Answer
$$s = \frac{\pi }{4}\sqrt {{\pi ^2} + 1} + \frac{1}{4}\ln \left| {\pi + \sqrt {{\pi ^2} + 1} } \right|$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {2\left( {\sin t - t\cos t} \right),2\left( {\cos t + t\sin t} \right),t} \right\rangle ,{\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[ {\left\langle {2\left( {\sin t - t\cos t} \right),2\left( {\cos t + t\sin t} \right),t} \right\rangle } \right] \cr
& {\bf{r}}'\left( t \right) = \left\langle {2\left( {\cos t - \cos t + t\sin t} \right),2\left( { - \sin t + t\cos t + \sin t} \right),1} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {2t\sin t,2t\cos t,1} \right\rangle \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {2t\sin t} \right)}^2} + {{\left( {2t\cos t} \right)}^2} + {{\left( 1 \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2}{{\sin }^2}t + 4{t^2}{{\cos }^2}t + 1} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {4{t^2} + 1} \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 {4{t^2} + 1} } dt \cr
& s = \frac{1}{2}\int_0^{\pi /2} {\sqrt {{{\left( {2t} \right)}^2} + 1} \left( 2 \right)} dt \cr
& {\text{Integrate by tables formula 26}}{\text{.}} \cr
& \int {\sqrt {{u^2} + {a^2}} du} = \frac{1}{2}\left( {u\sqrt {{u^2} + {a^2}} + {a^2}\ln \left| {u + \sqrt {{u^2} + {a^2}} } \right|} \right) + C \cr
& s = \frac{1}{4}\left[ {2t\sqrt {4{t^2} + 1} + \ln \left| {2t + \sqrt {4{t^2} + 1} } \right|} \right]_0^{\pi /2} \cr
& {\text{Integrate }} \cr
& s = \frac{1}{4}\left[ {2\left( {\frac{\pi }{2}} \right)\sqrt {4{{\left( {\frac{\pi }{2}} \right)}^2} + 1} + \ln \left| {2\left( {\frac{\pi }{2}} \right) + \sqrt {4{{\left( {\frac{\pi }{2}} \right)}^2} + 1} } \right|} \right] \cr
& - \frac{1}{4}\left[ {2\left( 0 \right)\sqrt {4{{\left( 0 \right)}^2} + 1} + \ln \left| {2\left( 0 \right) + \sqrt {4{{\left( 0 \right)}^2} + 1} } \right|} \right] \cr
& s = \frac{1}{4}\left[ {\pi \sqrt {{\pi ^2} + 1} + \ln \left| {\pi + \sqrt {{\pi ^2} + 1} } \right|} \right] - \frac{1}{4}\left[ {\ln 1} \right] \cr
& s = \frac{\pi }{4}\sqrt {{\pi ^2} + 1} + \frac{1}{4}\ln \left| {\pi + \sqrt {{\pi ^2} + 1} } \right| \cr
& \cr
& {\text{Graph}} \cr} $$
