Lab 1b for Math 17 Curious Cookie Questions

Spring 2012 Math 235 Lab 2: Clairaut’s Theorem A famous theorem is that the mixed partial derivatives of certain nice functions are the sameThis is Clairaut’s Theorem. Because most functions we work with are nice, it is easy to think that Clairaut’s Theorem applies to every function- In this lab, we will see that it does not. Go through the questions in this lab, using Maple for limits and graphs. Try to incorporate the answers to the questions in a narrative form. The grading criteria are listed on our class website. Goals for this Lab • Be able to graphically determine if a surface is continuous and/or differentiable. • Be able to compute the partial derivatives using Maple (and occasionally by hand) using the definition. • Understand how a given function that fails the hypotheses of Clairaut’s Theorem may NOT have equal mixed partials. • For this lab, you are writing up one particular example in detail. To get you started, think about how you would describe the problem to another student in class. • It is easiest to write it up if you already know what the answers are, so try to answer the outline questions below before you get started. Then think about how you can put the answers into a narrative. • Grading will be based on the following: – (5 pts) Put into narrative form. The structure of the paper follows a logical path to the conclusions. – (5 pts) Spelling, grammar, punctuation and typesetting (like parentheses, equation labeling, etc) – (10 pts) The mathematics are clearly explained, correct and complete. – (5 pts) You have (correctly, with the figure command) used at least one figure, and not too many figures. – (5 pts) A bibliography with at least one item. For the bibliography, there’s a nice web reference from our class web site. You should include a reference to some book (this is an exercise in how to do a short bibliography, so I don’t really care what you use, just how you put it into the paper). 1 Lab Details The overall goal of this lab is, given the particular function below, show that it does not satisfy the conditions of Clairaut’s Theorem. To assist you, there are several items below that will guide your thinking. In particular, you might first think about whether the function is continuous. Then look at if it is differentiable, and twice differentiable. The theorem is included below for your convenience. Theorem (Clairaut). Suppose f is defined on a disk D that contains the point (a, b). If the functions fxy and fyx are both continuous on D, then fxy (a, b) = fyx (a, b). Consider the function ( f (x, y) = xy(x2 −y 2 ) x2 +y 2 (x, y) 6= 0 0 (x, y) = 0 1. As an introduction to the lab, you might do a couple of examples that will satisfy the conditions of the theorem. See any calculus text or you may use an online resource (this might be a good place for the reference item needed for the lab- See comments below). 2. First get a feeling for what f is by plotting it (for the plot, Maple will ignore the origin so you can ignore the possible point of discontinuity at the origin). 3. Is f continuous at the origin? (a) Try using the definition (limits) in Maple. (b) Try using the squeeze theorem. (Hint: You might look to see if −|xy| ≤ f (x, y) ≤ |xy|. The absolute value function in Maple is abs()) 4. Is f differentiable? (a) Compute fx , fy using Maple (You might also use simplify here). (b) Compute fx (0, 0) and fy (0, 0) by using the definition of the derivative (by hand). (c) Show that fx and fy are continuous at the origin by seeing that −2|y| ≤ fx (x, y) ≤ 2|y|, and similarly, −2|x| ≤ fy (x, y) ≤ 2|x|. Can you show these algebraically? Do they help prove continuity at the origin? (How?) 5. Show that fxy (0, 0) 6= fyx (0, 0). (Note: Compute fxy (0, 0) using the definition of the derivative). 6. Discuss why Clairaut’s Theorem does not apply here. 2 Some helpful definitions: • The function z = f (x, y) is continuous at x = a, y = b if lim (x,y)→(a,b) f (x, y) = f (a, b) • Given z = f (x, y), the partial derivative with respect to x at the point x = a, y = b is: f (a + h, b) − f (a, b) h→0 h fx (a, b) = lim Similarly, the partial derivative with respect to y: fy (a, b) = lim h→0 f (a, b + h) − f (a, b) h 3