# Proving the ratio of curvature and torsion is constant.

This question has been asked slightly differently in a few different forums, but I wanted to discuss my approach and see if I was on the right track:

Prove that if the tangent lines of a curve make a constant angle with a fixed direction, then the ratio of its curvature to its torsion is constant.

So, I started by letting the curve be parameterized by arclength for convenience. Then, I let the fixed direction be the principal normal of the curve (as suggested by my professor). I know that the ratio of curvature to torsion is constant for a helix, so I was thinking of trying to prove that the assumptions imply that the curve must be a helix.

I tried using the cosine similarity formula as follows (with $T$ being the tangent vector, and $u$ being my fixed principal normal direction:

$cos(\theta)=\frac{T\cdot u}{\Vert{T}\Vert \Vert{u}\Vert}$ is constant

I think I can say that both $T$ and $u$ are unit, so then I'd have that $cos(\theta) =T\cdot u$ is constant.

Then, I was thinking if I showed $\frac{d}{ds}(T\cdot u)=0$, then I could somehow relate that back to curvature.

Am I on the right track?

Thank you very much for any help!

Suppose $(T,N,B)$ is a given Frenet-Serret frame for the curve and suppose $T\cdot u=\cos(\theta)$ for some constant vector $u$. It is sufficient to show $\tau/\kappa=\cot(\theta)$.

Differentiating $T\cdot u=\cos(\theta)$ yields $N\cdot u=0$. This implies $u=\cos(\theta)T+\sin(\theta)B$ because we can assume $u$ has unit length.

Differentiate this equation to obtain $0=\kappa cos(\theta)N-\tau \sin(\theta)N$ and so $\tau/\kappa=\cot(\theta)$.

To show the converse, first find a $\theta$ such that $\tau/\kappa=\cot(\theta)$ and work backwards through the proof above.

• Thank you, and thank you for correcting the notation. I am having trouble following one piece. Why does $N \cdot u = 0$ imply $u=cos(\theta)T+sin(\theta)B$? I can follow the steps after and before. @ArtW Commented Jun 11, 2016 at 22:26
• $\{T,N,B\}$ is an orthonormal basis, so we have the orthonormal expansion $u=(T\cdot u)T+(N\cdot u)N+(B\cdot u)B$. you're welcome :)
– ArtW
Commented Jun 11, 2016 at 22:29

To begin with, we should name the fixed direction some other way, say $v$. You already have that $T\cdot v$ is constant, so its derivative is zero. As $\dot v= 0$, we get $\kappa N \perp v$, and $v$ belongs to the vector space generated by $T$ and $B$. Since these are orthonormal, we get $$v= \langle v,T\rangle T+ \langle v, B\rangle B.$$ Now you can differentiate this equality on both sides and use Frenet-Serre equations to get the desired result.