Please help. The minute hand of a certain clock is 4 in long. Starting from the moment when the hand is pointing straight up, how fast is the area of the sector that is swept out by the hand increasing at any instant during the next revolution of the hand? Why is the answer 4/15pi sq inches.
Are we differentiating this with respect to time? I think that information might be missing from the problem (or is it assumed).
For a circular sector
A = 1/2 r^2 theta
dA/dt =1/2r^2 d(theta)/dt
=1/2 16 * 2pi rad/60sec