Is a function of velocity a function of time? Let $U:\mathbb{R}^6\to\mathbb{R}$ be a function which variable dependence is
$$
U(x,y,z,\dot{x},\dot{y},\dot{z}),
$$
where $x,y,z\equiv x(t),y(t),z(t)$ and $\dot{x}=\frac{dx}{dt}$. Is it correct to simplify the previous dependence to a function $U:\mathbb{R}^4\to\mathbb{R}$, such that
$$
U\equiv U(x,y,z,t)?
$$
My goal is to show that a function depending on the velocity is a function depending on time. Is this true?
Edit: This is related to this question, where $U$ is the potential energy associated with a non-conservative force. My goal is to simplify the following expression
$$
\frac{d}{d t}U (x,y,z,\dot{x},\dot{y},\dot{z}).
$$
Any ideas?
 A: There is a difference between writting $U(x,y,z,\dot{x},\dot{y},\dot{z})$ and $U(x,y,z,t)$. When you write $U(x,y,z,t)$ you mean that there is a possible explicit dependence on $t$, while in $U(x,y,z,\dot{x},\dot{y},\dot{z})$ this dependence is implicit. This will make a difference in the Lagrangian/Hamiltonian formalism since, for the first case, the partial derivative $\partial U(x,y,z,t)/\partial t$ can be evaluated and is not usually zero but in the second case $\partial U(x,y,z,\dot{x},\dot{y},\dot{z})/\partial t$ is always zero, since there is no explicit dependence on $t$. 
A: When we write $U:\mathbb{R}^6\to\mathbb{R}$ we are saying that every point $P$ in the $\mathbb{R}^6$ (or, at least, in some sub-domain of $\mathbb{R}^6$) is mapped to a value $U(P) \in \mathbb{R}$. There is no suggestion of a trajectory or flow of time here. The fact that we may interpret some of the co-ordinates in $\mathbb{R}^6$ as a position vector and others as a velocity vector is irrelevant.
If we now define a curve $Q(\lambda)$ in $\mathbb{R}^6$ as a sequence of points $Q: \mathbb{R} \to \mathbb{R}^6$ then we have an implicit function $V: \mathbb{R} \to \mathbb{R}$ defined by $V(\lambda) = U(Q(\lambda))$. If we now identify the parameter $\lambda$ with time and say that 
$Q(t) = (x(t), y(t), z(t), \dot x (t), \dot y (t), \dot z (t)) \\
V(t) = U(x(t), y(t), z(t), \dot x (t), \dot y (t), \dot z (t))$
then we have
$\frac {dV}{dt} = \frac {\partial U}{\partial x} \frac {dx}{dt} + \dots + \frac {\partial U}{\partial \dot z} \frac {d \dot z}{dt}$
By a conventional but somewhat confusing abuse of notation, we often write $\frac {dU}{dt}$ instead of $\frac{dV}{dt}$, glossing over the fact that the function $U: \mathbb{R} \to \mathbb{R}$ in $\frac {dU}{dt}$ is not the same as the function $U: \mathbb{R}^6 \to \mathbb{R}$.
