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    x = 15 - 12e^-t/14

    show that dx/dt = 1/14(15-x)

    im so stuck where the 15-x comes from...

    someone help out pls
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    (Original post by Gary)
    x = 15 - 12e^-t/14

    show that dx/dt = 1/4(15-x)

    im so stuck where the 15-x comes from...

    someone help out pls
    It comes from the original equation for x.

    Show your working.
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    (Original post by Gary)
    x = 15 - 12e^-t/14

    show that dx/dt = 1/4(15-x)

    im so stuck where the 15-x comes from...

    someone help out pls
    What's 15-15?
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    There is an error in the question by the way but I presume it is a typo ...?
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    The question is on AQA June 2007, question 4ci) if you want a better look at it


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    (Original post by Gary)
    The question is on AQA June 2007 / 4ci) if you want a better look at it
    Can't say that I do.
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    (Original post by Mr M)
    Can't say that I do.
    Well I don't think there's an error in the question as I checked and there's no typo


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    (Original post by Gary)
    Well I don't think there's an error in the question as I checked and there's no typo


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    Yes there is.
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    Oh yeah... Should be dy/dx = 1/14(15-x)


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    (Original post by Gary)
    Oh yeah... Should be dy/dx = 1/14(15-x)


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    So have you solved it yet or are you still struggling?
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    Nahh I'm stuck..


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    (Original post by Gary)
    Nahh I'm stuck..


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    Differentiate the equation for x to find dx/dt.

    Then replace the \displaystyle e^{-\frac{t}{14}} bit by changing the subject of the equation for x.
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    So I get dx/dt = 6/7e^-t/14

    But I don't get how you replace it by changing the subject of the equation for x

    Is it possible you post the soltuion step by step?

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    x= 15 -12e^(t/14)
    dx/dt= -12e^(t/14) * -1/14
    Rearrange origininal eq to get (15-x)=12e^(t/14)
    Sub in
    dx/dt=-(15-x)*-1/14
    =1/14(15-x)

    Pretty much what guy above said. Hope it helps.
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    So it was that simply all along.... Sigh

    Thank you very much for the help!


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