# nebusresearch

Joseph Nebus's work in progress.

## On Peeking At Cedar Point

by Joseph Nebus

I hadn’t intended to leave unanswered my little question about getting the best view of an obstructed attraction at Cedar Point, and apologize for that. Matters got in my way. And I really want to commend people to Geoffrey Brent’s solution, which avoids calculus in favor of geometric reasoning and so has that nice satisfying nature to it. (The part that turns into gibberish is rot13′d, so as not to spoil people: copy it to the box on Rot13.com and hit ‘Cypher’ to read it if you aren’t able to do the rot13 stuff in your head somehow.)

I do want to work out the solution by calculus methods, though, partly because that was actually easier for me, and partly to see whether my audience will put up with such. I’m trying to figure out how to present a more complicated subject which sure looks like it needs calculus to explain, and I’d like to have some sense whether I can write coherently on that topic so.

To set the stage: the problem was about where to stand, behind a tall obscuring fence, so as to see the greatest view of a building hidden behind the fence. To make for simple enough numbers, the viewer is assumed to have eyes six feet off the ground, the fence is eight feet tall, and the building, four feet beyond the fence, is twelve feet tall. Trusting that the ground is level — the reality isn’t quite, as it is at an amusement park — and that you can get as near or as far from the fence as you like, when does the angle between the top of the building and the top of the fence get its biggest?

I’m going to use x to represent how far the viewer is from the fence, and to use θ as the angle between the top of the building and the top of the fence. x is a traditional variable for the unknown quantity, and θ is almost as traditional for the angle. Here, exactly what the viewing angle θ is will depend on what x is, so while we’ll be solving for what x is, it will be through information about how θ depends on x. That sort of problem very often calls out for differential calculus. You can out-think this — Geoffrey Brent did — but thinking is hard work, and the tools are powerful enough that they’ll often work.

When I have a description of how the angle of what’s visible, θ, depends on how far the eye is from the fence, x, then I can find where the angle is biggest (or smallest) by taking the derivative of θ with respect to x. That derivative says how much a tiny change in x would make in θ; at the largest or the smallest angles, a tiny change in x produces no change in θ. If you don’t believe me, think of a marble rolling around in a bowl. If the marble is far away from the base, then, moving the marble a little bit sideways forces it up, or down, pretty far. If the marble is near the middle of the bowl, moving it sideways lets it move up or down not much at all.

We do need to know what the angle θ is. That’ll be the difference between the angle to the top of the building and the angle to the top of the fence. The viewer is, by definition, x feet away from the fence, and the top of the fence is two feet above the eye. Trusting that the fence is vertical, then, there are a couple of ways to describe the angle to the top of the fence, but the probably easiest is that it’s the arc-tangent of two divided by x. The building, meanwhile, is x + 4 feet away, and the top of the building is six feet above the eye. So the angle to that is the arc-tangent of six divided by the quantity x + 4.

This means that from the point x feet away from the fence, the angle of viewable building is:

$\theta = \arctan\left(\frac{6}{x + 4}\right) - \arctan\left(\frac{2}{x}\right)$

Taking the derivative — how much that expression changes as x changes — all at once looks bad, and is the sort of thing that makes me wonder why I don’t have a program which will do it automatically too, but it gets simpler if you break it down into parts. The first thing is that the derivative of the sum of two things is the same as the derivative of the first thing plus the derivative of the second, so we can take the derivative of those arctangent expressions separately. (Yes, that’s a subtraction rather than an addition, but once we reach introductory algebra we start treating subtraction as just being a special case of addition, which is coincidentally about when people really start grumbling about having to take math class.)

The next thing is that the derivative of the arctangent sounds bad, but, it isn’t: the derivative of arctan(x) with respect to x is the almost friendly-looking expression $\frac{1}{1 + x^2}$. That’s easy to deal with. Unfortunately, neither of the expressions we have is the arctangent of x; instead, they’re arctangents of expressions that depend on x.

But here again there’s a rule which makes things simpler; this one’s known as the chain rule. For this, it means the derivative, with respect to x, of $\arctan\left(\frac{2}{x}\right)$ is equal to $\frac{1}{1 + \left(\frac{2}{x}\right)^2}$ — using the formula for the derivative of arctan(x), with the thing in parentheses used in place of x — times the derivative of $\frac{2}{x}$ — which was the thing inside parentheses to start with. That derivative is an easy one, $-\frac{2}{x^2}$. So the derivative, with respect to x, of $\arctan\left(\frac{2}{x}\right)$ is $\frac{1}{1 + \left(\frac{2}{x}\right)^2}\cdot\frac{-2}{x^2}$.

Similarly again using the chain rule, the derivative of $\arctan\left(\frac{6}{x + 4}\right)$ is going to be the not-terribly-appealing-looking $\frac{1}{1 + \left(\frac{6}{x + 4}\right)^2}$ times the derivative of $\frac{6}{x + 4}$ with respect to x, which is $-\frac{6}{\left(x + 4\right)^2}$.

