# Hard but fun math problem

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• Feb 15th 2009, 07:55 PM
Melissa_Smith
Hard but fun math problem
A tank contains 60kg of salt and 2000L of water. A solution of a concentration .015kg of salt per liter enters a tank at the rate 6L/min. The solution is mixed and drains from the tank at the same rate.
Helpful Hints:
A differential Equation has to be used
Net Rate= (Rate In - Rate Out)
Initial Concentration is .03 kg/L
My answer was y=(((e^(6/-2000)t+ln(.08991))-.09)(-2000/6) but it was wrong
I dont know if y(0)=.03 or y(0)=60
Any help will be greatly appreciated.
• Feb 15th 2009, 09:31 PM
nmatthies1
okay then, this is not as hard as it might look.

Let Q(t) be the amount of salt in the tank at any time.
Now we assume that salt is neither created nor destroyed in the tank, therefore variations in the amount of salt are due solely to the flows in and out of the tank. And so we can say that dQ/dt, the rate of change of salt in the tank, is equal to the rate at which salt is flowing in minus the rate at which it is flowing out: dQ/dt= rate in - rate out.
Now the rate at which salt enters the tank is 0.015 kg/L x 6L/min= 0.09 kg/min.
To find the rate at which salt leaves the tank we multiply the concentration of salt in the tank by the rate of outflow. Since the rates of flow in and out are the same, the amount of water in the tank stays the same ar 2000 L and since the mixture is well stirred the concentration throughout the tank is the same, Q(t)/2000 kg/L.
Therefore the rate at which salt leaves the tank is Q(t)/2000 kg/L x 6L/min = 3Q(t)/1000 kg/min.
Therefore the differential equation modelling this system is
$\frac {dQ} {dt} = 0.09 - \frac {3Q} {1000}$ (1)
Now the initial condition is Q(0)=60
Now to solve the problem, note that (1) is both linear and separable. Rewriting we get $\frac {dQ} {dt} + \frac {3Q} {1000} = 0.09$ (2)
Thus the integrating factor is $e^{\int \frac {3} {1000} } = e^{\frac {3t} {1000}}$
Multiplying both sides of (2) by the IF we have $e^{\frac {3t} {1000}} \frac {dQ} {dt} + e^{\frac {3t} {1000}} \frac {3Q} {1000} = 0.09 e^{\frac {3t} {1000}}$ and we can now rewrite the left hand side as a derivative:
$\frac {d} {dt} (Q e^{\frac {3t} {1000}}) = 0.09 e^{\frac {3t} {1000}}$
Integrating this equation wrt to t between 0 and t we obtain $Q(t) e^{\frac {3t} {1000}} - Q(0) = \int_{0}^{t} 0.09 e^{\frac {3t} {1000}} dt$ = $[ \frac {1000} {3} * 0.09 e^{\frac {3t} {1000}}]_{0}^{t} = 1000 * 0.03 e^{\frac {3t} {1000}} -1000*0.03$
and so since Q(0)=60, $Q(t)=30 + 30 e^{\frac {-3t} {1000}}$