Name
ID
MATH 251
Exam 1 Version A
Sections 506
Solutions
1-12
/60
Spring 2013
13
/10
P. Yasskin
14
/10
15
/10
16
/10
Total
/100
Multiple Choice: (5 points each. No part credit.)
1. Find the line through P
3, 1, 2
and b
a.
x, y, z
2
t, 8
b.
x, y, z
2
t, 8
2t, 1
3t
c.
x, y, z
2
t, 8
2t, 1
3t
d.
x, y, z
1
2t, 2
8t, 3
t
e.
x, y, z
1
2t, 2
8t, 3
t
a
which is perpendicular to both of the vectors
1, 2, 3
1, 0, 2 .
2t, 1
3t
Correct Choice
î
SOLUTION:
v
a
b
k
3
1
2
1
0
2
2, 8, 1
X
P
tv
x, y, z
1
2t, 2
8t, 3
t
2. A triangle has vertices at A
1, 1, 1 , B
3, 4, 3 and C
3, 3, 2 .
Drop a perpendicular from B to the side AC.
Find the point P where the perpendicular intersects the side AC.
a.
b.
c.
d.
e.
7, 7, 5
3 3 3
4, 4, 2
3 3 3
5, 5, 4
3 3 3
41 , 47 ,
29 29
12 , 18 ,
29 29
SOLUTION:
proj AC AB
P
Correct Choice
5
29
24
29
AB B A
AB AC AC
4
2
4
AC
A proj AC AB
1, 1, 1
2, 3, 4
6
4
AC
4 2, 2, 1
1
4, 4, 2
3 3 3
C
A
2 2, 2, 1
3
2, 2, 1
4, 4, 2
3 3 3
7, 7, 5
3 3 3
1
3. If u points NorthEast and v points Down, then u
v points
a. NorthEast
b. NorthWest
Correct Choice
c. SouthEast
d. SouthWest
e. Up
SOLUTION:
Fingers of right hand point NorthEast with palm Down. The thumb points NorthWest.
4. Identify the quadratic surface for the equation
2x
2
2
y
2
3
z
2
2
x
2
2
2y
2
3
z
2
2
a. hyperboloid of 1 sheet
b. hyperboloid of 2 sheets
c. cone
d. hyperbolic paraboloid
Correct Choice
e. hyperbolic cylinder
SOLUTION:
x
2
2
x
2
2
Subtract the right side from the left side, expand the z terms and then solve for z:
y
3
2
z
y
3
2
4z
2
2
0
z
2
z
2
0
x 2
4
2
y
3
2
4
h x, y
10 x x 2 y 2 . If she is
1, 2 , in what unit vector direction should she walk to go up hill as fast
5. A girl scout is hiking up a mountain whose attitude is given by z
currently at the point
as possible?
a.
b.
c.
d.
e.
3,
5
3, 4
5 5
4,
5
4, 3
5 5
4
5
Correct Choice
3
5
4, 3
SOLUTION:
|v |
x, y
9
16
h
5
1
v
2x, 2y
v
h
1,2
1
2, 4
3, 4
3, 4
5 5
2
6. Find the arclength of 4 revolutions around the helix r t
2 cos 2t, 2 sin 2t,
NOTE: Each revolution covers an angle of 2 . How much does t change?
3t .
a. 2
b. 4
c. 5
d. 15
Correct Choice
e. 20
SOLUTION:
v
4 sin 2t,
4 cos 2t,
3
16 sin 2 2t
|v |
We cover 1 revolution as t runs from 0 to
4
L
|v | dt
ds
5 dt
5t
0
4
0
a. M
24
3
b. M
36
2
c. M
120
3
d. M
180
2
e. M
240
2
ds
20
2 cos 2t, 2 sin 2t,
0 and t 2 .
z2
z 2 |v | dt
9t 2
2
has linear mass density
3t
z2.
5
|v |
3
45 t
3
9t 2 5 dt
0
2
15 8
3
0 and t
2
120
3
0
8. Find the work done to move an object along the helix r t
between t
5
Correct Choice
SOLUTION:
M
9
.
7. A wire in the shape of the helix r t
Find its total mass between t
16 cos 2 2t
by the force F
yz,
xz,
2 cos 2t,
z.
2 sin 2t,
3t
a. 33
2
b. 33
c. 33
2
2
d. 33
2
e. 66
2
SOLUTION:
Correct Choice
Frt
6t sin 2t,
2
W
F ds
2
Frt
0
v dt
0
6t cos 2t,
3t
v
24t sin 2 2t
24t cos 2 2t
4 sin 2t,
4 cos 2t,
2
9t dt
33t dt
0
3
33 t 2
2
2
0
66
2
3
9. Find the tangent line to the helix r t
2 cos 2t,
2 sin 2t,
Where does it intersect the xy-plane?
HINT : What are the position and tangent vector at t
a.
x, y
1,
b.
x, y
1,
c.
x, y
2, 2
d.
x, y
2, 2
e.
x, y
Tangent Line:
r
2 cos ,
2
4 cos 2t, 3
2
?
X
P
3
2, 0, 3
2
2
4 sin , 4 cos , 3
0, 4, 3
2, 4t, 3
3t
2
3
3t 0 or t
. So x, y
2
2
v
2
x, y, z
tv
10. Find the plane tangent to the graph of z
b.
