Answer
$$s = 20\pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 10\cos t{\bf{i}} + 10\sin t{\bf{j}},{\text{ }}\left[ {0,2\pi } \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {10\cos t{\bf{i}} + 10\sin t{\bf{j}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 10\sin t{\bf{i}} + 10\cos t{\bf{j}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 10\sin t} \right)}^2} + {{\left( {10\cos t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {100{{\sin }^2}t + 100{{\cos }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {100} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = 10 \cr
& {\text{Find the arc length }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^{2\pi } {10} dt \cr
& {\text{Integrate}} \cr
& s = \left[ {10t} \right]_0^{2\pi } \cr
& s = 10\left[ {2\pi - 0} \right] \cr
& s = 20\pi \cr
& \cr
& {\text{Graph}} \cr} $$
