Solving the Example Problem: Tank w/Salt & Brine

AI Thread Summary
The discussion revolves around solving a problem involving a tank with salt and brine, where the initial conditions include 1000 gallons of water and 200 pounds of dissolved salt. The inflow rate of brine is 50 gallons per minute, with each gallon containing (1+cost) pounds of salt. The outflow rate is determined by the concentration of salt in the tank, which is y(t)/1000 pounds per gallon, leading to an outflow of y(t)/20 pounds per minute. A clarification was made regarding the calculation of the outflow rate, correcting an initial error in the setup. The conversation highlights the importance of unit consistency in solving the problem accurately.
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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."
 
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HINT: \frac{dA}{dt} = r_{in} - r_{out}

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 ...
 
Why's rate out = y(t)/1000 gal?
 
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.
 
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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