Suppose there is a differentiable vector field in 2d, $$\mathbf{F} = \mathbf{F}(x,y) \in \mathbb{R}^2.$$ I draw a curve through the $(x,y)$, parametrized in terms of $t$. So I have $(x(t), y(t))$. It follows that $$\mathbf{F}(t) = \mathbf{F}(x(t),y(t)).$$
I only know the parametric equations $(x(t),y(t))$, and the vector field $\mathbf{F}(t)$. I do not know the full vector field, $\mathbf{F}(x,y)$. My question is, given that I do not known the full vector field, but only a slice through the vector field through some parametric curve, am I able to calculate derivatives of the full vector field at each point along the curve?
In other words, I would like to calculate the derivatives $$\frac{d\mathbf{F}}{dx} \Big\vert_{(x(t),y(t))}, \frac{d\mathbf{F}}{dy}\Big\vert_{(x(t),y(t))}.$$
My thinking is that the following equation holds:$$\frac{d\mathbf{F}}{dx} \Big\vert_{(x(t_*),y(t_*))} = \frac{d\mathbf{F}}{dt} \Big\vert_{t_*} \frac{dt}{dx} \Big\vert_{x(t_*)},$$ and a similar expression for $y$. Am I misguided here? My intuition is telling me that if I only have access to the parametrized vector field $\mathbf{F}(t)$, than I can't calculate the derivative $d\mathbf{F}/dx$ along the parametric curve, because I only know how the vector field $\mathbf{F}$ changes along the curve, rather than some arbitrary direction in the (x,y) plane. However, the above equations seem to suggest otherwise. Which is right?
No, you can't.
Let $\boldsymbol u\in C^1(\mathbb R^2,\mathbb R^2)$ be a smooth vector field and let $\boldsymbol x\in C^1(\mathbb R,\mathbb R^2)$ be a parametric curve and let $\boldsymbol v=\boldsymbol u\circ\boldsymbol x$, i.e $\boldsymbol v(t)=\boldsymbol u(\boldsymbol x (t))$.
Now suppose that we know $\boldsymbol v$ (which is essentially a "slice" of $\boldsymbol u$) and we know $\boldsymbol x$, but not $\boldsymbol u$. I.e, we only know the part of the vector field that coincides with our parametric curve. The question is, using only the information we know, can we calculate $\nabla \boldsymbol u$, where $$\nabla\boldsymbol u=\begin{bmatrix}\frac{\partial u_1}{\partial x_1}&\frac{\partial u_1}{\partial x_2}\\ \frac{\partial u_2}{\partial x_1}&\frac{\partial u_2}{\partial x_2}\end{bmatrix}$$ ??
The answer is no. We can say, via the chain rule, that
$$\underbrace{\dot{\boldsymbol v}(t)}_{\text{vector}}=\underbrace{(\nabla\boldsymbol u)(\boldsymbol x(t))}_{\text{matrix}}\cdot\underbrace{\dot{\boldsymbol x}(t)}_{\text{vector}}$$
So the question basically boils down to this: Given known vectors $\boldsymbol x,\boldsymbol b$, can we uniquely determine a matrix $\mathbf A$ such that $$\mathbf{A}\boldsymbol x=\boldsymbol b$$ ??
Generally, the answer is no. For instance if $\boldsymbol b=\begin{bmatrix}1&-1\end{bmatrix}^\intercal$ and $\boldsymbol x=\begin{bmatrix}2&1\end{bmatrix}^\intercal$ then $$\begin{bmatrix}1&-1 \\ 1&-3\end{bmatrix}\boldsymbol x=\boldsymbol b$$ But also $$\begin{bmatrix}2&-5 \\ -1&1\end{bmatrix}\boldsymbol x=\boldsymbol b$$ As well as infinitely many more.