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I want to find the equation of the curve that minimizes the distance between two points $P \equiv (x_1,y_1)$ and $Q \equiv (x_2,y_2)$. Thus, with the curve $\gamma(x)=(t,y(x))$, we have that the distance is the integral between $x_1$ and $x_2$ of $||\gamma'(x)||$ which is: $$ d(P,Q) = \int_{x_1}^{x_2} \sqrt{1+y'(x)^2} \: dx $$ Then, we just need to apply the Euler equation $\frac{\partial f}{\partial y} - \frac{d}{dx}(\frac{\partial f}{\partial y'})=0$, with $f(x,y,y')=\sqrt{1+y'(x)^2}$.

My problem comes when doing the derivatives $$ \frac{\partial f}{\partial y} = \frac{(1+y'(x)^2)^{\frac{1}{2}-1}}{2} \cdot \frac{\partial y'}{\partial y} \stackrel{!!}{=} \frac{1}{2\sqrt{1+y'(x)^2}} \cdot 0=0 $$ But $\frac{\partial y'}{\partial y} \stackrel{!!}{=} 0$ because $y$ and $y'$ are independent. The other part is $$ \frac{\partial f}{\partial y'} = \frac{(1+y'(x)^2)^{\frac{1}{2}-1}}{2} \cdot \frac{\partial (y')^2}{\partial y'} = \frac{y'}{\sqrt{1+y'(x)^2}} $$ And now the last step would be $$ \frac{d}{dx}\left(\frac{\partial f}{\partial y'}\right) = \frac{y'' \sqrt{1+y'(x)^2} - \frac{y'}{\sqrt{1+y'(x)^2}} y' y'' }{1+y'(x)^2} = \frac{y'' + y''y'^2-y''y'^2}{\sqrt{1+y'(x)^2}^{3/2}} $$ Therefore as $\frac{\partial y'}{\partial y}=0$ the Euler equation is $$ \frac{-y''(x)}{(1+y'(x)^2)^{\frac{3}{2}}} = 0 $$ what happens when $y''(x)=0 \implies y'(x)=a \implies y(x)=ax+b$. Where $a=\frac{y_1-y_2}{x_1-x_2}$ and $b=\frac{-y_1 x_2+x_1 y_2}{x_1-x_2}$ the straight line that joins the points.

baristocrona
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    This question is similar to: How can $y$ and $y'$ be independent in variational calculus?. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. See this question also – whpowell96 Jun 27 '24 at 23:13
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    Watching some of your edits you have a strange way of showing $\frac{\partial f}{\partial y}=0,.$ This simply holds because your $f$ only depends on $y',.$ – Kurt G. Jun 29 '24 at 16:10
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    Reposting a comment from one of the linked posts by echinodermata, where here $f$ denotes the Lagrangian potential inside the integral of the functional: "Imo this is one of the cases wheere the traditional notation conflating the function $x\to y(x)$ and the current values $y(x)$ is most disastrous. They're both just written $y$. So it would sure seem plausible that $f$ takes the function as input right? But no, it actually takes only the current value. So $f$ needs to be told the current values $y$ and $y'$ separately. " I believe this is a valuable point of view. – whpowell96 Jun 29 '24 at 19:32
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    The functional takes in the function $y:x\mapsto y(x)$, but the Lagrangian potential that is integrated depends only on the values $y(x)$ and $y'(x)$. Obviously at a specific point devoid of context, $y(x)$ and $y'(x)$ can be varied separately, but they are eventually related when considering the entire functional – whpowell96 Jun 29 '24 at 19:34

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Fixing a mistake we have $f(x,y,y')=\sqrt{1+y'^{\,\color{red}2}}$ which does not depend explicitly on $x$ or $y\,.$ You have $$ \frac{\partial f}{\partial y'}=\frac{y'}{\sqrt{1+y'^{\,2}}}\,. $$ So your Euler-Lagrange equation is $$ 0=\frac d{dx}\frac{\partial f}{\partial y'} =\frac{y''}{(1+y'^{\,2})^{3/2}} $$ which is equivalent to $$ 0=y'' $$ having the expected straight lines as solutions.

Kurt G.
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  • Thanks for the appreciation. But in $\frac{\partial f}{\partial y'}$ must have a 1 in the numerator instead of $y'$? – baristocrona Jun 28 '24 at 08:52
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    @baristocrona I don't think so. Many people do not like that physicists notation. But it is safe when we consider $y'$ to be an argument of $f,,$ that is, a variable. Differentiate $\sqrt{1+z^2}$ with respect to $z,.$ It should be $z/\sqrt{1+z^2},.$ Now put back $y'$ for $z,.$ – Kurt G. Jun 28 '24 at 12:41
  • That's right, now I see thanks. – baristocrona Jun 28 '24 at 19:51