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A 12 hour water clock is designed in the shape of a paraboloid of height 4 ft and radius of the circular opening at the top as 1 ft with an opening for water flow at the bottom. The water clock is obtained by revolving the curve y= f(x) around y-axis.What should be this curve and what should be the radius of the circular hole at the bottom in order that the water level will fall at the constant rate of 4 inches per hour?

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This is obviously a homework, please show how you approached the problem. – Sasha Aug 26 '11 at 18:20

Hint: when the depth of the water is $y$ the rate of flow $\frac{dV}{dt}$ should be proportional to $\sqrt{y}$. If the cross-section area is $A(y)$ then $\frac{dV}{dt} = A \frac{dy}{dt}$.

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