calculus
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 $L^{-1}(x)$.
Divide the interval into $ n $ equal parts, $ a = x_0 \lt x_1 \lt \cdots \lt x_n = b $, with $ x_{i+1} = x_i + \Delta x_i $.
Suppose you want to approximate the curve between $(x_i,f(x_i))$ and $(x_i+\Delta x,f(x_i+\Delta x))$. You could simply approximate it with the straight line between the two points, whose length is
$$\sqrt{\left( f(x_i+\Delta x) - f(x_i)\right)^2 + (\Delta x)^2}.$$
In the picture below, the black line is the graph $y=f(x)$, and the green line is the line that joints $(x_i,f(x_i))$ on the bottom left and $(x_1+\Delta x,f(x_1+\Delta x))$ on the top right.
Then you would have that the arc length is approximated by the sum of the lengths
$$\text{Arc Length} \approx \sum_{i=1}^n \sqrt{(f(x_i+\Delta x) - f(x_i)))^2 + (\Delta x)^2}$$
and take the limit as $n\to \infty$. Unfortunately, the expression in the sum is not of the form necessary to view it as a Riemann sum, so you cannot turn that limit into a limit of Riemann sums, and from there to an integral.
So we take a slightly different approach. Instead of approximating the length of the curve from $(x_i,f(x_i))$ to $(x_i+\Delta x, f(x_i+\Delta x))$ with the straight line between the two points, we will approximate it with the tangent line to the graph of $f$ at $x_i$, from $(x_i,f(x_i))$ to the point $x_i+\Delta x$. This is the blue line in the picture above.
If $\Delta x$ is small, then we know the tangent line is a very good approximation for the curve on $[x_i,x_i+\Delta x]$, so the line will be a good approximation to the length of the curve.
Now, the tangent line to $y=f(x)$ through the point $x_i$ is given by $$y = f(x_i) + f'(x_i)(x-x_i).$$ At $x_i+\Delta x$, the line goes through $f(x_i) + f'(x_i)\Delta x$. So this tangent line goes from $(x_i,f(x_i))$ to $(x_i+\Delta x ,f(x_i)+f'(x_i)\Delta x)$. The length of the line between those two points is \begin{align*} &\sqrt{\Bigl( (x_i+\Delta x) - x_i\Bigr)^2 + \Bigl((f(x_i)+f'(x_i)\Delta x) - f(x_i)\Bigr)^2}\\\ &\quad = \sqrt{ (\Delta x)^2 + \left(f'(x_i)\Delta x\right)^2} \\\ &\quad = \sqrt{\left(1 + \left(f'(x_i)\right)^2\right)\Delta x^2} = \left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x. \end{align*}
Adding all of these, we get an approximation to the arc length: $$\text{Arc Length} \approx \sum_{i=1}^n \left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x.$$ Now, these can be seen as Riemann sums. So if we take the limit as $n\to\infty$, the approximation gets better and better (because the tangent gets closer and closer to the curve, giving a better approximation). At the limit, we get the exact arc length, and the limit of the Riemann sums becomes the integral. So \begin{align*} \text{Arc Length} &= \lim_{n\to\infty}\sum_{i=1}^n\left(\sqrt{1 + (f'(x_i))^2}\right)\Delta x\\\ &= \int_a^b \sqrt{1+(f'(x))^2}\,dx.{}{}{} \end{align*}
Best Answer
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!