$Q)$ In $\mathbb{R}^3$, There is a curve $\alpha(t)$ whose curvature $\kappa_{\alpha} = 3$ and torsion $\tau_{\alpha}=-4$ respectively. Here, $v (= \Vert a'(t) \Vert )= 2$
Find the curvature $\kappa_{\beta}$ of the curve $\beta(s) = \int_{0}^{s} N(t) dt$ (Here the $N(t)$ is a normal vector of the curve, $\alpha$)
Here is my solution.
By fundamental thm of Calculus, $\beta '(s) = N(s)$. Then the $\beta$ is a unit speed curve.
So, By Frenet-Serret, $\kappa_{\beta} = \vert \beta ''(s) \vert = \vert N'(s) \vert = \vert -3T-4B \vert =5$
(Here the $T$ and $B$ is a tangent vector and binormal vector respectively for the curve $\alpha$)
But the answer was $10 $.
What did I wrong in my solution?
Any help would be appreciated. Thanks.