20_new - HW08 gilbert (55035) This print-out should have 21...

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HW08 – gilbert – (55035) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Classify the quadric surface z y x 1. one-sheeted hyperboloid 2. elliptic paraboloid 3. two-sheeted hyperboloid 4. hyperbolic paraboloid 5. elliptic cone correct 6. ellipsoid Explanation: The trace on any horizontal plane xy-plane, i.e., any plane z = c, is always an ellipse, while the trace on the coordinate planes x = 0 and y = 0 is always a pair of straight lines. Consequently the quadric surface is an elliptic cone . keywords: quadric surface, ellipsoid, ellip-tic paraboloid, hyperbolic paraboloid, one-sheeted hyperboloid, two-sheeted hyper-boloid, 002 10.0 points Identify the quadric surface (all axes drawn to same scale). 1. z 2 = x 2 + y 2 2. x 2 + y 2 + z 2 = 1 correct 3. z 2 − x 2 − y 2 = 1 4. z = x 2 + y 2 5. 2x 2 + y 2 + 3z 2 = 1 6. x 2 + y 2 − z 2 = 1 Explanation: Because the axes are drawn to the same scale, the trace in any horizontal plane close to the xy-plane is always a circle, not an ellipse, and the same is true for the trace on both the xy-plane and the yz-plane. Consequently, the quadric surface is is the graph of a sphere, and the only equation whose graph has these properties is x 2 + y 2 + z 2 = 1 . keywords: Surfaces, SurfacesExam,
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HW08 – gilbert – (55035) 2 y 003 10.0 points Which one of the following equations has x graph when the circular cylinder has radius 1. 1. x 2 + y 2 − 2x = 0 correct 2. x 2 + y 2 + 2y = 0 3. y 2 + z 2 − 2y = 0 4. y 2 + z 2 − 4y = 0 5. x 2 + y 2 + 4y = 0 6. x 2 + y 2 − 4x = 0 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 y 2 + z 2 − 2y = 0, y 2 + z 2 − 4y = 0 . On the other hand, the intersection of the graph with the xy-plane, i.e. the z = 0 plane, is a circle centered on the x-axis and passing through the origin as shown in But this circle has radius 1 because the cylin-der has radius 1, and so its equation is (x − 1) 2 + y 2 = 1 as a circle in the xy-plane. Consequently, after expansion we see that the cylinder is the graph of the equation x 2 + y 2 − 2x = 0 . keywords: quadric surface, graph of equation, cylinder, 3D graph, circular cylinder, trace 004 10.0 points Which one of the following equations has graph when the circular cylinder has radius 1? 1. y 2 + z 2 + 4y = 0 2. y 2 + z 2 − 4z = 0
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HW08 – gilbert – (55035) 3 3. x 2 + y 2 − 4y = 0 005 10.0 points 4. y 2 + z 2 + 2y = 0 5. x 2 + y 2 − 2y = 0 correct 6. y 2 + z 2 − 2z = 0 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 y 2 + z 2 − 2z = 0, y 2 + z 2 − 4z = 0, y 2 + z 2 + 2y = 0, y 2 + z 2 + 4y = 0 . On the other hand, the intersection of the
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20_new - HW08 gilbert (55035) This print-out should have 21...

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