Find the curvature of the curve $\beta(s) = \int_{0}^{s} N(t) dt$

$$Q)$$ 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.

• 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. Jul 16 '20 at 18:23
• Dear @TedShifrin, thanks for your kind sharing your free textbook. It has been surely helpful for studying deifferential geometry. 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, $$|N'_\alpha|=v_\alpha|-3T_\alpha-4 B_\alpha|=10.$$