The surface area of a closed cylinder is given by A = 2*pi*r^2 + 2*pi*r*h, where h is the height and r is the radius of the base. At the time when the surface area is increasing at the rate of 20*pi cm^2 s^-1, the radius is 4 cm and the height is 1cm and is decreasing at the rate of 2cm s^-1. Find the rate of change of radius at this time.
My working so far:
I know that we need to work out dr/dt
So far I have dA/dt =20pi
Not sure where to go from here.
My working so far:
I know that we need to work out dr/dt
So far I have dA/dt =20pi
Not sure where to go from here.