Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Exam 2 due 09/22/2021 at 10:12pm MST
Problem 1. (1 point)
The radius of a right circular cone is increasing at a rate of 3 inches
per second and its height is decreasing at a rate of 4 inches per
second. At what rate is the volume of the cone changing when the
radius is 50 inches and the height is 10 inches?
NOTE: The volume of a cone with base radius rand height his
given by V=1
3πr2h.
cubic inches per second
Solution:
SOLUTION:
dV
dt =∂V
∂r
dr
dt +∂V
∂h
dh
dt
=2πrh
3(3) + πr2
3(−4)
=1000π−10000
3π
=−7000
3πin3/sec
Answer(s) submitted:
•
(incorrect)
Correct Answers:
•-7330.38285837618
Problem 2. (1 point)
Evaluate the integral by reversing the order of integration.
Z1
0Z2
2y
ex2dxdy =
Solution:
SOLUTION
The region of integration is shown below.
The region is bounded below by y=0 and above by y=x
2, with
0≤x≤2.
Thus
Z1
0Z2
2y
ex2dxdy =R2
0R
x
2
0ex2dydx
=R2
0hyex2iy=x
2
y=0dx
=R2
0
x
2ex2dx
Using the substitution u=x2,du =2xdx, yields
=1
4R4
0eudu
=1
4e4−1
Answer(s) submitted:
•13.3995
(correct)
Correct Answers:
•13.3995
1
Problem 3. (1 point)
Match the functions with the graphs of their domains.
1. f(x,y) = 3x−y
2. f(x,y) = e1
3x−y
3. f(x,y) = ln(3x−y)
4. f(x,y) = px5y3
A.
B.
C.
D.
Solution:
Solution:
The domain of ln(3x−y)is the region 3x−y>0, which is the
half-plane with boundary 3x−y=0.
The domain of 3x−yis the entire plane.
The domain of px5y3is the region x5y3≥0. The domain will
include the lines x=0 and y=0. The factor x5is positive for all
positive xand negative for all negative x. The factor y3is positive
for all positive yand negative for all negative y. The product is
positive if both factors are positive, or both factors are negative.
This occurs in Quadrant I and Quadrant III.
The domain of e1
3x−yis the same as the domain of the exponent.
This means that the domain is the same as the domain of 1
3x−y,
which is the set line 3x−y6=0. Thus, the domain is entire plane
except for the line 3x−y=0.
Answer(s) submitted:
•a
2
•d
•b
•c
(correct)
Correct Answers:
•A
•D
•B
•C
Problem 4. (1 point)
Find the linearization of the function f(x,y) = p137 −4x2−4y2
at the point (4,−4).
L(x,y) =
Use the linear approximation to estimate the value of f(3.9,−3.9)
f(3.9,−3.9)≈
Solution:
SOLUTION
f(x,y) = p137 −4x2−4y2⇒f(4,−4) = 3
fx(x,y) = −1
2p137 −4x2−4y2·4·2x⇒fx(4,−4) = −5.33333
fy(x,y) = −1
2p137 −4x2−4y2·4·2y⇒fy(4,−4) = 5.33333
Both fxand fyare continuous functions for 137−4x2−4y2>0, so
fis differentiable at (4,−4)and the linearization of fat (4,−4)
is
L(x,y) = 3−5.33333(x−4) + 5.33333(y+4)
We use the linear approximation to estimate the value of
f(3.9,−3.9):
f(3.9,−3.9)≈L(3.9,−3.9) = 3−5.33333(3.9−4)+5.33333(−3.9+4) = 4.06667
Answer(s) submitted:
•((137)/3)-((4x)/3)+4y
•-24.7333
(incorrect)
Correct Answers:
•3-5.33333*(x-4)+5.33333*(y+4)
•4.06667
3
Problem 5. (1 point)
Consider the following integral. Sketch its region of inte-
gration in the xy-plane.
Z3
0Z9
y2ysinx2dx dy
(a) Which graph shows the region of integration in the
xy-plane? [?/A/B/C/D]
(b) Write the integral with the order of integration re-
versed:
Z3
0Z9
y2ysinx2dx dy =ZB
AZD
C
ysinx2dy dx
with limits of integration
A =
B =
C =
D =
(c) Evaluate the integral.
A B
C D
(Click on a graph to enlarge it)
Solution:
SOLUTION
(a) The region is bounded on the left by the function x=y2
and on the right by the vertical line x=9. The bounds for yare
0≤y≤3. Thus the region corresponds to graph A.
