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Mixing problem

  • Thread starter asdf1
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there's an example problem in my textbook, but i'm stuck on how to make the first move~
"a tank contains 1000gal of water in which 200lb of salt are dissolved. Fifty gallonw of brine, each containing (1+cost)lb of dissolbed salt, run into the tank per minute. The mixture, kept uniform by stirring, runs out the same rate. Find the amount of salt y(t) in the tank at any time t."
 

Tide

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HINT: [tex]\frac{dA}{dt} = r_{in} - r_{out}[/tex]

where A is the amount of salt in the tank and r is the rate of salt flow in or out of the tank.
 
rate in = 50 lbs/min * (1+cost)lb / gal

rate in = y(t) / 1000 gal

dv /dt = rin - rout

find the intergating factor and ....
 
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Why's rate out = y(t)/1000 gal?
 

HallsofIvy

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asdf1 said:
Why's rate out = y(t)/1000 gal?
It's not- there's a slight error. If y(t) is the amount of salt in the entire tank then y(t)/1000 is the amount of salt in each gallon. (Notice that that is now in "pounds per gallon". Since the solution is going out of the tank at 50 gallons per minute, there will be (y(t)/1000 pounds/gallon)(50 gallon/minute)= y(t)/20 pounds/min

Since mathmike got the inflow right, I suspect that was just a typo.
 
734
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crummy.... I just noticed I have a lot of typos in my original question! sorry about that! :P
thanks! i didn't consider the different units... :P
 

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