Find the distance traveled by a particle with position (x, y) as t varies in the

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) t = 0 --> x(t) = cos²(0) = 1, y(t) = cos(0) = 1 starts @ (1,1) 2) Since the cosine starts @ it's maximum (1) it will now decrease to : t = p/2 --> x(p/2) =cos²(p/2) = 0, y(p/2) = cos(p/2) = 0 So it moves down and to the left towards (0,0) (the origin) It then moves further down to y = -1, but travels to the right: t = p --> x(p/2) =cos²(p) = +1, y(t) = cos(p) = -1 So it moves to (1, -1) 3) Now it will travel back up towards y = 0 then to y = 1...it should just retrace it's steps back to the original position... I don't know why they told you 4p, the period is 2p because the period of the cosine is 2p whereas the cos²() actually has as SMALLER period of just p...this is why the particle retraces its steps. The x-motion is identical from 0 -> p AND from p -> 2p, whereas the y-motion is SIMILAR from 0 -> p and from p -> 2p...but it's in the opposite direction. The first half it is moving down, the second half it does the same thing but moves upward (back to the original position). The cos²(t) is actually just (cos(2t) + 1)/2 = 1/2cos(2t) + 1/2

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