(b) The region is bounded below by y=0 and above by y=√x,
while 0 ≤x≤9. Thus
Z3
0Z9
y2ysinx2dx dy =Z9
0Z√x
0
ysin(x2)dy dx
(c)
R9
0R√x
0ysin(x2)dy dx =R9
0hy2
2i√x
0sin(x2)dx
=1
2R9
0xsin(x2)dx [substitution: u=x2,du =2xdx]
=1
4R81
0sin(u)du
=1
4[−cos(u)]81
0
=1−cos(81)
4
Answer(s) submitted:
•B
•0
•3
•yˆ(2)
•9
•.05582
(score 0.65)
Correct Answers:
•A
•0
•9
•0
•sqrt(x)
•[1-cos(3ˆ4)]/4
4
Problem 6. (1 point)
Consider the function f(x,y) = 2x3+y4on the region {(x,y)|x2+
y2≤36}.
Find the absolute minimum value:
Find the point(s) at which the absolute minimum is attained.
List your answer as comma separated list, e.g. (1, 1), (2,3)
).
Find the absolute maximum value:
Find the point(s) at which the absolute maximum is attained.
List your answer as comma separated list, e.g. (1,1), (2,3)
).
Solution:
SOLUTION
f(x,y) = 2x3+y4⇒fx(x,y) = 6x2,fy(x,y) = 4y3. Thus the only
critical point is (0,0), which is in the interior of the domain, and
f(0,0) = 0
The boundary of the region is x2+y2=36 ⇒y=
±√36 −x2,−6≤x≤6. Substituting in the function fyields
g(x) = f(x,±√36 −x2) = 2x3+ (36 −x2)2,−6≤x≤6.
Plotting the function g(x)in the given interval, we can see that it
attains its minimum at x=−6 and its maximum at x=0.
We have f(−6,0) = −432 and f(0,−6) = f(0,6) = 1296
Thus the absolute minimum value is −432 attained at (−6,0)and
the absolute maximum value is 1296 attained at (0,−6)and (0,6)
.
Answer(s) submitted:
•-432
•(-6,0)
•5616
•(0,-6),(0,6)
(score 0.75)
Correct Answers:
•-432
•(-6,0)
•1296
•(0,-6), (0,6)
Problem 7. (1 point)
Suppose that you are climbing a hill whose shape is given by z=
1062 −0.1x2−0.08y2, and that you are at the point (50,80,300).
In which direction should you proceed initially in order to reach
the top of the hill fastest?
If you climb in that direction, at what angle above the horizontal
will you be climbing initially (radian measure)?
Solution:
SOLUTION
z=f(x,y) = 1062 −0.1x2−0.08y2⇒∇f(x,y) =
h−0.2x,−0.16yiand
∇f(50,80) = h−10,−12.8iis the direction of largest slope at the
point (50,80).
If you climb in this direction, you will ascend at a rate of
|∇f(50,80)|=p(−10)2+ (−12.8)2=√263.84 vertical meters
per horizontal meters.
Thus the angle above the horizontal in which the path begins is
given by arctan(√263.84)≈1.50931 radians.
Answer(s) submitted:
•<-5,-6.4>
•.141748
(score 0.5)
Correct Answers:
•<-10,-12.8>
•1.50931
5
Problem 8. (1 point)
Suppose f(x,y) = xy(1−4x−9y).
f(x,y)has 4 critical points. List them in increasing lexographic
order. By that we mean that (x, y) comes before (z, w) if x<zor
if x=zand y<w. Also, determine whether the critical point a
local maximum, a local minimim, or a saddle point.
First point ( , ).
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Second point ( , ).
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Third point ( , ) .
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Fourth point ( , ).
Classification:
•
• local minimum
• local maximum
• saddle point
• cannot be determined
(local minimum, local maximum, saddle point, cannot be deter-
mined).
Solution:
SOLUTION
fx(x,y) = y(1−8x−9y),fy(x,y) = x(1−4x−18y).
Then fx=0 implies y=0 or y=1
9−8
9x.
If y=0, then substitution into fy=0 gives x(1−4x) = 0⇒x=0
or x=1
4, so we have critical points (0,0)and 1
4,0.
If y=1
9−8
9x, then substitution into fy=0 gives x(12x−1) =
0⇒x=0 or x=1
12 .
