Answer
$$s = \frac{{\sqrt {65} \pi }}{2}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = 8\cos t{\bf{i}} + 8\sin t{\bf{j}} + t{\bf{k}},{\text{ }}\left[ {0,\frac{\pi }{2}} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {8\cos t{\bf{i}} + 8\sin t{\bf{j}} + t{\bf{k}}} \right] \cr
& {\bf{r}}'\left( t \right) = - 8\sin t{\bf{i}} + 8\cos t{\bf{j}} + {\bf{k}} \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( { - 8\sin t} \right)}^2} + {{\left( {8\cos t} \right)}^2} + {{\left( 1 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {64{{\sin }^2}t + 64{{\cos }^2}t + 1} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {65} \cr
& {\text{Find the arc length }}s{\text{, using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^{\pi /2} {\sqrt {65} } dt \cr
& {\text{Integrate }} \cr
& s = \sqrt {65} \left[ t \right]_0^{\pi /2} \cr
& s = \frac{{\sqrt {65} \pi }}{2} \cr
& \cr
& {\text{Graph}} \cr} $$