So the derivative of θ with respect to x is:

$\frac{d\theta}{dx} = \frac{1}{1 + \left(\frac{6}{x + 4}\right)^2}\cdot\frac{-6}{\left(x + 4\right)^2} - \frac{1}{1 + \left(\frac{2}{x}\right)^2}\cdot\frac{-2}{x^2}$ which looks horrible. But those denominators work out pretty nicely when you actually try evaluating them, as they turn out to be a lot simpler once you get over the first impression. The derivative is:

$\frac{d\theta}{dx} = \frac{-6}{\left(x + 4\right)^2 + 6^2} + \frac{2}{x^2 + 4}$ and we want to find all the values of x for which this is zero. Again I start to wonder why I don’t have a computer program that will work out symbolic algebra like this — there are many such programs available and they’re wonderful — but it’s actually quicker to do this by hand rather than wait for a package of some open-source symbolic mathematics program to download and to find what hilarious level of deranged semi-functionality is billed as “works for Mac OS”. Relying on pencil and paper, and the general idea that it’s nice to get variables out of denominators as soon as that can be done without being confusing we get:

$\frac{-6}{\left(x + 4\right)^2 + 6^2} + \frac{2}{x^2 + 4} = 0 \\ \frac{-6}{\left(x + 4\right)^2 + 6^2} = - \frac{2}{x^2 + 4} \\ 6\left(x^2 + 4\right) = 2\left(\left(x + 4\right)^2 + 6^2\right) \\ 3\left(x^2 + 4\right) = \left(x + 4\right)^2 + 6^2 \\ 3x^2 + 12 = x^2 + 8x + 16 + 36 \\ 2x^2 - 8x - 40 = 0 \\ x^2 - 4x - 20 = 0$

That last line is one of finding the roots of a quadratic polynomial; we’re good at that. We have the quadratic formula for that, one that gives us two possible solutions:

$x = \frac{-\left(-4\right) \pm \sqrt{(-4)^2 - 4\cdot1\cdot(-20)}}{2\cdot 1} \\ x = \frac{4 \pm \sqrt{96}}{2}$

The square root of 96 has to be pretty close to the square root of 100, which is 10. So, four minus ten is a negative number, putting the viewer somewhere behind the fence, where we can’t get. The only answer that’s useful for us has the positive x, that is, $x = \frac{1}{2}\left(4 + \sqrt{96}\right)$. This is $2 + \sqrt{24}$, or just about 6.9 feet in front of the fence and I am so grateful that it’s exactly what Geoffrey Brent worked out as I like the confirmation that I probably didn’t do it wrong.

Just finding this answer doesn’t properly finish our work off, though. The maximum angle has to be found at some x where the derivative of θ with respect to x is zero, or at a boundary where the possible values of x are limited; however, just because the derivative is zero does not mean we have necessarily found a maximum. We might have found a minimum instead, or we might have found an inflection point, where the rate at which the angle changes with x changes how quickly it grows.

There are a couple of tests that might be done to see whether it is a maximum or minimum or an inflection point that we’ve found. The simplest, and practical for this question, is to just test it out: is the angle at the distance $x = 2 + \sqrt{24}$ smaller than the angle at other distances? At my claimed maximum the angle separating the top of the building from the top of the fence is about 0.2211 (this happens to be radians; it’s about 25.3 degrees). For positive values of x, ones that put us on the public side of the fence, we don’t see anything greater.

That’s not exactly rigorous, though we can get away with it in this case. A more convincing proof that we’re right would be to take the derivative with respect to x of this derivative a second time. If this derivative is a negative number when $x = 2 + \sqrt{24}$ then this point is a local maximum. This is known as the “second derivative test”, for the obvious reason that it is a test. As I work it out, this second derivative is $\frac{d^2\theta}{dx^2} = \frac{12\left(x + 4\right)}{\left(\left(x + 4\right)^2 + 6^2\right)^2} - \frac{4x}{\left(x^2 + 4\right)^2}$ which, at $x = 2 + \sqrt{24}$, is roughly -0.005 — not a huge number, but we don’t care about its size. It’s a negative number, and that’s all we needed. The point we found was a maximum. (And by now you see why I so appreciate Brent’s approach: mine took less insight, but paid for it with more drudge work. This is the usual tradeoff.)

For all that, however, we wouldn’t actually get the best peek over the fence by walking just under seven feet back from the fence. We’d get the best peek over it by going to one of the adjacent rides, such as the Ferris wheel, the Wicked Twister roller coaster, the Troika, or the giant Frisbee ride maXair, all of which carry the rider well above the height of the fence where it was clear that — as of late September — the Transport Refreshments buildings were intact and in good-looking shape. (I haven’t any pictures of this because I am not so foolish as to take photographs during a ride. I might have on the Ferris wheel, I admit, but I didn’t happen to go on it.) Of course, that doesn’t say much about what might be around after the new roller coaster is fully built.

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Posted on Tuesday, 12 March, 2013 at 11:49 pm in Geometry, Recreational Mathematics   |  RSS feed |   Respond   |   Trackback URL

Tags: amusement parks, calculus, Cedar Point, geometry, math, mathematics, maths, roller coasters

### One Comment			to “On Peeking At Cedar Point”

1. “You can out-think this — Geoffrey Brent did”

…I should probably mention that I saw a very similar problem in a book, yea these many years ago, so it only took very very minor tweaking.

But I will take credit for hanging onto that knowledge for a couple of decades!

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