Name
ID
MATH 251
Final Exam
Sections 506
1-13
/52
Spring 2013
14
/15
P. Yasskin
15
/15
16
/20
Total
/102
Solutions
Multiple Choice: (4 points each. No part credit.)
1. How much work is done to push a box up an incline from
2, 1
to
4, 2
by the force F
3, 2 ?
a. 65
b.
65
c. 2 65
d. 64
e. 8
Correct Choice
SOLUTION:
D
2. The graph of 2x 2
4, 2
4x
2, 1
3y
3z 2
2, 1
6z
W
F D
6
2
8
4 is a
a. hyperboloid of 1-sheet
b. hyperboloid of 2-sheets
c. elliptic paraboloid
d. hyperbolic paraboloid
Correct Choice
e. cone
SOLUTION:
y is linear, x 2 and z 2 have opposite signs
3. Find the tangential acceleration, a T , or the curve r t
a. 2
12t 2
b. 2
12t 2
c. 2t
4t 3
d. 2t
4t 3
e. 2
16t 2
SOLUTION:
hyperbolic paraboloid
t 2 , 4 t 3 , t 4 . HINT: Perfect square.
3
Correct Choice
v
2t, 4t 2 , 4t 3
|v |
4t 2
16t 4
16t 6
2t
4t 3
aT
d|v |
dt
2
12t 2
1
4. Find the equation of the plane through the points
1, 2, 3 ,
and
2, 4, 2
1, 2, 4 .
Its z-intercept is
a. 1
b. 2
Correct Choice
c. 4
d. 8
e. 16
SOLUTION:
P
1, 2, 3
î
N
u
v
1
u
2, 4, 2
1, 2, 3
1, 2, 1
v
1, 2, 4
1, 2, 3
2, 0, 1
k
2
1
2 0
1
N X N P
2x y 4z
The z-intercept is when x
î2
0
1
2
k0
4
2, 1, 4
21
2 43
16
y 0 and z 4.
5. Find the plane tangent to the graph of the function z
2x 2 y at the point
3, 2 .
x, y
Its z-intercept is
a.
72
b.
36
Correct Choice
c. 0
d. 36
e. 72
SOLUTION:
f x, y
2x 2 y
f 3, 2
f x x, y
4xy
f x 3, 2
24
f y x, y
2x 2
f y 3, 2
18
36
z
f tan x, y
36
24 x
z-intercept
f 3, 2
3
f x 3, 2 x
18 y
24x
f y 3, 2 y
18y
2
72
72
6. Find the plane tangent to the level set of the function F x, y, z
x, y, z
2
3
2x 2 yz 3 at the point
3, 2, 1 . Its z-intercept is
a. 1
b. 2
Correct Choice
c. 3
d. 108
e. 216
SOLUTION:
F
4xyz 3 , 2x 2 z 3 , 6x 2 yz 2
N X N P
24x 18y 108z 24 3
z-intercept 216/108 2
N
18 2
F 3, 2, 1
24, 18, 108
108 1 216
2
1 r 2 h.
3
If the radius r is currently 3 cm and decreasing at 2 cm/sec
7. The volume of a cone is V
while the height h is currently 4 cm and increasing at 1 cm/sec,
is the volume increasing or decreasing and at what rate?
a. increasing at 19 cm 3 / sec
b. increasing at 13 cm 3 / sec
c. neither increasing nor decreasing
d. decreasing at 13 cm 3 / sec
Correct Choice
e. decreasing at 19 cm 3 / sec
V dr
V dh
dV
dt
r dt
h dt
This is negative and so decreasing.
SOLUTION:
2 rh dr
3
dt
1 r 2 dh
3
dt
2
3
3 4
1
3
2
3
2
1
13
8. Duke Skywater is traveling in the Millenium Eagle through a dangerous galactic politon field whose
density is
2x 2 yz 3 . If Duke’s current position and velocity are r
3, 2, 1 and
v
. 25, . 5, . 25 , what is the current time rate of change of the politon field as seen by Duke?
a.
12
b.
