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