Question 1083191
A large cone of given size is being drained of water at the constant rate of 15 cubic cm each second. The water's surface level in the cone falls as a result. At what rate is the water level falling when the water is 4cm deep?
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No dimensions given.
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{{{V = pi*r^2*h/3}}}
r = h/k where k is a constant = height/radius of the cone.
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{{{V = pi*k^2*h^3/3}}}
{{{dV/dt = pi*k^2*h^2*(dh/dt)}}}
{{{15 = pi*k^2*16*(dh/dt)}}}
{{{dh/dt = 15/(16pi*k^2)}}}