Answer
$${\bf{T}}\left( t \right) = \frac{{\left\langle {1,t} \right\rangle }}{{\sqrt {1 + {t^2}} }}{\text{ and }}{\bf{N}}\left( t \right) = \frac{{\left\langle { - t,1} \right\rangle }}{{\sqrt {1 + {t^2}} }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {\frac{{{t^2}}}{2},\frac{{{t^3}}}{3}} \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 {t,{t^2}} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {{t^2} + {t^4}} \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|}} = \frac{{\left\langle {t,{t^2}} \right\rangle }}{{\sqrt {{t^2} + {t^4}} }} \cr
& {\bf{T}}\left( t \right) = \frac{{\left\langle {t,{t^2}} \right\rangle }}{{t\sqrt {1 + {t^2}} }} = \frac{{\left\langle {1,t} \right\rangle }}{{\sqrt {1 + {t^2}} }} \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{1}{{\sqrt {1 + {t^2}} }},\frac{t}{{\sqrt {1 + {t^2}} }}} \right\rangle \cr
& {\bf{T}}'\left( t \right) = \left\langle { - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}},\frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}} \right\rangle \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {\frac{{{t^2}}}{{{{\left( {1 + {t^2}} \right)}^3}}} + \frac{1}{{{{\left( {1 + {t^2}} \right)}^3}}}} \cr
& \left| {{\bf{T}}'\left( t \right)} \right| = \sqrt {\frac{{{t^2} + 1}}{{{{\left( {1 + {t^2}} \right)}^3}}}} = \sqrt {\frac{1}{{{{\left( {1 + {t^2}} \right)}^2}}}} = \frac{1}{{1 + {t^2}}} \cr
& {\bf{N}}\left( t \right) = \frac{{{\bf{T}}'\left( t \right)}}{{\left| {{\bf{T}}'\left( t \right)} \right|}} = \left( {1 + {t^2}} \right)\left\langle { - \frac{t}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}},\frac{1}{{{{\left( {1 + {t^2}} \right)}^{3/2}}}}} \right\rangle \cr
& {\bf{N}}\left( t \right) = \left\langle { - \frac{t}{{\sqrt {1 + {t^2}} }},\frac{1}{{\sqrt {1 + {t^2}} }}} \right\rangle \cr
& {\bf{N}}\left( t \right) = \frac{{\left\langle { - t,1} \right\rangle }}{{\sqrt {1 + {t^2}} }} \cr} $$
