I have a function $f(x,y)$ whose closed-form expression is elusive. However, the steepest direction $$v= \begin{bmatrix} v_1 \\ v_2 \\ \end{bmatrix} $$ is known. $||v||=1$ holds.
At a particular pooint $f(x_c, y_c)$, $$ f(x_c+v_1, y_c+v_2) - f(x_c, y_c) $$ is also known. Then how to find (or to approximate) the gradient of $f(x,y)$ at $(x_c, y_c)$?
The gradient of $f$ is given by $$\nabla f=\begin{bmatrix}\dfrac{\partial f}{\partial x}\\[6pt] \dfrac{\partial f}{\partial y}\end{bmatrix}$$ We also know that $\nabla f$ points in the same direction as $v$, i.e. that $\nabla f=\lambda v$ for some $\lambda > 0$. Thinking of $\nabla f$ as the change in $f$ as we move towards $v$, we get that $\nabla f\cdot v\approx f\left(x_c+v_1,y_c+v_2\right)-f\left(x_c,y_c\right)$. Since $||v||=1$ and $v$ and $\nabla f$ point in the same direction, $\nabla f\cdot v=||\nabla f||\approx f\left(x_c+v_1,y_c+v_2\right)-f\left(x_c,y_c\right)$. So, we conclude that $$\nabla f\approx \left[f\left(x_c+v_1,y_c+v_2\right)-f\left(x_c,y_c\right)\right]v$$