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

For a parametric curve, the arc length from \(t=0\) to \(t=\pi/2\) is

\(s=\int_0^{\pi/2}\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt\).

Given \(x=e^t\cos t\) and \(y=e^t\sin t\),

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

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

So

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

Hence

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

Integrating,

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