Answer
$$s = \sqrt {29} \pi $$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {2\sin t,5t,2\cos t} \right\rangle ,{\text{ }}\left[ {0,\pi } \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {2\sin t,5t,2\cos t} \right\rangle } \right] \cr
& {\bf{r}}'\left( t \right) = \left\langle {2\cos t,5, - 2\sin t} \right\rangle \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {2\cos t} \right)}^2} + {{\left( 5 \right)}^2} + {{\left( { - 2\sin t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{{\cos }^2}t + 25 + 4{{\sin }^2}t} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4 + 25} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {29} \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^\pi {\sqrt {29} } dt \cr
& {\text{Integrate }} \cr
& s = \left[ {\sqrt {29} t} \right]_0^\pi \cr
& s = \sqrt {29} \pi \cr
& \cr
& {\text{Graph}} \cr} $$
