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[SOLVED] Volume of liquid flowing into a tank
A tank in the shape of an inverted cone has liquid flowing into it at constant rate.
If the rate at which the water rises in the tank is 3x10^2 m s^1 when the liquid is at a depth of 2 m, calculate the volume of liquid flowing into the tank at m^3 s^1.
The cone has a maximum diameter of 4 m and height 3m.
i know that the vol. of a cone = [tex]\frac{\Pi r^{2}h}{3}[/tex]
I think this involves the volumes of revolution, where I have [tex] dy [/tex] as the thickness which would be [tex] 3 X 10^{2} m s^{1}[/tex] at a vol. of 2 m in height and then I have an integral of the area and limits of height. But because it's a rate i must need to differentiate it whereby [tex] dv/dt = vol. m^{3} s^{1} [/tex]???
Any clues would be appreciated, thanks.
A tank in the shape of an inverted cone has liquid flowing into it at constant rate.
If the rate at which the water rises in the tank is 3x10^2 m s^1 when the liquid is at a depth of 2 m, calculate the volume of liquid flowing into the tank at m^3 s^1.
The cone has a maximum diameter of 4 m and height 3m.
i know that the vol. of a cone = [tex]\frac{\Pi r^{2}h}{3}[/tex]
I think this involves the volumes of revolution, where I have [tex] dy [/tex] as the thickness which would be [tex] 3 X 10^{2} m s^{1}[/tex] at a vol. of 2 m in height and then I have an integral of the area and limits of height. But because it's a rate i must need to differentiate it whereby [tex] dv/dt = vol. m^{3} s^{1} [/tex]???
Any clues would be appreciated, thanks.
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