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Calc 2 differentials

  1. Apr 27, 2009 #1
    The air in a room with volume 180 m^3 contains 0.20% carbon dioxide initially. Fresher air with only 0.05% carbon dioxide flows into the room at a rate of 3 m^3/min and the mixed air flows out at the same rate.

    (a) Find the percentage of carbon dioxide in the room as a function of time.




    attempt :

    dy/dx = rate in - rate out

    rate in = 0.05 * 3

    rate out = 3 * y/180

    from that my answer turned out to be :

    y(t) = - (e^ (-t/60) * e ^(ln(9) + 1/300 ) - 9 )

    any help?
     
  2. jcsd
  3. Apr 28, 2009 #2

    HallsofIvy

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    What help are you asking for? If you want to know if that is the correct answer, no, it is not. For one thing, when t= 0, y should be 0.2(180)= 36 but your formula gives y(0)= eln(9)+ 1/300- 9. If you want to know what you did wrong, we will have to see what you did first!
     
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