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A water tank is in the shape of an inverted cone with depth 10 meters and top radius 8 meters. Water is flowing into the tank at 0.1 cubic meters/min but leaking out at a rate of 0.001[tex]h^2[/tex] cubic meters/min, where h is the depth of the water in meters. Will the tank ever overflow?

__Thoughts:__[tex]V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\pi (\frac{4}{5}h)^{2}h[/tex]

[tex]\frac{dV}{dt}=\frac{16}{25}\pi h^{2}\frac{dh}{dt}[/tex]

Now I replace [tex]\frac{dV}{dt}[/tex] with 0.1-0.001[tex]h^2[/tex]. This is where I am stuck. Any suggestions?

Thanks.