Answer: The arc length is \(\sqrt{2}\left(e^{\pi/2}-1\right)\).

For a parametric curve, the arc length is

\(s=\int_a^b \sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Here \(x=e^t\cos t\) and \(y=e^t\sin t\), for \(0\le t\le \frac{\pi}{2}\).

Differentiate:

\(\frac{dx}{dt}=e^t\cos t-e^t\sin t\),

\(\frac{dy}{dt}=e^t\sin t+e^t\cos t\).

Then

\(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2=e^{2t}(\cos t-\sin t)^2+e^{2t}(\sin t+\cos t)^2\)

\(=2e^{2t}(\cos^2 t+\sin^2 t)=2e^{2t}.\)

So

\(s=\int_0^{\pi/2} \sqrt{2e^{2t}}\,dt=\int_0^{\pi/2} \sqrt{2}\,e^t\,dt\)

\(=\left[\sqrt{2}e^t\right]_0^{\pi/2}=\sqrt{2}\left(e^{\pi/2}-1\right).\)