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Forming differential equationsfrom words to formula

  1. Oct 2, 2007 #1

    rock.freak667

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    1. The problem statement, all variables and given/known data

    The rate of increase of thickness of ice on a pond is inversely proportional ro the thickness of ice already present. It is known that,when the thickness is x cm and the temperature is of the air above the ice is -a degrees celcius, the rate is [itex]\frac{ax}{14400} cms^{-1}[/itex] . Form an appropriate diff. eq'n and show that, if the air temp is -10 degrees Cel. , the time taken for the thickness to increase from 5 com to 6 cm is a little more than 2 hours

    2. Relevant equations



    3. The attempt at a solution

    "The rate of increase of thickness of ice on a pond is inversely proportional or the thickness of ice already present."

    [tex] \frac{dx}{dt} \alpha x => \frac{dx}{dt}=kx[/tex]

    therefore [tex] lnx=Ae^{kt}[/tex] and that looks wrong as I have yet to use the given rate...I am confused about the thickness and the given rate
     
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  3. Oct 2, 2007 #2

    EnumaElish

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    How do you get [tex] \frac{dx}{dt} \alpha x => \frac{dx}{dt}=kx[/tex]? Is that really inversely proportional?
     
  4. Oct 2, 2007 #3

    Kurdt

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    The rate at a given point is merely there so you can determine your constant.
     
  5. Oct 2, 2007 #4

    rock.freak667

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    oh my bad..i did the question wrong.supposed to be [tex]x^2=Kt+C[/tex]..but I only have one piece of info and 2 constants 'a' and 'k'
     
  6. Oct 2, 2007 #5

    Kurdt

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    You've missed a factor of a half. What you do know is that:

    [tex] \frac{dx}{dt}=kx=\frac{ax}{14400} cms^{-1}[/tex] at some thickness x and temperature -a degrees C.
     
  7. Oct 2, 2007 #6

    rock.freak667

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    I multiplied through out by 2..2k=K and 2c=C ....so then k=a/14400 and then sub that in the eq'n
     
  8. Oct 3, 2007 #7

    HallsofIvy

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    And you have missed the point that EnumaElish raised: the rate is inversely proportional to the thickness, not directly proportional. Assuming x is the thickness of the ice at time t (I personally would insist upon that being explicitely stated), then
    [tex]\frac{dx}{dt}= \frac{k}{x}[/tex]
    That gives a whole different situation!
     
  9. Oct 3, 2007 #8

    Kurdt

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    Argh! I apologise. I did this myself at 3 am and was just about to post that he hadn't written it inversely proportional when enuma elish beat me to it, then I go and make the same mistake. Can't believe I did that. Anyway sorry to the OP for any confusion.
     
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