Version of non-linear functions with linearly increasing arc-length (proof)

Question

I am trying to find a function "f" such that: \begin{align}{ \int_{a}^{a+b} \sqrt{(\frac{d}{dt}(x(f(t))))^2+(\frac{d}{dt}(y(f(t))))^2 } \,dt}=B \hspace{3cm}a,b,B\in{\mathbb{R}}\end{align} This is easy for linear functions because the derivatives are constant. I have no idea how to prove how to do it generally. Is there a proof of a way to find the function "f"? I have tried expanding b into the integral of one. \begin{align}\int_{a}^{a+b}\frac{B}{b}dt=B \end{align}I have tried integrating but the integrals always has “f” multiplied by an irreducible function.\begin{align}{ \sqrt{(\frac{d}{dt}(x(f(t))))^2+(\frac{d}{dt}(y(f(t))))^2 } }=1 \end{align}\begin{align}{ (\frac{d}{dt}(x(f(t))))^2+(\frac{d}{dt}(y(f(t))))^2 }=1 \end{align}\begin{align}{ -\sqrt{(\frac{d}{dt}(x(f(t))))^2+(\frac{d}{dt}(y(f(t))))^2 } }\neq1 \end{align} From there I have tried using indefinite integrals but, It just introduces unknowns. I have also tried integrating on the interval [0,t] but it quickly becomes unintegrable because the integrand ends up having the following integral:\begin{align}\int_{f^{-1}(f^{-1}(y^{-1}(0)))}^{f^{-1}(t)}f(t)f^{-1}(t)dt\end{align}The most promising solution is inverting the line integral from [a,t] for f(t)=t. The result has no elementary inverse but, I can use root finding methods then concatenate over the lines from a to t to create a piecewise equation that is a discrete version of r(f(t)) where r(t)=(x(t),y(t)).Then I can shorten the lengths to find a better approximation. The trouble is proving that it works. \begin{align}\end{align}Edit: This question is about the reparameterization of parametric functions (in t) such that the change in t equals the change in arclength. The question is how to calculate that reparameterization and a proof that it works.

Answer 1

A couple of suggestions …

If you’re working with polynomial (i.e. Bézier) curves, they have a subset called Pythagorean Hodograph curves — the cool kids call them PH curves. On a PH curve, the troublesome square root goes away and the norm of the first derivative vector is a polynomial function. So, it’s easy to integrate to get arclength. Not quite what you were asking, but related, I think. Rida Farouki wrote an entire book about PH curves, if you’re interested.

More generally, presumably you can calculate a few values of $t$ and $s$, and do some simple curve fitting that gives you (approximately) $t$ as a function of $s$, or vice versa.

You asked for a proof that this works. If you do the reparameterisation curve fitting using polynomials, then the Stone-Weierstrass theorem tells you that the approximation can be made as good as you like.