$Q)$ 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.

  • $\begingroup$ You might find my free differential geometry text (linked in my profile) helpful. I have lots of examples (and exercises) of the application of the chain rule for non-arclength-parametrized curves. $\endgroup$ Jul 16 '20 at 18:23
  • $\begingroup$ Dear @TedShifrin, thanks for your kind sharing your free textbook. It has been surely helpful for studying deifferential geometry. $\endgroup$ Jul 16 '20 at 23:31

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, $$|N'_\alpha|=v_\alpha|-3T_\alpha-4 B_\alpha|=10.$$


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