Saturday, April 02, 2016

using calculus to understand the world

In my last post, I wrote about how I returned to teaching related rates in my calculus class and ranted a bit about the inanity of most related rates problems. There I mainly discussed the difficulty in reading the statement of such problems and how to make the questions they raise seem more natural. I’d like to expand on this theme with some more examples.

One feature of mathematics that doesn’t get emphasized enough, IMHO, is that it is a science, and as such is based in observation. Often, either we lead students through abstract reasoning to a previously unanticipated result, or we prove things that are so self-evident that the notion they need proof is itself baffling. Now, in the world of professional mathematics, it is true that even apparently obvious facts need proving (remember that “to prove” just means “to test”), and we often do get excited when we are led to something unexpected and beautiful. That is because we have learned how to use and trust our logical skills to examine the truth of something, and we delight in the uncovering of new truth by means of those skills. But even when a result is surprising to an audience, and even if it was at first surprising to the speaker, it is no longer so. A mathematical researcher plays with ideas until she notices something interesting, and then she tries to understand why it is so. That’s the exciting part of math, and that is what I believe we can share with our students through the process of modeling.

My goal in teaching related rates has become to ground as many questions as possible in direct observation. When I ask about the sliding ladder, as I described in my last post, before setting up the math but after asking students what they think will happen, I demonstrate by leaning a ruler against a book and slowly pulling the bottom end away. What one notices in this experiment is that the top end of the ruler moves very slowly at first, and very quickly just before reaching the ground. The speed in the final moment is so great that one is tempted to think the person pulling the bottom end lost control, and gravity took over. (This is even more credible when using the much larger, heavier ladder in a demonstration.) But the math shows that even if the person keeps complete control and moves at precisely the same speed, the same effect will occur. Let’s see why.

The exact length of the ladder doesn’t matter, of course, so call it $L$. If $x$ measures the distance from the wall to the bottom end of the ladder and $y$ measures the distance from the floor to the top of the ladder, then we have $x^2 + y^2 = L^2$. Then we differentiate both sides with respect to time and get $2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$, or \[ \frac{dy}{dt} = -\frac{x}{y} \frac{dx}{dt}. \] At this point most related rates problems would ask you about the size of $dy/dt$ for some particular values of $x$, $y$, and $dx/dt$, but look at how much we can determine just from this related rates equation: when $y$ is larger than $x$, the top end is moving more slowly than the bottom end, and conversely when $x$ is greater than $y$, the top end is moving more quickly than the bottom end. There is just one moment when the two ends are moving at the same speed, which is when $y = x$, or in other words, when the ladder is at a 45 degree angle. And as the distance between the top end and the floor approaches zero, the speed of the top end approaches infinity. (Not physically possible, of course, but it explains why there’s such a quick movement at the end of the process.) There; now I feel like I’ve learned something!

I have five more examples to illustrate how much more interesting I think related rates are when tied to direct observation. This will probably belabor the point, but unfortunately these examples are also stripped of any interest by focusing too much on a single moment in time, which is what every standard textbook does with them.

The next example involves inflating a balloon. This, again, is easy to demonstrate. I can’t take a deep enough breath to fill the whole balloon at once, but even with two puffs, exhaled at a near-constant rate, it’s obvious to students that the size (i.e., diameter, or radius) of the balloon grows more quickly at first, then more slowly. Anyone who’s worked with an air or helium tank has surely experienced this phenomenon. Why is this happening? And how much more slowly is the diameter increasing as time goes on? Here there’s very little modeling involved; essentially the entire model is provided by assuming the balloon is a sphere and using the formula for the volume of a sphere in terms of its radius, $V = \frac{4}{3}\pi r^3$. Differentiating with respect to time gives the relation $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$, so \[ \frac{dr}{dt} = \frac{1}{4\pi r^2}\frac{dV}{dt}. \] Some books reach this equation, but as with the ladder they again jump to plugging in values for a specific time, rather than noting the following: if $dV/dt$ is constant, then $dr/dt$ is inversely proportional to the square of the radius! And even more, $4\pi r^2$ is the surface area of a sphere with radius $r$, so this relationship between rates is directly related to the fact that the derivative of the volume of a sphere with respect to its radius is the surface area! That is, the size of the surface of the balloon is what determines, together with the rate the volume is increasing, how quickly the radius is increasing.

