Answer
${a_{\bf{T}}} = \frac{{4 + 18{t^2}}}{{\sqrt {4 + 9{t^2}} }}$
${a_{\bf{N}}} = \frac{{6t}}{{\sqrt {4 + 9{t^2}} }}$
Work Step by Step
We have ${\bf{r}}\left( t \right) = \left( {{t^2},{t^3}} \right)$. The velocity and acceleration vectors are ${\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) = \left( {2t,3{t^2}} \right)$ and ${\bf{a}}\left( t \right) = {\bf{r}}{\rm{''}}\left( t \right) = \left( {2,6t} \right)$, respectively.
By Eq. (2) of Theorem 1 we have
${a_{\bf{T}}} = \frac{{{\bf{a}}\cdot{\bf{v}}}}{{||{\bf{v}}||}} = \frac{{\left( {2,6t} \right)\cdot\left( {2t,3{t^2}} \right)}}{{\sqrt {\left( {2t,3{t^2}} \right)\cdot\left( {2t,3{t^2}} \right)} }}$
${a_{\bf{T}}} = \frac{1}{{t\sqrt {4 + 9{t^2}} }}\left( {4t + 18{t^3}} \right) = \frac{{4 + 18{t^2}}}{{\sqrt {4 + 9{t^2}} }}$
By Eq. (2) of Theorem 1 we have
${a_{\bf{N}}} = \sqrt {||{\bf{a}}|{|^2} - |{a_{\bf{T}}}{|^2}} $
${a_{\bf{N}}} = \sqrt {\left( {2,6t} \right)\cdot\left( {2,6t} \right) - {{\left( {\frac{{4 + 18{t^2}}}{{\sqrt {4 + 9{t^2}} }}} \right)}^2}} $
${a_{\bf{N}}} = \sqrt {4 + 36{t^2} - {{\left( {\frac{{4 + 18{t^2}}}{{\sqrt {4 + 9{t^2}} }}} \right)}^2}} = \frac{{6t}}{{\sqrt {4 + 9{t^2}} }}$
