Derive $dt$ from $dx$ and $v$

Question

Let $p$ be a vector-valued function

$$p: {\bf N} \longrightarrow {\bf R}^3$$ $$p: t \longmapsto (p_x,p_y,p_z)$$

whose values are the 3D position of something moving through 3D space (at discrete time steps).

We can approximate the velocity vector function

$$v: {\bf N} \longrightarrow {\bf R}^3$$ $$v: t \longmapsto (v_x,v_y,v_z)$$

as

$${\Delta p \over \Delta t}.$$

Now we're given a sequence of vectors $(p_x,p_y,p_z)$ and vectors $(v_x,v_y,v_z)$ and we want to approximate the scalar $\Delta t$. How could we do it?

If $p$ is a scalar function, then

$${\Delta p \over \Delta t} = v$$

so you just compute scalar division

$${\Delta p \over v} = \Delta t$$

but if $p$ and $v$ are vectors, what can we do? Vector division? For which vector product?

Another option is to compute the "$\Delta t$'s" in each dimension:

$$\Delta t_x = {\Delta p_x \over v_x}$$ $$\Delta t_y = {\Delta p_y \over v_y}$$ $$\Delta t_z = {\Delta p_z \over v_z}$$

But this gives 3 different values of $\Delta t$: $\Delta t_x$, $\Delta t_y$, and $\Delta t_z$. You could compute the norm of the vector $(\Delta t_x, \Delta t_y, \Delta t_z)$, but I don't know what that would mean.

Yet another option is to compute the scalar division of the norm $p_r $of $p$ and the norm $v_r$ of $v$:

$${\Delta p_r \over v_r} = \Delta t$$

but this gives yet another quantity.

Answer 1

When you write $\frac{\Delta p}{\Delta t} = v \in\mathbb{R}^3$, you are saying that you are doing this division coordinate-wise: $\frac{\Delta p}{\Delta t} = \frac{1}{\Delta t}\begin{pmatrix}\Delta p_1\\ \Delta p_2\\ \Delta p_3 \end{pmatrix} = \begin{pmatrix}\frac{\Delta p_1}{\Delta t}\\ \frac{\Delta p_2}{\Delta t}\\ \frac{\Delta p_3}{\Delta t} \end{pmatrix} = \begin{pmatrix}v_1\\ v_2\\ v_3\end{pmatrix} \implies \begin{cases}\Delta p_1/\Delta t=v_1\\ \Delta p_2/\Delta t=v_2\\ \Delta p_3/\Delta t=v_3 \end{cases}\implies \begin{cases}\Delta p_1=v_1\Delta t\\ \Delta p_2=v_2\Delta t\\ \Delta p_3=v_3\Delta t \end{cases}\implies \begin{cases}\Delta p_1/v1=\Delta t\\ \Delta p_2/v_2=\Delta t\\ \Delta p_3/v_3=\Delta t \end{cases}$

There are three equations all involving $\Delta t$. So, to invert this, you should divide each coordinate of $\Delta p$ by the corresponding coordinate of $v$. You say this gives you three different values, but this is not possible if $\frac{\Delta p}{\Delta t} = v$ holds like you described. Either your model $\frac{\Delta p}{\Delta t} = v$ is wrong or your data is wrong.

Answer 2

If the $p$ values were a sequence of positions, and the $v$ vectors the discrete velocities, then the $\Delta t_x$, $\Delta t_y$, $\Delta t_z$ values would all be equal - so the problem is just that of estimating a fixed but unknown value from a sample of three estimates of it. You could, for instance use least-squares estimation, which would estimate $\Delta t$ as the arithmetic mean $\frac{1}{3}(\Delta t_x +\Delta t_y + \Delta t_z)$. Your two methods are just other ways of averaging the three values. If these averages are all very close, it doesn't much matter which you use; if they aren't, you probably don't have enough information to get a sensible answer to your problem.