# When does the normal vector of a Moebius-strip intersect?

In class the teacher was talking about normal vectors. $r = \langle x,y\rangle$ then the normal vector is

$$N\left(t\right) = \frac{T^{\prime}\left(t\right)}{||T^{\prime}\left(t\right)||}$$

where $T\left(t\right) = \frac{r^{\prime}}{||r^{\prime}||}$

Now, one of the naughty students asked the question in the title.

I thought it was a cool question so I am asking here.

ps. I have no idea, still trying to figure out how the unit norm vector needs the 2nd derivative of the tangent vector T.

-
The question makes no sense. You gave the definition of the normal vector for a parametrised curve in $\mathbb{R}^2$; this is not the same formula (but it is similar) as that which describes the normal vector for a surface. The problem with the Moebius strip is that depending on parametrisation (in your case $t\mapsto -t$) there are two distinct normal vectors, and for the Moebius strip there is no consistent choice of a global parametrisation that chooses exactly one of the two normal vectors. I don't see what this has to do with "the normal vector intersecting." –  Willie Wong Oct 31 '12 at 8:50
Also, what do you want the normal vector to intersect? Usually intersection requires two objects. And a vector by itself (thinking in terms of a directed line segment) quite clearly cannot self intersect (unlike a curve). –  Willie Wong Oct 31 '12 at 8:51
@WillieWong Thanks, I didn't realize that, you have a nice and informative response. –  yiyi Oct 31 '12 at 12:21

Note that your definition of the normal vector along a curve $\gamma$ does not work at points where the curvature is zero. In stretches of $\gamma$ where the curvature is nonzero your normal vector points inward, which might be to the right or to the left vs. the forward direction.
If $\gamma$ is given as $$\gamma:\ t\mapsto{\bf r}(t)=\bigl(x(t),y(t)\bigr)$$ one might also define the normal vector by $${\bf n}(t):={\bigl(-\dot y(t),\dot x(t)\bigr)\over\bigl|\dot{\bf r}(t)\bigr|}\ .$$ In this way the normal vector is defined at all points of $\gamma$ and points to the left. Now the curvature, when nonzero, has a sign.