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!
 A: 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.
A: 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.
