Euler Lagrange
Euler-Lagrange Equation
\begin{equation} \frac{\partial F}{\partial y(x)}-\frac{d}{dx}\left(\frac{\partial F}{\partial y'(x)}\right)=0 \label{eq:eulerlagrange} \end{equation}
The right path minimizes the integral. The wrong path does not.
Imagine a wrong curve, shown in the dotted red line.
We can call it \begin{equation} Y(x) = y(x) + \eta(x) \end{equation}
Introduce a parameter $\alpha$: \begin{equation} Y(x) = y(x) + \alpha \eta(x) \label{eq:wrongpathwithparameter} \end{equation}
Now show that \begin{equation} \frac{dI}{d\alpha} = 0 \; \textrm{for} \; \alpha = 0 \label{eq:samestartendpoints} \end{equation}
The integral $I(\alpha)$ becomes, when written out: \begin{equation} I(\alpha) = \int_{x_a}^{x_b} F(Y, Y', x) dx \end{equation} or \begin{equation} I(\alpha) = \int_{x_a}^{x_b} F(y+\alpha \eta,\; y'+\alpha \eta',\; x)\; dx \end{equation}
Now we evaluate $\frac{\partial F}{\partial \alpha}$
Thus, $dI/d\alpha$: \begin{equation} \frac{\partial I}{\partial \alpha} = \int_{x_a}^{x_b} \frac{\partial F}{\partial \alpha} dx = \int_{x_a}^{x_b} \left( \eta \frac{\partial F}{\partial y} + \eta' \frac{\partial F}{\partial y'} \right)dx= 0 \end{equation}
Integration by Parts
\begin{equation} \int_{x_a}^{x_b} \eta' \frac{\partial F}{\partial y'} dx = \left[ \frac{\partial F}{\partial y'} \eta(x)\right]_{x_a}^{x_b} - \int_{x_a}^{x_b} \frac{d}{dx} \left(\frac{\partial F}{\partial y'} \right)\eta(x) dx \label{eq:byparts} \end{equation}Fundamental Lemma of Calculus of Variations:
if \begin{equation} \int \eta(x)g(x) dx = 0 \end{equation} then it can be shown that $g(x)$ must be zero.
If $I$ is an extremum: \begin{equation} I = \int_{x_a}^{x_b} F \left[ y(x), y'(x), x \right] dx \end{equation} then \begin{equation} \frac{\partial F}{\partial y} - \frac{d}{dx} \frac{\partial F}{\partial y'} = 0 \end{equation}
How will we use this? \begin{equation} I = \int \left(\frac{1}{2}m v^2 - U \right) dt = \int \left(T - U \right) dt \end{equation} where \begin{equation} T = \frac{1}{2}mv^2 \end{equation} and (for gravity for example) \begin{equation} U = mgy \end{equation}
The Lagrangian
\begin{equation} L = T - U \end{equation}\begin{equation} \frac{\partial L}{\partial x} - \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} = 0 \end{equation} leads to: \begin{equation} \frac{\partial L}{\partial x} = - \frac{\partial U}{\partial x} = F_x \end{equation} and \begin{equation} \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} = \frac{d}{dt}m \dot{x} = m\ddot{x} \end{equation} so: \begin{equation} F_x = m\ddot{x} \end{equation}
Do it for all three dimensions and you have \begin{equation} -\nabla U = m \mathbf{a} \Rightarrow \mathbf{F} = m \mathbf{a} \end{equation}