HW09-solutions

# This already eliminates the equations to standard

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Unformatted text preview: has radius 1. 1. x2 + z 2 − 4x = 0 2. x2 + y 2 + 4x = 0 3. z 2 + x2 + 2z = 0 correct 4. z 2 + x2 − 2x = 0 keywords: quadric surface, graph of equation, cylinder, 3D graph, circular cylinder, trace 014 5. x2 + y 2 + 2x = 0 2 as a circle in the xz -plane. Consequently, the graph is that of the equation z 2 + x2 + 2 z = 0 . 10.0 points Reduce the equation 2 6. z + x + 4z = 0 x2 − y 2 + z 2 + 2 x + 2 y + 4 z + 2 = 0 Explanation: The graph is a circular cylinder whose axis of symmetry is parallel to the y -axis, so it will be the graph of an equation containing no y -term. This already eliminates the equations to standard form and then classify the surface. (x + 1)2 (y − 1)2 (z + 2)2 + + = 1, 1. 2 2 2 ellipsoid (x + 1)2 (y + 1)2 (z − 2)2 2. − + = 1, 2 2 2 hyperboloid x2 + y 2 + 2 x = 0 , x2 + y 2 + 4 x = 0 . (x + 1)2 (y + 1)2 (z + 2)2 − + = 1, 3. 2 2 2 hyperboloid On the other hand, the intersection of the graph with the xz -plane, i.e. the y = 0 plane, is a circle centered on the z -axis and passing through the origin as shown in (x − 1)2 (y − 1)2 (z + 2)2 + + = 1, 4. 2 2 2 ellipsoid geyer (dag2798) – HW09 – rusin – (55735) (x + 1)2 (y − 1)2 (z + 2)2 5. − + = 1, 2 2 2 hyperboloid correct Explanation: Completing squares in all three variables gives (x + 1)2 − (y − 1)2 + (z + 2...
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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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