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Thread: simple differentiation??????!a level

  1. #1
    Member i_zz_y_ill's Avatar
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    simple differentiation??????!a level

    The radius of a circular ink black is increasing at a rate inversely poportional to its area A. find an expression for dA/dt.

    k so A=(pie)r^2,
    A'=2(pie)r dA/dt=dA/dr x dr/dt dr/dt proportionl to 1/A
    dr/dt = k/A
    dr/dt = k/(pie)r^2
    therefore: dA/dt = 2(pie)r x k/(pie)r^2
    =2k/r (A=(pie)r^2 IMPLIES r=a/(pie) squarerooted
    =2k x squarroot of pie/squareroot of a
    NOW, sorry aboout bad writing, i have answers and i cant get from here to
    ,,,,,,,,,,,,,,,,,,,,,,,,ans: k'/squareroot of a


    as in the differential of k,if anyone could spare a moment to engage their brains i bit better than myself that wold be gr8.thnx
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  2. #2
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    Hello, i_zz_y_ill!

    The radius of a circular ink blot is increasing at a rate inversely poportional to its area $\displaystyle A.$
    Find an expression for $\displaystyle \frac{dA}{dt}$ . . . . In terms of what? . $\displaystyle {\color{blue}r? \hdots A? \hdots \text{both?}} $

    Answer: .$\displaystyle \frac{k'}{\sqrt{A}}$
    A strange way to write the answer . . .
    Since $\displaystyle k$ is a constant, $\displaystyle k'$ is zero.
    So I must assume $\displaystyle k'$ means "some other constant" . . . a sloppy answer!

    We are told: .$\displaystyle \frac{dr}{dt} \:=\:\frac{k}{A}$ .[1]

    Since $\displaystyle A \:=\:\pi r^2$, we have: .$\displaystyle r \:=\:\frac{A^{\frac{1}{2}}}{\sqrt{\pi}} \quad\Rightarrow\quad \frac{dr}{dt} \:=\:\frac{1}{2\sqrt{\pi}A^{\frac{1}{2}}}\cdot\fra c{dA}{dt} $ .[2]


    Substitute [2] into [1]: . $\displaystyle \frac{1}{2\sqrt{\pi}{A^{\frac{1}{2}}}}\cdot\frac{d A}{dt} \:=\:\frac{k}{A} \quad\Rightarrow\quad \frac{dA}{dt} \;=\;\frac{2\sqrt{\pi}A^{\frac{1}{2}}}{1}\cdot\fra c{k}{A} $

    Therefore: . $\displaystyle \frac{dA}{dt} \;=\;\frac{2k\sqrt{\pi}}{\sqrt{A}} $

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