Answer
$${\bf{N}}\left( 1 \right) = - \frac{3}{{\sqrt {10} }}{\bf{i}} + \frac{1}{{\sqrt {10} }}{\bf{j}}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 2t{\bf{i}} + 3{t^2}{\bf{j}},{\text{ }}t = 1 \cr
& {\text{Calculate }}{\bf{r}}{\text{'}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {2t{\bf{i}} + 3{t^2}{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = 2{\bf{i}} + 6t{\bf{j}} \cr
& {\text{Find the unit Tangent Vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{r}}'\left( t \right)}}{{\left\| {{\bf{r}}'\left( t \right)} \right\|}},{\text{ }}{\bf{r}}'\left( t \right) \ne 0 \cr
& {\bf{T}}\left( t \right) = \frac{{2{\bf{i}} + 6t{\bf{j}}}}{{\left\| {2{\bf{i}} + 6t{\bf{j}}} \right\|}} = \frac{{2{\bf{i}} + 6t{\bf{j}}}}{{\sqrt {4 + 36{t^2}} }} \cr
& {\text{Find }}{\bf{T}}'\left( t \right) \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left[ {\frac{2}{{\sqrt {4 + 36{t^2}} }}{\bf{i}} + \frac{{6t}}{{\sqrt {4 + 36{t^2}} }}{\bf{j}}} \right] \cr
& {\text{Differentiate by hand and replace}} \cr
& {\bf{T}}'\left( t \right) = - \frac{{9t}}{{{{\left( {9{t^2} + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{3}{{{{\left( {9{t^2} + 1} \right)}^{3/2}}}}{\bf{j}} \cr
& {\text{Evaluate }}{\bf{T}}'\left( t \right){\text{ at }}t = 1 \cr
& {\bf{T}}'\left( 1 \right) = - \frac{{9\left( 1 \right)}}{{{{\left( {9{{\left( 1 \right)}^2} + 1} \right)}^{3/2}}}}{\bf{i}} + \frac{3}{{{{\left( {9{{\left( 1 \right)}^2} + 1} \right)}^{3/2}}}}{\bf{j}} \cr
& {\bf{T}}'\left( 1 \right) = - \frac{9}{{{{10}^{3/2}}}}{\bf{i}} + \frac{3}{{{{10}^{3/2}}}}{\bf{j}} \cr
& {\text{Finding the Principal Unit Normal Vector}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left\| {{\bf{T}}'\left( t \right)} \right\|}} \cr
& {\text{Evaluate }}{\bf{N}}'\left( t \right){\text{ at }}t = 1 \cr
& {\bf{N}}\left( 1 \right) = \frac{{{\bf{T}}'\left( 1 \right)}}{{\left\| {{\bf{T}}'\left( 1 \right)} \right\|}} = \frac{{ - \frac{9}{{{{10}^{3/2}}}}{\bf{i}} + \frac{3}{{{{10}^{3/2}}}}{\bf{j}}}}{{\left\| { - \frac{9}{{{{10}^{3/2}}}}{\bf{i}} + \frac{3}{{{{10}^{3/2}}}}{\bf{j}}} \right\|}} \cr
& {\bf{N}}\left( 1 \right) = \frac{{ - \frac{9}{{{{10}^{3/2}}}}{\bf{i}} + \frac{3}{{{{10}^{3/2}}}}{\bf{j}}}}{{3/10}} \cr
& {\bf{N}}\left( 1 \right) = - \frac{3}{{\sqrt {10} }}{\bf{i}} + \frac{1}{{\sqrt {10} }}{\bf{j}} \cr} $$
