Name
MATH 251
1-12
/48
Fall 2014
13
/20
P. Yasskin
14
/10
15
/25
Total
/103
Final Exam
Sections 508
Solutions
Multiple Choice: (4 points each. No part credit.)
1. Find the line of intersection of the planes 3x
2y
4z
8 and 3x
2y
2z
4.
This line intersects the xy-plane at
a.
b.
c.
d.
e.
1, 7 , 3
3
4, 10, 6
1, 2, 0
2, 1, 0
Correct Choice
22 , 2, 4
9
3
Solution: The desired point satisfies the plane equations and z 0.
So we solve 3x 2y 8 and 3x 2y 4.
Adding the 2 equations: 6x 12
x 2
Subtracting: 4y 4
y
2. The radius of a cylinder is currently r
Its height is currently h
dr
dt
50 cm and is increasing at
100 cm and decreasing at
dh
dt
4
1
2
cm
.
min
cm
.
min
At what rate is the volume changing?
a. dV
b.
c.
d.
e.
dt
dV
dt
dV
dt
dV
dt
dV
dt
10 000
10 000
Correct Choice
20 000
30 000
40 000
Solution: V
r2h
dV
V dr
V dh
dt
r dt
h dt
2 rh dr
dt
r 2 dh
dt
2 50 100 2
50
2
4
10 000
1
1
xy
3. Find the tangent plane to the graph of z
at
1, 2 . The z-intercept is
a. 0
b. 1
2
c. 3
2
d. 2
e. 5
2
Correct Choice
1 f y x, y
1
x2y
xy 2
1 f x 1, 2
1 f y 1, 2
1
f 1, 2
2
2
4
f tan x, y
f 1, 2 f x 1, 2 x 1 f y 1, 2 y 2
1
1 x 1
1 y 2
1
1 1
1 2
c
2
2
4
2
2
4
Solution:
z
z
1
xy
f x, y
f x x, y
4. Find the tangent plane to the graph of the ellipsoid x 2
3
2
xy
4y 2
z2
20 at
3, 1, 2 . The
z-intercept is
Correct Choice
10
20
40
80
160
a.
b.
c.
d.
e.
Solution: F
N
F
4c
40
x2
xy
4y 2
7, 11, 4
P
c
z2
N X
P
3, 1, 2
N P
7x
11y
2x
y, x
8y, 2z
4z
73
11 1
x3
3xy
y 3 have?
y
x2
x
42
40
10
5. How many critical points does the function f x, y
0
1
2
Correct Choice
3
infinitely many
a.
b.
c.
d.
e.
Solution: f x
x
F
x4
If x
x4
3x 2
x
0 then y
3y
0
0
x x3
0. If x
fy
1
3y 2
3x
0
x
1 then y
0
y2
0 or 1
1. So 2 critical points.
2
6. Duke Skywater is flying the Centenial Eagle through a dangerous polaron field whose density is given
by
x 4 y 3 z 2 . If Duke is currently at the point P
1, 2, 3 and has velocity v
4, 3, 2 ,
what is the rate of change of the polaron density as seen by Duke at the current instant?
a.
b.
c.
d.
e.
16
24
32
48
64
Correct Choice
Solution:
d
dt
v
4x 3 , 3y 2 , 2z
4, 3, 2
4, 12, 6
4, 12, 6
1,2,3
16
36
12
64
7. Duke Skywater is flying the Centenial Eagle through a dangerous polaron field whose density is given
by
x 4 y 3 z 2 . If Duke is currently at the point P
1, 2, 3 in what unit vector direction
should he fly to REDUCE the polaron density as fast as possible?
2, 6, 3
7 7 7
2, 6, 3
7 7 7
2, 6, 3
7 7 7
2 , 6 , 3
98 98 98
2 , 6 , 3
98 98 98
a.
b.
c.
d.
e.
Solution:
Correct Choice
4x 3 , 3y 2 , 2z
1,2,3
4, 12, 6
1
Fastest DECREASE along: u
x
12 2
62
14
2, 6, 3
7 7 7
1,2,3
8. Find the maximum value of the function f
42
2y
2z on the sphere of radius 5 centered at the
origin.
a. 5
b. 25
3
c. 9
d. 27
5
e. 15
Correct Choice
y 2 z 2 25.
