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Homework Help: Differential equation question

  1. May 7, 2006 #1
    Kinda lengthy question but bare with me please

    A 1200 m3 treatment tank is used for treating effluent. The use of a particular form of algae is a major part of this process. It is important that the concentration, C, of algae not fall below 10 gm per litre as the process will not work. Above a concentration of 20 gm per litre the algae begin to die off.

    Call the effective range for the concentration.
    Note: We will work in terms of time in hrs, mass in gm and volume in litres.

    As time passes algae are used up in the process in such a way that the rate of decrease of concentration of algae is proportional to the concentration of algae. To deal with this the algae is topped up at regular intervals of T hours. There is a special tank for growing algae from which the algae is taken.

    The problem is to find the correct mass of algae to add and the time intervals at which to add it to keep the concentration as high as possible within the effective range.

    1) Write the differential equation that describes the decrease in concentration of the algae. (i.e. don’t worry about the top up every T hours). If the initial concentration of algae is g /l write down the initial condition.

    For this as

    dC/dt is proportional to C

    dc/dt = kC

    2) Solve your equation using separation of variables. You MUST show every step in your solution.

    Next using seperation of variables i get

    > dc * 1/c = k dt
    > ln |C| = kt
    > C = Ae^kt A = e^c

    This is where im stuck how exactly do i solve this with the information give? any help would be appreciated.
  2. jcsd
  3. May 9, 2006 #2


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    You need an initial condition to find A. However, the description doesn't seem to give an initial C.

    Therefore, put somehting like C=C_0 at t=0 which then gives the constant to be A=C_0

    I presume the question goes on past part 2...
  4. May 9, 2006 #3
    Since the concentration decreases, it should be dC/dt = -kt. Don't forget your constant of integration, and from this you can find out what it is since the initial concentration is g/l.
  5. May 9, 2006 #4


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    I completely missed where it gave this in the question :redface:
  6. May 11, 2006 #5
    cheers for the help guys, got C = Co*e^(-kt)

    Would anyone be able to explain the mathematical reason why plotting ln|c| vs t approximating a straight line with experimental data proves this relationship.. cheers.
    Last edited: May 11, 2006
  7. May 11, 2006 #6
    If you were to take the natural log of both sides, you end up with ln(c) = ln (Co) -kt. Since ln(co) is just a number (let it equal c1), if you let y = ln(c), then the equation becomes y = -kt + c1 which is a linear equation with slope -k and intercet c1. So, from experimental data, if you plot the ln(c) versus t, and you get a linear plot, then this shows that the you have exponential decay, and using interpolation, you can calculate what k and Co should be.
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