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Given two smooth functions $k(s)$ and $\tau(s)$, how can we construct a curve in $R^3$ that its curvature is $k$ and torsion is $\tau$? Is there any conditions that $k$ and $\tau$ must satisfy as the curvature and torsion of a curve? It seems to be a very natural question, but not an easy task.

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    Check any textbook in elementary differential geometry –  Mar 16 '14 at 09:25
  • @AlexDegtyarev I agree with the sentiment that the answer to this question is very well known, but I have half a dozen differential geometry textbooks and not one of them answers this question. Even the relevant wikipedia page (http://en.wikipedia.org/wiki/Fundamental_theorem_of_curves) contains no references. It seems that the geometry of curves has become so standard that modern books don't really bother with it any more and you have to consult older books. – Paul Siegel Mar 16 '14 at 11:50
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    @PaulSiegel Probably, because it's obvious nowadays, and the proof is given in Wikipedia: (1) Solve Frenet equations (existence & uniqueness for linear ODE); (2) prove that the solutions are orthonormal frames (uniqueness for ODE); (3) integrate $\tau$. E.g., Manfredo P. Do Carmo, Differential Geometry of Curves and Surfaces. –  Mar 16 '14 at 12:28
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    Alex is right, this is pretty easy as stated. I would just like to say that if we require the curve to be closed, then the problem is interesting. –  Mar 16 '14 at 14:18
  • There is some relevant discussion in the earlier MO question, Existence and Properties of 3D Curves with unusual 2D (κ(s),τ(s)) Trajectories. – Joseph O'Rourke Mar 16 '14 at 15:56

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