Modeling problem

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  • #1
nrm
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I was hoping that I could find help solving this problem. I've been working on it for a while and haven't been able to solve it correctly.

A tank originally contains 100 gal of fresh water. Then water containing 0.5 lb of salt per gallon is poured into the tank at a rate of 2 gal/min, and the mixture is allowed to leave at the same rate. After ten minutes the process is stopped, and fresh water is poured into the tank at a rate of 2 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min.
So far I have:
[tex]
\frac{dQ}{dt} = r - \frac{rQ}{100}
[/tex]

Where Q is the amount of salt, r is the rate in and out.
I think that this may work for the first 10 minutes while the salt is part of what is getting mixed in, but after that I'm lost. Any help will be appriciated.
Thank you.
 
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Answers and Replies

  • #2
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you first need an expression for dy/dt

dy / dt = rate_in - rate_out

rate_in = .5 lb/gal * 2 gal/min = 1lb/min

rate_out = y(t) / 100 lb/gal * 2 gal / min = y(t) / 50 lb/min

so dy/dt = 1 lb/min - y(t) / 50 lb/min

then you find the IF and go from there
 
  • #3
HallsofIvy
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nrm said:
I was hoping that I could find help solving this problem. I've been working on it for a while and haven't been able to solve it correctly.



So far I have:
[tex]
\frac{dQ}{dt} = r - \frac{rQ}{100}
[/tex]
Where Q is the amount of salt, r is the rate in and out.
No, not quite. There are 2 gallons of water coming in every minute (I don't know why you call that "r") and each gallon carries 1/2 pound of salt so 2*(1/2) = 1 pound of salt comes in every minute ((1/2)r, not r). Each gallon of water going out contains [itex]\frac{Q}{100}[/itex] pounds of salt so the amount going out is [itex]\frac{rQ}{100}[/itex] as you have (with, of course, r= 2). The differential equation is:
[tex]\frac{dQ}{dt}= 1- \frac{Q}{50}[/tex]
Also, of course, Q(0)= 0.

I think that this may work for the first 10 minutes while the salt is part of what is getting mixed in, but after that I'm lost. Any help will be appriciated.
Thank you.
Yes, so solve that differential equation for Q(t) and use it to find Q(10).
Now, start all over again. Now, there is NO salt coming in so the differential equation is
[tex]\frac{dQ}{dt}= -\frac{Q}{50}[/tex].
Q(10)= whatever you found before and then find Q(20) (ten more minutes).

(Since t does not appear explicitely in that equation, you could just "restart" the clock: Let Q(0)= whatever you found before and then find Q(10) from this new equation.)
 
  • #4
nrm
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Thanks to both of you for replying. I've got it all worked out.
 

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