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 October 31st, 2009, 03:06 AM #1 Newbie   Joined: Oct 2009 Posts: 7 Thanks: 0 A water tank has shape of an inverted circular cone of altit A water tank has shape of an inverted circular cone of altitude 12ft and base radius 6ft. if water is being pumped into the tank at a rate of 10lit/min, approximate the rate at which the water level is rising when it is 3 ft deep please solve this its answer is 1.89ft/min
 October 31st, 2009, 08:18 AM #2 Guest   Joined: Posts: n/a Thanks: Re: A water tank has shape of an inverted circular cone of altit Perhaps I am in error, but I get dh/dt=$\frac{40}{9\pi}\approx 1.415$ $\frac{dh}{dt}=\frac{\frac{dv}{dt}}{A(t)}$ $A(t)=\text{area of surface of water when it is 3 feet deep}$ Area of water surface when 3 feet deep is $A={\pi}(\frac{3}{2})^{2}=\frac{9\pi}{4}$
November 1st, 2009, 03:06 AM   #3
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Quote:
 Originally Posted by urduworld . . . if water is being pumped into the tank at a rate of 10lit/min, . . .
Are you sure that the fill rate is being given in these units, when all the other measurements given use feet? If the fill rate is 10 ft³/min, I get 40/(9pi) ft/min = 1.41471... ft/min, as per galactus.

If the fill rate is 10 l/min, you'd need to know that 10 l/min = 0.353146667 ft³/min, which gives a numerically much smaller answer (0.04996... ft/min).

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# a water tank is in the shape of an inverted right circular cone of altitude 12 ft

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