Second derivative as a quotient

Posted on 9 February, 2015 by jesperg

The second derivative of a function y=f(x) can be expressed as

\displaystyle f''(x)=\frac{d}{dx}\frac{dy}{dx}=\frac{d^2ydx-d^2xdy}{dx^3}=\frac{d^2y}{dx^2}-\frac{d^2xdy}{dx^3}

So there is an extra correction term needed to be able to interpret the terms as quotients. To evaluate the second derivative we need to specify the velocity (dx,dy) and acceleration (d^2x,d^2y) at the point (x,y).

Example 1. Let (x,y)=(\cos\lambda,\sin\lambda) be a curve in \mathbf{R}^2. At the point (1/2,\sqrt{3}/2) we have (dx,dy)=(-\sqrt{3}/2,1/2) and (d^2x,d^2y)=(-1/2,-\sqrt{3}/2). We get

\displaystyle f''\left(\frac{1}{2}\right)=\frac{-\sqrt{3}/2}{(-\sqrt{3}/2)^2}-\frac{-1/2\cdot 1/2}{(-\sqrt{3}/2)^3}=-\frac{8}{3\sqrt{3}}.

Example 2. If we have a symmetric connection

\displaystyle \begin{aligned}d^2x&=-(\Gamma_{11}^1dx^2+2\Gamma_{12}^1dxdy+\Gamma_{22}^1dy^2)\\d^2y&=-(\Gamma_{11}^2dx^2+2\Gamma_{12}^2dxdy+\Gamma_{22}^2dy^2),\end{aligned}

we can combine the two acceleration expressions into

\displaystyle \frac{d^2ydx-d^2xdy}{dx^3}=\Gamma^1_{22}\left(\frac{dy}{dx}\right)^3-(\Gamma^2_{22}-2\Gamma^1_{12})\left(\frac{dy}{dx}\right)^2+(\Gamma^1_{11}-2\Gamma^2_{12})\left(\frac{dy}{dx}\right)-\Gamma^2_{11}

which is the ODE for geodesic paths

y''=\Gamma^1_{22}(y')^3-(\Gamma^2_{22}-2\Gamma^1_{12})(y')^2+(\Gamma^1_{11}-2\Gamma^2_{12})y'-\Gamma^2_{11}

This entry was posted in differential geometry and tagged , . Bookmark the permalink.

Leave a comment