.5 2.0 2.5
x
13. Duke Skywater is flying the Millenium Eagle through the galaxy. His current galactic position is
P
4, 4, 1 lightyears. He is passing through a deadly polaron field whose density is
xy z 2
3
polarons/lightyear .
a. If his current velocity is v
0. 3, 0. 2, 0. 1 lightyears/year, at what rate does he see the polaron
density changing?
Solution:
y, x, 2z
4, 4, 2
P
d
v
0. 3 4 0. 2 4 0. 1 2 1. 8 polarons/lightyear 3 /year.
P
dt
b. Duke decides to change his velocity to get out of the polaron field. If the Millenium Eagle’s
maximum speed is 0. 9 lightyears/year, with what velocity should Duke travel to reduce the
polaron’s density as fast as possible?
Solution:
The direction of maximum increase is
4, 4, 2 .
P
The direction of maximum decrease is
The length is
P
16
16
4
P
4, 4, 2 .
6.
So the unit vector of maximum decrease is v
P
1
6
4, 4, 2
2, 2, 1 .
3 3 3
P
Since the maximum speed is |v | 0. 9, Duke’s velocity should be
v |v |v 0. 9 2 , 2 , 1
0. 6, 0. 6, 0. 3 .
3 3 3
8