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