I'm just a simple high school math student, so please don't eat me =)

In my calculus text, I have the formula:

$$L(x) = \int_{c}^{x} \sqrt{[f'(t)]^2 + 1}\,dt$$

Where $L(x)$ is the arc length of a curve $f(x)$ from $c$ to $x$.

How can I invert this function so that I can find valid values of $x$ to satisfy a given arc length? Something like $L^{-1}(x)$.


Nice idea!

As long as the function $g(x)$ is well-behaved, we have the following very important result. Let $$G(x)=\int_c^x g(t)dt$$. Then $G'(x)=g(x)$.

This result (and some related ones) is called the Fundamental Theorem of (Integral) Calculus.

Now let us apply that to your problem. We obtain $$L'(x)=\sqrt{1+(f'(x))^2}$$

Use the above equation to solve for $f'(x)$ in terms of $L'(x)$. If you take $L(x)$ as known, you have found an explicit formula for $f'(x)$, and all you need to do is to integrate.

Now comes the unfortunate part. For most pleasant functions $L(x)$, the resulting integration problem will be either difficult or more often impossible (in terms of standard functions).

I hope that this gives you something to play with. You will find out why there is such a limited number of different arclength problems in calculus books!

  • $\begingroup$ I'm afraid I still don't understand. Now I have a function of $L'$, when what I need is a function of $L$. Can you show me an example with $f(x) = x^2$? $\endgroup$ – Clark Gaebel Apr 7 '11 at 23:19
  • $\begingroup$ @Clark: $L'$ is a function of $L$. If you know $L(x)$, then you can compute $L'(x)$ from it. $\endgroup$ – Andrés E. Caicedo Apr 7 '11 at 23:30
  • $\begingroup$ In your post, you wrote "given arc length". So I will take $L(x)=x^2$ as an example. Then, using the Fundamental Theorem of Calculus, we get $2x=\sqrt{1+(f'(x))^2}$. Solve for $f'(x)$. We get $f'(x)=\sqrt{4x^2-1}$ (I took the positive square root). To find $f(x)$, integrate. The constant of integration can be found by using the fact that the arc length from $c$ to $c$ is $0$. The integration in this case is slightly messy but doable. If you have trouble with it, ask Wolfram Alpha. $\endgroup$ – André Nicolas Apr 7 '11 at 23:35
  • $\begingroup$ Nono, what I meant was given a value for the arclength (let's say 10), a starting point (let's say 1), and a function $f(x) = x^2$, how do I find a value of $x$ such that $L(x) = 10$? Phrasing is difficult. $\endgroup$ – Clark Gaebel Apr 7 '11 at 23:46
  • 2
    $\begingroup$ @Gaebel: Indeed that is a different question! You can find an explicit formula for $L(x)$, and solve $L(x)=10$. Typically there should be two answers. And usually there will not be a pleasant formula for the inverse function, or indeed for $L(x)$, so you will need to use a numerical procedure. $\endgroup$ – André Nicolas Apr 8 '11 at 0:25

You can do it by a differential equation without getting $L(x)$ explicitly. If $\frac{dL}{dx} = \sqrt{1 + f'(x)^2}$, then $\frac{dx}{dL} = \frac{1}{\sqrt{1 + f'(x)^2}}$. Numerical methods can be used to solve this differential equation.


You are basically asking how can one find the "arclenght parametrisation" for the graph of a function (arclenght parametrisation is usually studied for parametric curves).

I might be wrong but from what I remember this Question is usually hard, the only way I know to solve it is by simply computing $L(x)$ and try to find its inverse function.


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