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