Austin Cholley Zhu MAT 267 ONLINE A Spring 2021
Assignment Midterm 2 due 02/12/2021 at 10:02pm MST
Problem 1.
1. (1 point)
Find the partial derivatives of the function
f(x,y) = −9x+6y
1x−5y
fx(x,y) =
fy(x,y) =
Answer(s) submitted:
•((39y)/((x-5y)ˆ(2)))
•-((39x)/((x-5y)ˆ(2)))
(correct)
Correct Answers:
•((1*x + -5*y)*-9 - (-9*x - -6*y)*1)/(1*x + -5*y)**2
•((1*x + -5*y)*(- -6) - (-9*x - -6*y)*-5)/(1*x + -5*y)**2
Problem 2.
2. (1 point)
Use differentials to estimate the amount of material in a
closed cylindrical can that is 60 cm high and 24 cm in diam-
eter if the metal in the top and bottom is 0.2 cm thick, and the
metal in the sides is 0.1 cm thick. Note, you are approximating
the volume of metal which makes up the can (i.e. melt the can
into a blob and measure its volume), not the volume it encloses.
The differential for the volume is
dV =dr+dh
dr =and dh =(be careful)
The approximate volume of material is cm3.
Answer(s) submitted:
•2pi rh
•pi rˆ(2)
•.1
•.4
•633.34507
(correct)
Correct Answers:
•2*3.14159265358979*r*h
•3.14159265358979*r**2
•0.1
•0.4
•633.345078963702
Problem 3.
3. (1 point)
Find the maximum rate of change of f(x,y) = ln(x2+y2)at
the point (-2, -3) and the direction in which it occurs.
Maximum rate of change:
Direction (unit vector) in which it occurs: h,
i
Answer(s) submitted:
•.55470019
•(4/(13))
•(6/(13))
(score 0.333333333333333)
Correct Answers:
•0.554700196225229
•-0.554700196225229
•-0.832050294337844
Problem 4.
4. (1 point) At a certain point on a heated metal plate, the
greatest rate of temperature increase, 5 degrees Celsius per me-
ter, is toward the northeast. If an object at this point moves
directly north, at what rate is the temperature increasing?
degrees Celsius per meter
Solution:
SOLUTION
Let Tbe the temperature function and Pthe point. The value
of the greatest rate of increase is the magnitude of the gradient.
Thus |∇T(P)|=5.
The gradient points in the northeast direction, so ∇T(P) =
h5cos(45o),5sin(45o)i=D5
√2,5
√2E.
A unit vector in the north direction is u=h0,1i, so the rate of
change of increase of the temperature in that direction is
DuT(P) = ∇T(P)·u=5
√2
Answer(s) submitted:
•(5/(sqrt(2)))
(correct)
Correct Answers:
•5/[sqrt(2)]
Problem 5.
5. (1 point)
The function fhas continuous second derivatives, and a critical
point at (-2, -7).
Suppose fxx(−2,−7) = −4,fxy(−2,−7) = −6,fyy(−2,−7) =
−9.
Then the point (-2, -7):
•A. is a local minimum
•B. cannot be determined
•C. is a saddle point
1
•D. is a local maximum
•E. None of the above
Answer(s) submitted:
•A
(incorrect)
Correct Answers:
•B
Problem 6.
6. (1 point) Consider the function f(x,y) = 2x3+y4on the
region {(x,y)|x2+y2≤16}.
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=16 ⇒y=
±√16 −x2,−4≤x≤4. Substituting in the function fyields
g(x) = f(x,±√16 −x2) = 2x3+ (16 −x2)2,−4≤x≤4.
Plotting the function g(x)in the given interval, we can see that
it attains its minimum at x=−4 and its maximum at x=0.
We have f(−4,0) = −128 and f(0,−4) = f(0,4) = 256
Thus the absolute minimum value is −128 attained at (−4,0)
and the absolute maximum value is 256 attained at (0,−4)and
(0,4).
Answer(s) submitted:
•-128
•(-4,0)
•256
•(0,-4),(0,4)
(correct)
Correct Answers:
•-128
•(-4,0)
•256
•(0,-4), (0,4)
Problem 7.
7. (1 point) Consider the curve x6+2xy +y4=4
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) = x6+2xy +y4−4=0. Then
dy
dx =−Fx
Fy
=−6x5+2y
2x+4y3
The slope of the tangent at (1,1)is then
m=−6+2
2+4=−4
3
The equation of the tangent line at (1,1)is y−1=−4
3(x−1)or
y=−4
3x+7
3.
Thus the y-intercept is b=7
3.
Answer(s) submitted:
•-(8/6)
•2.333333333333333333333
(correct)
Correct Answers:
•-1.33333333333333
•2.33333333333333
Problem 8.
8. (1 point) Evaluate the iterated integral I=R1
0R1+y
1−y(21y2+
8x)dxdy
Answer(s) submitted:
•((37)/2)
(correct)
Correct Answers:
•18.5
Problem 9.
9. (1 point)
Suppose Ris 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 =ZB
AZD
C
f(x,y)dydx
A =
B =
C =
D =
2
Solution:
SOLUTION
The region is bounded below by the line through the points
(−4,−1),(4,1). This line has equation y=1
4(x+4)−1.
The upper bound is the line y=2, while −4≤x≤4.
Thus
ZZR
f(x,y)dA =Z4
−4Z2
1
4(x+4)−1
f(x,y)dydx
Answer(s) submitted:
•-4
•4
•-1
•2
(score 0.75)
Correct Answers:
•-4
•4
•0.25*(x+4)-1
•2
Problem 10.
10. (1 point)
Consider the following integral. Sketch its region of
integration in the xy-plane.
Z1
0Zy
√y
180x2y3dx dy
(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)
Solution:
SOLUTION
(a) The given integral is equivalent to −Z1
0Z√y
y
180x2y3dx dy.
Thus the region is bounded on the by left by x=yand on the
the right by x=√y, or, equivalently, y=x2,x>0.
The bounds for the variable yare 0 ≤y≤1. Thus the graph of
the region of integration matches A.
(b)
Z1
0Zy
√y
180x2y3dx dy =180Z1
0
y3x3
3y
√y
dy
=180
3Z1
0
y3hy3−y3/2idy
=60Z1
0y6−y4.5dy
=60y7
7−y5.5
5.51
0
=601
7−1
5.5
=−180
77
Answer(s) submitted:
•A
•((-180)/(77))
(correct)
Correct Answers:
•A
•-2.33766
Problem 11.
11. (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:
•((eˆ(4)-1)/4)
(correct)
Correct Answers:
•13.3995
3
Problem 12.
12. (1 point)
Find the partial derivatives of the function
f(x,y) = xye−1y
fx(x,y) =
fy(x,y) =
fxy(x,y) =
fyx(x,y) =
Answer(s) submitted:
•yeˆ(-1y)
•x(1*eˆ(-1*y)+(-eˆ(-y))y)
•eˆ(-y)y+eˆ(-y)
•eˆ(-y)-eˆ(-y)y
(score 0.75)
Correct Answers:
•y*exp(-1*y)
•x*(-1*y*exp(-1*y) + exp(-1*y))
•-1*y*exp(-1*y) + exp(-1*y)
•-1*y*exp(-1*y) + exp(-1*y)
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