I was trying to find the curvature of an involute on the plane, given a curve which is parametrized by it's arc-length.
Suppose that we a have a curve, $\alpha(s)$ which is parametrized by it's arc-length, then we know that an involute of $\alpha$ is parametrized by $$\beta(s)=\alpha(s)+(c-s)\alpha'(s)=\alpha(s)+(c-s)T_{\alpha}(s)$$ where $(T_{\alpha}(s),N_{\alpha}(s))$ is the Frenet reference of the curve $\alpha$ at the point $s$.
Since we know that $||\beta''(s)||=k_{\beta}(s)$, I suppose that we need to differentiate $\beta(s)$ two times and then take the norm of it, right?
Differentiating we get, $$\beta'(s)=T(s)+(c-s)k_{\alpha}(s)N_{\alpha}(s)-T_{\alpha}(s)=(c-s)k_{\alpha}(s)N_{\alpha}(s)$$
because $T_{\alpha}(s)=k_{\alpha}(s)N_{\alpha}(s)$. Differentiating a second time,
$$\beta''(s)=k_{\alpha}(s)\big(-N_{\alpha}(s)+(c-s)(-k_{\alpha}(s)T_{\alpha}(s))\big)+(c-s)N_{\alpha}(s)k'_{\alpha}(s)=$$
$$-k_{\alpha}(s)N_{\alpha}(s)-(c-s)k^2_{\alpha}(s)T_{\alpha}(s)+(c-s)N_{\alpha}(s)k'_{\alpha}(s)$$ the problem comes whenever I have to take the norm of this vector, which I don't know how to do. Can anyone help me? Any help will be really much appreciated.
Best Answer
Note that $\|\beta''(s)\|=k_\beta(s)$ doesn't hold here, because the involute $\beta(s)$ is not parametrized by arc length: $\|\beta'(s)\| = (c-s)k_\alpha$. We need to use the formula $$ k_\beta = \frac{\det(\beta'(s)\; \beta''(s))}{\|\beta'(s)\|^3}. $$ Filling in $\beta'(s)$ and $\beta''(s)$ in this formula gives $k_\beta(s) = \frac{1}{s}$.