what is the second derivative of x^2 + y^2 = s^2

thanks

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- Dec 4th 2007, 06:12 PMdoctorgkderivative
what is the second derivative of x^2 + y^2 = s^2

thanks - Dec 4th 2007, 06:22 PMtopsquark
Ummm... You are looking for $\displaystyle \frac{d^2y}{dx^2}$? And I presume s is a constant?

$\displaystyle 2x + 2y \frac{dy}{dx} = 0$

$\displaystyle \frac{dy}{dx} = -\frac{x}{y}$

$\displaystyle \frac{d^2y}{dx^2} = -\frac{y - x \frac{dy}{dx}}{y^2}$

Insert the equation for $\displaystyle \frac{dy}{dx}$:

$\displaystyle \frac{d^2y}{dx^2} = -\frac{y - x \cdot -\frac{x}{y}}{y^2}$

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

Inserting the original equation:

$\displaystyle \frac{d^2y}{dx^2} = -\frac{s^2}{y^3}$

I'd probably leave it like that.

-Dan