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Math Help - Application of Differential Equations.

  1. #1
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    Application of Differential Equations.

    A 25% nitric acid solution flows at a rate of 6 litres/min into a large tank. Initially, the tank holds 200 litres of 5% nitric acid solution. The solution inside the tank is kept well stirred and flows out of the tank at a rate of 8 litres/min.

    (a) If x(t) is the amount (in litres) of nitric acid in the tank after t minutes, prove that

    \frac{dx}{dt} + \frac{4x}{100-t} = \frac{3}{2}

    Okay .. I have been trying to do it, I'm close to the answer but .. not getting there. Please help me out ?

    I know that

    \frac{dx}{dt} = (Inflow Rate x Concentration) - (Outflow Rate) x \frac{x}{v(t)}

    v(t) is the amount of solution in the tank at any minute which is defined as,

    v(t) = 200 - (Inflow Rate - Outflow Rate)(t)
    v(t) = 200 - 2t

    so putting that back up to the dx/dt equation..

    \frac{dx}{dt} = (6 x 25%) - 8 x \frac{x}{200 - 2t}

    Hmm, I don't know what went wrong.

    Anyway, does it matter that they told us that there was 5% of the nitric acid in the solution? Or can I just ignore that?

    Thank youu.
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  2. #2
    Member sinewave85's Avatar
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    Quote Originally Posted by panda* View Post
    Okay .. I have been trying to do it, I'm close to the answer but .. not getting there. Please help me out ?

    \frac{dx}{dt} = (6 x 25%) - 8 x \frac{x}{200 - 2t}

    Thank youu.

    Yes, you are very close! In fact, you are only a few simple steps away from completing your proof. Watch this:

    \frac{dx}{dt} = (6)(0.25) - (8)\frac{x}{200 - 2t}

    \frac{dx}{dt} = (6)(1/4) - \frac{8x}{200 - 2t}

    \frac{dx}{dt} = \frac{6}{4} - \frac{8x}{200 - 2t}

    \frac{dx}{dt} = \frac{3}{2} - \frac{4x}{100 - t}

    \frac{dx}{dt} + \frac{4x}{100 - t} = \frac{3}{2}

    You had the answer, you just needed to put it in the same format as the equation provided.
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  3. #3
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    OH! My bad. I didn't notice it was 6/4, I immediately simplified it to 3/2!
    Thank you so much!
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