Answer
$$L = \sqrt 2 \left( {{e^{\pi /2}} - 1} \right)$$
Work Step by Step
$$\eqalign{
& {\text{Let }}x = {e^t}\cos t,{\text{ }}y = {e^t}\sin t,{\text{ }}\left( {} \right) \cr
& {\text{Find the arc length using }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} } dt \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}\cos t} \right] = {e^t}\cos t - {e^t}\sin t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {{e^t}\sin t} \right] = {e^t}\sin t + {e^t}\cos t \cr
& L = \int_0^{\pi /2} {\sqrt {{{\left( {{e^t}\cos t - {e^t}\sin t} \right)}^2} + {{\left( {{e^t}\sin t + {e^t}\cos t} \right)}^2}} dt} \cr
& {\text{Simplifying}} \cr
& L = \int_0^{\pi /2} {\sqrt {{e^{2t}} - 2{e^{2t}}\sin t\cos t + {e^{2t}} + 2{e^{2t}}\sin t\cos t} dt} \cr
& L = \int_0^{\pi /2} {\sqrt {2{e^{2t}}} dt} \cr
& L = \sqrt 2 \int_0^{\pi /2} {{e^t}dt} \cr
& {\text{Integrating}} \cr
& L = \sqrt 2 \left[ {{e^t}} \right]_0^{\pi /2} \cr
& L = \sqrt 2 \left( {{e^{\pi /2}} - 1} \right) \cr} $$
