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a tank contains 100 gallons of water. in error, 300 lbs of salt are poured into the tank instead of 200 lbs. to correct this condition, a stopper is removed from the bottom of the tank allowing 3 gal of brine to flow out per minute. at the same time, 3 gallons of fresh water per minute are pumped into the tank. if the mixtire is kept uniform by constant stirring, how long will it take for the brine to contain the desired amount of salt?

MIT 18.03SC Differential Equations
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You might want to look at Practice Problems 4. There is a mixing problem there. I know there have also been mixing problems in the lecture, though I am not sure exactly where. Perhaps you can find some just before Practice Problems 4. Here is one way to approach the question: v= 100 gallons, s=amount of salt in tank, t= time in min \[ds/dt = -[s(t)/100 gallons]\times (3 gallons) \]Just solve the ODE. I won't show all the steps. \[\int\limits_{}^{}1/s ds = -\int\limits_{}^{} 3/100 dt\]\[s=ce ^{3/100t}\]\[s=300e ^{-3/100t}\]Then we want to solve \[200=300e ^{-3/100t}\]. We find t=-100/3ln(2/3). (Don't worry, t is not negative since ln(2/3)<0.)
http://ocw.mit.edu/courses/mathematics/18-03sc-differential-equations-fall-2011/unit-i-first-order-differential-equations/first-order-linear-odes/MIT18_03SCF11_rec_03s4.pdf
okay thanks..i'll try this

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Okay, I hope it makes sense. The ODE ds/dt =-s(t)/100x3 is just saying that the change in the amount of salt over time is, first of all, negative, because the amount of salt in the tank is decreasing. How much is the amount of salt in the tank decreasing by? Well, at time t, the concentration of salt in the tank is the amount of salt in the tank divided by the volume, i.e. s(t)/100gallons. And, in 1 minute, 3 gallons leave the tank. So, we want to multiply s(t)/100gallons times 3 gallons to get the total amount of salt leaving the tank per minute. Thus ds/dt = -s(t)/100gallons x 3 gallons. If the ODE is right, I think it should make sense to you.

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