# Curvature of implicit equation

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• Jan 4th 2013, 08:56 AM
PStudent175
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.
• Jan 4th 2013, 09:17 AM
skeeter
Re: Curvature of implicit equation
Quote:

Originally Posted by PStudent175
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 ...
• Jan 4th 2013, 09:21 AM
PStudent175
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
• Jan 4th 2013, 09:30 AM
skeeter
Re: Curvature of implicit equation
implicit derivative ...

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

$2y \cdot y' = -2x$

$y' = \frac{-2x}{2y} = -\frac{x}{y}$
• Jan 4th 2013, 11:06 AM
PStudent175
Re: Curvature of implicit equation
Ok great thanks, could you also show how you got y'' = -R^2 / y^3 ?
• Jan 4th 2013, 11:18 AM
skeeter
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 ...

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

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