Prove that $\kappa(s) = \lim\limits_{h \to 0} \left|\frac{\theta (s,h)}{h}\right|$

Question

$\newcommand{\Reals}{\mathbf{R}}$Let $\gamma:I\subset \Reals \to \Reals^{n}$ the arc length parameterization of a curve $C\subset \Reals^{n}$.

If $\theta(s, h)$ is the angle between $T(s)$ and $T(s + h)$, show that $$ \kappa(s) = \lim_{h \to 0} \left|\frac{\theta (s, h)}{h}\right|. $$

Indeed, I can see that geometrically this is what happens, but how can I prove it? I have tried using that $$ \cos\theta = \langle T(s), T(s + h)\rangle, $$ but I can't get to anything, any help will be appreciated, thanks.

Answer 1

$\newcommand{\Brak}[1]{\langle#1\rangle}$As you noted, $$ \cos\theta(s, h) = \Brak{T(s + h), T(s)}. $$ Here, however, it may be more convenient to note that if $T' = \kappa N$, then $$ \sin\theta(s, h) = \Brak{T(s + h), N(s)}. \tag{1} $$ (Why is this true?)

Now, by definition $$ \kappa(s)N(s) = T'(s) = \lim_{h \to 0} \frac{T(s + h) - T(s)}{h}. \tag{2} $$ Dot both sides with $N$ and see what you get.