PX285 - C6b - a particle in a cartesian plane

consider a particle moving in a cartesian plane with a potential, $V (x, y)$
the lagrangian:

L = \frac{1}{2} m v^{2} - V (x, y)

where, $v^{2} = {\dot{x}}^{2} + {\dot{y}}^{2}$

here, there are two position coordinates, ie: $d = 2$ , so expect two euler-lagrange equations:

\begin{aligned} - \frac{\partial V}{\partial x} & = \frac{d}{d t} m \dot{x} = \frac{d}{d t} p_{x} \\ - \frac{\partial V}{\partial y} & = \frac{d}{d t} m \dot{y} = \frac{d}{d t} p_{y} \end{aligned}

define the momentum vector:

\vec{p} = (\begin{matrix} p_{x} \\ p_{y} \end{matrix})

\frac{d}{d t} (\begin{matrix} p_{x} \\ p_{y} \end{matrix}) = (\begin{matrix} - \frac{\partial V}{\partial x} \\ - \frac{\partial V}{\partial y} \end{matrix}) = \vec{F}

therefore, a fully vectorized form of newton's second law is obtained:

\frac{d}{d t} \vec{p} = \vec{F} = - \vec{\nabla} V