Since #x=cos^{2}(t)# and #y=sin^{2}(t)#, it follows that #x+y=cos^{2}(t)+sin^{2}(t)=1# for all values of #t#. Therefore, the motion is always on the straight line with #xy#-equation #x+y=1#, which is equivalent to #y=-x+1# (a straight line with a slope of #-1# and a #y#-intercept of #1#).

Also, since #cos^{2}(t)\geq 0# and #sin^{2}(t)\geq 0# for all #t#, this motion is always in the 1st quadrant of the plane where #x\geq 0# and #y\geq 0#.

Now think about, for #0\leq t\leq 2pi#, how the values of #cos^{2}(t)# oscillate from 1 to 0 to 1 to 0 and back to 1 again, while the values of #sin^{2}(t)# oscillate from 0 to 1 to 0 to 1 and back to 0 again. In other words, the motion traverses the line segment from #(1,0)# to #(0,1)# four times. By the Pythagorean Theorem (draw an appropriate right triangle), this line segment has length #sqrt(1^{2}+1^{2})=sqrt(2)#.

This leads us to conclude that the total distance traveled (arc length) is #4sqrt(2)#.