Finding the inverse of the arc length function 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 $L^{-1}(x)$.
 A: 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!
A: 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.
A: 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.