120
Correct Choice
c. 12
d. 120
e. 12, 564
SOLUTION:
d
v
dt
4xyz 3 , 2x 2 z 3 , 6x 2 yz 2
. 25, . 5, . 25
24, 18, 108
3, 2, 1
6
9
27
24, 18, 108
12
9. Duke Skywater is traveling in the Millenium Eagle through a dangerous galactic politon field whose
density is
2x 2 yz 3 . If Duke’s current position is r
3, 2, 1 , in what unit vector direction should
he travel to reduce the politon field as fast as possible?
a.
b.
c.
d.
1
4, 3, 18
349
1
4, 3, 18
349
1
4, 3, 18
349
1
4, 3, 18
349
Correct Choice
SOLUTION:
4xyz 3 , 2x 2 z 3 , 6x 2 yz 2
3, 2, 1
24, 18, 108
The politon field decreases fastest in the direction of
6 4, 3, 18 .
1
1
So the unit vector direction is û
4, 3, 18
349
42
32
18 2
349
3
Compute
10.
F ds where F
along the curve r t
2x
2y, 2x
y3
2y
4 2 t cos t , 4 2 t sin t
2
for
t
.
4
4
HINT: Find a scalar potential.
1
a. 2
-3
-2
-1
0
1
2
3
x
b. 4
c. 8
d. 4
2
e. 8
2
Correct Choice
SOLUTION:
x2
f
y2
2xy
f ds
,
Compute
F
r
,
F ds
11.
f
2x
2y, 2x
4 2
4
f ,
F ds for F
2y
2y
f
x2
2xy
gy
yf
2x
2y
f
y2
2xy
hx
2
y, x
the one leaf of the 3 leaf rose r
2x
1 , 4 2
4
2
4
f ,
x
xf
y
1
2
2
2
By the F.T.C.C.
,
2
2
2
2
2
4
counterclockwise around
with x
cos 3
2
y
0.
HINT: Use Green’s Theorem.
a.
b.
c.
d.
e.
x
2
3
4
Correct Choice
6
12
SOLUTION:
cos 3
0 when 3
or
2
Let P
6
x
y and Q
x
y.
By Green’s Theorem,
/6
F ds
P dx
/6
/6
r2
cos 3
0
Q dy
xQ
/6
d
/6
yP
cos 2 3
dx dy
1
2r dr d
/6
/6
d
/6
1
cos 3
1 dx dy
cos 6
2
d
1
2
0
sin 6
6
/6
/6
6
4
12.
Compute
F dS for F
xy 2 , yx 2 , z x 2
y2
over the complete surface of the solid
x2
above the paraboloid z
below the plane z
a.
64
5
b.
64
3
c.
64
15
d.
256
15
e.
256
3
y2
By Gauss’ Theorem
x2
x2
2
2
y2
4
F dS
0
2
4, oriented outward.
Correct Choice
SOLUTION:
F
y2
0
x3
dV
r dr d dz
2
3
2r dz dr d
2
64
3
32
13. Compute
r
2r 2
F dV Use cylindrical coordinates.
F dS
2
2r 3 z
0
2
4
r
dr
2
2
8r 3
0
2r 5 dr
2
2r 4
r6
3
2
0
64
3
xy 2 dx dy over the quarter circle x 2
y2
4 in the first quadrant.
a. 4
b. 8
5
c. 16
5
d. 64
5
e. 32
5
Correct Choice
SOLUTION:
x3
x3
xy 2
/2
xy 2 dx dy
0
2
0
x x2
y2
r 4 cos dr d
r 3 cos
sin
/2
0
dx dy r dr d
r5 2
32
5 0
5
5
Work Out: (Points indicated. Part credit possible. Show all work.)
