Austin Cholley Zhu MAT 267 ONLINE A Spring 2021
Assignment Section 11.1 due 01/31/2021 at 11:59pm MST
1. (1 point)
Evaluate the function at the specified points.
h(x,y,z) = x1
y4z, , ,(−3 3, ,−2) (2 4,2)
At (−3 2 :,3,−)
At (2 4 :, ,2)
Solution:
Solution: Substitute (x,y,z) = (−3, ,3−2)to find
h(− −3 3, , 2) = −3·1
34·−2=6
81 =0.0740741.
For the second point, substitute (x,y,z) = (2, ,4 2)to find
h(2 4, ,2) = 2·1
44·2=4
256 =0.015625.
Answer(s) submitted:
•(2/(27))
•(1/(64))
(correct)
Correct Answers:
•0.0740741
•0.015625
2. (1 point) The domain of the function isf(x,y) = √x+√y
A. The union of two intervals
B. The first quadrant
C. The first and third quadrants
D. All of the xy-plane
E. The area inside a parabola
Solution:
SOLUTION:
√xis defined for x≥0 and √yis defined for 0.y≥
Thus the domain is {(x,y)|x≥0 0,y≥ }, the first quadrant in
the -plane.xy
Answer(s) submitted:
•B
(correct)
Correct Answers:
•B
3. (1 point) The domain of the function f(x,y) = 3x+5y
x2+y2−4
is
A. The union of two intervals
B. The area inside a circle (including the circle)
C. The area inside a circle (not including the circle)
D. The first quadrant
E. All the xy-plane except a circle
Solution:
SOLUTION:
f(x x,y)is defined when 2+y2−46=0. Thus the domain is
(x x,y)|2+y26=4 , that is, all the -plane except the circlexy
x2+y2=4.
Answer(s) submitted:
•E
(correct)
Correct Answers:
•E
4. (1 point)
Match the functions with the graphs of their domains.
1. f(x,y) = ln 2(2y−x)
2. f(x,y) = 2 2y−x
3. f(x,y) = px3y3
4. f(x,y) = e1
2 2y−x
A.
B.
1
C.
D.
Solution:
Solution:
The domain of ln is the region 2(2y−2x)y−2x>0, which
is the half-plane with boundary 2y−2 0.x=
The domain of 2 is the entire plane.y−2x
The domain of is the region 0. The domain
px3y3x3y3≥
will include the lines 0 and 0. The factor is positivex=y=x3
for all positive and negative for all negative . The factorx x y3
is positive for all positive and negative for all negativey 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
2 2y−xis the same as the domain of the expo-
nent. This means that the domain is the same as the domain of
1
2y−2x, which is the set line 2 0. Thus, the domain isy−2x6=
entire plane except for the line 2 0.y−2x=
Answer(s) submitted:
•d
•c
•b
•a
(correct)
Correct Answers:
•D
•C
•B
•A
5. (1 point) Match the functions with the graphs labeled A -
G. As always, you may click on the thumbnail image to produce
a larger image in a new window (sometimes exactly on top of
the old one). Just take your time; process of elimination will
help with ones that are not obvious.
1. f(x,y) = (x2−y2)2
2. f(x,y) = sin sin(x) ( )y e−x2−y2
3. f(x,y) = sin( )y
4. f(x,y) = cos(x2+y2) (/1+x2+y2)
5. f( )x,y) = (x−y2
6. f(x,y) = 1/(1+x2+y2)
7. f(x,y) = 3−x2−y2
A B C D E
Solution:
SOLUTION:
1. The function is 0 along the lines x=yand x=−y. Also,
