$Q)$ 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.
Best Answer
The mistake is in your expression for $N'$. There you use a Frenet-Serret equation that only holds for arc length parametrized curves. But $\alpha$ is not arc length parameterized; its speed is equal to $2$.
In general, for a regular curve $\alpha$, the usual Frenet-Serret formulas pick up an extra factor $v_\alpha$ (the speed of the curve), because of the chain rule. So $$ \begin{cases} T'_\alpha=v_\alpha\kappa_\alpha N_\alpha\\ N'_\alpha=-v_\alpha\kappa_\alpha T_\alpha+v_\alpha \tau_\alpha B_\alpha\\ B'_\alpha=-v_\alpha \tau_\alpha N_\alpha \end{cases}. $$ Consequently, $$|N'_\alpha|=v_\alpha|-3T_\alpha-4 B_\alpha|=10.$$