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7 April, 10:29

A tank contains 1,000 L of brine with 12 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes

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  1. 7 April, 13:29
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    y = 12*e^[-t/100]

    Step-by-step explanation:

    We have to mix problems we use:

    dy / dt = (rate in) - (rate out)

    The water entering the tank at a speed of 10 L / min but it has no salt, therefore

    rate in = 0

    The tank has 1000 liters of brine with 12 kg of salt initially, therefore the concentration of salt at time "t" is:

    y (t) / 1000

    then rate out would be:

    rate out = y (t) / 1000 * 10 = y (t) / 100

    The difference equation would then be:

    dy / dt = 0 - y (t) / 100

    1 / y (t) * dy = 1/100 * dt

    We integrate from both sides and we have:

    ln y = - (1/100) * t + C1

    y = e ^ [ - (1/100) * t + C1]

    y = e ^ [ - (1/100) * t] * e ^ [C1]

    We assume that C = e ^ [C1]

    Thus:

    y = C * e ^ [ - (1/100) * t]

    now y = 12 to t = 0, replacing:

    12 = C * e ^ [ - (1/100) * 0]

    12 = C, therefore we would have:

    y = 12 * e ^ [ - t / 100]
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