Answer
$$s = \frac{{\sqrt 5 }}{{16}}{\pi ^2}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {\cos t + t\sin t,\sin t - t\cos t,{t^2}} \right\rangle ,{\text{ }}\left[ {0,\frac{\pi }{2}} \right] \cr
& {\text{Differentiate }}{\bf{r}}\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left[ {\left\langle {\cos t + t\sin t,\sin t - t\cos t,{t^2}} \right\rangle } \right] \cr
& {\bf{r}}'\left( t \right) = \left\langle { - \sin t + \sin t + t\cos t,\cos t - \cos t + t\sin t,2t} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {t\cos t,t\sin t,2t} \right\rangle \cr
& {\text{Find }}\left\| {{\bf{r}}'\left( t \right)} \right\| \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{{\left( {t\cos t} \right)}^2} + {{\left( {t\sin t} \right)}^2} + {{\left( {2t} \right)}^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{t^2}{{\cos }^2}t + {t^2}{{\sin }^2}t + 4{t^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {{t^2}\left( {{{\cos }^2}t + {{\sin }^2}t} \right) + 4{t^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt {5{t^2}} \cr
& \left\| {{\bf{r}}'\left( t \right)} \right\| = \sqrt 5 t \cr
& {\text{Find the arc length }}s{\text{ using the formula }}s = \int_a^b {\left\| {{\bf{r}}'\left( t \right)} \right\|} dt \cr
& s = \int_0^{\pi /2} {\sqrt 5 t} dt \cr
& {\text{Integrate }} \cr
& s = \sqrt 5 \left[ {\frac{{{t^2}}}{2}} \right]_0^{\pi /2} \cr
& s = \frac{{\sqrt 5 }}{2}\left( {\frac{{{\pi ^2}}}{8}} \right) \cr
& s = \frac{{\sqrt 5 }}{{16}}{\pi ^2} \cr} $$
