The velocity vector field can be described as follows:
$$\vec{V} = u(x,y,z,t)\hat{i} + v(x,y,z,t)\hat{j} + w(x,y,z,t)\hat{k}$$
What I understand from this is that for ALL particles at some random arbitrary location $(x_0, y_0, z_0)$ and some random arbitrary time $(t_0)$, we have
$$\vec{V_0} = u(x_0,y_0,z_0,t_0)\hat{i} + v(x_0,y_0,z_0,t_0)\hat{j} + w(x_0,y_0,z_0,t_0)\hat{k}$$
This makes sense to me!
However, I start to think about the position function. Recall that we are also using Eulerian frame.
For example, say we have a position vector for particle A
$$\vec{r_A}=x(t)\hat{i} + y(t)\hat{j} + z(t)\hat{k}$$
taken with respect to the origin of our Cartesian coordinate frame.
Meaning that
$$\vec{r_A}=\vec{r_A}(x(t), y(t), z(t))=\vec{r_A}(t)$$
It also follow that, to get the velocity vector for particle A,
$$\frac{d\vec{r_A}}{dt} = \vec{V_A}=\vec{V_A}(t)$$
QUESTION: How does one go about, mathematically, of describing a position vector field for ALL particles?
Thinking about it further, it would make sense to describe the position vector field for ALL particles as
$$\vec{r}=x\hat{i} + y\hat{j} + z\hat{k}$$
since the position of ALL particles is, well, its position. My guess is that this would not be a function of time since we are only looking at that location. Much like $\vec{V}$, we are only describing a certain location at a certain time. However, the location we are looking at is the position
However, to get the velocity vector field, we have to differentiate position,
$$\frac{d\vec{r}(x, y, z)}{dt} = \vec{V}(x,y,z,t)$$
Here is where I get stuck again...
QUESTION: If my assumption is correct, what would the chain rule look like for this equation?
QUESTION: If the velocity field for ALL particles is a function of position and time, $\vec{V}=\vec{V}(x,y,z,t)$, would that mean that the position is also a function of time? How can that be?
Thank you! Feel free to suggest reading material that can also help! I am very curious!