The volume of a come (V=1/3r^2h) is increasing at the rate of 28pi cubic units per second. At the instant when the radius r of the cone is 3 units, its volume is 12pi cubic units and the radius is increasing at 1/2 unit per second. When the radius of the come is 3 units, what is the rate of change of the area of its base? Of its height? Three instantaneous rate of change of the area of its base with respect to its height?

Question

anonymous · Answer

Not three at the end, I meant the!

anonymous · Answer

do you know the formula for the area of the base?