Differentiation

(a) The position vector r(t) is given by r(t) = R cos(wt)i + R sin(wt)j.


To find the velocity vector v(t), we differentiate r(t) with respect to time t.


v(t) = dr(t)/dt = -Rw sin(wt)i + Rw cos(wt)j.


Similarly, to find the acceleration vector a(t), we differentiate v(t) with respect to time t.


a(t) = dv(t)/dt = -Rw^2 cos(wt)i - Rw^2 sin(wt)j.


(b) To show that a(t) is perpendicular to v(t) for all t, we take the dot product of a(t) and v(t). If the dot product is zero, then the vectors are perpendicular.


a(t) . v(t) = (-Rw^2 cos(wt)i - Rw^2 sin(wt)j) . (-Rw sin(wt)i + Rw cos(wt)j)
= Rw^3 cos(wt) sin(wt) - Rw^3 sin(wt) cos(wt)
= 0.


Since the dot product is zero, a(t) is perpendicular to v(t) for all t.


(c) To show that a(t) . r(t) = -w^2R^2 for all t, we take the dot product of a(t) and r(t).


a(t) . r(t) = (-Rw^2 cos(wt)i - Rw^2 sin(wt)j) . (R cos(wt)i + R sin(wt)j)
= -R^2w^2 cos^2(wt) - R^2w^2 sin^2(wt)
= -R^2w^2 (cos^2(wt) + sin^2(wt))
= -R^2w^2 * 1
= -w^2R^2.


Therefore, a(t) . r(t) = -w^2R^2 for all t.

Does this help?