I've been looking a bit into differential geometry and have gotten stuck on a question:
Given a function $f,$ is there a way to find the explicit space curve which has $f$ as both it's curvature and torsion?
I've been able to find a formula for a plane curve with curvature $k(s),$ but extending to what seems to be the next simplest case (curvature = torsion) has been difficult.
Any hints on how to proceed?
If $\kappa = \tau$, let $U = T + B$ and $W = T - B$. Derivatives in the following are with respect to arclength $s$. From the Frenet-Serret formulas, $U'= 0$ so $U$ is constant. WLOG we can take it to be $[0,0,\sqrt{2}]$. Now we have $W$ and $N$ orthogonal to $U$ with $U \times W = 2 N$, so $W = [\sqrt{2} \cos(w), \sqrt{2} \sin(w), 0]$ and $N = [-\sin(w),\cos(w),0]$. Then $W' = 2 \kappa N$ says $w' = \kappa$. So the first thing you need is an antiderivative of $w$ of $\kappa$. Then $T = (U+W)/2 = [\sqrt{2} \cos(w)/2, \sqrt{2} \sin(w)/2, \sqrt{2}/2]$, and integrating this (which requires antiderivatives of $\cos(w)$ and $\sin(w)$) will give you your curve.
Using the curvature and torsion, you can determine the moving trihedral of the space curve by solving a certain differential equation. If you wish, you can then use this to get a parameterization of the curve. All the necessary information can be found here, although they refer to the moving trihedral as the Frenet-Serret frame.
