# Proof that the shortest curve joining two points is a straight line.

Let the two points be labelled $$A$$ and $$B$$ and have coordinates $$(a,y(a))$$ and $$(b,y(b))$$. The length of an element of path $$ds$$ is given by $ds=\sqrt{dx^2+dy^2} = \sqrt{1+y'^2}\;dx$ and hence the total path length is given by $L=\int_a^b{ \sqrt{1+y'^2}} \;dx.$ The Euler-Lagrange equation $\frac{\partial F}{\partial y} = \frac{d}{dx}\left(\frac{\partial F}{\partial y'} \right)$ can be rearranged to $\frac{\partial F}{\partial y'} = \mathrm{const.}$ since $$F= \sqrt{1+y'^2}$$ does not contain $$y$$ explicity. We can now differentiate $$F$$ and set it to a constant value. $k=\frac{\partial F}{\partial y'} = \frac{y'}{\sqrt{1+y'^2}}$ $k^2(1+y'^2) = y'^2$ $k^2 = y'^2-k^2y'^2 = y'^2(1-k^2)$ $\Rightarrow y'=\frac{k}{\sqrt{1-k^2}}$ Integration gives $y(x) = \frac{k}{\sqrt{1-k^2}}x+c\$ which, as expected, is the equation of a straight line in the form $$y=mx+c$$. $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Box.$

Note by Samuel Braun
1 year, 4 months ago

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## Comments

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You should mention that the curve is confined to a plane. Such curves are called geodesics, and have different shapes on different surfaces. On the surface of a sphere for example, it is the great circle of the sphere. As you are including $x$ and $y$ only, it refers to a plane curve.

- 1 year, 4 months ago

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