Answer
$$L = \frac{{2\left( {73\sqrt {73} - 1} \right)}}{{27}}$$
Work Step by Step
$$\eqalign{ & {\bf{r}}\left( t \right) = \left\langle {2{t^{9/2}},{t^3}} \right\rangle ,\,\,\,\,\,for\,\,\,\,\,0 \leqslant t \leqslant 2 \cr & {\text{Calculate }}{\bf{r}}'\left( t \right){\text{ and }}\left| {{\bf{r}}'\left( t \right)} \right| \cr & {\bf{r}}'\left( t \right) = \left\langle {2\left( {\frac{9}{2}} \right){t^{7/2}},3{t^2}} \right\rangle \cr & {\bf{r}}'\left( t \right) = \left\langle {9{t^{7/2}},3{t^2}} \right\rangle \cr & \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {{{\left( {9{t^{7/2}}} \right)}^2} + {{\left( {3{t^2}} \right)}^2}} \cr & \left| {{\bf{r}}'\left( t \right)} \right| = \sqrt {81{t^7} + 9{t^4}} \cr & {\text{The Arc Length is }}\int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr & L = \int_0^2 {\sqrt {81{t^7} + 9{t^4}} } dt \cr & L = \int_0^2 {3{t^2}\sqrt {9{t^3} + 1} } dt \cr & L = \frac{1}{9}\int_0^2 {{{\left( {9{t^3} + 1} \right)}^{1/2}}} \left( {27{t^2}} \right)dt \cr & {\text{Integrate}} \cr & L = \frac{1}{9}\left[ {\frac{{{{\left( {9{t^3} + 1} \right)}^{3/2}}}}{{3/2}}} \right]_0^2 \cr & L = \frac{2}{{27}}\left[ {{{\left( {9{{\left( 2 \right)}^3} + 1} \right)}^{3/2}} - {{\left( {9{{\left( 0 \right)}^3} + 1} \right)}^{3/2}}} \right] \cr & L = \frac{2}{{27}}\left[ {{{\left( {73} \right)}^{3/2}} - {{\left( 1 \right)}^{3/2}}} \right] \cr & L = \frac{{2\left( {73\sqrt {73} - 1} \right)}}{{27}} \cr} $$
