Answer
The normal vector at $t = \sqrt \pi $ is
${\bf{N}}\left( {\sqrt \pi } \right) = \left( {0, - 1} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$, where
$x\left( t \right) = \mathop \smallint \limits_0^t \sin \frac{{{u^2}}}{2}{\rm{d}}u$, ${\ \ \ }$ $y\left( t \right) = \mathop \smallint \limits_0^t \cos \frac{{{u^2}}}{2}{\rm{d}}u$
By the Fundamental Theorem of Calculus:
$x'\left( t \right) = \sin \frac{{{t^2}}}{2}$, ${\ \ \ }$ $y'\left( t \right) = \cos \frac{{{t^2}}}{2}$
Thus, the tangent vector is
${\bf{r}}'\left( t \right) = \left( {x'\left( t \right),y'\left( t \right)} \right) = \left( {\sin \frac{{{t^2}}}{2},\cos \frac{{{t^2}}}{2}} \right)$
We find the unit tangent vector ${\bf{T}}\left( t \right)$:
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {\sin \frac{{{t^2}}}{2},\cos \frac{{{t^2}}}{2}} \right)}}{{\sqrt {\left( {\sin \frac{{{t^2}}}{2},\cos \frac{{{t^2}}}{2}} \right)\cdot\left( {\sin \frac{{{t^2}}}{2},\cos \frac{{{t^2}}}{2}} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {\sin \frac{{{t^2}}}{2},\cos \frac{{{t^2}}}{2}} \right)}}{{\sqrt {{{\sin }^2}\frac{{{t^2}}}{2} + {{\cos }^2}\frac{{{t^2}}}{2}} }}$
Since ${\sin ^2}\frac{{{t^2}}}{2} + {\cos ^2}\frac{{{t^2}}}{2} = 1$, so ${\bf{T}}\left( t \right) = \left( {\sin \frac{{{t^2}}}{2},\cos \frac{{{t^2}}}{2}} \right)$.
Taking the derivative, we obtain
${\bf{T}}'\left( t \right) = \left( {t\cos \frac{{{t^2}}}{2}, - t\sin \frac{{{t^2}}}{2}} \right) = t\left( {\cos \frac{{{t^2}}}{2}, - \sin \frac{{{t^2}}}{2}} \right)$
So,
$||{\bf{T}}'\left( t \right)|| = \sqrt {{t^2}\left( {\cos \frac{{{t^2}}}{2}, - \sin \frac{{{t^2}}}{2}} \right)\cdot\left( {\cos \frac{{{t^2}}}{2}, - \sin \frac{{{t^2}}}{2}} \right)} $
$||{\bf{T}}'\left( t \right)|| = t\sqrt {{{\cos }^2}\frac{{{t^2}}}{2} + {{\sin }^2}\frac{{{t^2}}}{2}} = t$
By Eq. (6), the normal vector ${\bf{N}}\left( t \right)$ is given by
${\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{||{\bf{T}}'\left( t \right)||}}$
${\bf{N}}\left( t \right) = \frac{{t\left( {\cos \frac{{{t^2}}}{2}, - \sin \frac{{{t^2}}}{2}} \right)}}{t} = \left( {\cos \frac{{{t^2}}}{2}, - \sin \frac{{{t^2}}}{2}} \right)$
At $t = \sqrt \pi $, we get ${\bf{N}}\left( {\sqrt \pi } \right) = \left( {0, - 1} \right)$.
