A point moves in the plane `xy` according to the law `x=a sin omega t, y=b cos omegat , ` where `a,b` and `omega` are positive constants.
Find `:`
`(a)` the trajectory equation `y(x)` of the point and the direction of its motion along this trajectory ,
`(b)` the acceleration `w` of the point as a function of its radius vector `r` relative to the orgin of coordinates.

Explanation

From the Eqn `: x= a sin omega t`
`sin^(2)omegat=x^(2)//a^(2)` or `cos^(2) omegat =1-(x^(2))/(a^(2)) ....(1)`
And from the equation `: y=b cos omegat `
`cos^(2)omegat =y^(2)//b^(2) ...(2)`
From Eqns `(1)` and `(2)`, we get `:`
`1-(x^(2))/(a^(2))=(y^(2))/(b^(2))` or `(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
Which is the standard equation of the ellipse shown in the figure.
we observe that,
at `t=0, x=0` and `y=b`
and at `t=(pi)/(2 omega), x=+a` and `y=0`
Thus we observe that at `t=0`, the point is at point (figure) and at the following moments, the co- ordinat `y` diminishes and `x` beomes positive. Consequenctly the motion is clock-wise
`(b)` As `x=a sin omegat ` and `y=b cos omegat`
so we amy write `vec(r)= a sin omega t vec(i+)b cos omegat vec(j)`
Thus `ddot(r)=vec(w)=-omega^(2)vec(r)`