Lagrange equations:
f
g
1
2x y z
5
10
10
g x 2 4x 2 4x 2 25
9x 2 25
x
y
z
3
3
3
5
10
10
For a maximum, we need them positive.
f
2
2
3
3
3
Solution: The constraint is g
f
1, 2, 2
g
2x, 2y, 2z
1
2x
2
2y
2
2z
x2
15
3
x 2 which may be parametrized as r t
6 , 6 . Find it mass if its linear density is
x.
9. A wire has the shape of the parabola y
0, 0
to
t, t 2
from
a. 1 ln 2 6
b.
c.
d.
e.
5 6
5
4
2
1 ln 2 6 5
5 6
4
2
125
12
31
Correct Choice
3
21
2
Solution: v
1, 2t
12
|v |
6
M
6
|v | dt
ds
t 1
0
M
1
8
25
u du
1
10. Compute
1
8
0
3/2
2u
3
F ds for F
25
1
2t
2
4t 2 dt
1 125
12
2xy 2 , 2x 2 y
1
4t 2
u
1
x
4t 2
du
t
8t dt
1 du
8
t dt
x 3 from
1, 1
to
31
3
1
along the cubic y
2, 8 .
Hint: Use a theorem.
a.
b.
c.
d.
e.
255
256
510
60
66
Correct Choice
Solution: To use the Fundamental Theorem of Calculus for Curves, we need a scalar potential.
F
f
2xy 2
2x 2 y
f x2y2
xf
yf
2,8
F ds
f ds
f 2, 8
f 1, 1
2282
1212
255
1,1
4
11.
Compute
2x 2 dx
3y
3y 2
y6
2x dy
counterclockwise over the complete boundary of the
4
shape at the right, which is a square of side 4
2
under an isosceles triangle with height 3.
Hint: Use a theorem
a.
110
b.
22
-2
Correct Choice
0
2
x
c. 0
d. 22
e. 110
Solution: Green’s Theorem says
P dx
R
Here P
3y
2x 2 dx
3y
2x 2 , Q
3y 2
3y 2
2x dy
dx dy
R
12.
x
R
3
R
P
y
dx dy
2x and R is the region inside the square and triangle.
R
2
Q
x
Q dy
5
3y 2
1 dx dy
2x
y
5 Area
3y
2x 2
5 42
dx dy
14 3
2
110
R
Compute
F dS over the piece of the hyperbolic
P
paraboloid z
x2
y2
xy oriented upward inside the cylinder
3y, 3x, x 2
9 for the vector field F
y2 .
Note, the paraboloid may be parametrized by
R r,
r cos , r sin , r 2 sin cos
.
Hint: Use a theorem.
a. 27
b.
c.
d.
e.
2
27
54
108
216
Correct Choice
Solution: Stokes’ Theorem says
F ds.
F dS
P
P
The boundary of the paraboloid is its intersection with the cylinder where r 3.
So the boundary is the curve r
3 cos , 3 sin , 9 sin cos .
2
2
v
3 sin , 3 cos , 9 cos
9 sin
On the curve F
9 sin , 9 cos , 3 . So F v 27 sin 2
27 cos 2
27 cos 2
27 sin 2
2
2
2
sin 2 2
F ds
F v dt
54 cos 2 dt
27 1 cos 2 dt 27
54
2
0
0
0
0
54 cos 2
P
5
Work Out: (Points indicated. Part credit possible. Show all work.)
13. (20 points) For each integral, plot the region of integration and then compute the integral.
2
4
0
y2
a. I
cos x 3/2 dx dy
y 2.0
Solution:
4
x 1/2
0
0
I
1.5
4
1.0
0.0
0
4 y2
2
0
Solution:
y
cos x 3/2 y
0
x 1/2
y 0
dx
cos x 3/2 x 1/2 dx
0
0.5
b. J
4
cos x 3/2 dy dx
1
y
arctan x
2
3
u
x 3/2
I
2
3
4
3 x 1/2 dx
2 du
x 1/2 dx
2
3
8
8
2 sin u
2 sin 8
cos u du
3
3
0
0
du
x
dx dy
/4
y
2
J
0
1.0
0
/4
arctan r sin
r cos
r dr d
2
r dr d
0.5
0
0
/4
2
0.0
0
1
2
x
2
0
r2
2
2
2
0
16
6
14.
