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Math 221 Quiz 3 Spring 2013 Several Variable Calculus 1. Suppose f .x; y/ D xy. Which of the following pictures is a contour map of this function? Explain how you know. 1.0 1.0 1.0 0.5 0.5 0.5 0.0 0.0 0.0 -0.5 -0.5 -0.5 -1.0 -1.0 -0.5 0.0 (A) 0.5 1.0 -1.0 -1.0 -0.5 0.0 0.5 1.0 -1.0 -1.0 -0.5 (B) 0.0 0.5 1.0 (C) Solution. Setting xy equal to a constant produces a hyperbola with the two coordinate axes as asymptotes (see page A21 in Appendix C to the textbook), so the answer is plot (B). (The special case when xy D 0 produces the coordinate axes themselves.) Alternatively, you can get the answer without using any knowledge of conic sections. Consider a typical level curve, say xy D 1. When x gets close to 0, the value of y has to get large in magnitude (since y D 1=x). In picture (A), the value of y gets close to 0 when x does, and in picture (C), the value of y approaches a finite value when x gets close to 0. Therefore the answer has to be (B). .x C y/2 exist? Explain why or why not. 2. Does lim .x;y/!.0;0/ x 2 C y 2 Solution. (This problem is Exercise 12 on page 739 of the textbook.) This two-dimensional limit does not exist, for there are different limits along different paths approaching the point .0; 0/. Indeed, on the x-axis (where y D 0), the function equals x 2 =x 2 , or 1, so the limit as x ! 0 equals 1. But along the line where y D x, the function equals 0, so the limit at .0; 0/ along this line equals 0. And along the line where y D x, the function equals .2x/2 =.2x 2 /, or 2, so the limit at .0; 0/ along this line equals 2. Since there are different limits at .0; 0/ along different paths, the two-dimensional limit does not exist. January 31, 2013 Page 1 of 2 Dr. Boas Math 221 Quiz 3 Spring 2013 Several Variable Calculus Another way to analyze the problem is to expand the numerator to simplify the function: .x C y/2 x 2 C 2xy C y 2 xy D D1C2 2 : 2 2 2 2 x Cy x Cy x C y2 And now the problem reduces to Example 2 on page 734 of the textbook. 3. Suppose f .x; y; ´/ D p x 2 C y 2 C ´2 . Compute @f .0; 3; 4/. @´ Solution. (This problem is Exercise 14 on page 746 of the textbook.) To compute the ´ partial derivative, treat the variables x and y as constants. Then ˇ ˇ @f 1 2 4 2 2 1=2 .0; 3; 4/ D .x C y C ´ / .0 C 0 C 2´/ ˇˇ D : @´ 2 5 .x;y;´/D.0;3;4/ Alternatively, since x and y are treated as constants, youpcould plug in their values at the beginning of the calculation: f .0; 3; ´/ D 9 C ´2 , so ˇ ˇ ˇ ˇ @f 1 d p 4 2 1=2 2 ˇ 9 C ´ ˇ D .9 C ´ / .0; 3; 4/ D .2´/ ˇˇ D : @´ d´ 2 5 ´D4 ´D4 January 31, 2013 Page 2 of 2 Dr. Boas