# 12-6 - y-axis It matches with III 26 This is a cone with...

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Solutions to Homework Assignment 7 1. (a) It represents a parabola. (b) It represents a vertical parabolic cylinder. (c) This is also a parabolic cylinder with its vertex along the x -axis. 3. This is independent of x, so it is a cylinder along the x -axis. In each cross-section x = k, the trace is an ellipse, so this is an elliptical cylinder. -4 -2 0 x -4 -4 2 -2 -2 0 0 2 4 z y 2 4 4 4. Since this is independent of y, it is a cylinder along the y -axis. In the xz -plane, it is a parabola, so the graph is a parabolic cylinder. -4 -2 0 y -4 4 2 -2 2 z x 0 0 2 -2 4 4 -4 6. This is a hyperbolic cylinder along the x -axis. -4 -2 0 x 2 -4 -4 -2 -2 y 4 0 z 0 2 2 4 4 11. In each of the planes x = k, y = k, z = k, the trace is an ellipse. This surface is an ellipsoid.

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-4 -4 -2 4 -2 2 0 0 y z 0 2 2 x -2 4 4 -4 17. When y is a constant, the trace is an ellipse. When z is constant, the trace is the parabola y = x 2 +4 k 2 . When x is constant, the trace is also a parabola: y = 4 z 2 + k 2 . This gives an elliptic paraboloid. -4 -2 -4 0 0 1 x -2 2 2 y 3 4 0 4 z 2 4 22. This is an ellipsoid. Its longest axis is along the z -axis, so this matches with IV. 24. This is a hyperboloid of two sheets whose axis is the
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Unformatted text preview: y-axis. It matches with III. 26. This is a cone with elliptical cross sections and the y-axis for its axis; it matches with I. 28. This is a hyperbolic paraboloid. The only one graphed is V. 33. We complete the square: 4 x 2 + ( y-2) 2 + 4( z-3) 2 =-36 + 4 + 36 = 4 . We then get x 2 + ( y-2) 2 2 2 + ( z-3) 2 = 1 . This is an ellipsoid whose center is at (0 , 2 , 3) .-4-2 y 2 4-4-2 x 2 4-4-2 z 2 4 45. This is very similar to the focus-directrix deFnition of a parabola, so I’m thinking it’s probably a paraboloid. Let’s check it out. The square of the distance from the generic point ( x, y, z ) to the point (-1 , , 0) is ( x + 1) 2 + y 2 + z 2 . The square of the distance from ( x, y, z ) to the plane x = 1 is ( x-1) 2 . These two quantities are supposed to be equal. We get x 2 + 2 x + 1 + y 2 + z 2 = x 2-2 x + 1 , so-4 x = y 2 + z 2 . This is indeed a circular paraboloid....
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