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...

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MATH2321, Calculus III for Science and Engineering, Fall 2015 1 Exam 1 Name (Printed) Date Signature Instructions STOP. Print your name, the date, and then sign the exam on the line above. This exam consists of 5 problems, each worth 20 points apiece, for a total of 100 points. Work as many problems as you can within 65 minutes. You may work the problems in any order, so use your time wisely. If you finish early, I encourage you to check your work before you hand in the exam. Work each problem in the space provided, or on the back of the preceding page. Be sure to indicate if your work runs to the back of a page. Circle or box your final answer to each problem, and show all work. I will assign partial credit for incorrect answers based upon the work that you submit. You may use your calculator on this exam. The use of any other electronic device is NOT permitted. This exam is closed book. If you brought your textbook or any notes with you today, keep them out of sight for the duration of the exam. If the wording of any problem is unclear, raise your hand. I will come to your desk and attempt to clarify. Do not turn the page until I give the signal to begin. Good luck!
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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.
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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.
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