Möbius strip parametrization

Question

This is from Pressley’s book. How to arrive at the parametric form using rotation matrices? The rotation about $z-axis$ can be obtained from basic rotation matrix but what about the rotation about the plane containing the point $P$ and the $z-axis$? Is second rotation a basic rotation about $y-axis$? EDIT: After applying rotation of $\theta$ about $z-axis$ and $\theta/2$ about $y-axis$ for the point $\left(1,0,t\right)$ I got $\left(cos(\theta)cos(\theta/2)-tsin(\theta/2), -sin(\theta), cos(\theta)sin(\theta/2)+tcos(\theta/2)\right)$. What am I missing?

Answer 1

We have a combination of two rotations: In time $\Theta = 2\pi$ the segment $\mathcal L$ rotates with speed $1/2$ in the $xz$-plane (which is spanned by the $z$-axis and $\mathcal L$) around its midpoint $P = (1,0,0)$, and the $xz$-plane rotates with speed $1$ around the $z$-axis.

Note that we must assume that the segment $\mathcal L$ has the form $\{(1,0)\} \times [-r,r]$ with $r < 2$ (otherwise $\mathcal L$ would intersect $\mathcal C$ in two points at rotation time $\theta = \pi$).

A plane rotation by an angle $\phi$ around the origin is given by the matrix $$A(\phi) = \begin{pmatrix}\cos \phi & -\sin \phi\\\ \sin \phi & \cos \phi \end{pmatrix}$$ A rotation in the $xz$-plane only affects the first and third coordinate of a vector, in particular we see that the vector $(0,0,t)$ is rotated to $(-t\sin \phi,0,t\cos \phi)$.

At time $\theta \in [0,2\pi]$ the point $P_t = (1,0,t) = P + (0,0,t) \in \mathcal L$ has reached the position $P_t(\theta) = P + (-t\sin(\theta/2),0,t\cos(\theta/2) ) = (1 - t\sin(\theta/2),0,t\cos(\theta/2))$.

The rotation of the point $P_t(\theta) = (P_1,0,P_3)$ around the $z$-axis only affects the first two coordinates and at time $\theta \in [0,2\pi]$ we reach $$P_t'(\theta) = (P_1\cos \theta,P_1\sin \theta ,P_3) = ((1 - t\sin(\theta/2))\cos \theta,(1 - t\sin(\theta/2))\sin \theta,t\cos(\theta/2)) .$$ In particular you see that $P_t'(2\pi) = (1,0,-t)$.