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 * Tank Example 2**

A holding tank is used to control the release of pollutants into a sewerage system. The tank initially contains 400 litres of water with 800kg of pollutants disolved in it. Polluted water containing 4 kg of pollutants per litre flows into the tank at a rate of 3 litres/minute. The mixture is kept uniform by stirring and it flows out of the tank at the same rate. a) Set up the differential equation describing the rate of change of the quantity of pollutants (Q kg) in the tank. b) Solve using calculus to find an expression for Q(t) c) Find the amount of pollutants after 20 minutes.

From the question ... ... V = 400 litres (constant) ... ... Q(0) = 800 kg
 * a) Set up the DE **

... ... C(in) = 4 kg/litre

math \\ . \qquad \dfrac{dV(in)}{dt} = 3 \; \text{litres/minute} \\. \\ . \qquad \dfrac{dV(out)}{dt} = 3 \; \text{litres/minute} math

... ... **Rate In = (concentration of inflow) × (rate of inflow)**

math . \qquad \dfrac{dQ(in)}{dt} = 4 \times 3 = 12 \; \text{kg/minute} math

... ... **Rate Out = (concentration within tank) × (rate of outflow)**

math \\ . \qquad \dfrac{dQ(out)}{dt} = \dfrac{Q}{400} \times 3 \\ .\\ . \qquad \qquad \quad \; = \dfrac{3Q}{400} \; \text{kg/minute} math

... ... **DE = Rate In – Rate Out**

math . \qquad \dfrac{dQ}{dt}=12-\dfrac{3Q}{400} \; \text{kg/minute} math

**b) Solve to find Q(t)**

math \\ . \qquad \dfrac{dQ}{dt}=12-\dfrac{3Q}{400} \\. \\ . \qquad \dfrac{dQ}{dt}=\dfrac{4800-3Q}{400} math

Take the reciprocal of both sides math . \qquad \dfrac{dt}{dQ}=\dfrac{400}{4800-3Q} math

integrate math \\ . \qquad t=-\dfrac{400}{3} \displaystyle{\int}\dfrac{-3}{4800-3Q} \, dQ \\. \\ . \qquad t=-\dfrac{400}{3} \log_e \big| 4800-3Q \big| +c math

Make **Q** the subject math \\ . \qquad \dfrac{-3t}{400}=\log_e \big| 4800-3Q \big| - \dfrac{3c}{400} \\. \\ . \qquad \log_e \big| 4800-3Q \big| = \dfrac{3c}{400} - \dfrac{3t}{400} math

Use log laws math \\ . \qquad \big| 4800-3Q \big| =e^{\frac{3c}{400}} \times e^{\frac{-3t}{400}} \\. \\ . \qquad 4800-3Q =e^{\frac{3c}{400}} \times e^{\frac{-3t}{400}} \qquad \text{for } Q < 1600 math

Introduce the constant, **A** math \\ . \qquad \text{Let } A=e^{\frac{3c}{400}} \\. \\ . \qquad 4800-3Q=Ae^{\frac{-3t}{400}} \\. \\ . \qquad 3Q=-Ae^{\frac{-3t}{400}}+4800 \\. \\ . \qquad Q=-\frac{A}{3}e^{\frac{-3t}{400}}+1600 math

{Notice that because of the constant, Q must always be less than 1600}

To find **A**, use **Q(0) = 800** math \\ . \qquad 800=-\frac{A}{3}+1600 \\. \\ . \qquad A = 2400 math Hence solution is math . \qquad Q=-800e^{\frac{-3t}{400}}+1600 \; \text{kg} math

This can be solved on your calculator, by entering **dSolve**(q'=12 – 3q/400, t, q, t=0, q=800)

**c) Find Q(20)** math . \qquad Q(20)=-800e^{\frac{-3 \times 20}{400}}+1600 math

... ... Q(20) = 911 kg.