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First-Order Diff. Eq.

  1. Apr 8, 2007 #1
    1. The problem statement, all variables and given/known data
    A 5-gallon bucket is full of pure water. Suppose we begin dumping salt into the bucket at a rate of .25 pounds per minute. Also, we open the spigot so that .5 gallons per minute leaves the bucket, and we add water to keep the bucket full. If the saltwater solution is always well mixed, what is the amount of salt in the bucket after
    a) 1 minute?
    b) 10 minutes?
    c) 60 minutes?
    d) 1000 minutes?
    e) a very, very long time?

    2. Relevant equations
    16 ounces = 1 pound
    1 gallon = 128 oz
    dSalt/dt = .25 lbs/min
    dSolution/dt = -.5 gal/min

    3. The attempt at a solution

    At first, I just set this up like a simple ratio problem...
    .25 pounds salt = 4 ounces
    5 gallons water = 640 ounces
    4 oz salt / 640 oz water = x oz salt / 576 oz water
    x = (576*4)/(640) = 3.6 oz salt = .225 pounds salt

    However, I checked in the back of the book and the answer listed is .238 pounds of salt. I don't understand how they came up with that answer?

    Also, I'm a little confused as to how to incorporate the diff. eqs. into the solution. I don't think the way I solved it was right because I mostly just used alegbra. I think there should be a way to use both the rate that the salt is being poured in and the rate at which the solution is leaking out, but I don't know exactly how. If I can come up with the equation, I think I could do the rest of the integration and stuff, but it seems like I'm just stuck on the first part.

    Any help would be greatly appreciated. Thanks in advance :)
  2. jcsd
  3. Apr 9, 2007 #2


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    Staff Emeritus
    Science Advisor

    "Set it up as a simple ratio problem"?? You titled this "First Order Diff. Eq."! I doubt they are going to give you a "simple ratio problem" in Differential Equations!

    Let X(t) be the amount of salt, in pounds, in the bucket at t minutes. Since the bucket is kept full and the concentration is kept uniform, the concentration at t minutes is X(t)/5 pounds per gallon. Since 1/2 gallon of mixture is leaving the bucket every minute, (1/2)(X(t))/5 = X(t)/10 pounds of salt are leaving every minute. Also, there are .25 pounds of salt going into the bucket every minute. What is the net change in salt in the bucket every minute? In other words, what is dX/dt? Of course, X(0)= 0.
  4. Apr 9, 2007 #3
    Thank you very much for your help, HallsofIvy :) I think I figured it out now. I don't know what I was thinking before. I guess I just got desperate after spending so long stuck on this one problem and started trying anything I could think of.

    dX/dt = .25 - t/10
    dX / (t - 2.5) = -(1/10) dt
    ln |t - 2.5| = -(1/10)t + c
    t = 2.5 + ce^(-t/10)

    Substituting initial value in...
    0 = 2.5 + c
    c = -2.5

    So, the equation is 2.5 - 2.5e^(-t/10). I can use that to find the rest of the answers I think.
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