The volume V of a cone (V = 1/3*pi*r^2*h) 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. At the instant when the radius of the cone is 3 units, what is the rate of change of the area of its height h? Know: dV/dt = 28pi cubic units per second, dr/dt = 1/2 unit per second, and dA/dt = 3pi units square per second. Need: dh/dt 12pi = 1/3pi*3^2*h h=4 V = 1/3*pi*r^2*h dV/dt = 1/3*pi*2r*dr/dt*h + r^2*dh/dt (use the product rule. do you multiply 1/3 pi times everything like this 1/3*pi(2r*dr/dt*h + r^2*dh/dt) or how I did it?) 28pi = 1/3pi*2(3)*1/2*4 + 6*dh/dt 24pi = 6*dh/dt dh/dt = 4pi units per second Is this correct?