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A textbook example asks me:

A large tank is filled to capacity with 100 gallons of pure water. Brine containing 3 pounds of salt per gallon is pumped into the tank at a rate of 4 gal/min. The well-mixed solution is pumped out of the tank at the rate of 5 gal/min. Find the amount of salt after 30 minutes.

I took the following apparently incorrect approach to solving it:

I first calculated the total amount of liquid in the tank to be $100+4t-5t$ with $t$ denoting time in minutes.

I then observed that the amount of salt coming in was $\frac{3 \ \text{lbs}}{\text{min}}$. I determined that the amount of salt going out was $ \frac{\text{Salt}\frac{\text{lbs}}{\text{gal}}}{100-t}$.

To calculate the total amount of salt, I simply took the amount of incoming salt and subtracted the outgoing salt, plugged in a value of 30 for $t$ and arrived at 87.85 pounds of salt at 30 minutes. However, the correct answer in the answer section is given as 209.97 lbs.

Can anyone suggest what I did wrong?

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For starters, how did you arrive at three pounds per four gallons? The text says 3 pounds per gallon. –  joriki Oct 4 '12 at 16:00
Sorry, that should say $\frac{3 \ gallons}{4 \ minute}$. I'll fix that now... –  Imray Oct 4 '12 at 16:06
Now I'm really confused. You announced that you'd change it to 3 gallons/4 minutes, and now you've changed it to 3 pounds/4 minutes. I think the most important thing you probably need to solve the problem is to be more careful. –  joriki Oct 4 '12 at 16:19
Sorry Sorry - silly mistake!! That might have been my issue, I'll try it again and let you know if I got it... –  Imray Oct 4 '12 at 16:25
I fixed it and solved it properly. Thank you for pointing out my mistake. I'll reread a few more times before posting it up next time :) –  Imray Oct 4 '12 at 16:42

2 Answers 2

up vote 0 down vote accepted

I misread the question and misinterpreted the amount of incoming salt.

There are 3 pounds per gallon of incoming brine, and 4 gallons being pumped in per minute, in other words, 12 pounds incoming.

If $A(t)$ is the function describing the total amount of salt as a function of time, then $\frac{dA}{dt} = 12 - \frac{5A(t)}{100-t}$.

Solving the differential equation gave me $A(t) = [3(100-t)^{-4}+C](100-t)^5 $ and then I solved for C by using $A(0) = 0$ as an initial condition (since the tank has no brine at the beginning). My final function for $A(t)$ is $ = 3(100-t)-\frac{3}{100^4}(100-t)^5$ and $A(30) = 209.97$


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Finally you got a differential equation into the solution of your differential equation problem :) –  rschwieb Oct 4 '12 at 20:03
lol that should've given it away :) –  Imray Oct 4 '12 at 20:09

this is how i interpret the problem..

Let x= be the amount of the water after (t) minutes Thus: volume of the mixture of the tank after (t) minutes =100+(4-5)t =100-t

dx/dt=rate of gain - rate of loss dx/dt=3lb/gal)(4gal/min)-(xlb/100gal-t gal/min) dx/dt=(12lb/min)-[x lb/gal(100-t/min)][5gal/min] dx/dt=(12lb/min)-(5x lb/100min-t) dx=12dt-(5x/100-t)dt soving the d.e dx+5x/100-t=12dt integrate. linear in x

e^integral of (5/100-t)dt =e^-5ln(100-t) =(100-t)^-5 integral of [ (100-t)^-5(dx+5x/(100-t)]= integral of [ (100-t)^-5(12dt) (100-t)^-5 x = 12(100-t)^-4/-4 + c (100-t)^-5 x=-3(100-t)^-4 + c where t=0 x=0 (100-0)^-5 (0) = -3(100-0)^-4 0=-3(100)^-4 + c c=3(100)^-4 thus: [(100-t)^-5 x = -3(100-t)^-4 + 3(100)^-4] divide both side by (100-t)^-4

where t=30


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