Answer
$$L = \frac{{2\sqrt 2 - 1}}{3}$$
Work Step by Step
$$\eqalign{
& x = \frac{1}{3}{t^3},\,\,\,\,\,y = \frac{1}{2}{t^2},\,\,\,\,\left( {0 \leqslant t \leqslant 1} \right) \cr
& {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{3}{t^3}} \right] = {t^2} \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\frac{1}{2}{t^2}} \right] = t \cr
& {\text{Use the arc length formula }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr
& L = \int_0^1 {\sqrt {{{\left( {{t^2}} \right)}^2} + {{\left( t \right)}^2}} dt} \cr
& L = \int_0^1 {\sqrt {{t^4} + {t^2}} dt} \cr
& L = \int_0^1 {t\sqrt {{t^2} + 1} dt} \cr
& L = \frac{1}{2}\int_0^1 {\left( {2t} \right)\sqrt {{t^2} + 1} dt} \cr
& {\text{Integrating}} \cr
& L = \frac{1}{2}\left[ {\frac{{2{{\left( {{t^2} + 1} \right)}^{3/2}}}}{3}} \right]_0^1 \cr
& L = \frac{1}{3}\left[ {{{\left( {{t^2} + 1} \right)}^{3/2}}} \right]_0^1 \cr
& L = \frac{1}{3}\left[ {{{\left( {{1^2} + 1} \right)}^{3/2}} - {{\left( {{0^2} + 1} \right)}^{3/2}}} \right] \cr
& L = \frac{1}{3}\left[ {{2^{3/2}} - 1} \right] \cr
& L = \frac{{2\sqrt 2 - 1}}{3} \cr} $$
