Answer
$${\bf{T}}\left( t \right) = \frac{{\left\langle {2t,1} \right\rangle }}{{\sqrt {4{t^2} + 1} }}{\text{ and }}{\bf{N}}\left( t \right) = \frac{{\left\langle {1, - 2t} \right\rangle }}{{\sqrt {4{t^2} + 1} }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{t^2},t} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right){\text{ and }}\left| {{\bf{v}}\left( t \right)} \right| \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \left\langle {2t,1} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {4{t^2} + 1} \cr
& \cr
& {\text{Find the unit tangent vector }}{\bf{T}}\left( t \right) \cr
& {\bf{T}}\left( t \right) = \frac{{{\bf{v}}\left( t \right)}}{{\left| {{\bf{v}}\left( t \right)} \right|}} \cr
& {\bf{T}}\left( t \right) = \frac{{\left\langle {2t,1} \right\rangle }}{{\sqrt {4{t^2} + 1} }} \cr
& \cr
& {\text{Find the principal unit normal vector }}{\bf{N}}\left( t \right) \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} \cr
& {\bf{T}}'\left( t \right) = \frac{d}{{dt}}\left\langle {\frac{{2t}}{{\sqrt {4{t^2} + 1} }},\frac{1}{{\sqrt {4{t^2} + 1} }}} \right\rangle \cr
& {\bf{T}}'\left( t \right) = \left\langle {\frac{2}{{{{\left( {4{t^2} + 1} \right)}^{3/2}}}}, - \frac{{4t}}{{{{\left( {4{t^2} + 1} \right)}^{3/2}}}}} \right\rangle \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {\frac{{4 + 16{t^2}}}{{{{\left( {4{t^2} + 1} \right)}^3}}}} = \sqrt {\frac{{4\left( {4{t^2} + 1} \right)}}{{{{\left( {4{t^2} + 1} \right)}^3}}}} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = 2\sqrt {\frac{1}{{{{\left( {4{t^2} + 1} \right)}^2}}}} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \frac{2}{{4{t^2} + 1}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} = \frac{{4{t^2} + 1}}{2}\left\langle {\frac{2}{{{{\left( {4{t^2} + 1} \right)}^{3/2}}}}, - \frac{{4t}}{{{{\left( {4{t^2} + 1} \right)}^{3/2}}}}} \right\rangle \cr
& {\bf{N}}\left( t \right) = \left\langle {\frac{1}{{\sqrt {4{t^2} + 1} }}, - \frac{{2t}}{{\sqrt {4{t^2} + 1} }}} \right\rangle \cr
& {\bf{N}}\left( t \right) = \frac{{\left\langle {1, - 2t} \right\rangle }}{{\sqrt {4{t^2} + 1} }} \cr} $$
