HW06-solutions - qazi (kaq87) HW06 berg (55290) 1 This...

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Unformatted text preview: qazi (kaq87) HW06 berg (55290) 1 This print-out should have 19 questions. Multiple-choice 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 xz-planes 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 y-axis, so it will be the graph of an equation containing no y-term. 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 xz-plane, i.e. the y = 0 plane, is a circle centered on the x-axis 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 xz-plane. 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 y-term. 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 xz-plane, 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.

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HW06-solutions - qazi (kaq87) HW06 berg (55290) 1 This...

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