HW09-solutions

# 12 x y2 z2 36 z k ry 6 6 thus the equation is 016

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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 x-axis 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.

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