Name
Sec
MATH 253
Exam 1
Sections 501,503
1-11
/66
13
/10
12
/20
14
/10
Total
/106
Fall 2009
Solutions
P. Yasskin
Multiple Choice: (6 points each. No part credit.)
1. If fx, y = x 2 cosy 2 , which of the following is FALSE?
a. f x x, y = 2x cosy 2 
b. f y x, y = −2x 2 y siny 2 
c. f xx x, y = 2 cosy 2 
d. f yy x, y = −4x 2 y cosy 2 
Correct Choice
e. f xy x, y = −4xy siny 2 
f yy x, y = d −2x 2 y siny 2  = −2x 2 siny 2  − 4x 2 y 2 cosy 2 
dy
2. The quadratic surface x 2 − y 2 + z 2 − 4x − 6y − 10z + 16 = 0 is a
a. hyperboloid of 1 sheet and center 2, 3, 5
b. hyperboloid of 1 sheet and center 2, −3, 5
Correct Choice
c. hyperboloid of 2 sheets and center 2, 3, 5
d. hyperboloid of 2 sheets and center 2, −3, 5
e. cone with vertex 2, 3, 5
x 2 − 4x − y 2 + 6y + z 2 − 10z + 16 = 0
x 2 − 4x + 4 − y 2 + 6y + 9 + z 2 − 10z + 25 + 16 = 4 − 9 + 25
x − 2 2 − y + 3 2 + z − 5 2 = 4
x − 2 2 + z − 5 2 = 4 + y + 3 2
hyperboloid with center 2, −3, 5
x − 2 2 + z − 5 2 ≥ 4
1 sheet
3. An airplane is travelling due North at constant speed and a constant altitude as it crosses the
equator. In what direction does the B̂ vector point?
HINTS: Remember the Earth is curved. Ignore the rotation of the Earth.
a. East
b. West
Correct Choice
c. South
d. Up
e. Down
T̂ points North. N̂ points Down. So B̂ = T̂ × N̂ points West.
1
4. A triangle has edge vectors AB = 2, 1, −2 and AC = −2, −2, 4.
Find the altitude of the triangle if AB is the base.
a.
2 5
3
b.
5
3
Correct Choice
c. 2 5
5
d.
e. 3 5
AB × AC =
î
̂
k̂
2
1
−2
−2 −2
4
= î4 − 4 − ̂ 8 − 4 + k̂ −4 + 2 = 0, −4, −2
Area = 1 AB × AC = 1 −4 2 + −2 2 = 1 20 = 5
2
2
2
1
Area = Base ⋅ Altitude
Base = AB = 4 + 1 + 4 = 3
2
2 5
Altitude = 2Area =
Base
3
5. A box slides down the helical ramp ⃗
rt = 4 cos t, 4 sin t, 9 − 3t starting at height z = 9 and
ending at height z = 0. How far does the box slide?
a. 3
b. 5
c. 15
Correct Choice
d. 25
e. 75
⃗
v = −4 sin t, 4 cos t, −3
L=
|v⃗| =
∫ z=9 ds = ∫ 0 |v⃗| dt = ∫ 0 5 dt =
z=0
3
16 sin 2 t + 16 cos 2 t + 9 = 5
3
5t
3
0
= 15
6. A box slides down the helical ramp ⃗
rt = 4 cos t, 4 sin t, 9 − 3t starting at height z = 9 and
⃗ = −yz, xz, 5z.
ending at height z = 0 under the action of the force F
Find the work done on the box.
a. 9
2
b. 9
c. 25
2
d. 27
2
Correct Choice
e. 27
⃗
v = −4 sin t, 4 cos t, −3
⃗ = −yz, xz, 5z = −4 sin t 9 − 3t, 4 cos t 9 − 3t, 5 9 − 3t
F
⃗ ⋅⃗
F
v = 16 sin 2 t 9 − 3t + 16 cos 2 t 9 − 3t − 15 9 − 3t = 9 − 3t
z=0
3
3
3
⃗ ⋅ ds⃗ = ∫ F
⃗ ⋅⃗
v dt = ∫ 9 − 3t dt = 9t − 3 t 2
= 27
W=∫ F
2
2
0
z=9
0
0
2
7. The diameter and height of a cylindrical trash can (no lid) are measured as D = 30
cm and h = 40 cm. The metal is 0. 2 cm thick. Use differentials to estimate the volume of
metal used to make the can.
