Answer
$$s = 60$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 10{\cos ^3}t{\bf{i}} + 10{\sin ^3}t{\bf{j}},{\text{ }}\left[ {0,2\pi } \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {10{{\cos }^3}t{\bf{i}} + 10{{\sin }^3}t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 30{\cos ^2}t\sin t{\bf{i}} + 30{\sin ^2}t\cos t{\bf{j}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 30{{\cos }^2}t\sin t} \right)}^2} + {{\left( {30{{\sin }^2}t\cos t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {900{{\cos }^4}t{{\sin }^2}t + 900{{\sin }^4}t{{\cos }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {900{{\cos }^2}t{{\sin }^2}t\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = 30\left| {\cos t\sin t} \right| \cr
& {\text{Find the arc length }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& {\text{From the graph we can set the arc length from }}\left[ {0,\frac{\pi }{2}} \right]{\text{ as}} \cr
& s = 4\int_0^{\pi /2} {30\cos t\sin t} dt \cr
& {\text{Integrate}} \cr
& s = 120\left[ {\frac{{{{\sin }^2}t}}{2}} \right]_0^{\pi /2} \cr
& s = 60\left[ {{{\sin }^2}\left( {\frac{\pi }{2}} \right) - {{\sin }^2}\left( 0 \right)} \right] \cr
& s = 60 \cr
& \cr
& {\text{Graph}} \cr} $$
