I'm trying to see how we use the E-L equation
\begin{equation} L_x(t,q(t),q'(t))-\dfrac{d}{dt}L_v(t,q(t),q'(t))=0 \end{equation}
to find the shortest distance between two points in the Euclidean Plane (i.e. equation of a straight line?)
This is the proof I have done but I'm unsure as to whether I have executed it correctly? It needs to be precise
Let $\ell(f)$ be the length of the graph, such that:
\begin{equation} \ell(f)= \int_b^a \sqrt{1+(f'(x))^2}\,dx \end{equation}
Let $L(x, f(x), f'(x)) = L(x, y, y')$, such that:
\begin{equation} \int_b^a \sqrt{1+(f'(x))^2}\,dx =\large \int_b^a \sqrt{1+(y')^2}\,dx \end{equation}
Evaluating: \begin{equation} \quad \dfrac{\partial L}{\partial y'}=\dfrac{y'}{\sqrt{1+(y')^2}} \quad and \quad \dfrac{\partial L}{\partial y}=0 \end{equation}
By substituting these into the E-L equation, we now have:
\begin{align} \dfrac{d }{dx} \dfrac{y'}{\sqrt{1+(y')^2}} &= 0\\ \Rightarrow \dfrac{d }{dx} \dfrac{f'(x)}{\sqrt{1+(f'(x))^2}} &=0\\ \Rightarrow \displaystyle\int_b^a \dfrac{d }{dx} \dfrac{f'(x)}{\sqrt{1+(f'(x))^2}} \,dx &= constant = C\\ \Rightarrow f'(x) = \dfrac{C}{\sqrt{1-C^2}} &= m\\ \Rightarrow \quad f(x) &= mx+ \text{constant} \end{align}
Suppose we have the two points $(x,y)=(0,a)$ and $(x,y) = (1,b)$. Suppose that the solution can be expressed as a function of one variable so that $y= y(x)$. We seek $y$ such that
$$D = \int_{0}^1 \sqrt{1+ (\frac{dy}{dx})^2} dx$$ such that y(0)=a, y(1)=b, is minimized. Of course what this equation represents is the arclength formula in 1D.
Then for the Functional D, the EL equation is :
$$ \frac{d}{dx} ( \frac{y'}{\sqrt{1+y'^2}})=0$$
Integrating this equation se have that $y' = $ constant. and so this implies that
$$y(x) = a +(b-a)x$$