# Showing a characterisation of curves with constant slope.

Here is what I would like to prove:

Let $\alpha:I\rightarrow\mathbb{R}^3$ be a curve. Assume $\alpha\in C^3$ and that $\dot\alpha(t)$ and $\ddot\alpha(t)$ are linearly independent $\forall t\in I$. Then:

$\alpha$ is a curve with constant slope $\iff \vec{T}(\alpha,t)$ makes a constant angle with the Darboux vector.

Notation:

$\tau=\frac{\langle \vec{B},\vec{N} \rangle}{\|\dot\alpha(t)\|}$ is the torsion.

$\kappa=\|\vec{K}\|=\|\frac{\dot{T}}{\|\dot\alpha(t)\|}\|$ is the curvature.

The Darboux vector is : $\vec{D}(\alpha,t)=\tau(\alpha,t)\vec{T}(\alpha,t)+\kappa(\alpha,t)\vec{B}(\alpha,t)$

$\vec{T}$ is the tangent vector (i.e. $\vec{T}(\alpha,t)=\frac{\dot\alpha(t)}{\|\dot\alpha(t)\|}$), $\vec{N}$ the normal vector and $\vec{B}$ is the binormal vector (all three from the Frenet–Serret frame).

What I know:

I know that a curve has a constant slope if and only if the ratio between the torsion and the curvature is constant, i.e. $\frac{\tau}{\kappa}$ is constant.

For the $\implies$ part of the proof I wanted to write :

$$\cos(\angle(\vec{T},\vec{D}))=\frac{\langle\vec{T},\vec{D}\rangle}{\|\vec{T}\|\|\vec{D}\|}=\frac{\langle\vec{T},\tau\vec{T}+\kappa\vec{B}\rangle}{\|\vec{T}\|\|\tau\vec{T}+\kappa\vec{B}\|}$$

to show that the angle is constant but I can't get anywhere. As for the other part of the proof ($\Longleftarrow$) I do not know where to begin.

Any hint or help would be appreciated, thanks in advance.

Note that if $\tau=c\kappa$, then the Darboux vector is parallel to $c\vec T+\vec B$, which in turn has constant length.
For the converse, just use high school algebra with your formula and solve for $\tau/\kappa$.