14. (15 points) The half cylinder x 2
y 2 9 for y 0 and 0 z 4 has mass surface density
y . Find the total mass and the center of mass. Follow these steps:
2
Parametrize the surface:
R ,z
3 cos , 3 sin , z
Find the tangent vectors, the normal vector and its length:
î
k
N
e
3 sin
3 cos ,
0
ez
0,
0,
1
N
9 cos 2
9 sin 2
er
e
î 3 cos
3 sin
k0
3 cos , 3 sin , 0
3
Evaluate the density on the surface and compute the total mass:
y2
|R
9 sin 2
,z
4
M
dS
|R
,z
sin 2
2
54
9 sin 2 3 d dz
N d dz
0
1
27 4
0
0
cos 2
2
d
54
0
Use symmetry to determine the x and z components fo the center of mass and then compute the y
component of the center of mass.
x
0
z
2
4
M xz
27 sin 3 3 d dz
y dS
0
1
324
1
u 2 du
1
y
M xz
M
432
54
81 4
0
1
cos 2
sin d
u
1
3
324 1
cos
du
sin d
0
324 u
u3
3
1
324
1
1
1
3
648
2
3
432
8
6
15. (15 points) A rectangular box sits on the xy-plane with its top 4 vertices on the paraboloid
2x 2
z
8y 2
8. Find the dimensions and volume of the largest such box.
NOTE: Full Credit for solving by Lagrange Multipliers, Half Credit for Eliminating a Variable.
SOLUTION: Let the corner in the first octant be
The dimensions are L
2x, W
So we need to maximize V
and the volume is V
z
4xyz subject to the constraint g
Lagrange Multiplier Method:
V
4yz, 4xz, 4xy
4x
4xyz
4x
4xz
16y
4xyz
16y 2
4xyz
z
4x 2
16y 2
z
x
8
z
8y 2
z
z
2
z
2
Eliminate a Variable Method: z
8
2x 2
8y 2
Use the constraint: z
V
4xyz
4xy 8
2x 2
Vx
32y
24x 2 y
32y 3
Vy
32x
8x 3
96xy 2
3
1
2 :
8x 2
3
2
1 :
32y 2
Conclusions: L
2x
2x 2
g
2x 2y z
2x 2
z
8y 2
4x, 16y, 1
4xyz.
8.
V
g
2
4yz
4xy
2y, H
x, y, z . So x, y and z are positive.
8y 2
32xy
3x 2
8y 4
8x 4
x2
8
2. W
V is positive on the region 2x 2
32xy 3
4y 2
0
12y 2
x
8
8x 3 y
0
z
and y
2
z
4
z
2
4, x
z
4
1, y
Since y
0, 3x 2
Since x
0, x 2
1
2
4y 2
4
1
12y 2
4
2
1
y
2y
8y 2
1
2
z
1, H
z
4
8 with x
8
2x 2
8y 2
V
0 and y
8
2 1 4
2
2
4
8
0 and V
0 on the boundary.
Since there is only one critical point, it must be a maximum.
7
16.
(20 points) Verify Stokes’ Theorem
F dS
F ds
C
C
for the cone C given by z
x
2
y
2
for z
3
oriented down and out, and the vector field F
yz, xz, z 2 .
Note: The boundary of the cone is the circle, x 2
y2
9,
Be sure to check the orientations. Use the following steps:
a. The cone, C, may be parametrized as R r,
r cos , r sin , r
Compute the surface integral by successively finding:
e r , e , N,
F,
F
R r,
,
F dS
C
î
k
er
cos
sin ,
e
r sin , r cos ,
Reverse N
1
F
e
0
k r cos 2
r sin
r sin 2
r cos , r sin , r
so it points down and out.
k
x
y
z
yz,
xz,
z2
î0
x
0
y
kz
z
x, y, 2z
r cos , r sin , 2r
R r,
F dS
2
3
0
0
C
2
3
0
3r 2 dr d
0
b. Parametrize the circle,
, v, F
r
2
,
r 2 sin 2
r 2 cos 2
F N dr d
C
r
er
î r cos
r cos , r sin , r
î
F
N
r3
3
2r 2 dr d
54
0
C, and compute the line integral by successively finding:
F ds
C
r
F
3 cos , 3 sin , 3
r
yz, xz, z 2
2
F ds
S
v
Reverse v
3 sin , 3 cos , 0
9 sin , 9 cos , 9
2
F vd
0
3 sin , 3 cos , 0
27 sin 2
27 cos 2
d
54
0
8