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Math Help - Differiential Equation, Mixing problem

  1. #1
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    Differiential Equation, Mixing problem

    A 50 litre tank is initially filled with 10 litres pf brine solution containing 20 kg of salt. Starting from time t=0, distilled water is poured into the tank at a constant rate of 4 litres per minute. At the same time, the mixture leaves the tank at a constant rate of k^(1/2) litre per minute, where k^(1/2) >0. The time taken for overflow to occur is 20 minutes.

    (a) Let Q be the amount of salt in the tank at time t minutes. Show that the rate of change of Q is given by:

    dQ/dt= (-Qk^(1/2))/(10+(4-k^(1/2))t)

    * k^(1/2) means square root of k,

    Hence, express Q in term of t,

    (b) Show that k = 4, and calculate the amount of salt in the tank at the instant outflow occurs.

    (c) Sketch the graph of Q against t for 0 < t < 20








    Please help me to solve. Thank you....
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  2. #2
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    a.)

    \frac{dQ}{dt} = C(t) \times (- \sqrt{k})

    where C(t) is the concentration of salt in the water at time t. Hence this states that the rate of salt loss is the concentration at time t multiplied by the rate of loss of salty water volume. Now the concentration is simply the amount of salt Q at time t divided by the total volume of water in the tank at time t so:
    <br />
C(t) = \frac{Q}{V(t)} = \frac{Q}{10 +(4-\sqrt{k}) t}

    hence

    \frac{dQ}{dt} = -\frac{Q \sqrt{k}}{10 +(4-\sqrt{k}) t} .

    b.)

    Here you know the time taken for overflow is 20 minutes and you know the tank has total volume 50 litres so use the relationship for the volume that:

    V(t) = 10 +(4-\sqrt{k}) t

    put V = 50 and t = 20 and then solve for k.

    Then substituting back into your original differential equation you should find

    \frac{dQ}{dt} = -\frac{Q}{5 + t}

    so

    \int_{Q_0}^{Q_f} \frac{1}{Q} \, \mathrm{d}Q = - \int_{0}^{20} \frac{1}{5+t} \, \mathrm{d}t

    where Q_0 is the initial amount of salt (20 kg) and Q_f is the amount of salt when overflow occurs.

    I think you can take it from there, right?
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