Problem 1I-3 (a) says $\vec{r} = < 2\cos^2 t , \ \sin^2 t>$ in other words, r is a position vector with x component = 2 cos^2 t and a y component = sin^2 t The attached graph show the locus of points we create as we increment t notice if we start with $x= 2 \cos^2 t \\ y= \sin^2 t$ and (because we want to use sin^2( t) + cos^2(t) = 1 to simplify the equation) we multiply the 2nd equation by 2 and add the two equations: $x + 2y = 2\cos^2 t + 2 \sin^2 t \\ x+2y= 2(\cos^2 t + \sin^2 t) \\ x+2y= 2$ that is the equation of a line. In slope-intercept form $y = - \frac{1}{2} x +1$ so we should expect the locus of points to lie on this line (and it does) because sin and cos are cyclical, we will get cyclical behavior... the point moves back and forth between (0,1) and (2,0)