soln.exam1.MATH2321-F15 - MATH2321 Calculus III for Science and Engineering Fall 2015 1 Exam 1 Name(Printed Date Signature Instructions STOP above Print

# soln.exam1.MATH2321-F15 - MATH2321 Calculus III for Science...

This preview shows page 1 - 4 out of 12 pages. MATH2321, Calculus III for Science and Engineering, Fall 2015 2 Problem 1. [20 points] Consider the surface in R 3 which is described by the equation - z 2 + 4 = x 2 + 2 y 2 . (a.) [4 points] Graph the cross-section of the surface in the plane where y = 0. A quick sketch is sufficient, but be sure to label the axes of your graph. Answer : The cross-section with y = 0 consists of those points in the xz -plane which satisfy x 2 + z 2 = 4 . (1) The graph is thus a circle centered at the origin with radius r = 2, as sketched in Figure 1. z x y=0 Figure 1: Cross-section for y = 0 in xz -plane. (b.) [4 points] Graph the cross-section of the surface in the plane where x = 0. A quick sketch is sufficient, but be sure to label the axes of your graph. Answer : The cross-section with x = 0 consists of those points in the yz -plane which similarly satisfy 2 y 2 + z 2 = 4 . (2) In this case, the graph is an ellipse at the origin with minor and major axes along the y - and z -axes, respectively. The y -intercepts of the ellipse are y = ± 2, and the z -intercepts are z = ± 2. See Figure 2 for a sketch. (c.) [8 points] Graph the contours of the surface in the planes where z = 0 , ± 1 , ± 2. Label each contour with the corresponding value of z . Be sure to label the axes of your graph. MATH2321, Calculus III for Science and Engineering, Fall 2015 3 x=0 z y Figure 2: Cross-section for x = 0 in yz -plane. Answer : In the horizontal planes where z = 0, z = ± 1, and z = ± 2, the corresponding contour lines in the xy -plane satisfy z = 0 : x 2 + 2 y 2 = 4 , z = ± 1 : x 2 + 2 y 2 = 3 , z = ± 2 : x 2 + 2 y 2 = 0 . (3) For z = 0 or z = ± 1, the contour line is an ellipse centered at the origin, with major and minor axes in the x - and y -directions, respectively. As the absolute value of z increases from zero, the size of the corresponding ellipse decreases. For z = ± 2, the only solution is ( x, y ) = (0 , 0), so the contour is simply the point at the origin. The contour diagram is graphed in Figure 3.  #### You've reached the end of your free preview.

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