I'm having trouble understanding an argument: the modulus of curvature it is the speed at which tangents change direction.
Let $\alpha:I \rightarrow \mathbb{R}^2$ a plane curve arc length parametrized and $k(s)$ the curvature of $\alpha$ is $s$. Consider the tangent vectors $\alpha'(s_0)$ and $\alpha'(s_0+h)$, where $s_0 \in I$ is fixed and $s_0 + h \in I$. Denote by $\phi(h)$ the angle between these two vectors, ie,
\begin{equation}
\cos\phi(h) = \langle \alpha'(s_0) ,\alpha'(s_0+h) \rangle.
\end{equation}
(because $|\alpha'(s_0)| = |\alpha'(s_0+ h)| = 1$).
The limit $\displaystyle \lim_{h \rightarrow 0} \frac{\phi(h)}{h}$ is the speed at which tangents change direction. We have
\begin{equation}
|\alpha'(s_0+h) – \alpha'(s_0)| = 2 \sin\frac{\phi(h)}{2}
\end{equation}
for all $h$ and so
\begin{equation}
|k(s_0)| = |\alpha''(s_0)| = \lim_{h \rightarrow 0} \frac{\phi(h)}{h}.
\end{equation}
Comments:
I'm not able to verify the last two equations:
$$|\alpha'(s_0+h) – \alpha'(s_0)|^2 = |\alpha'(s_0+h)|^2 – 2 \langle \alpha'(s_0+h) , \alpha'(s_0) \rangle + |\alpha'(s_0)|^2 = 2 – 2\cos{\phi(h)}.$$
I don't know which trigonometric identity is being used.
The last equation is also not being able to verify.
Thank you for your help.
Best Answer
The first equation uses cosine double-angle formula: $$ \cos\phi=1-2\sin^2{\phi\over2}. $$ The second equation features a standard trick to compute the limit: $$ |\alpha''(s_0)| =\lim_{h\to0}{|\alpha'(s_0+h) - \alpha'(s_0)|\over h} = \lim_{h\to0}\left({2\over h} \sin\frac{\phi(h)}{2}\right) \\ = \lim_{h\to0}\left({\phi(h)\over h}\cdot {\sin(\phi(h)/2)\over\phi(h)/2}\right) = \lim_{h\to0}\left({\phi(h)\over h}\right), $$ because $\displaystyle{\sin\theta\over\theta}\to1$ for $\theta\to0$.