The area of a circle is decreasing at the rate of 5m^2/s. At what rate is the radius of the circle decreasing at that moment. Also, at what rate is the diameter decreasing at that moment?

Answer 1

Area = A = pi(r)^2

A'(t) = 2pirr'

r' = -5/2pir. depending on the radius the rate of change is different for a given rate of change of Area

for d' rewrite the Area equation in terms of d

A=pi(r)^2 = pi(d/2)^2 = pid^2/4 = (pi/4)d^2

A' = (pi/2)dd'

d' = -10/pid = -10/2rpi = -5/rpi = twice the rate of change for the radius