Graphs for which a calculus student can reasonably compute the arclength

Solution 1:

Ferdinands, in his short note "Finding Curves with Computable Arc Length", also comments on the difficulty of coming up with suitable examples of curves with easily-computable arclengths. In particular, he gives a simple recipe for coming up with examples: let

$$f(x)=\frac12\int \left(g(x)-\frac1{g(x)}\right)\,\mathrm dx$$

for some suitably differentiable $g(x)$ over the desired integration interval for the arclength. The arclength over $[a,b]$ is then given by

$$\frac12\int_a^b\left(g(x)+\frac1{g(x)}\right)\,\mathrm dx$$

$g(x)=x^{10}$ and $g(x)=\tan x$ are some of the example functions given in the article that are amenable to this recipe.

Solution 2:

Another example: you can get $$ \sqrt{1 + [f'(x)]^2} = ax + \frac 1{2ax} $$ by taking $f(x) = \frac 12 a x^2 - \frac 1{4a} \ln(x)$ for any constant $a$.

A possibly helpful way of reframing the question: we would like to know for which "nicely integrable" functions $g(x)$ is there a "reasonable" $f(x)$ satisfying $\sqrt{1 + [f'(x)]^2} = g(x)$. In other words, for which nicely integrable $g(x)$ does the function $\sqrt{[g(x)]^2 - 1}$ have a closed-form integral?

Solution 3:

I will dissent here (as often) and say: DON'T.

The problem here is looking at everything as something that needs to be computed, "solved", or otherwise manipulated into some set, pat form.

As it is well-known, few of these integrals are amenable to exact representation in terms of anything encountered at this point (if anything at all). Any exercise you can give effectively amounts to little more than an exercise in symbolic integration, and it wouldn't be particularly meaningful. If you want to exercise symbolic integration, then you should have done that already for its own sake.

What would be much better to do is to give exercises to set up the arc length integral in a variety of scenarios where it may be required - NOT to solve it. To recognize what is being asked for is an arc length, and then show understanding of the integral definition by writing that specific case down. Many people get a lot of notions like that "this integral doesn't exist" because you can't write down a formula, or that somehow, if you don't have "a formula", you don't or can't really "understand" the problem. And the fact is: most real-life integrals just don't have a simple formula or - perhaps a better way to look at it is, the integral is the formula.

People need to be disabused of the notion that there is one "true" or "correct" representation for a mathematical object, whether it's a number, a function, a space of some sort, or anything else, and instead understand and get comfortable with the merits of working with different objects. And it doesn't stop here - if anything, this is already too late, because too many think things like "$\pi$ is infinite", which is not the case: a particular representation is infinite (but not all need be - I just gave you one! $\pi$.), and that representation is actually a pretty useless one insofar as an exact representation is concerned because it has no discernible pattern, while on the other hand, other infinite representations, like

$$\pi = 4\left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)$$

are far more transparent. (And the $4$ even has a meaning: the right-hand bit is the area of a quarter of a unit circle. 4 of those make the whole thing, which has area $\pi$.) $\pi$ itself, though, is a finite number: just a little more than 3.

So give them realistic, interesting cases. Tell them that they don't need to solve it, but to understand the formula. You can also give a numerical check as to an approximate value for the arc length, so one can use a computer to verify the correctness. For example, we might suggest something like this - a very natural, real-life problem:

Over the course of its annual journey, the Earth travels around the Sun in an orbit that is, to a close approximation, an ellipse, with an eccentricity of $e_E = 0.016\ 7086$, and a semi-major axis of $a_E = 149.598\ \mathrm{Gm}$. Let this ellipse lie in the $xy$-plane, and write, from first principles:

1. the equation of the ellipse in standard form in terms of $e_E$ and $a_E$, with coordinates being distances in gigameters (Gm),
2. the integral for the arc length of a quarter-orbit,
3. the integral for the arc length of a full orbit, i.e. the distance the Earth travels in one year,
4. Use a computer, Wolfram Alpha, or other calculation tool to numerically approximate the integral with the given values, and check that the length of the quarter-orbit is approximately 234.0 Gm, and the full orbit is likewise approximately 936.0 Gm long.

And I'm sure you could find many, many exciting examples this way. And a few might just have a solution - you could mark those, e.g. give a catenary (hanging chain), and point that out ("This one actually can be reduced to an elementary formula! Do so.").

Solution 4:

This example $$ y = a\cosh \frac{x}{a} $$ is quite simple for computations.

Solution 5:

You can try $f(x)=\dfrac{\sqrt{a^2e^{2ax}-1}-\tan^{-1}\sqrt{a^2e^{2ax}-1}}{a}$, which has arc-length $e^{ax}-1$ and isn't too hard to work with as long as you remember $\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}x$.

But other than that, you could always define your function as an unsolved integral, $f(x)=\int\sqrt{L'(x)^2-1}\ \mathrm{d}x$. Then even when the function itself has no closed form, you can define a closed form for the arc-length, $L(x)$. Students can then use their knowledge of integration rules and the fundamental theorem of calculus to compute the arc-length.

For example, take $f(x)=\int \sqrt{\sec^4x - 1} \ \mathrm{d}x$, which has a horribly unwieldy closed form when the integral is solved. Students could compute the arc length as

$$\begin{aligned}L(x)&=\int\sqrt{1+{\left({\int \sqrt{\sec^4x - 1}\ \mathrm{d}x}'\right)}^2}\ \mathrm{d}x\\ &=\int\sqrt{1+\left[\sqrt{\sec^4x - 1}\right]^2}\ \mathrm{d}x\\ &=\int{\sec^2x}\ \mathrm{d}x\\ &=\tan x+C \end{aligned}$$

Which has the added gratification of reducing the integral into a satisfyingly neat conclusion. Behind the scenes, this works because we chose $L(x)=\tan(x)$, when we defined $f(x)=\int \sqrt{\left(\frac{\mathrm{d}}{\mathrm{d}x}\tan x\right)^2-1}\ \mathrm{d}x$.

The problem with that tactic is that you could only do it for a couple of problems since the students would soon see that your choice of $L(x)$ is the arc length. You'd probably also want to put a note in the question that students don't need to evaluate the integral form of $f(x)$, otherwise they'd get lost in it.