Solving the Example Problem: Tank w/Salt & Brine

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Homework Help Overview

The discussion revolves around a problem involving a tank containing water and salt, where brine flows in and out at a constant rate. Participants are exploring how to model the amount of salt in the tank over time, given the rates of inflow and outflow of the brine solution.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to establish the rates of salt flow into and out of the tank. There are questions regarding the correct formulation of the outflow rate based on the concentration of salt in the tank.

Discussion Status

Some participants have provided hints regarding the differential equation that models the situation, while others are questioning the assumptions made about the outflow rate. There is acknowledgment of potential errors in the original problem setup, and participants are clarifying the units involved.

Contextual Notes

There are indications of typos in the original problem statement, which may affect the interpretation of the rates and units involved. Participants are also navigating the constraints of the problem as presented in the textbook.

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."
 
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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 ...
 
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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