Derivative with respect to a derivative

Question

Let $q=q(t)\in C^1(\mathbb{R})$ and $V=V(x)\in C^1(\mathbb{R})$. My book uses the following fact over and over again $$\frac{\partial V(q)}{\partial \dot{q}}=0.$$ Why is this true?

Answer 1

Don't think of $\dot{q}$ as a derivative, but as an independent variable. This equation shows up in mechanics where $V(q)$ describes the potential of a field. It states that $V$ represents a field depending on positions $q$ only and not velocities $\dot{q}$.

If you had a dissipative field $V(q,\dot{q})$ then $\frac{\partial V}{\partial \dot{q}} \neq 0$.

Example of a conservative field would be gravity or electrostatic potential

$$ V(q) = -\frac{K}{q} $$