If a path has both curvature and torsion constant why it is necessarily a helix? How to prove it?
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2you mean the Frenet-Serret' system of differential equations has a unique solution? – user180259 Oct 01 '14 at 18:04
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and how to prove it has a unique solution? – user180259 Oct 01 '14 at 18:08
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1I may miss a detail or two (it's been over 30 years since I read the book), but IIRC in Lipschutz' Differential Geometry there is a theorem that curvature and torsion, as functions of the arc length, determine a 3D-curve uniquely to the extent possible, i.e. up to translation and rotation. This implies that the answer to your question is affirmative. – Jyrki Lahtonen Oct 01 '14 at 18:44
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If both curvature and torsion are constant, the curve is in fact a (right) circular helix. You can, as @JyrkiLahtonen suggests, deduce this from the Fundamental Theorem of Curve Theory: By direct computation, a circular helix has constant curvature and torsion, and the curvature and torsion functions determine the curve up to rigid motion (isometry of $\Bbb R^3$).
Alternatively, with a bit of work and cleverness, you can actually explicitly integrate the Frenet equations \begin{align*} T'(s) &= &\kappa N(s) \\ N'(s) &= -\kappa T(s) + &&&\tau B(s) \\ B'(s) &= &-\tau N(s) \end{align*} to show that $\alpha(s)$ is a right circular helix. (Hint: Show that $N''(s) = -(\kappa^2+\tau^2)N(s)$.)
A generalized helix is a curve whose tangent vector makes a constant angle with a fixed vector. Such curves are characterized by the weaker condition $\tau/\kappa = \text{constant}$.
Ted Shifrin
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