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Homework Help: Differential Equations - Word Problem

  1. Sep 5, 2010 #1
    1. The problem statement, all variables and given/known data
    A pond initially contains 1,000,000 gal of water and an unknown amount
    of an undesirable chemical. Water containing 0.01 gram of this chemical per
    gallon flows into the pond at a rate of 300 gal/hour. The mixture flows out at
    the same rate, so the amount of water in the pond remains constant. Assume
    that the chemical is uniformly distributed throughout the pond.
    a) Write a di erential equation for the amount of the chemical at any time.
    b) How much of the chemical will be in the pond after a very long time?
    Does this limiting amount depend on the amount that was present initially?


    2. Relevant equations



    3. The attempt at a solution

    t (hours)
    y (grams)

    y(0) = y0

    dy/dt = 3g/hr - (??) -- How do I determine the rate at which it's flowing out if I don't know the initial value?
     
  2. jcsd
  3. Sep 6, 2010 #2
    It says the mixture flows out at the same rate as it flows in such that the volume remains constant (1,000,000 gal). Now just remember that (like you did with the flow in) the flow out is going to be equal to the rate (300gal/hour) multiplied by the concentration of the unknown chemical. Since the volume is constant and the mixture is perfectly mixed you can conclude that the concentration flowing out would be equal to y/V.
     
  4. Sep 6, 2010 #3
    Thanks. How about this one?

    1. The problem statement, all variables and given/known data

    1. In the theory of learning, the rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized. If M denotes the total amount that is to be memorized and A(t) the amount memorized in time t, a nd the di fferential equation for A.

    2. In Problem 1, assume that the amount of material forgotten
    is proportional to the amount memorized at time t. What is the differential
    equation for A when forgetfulness is taken into account?

    3. The attempt at a solution

    1. dA/dt = 1/M -- should I include an arbitrary rate constant?

    2. dA/dt = 1/M - A
     
  5. Sep 6, 2010 #4
    #1:
    It says that the rate at which the material is memorized is proportional to the amount that is left to be memorized. So this could be restated as

    (Rate of Memorizing Material) [tex]\propto[/tex] (Total Amount to Be Memorized) - (Amount Memorized)

    So, since A(t) is the amount memorized at time t, and M is the total amount to be memorized, you could say that

    [tex]\frac{dA(t)}{dt}[/tex] [tex]\propto[/tex] M - A(t)

    or like you mentioned we could use a proportionality constant, lets say "c", so

    [tex]\frac{dA(t)}{dt}[/tex] = c(M - A(t))

    That is my guess for an answer for this question. I solved the differential equation and got a solution that made sense, you could do the same to see what I'm talking about.

    #2:

    For this one we take forgetfulness into account. This means that to our previous differential equation,
    [tex]\frac{dA(t)}{dt}[/tex] = c(M - A(t))

    ,we must introduce a negative value on the right hand side. That is because forgetting decreases the rate of memorization. If we denote the forgetfulness as F, then using what they gave us we know that F [tex]\propto[/tex] A ,or, if we again introduce a constant of proportionality, F = kA

    So the new differential equation we would get is
    [tex]\frac{dA(t)}{dt}[/tex] = c(M - A(t)) - kA

    I also solved this one and it made sense. If you want to try solving these go for it (it's good practice and a great way to check your answer).
     
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