Because the last two lectures were a review day and an exam, I am going to once again break from the mold a bit, and discuss both events in one entry.

### The Review

Over the weekend, I produced a rough draft of a study guide, cribbed from that to create a rough exam, then pared the guide down a little and changed some wording to emphasize the material that (1) I felt was most important and (2) I was actually going to test them on (with some important red herrings thrown in to make sure that they studied those, as well). I handed out the study guide on Monday, and told them to prepare to go over it on Thursday.

When Thursday rolled around, I had two quick corollaries to go over, and a quiz prepared. We were then able to spend the rest of class answering questions. Unfortunately, I did an embarrassingly poor job with the very first question, which I think spoiled the mood and morale for the day.

This question was something like this: find the critical numbers of the function \(f(x) = |3x-5|\). Intuitively, this is the absolute value of a linear function, thus there should be one critical number, which will occur when the function is 0 (i.e. the vertex of the absolute value function). In fact, if we know anything about transformations of graphs, it isn’t difficult to see that the critical number occurs at \(x=\frac{5}{3}\).

However, I wanted to work out the derivative, and demonstrate that it didn’t exist at this point, hence it was a critical number. This was a catastrophe. It started out okay: I tried to do too many steps at once. By definition, we have \[

f(x) = \begin{cases}

3x-5 & \text{if $3x-5\ge 0$,} \\

-3x+5 & \text{if $3x-5< 0$.} \\
\end{cases}
\] Except that I was already trying to figure out the derivative, so I wrote\[
\eqalign{
\frac{d}{dx}f(x)
&= \begin{cases}
(3x-5)\frac{d}{dx}(3x+5) & \text{if $3x-5\ge 0$,} \\
(-3x+5)\frac{d}{dx}(3x+5) & \text{if $3x-5< 0$.} \\
\end{cases} \\
&= \begin{cases}
9x-15 & \text{if $3x-5\ge 0$,} \\
-9x+15 & \text{if $3x-5< 0$.} \\
\end{cases} \\
}
\] Setting this equal to zero and solving, we get a critical point at \(x = \frac{15}{9} = \frac{5}{3}\), which is the correct answer, but for the wrong reasons. This *looked* wrong to me on the board, and felt wrong to the students (go them!), but for the life of me, I couldn’t find the error, and ended up confusing the hell out of everyone (myself included) trying to track it down.

Collectively, we eventually managed to go back to the beginning of the problem and work it out, but we ended up wasting a lot of time in (what I think was) an unproductive activity. By the end of it, everyone seemed somewhat dazed and demoralized, ready to go home.

I pushed on until the end of class, and we did get some productive work done, but it was not a happy day. Of course, this comes back to my constant refrain: slow down, slow down, slow down. Do one step at a time.

### The Exam

Again, it seems, the exam was too long. Half of the students took more than the 95 minutes allotted for the class. This despite the fact that the exam was far shorter (no multiple choice questions, two fewer true/false questions (six, compared to eight), and what I thought was a set of relatively straight-forward free response computations).

I will need to discuss the exam with the students this evening in lecture to confirm this, but I suspect that the biggest problems were the related rates and implicit differentiation problems in the free response section.

The related rates problem involved two boats passing a buoy at a different times. The goal was to determine how fast they were converging/diverging at a specific time. We have done several such problems, but this example was slightly different, in that the boats did not pass the buoy at the same time. This difference made the set-up a little more difficult, and introduced a really subtle sign error. The set-up killed half the class, and the sign error wiped out almost all of the rest—out of 27 students who took the exam, there were only two correct answers to this question.

In principle, this isn’t a bad ratio—those two students are the A students, those that managed to solve it with the sign error are B students, those that got the set-up more or less correct are the C students, and those that made no progress fall into the D/F range. However, I had not intended this question to be a discriminating question—I just wanted to know if they could solve a related rates problem. For that purpose, it probably was not the best possible question

The other difficult question involved an implicit differentiation, i.e. find the slope of the line tangent to the graph of \(x^{2/3} + y^{2/3} = 1\) at the point \(\left(\frac{3}{8}\sqrt{3},\frac{1}{8}\right)\). Most of the class managed the calculus, but spun their wheels on the algebraic simplification. I am sure it was quite frustrating for them, and I was disappointed by the results. The most annoying aspect of it is that the could have pulled out their calculators right at the beginning, forgone most of the algebra, and just given a decimal answer. In fact, that is what I *told* them to do.

Overall, the distribution of scores was quite strange. There were four scores above 85%. These scores represented my A range. There were three scores between 80% and 85%, then a trickle of scores down to 70% without any obvious gaps or clusters. In general, I would have liked to award some of those in the 70-80% range with Bs, but there was no fair way to draw a line, so I went by the syllabus and gave a B to the 80% and Cs those that scored between 65% and 70% (actually, the lowest C ended up being at 63.3%). Ds and Fs were also awarded exactly as the syllabus describes.

Thus there were a few students in the B and D ranges who were helped by the curve, but almost everyone was completely unaffected by it, and the overall distribution seemed heavily weighted to the bottom (lots of low Cs, Ds, and Fs). I’m not entirely sure what to make of it.