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Math Help - Slope of a Circle

  1. #1
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    Slope of a Circle

    ...is y'=\frac{-x}{y}

    My book says the second derivative of a circle is also y''=\frac{-x}{y}
    Every time I do it, I get y''=\frac{-y}{x}

    So what is the 2nd derivative of a circle?
    Por favor show some steps or tell me how you did it?

    Ty
    Editops! wrong secition. This isn't homework.
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  2. #2
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    Krizalid's Avatar
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    Quote Originally Posted by Truthbetold View Post
    My book says the second derivative of a circle is also y''=\frac{-x}{y}
    Every time I do it, I get y''=\frac{-y}{x}
    Show your procedure.
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  3. #3
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    Quote Originally Posted by Krizalid View Post
    Show your procedure.
    Got it solved by help from somewhere else.

    The book doesn't say y''= -x/y
    It doesn't say anything about the derivative of -x/y, even thouhg they ask you to do it and its an odd problem. Very strange this book.

    y'' solved by complicated methods i've never heard before = \frac{36}{y^3}
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  4. #4
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    Quote Originally Posted by Truthbetold View Post
    ...is y'=\frac{-x}{y}

    ...
    Hi, obviously you are refering to a circle with the equation

    x^2 + y^2 = r^2

    Then the first derivative (using implicite derivation) is

    y' = \frac{-x}{y}

    To get the second derivative use quotient rule and chain rule:

    y''=\frac{y \cdot (-1) - (-x) \cdot y \cdot y'}{y^2}

    Substitute y':

    y''=\dfrac{y \cdot (-1) - (-x) \cdot y \cdot \frac{-x}{y}}{y^2} = \frac{-x^2 - y^2}{y^3}

    Use the first equation and you'll get:

    y''=\frac{-r^2}{y^3}

    Actually you have to consider 2 cases:

    1. The conditions of the upper hemisphere
    2. The conditions of the lower hemisphere
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