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Unformatted text preview: patino (mp25752) – Assignment6 – luecke – (57510) 1 This printout should have 10 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points This question is from section 13.6. Find an equation for the surface obtained by rotating the parabola x = y 2 about the xaxis. 1. z 2 − y 2 = x 2. y 2 + z 2 = x correct 3. y 2 − z 2 = x 4. x 2 − z 2 = y 5. x 2 + y 2 = z Explanation: The surface of paraboloid of revolution (cir cular paraboloid) with vertex at the origin, axis the xaxis and opens to the right. Thus the trace in the xz plane is also a parabola: x = z 2 , and equation for surface is y 2 + z 2 = x . 002 10.0 points Hint: This is from Chapter 13. Use the distance formula to write down the conditions described on a point (x,y,z). This will give you the equation for such points. Find an equation for the surface consisting of all points P ( x, y, z ) equidistant from the point P ( − 3 , , 0) and the plane x = 3. 1. 12 z 2 + x 2 = 12 y 2. y 2 − z 2 + 12 x = 0 3. y 2 − x 2 = 6 z 4. y 2 + z 2 + 12 x = 0 correct 5. y 2 + 12 x 2 = 1 Explanation: The distance from P ( x, y, z ) to the point P ( − 3 , , 0) is radicalBig ( x − − 3) 2 + ( y − (0)) 2 + ( z − 0) 2 , while the distance from P ( x, y, z, ) to the plane x = 3 is  x − 3  . When these distances are equal, therefore,  x − 3  = radicalBig ( x − − 3) 2 + ( y − (0)) 2 + ( z − 0) 2 . Consequently, after squaring both sides and simplifying we see that the set of all equi distant points P ( x, y, z ) is the surface y 2 + z 2 + 12 x = 0 . keywords: plane, locus, equidistant from point and plane 003 10.0 points Find a vector function whose graph is the curve of intersection of the sphere x 2 + y 2 + z 2 = 8 and the plane y = 2 ....
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This note was uploaded on 10/21/2009 for the course M 53215 taught by Professor Lueke during the Spring '09 term at University of Texas at Austin.
 Spring '09
 Lueke

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