Let's call omega w until LaTeX is back up...

a) Suppose the particle is at the point (x, y). Then

x = a cos(wt)

y = sin(wt)

The acceleration is s'' where s is the displacement vector and each derivative is taken with respect to time. The displacement is a vector from the origin to the point in question, so in terms of the unit vectors i and j:

s = x*i + y*j = a cos(wt)*i + sin(wt)*j

Then taking the derivatives:

s' = -aw sin(wt)*i + w cos(wt)*j

s'' = -aw^2 cos(wt)*i - w^2 sin(wt)*j

Let's play with this a bit and factor a -w^2:

s'' = -w^2 (a cos(wt) + sin(wt)) = -w^2*s

So s'' is a vector that is w^2 longer than s and, most importantly, points in exactly in the opposite direction to s. Since s points from the origin to the point, s'' points from the point to the origin.

b) The magnitude of the acceleration is |s''|, so

|s''| = w^4 * |s|

Thus the two are proportional.

-Dan

Note carefully that the centripetal acceleration does NOT point toward the origin in general. The centripetal acceleration is the accelerationcomponentin a direction perpendicular to the velocity.