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Math Help - Some help

  1. #1
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    Some help

    The rate at which the temperaturex of a liquid decreases is inversely proportional ites temperature at time t. write down a differential eq and solve it. given that the temp of the liquid increases from 20 to 20.5 in 2 seconds find temp of body after 10 seconds...

    Inversely proportional = 1/k

    so eq would be

    dx/dt = 1/kx

    kx dx = dt

    integrating

    kx^2 = t + c

    ? 2 unknowns ? or what am i doing wrong here
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  2. #2
    Super Member Matt Westwood's Avatar
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    \frac {dx}{dt} = \frac 1 {kt} is more like it!

    Mind, now I look at it I'm not sure why ...
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  3. #3
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    Quote Originally Posted by ramzel View Post
    The rate at which the temperaturex of a liquid decreases is inversely proportional ites temperature at time t. write down a differential eq and solve it. given that the temp of the liquid increases from 20 to 20.5 in 2 seconds find temp of body after 10 seconds...
    \frac{dx}{dt} = \frac{k}{x}
    Last edited by skeeter; September 2nd 2009 at 03:09 PM. Reason: error in set up
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  4. #4
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    \frac{dx}{dt} = \frac{k}{x}

    separate variables ...

    \frac{dx}{x} = k \, dt

    huh when you cross multiply doesn't the x goes up ?
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  5. #5
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    Quote Originally Posted by ramzel View Post
    \frac{dx}{dt} = \frac{k}{x}

    separate variables ...

    \frac{dx}{x} = k \, dt

    huh when you cross multiply doesn't the x goes up ?
    you're right ... my error. was thinking of something else.

    \frac{dx}{dt} = \frac{k}{x}

    x \, dx = k \, dt
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  6. #6
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    Thanks i got the right answer
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