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    The radius of a circular ink blot is increasing at a rate inversely proportional to its area A. Find an expression for dA/dt.

    So
    dr/dt = k/A

    A= pir^2
    dA/dr = 2pi(r)

    Then what? Are we supposed to use the chain rule? Becuase I tried that and it doesn't work, the answer in the back says-
    k'/rootA

    HELP!!
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    (Original post by Hippysnake)
    The radius of a circular ink blot is increasing at a rate inversely proportional to its area A. Find an expression for dA/dt.

    So
    dr/dt = k/A

    A= pir^2
    dA/dr = 2pi(r)

    Then what? Are we supposed to use the chain rule? Becuase I tried that and it doesn't work, the answer in the back says-
    k'/rootA

    HELP!!
    you've got

    \displaystyle \frac{dr}{dt} = \frac{k}{A}

    \displaystyle A= \pi r^2

    \displaystyle \frac{dA}{dr} = 2\pi r

    so \displaystyle \frac{dr}{dt}  \frac{dA}{dr} = \frac{dA}{dt} = \frac{2k}{r}

    but \displaystyle A= \pi r^2 so \displaystyle \sqrt{\frac{A}{\pi}}= r

    see what you can do from there
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    The radius of a circular ink blot is increasing at a rate inversely proportional to its area A. Find an expression for dA/dt.

    So
    dr/dt = k/A

    A= pir^2
    dA/dr = 2pi(r)

    Then what? Are we supposed to use the chain rule? Becuase I tried that and it doesn't work, the answer in the back says-
    k'/rootA

    HELP!!
    Use that

     \frac{dA}{dt} = \frac{dA}{dr} \times \frac{dr}{dt}

    and then mess about with some constants until you get the answer
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    (Original post by Chaoslord)
    you've got

    \displaystyle \frac{dr}{dt} = \frac{k}{A}

    \displaystyle A= \pi r^2

    \displaystyle \frac{dA}{dr} = 2\pi r

    so \displaystyle \frac{dr}{dt}  \frac{dA}{dr} = \frac{dA}{dt} = \frac{2k}{r}

    but \displaystyle A= \pi r^2 so \displaystyle \sqrt{\frac{A}{\pi}}= r

    see what you can do from there
    Ahh thank you, I wasn't getting the answer stated by the book.

    However, why is answer \frac{2k\sqrt{\pi}}{\sqrt{A}} better than \frac{2k}{r} ?
 
 
 
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