Diff EQ applications problem

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[SOLVED] Diff EQ applications problem

A tank initially holds 25gallons of water. Alcohol enters at the rate of 2gallons/minute and the mixture leaves at a rate of 1 gallon/min. What will the concentration of alcohol be when there is 60gallons of fluid in the tank



The Attempt at a Solution


I don't know how to set up the equation to the problem, this is where I'm stuck-
 

Answers and Replies

  • #2
HallsofIvy
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Let A(t) be the amount of alcohol in the tank, in gallons, after t minutes. There are two gallons of alcohol coming in every minute, one gallon of mixture leaving it: the net amount of liquid coming in is 1 gal per minute. The total amount of liquid in the tank after t minutes is 25+ t. The concentration of alcohol after t minutes is A(t)/(25+t).

Now, dA/dt is the rate at which alcohol is coming in/going out in gal/min. You are told there are 2 gal/min of alcohol coming in and each gallon going out contains A(t)/(25+ t) gallions of alcohol. dA/dt= what?
 
  • #3
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So A(t)=#gal x time=at
Alcohol concentration of mixture leaving the tank at/(25+t), rate leaving 1gal/min=t
dA/dt=rate coming in - rate going out

would the initial equation be dA/dt=2t - t(at/(25+t))?
 
  • #4
HallsofIvy
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The rate at which alcohol is coming in is 2 gal/min, not "2t gallons". The rate at which liquid is leaving is -1 gal/min so the rate at which alcohol is leaving is -A(t)/(25+ t).

dA/dt= 2- A(t)/(25+ t). (A(t) is not "at".)
 

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