A similar phenomenon happens with the standard filling-an-inverted-cone problem. The demonstration here involves a martini glass and some colored water. (As I promise my students when doing this experiment, it’s just water.) My martini glass is about 12 cm across on top, and about 8 cm deep, giving it a volume of 300 milliliters. (That’s about 10 ounces, the size of two regular martinis—you don’t want to fill this glass with gin and drink it too quickly.) Having a nice big glass is useful for this demonstration: if I pour at a constant rate, the water level rises much more slowly near the top than near the bottom. The math shows just how much more slowly. The volume of a cone with height $h$ and base radius $r$ is $V = \frac{1}{3}\pi r^2 h$. From the geometry of this situation (using similar triangles, for instance), for my glass the radius of the surface of the water is always three-quarters of the water’s depth (here interpreted as height). We could use the relation $r = \frac{3}{4}h$ and substitute into the volume formula to get rid of the variable $r$, but there’s also no harm (as my students taught me) in differentiating first, using the product rule: \[ \frac{dV}{dt} = \frac{\pi}{3} \left( 2rh\frac{dr}{dt} + r^2\frac{dh}{dt}\right). \] Notice that this formula is valid for all cones varying in height, radius, and volume, whether or not the height and radius are linearly related at all times. The most obvious quantity of interest (assuming constant $dV/dt$) is $dh/dt$. From $r = \frac{3}{4}h$ we get $\frac{dr}{dt} = \frac{3}{4}\frac{dh}{dt}$, and also $h = \frac{4}{3}r$. The reason to solve for both of these quantities is that, by keeping both $dh/dt$ and $r$ in the equation and substituting out $h$ and $dr/dt$, we get $\frac{dV}{dt} = \frac{\pi}{3}\left(2r^2\frac{dh}{dt}+r^2\frac{dh}{dt}\right)$, or, after solving for $dh/dt$, \[ \frac{dh}{dt} = \frac{1}{\pi r^2} \frac{dV}{dt}. \] First of all, the ratio between the height and the radius has disappeared, so this formula now works for any inverted cone, not just my martini glass. And second of all, just as with the balloon, the rate at which the height increases depends on the “surface area” that is expanding, which in this case is just the base of the cone! Thus, again, the reason the water level rises more slowly near the top of the glass has a clear geometric interpretation. (Here’s a real-world application: I argue this works to the benefit of bartenders, who can pour into a martini glass fairly quickly without risk of overflowing, because the beverage level rises slowly near the top of the glass.)

I took the next two examples from Cornell’s Good Questions Project; come to think of it, it may be these questions that first planted in my head the idea of looking at related rates problems over time, without numbers. The situations are again standard for related rates problems, but the conclusions are much more interesting than a single rate at a single moment.

Consider an actor (say, Benedict) on a stage, illuminated by a light at the foot of the stage. Benedict casts a shadow on the back wall; how does the length of his shadow vary if he walks towards the light at a constant speed? The demonstration of this situation is particularly exciting, because you get to turn off the classroom lights, pull out a flashlight and a doll or figurine, and watch what happens to the shadow of the doll/figurine/actor on the wall as it moves towards the flashlight. Students observe that at first the shadow grows slowly (when the figure is close to the wall), then more quickly as he approaches the light. Modeling this situation generally provides the first major geometric hurdle for my students, because it involves the imagined line that emanates from the light, passes by Benedict’s head, and finally reaches the back wall, thereby determining the height of the shadow. (I wonder if many of them have never thought about the geometry of how shadows relate to the objects that cast them.) I’ll let the reader work out the fact that, if Benedict’s height is $h$, the distance from the light to the back wall is $D$, the distance from Benedict to the light is $x$, and the height of the shadow is $s$, then $\frac{s}{D} = \frac{h}{x}$. (Hint: use similar triangles.) Here the only variables are $x$ and $s$, so the related rates equation is \[ \frac{ds}{dt} = -\frac{hD}{x^2}\frac{dx}{dt}. \] Students are at first perplexed by the negative sign: shouldn’t the shadow be increasing? If so, why does its derivative appear to be negative? Then they realize: ah, if Benedict is walking towards the light, then $dx/dt$ is negative, so $ds/dt$ is in fact positive! And so it becomes clear that the height of the shadow increases much more rapidly when Benedict is near the light than when he is near the wall. (I generally give specific values for the height of the actor and the distance from the wall to the light in this question, so that it’s more obvious which values are constant.)

I don’t have a standard demonstration for this next problem, because I use it as a quiz question (although maybe not anymore, now that I’ve written about it here), but it’s easy enough to devise an experiment. This situation is similar enough to the previous one that its result is a bit surprising. Suppose a streetlight at height $L$ is the only source of illumination nearby, and a woman (say, Agatha) of height $h$ walks at a constant speed away from the light. As she gets farther away from the light, does her shadow grow more quickly, more slowly, or does it grow at a constant rate? If $x$ again denotes the distance to the light (well, really from Agatha’s feet to the base of the lamp, which is not the same as her distance to the source of illumination), and $s$ is the length of Agatha’s shadow, then similar triangles produce the relation $\frac{s}{h} = \frac{s + x}{L}$. We can rearrange this into a simple proportion between $s$ and $x$: $s = \frac{h}{L-h} x$. (Here’s an interesting feature of this equation already: it only makes sense if $h < L$, that is, if Agatha is shorter than the lamppost!) Now we differentiate to get \[ \frac{ds}{dt} = \frac{h}{L - h} \frac{dx}{dt}. \] So if Agatha’s speed is constant, then her shadow’s length is also increasing at a constant rate. This example shows especially well why it’s dumb to look at related rates at a single moment in time. Most book exercises of this sort ask how quickly the shadow is growing when Agatha is at a particular distance from the lamp. But it doesn’t matter how far away she is, and the math proves that it doesn’t matter.

