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