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Math Help - inhibited growth

  1. #1
    Junior Member
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    inhibited growth

    There is a problem in my textbook that is exactly like one of the examples, but just uses different numbers.

    Problem: A tank initially contains 400 gal of brine in which 100 lbs of salt are dissolved. Brine containing 1/10 lb of salt per gallon is run into the tank at 20 gal/min, the mixture being drained off at the same rate. How many lbs of salt remain in the tank after 30 minutes?

    In the example, we separate variables and integrate. I use the same method to solve this problem, it just has different numbers and I get: 40-60e^(-t/20). The books's answer is 40 + 60e^(-t/20).

    What am I doing wrong?
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  2. #2
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    I got it.

    When going through the problem you deal with the quantity ln(40-x). But 40-x would be negative at t=0, so we must be dealing with -(40-x)=(x-40)
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  3. #3
    A Plied Mathematician
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    Looks to me like you probably integrated

    \dfrac{1}{40-x} and got \ln|40-x|, whereas you should have gotten -\ln|40-x|. You have to do a u substitution in order to use that rule!
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