(10 points) An aquarium in the shape of a rectangular solid has
a slate bottom costing $6 per ft 2 , a glass front costing $2 per ft 2 ,
and aluminum sides and back costing $1 per ft 2 . There is no top.
Let L be the length front to back, W be the width side to side and
H be the height. Write a formula for the total cost, C, and find the
dimensions and cost of the cheapest such aquarium if the volume
36 ft 3 . Do not use decimals.
is V
Solution: The Cost is C
6LW
The Volume constraint is V
2WH
LWH
1 2LH
WH
6LW
3WH
2LH.
36.
Method 1: Lagrange Multuipliers:
C
6W
2H, 6L
3H, 3W
2L
V
WH, LH, LW
3W
2L
The Lagrange equations are:
6W
2H
WH
6L
3H
LH
LW
Multiply the 1 st equation by L, the 2 nd equation by W, and the 3 rd equation by H so the right
sides are all LWH.
Then equate them:
LWH
6LW
2LH
6LW
3WH
3WH
2LH
The 1 st and 2 nd say 2LH 3WH or 2L 3W. The 2 nd and 3 rd say 6LW 2LH or 3W
3 W and H 3W into the volume constraint:
Substitute L
2
3 W W 3W
9 W 3 36
36
W3 8
W 2
2
2
Substituting back:
L 3
H 6
and
C 6 3 2 3 2 6 2 3 6 $108
H.
Method 3: Eliminate a Variable:
108
72
L
W
108
CL
0
C W 6L 722
L2
W
4
18
1 W4
W
18 W
W3
2
144
8
12
W2
12
Substituting back:
L
3
H
22
H
36
LW
6W
C
6LW
0
W
18
L2
8
W
2
36
3 2
6
L
and
12
W2
C
6 3 2
108
3
72
2
$108
7
15.
(25 points) Verify Gauss’ Theorem
F dV
F dS
V
V
2
for the vector field F
x2
cone
y2
z
2
yz , xz , x
2
2
and the solid
y z
4.
Be careful with orientations. Use the following steps:
First the Left Hand Side:
a. Compute the divergence:
F
0
x2
0
y2
x2
y2
b. Name your coordinate system:
cylindrical
c. Express the divergence and the volume element in those coordinates:
r2
F
dV
r dr d dz
d. Compute the left hand side:
2
4
4
F dV
0
V
4
2
4r 3
0
r 4 dr
4
r 2 r dz dr d
2
r
r5
5
r4
2
0
4
r3z
4
0
z r
2 44 1
4
5
0
dr
512
5
Second the Right Hand Side:
The boundary surface consists of a cone C and a disk D with appropriate orientations.
e. Parametrize the disk D:
R r,
r cos , r sin , 4
f. Compute the tangent vectors:
er
cos ,
e
r sin ,
sin ,
0
r cos ,
0
g. Compute the normal vector:
N
î0
h. Evaluate F
F
R r,
k r cos 2
r sin 2
0, 0, r
yz 2 , xz 2 , x 2
y2 z
on the disk:
0
This is up. Oriented correctly.
16r sin , 16r cos , 4r 2
i. Compute the dot product:
F N
4r 3
j. Compute the flux through D:
2
4
F dS
D
0
0
4r 3 dr d
2
r4
4
0
2 44
512
8
The cone C may be parametrized by:
R r,
r cos , r sin , r
k. Compute the tangent vectors:
er
cos ,
e
r sin ,
sin ,
1
r cos ,
0
l. Compute the normal vector:
N
î0
r cos
0
r cos , r sin , r
m. Evaluate F
F
r sin 2
This is oriented up and in. We need down and out.
r cos , r sin , r
Reverse N
k r cos 2
r sin
yz 2 , xz 2 , x 2
y2 z
on the cone:
r 3 sin , r 3 cos , r 3
R r,
n. Compute the dot product:
F N
r 4 cos sin
r 4 sin cos
r4
2r 4 cos sin
r4
r 4 sin 2
r4
o. Compute the flux through C:
2
4
0
0
F dS
C
r 4 sin 2
r 4 dr d
cos 2
2
2
2048
5
512
5
0
r5
5
4
0
5
2 4
5
2048
5
p. Compute the TOTAL right hand side:
F dS
V
F dS
D
F dS
C
512
which agrees with (d).
9