Austin Cholley Zhu MAT 267 ONLINE A Fall 2021
Assignment Section 10.6 due 08/29/2021 at 11:59pm MST
Problem 1. (1 point)
Match the surfaces with the appropriate descriptions.
1. z=2x2+3y2
2. z=2x+3y
3. x2+2y2+3z2=1
4. z=4
5. z=y2−2x2
6. x2+y2=5
7. z=x2
A. ellipsoid
B. circular cylinder
C. nonhorizontal plane
D. horizontal plane
E. hyperbolic paraboloid
F. elliptic paraboloid
G. parabolic cylinder
Solution:
SOLUTION
1. The traces in x=kare 2x2+3y2=k.When k>0 we have a
family of ellipses. When k=0 we have just a point at the origin,
and trace is empty for k<0. The traces in y=kare z=2x2+3k2,
a family of parabolas opening in the positive z direction. Simi-
larly, the traces in x=kare z=2k2+3y2, a family of parabolas
opening in the positive z-direction. Thus the surface is an elliptic
paraboloid.
2. The equation is linear in the variables x,yand zand it represents
a non horizontal plane.
3. All the traces are family of ellipses, thus the surface is an ellip-
soid.
4. The equation represents a horizontal plane.
5. The traces in x=kare the parabolas z=y2−2k2; the traces
in z=kare k=y2−2x2, which are hyperbolas; and the traces in
y=kare the parabolas z=k2−2x2. Thus the surface is a hyper-
bolic paraboloid.
6. Since zis missing from the equation, the horizontal traces
x2+y2=5,z=k, are copies of the same circle in the plane z=k.
Thus the surface is a circular cylinder .
7. The equation reprents a parabolic cylinder. Since the ydoes not
appear, the graph is formed by moving the parabola z=x2in the
direction of the y-axis.
Answer(s) submitted:
•f
•c
•a
•d
•e
•b
•g
(correct)
Correct Answers:
•F
•C
•A
•D
•E
•B
•G
1
Problem 2. (1 point)
Match the equation with its graph labeled A-F. You may click on
any image to get a larger view.
A.
B.
C.
D.
2
E.
F.
1. y2=x2+2z2
2. x2+2z2=1
3. 9x2+4y2+z2=1
4. x2+4y2+9z2=1
5. x2−y2+z2=1
6. −x2+y2−z2=1
Answer(s) submitted:
•e
•f
•b
•a
•d
•c
(correct)
Correct Answers:
•E
•F
•B
•A
•D
•C
Problem 3. (1 point)
State the type of the quadratic surface:
x
62+y
62+z
52=1
1. Ellipsoid
2. Hyperboloid of one sheet
3. Hyperboloid of two sheets
4. None of these
Describe the trace obtained by intersecting with the plane z=1:
1. Ellipse
2. Hyperbola
3. Circle
4. Empty set
Solution:
Solution:
This is the equation of an ellipsoid. Substituting z=1 we get:
x2
36 +y2
36 +1
25 =1
x2
36 +y2
36 =24
25
x
6q24
25 !2
+ y
6q24
25 !2
=1
Thus, the trace on the plane z=1 is an ellipse.
Answer(s) submitted:
•1
•1
(correct)
Correct Answers:
•1
•1
3
Problem 4. (1 point)
Match the ellipsoids shown in the figure above with the equations:
1) x2+16y2+16z2=256
2) 16x2+y2+16z2=256
3) 16x2+16y2+z2=256
Solution:
Solution:
1) We rewrite the equation in the form:
x
16 2+y
42+z
42=1
The ellipsoid intersects the x,yand zaxes at the points
(±16,0,0),(0,±4,0)and (0,0,±4), hence Bis the correspond-
ing figure.
2) We rewrite the equation in the form:
x
42+y
16 2+z
42=1
The ellipsoid intersects the x,yand zaxes at the points
(±4,0,0),(0,±16,0)and (0,0,±4), hence Cis the correspond-
ing figure.
3) We rewrite the equation in the form:
x
42+y
42+z
16 2=1
The ellipsoid intersects the x,yand zaxes at the points
(±4,0,0),(0,±4,0)and (0,0,±16), hence Ais the correspond-
ing figure.
Answer(s) submitted:
•b
•c
•a
(correct)
Correct Answers:
•B
•C
•A
Problem 5. (1 point)
State whether the equation
z=x
82+y
32
defines: ?
Solution:
Solution:
This equation is the equation of an elliptic paraboloid.
Answer(s) submitted:
•An elliptic paraboloid
(correct)
Correct Answers:
•AN ELLIPTIC PARABOLOID
Problem 6. (1 point)
State whether the equation
16x2−36y2−36z2=1
defines (enter number of statement):
1. A hyperboloid of two sheets
2. A hyperboloid of one sheet
3. An ellipsoid
4. None of these
Solution:
Solution:
We rewrite the equation in the form
x
1
42
−y
1
62
−z
1
62
=1
This equation is the equation of a hyperboloid of two sheets. The
correct answer is 1.
Answer(s) submitted:
•2
(incorrect)
Correct Answers:
•1
4
Problem 7. (1 point)
Find the equation of the ellipsoid passing through the points
(±8,0,0),(0,±5,0)and (0,0,±5)
=1
Solution:
Solution:
The desired ellipsoid is x
82+y
52+z
52=1
Answer(s) submitted:
•((x/8))ˆ(2)+((y/5))ˆ(2)+((z/5))ˆ(2)
(correct)
Correct Answers:
•(x/8)ˆ2+(y/5)ˆ2+(z/5)ˆ2
Problem 8. (1 point)
State the type of the quadratic surface:
x2+y
32+z2=1
1. Hyperboloid of two sheets
2. Hyperboloid of one sheet
3. Ellipsoid
4. None of these
Describe the trace obtained by intersecting with the plane y=0:
1. Ellipse
2. Hyperbola
3. Circle
4. Empty set
Solution:
Solution:
This equation is the equation of an ellipsoid.
The xz-trace is obtained by substituting y=0 in the equation. This
gives the equation x2+z2=1 which defines a circle in the xz-
plane.
Answer(s) submitted:
•3
•3
(correct)
Correct Answers:
•3
•3
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