Editing Calculus of variations (section)

=== Example ===
In order to illustrate this process, consider the problem of finding the extremal function <math>y = f(x),</math> which is the shortest curve that connects two points <math>\left(x_1, y_1\right)</math> and <math>\left(x_2, y_2\right).</math> The [[arc length]] of the curve is given by
<math display="block">A[y] = \int_{x_1}^{x_2} \sqrt{1 + [ y'(x) ]^2} \, dx \, ,</math>
with
<math display="block">y'(x) = \frac{dy}{dx} \, , \ \ y_1=f(x_1) \, , \ \ y_2=f(x_2) \, .</math>
Note that assuming {{mvar|y}} is a function of {{mvar|x}} loses generality; ideally both should be a function of some other parameter. This approach is good solely for instructive purposes.

The Euler–Lagrange equation will now be used to find the extremal function <math>f(x)</math> that minimizes the functional <math>A[y].</math>
<math display="block">\frac{\partial L}{\partial f} -\frac{d}{dx} \frac{\partial L}{\partial f'}=0</math>
with
<math display="block">L = \sqrt{1 + [ f'(x) ]^2} \, .</math>

Since <math>f</math> does not appear explicitly in <math>L,</math> the first term in the Euler–Lagrange equation vanishes for all <math>f(x)</math> and thus,
<math display="block">\frac{d}{dx} \frac{\partial L}{\partial f'} = 0 \, .</math>
Substituting for <math>L</math> and taking the derivative,
<math display="block">\frac{d}{dx} \ \frac{f'(x)} {\sqrt{1 + [f'(x)]^2}} \ = 0 \, .</math>

Thus
<math display="block">\frac{f'(x)}{\sqrt{1+[f'(x)]^2}} = c \, ,</math>
for some constant <math>c.</math> Then
<math display="block">\frac{[f'(x)]^2}{1+[f'(x)]^2} = c^2 \, ,</math>
where
<math display="block">0 \le c^2<1.</math>
Solving, we get
<math display="block">[f'(x)]^2=\frac{c^2}{1-c^2}</math>
which implies that
<math display="block">f'(x)=m</math>
is a constant and therefore that the shortest curve that connects two points <math>\left(x_1, y_1\right)</math> and <math>\left(x_2, y_2\right)</math> is
<math display="block">f(x) = m x + b \qquad \text{with} \ \ m = \frac{y_2 - y_1}{x_2 - x_1} \quad \text{and} \quad b = \frac{x_2 y_1 - x_1 y_2}{x_2 - x_1}</math>
and we have thus found the extremal function <math>f(x)</math> that minimizes the functional <math>A[y]</math> so that <math>A[f]</math> is a minimum. The equation for a straight line is <math>y = mx+b.</math> In other words, the shortest distance between two points is a straight line.{{efn|name=ArchimedesStraight| As a historical note, this is an axiom of [[Archimedes]]. See e.g. Kelland (1843).<ref>{{cite book |last=Kelland |first=Philip |author-link=Philip Kelland| title=Lectures on the principles of demonstrative mathematics |year=1843 |page=58 |url=https://books.google.com/books?id=yQCFAAAAIAAJ&pg=PA58 |via=Google Books}}</ref>}}