Water is poured into an inverted cone at a rate of π-units per second. If the radius of the base of the cone is r and its height is 2r, what is the rate at which the depth of the water is changing when the height of the water is 1/2r?
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$\begingroup$ $\frac{dV}{dt} = \pi$. Further from the given information, $dV = \frac{\pi h^2}{4} dh$ $\endgroup$ – Sandeep Thilakan Mar 12 '14 at 4:42
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Let h, and R be the height of the water and the radius of the water surface. Then:
h/2r = R/r ==> R = h/2.
The volume V of the water in the cone is: V = 1/3*pi*R^2*h = 1/12*pi*h^3.
So: pi = dV/dt = 1/4*pi*h^2*dh/dt = 1/4*pi*(1/(4*r^2))*dh/dt ==> dh/dt = 16r^2
