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 \( y=mx+c \). \[\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Box.\]

#Calculus

Note by Samuel Braun
1 year, 9 months ago

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Comments

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 xx and yy only, it refers to a plane curve.

A Former Brilliant Member - 1 year, 9 months ago
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