Expert: Sherry Wallin - 3/27/2011

Question
The minute hand of  the 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 nest revolution of the hand?

Answer
Hi~


Area of a sector = (1/2)r^2*theta where theta is measured in radians so we want DA/dt

But we don't have time yet...but we can relate degrees with time in this way:

When the minute hand sweeps out the area of the whole circle then it will have moved 2pi radians but this will take 12 hours, so how far will the minute hand travel in 1 hour? The minute hand actually moves at a rate of
(2pi radians)/(12 hour) or pi/6 radians per hour, this will be your Dtheta/dt
(another way to think of this is that 1/12 of the circle is swept out in 1 hour)

Now for the calculus:
DA/dt = (1/2)r^2 Dtheta/dt  but we know what r is, it doesn't change, so instead, do this:

DA/dt = (1/2)4^2 Dtheta/dt = 8 Dtheta/dt but we know Dtheta/dt is pi/6 so DA/dt = 8(pi/6)
= 4pi/3, the rate at which the area of the sector of circle is changing at any instant.

Math Prof