Answer
(a) See the figure attached
(b) We verify:
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\sin u\sin \frac{u}{2}, - \cos \frac{u}{2}} \right)$
(c) We show that ${\bf{N}}\left( {u,0} \right)$ varies continuously, however $S$ is not orientable.
Work Step by Step
(a) We plot the parametrization
$G\left( {u,v} \right) = \left( {\left( {1 + v\cos \frac{u}{2}} \right)\cos u,\left( {1 + v\cos \frac{u}{2}} \right)\sin u,v\sin \frac{u}{2}} \right)$
for $0 \le u \le 2\pi $, $ - \frac{1}{2} \le v \le \frac{1}{2}$.
This is a Möbius strip as is seen in the figure attached.
(b) We have
$G\left( {u,v} \right) = \left( {\left( {1 + v\cos \frac{u}{2}} \right)\cos u,\left( {1 + v\cos \frac{u}{2}} \right)\sin u,v\sin \frac{u}{2}} \right)$
Using a computer algebra system, we obtain
${{\bf{T}}_u} = ( - \frac{1}{2}v\cos u\sin \frac{u}{2} - \left( {1 + v\cos \frac{u}{2}} \right)\sin u,$
$ - \frac{1}{2}v\sin u\sin \frac{u}{2} + \left( {1 + v\cos \frac{u}{2}} \right)\cos u,\frac{1}{2}v\cos \frac{u}{2})$
${{\bf{T}}_v} = \left( {\cos \frac{u}{2}\cos u,\cos \frac{u}{2}\sin u,\sin \frac{u}{2}} \right)$
${\bf{N}}\left( {u,v} \right) = {{\bf{T}}_u} \times {{\bf{T}}_v} = (\frac{1}{2}\left( { - v\cos \frac{u}{2} + 2\cos u + v\cos \frac{{3u}}{2}} \right)\sin \frac{u}{2},$
$\frac{1}{4}\left( {v + 2\cos \frac{u}{2} + 2v\cos u - 2\cos \frac{{3u}}{2} - v\cos 2u} \right),$
$ - \cos \frac{u}{2}\left( {1 + v\cos \frac{u}{2}} \right))$
Setting $v=0$, we obtain the unit circle $G\left( {u,0} \right) = \left( {\cos u,\sin u,0} \right)$. Thus,
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\left( {\cos \frac{u}{2} - \cos \frac{{3u}}{2}} \right), - \cos \frac{u}{2}} \right)$
We can write $\cos \frac{{3u}}{2} = \cos \left( {u + \frac{u}{2}} \right)$. Using the trigonometric identity:
$\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$
we get
$\cos \left( {u + \frac{u}{2}} \right) = \cos u\cos \frac{u}{2} - \sin u\sin \frac{u}{2}$
So,
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\left( {\cos \frac{u}{2} - \cos u\cos \frac{u}{2} + \sin u\sin \frac{u}{2}} \right), - \cos \frac{u}{2}} \right)$
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\left( {\cos \frac{u}{2}\left( {1 - \cos u} \right) + \sin u\sin \frac{u}{2}} \right), - \cos \frac{u}{2}} \right)$
Using the trigonometric identity:
$\cos 2A = 1 - 2{\sin ^2}A$
we get
$\cos u = 1 - 2{\sin ^2}\frac{u}{2}$
So,
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\left( {\cos \frac{u}{2}\left( {2{{\sin }^2}\frac{u}{2}} \right) + \sin u\sin \frac{u}{2}} \right), - \cos \frac{u}{2}} \right)$
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\left( {2\cos \frac{u}{2}{{\sin }^2}\frac{u}{2} + \sin u\sin \frac{u}{2}} \right), - \cos \frac{u}{2}} \right)$
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\sin \frac{u}{2}\left( {2\cos \frac{u}{2}\sin \frac{u}{2} + \sin u} \right), - \cos \frac{u}{2}} \right)$
Since $\sin u = 2\cos \frac{u}{2}\sin \frac{u}{2}$, so
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\frac{1}{2}\sin \frac{u}{2}\left( {2\sin u} \right), - \cos \frac{u}{2}} \right)$
Finally, the normal vector along this circle is
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\sin u\sin \frac{u}{2}, - \cos \frac{u}{2}} \right)$
(c) From part (b), we obtain the normal vector along the unit circle $G\left( {u,0} \right) = \left( {\cos u,\sin u,0} \right)$:
${\bf{N}}\left( {u,0} \right) = \left( {\cos u\sin \frac{u}{2},\sin u\sin \frac{u}{2}, - \cos \frac{u}{2}} \right)$
Since sine and cosine functions are continues, ${\bf{N}}\left( {u,0} \right)$ varies continuously.
However,
${\bf{N}}\left( {0,0} \right) = \left( {0,0, - 1} \right)$
and
${\bf{N}}\left( {2\pi ,0} \right) = \left( {0,0,1} \right)$,
we get ${\bf{N}}\left( {2\pi ,0} \right) = - {\bf{N}}\left( {0,0} \right)$. This shows that $S$ is not orientable.
