Why does the partial derivative with respect to $x$ of a function depending only on $\dot{x}$ vanish? [duplicate]

In Classical Mechanics by Goldstein it says:

$$\sum \left\{ \left[ \frac{d}{dt} \left( \frac{\partial T}{\partial \dot q_j} \right) - \frac{\partial T}{\partial q_j} \right] - Q_j \right\} \delta q_j = 0.$$ Note that in a system of Cartesian coordinates the partial derivative of $T$ with respect to $q_j$ vanishes. Thus, speaking in the language of differential geometry, this term arises from the curvature of the coordinates $q_j$. In polar coordinates, e.g., it is the partial derivative of $T$ with respect to an angle coordinate that the centripetal acceleration term appears. I don't understand what's said here that the partial derivative of $T$ vanishes when differentiating with respect to a Cartesian coordinate. How is that possible? Isn't $\dot x$ is a function of $x$?

This is a common point of confusion when one gets started in Lagrangian mechanics. The important thing to notice is that we are taking partial derivatives, not full derivatives with respect to $x$.
From the point of view of our partial derivatives, $x$ and $\dot{x}$ are completely separate variables with no relation to each other.
• The point of the intuitive argument I gave was solely to remind the asker that they already knew that $\frac{\partial \dot{x}}{\partial x}$ must vanish because they know that the form kinetic energy they are most familiar with, $\frac{1}{2} m \dot{x}^2$ does not depend on position. Oct 10, 2016 at 4:02