Unformatted text preview: )2 = 2 or
(x + 1)2 (y − 1)2 (z + 2)2
−
+
= 1, a hyper2
2
2
boloid.
015 10.0 points Find an equation for the surface obtained
by rotating the line z = 6y about the z axis. 8 2. x2 + y 2 − 2x − 6z − 8 = 0
3. x2 + z 2 + 6y + 2z − 8 = 0
4. y 2 + z 2 − 6x − 2y − 8 = 0 correct
5. x2 + z 2 − 6y − 2z − 8 = 0
6. x2 + y 2 + 2x + 6z − 8 = 0
Explanation:
The distance from P (x, y, z ) to P (0, 1, 0)
is
x2 + (y − 1)2 + z 2 ,
while the distance from P (x, y, z ) to the
plane x = −3 is x + 3. Thus P lies on
the graph of 1. x2 + 36y 2 = z 2
2. x2 + y 2 = 36z 2  x + 3 =
3. x2 + y 2 = 12
z
6 4. x2 + y 2 = 12
z correct
36 5. 12
x + y2 = z2
36 z
k
r=y=
=.
6
6
Thus the equation is 016 After squaring and expanding both sides this
becomes
x2 + 6 x + 9 = x2 + y 2 − 2 y + 1 + z 2 .
Consequently, an equation for the surface is Explanation:
The surface is a right circular cone with
vertex at (0, 0, 0) and axis the z axis. For
z = k = 0, the trace is a circle with center on
the z axis and radius x2 + y 2 = x2 + (y − 1)2 + z 2 . 12
z.
36 10.0 points Find an equation for the surface consisting
of all points P (x, y, z ) equidistant from the
point P (0, 1, 0) and the plane x = −3.
1. y 2 + z 2 + 6x + 2y − 8 = 0 y 2 + z 2 − 6x − 2y − 8 = 0 .
017 10.0 points Find an equation for the surface consisting
of all points P (x, y, z ) whose distance from
the y axis is twice its distance from the xz plane.
1. 4x2 − y 2 + 4z 2 = 0
2. x2 − 4y 2 − 4z 2 = 0
3. x2 + y 2 − 4z 2 = 0
4. 4x2 − y 2 − z 2 = 0
5. x2 − 4y 2 + z 2 = 0 correct
6. 4x2 + 4y 2 − z 2 = 0 geyer (dag2798) – HW09 – rusin – (55735)
Explanation:
The distance from P (x, y, z ) to the xaxis
is
y2 + z2 ,
while the distance from P (x, y, z ) to the xz plane is y . Thus P lies on the graph of
y2 + z2 . 2 y  = Explanation:
The graph is a circular cylinder whose axis
of symmetry is parallel to the z axis, so it
will be the graph of an equation containing no
z term. This already eliminates the equations
x2 + z 2 + 2 x = 0 ,
y 2 + z 2 − 2z = 0 . Consequently, after squaring both sides and
rearranging we see that an equation for the
surface is On the other hand, the intersection of the
graph with the xy plane, i.e. the z = 0 plane
is
y x2 − 4 y 2 + z 2 = 0 .
018 9 10.0 points
x Which one of the following equations has
graph
z Consequently, the graph is that of the equation
x2 + y 2 + 2 x = 0 . z
xy keywords: quadric surface, graph of equation,
cylinder, 3D graph, circular cylinder, trace x
y 1. x2 + y 2 − 2y = 0
2. x2 + y 2 − 2x = 0
3. x2 + y 2 + 2x = 0 correct
4. x2 + y 2 + 2y = 0
5. y 2 + z 2 − 2z = 0
6. x2 + z 2 + 2x = 0...
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This note was uploaded on 05/12/2013 for the course M 408d taught by Professor Sadler during the Spring '07 term at University of Texas.
 Spring '07
 Sadler
 Determinant, Multivariable Calculus

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