# Curvature of implicit equation

• 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.

$\displaystyle y^2 = R^2 - x^2$

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

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

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

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 ...

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

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

$\displaystyle y' = \frac{-2x}{2y} = -\frac{x}{y}$
• 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
$\displaystyle y' = -\frac{x}{y}$

quotient rule ...

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

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

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

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

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

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