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

Let the two points be labelled AA and BB and have coordinates (a,y(a))(a,y(a)) and (b,y(b))(b,y(b)). The length of an element of path dsds is given by ds=dx2+dy2=1+y2  dxds=\sqrt{dx^2+dy^2} = \sqrt{1+y'^2}\;dx and hence the total path length is given by L=ab1+y2  dx. L=\int_a^b{ \sqrt{1+y'^2}} \;dx. The Euler-Lagrange equation Fy=ddx(Fy)\frac{\partial F}{\partial y} = \frac{d}{dx}\left(\frac{\partial F}{\partial y'} \right) can be rearranged to Fy=const.\frac{\partial F}{\partial y'} = \mathrm{const.} since F=1+y2 F= \sqrt{1+y'^2} does not contain yy explicity. We can now differentiate FF and set it to a constant value. k=Fy=y1+y2 k=\frac{\partial F}{\partial y'} = \frac{y'}{\sqrt{1+y'^2}} k2(1+y2)=y2 k^2(1+y'^2) = y'^2 k2=y2k2y2=y2(1k2) k^2 = y'^2-k^2y'^2 = y'^2(1-k^2) y=k1k2 \Rightarrow y'=\frac{k}{\sqrt{1-k^2}} Integration gives y(x)=k1k2x+c  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 y=mx+c .                                                                                                                                                                                                                                                                                                                                 .\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\Box.

Note by Samuel Braun
6 months ago

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

Alak Bhattacharya - 6 months ago

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