Answer
$$L = \frac{{{\pi ^2}}}{8}$$
Work Step by Step
$$\eqalign{
& {\bf{r}}\left( t \right) = \left\langle {\cos t + t\sin t,\sin t - t\cos t} \right\rangle ,\,\,\,\,\,{\text{for }}0 \leqslant t \leqslant \pi /2 \cr
& {\text{find }}{\bf{r}}'\left( t \right) \cr
& {\bf{r}}'\left( t \right) = \frac{d}{{dt}}\left\langle {\cos t + t\sin t,\sin t - t\cos t} \right\rangle \cr
& {\text{use product rule for }}t\sin t{\text{ and }}t\cos t.{\text{ then}} \cr
& {\bf{r}}'\left( t \right) = \left\langle { - \sin t + t\cos t + \sin t,\cos t + t\sin t - \cos t} \right\rangle \cr
& {\bf{r}}'\left( t \right) = \left\langle {t\cos t,t\sin t} \right\rangle \cr
& {\text{use the Definition of Arc Length for Vector Functions }}\left( {{\text{see page 832}}} \right) \cr
& {\text{for a vector }}{\bf{r}}'\left( t \right) = \left\langle {f'\left( t \right),g'\left( t \right),h'\left( t \right)} \right\rangle \cr
& L = \int_a^b {\sqrt {f'{{\left( t \right)}^2} + g'{{\left( t \right)}^2} + h'{{\left( t \right)}^2}} dt} = \int_a^b {\left| {{\bf{r}}'\left( t \right)} \right|} dt \cr
& {\text{then}} \cr
& \,{\text{for }}0 \leqslant t \leqslant \pi /2 \to a = 0{\text{ and }}b = \pi /2.{\text{ then}} \cr
& L = \int_0^{\pi /2} {\left| {\left\langle {t\cos t,t\sin t} \right\rangle } \right|} dt \cr
& L = \int_0^{\pi /2} {\sqrt {{{\left( {t\cos t} \right)}^2} + {{\left( {t\sin t} \right)}^2}} } dt \cr
& {\text{simplifying}} \cr
& L = \int_0^{\pi /2} {\sqrt {{t^2}{{\cos }^2}t + {t^2}{{\sin }^2}t} } dt \cr
& L = \int_0^{\pi /2} {\sqrt {{t^2}\left( {{{\cos }^2}t + {{\sin }^2}t} \right)} } dt \cr
& L = \int_0^{\pi /2} {\sqrt {{t^2}\left( 1 \right)} } dt \cr
& L = \int_0^{\pi /2} t dt \cr
& {\text{integrating}} \cr
& L = \left( {\frac{{{t^2}}}{2}} \right)_0^{\pi /2} \cr
& L = \frac{{{{\left( {\pi /2} \right)}^2}}}{2} - \frac{{{{\left( 0 \right)}^2}}}{2} \cr
& {\text{simplifying}} \cr
& L = \frac{{{\pi ^2}}}{8} \cr} $$
