In Goldstein, Classical Mechanics, Chap. 1.4 we derive Lagrange's equations from D'Alembert's Principle. My question is regarding the last part of the derivation, specifically the part where he introduces the Lagrangian $L$ defined as $T - V$: $$ \frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial T}{\partial q_j} - Q_j = 0 ,\tag{1.53} $$ $$ Q_j = - \frac{\partial V}{\partial q_j}. \tag{1.54}$$ Which when substituting yields: $$ \frac{d}{dt} \left(\frac{\partial T}{\partial \dot{q}_j} \right) - \frac{\partial (T-V)}{\partial q_j} = 0\tag{1.55}. $$
The next step is what confuses me. He states that the potential $V$ does not depend on the generalized velocities so he makes the following substitution which "has no effect on differentiation with respect to $ \dot{q}_j $": $$ \frac{d}{dt} \left(\frac{\partial (T - V)}{\partial \dot{q}_j} \right) - \frac{\partial (T-V)}{\partial q_j} = 0. $$ Which leads to the familiar: $$ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}_j} \right) - \frac{\partial L}{\partial q_j} = 0.\tag{1.57} $$
It makes sense intuitively. After all, you can vary a particles position without effecting its instantaneous velocity at the end of its path, therefore causing a change in potential without a corresponding change in velocity.
My confusion arises when you know the equations of motion as a priori: $V$ is dependent on position, position is dependent on time (from equations of motions), and velocity is dependent on time. Doesn't this means there exists some relationship between the potential function and velocity?
