The derivative of the normal unit vector

Question

I want to derive the unit vector $N(s)=\dfrac{T'(s)}{\|T'(S)\|}$, where $T'(s)$ is the derivative of the tangent unit vector, how can I derive this vector? Thanks!

Answer 1

I'm not sure Omar the OP meant "differentiate" or simply "derive" as in "show that such-and-such is the case. I assumed the former:

If the context is $\Bbb R^2$, i.e., the curves lie in the plane, then since $N(s)$ and $\kappa$ are defined via

$T'(s) = \kappa N(s), \tag{1}$

together with the requirement that $\Vert N(s) \Vert = 1$, where $\kappa$ is the curvature of the curve, it follows that

$N'(s) = -\kappa T(s); \tag{2}$

you don't have to do any more work once you've got $T'(s)$; you're done. A complete explanation may be found in my answer to this question.

If you're in $\Bbb R^3$, or even $\Bbb R^n$, $n > 3$ (1) still defines $\kappa$ and $N(s)$ in this case, and we can use it to write $N'(s)$ via

$T''(s) = \kappa' N(s) + \kappa N'(s), \tag{3}$

provided we can compute $\kappa'(s)$. But

$\kappa(s) = \Vert T'(s) \Vert = \langle T'(s), T'(s) \rangle^{\frac{1}{2}}, \tag{4}$

so that

$\kappa'(s) = \frac{1}{2}\langle T'(s), T'(s) \rangle^{-\frac{1}{2}}\langle T'(s), T'(s) \rangle' = \frac{1}{2}\kappa^{-1}\langle T'(s), T'(s) \rangle'. \tag{5}$

Now

$\langle T'(s), T'(s) \rangle' = \langle T''(s), T'(s) \rangle + \langle T'(s), T''(s) \rangle = 2\langle T'(s), T''(s) \rangle, \tag{6}$

so combining this with (5) yields

$\kappa'(s) = \langle T'(s), T'(s) \rangle^{-\frac{1}{2}}\langle T'(s), T''(s) \rangle, \tag{7}$

and now using (1) and (4) we see that

$\kappa'(s) = \kappa^{-1}\langle \kappa N(s), T''(s) \rangle = \langle N(s), T''(s) \rangle. \tag{8}$

(8) may be used in (3) which, after a little re-arranging, becomes

$N'(s) = \kappa^{-1}(T''(s) - \langle N(s), T''(s) \rangle N(s)). \tag{9}$

That's most likely how I'd do it, in any event. A "recipe" or set of directions for obtaining $N'(s)$ based on (9) might read:

Given $T(s)$,

1.) Differentiate $T(s)$ to obtain $T'(s) = kN$; this gives $\kappa$ and $N$;

2.) Differentiate $T'(s)$ to obtain $T''(s)$;

3.) From the data derived in steps (1) and (2) above, algebraically calculate, in accordance with equation (9), $N'(s) = \kappa^{-1}(T''(s) - \langle N(s), T''(s) \rangle N(s))$.

4.) Halt.

Of course, it is also possible to compute $N'(s)$ directly from (1) by writing

$N(s) = \kappa^{-1}T'(s), \tag{10}$

reducing it to simplest terms "algebraically", and then differentiate the resulting expression directly. But it's not clear that it would be less work the above prescpition; indeed, my personal experience shows it would often be more. Addionally but not extraneously, the nice thing about (9) is that it breaks out everything in geometric terms; indeed, $T''(s) - \langle N(s), T''(s) \rangle N(s)$ is the projection of $T''(s)$ onto $N^\perp(s)$, the subspace normal to $N(s)$. So (9) has a nice interpretation, geometrically speaking: it says that the derivative of $N(s)$ with respect to $s$ is the component of $T''(s)$ normal to $N(s)$, divided by $\kappa$. One can of course assign similar geometric interpretations to most of the other equations occurring in the above discussion.

Of course all of the above is subject to the usual caveats concerning the vanishing of $\kappa$; we must generally assume we are working in a neighborhood where $\kappa(s) \ne 0$, since $N(s)$ can be defined, and the ensuing divisions by $\kappa$ are permitted.

Hope this helps. Cheerio,

and of course,

Fiat Lux!!!