.
2 sin ,
The line intersects the xy-plane when z
e
2
Correct Choice
P
4 sin 2t,
a.
2
at the point t
2,
SOLUTION:
vt
3t
y ln x at the point
2, 2
e, 2 . Its z-intercept is
Correct Choice
c. 0
d. 2
e. e
SOLUTION:
f
y ln x
f e, 2
2
z
fx
y
x
f x e, 2
2
e
fy
ln x
f y e, 2
1
f e, 2
2
When x
f x e, 2 x
2 x
e
y
e
e
1y
f y e, 2 y
2
2
0, we have z
2
2
2
2.
4
11. Find the plane tangent to the graph of x 2 z 2
2zy 2
yx 3
71 at the point
2, 1, 0 . Its z-intercept is
a. 2
b. 4
c. 8
Correct Choice
d. 16
e. 32
SOLUTION:
F x, y, z
x2z2
N
12, 8, 2
N X
F
2,1,0
When x
y
0, we have z
2zy 2
yx 3
N P
2xz 2
F
12x
8y
3yx 2 , 4zy
2z
12 2
x 3 , 2x 2 z
8 1
2 0
2y 2
32
16.
x, y
1, 1
is a critical point of the function f x, y
2
Second Derivative Test to classify this critical point.
12. The point
4xy
x3y
4xy 3 . Use the
a. local minimum
b. local maximum
Correct Choice
c. saddle point
d. TEST FAILS
SOLUTION:
fx
fy
f xx
f yy
f xy
D
f x 1, 1
4 1
3 1
2
2
2
1
1
3
2
4x x 12xy
f y 1,
4 1 12
2
2
1
6xy
f xx 1,
3
2
24xy
f yy 1, 1
12
2
4 3x 2 12y 2
f xy 1, 1
4 3 12 1
2
2
2
2
f xx f yy f xy
3
12
2
32
4y
3x 2 y
Since D
3
4 1
0
Checked
2
0
Checked
4y 3
0 and f xx
2
2
2
0 it is a local maximum.
Work Out: (10 points each. Part credit possible. Show all work.)
13. Find the point where the line
x, y, z
4
3t, 3
2t, 2
t
intersects the plane x
2y
3z
20, or
explain why they are parallel.
SOLUTION:
Substitute the line into the plane and solve for t:
20 x 2y 3z
4 3t 2 3 2t 3 2 t
2t
Substitute back into the line:
x, y, z
4 3 2 ,3 2 2 ,2 2
10, 1, 4
Check by substituting into the plane:
x 2y 3z
10 2 1 3 4
20
16
20
t
2
5
14. Find the line where the planes
2x
6y
4z
7 and 3x
9y
6z
5 intersect, or explain why
they are parallel.
SOLUTION:
2, 6, 4 .
The normal to the first plane is N 1
The normal to the second plane is N 2
3, 9, 6 .
3
Notice that N 2
N , so the normals are parallel.
2 1
Alternatively, compute N 2 N 1 0, so the normals are parallel.
In either case the planes are parallel and do not intersect.
15. If two adjustable resistors, with resistances R 1
and R 2 , are arranged in parallel, the total
resistance R is given by
R
R1R2
R1 R2
Currently, R 1 3
and R 2 7
and they are changing according to dR 1
0. 1
dt
dR 2
0. 2 sec . Find R and dR . Is R increasing or decreasing?
sec and
dt
dt
SOLUTION:
dR
dt
R
3
3
R
R2
R dR 1
R 1 dt
R2 2
dR 1
2 dt
R1 R2
0. 031 sec
7
2. 1 .
7
R 1 R 2 R 2 R 1 R 2 1 dR 1
dR 2
dt
dt
R1 R2 2
2
R1
dR 2
72
0. 1
2 dt
R1 R2
3 7 2
So R is decreasing.
R1
R 2 R 1 R 1 R 2 1 dR 2
dt
R1 R2 2
32
4. 9 1. 8
.2
2
100
3 7
16. A rectangular box sits on the xy-plane with its top 4 vertices in the paraboloid z
8
2x 2
8y 2 .
Find the dimensions and volume of the largest such box.
SOLUTION: Let the corner in the first quadrant be x, y, z . The dimensions are L 2x, W 2y,
H z. So x, y and z are positive. So the volume is
V
2x 2y z 4xyz 4xy 8 2x 2 8y 2
32xy 8x 3 y 32xy 3
V x 32y 24x 2 y 32y 3 8y 4 3x 2 4y 2
0
Since y 0, 3x 2 4y 2 4
1
3
2
2
2
2
2
V y 32x 8x 96xy
8x 4 x 12y
0
Since x 0, x 12y
4
2
3 1
2 :
8x 2 8
x 1
1
3 2
1 :
32y 2 8
y
z 8 2x 2 8y 2 8 2 2 4
2
L 2x 2. W 2y 1, H z 4
V 2 1 4 8
2
2
V is positive on the region 2x 8y
8 with x 0 and y 0 and V 0 on the boundary.
Since there is only one critical point, it must be a maximum.
Note: Problem 12 shows V is a local maximum.
6