Calculus
posted by Luis .
The minute hand of a clock is 6in long. Starting from noon, how fast is the area of the sector swept by the minute hand increasing in in^2/min at any instant?

The area of the sector is proportional to the central angle
Let the area of the sector be A
A/(pir^2) = theta/(2pi)
piA = 36theta
A = (36/pi)theta
dA/dt = (36/pi)d(theta)/dt
but d(theta)/dt = 2pi radians/60 min = pi/30 radians/min
so dA/dt = (36/pi)(pi/30) = 6/5 in^2/min 
The answer given by Reiny is not correct.
Let the central angle be theta. The area of the sector is proportional to the ratio of the central angle to the angle which goes the whole way around the clockwhich is theta/2pi.
If we multiply the formula for the area of a circle by this ratio (theta/2pi), we get the area of the sector. The area of the sector is, therefore, pi(r^2)(theta/2pi). Since we know our radius to be 6, we get 36pi(theta)/2pi, or simplified, 18theta.
Taking the derivative of A=18theta, we have dA/dt = 18(dtheta/dt). We know dtheta/dt because we know that it takes 1 hour for the minute hand to travel all the way around the clock. It will have traveled 2pi radians in 1 hour, or pi/30 radians in a minute.
Plugging into the previous equation, we have dA/dt = 18(pi/30) =
3pi/5 in^2/min,
which is the correct answer to this problem.
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