a. 165π cm 3
b. 210π cm 3
c. 285π cm 3
Correct Choice
d. 330π cm 3
e. 525π cm 3
V = πr 2 h
r = 15
h = 40
dr = 0. 2
dh = 0. 2
dV = ∂V dr + ∂V dh = 2πrhdr + πr 2 dh = 2π ⋅ 15 ⋅ 40 ⋅ 0. 2 + π15 2 ⋅ 0. 2 = 285π
∂r
∂h
V = 1 πD 2 h
D = 30
4
dV = ∂V dD + ∂V dh =
∂D
∂h
h = 40
dD = 0. 4
1 πDh dD +
2
dh = 0. 2
1 πD 2 dh = 1 π ⋅ 30 ⋅ 40 ⋅ 0. 4 + 1 π30 2 ⋅ 0. 2 = 285π
4
2
4
8. Find the equation of the plane tangent to the surface z = x 3 y 2 at the point 2, 1.
Then the z-intercept is z =
a. −40
b. 8
c. −8
d. 32
e. −32
Correct Choice
fx, y = x 3 y 2
f2, 1 = 8
The tangent plane is
f x x, y = 3x 2 y 2
f x 2, 1 = 12
z = f2, 1 + f x 2, 1x − 2 + f y 2, 1y − 1
f y x, y = 2x y
f y 2, 1 = 16
3
= 8 + 12x − 2 + 16y − 1
= 12x + 15y − 32
The intercept is z = −32.
9. Find the equation of the plane tangent to the surface 12xyz − z 3 = 45 at the point 1, 2, 3.
Then the z-intercept is z =
a. 135
b. 45
c. − 3 6
d. −45
Correct Choice
e. −135
Fx, y, z = xyz − z 3
P = 1, 2, 3
⃗
∇F = 12yz, 12xz, 12xy − 3z 2 
⃗⋅X = N
⃗⋅P
N
z-intercept when:
⃗ = ⃗
N
∇F
1,2,3
= 72, 36, −3
72x + 36y − 3z = 72 ⋅ 1 + 36 ⋅ 2 − 3 ⋅ 3 = 135
x = 0, y = 0,
−3z = 135
z = −45
3
10. Starting from the point 1, −2, find the maximum rate at which the function fx, y = x 2 y 3
increases.
a. 20
Correct Choice
b. 25
c. 400
d. −16, 12
e. 16, −12
⃗
∇f = 2xy 3 , 3x 2 y 2 
⃗
∇f1, −2 = −16, 12
The maximum rate of increase is ⃗
∇f1, −2 =
16 2 + 12 2 =
256 + 144 =
400 = 20
11. Which of the following is the plot of the vector field Fx, y = x + y, x − y?
y
y
4
2
4
2
d.
a.
-4
-2
2
4
-2
-4
x
-2
-4
y
y
4
Correct
2
4
-2
x
-4
-2
-2
x
-4
4
c.
-2
2
-2
⃗ is horizontal.
F 2 = x − y = 0 and F
⃗ is vertical.
When y = −x, F 1 = x + y = 0 and F
⃗ is (b).
So F
When y = x,
2
-4
4
4
e.
Choice
-4
y
2
x
2
b.
-2
4
-4
2
-4
2
-2
4
x
-4
4
Work Out: (Points indicated. Part credit possible. Show all work.)
12. (20 points) Find the point on the curve ⃗
rt =
e t , 2 t, e −t
where the curvature is a local
maximum or local minimum. Is it a local maximum or local minimum?
a|
|v⃗ × ⃗
HINTS: First find the curvature κ =
. Then find the critical point and apply the first or
3
|v⃗|
second derivative test.
