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    $\begingroup$ It seems to be the same answer, just expressed very differently and very much more difficult to get the answer from the method avoiding Beltrami identity. Why should textbook authors waste paper talking about solution methods that are unnecessarily difficult? $\endgroup$
    – naturallyInconsistent
    Commented Jul 2, 2023 at 17:37
  • $\begingroup$ How do you get your "first integraL". As you have two variables $x(t)$, $y(t)$ the first integral should be $$ f- \dot x \frac{\partial f}{\partial \dot x} -\dot y \frac{\partial f}{\partial \dot y}=c $$ Also you have a constraint $v= \sqrt{\dot x^2+\dot y^2}$ which makes your functional $$ T= \int \frac v v dt= \int dt, $$ which, being identically satisfied, does not seem much use. $\endgroup$
    – mike stone
    Commented Jul 2, 2023 at 17:40
  • $\begingroup$ Related : Another Solution To Brachistochrone Problem. My notes on brachistochrone $\endgroup$
    – Frobenius
    Commented Jul 2, 2023 at 21:08
  • $\begingroup$ @Eli : Precisely. It's not wrong and identical to my notes equation (1.13). Note that for your constant we have $\;k=1/D=1/(2R)\;$ where $\;R=D/2\;$ the radius of the circle. $\endgroup$
    – Frobenius
    Commented Jul 3, 2023 at 8:47
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    $\begingroup$ @mikestone If I'm not mistaken the first integral comes from the fact that $\partial f/\partial x = 0$, implying that $d/dt(\partial f/\partial \dot{x}) = 0$. (b) This sort of "identical" variational principle can still be of use; for example, one can derive the geodesic equation on a manifold by defining an arbitrary curve $\vec{q}(\lambda)$ and then varying the arc length $S = \int ds = \int \sqrt{ g_{ij} \dot{q}^i(\lambda) \dot{q}^j(\lambda)} d \lambda$. $\endgroup$
    – Michael Seifert
    Commented Jul 3, 2023 at 15:03