The Mixing Problem: Understanding the Dilution of Brine in a Tank

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The discussion revolves around solving a differential equation related to the dilution of brine in a tank. The initial problem involves a tank with 100 gallons of brine containing 30 pounds of salt, with fresh water entering at 3 gallons per minute and brine exiting at the same rate. A subsequent problem changes the inflow rate to 2 gallons per minute, leading to confusion about how this affects the salt concentration. Participants clarify that while fresh water dilutes the brine, the rate of salt entering the tank remains zero since only fresh water is added. Ultimately, the fresh water's inflow rate directly impacts the dilution process, leading to a gradual reduction in salt concentration over time.
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Hi all. I'm trying to help a friend work a differential equation problem since I've already taken this course and gotten an A in it, but this other problem has me a little confused.

The book is ODE's by Tenenbaum and Pollard, chapter 3 lesson 15a number 2.

#1) Tank initially holds 100 gal of brine containing 30 lb of dissolved salt. Fresh water flows into the tank at a rate of 3 gal/min and brine flows out at the same rate. (a) Find the salt content of the brine at the end of 10 minutes and (b) When will the salt content be 15 lb?

This one I've solved easily but number 2 says,

#2) Solve problem 1 if 2gal/min of fresh water enter the tank instead of 3 gal/min.

The solution on the next page says it should be (a) x=30(1-.01t)^3 which, when 10 is plugged for t gives 21.87 lb.

Can someone explain what I'm missing? Shouldn't the rate in still be zero?
 
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It is very difficult to explain what you are missing when you haven't explained what you did to try to solve the problem. Can you tell us what your solution is for both parts?

And it is unclear what you mean when you say "the rate in should still be zero". The rate of water flowing into the tank? This is clearly not zero.
 
Well for #1 I got a) x=30e^(-.03t), x(10)= 22.2 lb. b) 23.1 minutes

For number I've tried finding the integrating factor but that still leaves me with the question of whether the rate of fresh water in changing from 3gal/min to just 2gal/min has anything to do with the amount of solution in the tank. When I use rate in= 2gal/min and try the integrating factor method I get a solution that looks like this: x=-2/.03+(30+2/.03)e^(-.03t) which is a ridiculous result. Any clues?

And the rate of solution coming in is zero because the problem says it is fresh water flowing in. So rate of solution flowing in is zero but fresh water flowing in is 2 gal/min in problem 2 and 3 gal/min in problem 1.
 
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I guess my real question isn't "can someone help me solve this?" It's more "can someone explain how the rate of fresh water coming in affects the solution inside the tank?"
 
The fresh water flowing into the tank dilutes the brine solution in the tank. Brine is flowing out at the same rate as fresh water is flowing into the tank, so in time, what's in the tank will be just water with no salt.
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

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