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. They can be found here.

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 $t = 1/2$.

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.

- 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?
- 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.

#### Final Comments

I've said it in class, but it merits resaying: related rates is an application of the chain rule, and it comes up all the time. In many ways, it is one of the bread-and-butter questions of the course. A related rates problem will certainly appear on the next midterm, and one will appear on the final.

They're also one step further removed from standard arithmetic, in the sense that they're often given as word problems without a designated path to the solution. Often, you must relate the various values yourself (as if it were a real problem). This can be hard to set up or to visualize. If you have any trouble, let me know, and we'll see what we can do.

It was a pleasure to lecture to you all on Wednesday, but I'm sure you're happy to be back in the capable hands of Tom.

Finally, I will be in the MRC on both Monday and Tuesday of next week. But I won't be in the MRC at all on the following Tuesday.

Good luck, and have a good weekend.

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