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DE word problem

  1. Feb 2, 2006 #1


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    Ok, i think this is the last DE problem i do before i strategically place myself between a wall and a speeding train.

    The question is:

    A tank initially contains 60 gal. of pure water. Brine containing 1 lb of salt per gallon enters the tank at a rate of 2 gal./min. The well-mixed solution flows out of a hole in the tank at a rate of 3 gal./min. What is the maximum amount of salt the tank will ever hold? (Hint: First determine the amount of salt in the tank at any given time.)

    Now i think i'm suppose to start here but im not sure.:

    r_i = 2 \\
    c_i = 1 \\
    r_o = 1.5 \\
    V = 60 \\
    \frac{{dx}}{{dt}} = r_i c_i - \frac{{r_o }}{v}x \\

    ri = rate in
    ci = concentration in lb/gal
    ro = rate out
    V = volume

    I'm very lost and starting to looooose my mind over this course! Any help? tips? suggestions? death threats?
  2. jcsd
  3. Feb 2, 2006 #2


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    I like the fact that you carefully write out what each parameter means- not only does it help us, it helps you keep your thoughts straight. That's why I comment on the fact that "x" is not in that list! Also "rate in" and "rate out" are not specific enough. Rate of water or salt in and out? And what units. Oh, and why is ro= 1.5? You problem says the water flows out at 3 gal/min.

    From the equation, I guess that x(t) is the amount of salt, in lbs, in the tank at time t, in minutes, so that [itex]\frac{dx}{dt}[/itex] is the rate at which that amount is changing, in lbs/min. Okay, that amount is changing because there is some salt coming in and some going out.

    "Brine containing 1 lb of salt per gallon enters the tank at a rate of 2 gal./min." Okay so salt is coming in at (1 lb/gal)(2 gal/min)= 2 lb/min.

    " The well-mixed solution flows out of a hole in the tank at a rate of 3 gal./min." With x the amount of salt in the tank, in lbs, and V the volume of water in the tank in gal., x/v is the concentration of salt in the water in lb/gal. Water at the concentration is leaving the tank at 3 gal/min so salt is leaving the tanks at (x/v lb/gal)(3 gal/min)= 3x/v lb/min

    dx/dt= 2- (3/v)x. That's exactly what you get when you put the numbers into your formula (and why didn't you?). It's much better to be able to derive the d.e. from the situation that just copy a formula.

    One complication here- in most of these problems the amount of water flowing in and out are the same so the volume of water is a constant. Here there are 2 gal/min coming in and 3 gal/min going out. The tank is losing a net 1 gal/min. Can you see that V= 60gal- (1 gal/min)(t min)=
    60- t?

    Your differential equation is
    [tex]\frac{dx}{dt}= 2- \frac{3}{60-t}x= \frac{120- 2t- 3x}{60- t}[/tex]
    with x(0)= 0

    That equation is not separable nor is it exact. But it is "linear" and it's not difficult to find an integrating factor.

    Once you've solved the equation, find the maximum value of x. Don't forget to check the endpoints. One endpoint is obviously at t= 0, when the amount of salt in the tank is 0- obviously a minimum, not a maximum. The other endpoint is when the tank runs dry. When will that be?

    Oh, and keep careful track of "signs". Remember that the integral of 1/x is ln |x| and that t is always less than ___.
    Last edited: Feb 3, 2006
  4. Feb 2, 2006 #3


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    God! I miss one lecture and i might as well have dropped the class haha. Every method i seem to have needed to use was covered the 1 day i missed a lecture... I'll try to figure out how to do this when i get home
  5. Feb 2, 2006 #4


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    A linear first order differential equation is an equation of the form
    dy/dx+ p(x)y= f(x), where p(x) and f(x) are functions of x. That is, there is no "function" of y, like y2 or sin(y) or such.

    Such an equation is "exact" if the left side can be written as a single derivative: d(u(t)y)/dx= u(t)(dy/dx)+ (du/dx)y.

    An integrating factor is a function m(x) such that multiplying the entire equation by it make the equation "exact":
    m(x)(dy/dx)+ m(x)p(x)y= d(m(x)y)/dx= m(x)(dy/dx)+ (dm/dx)y.

    Notice that the m(x)(dy/dx) terms cancel and we can divide through then by y to get m(x)p(x)= dm/dx, a separable differential equation for m(x).

    In your equation
    [tex]\frac{dx}{dt}+\frac{3}{60-t}= 2[/tex]
    p(t) (y= x and x= t in what I wrote above!) so the equation for the integrating factor is
    [tex]\frac{dm}{dt}= \frac{3m}{60- t}[/itex]
    an easy separable equation for m.
  6. Feb 3, 2006 #5


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    Same thing happens to me!

    Very annoying because it not means I can't miss class period.

    I rarely ever skip unless I have to, but when that happens, it's actually valuable!
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