Name
1-14
/70
Fall 2014
15
/10
Solutions P. Yasskin
16
/10
17
/10
Total
/100
MATH 251/253 (circle one)
Exam 1
Sections 508/201/202(circle one)
Multiple Choice: (5 points each. No part credit.)
1. The vertices of a triangle are A
2, 1, 2 , B
and C
3, 2, 2 2
4, 3, 2 .
Find the angle at A.
a.
b.
c.
d.
e.
30°
45°
60°
120°
135°
Correct Choice
Solution: AB
AB AC
2
B
2
A
4
AB
AB AC
AB AC
cos
1, 1, 2
AC
1
4
2 2 2
1
2
C
2
1
2
A
2, 2, 0 .
AC
arccos
4
1
2
2. The vertices of a triangle are A
2, 1, 2 , B
3, 2, 2 2
Find a vector perpendicular to the plane of this triangle.
a.
b.
c.
d.
e.
AC
2 2
45°
and C
4, 3, 2 .
Correct Choice
1, 1, 0
1, 1, 0
1, 1, 1
1, 1, 1
1, 1, 1
Solution: AB
AB
4
1, 1, 2
î
k
1 1
2
2 2
0
2 2 ,2 2 ,0
AC
î 0
2, 2, 0 .
2 2
0
2 2
k0
2 2 1, 1, 0
1
3. Which of the following points lies on the line
and on the plane 2x
1, 1, 1
4, 3, 2
2, 3, 2
3, 1, 3
2, 2, 2
a.
b.
c.
d.
e.
3y
4z
x, y, z
2
t, 3
2t, 4
t
21?
Correct Choice
Solution: We are looking for the intersection. Plug the line into the plane:
2 2 t 3 3 2t 4 4 t
21 So 29 8t 21 or 8t
8 or t
1.
So the point of intersection is x, y, z
2 1, 3 2, 4 1
3, 1, 3
4. The quadratic surface x 2
a.
b.
c.
d.
e.
y2
6x
4y
2
0 is a
hyperboloid
hyperbolic ellipsoid
hyperbola
hyperboic paraboloid
hyperbolic cylinder
Correct Choice
Solution: Since there are no z’s, this surface is a cylinder. Since the equation is a hyperbola in the
xy-plane, this surface is a hyperbolic cylinder.
5. For the "twisted cubic" curve r t
2t 2 ,
2t 2 1
1 ,
2t 2 1
2t 2 ,
2
2t 1
1 ,
2t 2 1
2t 2 ,
2t 2 1
a.
b.
c.
d.
e.
Solution: v
î
v
|v
a
a|
2t ,
2t 2 1
2t ,
2t 2 1
2t ,
2t 2 1
2t ,
2t 2 1
2t ,
2t 2 1
1
2t 2
2t 2
2t 2
1
2t 2
2t 2
2t 2
1
2t 2
1, 2t, 2t 2
t, t 2 , 2 t 3 , find the binormal vector B.
3
Correct Choice
1
1
1
1
1
a
0, 2, 4t
k
î 8t 2
1 2t 2t 2
0 2
4t
16t 4
16t 2
4
4t 2
4t 2
2
4t
B
0
v
|v
k2
a
a|
0
4t 2 , 4t, 2
2t 2 ,
2t ,
1
2t 2 1 2t 2 1 2t 2 1
2
t, t 2 , 2 t 3
3
6. Find the mass of the "twisted cubic" curve r t
y2
if the linear density is
between t
0 and t
1
6xz.
a. 1
b. 1
5
c. 7
5
d. 20
7
e. 17
7
Correct Choice
y2
Solution:
1, 2t, 2t 2
v
6xz
1
|v |
4t 2
4t 4
1
M
ds
rt
t2
rt
|v | dt
5t 4 1
2
7
10 t
7
1
1
2t 2 dt
5t 4
10t 6 dt
0
10
7
1
0
5t 4
2t 2
1
0
t5
2 t3
3
6t
17
7
7. Find the work done when a bead is pushed along the "twisted cubic" curve r t
between t
0 and t
1 if you apply the force F
t, t 2 , 2 t 3
3
3z, y, x .
a. 1
2
b. 1
c. 3
2
d. 2
e. 5
2
Correct Choice
Solution: F
3z, y, x
W
Frt
1
F ds
2t 3 , t 2 , t
Frt
v dt
2t 3
2t 3
v
2t 3 dt
0
1, 2t, 2t 2
1
0
6t 3 dt
6t 4
4
1
0
3
2
8. You are riding on a train which is currently travelling EAST but curving toward the SOUTH.
Where do B and N for the train currently point?
a.
b.
c.
d.
e.
B
B
B
B
B
points
points
points
points
points
SOUTH
SOUTH
UP
DOWN
DOWN
and
and
and
and
and
N
N
N
N
N
points
points
points
points
points
DOWN.
UP.
SOUTH.
SOUTH.
Correct Choice
SOUTHEAST.
Solution: v and T point EAST and a points horizontally to the right of East. So N points
SOUTH and B T N points DOWN.
3
9. For the function f
x sin yz , which of the following are correct?
2
II.
a.
b.
c.
d.
e.
