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Math Help - Curvature of implicit equation

  1. #1
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    Curvature of implicit equation

    Hi,

    I need to find the curvature k of y=sqrt(R^2 - x^2)

    I know k = |f''| / (1 + f' ^2) ^1.5

    But I think I have gone somewhere with my differentiation as I didn't end up with k = 1/R which is the answer :/

    Thanks in advance for the help.
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  2. #2
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    Re: Curvature of implicit equation

    Quote Originally Posted by PStudent175 View Post
    Hi,

    I need to find the curvature k of y=sqrt(R^2 - x^2)

    I know k = |f''| / (1 + f' ^2) ^1.5

    But I think I have gone somewhere with my differentiation as I didn't end up with k = 1/R which is the answer :/

    Thanks in advance for the help.
    y^2 = R^2 - x^2

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

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


    k = \frac{\frac{R^2}{y^3}}{\left(1 + \frac{x^2}{y^2}\right)^{\frac{3}{2}}}

    clean up the algebra ...
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  3. #3
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    Re: Curvature of implicit equation

    Sorry how did you get y' = -x/y ?

    I got y' = f' = -x(R^2 - x^2)^-0.5 from the chain rule
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  4. #4
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    Re: Curvature of implicit equation

    implicit derivative ...

    \frac{d}{dx}(y^2 = R^2 - x^2) ... note that R is a constant

    2y \cdot y' = -2x

    y' = \frac{-2x}{2y} = -\frac{x}{y}
    Thanks from PStudent175
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  5. #5
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    Re: Curvature of implicit equation

    Ok great thanks, could you also show how you got y'' = -R^2 / y^3 ?
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  6. #6
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    Re: Curvature of implicit equation

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

    quotient rule ...

    y'' = -\frac{y - xy'}{y^2}

    y'' = -\frac{y - x\left(-\frac{x}{y}\right)}{y^2}

    y'' = -\frac{y + \frac{x^2}{y}}{y^2}

    multiply numerator and denominator by y to clear the complex fraction ...

    y'' = -\frac{y^2 + x^2}{y^3}

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