### 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:

• Lots of people didn't evaluate $\cos \frac{\pi}{6}$ as $\frac{\sqrt{3}}{2}$, and instead rounded their answer. If you round an answer you lose information and therefore accuracy... try to avoid rounding wherever possible. It's especially amusing when people write the exact answer and then the rounded answer, and draw a box around the rounded answer... that's like saying "I've found the answer, now I'm going to make it slightly inaccurate and submit that instead"!
• Most people failed to really answer the question. It asks you at what rate the angle changes; writing the value of $\frac{d\theta}{dt}$ is all well and good, but in the end it's just a number. The interpretation of this number is what's important: in particular, you should include units, and even better, you should interpret the sign of the number you get. The fact that $\frac{d\theta}{dt} = -\frac{1}{12\sqrt{3}}$ means that "the angle is decreasing at $\frac{1}{12\sqrt{3}}$ radians per second" — this is the answer to the question, not a simple number $-\frac{1}{12\sqrt{3}}$.

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 $z=f(x,y)$:

• If $\mathbf{u}$ is a general vector, the rate of change of $f$ in the direction of $\mathbf{u}$ is $\nabla f \cdot \widehat{\mathbf{u}}$, where $\widehat{\mathbf{u}}$ is the unit vector in the direction of $f$. So for part (a), since $\langle 1,1 \rangle$ is not a unit vector, you need to divide $\nabla f \cdot \mathbf{u}$ by the magnitude of $\langle 1,1 \rangle$ (namely $\sqrt{2}$).
• The rate of most rapid increase of $f$ at $(a,b)$ is in the direction of $\nabla f(a,b)$; and the rate of most rapid decrease is in the direction of $-\nabla f(a,b)$. So the unit vector in the direction of most rapid decrease for this question was $-\frac{\nabla f(4,3)}{|\nabla f(4,3)|}$.
• The value of the function doesn't increase in the directions perpendicular to $\nabla f$. Here's a quick way of finding perpendicular vectors in $\mathbb{R}^2$: given a vector $\langle a,b \rangle$, a vector perprendicular to it is $\langle -b, a \rangle$. (To see why, dot them.) Now $\langle a,b \rangle$ and $\langle -b, a \rangle$ have the same magnitude, so if one is a unit vector then so is the other. So to go from part (b) to part (c), all you had to do is apply this to $-\frac{\nabla f(4,3)}{|\nabla f(4,3)|}$.

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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