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The Euler-Lagrange equations for a bob attached to a spring are $$ \frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial v} = \frac{\partial L}{\partial x} $$ But $v$ is a function of $x$. Is it or is it not. Because normal thinking says that $x$ is a function of $t$ and $v$ is a function of $t$ but it is not necessary that $v$ be a function of $x$. But mathematically
\begin{align} x&=f(t) \\ t&=f^{-1}(x) \\ v &= g(t) \\ &=g(f^{-1} (x)).\end{align} So chain rule should be applied in the Euler Lagrange equations. Then why is it not? Is it because on applying the chain rule on $f^{-1}$ and on applying the chain rule on both the sides of the equation we will get the same result. If not then what is the reason for not applying the chain rule? Or is it that $v$ is not a function of $x$ but then $v$ is related to $x$. Where am I going wrong in a such a basic thing is puzzling me a lot.

By Symmetry
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Shashaank
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  • This would be much easier to read with formulas; use MathJax. Otherwise we are forced to parse your text, and write it down as we go, errors and all. MathJax is described in the help. – Peter Diehr May 27 '16 at 21:12
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    Possible duplicates: http://physics.stackexchange.com/q/885/2451 and links therein. – Qmechanic May 27 '16 at 21:18
  • @Qmechanic thanks a lot , I was looking for something just like this , thanks – Shashaank May 27 '16 at 21:20
  • Position and velocity are algebraicly independent; given a velocity you cannot know the position unless you also have an algebraic constraint that is known in advance. Example: consider a car racing on a circular track. – Peter Diehr May 28 '16 at 12:32

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