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Math Help - Salt problem

  1. #1
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    Salt problem

    A tank with capacity of 400 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/ min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/ min. Let Q(t) lb. be the amount of salt in the tank, V(t) gal be the volume of water in the tank.
    Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Your answer should be a function in terms of t.




    Doesn't Q(T) = t +200?


    Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Round your answer to 3 decimal places.

    Doesn't this just equal 200 +200 = 400?
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  2. #2
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    Quote Originally Posted by lord12 View Post
    A tank with capacity of 400 gal of water originally contains 200 gal of water with 100 lb of salt in solution. Water containing 1 lb of salt per gallon is entering at a rate of 3 gal/ min, and the mixture is allowed to flow out of the tank at a rate of 2 gal/ min. Let Q(t) lb. be the amount of salt in the tank, V(t) gal be the volume of water in the tank.
    Find the amount of salt in the tank at any time prior to the instant when the solution begins to overflow. Your answer should be a function in terms of t.




    Doesn't Q(T) = t +200?


    Find the concentration (in pounds per gallon) of salt in the tank when it is on the point of overflowing. Round your answer to 3 decimal places.

    Doesn't this just equal 200 +200 = 400?
    Well if you think about it, initially you have 100 lbs of salt in 200 gallons or 1/2 lb/gal. The concentration of the incoming salt solution is 1 lb/gal. So if you assume perfect (and instanteous) mixing, the concentration will change from 1/2 to 1 (assuming an infinite tank). Here we must set up a differential equation for the process.

    First the tank in fill up V =(3-2)t + 200.

    Next, the change in the amount of salt A is

    \frac{dA}{dt} = r_i c_i - r_o c_o

    where ri - rate in, ro - rate out, ci - concentration in and co - concentration out.

    so

    \frac{dA}{dt} = 3 \cdot 1 - 2 \cdot \frac{A}{V} = 3 - \frac{2A}{t+200}

    the initial condition

    A(0) = 100. A linear ODE for A. Can you solve it?
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  3. #3
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    I solve the diffeq and get that A = t +200
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  4. #4
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    Quote Originally Posted by lord12 View Post
    I solve the diffeq and get that A = t +200
    Please, let me ask - how?
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  5. #5
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    is it t-100?
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  6. #6
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    Quote Originally Posted by lord12 View Post
    is it t-100?
    I got

    A = t + 200 - \frac{4,000,000}{(t+200)^2}.
    Last edited by Jester; January 29th 2009 at 07:23 AM. Reason: wrong answer
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