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Math Help - losing moisture at a rate proportional to moisture content

  1. #1
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    losing moisture at a rate proportional to moisture content

    A wet towel hung from a clothesline to dry loses moisture through evaporation at a rate proportional to its moisture content. If after hours the towel has lost % of its original moisture content, after how long will it have lost %?

    t=____ hours
    Last edited by mr fantastic; April 9th 2010 at 01:22 AM. Reason: Changed post title
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  2. #2
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ewkimchi View Post
    A wet towel hung from a clothesline to dry loses moisture through evaporation at a rate proportional to its moisture content. If after hours the towel has lost % of its original moisture content, after how long will it have lost %?

    t=____ hours
    \frac{dy}{dt} = k.y

    where k is the rate, and y is the proportion of the moisture.

    so,

     \frac{dy}{y} = k. dt

    Integrate both sides:

    \int \frac{dy}{y} = \int k. dt

    You will get:

    log(y) = kt + C..................(1)

    Use this equation to find out what your question is asking. First start with time t=0 to find out C. At time t=0, the towel

    has not lost any moisture. That means y=100

    So,

    log(100) = k \times 0 + C \rightarrow 2 = 0+C

     \therefore C=2

    Now find k. To find k, use the first part of your question.

    At time t=5 \text{hours}, the moisture lost is 10%. So the remaining moisture in the towel is 90.

    So, log(90) = (k \times 5) + 2

    \therefore k = -0.00915

    Now, that you have k and C, find t.

    Let t be the time when 70% moisture is lost, that is, the remaining moisture is 30. So, use equation (1) in a similar manner to find the time(t).
    Last edited by harish21; April 8th 2010 at 07:55 PM.
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    THanks

    but I got t= and it's wrong; what did you get?
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by ewkimchi View Post
    but I got t= and it's wrong; what did you get?
    I am sorry. I posted the wrong value of k.

    k = \frac{log(90)-2}{5} = -0.00915

    With this, and 70% moisture gone, the time should be

    log(30) = (-0.00915) \times t + 2

    which gives t = 57.14 hours
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