If x=0, then y=1
9, and if x=1
12 , then y=1
27 , so 0,1
9and
1
12 ,1
27 . are critical points.
Thus the four critical points are (0,0),0,1
9,1
12 ,1
27 ,1
4,0,
We use the second derivative test to classify the points: fxx(x,y) =
−8y,fyy(x,y) = −18x,fxy(x,y) = 1−8x−18y. Thus D(x,y) =
(−8y)(−18x)−(1−8x−18y)2.
D(0,0) = 0−12=−1<0.
D0,1
9=0−(−1)2=−1<0
Hence (0,0)and 0,1
9are saddle points.
D1
12 ,1
27 =4
9−−1
32=1
3>0
fxx 1
12 ,1
27 =−8
27 <0.
Thus 1
12 ,1
27 is a local maximum.
D1
4,0=0−(−1)2=−1<0
Thus 1
4,0is a saddle point.
Answer(s) submitted:
•0
•0
•saddle point
•0
•(1/9)
•saddle point
•(1/(12))
•(1/(27))
•local maximum
•(1/4)
•0
•saddle point
(correct)
Correct Answers:
•0
•0
•saddle point
•0
•0.111111
•saddle point
•0.0833333
•0.037037
•local maximum
•0.25
•0
•saddle point
6
Problem 9. (1 point)
Consider a function f(x,y)at the point (5,3).
At that point the function has directional derivatives:
4
√45 in the direction (parallel to) h6,3i, and
4
√41 in the direction (parallel to) h5,4i.
The gradient of fat the point (5,3)is
(,).
Solution:
SOLUTION
The unit vector in the direction of h6,3iis u1=D6
√45 ,3
√45 E,
while the unit vector in the direction of h5,4iis u2=D5
√41 ,4
√41 E
.
Let f1=fx(5,3)and f2=fy(5,3), then
f16
√45 +f23
√45 =4
√45 and
f15
√41 +f24
√41 =4
√41
Simplyfying yields the sytem:
6f1+3f2=4
5f1+4f2=4
Solving the system gives f1=4
9,f2=4
9, so the gradient of fat
the point (5,3)is
4
9,4
9
Answer(s) submitted:
•0
•0
(incorrect)
Correct Answers:
•0.444444444444445
•0.444444444444444
Problem 10. (1 point)
Consider the curve x2+6xy +y3=8
The equation of the tangent line to the curve at the point (1,1)has
the form y=mx +bwhere
m=and b=
Solution:
SOLUTION
Let F(x,y) = x2+6xy +y3−8=0. Then
dy
dx =−Fx
Fy
=−2x1+6y
6x+3y2
The slope of the tangent at (1,1)is then
m=−2+6
6+3=−8
9
The equation of the tangent line at (1,1)is y−1=−8
9(x−1)or
y=−8
9x+17
9.
Thus the y-intercept is b=17
9.
Answer(s) submitted:
•-(8/9)
•((17)/9)
(correct)
Correct Answers:
•-0.888888888888889
•1.88888888888889
7
Problem 11. (1 point)
Find the volume of the solid enclosed by the paraboloids z=
16x2+y2and z=32 −16x2+y2.
Solution:
SOLUTION
The two paraboloids intersect when
16x2+y2=32 −16 x2+y2or x2+y2=1. So
V=ZZx2+y2≤132 −16 x2+y2−16 x2+y2 dA
=Z2π
0Z1
0
(32 −32r2)r dr dθ
=Z2π
0
dθZ1
0
(32r−32r3)dr
= [θ]2π
016r2−8r41
0
=2π16(1)2−8(1)4
=2π(8) = 16π
Answer(s) submitted:
•144pi
(incorrect)
Correct Answers:
•50.2655
Problem 12. (1 point)
Find the partial derivatives of the function
f(x,y) = −8x+3y
9x+9y
fx(x,y) =
fy(x,y) =
Answer(s) submitted:
•((((-8x+3y)-8(9x+9y)))/((9x+9y)ˆ(2)))
•((3(9x+9y)-9(-8x+3y))/((9x+9y)ˆ(2)))
(score 0.5)
Correct Answers:
•((9*x + 9*y)*-8 - (-8*x - -3*y)*9)/(9*x + 9*y)**2
•((9*x + 9*y)*(- -3) - (-8*x - -3*y)*9)/(9*x + 9*y)**2
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