Graphs for which a calculus student can reasonably compute the arclength

Question

Given a differentiable real-valued function $f$, the arclength of its graph on $[a,b]$ is given by

$$\int_a^b\sqrt{1+\left(f'(x)\right)^2}\,\mathrm{d}x$$

For many choices of $f$ this can be a tricky integral to evaluate, especially for calculus students first learning integration. I've found a few choices of $f$ that make the computation pretty easy:

• Letting $f$ be linear is super easy, but then you don't even need the formula.
• Taking $f$ of the form $(\text{stuff})^{\frac{3}{2}}$ might work out nicely if $\text{stuff}$ is chosen carefully.
• Calculating it for $f(x) = \sqrt{1-x^2}$ is alright if you remember that $\int\frac{1}{x^2+1}\,\mathrm{d}x$ is $\arctan(x)+C$.
• Letting $f(x) = \ln(\sec(x))$ results in $\int\sec(x)\,\mathrm{d}x$, which classically sucks.

But it looks like most choices of $f$ suggest at least a trig substitution $f'(x) \mapsto \tan(\theta)$, and will be computationally intensive, and unreasonable to ask a student to do. Are there other examples of a function $f$ such that computing the arclength of the graph of $f$ won't be too arduous to ask a calculus student to do?

Answer 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.

Answer 2

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

Answer 3

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 $a$.

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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?

Answer 4

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.

Answer 5

Difficult, but very interesting problem covering very much of integral calculus.

Perhaps worth your while if you provide some of the steps where indicated as an integral table:

Prove that, like circles, all parabolas are similar.

Specifically, given the parabola $y=Ax^2$, prove the ratio of the length of the line passing through the focus, $y=1/4A$ and intersecting the parabola, divided by the arclength of the parabola between the points of intersection is a constant, i.e. independent of A. And find that constant.

First find the points of intersection: $$Ax^2=1/4A\implies x=+-(1/2A)$$

That gives a latus rectum length of 1/A.

Now we set up our integral:

$y=Ax^2$

$y'=2Ax$

So:

$$L=\int_{-1/2A}^{1/2A} \sqrt{1+4Ax^2} dx$$

Let $u=2Ax$. Then $du=2Adx$ and we can rewrite the integeral.

$$L=\frac{1}{2A}\int_{-1}^1 \sqrt{1+u^2} du$$

Already, we can see that the latus rectum length divided by this length cancels A, so it follows that this ratio is a constant for all parabolas.

Now you can use trig substitution combined with a halfing trick. This is the part where you might want to give them part of the answer.

Imagine a right triangle with base of length 1, height of length $u$ and hypotenuse therefor of length $\sqrt{1+u^2}$. Then $u=\tan{\theta}$ so $du=\sec^2{\theta} d\theta$ and $\sqrt{1+u^2}=\sec{\theta}$. And using trig substitution:

$$L=\frac{1}{2A}\int_{-\pi/4}^{\pi/4} \sec^3{\theta} d\theta$$

Here's where providing some steps could be useful.

$$\sec^3{\theta}=\sec{\theta}(1+\tan^2{\theta})=\sec{\theta}+\sec{\theta}\tan^2{\theta}$$

So the integral can be broken up into two integrals.

Now you can use integration by parts:

$$r=\tan{\theta}$$ $$ds=\sec{\theta}\tan{\theta}$$ $$s=\sec{\theta}$$ $$dr=\sec^2{\theta}$$

$$\int r ds= rs-\int s dr$$

So:

$$\int \sec^3{\theta} d\theta=\ln|\sec{\theta}+\tan{\theta}| + \sec{\theta}\tan{\theta}-\int \sec^3{\theta}$$

Rearranging we finally have:

$$\int \sec^3{\theta} d\theta = \frac{\ln|\sec{\theta}+\tan{\theta}|+\sec{\theta}\tan{\theta}}{2}$$

Ending with:

$$L=\frac{1}{A}(\ln|\sqrt{2}+1|+\sqrt{2})$$

Divide by $1/A$ to get the ratio.

Answer 6

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.").

Answer 7

Since people may find this question looking for specific functions that they can use in exercises for students, it's a good idea to have a community wiki answer to collect explicit examples. Cracking open Stewart's Calculus and looking over the "calculate this arclength" exercises, all the functions listed appear have a form similar to something already mentioned here.

With these functions, $1+f'(x)^2$ will be a square or just be a single summand, and so evaluating the arclength formula will be straightforward for a calculus student. Seeing that it's a square though, might not be so easy: \begin{align} f(x) = \frac{2}{3}(x^2-1)^\frac{3}{2}\quad &\implies \quad 1+f'(x)^2 = (2x^2-1)^2 \\ f(x) = \left(\frac{2}{3}x-1\right)^\frac{3}{2} \quad &\implies \quad 1+f'(x)^2 = \frac{2}{3}x \\ f(x) = \frac{1}{3}(x^2+2)^\frac{3}{2}\quad &\implies \quad 1+f'(x)^2 = (x^2+1)^2 \\ f(x) = \frac{1}{3}\sqrt{x}(x-3)\quad &\implies \quad 1+f'(x)^2 = \left(\frac{\sqrt{x}}{2}+\frac{1}{2\sqrt{x}}\right)^2 \\ f(x) = \frac{x^2}{2} - \frac{1}{4}\ln(x) \quad &\implies \quad 1+f'(x)^2 = \left(x+\frac{1}{4x}\right)^2 \\ f(x) = \ln(1-x^2) \quad &\implies \quad 1+f'(x)^2 = \left(\frac{1+x^2}{1-x^2}\right)^2 \end{align}

Then these ones, from what I can tell, are reasonable besides requiring trig-substitution.
$$ f(x) = e^x \qquad f(x) = \ln(x) \qquad f(x) = 2\sqrt{x} $$

Answer 8

Let $u(x) = f'(x)$ and $v(x) = \sqrt{1+u(x)^2}$. So we want $u$ to have an elementary integral (so that we can write down $f$ on the assignment sheet) and $v$ to have an elementary integral (so our students can solve it.) In other words, we want functions $u$ and $v$, both with elementary integrals, so that $v^2 = 1 + u^2$.

Rewrite this as $(v+u) (v-u) = 1$. If $v$ and $u$ have elementary integrals then so do $v+u$ and $v-u$. Conversely, $v = \tfrac{1}{2} \left( (v+u) + (v-u) \right)$ and $u = \tfrac{1}{2} \left( (v+u) - (v-u) \right)$ so, if $v \pm u$ have elementary integrals, then so do $u$ and $v$. So the problem reduces to finding $h$ where $h$ and $1/h$ both have elementary integrals.

Some candidates for $h$ from this perspective:

• Any rational function.
• Any rational funtion of $e^x$.
• Any rational function of $\sin x$ and $\cos x$.