Answer
The normal vector at $t = {\pi ^{1/3}}$ is
${\bf{N}}\left( {{\pi ^{1/3}}} \right) = \left( {\frac{1}{2}, - \frac{1}{2}\sqrt 3 } \right)$
Work Step by Step
We have the parametrization of the clothoid ${\bf{r}}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right)$, where
$x\left( t \right) = \mathop \smallint \limits_0^t \sin \frac{{{u^3}}}{3}{\rm{d}}u$, ${\ \ }$ $y\left( t \right) = \mathop \smallint \limits_0^t \cos \frac{{{u^3}}}{3}{\rm{d}}u$
By the Fundamental Theorem of Calculus:
$x'\left( t \right) = \sin \frac{{{t^3}}}{3}$, ${\ \ }$ $y'\left( t \right) = \cos \frac{{{t^3}}}{3}$
So, the tangent vector is
${\bf{r}}'\left( t \right) = \left( {x'\left( t \right),y'\left( t \right)} \right) = \left( {\sin \frac{{{t^3}}}{3},\cos \frac{{{t^3}}}{3}} \right)$
The unit tangent vector is
${\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{||{\bf{r}}'\left( t \right)||}} = \frac{{\left( {\sin \frac{{{t^3}}}{3},\cos \frac{{{t^3}}}{3}} \right)}}{{\sqrt {\left( {\sin \frac{{{t^3}}}{3},\cos \frac{{{t^3}}}{3}} \right)\cdot\left( {\sin \frac{{{t^3}}}{3},\cos \frac{{{t^3}}}{3}} \right)} }}$
${\bf{T}}\left( t \right) = \frac{{\left( {\sin \frac{{{t^3}}}{3},\cos \frac{{{t^3}}}{3}} \right)}}{{\sqrt {{{\sin }^2}\frac{{{t^3}}}{3} + {{\cos }^2}\frac{{{t^3}}}{3}} }}$
Since ${\sin ^2}\frac{{{t^3}}}{3} + {\cos ^2}\frac{{{t^3}}}{3} = 1$, we get ${\bf{T}}\left( t \right) = \left( {\sin \frac{{{t^3}}}{3},\cos \frac{{{t^3}}}{3}} \right)$.
We evaluate the derivative of ${\bf{T}}\left( t \right)$:
${\bf{T}}'\left( t \right) = \left( {{t^2}\cos \frac{{{t^3}}}{3}, - {t^2}\sin \frac{{{t^3}}}{3}} \right)$
At $t = {\pi ^{1/3}}$, we get
${\bf{T}}'\left( {{\pi ^{1/3}}} \right) = \left( {\frac{1}{2}{\pi ^{2/3}}, - \frac{1}{2}\sqrt 3 {\pi ^{2/3}}} \right)$
The normal vector at $t = {\pi ^{1/3}}$:
${\bf{N}}\left( {{\pi ^{1/3}}} \right) = \frac{{{\bf{T}}'\left( {{\pi ^{1/3}}} \right)}}{{||{\bf{T}}'\left( {{\pi ^{1/3}}} \right)||}}$
${\bf{N}}\left( {{\pi ^{1/3}}} \right) = \frac{{\left( {\frac{1}{2}{\pi ^{2/3}}, - \frac{1}{2}\sqrt 3 {\pi ^{2/3}}} \right)}}{{\sqrt {\left( {\frac{1}{2}{\pi ^{2/3}}, - \frac{1}{2}\sqrt 3 {\pi ^{2/3}}} \right)\cdot\left( {\frac{1}{2}{\pi ^{2/3}}, - \frac{1}{2}\sqrt 3 {\pi ^{2/3}}} \right)} }}$
${\bf{N}}\left( {{\pi ^{1/3}}} \right) = \frac{{\left( {\frac{1}{2}{\pi ^{2/3}}, - \frac{1}{2}\sqrt 3 {\pi ^{2/3}}} \right)}}{{\sqrt {\frac{1}{4}{\pi ^{4/3}} + \frac{3}{4}{\pi ^{4/3}}} }} = \frac{1}{{{\pi ^{2/3}}}}\left( {\frac{1}{2}{\pi ^{2/3}}, - \frac{1}{2}\sqrt 3 {\pi ^{2/3}}} \right)$
${\bf{N}}\left( {{\pi ^{1/3}}} \right) = \left( {\frac{1}{2}, - \frac{1}{2}\sqrt 3 } \right)$
