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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?

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What does explicit formula mean? – Jonas Meyer Mar 25 '11 at 2:50
You have specified a differential equation. Solve it! To get started, consider the case $\kappa = \text{constant}, \tau = \text{constant}$. – whuber Mar 25 '11 at 3:00
up vote 6 down vote accepted

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.

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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.

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Moving trihedron perhaps? – Jesse Madnick Mar 25 '11 at 8:32
@Jesse: an emphatic no. A trihedron would be a surface of 3 faces (which is impossible in Euclidean geometry, the same way that a two-sided polygon is not possible in the plane). A trihedral (the noun and the adjective forms are spelled the same way) is a figure formed by three non-coplanar lines intersecting at a point. (Yes, an -al ending can be a noun; the transversal is the noun, while two surfaces can be transverse to each other.) See, e.g. James and James, Mathematical Dictionary. Admittedly, the word is somewhat out of fashion nowadays. – Willie Wong Jul 30 '11 at 3:49
@Willie Wong: That's fair, but I'm still used to hearing trihedron. For example here or here. – Jesse Madnick Jul 30 '11 at 5:07
@Jesse: Well, in terms of the Greek roots, either is fine. εδρα means "base" or "facet" or "seat" (legal parlance). For me it is a personal preference to reserve -hedron for stuff like polyhedra. Cheers. – Willie Wong Jul 30 '11 at 13:44

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