(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Another related rates problem

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