A calculus of variations question

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The discussion centers on the calculus of variations, specifically the necessity of including the derivative of the function q, denoted as \dot{q}, as an argument in the Lagrangian function L(q, \dot{q}, x). The Euler-Lagrange equation is derived from the functional S[q, \dot{q}, x] = ∫(a to b) L(q(x), \dot{q}(x), x) dx. The inclusion of \dot{q} allows for the proper calculation of partial derivatives, which is essential for deriving conditions for extrema. The conversation concludes with a clear understanding of the role of \dot{q} in the context of functionals.

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Saketh
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I am trying to learn the calculus of variations, and I understand the mathematical derivation of the Euler-Lagrange equation.

As I understand it, the calculus of variations seeks to find extrema for functions of the form:
S[q,\dot{q}, x] = \int_{a}^{b} L(q(x),\dot{q}(x), x) \,dx.

Here is my question: Why is it necessary to have \dot{q} as an argument of L? Isn't the derivative of the function q "included" with the function itself?

Thanks in advance.
 
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L is a function of three variables L(z,y,x) where we consider the cases where z = q(x), y = q'(x). Now when we write L(z,y,x), then it makes sense to consider (dL/dy)(z,y,x). If we consider L a function of two variables, writing L(z,x), then how can we make sense of the derivative of L with respect to (dz/dx)(x)? So although the second argument of L ends up being the derivative of the first argument w.r.t. the third (and so, seems to already be "contained" in the other arguments), if we want to do things like take the partial of L with respect to q'(x), we need to have q'(x) occupy its own argument.
 
AKG said:
L is a function of three variables L(z,y,x) where we consider the cases where z = q(x), y = q'(x). Now when we write L(z,y,x), then it makes sense to consider (dL/dy)(z,y,x). If we consider L a function of two variables, writing L(z,x), then how can we make sense of the derivative of L with respect to (dz/dx)(x)? So although the second argument of L ends up being the derivative of the first argument w.r.t. the third (and so, seems to already be "contained" in the other arguments), if we want to do things like take the partial of L with respect to q'(x), we need to have q'(x) occupy its own argument.
Thanks for the explanation - I understand now.

I suppose this is something peculiar to functionals, but it's simple enough.
 

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