Derivative of Potential with respect to dq/dt Watch

QuantumOverlord
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You can derive the Euler Lagrange equation using D’Alembert’s principle. The algebra is annoying but after a lot of messing about you can get to this point fairly intuitively:

-dV/dq - d/dt (dT/d(dq/dt)) + dT/dq = 0

where q is a generalized coordinate with an implied index i (omitted for clarity), and dq/dt is similarly indexed and completely independant from q. T is kinetic energy, and V is potential energy.

Now if you define L = T - V (the Lagrangian formation)

the E-L equation should immediately follow, and it does, except it requires d/dt(dV/(dq/dt))) = 0 and I am not sure why that should necessarily be the case, why can't the potential vary with a particles velocity? (Isn't that what it implies), or is it the case that it indeed it is true but the assumption in the E-L equation is that V is a function of q only.

Note: I know how to prove the EL equation using calculus of variations with hamiltonians principle of least action, and I know its considerably less messy, and more general. However the good thing about this method is it starts from basic newtonian mechanics wheras hamiltonians principle is more of an axiom.
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