Answer
$$L = \frac{1}{9}\left( {\sqrt {148} - \sqrt 4 } \right)$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {{t^2},{t^3}} \right\rangle {\text{, for }}0 \leqslant t \leqslant 4 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {{t^2},{t^3}} \right\rangle \cr
& {\text{differentiating}} \cr
& {\bf{r}}'\left( t \right) = \left\langle {2t,3{t^2}} \right\rangle \cr
& {\text{Use the Definition of Arc Length for Vector Functions }} \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& L = \int_0^4 {\left| {\left\langle {2t,3{t^2}} \right\rangle } \right|} dt \cr
& L = \int_0^4 {\sqrt {{{\left( {2t} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^4 {\sqrt {4{t^2} + 9{t^4}} } dt \cr
& L = \int_0^4 {\sqrt {{t^2}\left( {4 + 9{t^2}} \right)} } dt \cr
& L = \int_0^4 {\sqrt {4 + 9{t^2}} \left( t \right)} dt \cr
& L = \frac{1}{{18}}\int_0^4 {\sqrt {4 + 9{t^2}} \left( {18t} \right)} dt \cr
& {\text{integrating}} \cr
& L = \frac{1}{{18}}\left( {\frac{{{{\left( {4 + 9{t^2}} \right)}^{1/2}}}}{{1/2}}} \right)_0^4 \cr
& L = \frac{1}{9}\left( {\sqrt {4 + 9{t^2}} } \right)_0^4 \cr
& L = \frac{1}{9}\left( {\sqrt {4 + 9{{\left( 4 \right)}^2}} - \sqrt {4 + 9{{\left( 0 \right)}^2}} } \right) \cr
& L = \frac{1}{9}\left( {\sqrt {148} - \sqrt 4 } \right) \cr} $$
