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Unformatted text preview: qazi (kaq87) HW06 berg (55290) 1 This printout should have 19 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 10.0 points For which of the following quadratic rela tions is its graph a hyperbolic paraboloid? 1. z 2 = x 2 4 + y 2 16 2. z = x 2 16 + y 2 25 3. x 2 + y 2 25 + z 2 4 = 1 4. z = x 2 25 y 2 4 correct 5. z 2 16 x 2 y 2 = 1 6. x 2 + y 2 z 2 16 = 1 Explanation: In every case, the graph of the given quadratic relation will be a quadric surface in standard position. Now if a hyperbolic paraboloid is in stan dard position and its trace on the yz and xzplanes are the parabolas z = y 2 b 2 , z = x 2 a 2 , respectively, then it will the graph of a quadratic relation z = x 2 a 2 y 2 b 2 , while its trace on horizontal planes z = c is the hyperbola x 2 a 2 y 2 b 2 = c , for all c negationslash = 0. Consequently, only the graph of the quadratic relation z = x 2 25 y 2 4 is a hyperbolic paraboloid. keywords: 002 10.0 points Which one of the following equations has graph when the circular cylinder has radius 1. 1. x 2 + y 2 + 2 x = 0 2. x 2 + z 2 4 x = 0 3. z 2 + x 2 + 2 z = 0 4. x 2 + y 2 + 4 x = 0 5. z 2 + x 2 + 4 z = 0 6. z 2 + x 2 2 x = 0 correct Explanation: The graph is a circular cylinder whose axis of symmetry is parallel to the yaxis, so it will be the graph of an equation containing no yterm. This already eliminates the equations x 2 + y 2 + 2 x = 0 , x 2 + y 2 + 4 x = 0 . On the other hand, the intersection of the graph with the xzplane, i.e. the y = 0 plane, is a circle centered on the xaxis and passing through the origin as shown in qazi (kaq87) HW06 berg (55290) 2 x z But this circle has radius 1 because the cylin der has radius 1, and so its equation is ( x 1) 2 + z 2 = 1 as a circle in the xzplane. Consequently, after expansion we see that the cylinder is the graph of the equation z 2 + x 2 2 x = 0 . keywords: quadric surface, graph of equation, cylinder, 3D graph, circular cylinder, trace 003 10.0 points Which one of the following equations has graph 1. y x 2 + 4 = 0 2. y + z 2 4 = 0 3. x z 2 + 4 = 0 correct 4. z y 2 + 4 = 0 5. x + y 2 4 = 0 6. z + x 2 4 = 0 Explanation: The graph is a parabolic cylinder that has constant value on any line parallel to the y axis, so it will be the graph of an equation containing no yterm. This already eliminates the equations x + y 2 4 = 0 , y + z 2 4 = 0 , y x 2 + 4 = 0 , z y 2 + 4 = 0 . On the other hand, the intersection of the graph with the xzplane, i.e. the y = 0 plane, is a parabola opening to the right on the x axis as shown in x z Consequently, the graph is that of the equa tion x z 2 + 4 = 0 ....
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This note was uploaded on 03/28/2012 for the course M 408 M taught by Professor Gilbert during the Spring '07 term at University of Texas at Austin.
 Spring '07
 Gilbert
 Multivariable Calculus

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