MAT 267 Midterm 2
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School
Arizona State University *
*We aren’t endorsed by this school
Course
267
Subject
Mathematics
Date
Feb 20, 2024
Type
Pages
3
Uploaded by Ovidis
Matthew Guynn
Ashbrook
MAT
267
ONLINE
A
Spring
2024
Assignment Midterm
2 due 02/08/2024 at 02:58pm MST
Problem 1.
(1 point)
Use the contour diagram of
f
to decide if the specified di-
rectional derivative is positive, negative, or approximately
zero.
?
1. At the point
(
1
,
0
)
in the direction of
−
⃗
j
,
?
2. At the point
(
0
,
−
2
)
in the direction of
(
⃗
i
−
2
⃗
j
)
/
√
5,
?
3. At the point
(
−
2
,
2
)
in the direction of
⃗
i
,
?
4. At the point
(
−
1
,
1
)
in the direction of
(
−
⃗
i
−
⃗
j
)
/
√
2,
?
5. At the point
(
0
,
2
)
in the direction of
⃗
j
,
?
6. At the point
(
−
1
,
1
)
in the direction of
(
−
⃗
i
+
⃗
j
)
/
√
2,
(Click graph to enlarge)
Correct Answers:
•
ZERO
•
POSITIVE
•
NEGATIVE
•
ZERO
•
POSITIVE
•
POSITIVE
Problem 2.
(1 point)
Calculate all four second-order partial derivatives of
f
(
x
,
y
) =
3
x
2
y
+
5
xy
3
.
f
xx
(
x
,
y
) =
f
xy
(
x
,
y
) =
f
yx
(
x
,
y
) =
f
yy
(
x
,
y
) =
Correct Answers:
•
6*y
•
3*2*x+5*3*yˆ2
•
3*2*x+5*3*yˆ2
•
5*x*3*2*y
Problem 3.
(1 point)
Find the directional derivative of
f
(
x
,
y
,
z
) =
z
3
−
x
2
y
at the point
(-1, -3, 3) in the direction of the vector
v
=
⟨−
4
,
5
,
5
⟩
.
Correct Answers:
•
18.9560896108172
Problem 4.
(1 point)
Find the maximum rate of change of
f
(
s
,
t
) =
te
st
at the point
(
0
,
4
)
.
Answer:
Correct Answers:
•
16.0312
Problem 5.
(1 point)
Find the partial derivatives of the function
f
(
x
,
y
) =
xye
−
5
y
f
x
(
x
,
y
) =
f
y
(
x
,
y
) =
f
xy
(
x
,
y
) =
f
yx
(
x
,
y
) =
Correct Answers:
•
y*exp(-5*y)
•
x*(-5*y*exp(-5*y) + exp(-5*y))
•
-5*y*exp(-5*y) + exp(-5*y)
•
-5*y*exp(-5*y) + exp(-5*y)
1
Problem 6.
(1 point)
Find the partial derivatives of the function
f
(
x
,
y
) =
−
8
x
−
2
y
−
4
x
+
5
y
f
x
(
x
,
y
) =
f
y
(
x
,
y
) =
Correct Answers:
•
((-4*x + 5*y)*-8 - (-8*x - 2*y)*-4)/(-4*x + 5*y)**2
•
((-4*x + 5*y)*(- 2) - (-8*x - 2*y)*5)/(-4*x + 5*y)**2
Problem 7.
(1 point)
The function
f
has continuous second derivatives, and a critical
point at (4, 3).
Suppose
f
xx
(
4
,
3
) =
−
20
,
f
xy
(
4
,
3
) =
−
10
,
f
yy
(
4
,
3
) =
−
5.
Then the point (4, 3):
•
A. is a saddle point
•
B. is a local minimum
•
C. is a local maximum
•
D. cannot be determined
•
E. None of the above
Correct Answers:
•
D
Problem 8.
(1 point)
Does the function
f
(
x
,
y
) =
x
2
2
+
5
y
4
+
8
y
2
−
2
x
have a global maximum and global minimum? If it does, identify
the value of the maximum and minimum. If it does not, be sure
that you are able to explain why.
Global maximum?
(Enter the value of the global maximum, or
none
if there is no
global maximum.)
Global minimum?
(Enter the value of the global minimum, or
none
if there is no
global minimum.)
Correct Answers:
•
none
•
-2*2/2
Problem 9.
(1 point)
Consider the following integral. Sketch its region of inte-
gration in the
xy
-plane.
Z
1
0
Z
y
√
y
180
x
2
y
3
dxdy
(a) Which graph shows the region of integration in the
xy
-plane? [?/A/B]
(b) Evaluate the integral.
A
B
(Click on a graph to enlarge it)
Correct Answers:
•
B
•
-2.33766
Problem 10.
(1 point)
Evaluate the following integral.
Z
3
1
Z
3
0
(
5
x
2
+
y
2
)
dxdy
=
Correct Answers:
•
116
2
Problem 11.
(1 point)
Suppose
R
is the shaded region in the figure, and
f
(
x
,
y
)
is
a continuous function on
R
. Find the limits of integration
for the following iterated integral.
(a)
ZZ
R
f
(
x
,
y
)
dA
=
Z
B
A
Z
D
C
f
(
x
,
y
)
dydx
A
=
B
=
C
=
D
=
Correct Answers:
•
-2
•
1
•
0.333333*(x+2)-1
•
2
Problem 12.
(1 point)
Calculate the volume under the elliptic paraboloid
z
=
x
2
+
5
y
2
and over the rectangle
R
= [
−
4
,
4
]
×
[
−
3
,
3
]
.
Correct Answers:
•
976
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3
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