Answer
$${a_N} = \frac{{6t}}{{\sqrt {9{t^2} + 4} }}{\text{ and }}{a_T} = \frac{{18{t^2} + 4}}{{\sqrt {9{t^2} + 4} }}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{t^3},{t^2}} \right\rangle \cr
& {\text{Calculate }}{\bf{v}}\left( t \right){\text{, }}\left| {{\bf{v}}\left( t \right)} \right|{\text{ and }}{\bf{a}}\left( t \right) \cr
& {\bf{v}}\left( t \right) = {\bf{r}}'\left( t \right) \cr
& {\bf{v}}\left( t \right) = \left\langle {3{t^2},2t} \right\rangle \cr
& \left| {{\bf{v}}\left( t \right)} \right| = \sqrt {9{t^4} + 4{t^2}} = t\sqrt {9{t^2} + 4} \cr
& \cr
& {\bf{a}}\left( t \right) = {\bf{v}}'\left( t \right) \cr
& {\bf{a}}\left( t \right) = \left\langle {6t,2} \right\rangle \cr
& \cr
& {\text{Find the components of acceleration: }}{a_N}{\bf{N}} + {a_T}{\bf{T}}, \cr
& {\text{Where }}{a_N} = \kappa {\left| {\bf{v}} \right|^2} = \frac{{\left| {{\bf{v}} \times {\bf{a}}} \right|}}{{\left| {\bf{v}} \right|}}{\text{ and }}{a_T} = \frac{{{d^2}s}}{{d{t^2}}} = \frac{{{\bf{v}} \cdot {\bf{a}}}}{{\left| {\bf{v}} \right|}} \cr
& {\text{Then}}{\text{,}} \cr} $$
\[{\mathbf{v}} \times {\mathbf{a}} = \left| {\begin{array}{*{20}{c}}
{\mathbf{i}}&{\mathbf{j}}&{\mathbf{k}} \\
{3{t^2}}&{2t}&0 \\
{6t}&2&0
\end{array}} \right|\]
$$\eqalign{
& {\bf{v}} \times {\bf{a}} = 0{\bf{i}} - 0{\bf{j}}\, + \left( {6{t^2} - 12{t^2}} \right){\bf{k}} \cr
& {\bf{v}} \times {\bf{a}} = - 6{t^2}{\bf{k}} \cr
& \left| {{\bf{v}} \times {\bf{a}}} \right| = 6{t^2} \cr
& \cr
& {a_N} = \frac{{\left| {{\bf{v}} \times {\bf{a}}} \right|}}{{\left| {\bf{v}} \right|}} = \frac{{6{t^2}}}{{t\sqrt {9{t^2} + 4} }} \cr
& {a_N} = \frac{{6t}}{{\sqrt {9{t^2} + 4} }} \cr
& \cr
& and \cr
& \cr
& {\bf{v}} \cdot {\bf{a}} = \left\langle {3{t^2},2t} \right\rangle \cdot \left\langle {6t,2} \right\rangle \cr
& {\bf{v}} \cdot {\bf{a}} = 18{t^3} + 4t \cr
& {a_T} = \frac{{{\bf{v}} \cdot {\bf{a}}}}{{\left| {\bf{v}} \right|}} = \frac{{18{t^3} + 4t}}{{t\sqrt {9{t^2} + 4} }} \cr
& {a_T} = \frac{{18{t^2} + 4}}{{\sqrt {9{t^2} + 4} }} \cr
& \cr
& {a_N} = \frac{{6t}}{{\sqrt {9{t^2} + 4} }}{\text{ and }}{a_T} = \frac{{18{t^2} + 4}}{{\sqrt {9{t^2} + 4} }} \cr} $$
