The radius of a circle is changing at the rate of 1/π inches per second. At what rate, in square inches per second, is the circle’s area changing when the radius is 5 inches?
Since area is \(A=\pi r^2\) then \(\frac{dA}{dt} = 2\pi r\frac{dr}{dt}\)
So the rate of change of area is \(2\pi r\) times the rate of change of radius.
I'll leave you to plug in the numbers.
+8 
