The Frenet-Serret Formulas involve a unit tangent vector,
$$\hat T = \frac{\vec r’(t)}{||\vec r’(t)||}$$
and a unit normal vector (differentiating with respect to $t$)
$$\hat N = \frac{\vec T’(t)}{||\vec T’(t)||}$$
and a unit binormal vector, the cross product of the two above:
$$\hat B = \hat T \times \hat N$$
Normally, they are written with respect to the arclength parameter $s$, but I’ve seen formulas with respect to $t$ that look way more complicated, such as these from wikipedia:
How do you derive these from the definitions above?
So far, I've gotten that
$$\hat N = \frac{||\vec r’(t)|| r''(t) – r'(t) \frac{r'(t) \cdot r''(t)}{||\vec r’(t)||}}{\Big|\Big|||\vec r’(t)|| r''(t) – r'(t) \frac{r'(t) \cdot r''(t)}{||\vec r’(t)||}\Big|\Big|}$$
just from quotient rule, but I'm not even sure I differentiated $||\vec r’(t)||=\sqrt{\vec r'(t) \cdot \vec r'(t)}$ right. I'm not sure how to proceed — any help with the $\hat N$ case or $\hat B$, or with the curvature $\kappa$ or torsion $\tau$ is much appreciated.
Best Answer
Working with the natural parameter $s$ we see that $T,N,B$ form a right-handed orthonormal set.
Now let $r=r(t)$ and use $r'$ for differentiation wrt $t$.
You have proved so far that $T$ is parallel to $r'$, and that $N$ is a linear combination of $r'$ and $r''$. (It's actually a lot easier to prove only this, never mind the precise coefficients.)
Recall that $B=T\times N$ to get that $B$ is parallel to $r'\times r''$, since $r'\times r'=0$. Hence $B$ being unit, is $\frac{r'\times r''}{||r'\times r''||}$ as required.
Now we know that $N=B\times T$; it is trivial to get the expression you want.