Answer
$1.120$
Work Step by Step
$x=e^{-t}cost$, $y=e^{-t}sint$, $0\leq t \leq \frac{\pi}{2}$
The arc length of a parametric curve is calculated using the formula:
$L=\int_{a}^{b}\sqrt ((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)dt$
$\frac{dx}{dt}=-e^{-t}cost-e^{-t}sint$
$\frac{dy}{dt}=-e^{-t}sint+e^{-t}cost$
$L=\int_{0}^{\frac{\pi}{2}}\sqrt ((-e^{-t}cost-e^{-t}sint)^2+(-e^{-t}sint+e^{-t}cost)^2)dt$
$L=\int_{0}^{\frac{\pi}{2}}\sqrt (e^{-2t}cos^2t+2e^{-2t}costsint+e^{-2t}sin^2t+e^{-2t}sin^2t-2e^{-2t}costsint+e^{-2t}cos^2t)dt$
$L=\int_{0}^{\frac{\pi}{2}}\sqrt (2e^{-2t}cos^2t+2e^{-2t}sin^2t)dt$
$L=\int_{0}^{\frac{\pi}{2}}\sqrt (2e^{-2t}(cos^2t+sin^2t))dt$
By the Pythagorean identity $sin^2x+cos^2x=1$:
$L=\int_{0}^{\frac{\pi}{2}}\sqrt (2e^{-2t})dt$
$L=\sqrt 2\int_{0}^{\frac{\pi}{2}}e^{-t}dt$
$L=\sqrt 2[-e^{-t}]_{0}^{\frac{\pi}{2}}$
$L=\sqrt 2[-e^{-\frac{\pi}{2}}+1]$
$L\approx1.120$
