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