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A water tank has the shape of an inverted right circular cone of altitude 12 ft and base radius 6 ft. if water is being pumped into the tank at a rate of 10 gal/min, approximate the rate at which the water level is rising when it is 3 ft deep

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sqdancefan
The radius of the water surface at that point is (3 ft)/(12 ft)*(6 ft) = 1.5 ft, so the surface area is
.. π(1.5 ft)^2 = 2.25π ft^2

The rate of rise is then
.. (10 gal/min)*(77/576 ft^3/gal)/(2.25π ft^2) ≈ 0.18911 ft/min

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The rate of change of depth is equal to the rate of change of volume divided by the surface area. Here's how you can get there from derivatives.

V = (1/3)Bh
V' = (1/3)(B'h +Bh')
Here we have B = π(h/2)^2 = (π/4)h^2, so B' = (π/2)h*h'
V' = (1/3)((π/2)*h^2*h' +(π/4)h^2*h')
V' = (π/4)h^2*h'
V' = B*h'

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