D.E mixing problem with rate out

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SUMMARY

The discussion focuses on a differential equation modeling the concentration of alcohol in a tank as it fills with pure alcohol at a rate of 2 gallons per minute while simultaneously allowing a mixture to exit at 1 gallon per minute. The correct formulation for the outflow rate is established as q/(40+t), leading to the differential equation dq/dt = 2 - q/(40+t). The concentration of alcohol is determined by dividing the amount of alcohol, q, by the total volume of fluid in the tank, which is 40 + 2t gallons.

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Homework Statement


A large tank initially hold 40 gallons of water. Pure alcohol enters at a rate of 2 Gal/min is well stirred and the mixture leaves at 1gal/min. What will be the concentration when of alcohol when the tank has 80 gallons of fluid in it.


Homework Equations


q= amount of alcohol
i have dq/dt=rate in -rate out
rate in=2gal/min of OH
it will take 40 mins to fill to 80 gal
and the thing I am wandering about is the rate out.
i currently have
rate out=q/(40+2t), but I am thinking maybee its q/(40+t)

as for the rest i think I am fine its just this outflow rate I am not sure on.
 
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Your second guess for the outflow rate appears correct to me, making your differential equation:
[tex]\frac{dq}{dt}=2-\frac{q}{40+t}[/tex]

Any reason you are not sure that you would like to discuss?

Don't forget that when you are answering the problem they are asking for concentration of alcohol; your state variable q is currently defined as the amount of alcohol in gallons.
 
well at first i was thinking of using (40+2t) since that was the initial volume + the flow in but then i thought about how the acutal vol change is just t when you account for the out flow.

as far as the concentration goes what i did was set up the D>E for the quantity of OH in the container then just divid q by (40+2t)

thanks for the help i sortof needed that conformation.
 

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