Interpolating with Hermite cubics in two dimensions I want to estimate the value of the function $f(x,y)$ at a particular point.  Suppose I am given two points, $(x_1,y_1)$ and $(x_2,y_2)$, along with the value of $f$ and its partial derivatives $f_x$ and $f_y$ at the two points.
Using cubic Hermite interpolation, I would like to find the value of any point $$(x_∗,y_∗)=(x_1 +t_∗(x_2 − x_1),y_1 +t_∗(y_2 − y_1))$$ where $t_∗$ parametrizes the line between the two points.
I have thought about the problem and wonder if it is possible to interpolate along the $x$-axis using the two points and the $x$-derivatives at both, do the same for $y$, and then somehow combine both answers to get the final answer?
EDIT: It appears as though using a directional derivative would be more appropriate.
 A: Hermite interpolation makes sense to me here too. Let $p_1=(x_1,y_1)$ and $p_2=(x_2,y_2)$. We have $f_1 = f(p_1)$ and $f_2 = f(p_2)$, as well as $g_1 = \nabla f(p_1)$ and $g_2 = \nabla f(p_2)$. 
Consider the line between the points: $\ell(t) = p_1t + (1-t)p_2$ for $t\in[0,1]=[t_1,t_2]$. 
What we can do is consider the space curve $(x(t),y(t),f(t))$ where $f(t)=f(x(t),y(t))$. Then, we have $$\ell(t) = \begin{bmatrix} 
x(t)\\y(t) \end{bmatrix}$$
Separately, we have $(t_1,f(t_1))$ and $(t_2,f(t_2))$ as the values we'd like to interpolate. But we need the derivatives.
By the chain rule you have
$$\partial_tf = \partial_x \,f\partial_t x + \partial_y \,f \partial_t y = \nabla f \cdot \begin{bmatrix} x_1 - x_2\\ y_1 - y_2 \end{bmatrix} = \nabla f \cdot \Delta $$
where 
$$ m_1 = \partial_t\, f(0) = g_1\cdot\Delta\;\;\;\text{and}\;\;\; m_2 = \partial_t \,f(1) = g_2\cdot\Delta $$
We can now apply the cubic Hermite polynomial interpolation formula:
$$
f(t) = (2t^3-3t^2+1)\,f_1 +(t^3-2t^2+t)m_1 +(-2t^3+2t^2)f_2 + (t^3-t^2)m_2
$$
