Answer
Distance = $6\sqrt 2$
$L$ = $\sqrt 2$
Work Step by Step
$x$ = $\sin^2t$
$y$ = $\cos^2t$
$\frac{dx}{dt}$ = $2\sin t\cos t$
$\frac{dy}{dt}$ = $-2\sin t\cos t$
$\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2$ = $(2\sin t\cos t)^2+(-2\sin t\cos t)^2$ = $8\sin^2t\cos^2t$ = $2\sin^2 2t$
Distance = $\int_0^{3\pi}{\sqrt {2\sin^2 2t}}dt$ = $6\sqrt {2}\int_0^{\frac{\pi}{2}}{\sin 2t}dt$ = $6\sqrt 2$
The full curve is traversed as t goes from $0$ to $\frac{\pi}{2}$ because the curve is the segment of $x+y$ = $1$ that lies in the first quadrant and this segment is completely traversed as $t$ goes from $0$ to $\frac{\pi}{2}$.
Thus
$L$ = $\int_0^{\frac{\pi}{2}}{\sin 2t}dt$ = $\sqrt 2$