There’s a risk in related rates exercises to always resort to problems that only involve differentiating polynomials, so here’s an example that uses trigonometric functions. The demonstration I use: I walk back in forth in front of the class and tell the students to be mindful of what their heads do as they follow my movement. After a couple of times, several of them observe that their heads must turn more quickly when I’m closer to them. I point out that this is something anyone who’s had to run a video camera at a race must be aware of. (It’s also apparent to someone riding in the passenger seat of a car, keeping their gaze fixed on a single tree or other immobile object: for a long time, your head turns little, but when you’re close to the object, you have to turn quickly to keep it in view.) I generally set up the problem on the board as though it is taking place at a racetrack. Suppose a runner is moving along a track (let’s assume it’s straight for simplicity) at $v$ feet per second. You’re watching from a position $D$ feet away from the track. How quickly does your head need to turn to keep following the runner? The answer depends on how far away the runner is. One has to introduce a reasonable coordinate system and some useful variables: good choices are the position $x$ of the runner relative to the point of the track closest to you, and the angle $\theta$ by which your head is turned from looking at this closest point. Then we get the relation $\tan\theta = \frac{x}{D}$, and differentiating with respect to time results in the equation $\sec^2\theta \frac{d\theta}{dt} = \frac{1}{D} \frac{dx}{dt}$, or \[ \frac{d\theta}{dt} = \frac{v}{D} \cos^2\theta \] (using the assumption that $dx/dt = v$). When $\theta = 0$, so that the runner is closest to you, the rate at which your head turns is $v/D$, which depends only on how fast the runner is going and how far away from the track you are. (Notice that the units work out: the radian measure of an angle is technically dimensionless, and so we expect its rate of change not to have any dimension other than 1/time. Since $v$ has dimension of distance/time and $D$ has the dimension of distance, $v/D$ has the dimension 1/time.) As $\theta$ increases (in this scenario, $\theta$ is never greater than a right angle), the change in the angle of your head to follow the runner happens more slowly, because $\cos^2\theta$ is closer to zero.

These are just a few examples of standard situations involving related rates that become much more interesting when the myopic attention to a single moment in time is removed. I’m sure most readers of this post can do the calculations I’ve shown on their own, but the tendency to hone in on a single rate at a single point in time is so entrenched that I wanted to show how much more interesting related rates become when that element is removed. I don’t know that my students are better at solving related rates problems than other students, but I have noticed that they’re much less likely to insert specific quantities into a relation before it’s necessary than when I taught the subject years ago. I haven’t had time to strip all such problems of the detritus that comes with wanting a numeric answer, but I believe our understanding (and our calculus students’ understanding) of the world will be much improved by making the effort to transform these problems into meaningful questions.

Here are two other examples that I won’t work out in detail. One scenario has a boat being pulled into a dock by a rope attached to a pulley elevated some distance above the boat. If the rope is pulled at a constant rate, the boat in fact speeds up as it approaches the dock! (I tried demonstrating this once with a string tied to a stuffed animal pulled across a desk, with moderate success.) Another common type of problem considers two boats moving in perpendicular directions (or cars moving along perpendicular roads), and asks at a certain point in time whether the distance between them is increasing or decreasing. That’s silly. Why not establish the relation between them, and ask at what times the distance is increasing, and at what times the distance is decreasing? If there’s a time when the rate of change in distance is zero, then the boats (or cars) are at their closest (or farthest) positions, which connects to the study of optimization, which has its own set of issues…

P.S. I should have known better than to look at Khan Academy’s treatment of related rates. His videos show all the marks of what is classically wrong with these problems: the irrelevant information of what variables equal at a single moment in time is presented up front along with everything that’s constant in the situation, and in the end the answer is a single, uninformative number. Even when an interesting equation is present on the screen, Khan rushes past it to get to the final number. How can we get our students to ask and answer more interesting questions than these, about the same situations?

1 comment:

John Golden said...

Just super. Calculus as a way of thought is my big content objective, and this is a bullseye. On Twitter, Lana connected this to Brian Bushart's numberless story problems, which is an excellent connection.