2
f
x y
2
f
y x
I.
f
x z
2
f
z x
III.
z cos yz
IV.
z cos yz
2
y cos yz
V.
y cos yz
VI.
f
y z
2
f
z y
x cos yz
xyz sin yz
x cos yz
xyz sin yz
I and II.
III and IV.
Correct Choice
V and VI.
I, II and III.
IV, V and VI.
Solution: Since mixed partial derivatives are equal, I and II cannot both be correct and V and VI
cannot both be correct.
10. Find the equation of the plane tangent to the graph of the function z
f x, y
xy 3 at
2, 1 . What is the z-intercept?
x, y
Correct Choice
a. 14
b. 6
c. 6
d. 14
e. 26
Solution:
x2y
f x, y
f x x, y
2xy
f y x, y
2
z-intercept is c
x
6
xy 3
f 2, 1
y3
3xy
5 2
2
6
z
f tan x, y
f x 2, 1
5
z
f 2, 1
f y 2, 1
10
z
6
10 1
5x
f x 2, 1 x
2
10 y
2
f y 2, 1 y
2, 1 ,
4
P
2
b. 1
1
1
14
11. Find the equation of the plane tangent to the graph of the equation x sin yz
a.
x2y
1 at
. What is the z-intercept?
4
2
c. 2
d. 4 2
Correct Choice
e. 2 2 2
Solution: The graph is a level set of the function F
F
sin yz , xz cos yz , xy cos yz
N
N X
F
P
sin
N P
z-intercept is c
, 2 cos
, 2 1 cos
4
4
4
4
1 x
1
1
y 1z
2
4
4
2
2
4 1
4 2
2
x sin yz .
1
4
1 , ,1
4
2
1
2
4
12. A fish is currently at the point
x, y, z
1, 2, 3 and has velocity v
1, 2, 1 .
dD
xyz 2 , find
, the time rate of change of the density as seen by the fish
dt
If the salt density is D
at the current instant.
a.
b.
c.
d.
e.
12
24
36
48
60
Correct Choice
yz 2 , xz 2 , 2xyz
18, 9, 12
Solution: D
dD
v D
1, 2, 1
18, 9, 12
18 18
dt
12
24
13. The equation z 3 sin x
z cos y 3 defines z as an implicit function of x and y. Notice that its
z at
graph passes through the point
, , 2 . Find
,
.
4 4
4 4
y
a.
b.
c.
d.
e.
2
5
2
6
2
7
1
6
1
7
Correct Choice
Solution: Apply
6 z 1
y 2
14.
to both sides:
y
z 1
y 2
2 1
2
0
z cos y
3z 2 z sin x
y
y
z
7 z
2
0
y
y
z sin y
0
2
7
The plot at the right is the contour plot
of which of these functions?
a. f x, y
sin x sin y
b. f x, y
x2
c. f x, y
sin
d. f x, y
sin x
e. f x, y
sin xy
y2
x2
y2
Correct Choice
sin y
Solution: Since the plot is circularly symmetric, the function must be a function of only the polar
coordinate r
x2 y2 .
5
Work Out: (10 points each. Part credit possible. Show all work.)
15. The pressure P, the temperature T, and the density
, of a certain ideal gas are related by
T. Currently, the temperature is T 300°K and is increasing at 2°K per minute while
gm
gm
the density is
4
and is decreasing at 0. 05
per minute. Consequently, the pressure
3
3
cm
cm
is currently P 10 3 T 10 3 4 300
1. 2 atm. At what rate is P changing and is it increasing
or decreasing?
10
P
3
d
P
P
0. 05
10 3
. 004
10 3 T 0. 3
dt
T
P dT
P d
By chain rule: dP
. 004 2 0. 3
0. 05 . 008 . 015
0. 007
dt
dt
T dt
So P is decreasing at 0. 007 atm per minute.
Solution:
dT
dt
2
1 r 2 h. If the radius and height are measured to be
3
3cm 0. 02cm and h 5cm 0. 03cm, then the volume is computed to be
1 3 2 5 15 cm 3 . Use differentials to estimate the error in this computed volume.
3
16. The volume of a cone is V
r
V
V dr
r
1 3
3
Solution: V dV
2 3 5 . 02
3
2
V dh
2 rh dr 1 r 2 dh
3
3
h
. 03
0. 29
0. 911 06
x2
17. Find the minimum value of the function f
2y 2
4z 2 on the plane x
y
z
14.
Solution:
METHOD I: Eliminate a Variable:
x
14
f
y
6y
y
z
2 14
2z
28
f
y
14
y
z
4y
0
2y
10z
28
z
2
2y 2
f
z
4z 2
2 14
y
y
4, z
z
8z
0
2
and
f 8, 4, 2
64 32 16 112
So
x 8
There is only one critical point and f can be arbitrarily large.
So the critical point must be the minimum.
METHOD II: Lagrange Multipliers
f
2x, 4y, 8z
2x
4y
g
8z
Use the constraint:
x
1, 1, 1
4z
4z
y
2z
z
f
g
14
7z
2x
4y
8z
2z
14
z
2
x
8
y
4
f 8, 4, 2
64 32 16 112
There is only one critical point and f can be arbitrarily large.
So the critical point must be the minimum.
6