Hi need to compute the curvature of a curve defined by:
\begin{cases} r_1(s) = a \cdot \cos\left[{\frac{s}{a} \cos\alpha}\right] \\ r_2(s) = a \cdot \sin\left[{\frac{s}{a} \cos\alpha}\right] \\ r_3(s) = s \cdot \sin\left({\alpha}\right) \end{cases}
With $a \in \mathbb{R}; a > 0$ and $\alpha \in \mathbb{R}$
but to do this first I need to test if the curve is parametrized by the arc complement then I cant compute the curvature as:
$k=||\frac{dT}{ds}||$
with $T(s)=\frac{dr}{ds}$
If we start with the curve $$x=a\cos(\frac{t\cos \alpha}{a}),y=a\sin(\frac{t\cos \alpha}{a}),z=t\sin\alpha$$ we calculate $ \mathbf v=[\dot x,\dot y,\dot z] \text { and }\dot s=\vert \vert \mathbf v \vert \vert=1$ so the curve really is parametrized by the arc length and there is no harm in taking $s=t.$ Since $\dot s=1,$ the formula for the curvature is just$$\kappa=\vert \vert \mathbf v \times \mathbf a\vert \vert$$. We already worked out $\mathbf v$. Differentiate its components to find $\mathbf a$ and then take the cross-product and calculate its magnitude.The result is $$\kappa=\frac{\cos^2\alpha}{\vert a \vert}.$$