Derive the Frenet equations I was looking for a derivation of the Frenet equations. I've been following this reference but I've been having problems in understanding this statement (found before Eq.$(2.19)$):

When r'($s+\Delta s$) is moved from $Q$ to $P$, then r'($s$), r'($s+\Delta s$) and r'($s+\Delta s$)-r'($s$) form an isosceles triangle, since r'($s+\Delta s$) and r'($s$) are unit tangent vectors. Thus we have $|\textbf{r}'(s+\Delta s)-\textbf{r}'(s)|=\Delta \theta \cdot 1=\Delta \theta=|\textbf{r}''(s)\Delta s| $ as $\Delta s \to 0$ [...]

My problem is in understanding the last string: I mean how can I prove mathematically that $|\textbf{r}'(s+\Delta s)-\textbf{r}'(s)|=\Delta \theta \cdot 1=\Delta \theta$. Is this taken from the dot product propriety:
$$|\textbf{a}-\textbf{b}|^2=a^2+b^2 + 2 ab \cos(\theta),$$
assuming $\textbf{a}$ and $\textbf{b}$ are unitary vectors and taking the angle between the vectors approximately zero, thus Taylor expanding the cosine to second order?
 A: You've mis-typed things a little -- in each place where you have $|r'(s + \Delta s) - r'(s + \Delta s)|$, you should have $|r'(s + \Delta s) - r'(s)|$; I suspect this is a cut-and-paste error. 
The claim the authors make -- that the difference vector EQUALS $\Delta \theta \cdot 1$ --  is false. But what's true is that if you have an arc of a unit circle subtending angle $\Delta \theta$, then the length of the arc is also $\Delta \theta$, so that the rightmost arrow in the Figure 2.4 that you refer to (the one pointing SE) has a length that's APPROXIMATELY the same as the length of a circle arc between its two endpoints. As $\Delta \theta$ gets small, this approximation gets better and better (basically, it's the approximation $\sin x \approx x$ for small $x$), so the rest of the argument is valid. 
I have to say, a quick look at that page doesn't give me a lot of confidence. You might want to look at Barrett O'Neill's book, "Elementary Differential Geometry", or Millman and Parker's book, which is a bit more sophisticated, or doCarmo, which sits somewhere in the middle. 
