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I am trying to find a proof for the 9th question of section 2.4, from the book Elementary Differential Geometry by Barrett O'Neill. I want to show that a curve $\alpha$ with curvature $\kappa$ and torsion $\tau$ both constant is a circular helix.

My thoughts: I know $\alpha$ must be a cylindrical helix since $\tau/\kappa$ is a constant. Remains to show that it is circular. I'm not sure how to do this. I found this question here : Is the helix the unique path with constant curvature and constant torsion? where the answer gives a hint to prove N′′(s)=−(κ2+τ2)N(s). I can do this, but where does it head me toward? I cannot see where this will lead me.

Random thought: It occurs to me that if I can project the curve $\alpha$ on the xz plane, it must be a circle. But I don't know how I would go about showing this.

Any help is appreciated. Thank you for your time.

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Of course the uniqueness theorem (Fundamental Theorem of Curves) will do it, since you know (or can easily show) that circular helices have constant curvature and torsion.

The intent of my hint was to then deduce that $\vec N(s)=\cos(ks)\vec c_1+\sin(cs)\vec c_2$ with $k^2=\kappa^2+\tau^2$ and $\vec c_1,\vec c_2$ fixed vectors. Show then that $\vec c_1$ and $\vec c_2$ must be orthogonal and show that you get circular motion in the plane they span. Can you finish from there?

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  • $\begingroup$ I will try and come back to comment. Thanks a lot for the answer. $\endgroup$ – user264750 Sep 27 '17 at 5:54
  • $\begingroup$ Sure thing. If you get very stuck and/or have further questions, let me know. When you figure it out, accept the answer so that the question doesn't stay on the unanswered list. $\endgroup$ – Ted Shifrin Sep 27 '17 at 6:01

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