⃗
vt =
|v⃗| =
e t , 2 , −e −t
e 2t + 2 + e −2t = e t + e −t
⃗
at = e t , 0, e −t 
⃗
v×⃗
a=
a| =
|v⃗ × ⃗
k̂
î
̂
et
2 −e −t
et
0
2e −2t + 4 + 2e 2t =
a|
|v⃗ × ⃗
=
3
|v⃗|
κ′ =
−2 2 e t − e −t 
=0
e t + e −t  3
κ ′′ =
=
2 e −t − ̂ 1 + 1 + k̂ − 2 e t
2 e −t , −2, − 2 e t
=
e −t
κ=
⃗r0 =
=î
2 e −2t + 2 + e 2t =
2 e t + e −t 
2 e t + e −t 
2
=
t
−t 3
t
e + e 
e + e −t  2
e 0 , 2 0, e −0
e t = e −t
e 2t = 1
t=0
= 1, 0, 1
−2 2 e t + e −t  3 e t + e −t  + 2 2 e t − e −t 3e t + e −t  2 e t − e −t 
e t + e −t  6
−2 2 e t + e −t  2 + 2 2 e t − e −t 3e t − e −t 
e t + e −t  4
κ ′′ 0 =
−2 2 1 + 1 2 + 2 2 1 − 131 − 1
−2 2
=
<0
4
4
1 + 1
local maximum
13. (10 points) The pressure, P, density, D, and temperature, T, of a certain ideal gas are related
by P = 4DT. A fly is currently at the point ⃗rt 0  = 3, 2, 4 and has velocity ⃗
vt 0  = 2, 1, 2.
At the point 3, 2, 4, the density and temperature and their gradients are
D = 50
⃗
∇D =
∂D , ∂D , ∂D
∂x ∂y ∂z
T = 300
⃗
∇T =
∂T , ∂T , ∂T
∂x ∂y ∂z
= 0. 1, 0. 4, 0. 2
= 2, 3, 1
Find the time rate of change of the pressure, dP , as seen by the fly.
dt
dP = dP dD + dP dT = dP
dt
dD dt
dT dt
dD
= 4T ⃗
∇D ⋅ ⃗
v + 4D ⃗
∇T ⋅ ⃗
v
dD dx + dD dy + dD dz
dx dt
dy dt
dz dt
+ dP
dT
dT dx + dT dy + dT dz
dx dt
dy dt
dz dt
= 4 ⋅ 300 0. 2 + 0. 4 + 0. 4 + 4 ⋅ 50 4 + 3 + 2 = 1200 + 1800 = 3000
5
14. (10 points) Suppose p = px, y, while x = xu, v and y = yu, v.
Further, you know the following information:
x2, 1 = 3
y2, 1 = 4
∂p
2, 1 = 5
∂x
∂p
2, 1 = 6
∂y
∂p
3, 4 = 7
∂x
∂p
3, 4 = 8
∂y
∂x 2, 1 = 9
∂u
∂x 2, 1 = 10
∂v
∂x 3, 4 = 11
∂u
∂x 3, 4 = 12
∂v
∂y
2, 1 = 18
∂u
∂y
2, 1 = 17
∂v
∂y
3, 4 = 16
∂u
∂y
3, 4 = 15
∂v
a. Write out the chain rule for
p2, 1 = 13
p3, 4 = 14
∂p
.
∂u
∂p
∂p ∂x
∂p ∂y
=
+
∂u
∂x ∂u
∂y ∂u
b. Then use it and the above information to compute
∂p
∂p
=
∂u
∂x
xu,v,yu,v
∂x + ∂p
∂u
∂y
xu,v,yu,v
∂p
2, 1.
∂u
∂y
∂u
∂p
∂p
∂p
∂y
2, 1 =
3, 4 ∂x 2, 1 +
3, 4
2, 1 = 7 ⋅ 9 + 8 ⋅ 18 = 207
∂u
∂x
∂u
∂y
∂u
6