Show that curvature over torsion is constant when curve dot constant vector is constant

[Dragonfall]

Homework Statement

Suppose $$\alpha$$ is a regular curve in $$\mathbb{R}^3$$ with arc-length parametrization such that the torsion $$\tau(s)\neq 0$$, and suppose that there is a vector $$Y\in \mathbb{R}^3$$ such that $$<\alpha',Y>=A$$ for some constant A. Show that $$\frac{k(s)}{\tau(s)}=B$$ for some constant B, where k(s) is the curvature of alpha.

The Attempt at a Solution

I think the Frenet formula in question that I can use is $$n'=-kt-\tau b$$, but I can't make it work.

 

[Dragonfall]

What about the converse? Suppose k/tau is constaint and tau is nonzero everywhere, show that there exists a nonzero vector Y such that <a', Y> is constant.

 

[Dick]

Ok, I'm pretty rusty at this so bear with me. Take k(s) nonzero (if it is zero, it's not even clear to me how to define the Frenet frame). Since you have arc-length parametrization, alpha'=t (the tangent). So <t,Y> is constant. Now you should be able to prove a bunch of stuff by differentiating <x,Y> where x is various vectors.

i) Show <n,Y>=0.
ii) Show <b,Y> is also constant.
iii) Differentiate <n,Y> and use your favorite Frenet formula.