Meaning of vector derivative of velocity with respect to position

Question

I would like to compute an expression for $\frac{\partial v}{\partial x}$, where $x$ is a position vector and $v = \dot{x}$ is a velocity vector. Using the Chain Rule, we should have:

$$\frac{\partial v}{\partial x} = \frac{\partial t}{\partial x}\frac{\partial v}{\partial t} = \frac{\partial t}{\partial x}\dot{v}^\intercal$$

Using differentials $dx$ and $dt$, we have

$$dx = (\frac{\partial x}{\partial t})^\intercal dt, dt = (\frac{\partial t}{\partial x})^\intercal dx$$

and by combining we get

$$I = \frac{\partial t}{\partial x}\frac{\partial x}{\partial t}$$

with some manipulation, we can get

$$v = \frac{\partial x}{\partial t}^\intercal = \frac{\partial t}{\partial x} ||v||^2$$

Thus we get

$$\frac{\partial t}{\partial x} = \frac{v}{||v||^2}$$

and finally

$$\frac{\partial v}{\partial x} = \frac{v\dot{v}^\intercal}{||v||^2}$$

This seems to make logical sense from the fundamentals of matrix calculus (which I am still trying to grasp), but I am not sure how to interpret this result.

Answer 1

Let $(x,v,a)$ denote the position, velocity, and acceleration vectors.

The Moore-Penrose inverse of the velocity vector is $$\eqalign{ v^+ &= \Bigg(\frac{v^T}{v^Tv}\Bigg) ^\implies v^+v=1\cr }$$ but please note that $\,\,vv^+\ne I$

The MP inverse allows you to write $$\eqalign{ a &= a\,(v^+v) \cr \frac{dv}{dt} &= av^+\,\frac{dx}{dt} \cr }$$ Multiplying the LHS and RHS by $dt$ yields the differential result $$\eqalign{ dv &= av^+\,dx \cr }$$ It is the interpretation of your next step $$\eqalign{ \frac{\partial v}{\partial x} &= av^+ = {\dot v}v^+ \cr }$$ that raises questions, since it implies that velocity (aka momentum) is a function of position.
But position and momentum are treated as independent variables in mechanics.