Calculus in Three Dimensions (21-259) — Feedback on Homework 4

Homework 4 was due on Tuesday 7th October 2014. Questions 2, 3, 4, 7(b) and 8(a) were graded.

Question 2. This was mostly done very well. Most errors arose either from not using the implicit function theorem (the "shortcut") and messing up the subsequent algebra, or from using the implicit function theorem and omitting the all-important minus sign.

Question 3. I had a few qualms with most answers to this question:

Question 4. This question was easy when you knew the results, but a lot of people were trying to do these things from scratch or just made up things they wanted to be true. Here are some facts you should know about functions of the form $z=f(x,y)$:

Question 7(b). Everyone had the right idea with this question: set the partial derivatives equal to zero and solve, and then use the second derivative test to classify the critical points. But people seemed to have problems solving the pair of equations you get when you set the partial derivatives equal to zero... because people kept diving by zero! If you did find all four critical points but classified them incorrectly, it's probably because you made an arithmetic error in computing the discriminant.

In particular, $f_y = 2xy+2y$. Setting this equal to zero doesn't mean you can cancel the $y$s and get $x=-1$; it means either $y=0$, or $y \ne 0$ and $x=-1$. You need to consider both cases. The solution $y=0$ corresponds with $(0,0)$ and $(-\frac{5}{3}, 0)$, and the solution $x=-1$ corresponds with $(-1,2)$ and $(-1,-2)$. By ignoring one case or the other, you omit two solutions.

Question 8(a). A few people seemed to simply have no idea how to do this kind of problem, in which case I refer you to my notes here. For those of you that did, the most common error was only checking the vertices of the triangle, and not also looking for the critical points of the restrictions of $f$ to the boundary of the triangle. That is, you needed to find the critical points of the functions $g(x) = f(x,0)$, $h(x) = f(1,y)$ and $k(x) = f(x,5-x)$. If you didn't do this, you will have missed the point $(3,2)$, which is a global maximum along with $(1,0)$.

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