Given a set of points in $\mathbb{R}^3$ $$x^\star(t), y^\star(t), z^\star(t)\text{,}\quad 0\le t\le 1$$ I am looking for a scalar potential $V(x, y): \mathbb{R^2} \rightarrow \mathbb{R}$, whose partial derivatives satisfy the following conditions, \begin{align*} \dfrac{\partial V}{\partial x}(x^\star(t),y^\star(t)) &= 0\\ \dfrac{\partial V}{\partial y}(x^\star(t),y^\star(t))&=z^\star(t)\text{,}\\ \forall 0\le t\le 1\text{.} \end{align*} Any idea on how to find $V$?
Less confusing formulation
Given a curve $x^\star, y(x^\star)$ in $\mathbb{R}^2$, I am looking for a scalar potential $V(x, y)$ having a specific gradient along that curve, i.e. \begin{align*} \dfrac{\partial V}{\partial x}(x^\star,y(x^\star)) &= 0\\ \dfrac{\partial V}{\partial y}(x^\star,y(x^\star))&=z(x^\star)\text{,}\\ \end{align*} where $z: \mathbb{R} \rightarrow \mathbb{R}$ is a given function.
Context
This problem arises when one tries to derive the potential elastic energy of an elastic structure with two degrees of freedom $x$ and $y$ given its equilibrium path. The equilibrium path is the set of all the static equilibrium states the structure can be in when a force acts on it. This path can be represented by a curve in a three-dimension space: two dimensions for the degrees of freedom ($x$ and $y$), and a third dimension for the force $z$ acting on one of the degree of freedom ($y$ in this case). Each triplet of points $(x^\star(t), y^\star(t), z^\star(t))$ corresponds to a static configuration of the structure.
When on tries to find the equilibrium path of an elastic structure whose elastic energy is given by $V(x, y)$, the classic approach consists of forming the total potential energy $\Pi$ defined as $$\Pi = V(x, y) - zy\text{,}$$ where $z$ represents the external force acting on the degree of freedom $y$. We consider the case where no external force acts on the degree of freedom $x$.
The static equilibrium states ($x^\star$, $y^\star$, $z^\star$) are the stationary points of the potential $\Pi$ (w.r.t the variables $x$ and $y$), which leads to the following set of equations \begin{equation*} 0=\dfrac{\partial V}{\partial x}(x^\star, y^\star),\quad 0=\dfrac{\partial V}{\partial y}(x^\star, y^\star) - z^\star\text{.} \end{equation*}
You can inject the two conditions inside the Taylor series of $V$ and... the job is basically done ! Concretely, let's recall that a bivariate Taylor expansion around a point $(x_0,y_0)$ is given by $$ V(x,y) = \sum_{n,m\ge0} \left.\frac{\partial^{n+m}V}{\partial x^n \partial y^m}\right|_{x=x_0,y=y_0} \frac{(x-x_0)^n}{n!} \frac{(y-y_0)^m}{m!}. $$ Choosing $x_0 = x^\star$ and $y_0 = y^\star$, we end up with $$ V(x,y) = a_{00} + z^\star(y-y^\star) + \sum_{n,m\ge2} a_{nm} \frac{(x-x^\star)^n}{n!} \frac{(y-y^\star)^m}{m!}, $$ where $a_{nm} = \partial_x^n \partial_y^m V(x^\star,y^\star)$ are free parameters.