Water Tank Problem

puff2978

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Joined
Mar 15, 2010
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2
A water tank has a circular base of
radius 4 meters and initally contains water to a depth
of 5 meters. Then water drains out at an increasing rate
so that after t hours the water is leaving at a rate of 6t m^3/hr
How many hours until the tank is empty ?

I know that Area*height is the volume of the cylinder...
 
\(\displaystyle V_{tank} \ = \ \pi r^{2} h \ = \ \pi(16)(5) \ = \ 80\pi\)

\(\displaystyle \frac{dV}{dt} \ = \ -6t \ \implies \ \int dV \ =\int -6t dt \ \implies \ V(t) \ = \ -3t^{2}+C\)

\(\displaystyle V(0) \ = \ 80\pi \ = \ -3(0)+C, \ \implies \ C \ = \ 80\pi\)

\(\displaystyle Hence, \ V(t) \ = \ -3t^{2}+80\pi\)

\(\displaystyle Now \ when \ the \ tank \ is \ empty, \ V(t) \ = \ 0, \ ergo, \ 3t^{2} \ = \ 80\pi \ \implies \ t \ \dot= \ 9.1529 \ hrs.\)

\(\displaystyle Check: \ V(9.1529) \ = \ -3(9.1529)^{2}+80\pi \ \dot= \ .00068, \ good \ enough \ for \ government \ work.\)
 
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