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Math Help - Euler Lagrange equation

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    Euler Lagrange equation




    i have no idea with the part under red line
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  2. #2
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    The path of the light ray is given by an equation y = y(x). The time taken for the ray to travel along the path is given in infinitesimal terms by dT = \frac{\sqrt{1+y'^2}\,dx}{c(x)} ("time = distance/speed"). Integrate that to get T = \int_{x_1}^{x_2}\frac{\sqrt{1+y'^2}}{c(x)}\,dx. The E–L equation tells you that the extremal path is given by 0 = \frac d{dx}\Bigl(\frac\partial{\partial y'}\Bigl(\frac{\sqrt{1+y'^2}}{c(x)}\Bigr)\Bigr) = \frac d{dx}\Bigl(\frac{y'}{c(x)\sqrt{1+y'^2}}\Bigr). In other words, \frac{y'}{c(x)\sqrt{1+y'^2}} is a constant, which for convenience we can call k^{-1}.

    Then ky' = c(x)\sqrt{1+y'^2}. Square both sides, solve for y'^2 and take the square root again, to get y' = \frac{c(x)}{\sqrt{k^2-c^2(x)}}. Integrate that to get y_2 - y_1 = \int_{x_1}^{x_2}\frac{c(x)}{\sqrt{k^2-c^2(x)}}\,dx.

    The last part is straightforward calculus.
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