I haven’t quite yet finished writing up the solutions to the problems we did in class yesterday. But I wanted to go ahead an make the solutions to the test available. If you ask me for them, I can send you a link to them.

But please note that there is an error in the key! In particular, on problem 7(b), I forgot that we only care about $t \geq 0$. So the final answer should not include $t = 1/2$.
The notes for the day are after the fold:We considered three basic questions today. Two were related rates problems, and one was a preview of thinking of the extrema of graph, the zeroes of derivatives, and the extreme-value theorem. Unless there are any questions, I’ll just go over the two related rates problems.

1. The surface area of a cube is increasing at $72$ cm^2/s. At what rate is the side length increasing when the surface area is $96$ cm^2? At what rate is the volume increasing at that time?
2. Charlie is testing a new candy: the ever-stretching Laffy-Taffy. So he steps into his glass elevator with one end of the Laffy taffy in his hand. An oompa-loompa stands outside the elevator holding the other end, 4 meters away from Charlie. Charlie hits the button, and rises at 100 m/s. At what rate does the angle of inclination (from the oompa-loompa’s perspective) change when the elevator is 4 m high? At what rate is the Laffy-Taffy stretching at that time?

#### Questions 1:

Although I don’t do it here – I still recommend that the first thing you do is draw a picture. Here, we have a cube. We know how quickly the surface area is changing. We want to know how quickly the side length is changing. How do we relate surface area to side length?

Well – each side of a cube is a square with side length $s(t)$, where $s(t)$ is the side length of the cube at time $t$. Since a cube has six faces, this means that at time $t$, a cube has surface area $A(t) = 6s(t)^2$. This formula relates side length to surface area, so it’s the exact formula that we are seeking. Differentiating both sides with respect to $t$, we get that $A'(t) = 12s(t)s'(t)$.

At our particular moment in time, we know that $A'(t) = 72$. We want to know $s'(t)$ But we also need to know $s(t)$. Do we know this? Well, we know that the surface area at this tie is $96$ cm^2. This means that $6s^2 = 96$, so $s = 4$ in our situation. Thus from $A'(t) = 12s(t)s'(t)$, we get $72 = 12 \cdot 4 s'$, or that $s' = 3/2$ cm/s.

Now we want to know how quickly the volume is changing. Well – the volume of a cube satisfies $V(t) = s(t)^3$, so that $V'(t) = 3s(t)^2s'(t)$. Since we just calculated $s(t)$ and $s'(t)$, we know that $V' = 3 \cdot 4 \cdot 3/2 = 18$ cm^3/s.

#### An aside

As an aside – there is a common mathematical fallacy that comes up related to this concept. We know that volume is 3-dimensional and length is 1-dimensional, but we are so accustomed to cubing length to get volume that we often expect that we should cube the rate of length-change to get the rate of volume-change. But this isn’t how it works! The ability to understand how to estimate the rate of change of something from a known or at least an estimate of another thing’s rate of change is a fundamental task that we do all the time. Understanding that this isn’t a naive process might lead to a greater grasp of numeracy (which I always emphasize is important).

#### Question 2

If someone asks me to, I will upload a picture describing this question. But at the moment, I don’t. Let’s let $h(t)$ describe the height of the elevator at time $t$, so that we know $h'(t)$ and we want to know things when $h(t) = 4$ . If $\theta$ denotes the angle of inclination, then we want to know $\theta'(t)$.

Since the oompa-loompa stands $4$ meters away from the base of the elevator at the start, we can set up a triangle with base $4$ meters, height $h(t)$, and with hypotenuse equal to the length of the Laffy-Taffy. Then $\tan \theta = \dfrac{h(t)}{4}$. Differentiating both sides with respect to $t$, we get that $\sec^2 (\theta(t)) \theta'(t) = h'(t)/4$. So at our time in question, we know that $h'(t) = 100$. What is $\theta(t)$? Well, since we’re interested in what happens when $h(t) = 4$, our $\theta$ is $\pi/4$, since in the triangle the height is equal to the width. And since $\cos \pi/4 = 1/(\sqrt 2)$, we know that $\sec^2 \pi/4 = 2$. So we have that $2 \theta ' = 100/4$, or rather $\theta' = 100/8$ radians per second. That’s pretty speedy.

We also want to know how quickly the Laffy-Taffy is stretching. Let $l(t)$ denote the length of the Laffy-Taffy. Then from our triangle, we know that $l(t) = \sqrt{ 16 + h(t)^2}$. Differentiating, we see that $l'(t) = \frac{1}{2} (16 + h(t))^{-1/2} 2h(t) h'(t)$. Plugging everything in that we know, we get that $l'(t) = \frac{1}{2} (16 + 16)^{-1/2} \cdot 2 \cdot 4 \cdot 100$. And simplifying yields the answer.