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Math Help - Mixing Problem - Deriving a differential equation

  1. #1
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    Unhappy Mixing Problem - Deriving a differential equation

    Hi everyone, first post on here In a bit of strife though and looking for a bit of guidance...

    The Problem:
    A 150 litre tank contains 40 litres of water and 20 litres of alcohol mixed together. Pure alcohol is run in at 5 litres/minute, and the mixture is stirred to keep it perfectly uniform. Simultaneously, the mixture is withdrawn at 3 litres/minute.

    Derive a di fferential equation for the amount of alcohol, x(t) litres, in the tank at time t minutes.

    My attempt so far:
    This is my first time dealing with two liquids in a mixture (L/L) as opposed to a solid (such as a pollutant in a solution which is measured in g/L)

    Let x be the amount of alcohol in the tank at time t.

    \frac{dx}{dt} = rate alcohol flows in - rate alcohol flows out

    Rate in = Concentration X Flow rate in = 1 L/L X 5 \frac{L}{min}

    This is where I've hit a stumbling block as I'm having trouble calculating what rate out comes to.

    Rate out = Concentration X Flow rate out = x/(60+2t) X 3 \frac{L}{min}

    The reason I've gone with (60 + 2t) as the volume is that there's 60L at the start, and the difference between the two rates of flow is 2t. Am I right in doing so?

    And then the final differential equation I get to is:

    \frac{dx}{dt} = 5 - x/(20 + t) (simplified from above)

    A premeditated thanks for the help!
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  2. #2
    MHF Contributor
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    Fine apart from arithmetic where you
    Quote Originally Posted by omega24 View Post
    (simplified from above)
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