In the following problem I am asked to find the curvature of a curve. The problem is the following: \
Given the curve $\alpha(t), \alpha: \mathbb{R} \rightarrow \mathbb{R}^3$, which is parametrised by arc-length, we define the curve $\beta(t) = \alpha'(t)$. Show that the curvature of $\beta,\kappa_{\beta} = \sqrt{1+\frac{\tau^2}{\kappa_{\alpha}^²}}$, where $\tau$ is the torsion of $\alpha$ and $\kappa_{\alpha}$ its curvature.
My attempt is the following:
I know that $\kappa_{\beta} = \frac{||\beta' \times \beta''||}{||\beta^3||}, ||\beta' \times \beta''|| = (||\beta'^2||||\beta''^2|| - (\beta'\cdot\beta'')^2)^\frac{1}{2}$. However, since torsion of $\alpha$ is $\tau = N'\cdot B$, I don't know could the torsion appear here. I also noticed that $N = T_{\beta}$, where $N$ is the normal vector of $\alpha$ and $T_{\beta}$ the tangent unit vector of $\beta$. I would be grateful if somebody could give me a hint in order to solve the exercise.
Let me write out the details for this one. We will use $T, N, B$ for the Frenet frame of $\alpha$, and I will write the curvature and torsion of $\alpha$ as $k$ and $\tau$ without subscripts. Then by the structure equation, \begin{align} \alpha'' &= k N,\\ \alpha''' &= (kN)' = k'N + kN' = k'N + k(-kT + \tau B). \end{align}
Then $\|\beta'\|=\|\alpha''\| = k$. Also $$ \|\beta'\times\beta''\| = \|\alpha''\times\alpha'''\| = \|kN\times (k'N+k(-kT+\tau B)\|= k^2\|k B + \tau T\| = k^2\sqrt{k^2+\tau^2}, $$ by $N\times N=0, N\times T=-B, N\times B=T$.
Putting these together, we see $$ k_\beta = \frac{\|\beta'\times\beta''\|}{\|\beta'\|^3} = \frac{k^2\sqrt{k^2+\tau^2}}{k^3} = \frac{\sqrt{k^2+\tau^2}}{k} = \sqrt{1+\frac{\tau_\alpha^2}{k_\alpha^2}}. $$