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C4 Differentiation Edexcel

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    (Original post by rubadubdub)
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    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A
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    (Original post by rubadubdub)
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    they have k' as a positive constant.

    they then have -k'/sqrtA times sqrt(h)

    so they say that k'/sqrtA is a positive constant k, leaving -k sqrt(h)

    i.e. k and k' are two different constants.
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    (Original post by rubadubdub)
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    They have mentioned that  k=\dfrac{k'}{\sqrt{A}}

    You are confusing  k with  k' , both are different constants.
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    (Original post by rubadubdub)
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    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





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    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A
    (Original post by just george)
    they have k' as a positive constant.

    they then have -k'/sqrtA times sqrt(h)

    so they say that k'/sqrtA is a positive constant k, leaving -k sqrt(h)

    i.e. k and k' are two different constants.
    (Original post by raheem94)
    They have mentioned that  k=\dfrac{k'}{\sqrt{A}}

    You are confusing  k with  k' , both are different constants.

    Thanks but why is k' equivalent to (-k' . A^-1/2)


    Sorry i understand now thanks
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    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





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    thanks
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    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





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     k = \dfrac{k'}{\sqrt{A}}

k \not= -\dfrac{k'}{\sqrt{A}}

    Probably a typo from you.
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    (Original post by raheem94)
     k = \dfrac{k'}{\sqrt{A}}

k \not= -\dfrac{k'}{\sqrt{A}}

    Probably a typo from you.
    No typo.

    There is nothing wrong with what I have typed.
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    (Original post by steve2005)
    There are two k's one is K dash and then other is k.

    k is defined by reference to k dash and the square root of A





    Uploaded with ImageShack.us

    (Original post by steve2005)
    No typo.

    There is nothing wrong with what I have typed.
     \frac {dh}{dt} = \frac{-k'\sqrt{Ah}}{A} = (\frac{-k'}{\sqrt{A}})\sqrt{h} = -k\sqrt{h}

     where  k = (\frac{k'}{\sqrt{A}})

    I think thats what raheem94 was trying to point out
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    (Original post by just george)
     \frac {dh}{dt} = \frac{-k'\sqrt{Ah}}{A} = (\frac{-k'}{\sqrt{A}})\sqrt{h} = -k\sqrt{h}

     where  k = (\frac{k'}{\sqrt{A}})

    I think thats what raheem94 was trying to point out
    I know what he was trying to say. But what I wrote is correct because k is defined, in this case, to be negative.

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