Answer
$$L = \sqrt {2} (e^{\pi} - 1)$$
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Work Step by Step
1. Write the formula for the length "$L$"of a curve:
$$L = \int_{\alpha}^{\beta} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \space dt$$
2. Find $\frac{dx}{dt}$ and $\frac{dy}{dt}$
$\frac{dx}{dt} = \frac{d(e^tcos(t))}{dt} = e^tcos(t) - e^tsin(t)$
$\frac{dy}{dt} = \frac{d(e^tsin(t))}{dt} = e^tsin(t) + e^tcos(t)$
3. Substitute the values into the definite integral:
$L = \int_{0}^{\pi} \sqrt{( e^tcos(t) - e^tsin(t))^2 + ( e^tsin(t) + e^tcos(t))^2} \space dt$
$L = \int_{0}^{\pi} \sqrt{( e^{2t}cos^2(t)- 2e^tcos(t)e^tsin(t) + e^{2t}sin^2(t)) + ( e^{2t}sin^2(t) + 2e^tsin(t)e^tcos(t) + e^{2t}cos^2(t))} \space dt$
$L = \int_{0}^{\pi} \sqrt{(2e^{2t}(cos^2(t) + sin^2(t)) )} \space dt$
$L = \int_{0}^{\pi} \sqrt{(2e^{2t} )} \space dt$
$L = \int_{0}^{\pi} \sqrt{2}(e^t) \space dt$
$L = \sqrt{2} \int_{0}^{\pi} (e^t) \space dt$
$L = \sqrt{2} [(e^t)]^{\pi}_0$
$L = \sqrt {2} (e^{\pi} - e^0)$
$L = \sqrt {2} (e^{\pi} - 1)$
