Answer
The normal vector ${\bf{N}}\left( t \right)$ at $t=1$ is
${\bf{N}}\left( 1 \right) = \left( { - \frac{3}{{\sqrt {13} }},\frac{2}{{\sqrt {13} }}} \right)$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$. The derivatives are ${\bf{r}}'\left( t \right) = \left( {2t,3{t^2}} \right)$ and ${\bf{r}}{\rm{''}}\left( t \right) = \left( {2,6t} \right)$.
Let ${\rm{v}}\left( t \right) = ||{\bf{r}}'\left( t \right)||$. So,
${\rm{v}}\left( t \right) = \sqrt {4{t^2} + 9{t^4}} = t\sqrt {4 + 9{t^2}} $
$v'\left( t \right) = \frac{{8t + 36{t^3}}}{{2\sqrt {4{t^2} + 9{t^4}} }} = \frac{{4 + 18{t^2}}}{{\sqrt {4 + 9{t^2}} }}$
Evaluate $v\left( t \right){\bf{r}}{\rm{''}}\left( t \right) - v'\left( t \right){\bf{r}}'\left( t \right)$ at $t=1$:
$v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right) = \sqrt {13} \left( {2,6} \right) - \frac{{22}}{{\sqrt {13} }}\left( {2,3} \right)$
$v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right) = \left( {2\sqrt {13} - \frac{{44}}{{\sqrt {13} }},6\sqrt {13} - \frac{{66}}{{\sqrt {13} }}} \right)$
By Eq. (12), the normal vector ${\bf{N}}\left( t \right)$ at $t=1$ is given by
${\bf{N}}\left( 1 \right) = \frac{{v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right)}}{{||v\left( 1 \right){\bf{r}}{\rm{''}}\left( 1 \right) - v'\left( 1 \right){\bf{r}}'\left( 1 \right)||}}$
${\bf{N}}\left( 1 \right) = \frac{{\left( {2\sqrt {13} - \frac{{44}}{{\sqrt {13} }},6\sqrt {13} - \frac{{66}}{{\sqrt {13} }}} \right)}}{{||\left( {2\sqrt {13} - \frac{{44}}{{\sqrt {13} }},6\sqrt {13} - \frac{{66}}{{\sqrt {13} }}} \right)||}}$
${\bf{N}}\left( 1 \right) = \frac{{\left( {2\sqrt {13} - \frac{{44}}{{\sqrt {13} }},6\sqrt {13} - \frac{{66}}{{\sqrt {13} }}} \right)}}{{\sqrt {\left( {2\sqrt {13} - \frac{{44}}{{\sqrt {13} }},6\sqrt {13} - \frac{{66}}{{\sqrt {13} }}} \right)\cdot\left( {2\sqrt {13} - \frac{{44}}{{\sqrt {13} }},6\sqrt {13} - \frac{{66}}{{\sqrt {13} }}} \right)} }}$
${\bf{N}}\left( 1 \right) = \left( { - \frac{3}{{\sqrt {13} }},\frac{2}{{\sqrt {13} }}} \right)$
