Answer
The normal vector ${\bf{N}}\left( t \right)$ at $t = \pi $ is
${\bf{N}}\left( \pi \right) = \left( {0, - 1} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {t - \sin t,1 - \cos t} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {1 - \cos t,\sin t} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {\sin t,\cos t} \right)$.
Let ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$. So,
${\rm{v}}\left( t \right) = \sqrt {{{\left( {1 - \cos t} \right)}^2} + {{\sin }^2}t} = \sqrt {2 - 2\cos t} $
$v'\left( t \right) = \frac{1}{{2\sqrt {2 - 2\cos t} }}2\sin t = \frac{{\sin t}}{{\sqrt {2 - 2\cos t} }}$
Evaluate $v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ at $t = \pi $:
$v\left( \pi \right){\bf{r}}{\rm{''}}\left( \pi \right) - v'\left( \pi \right){\bf{r}}'\left( \pi \right) = 2\left( {0, - 1} \right) - 0\left( {2,0} \right)$
$v\left( \pi \right){\bf{r}}{\rm{''}}\left( \pi \right) - v'\left( \pi \right){\bf{r}}'\left( \pi \right) = \left( {0, - 2} \right)$
By Eq. (12), the normal vector ${\bf{N}}\left( t \right)$ at $t = \pi $ is given by
${\bf{N}}\left( \pi \right) = \frac{{v\left( \pi \right){\bf{r}}{\rm{''}}\left( \pi \right) - v'\left( \pi \right){\bf{r}}'\left( \pi \right)}}{{||v\left( \pi \right){\bf{r}}{\rm{''}}\left( \pi \right) - v'\left( \pi \right){\bf{r}}'\left( \pi \right)||}}$
${\bf{N}}\left( \pi \right) = \frac{{\left( {0, - 2} \right)}}{{||\left( {0, - 2} \right)||}} = \frac{1}{2}\left( {0, - 2} \right) = \left( {0, - 1} \right)$
