# Strongest increase of a gradient $f(x,y)$ at the position $x_0$

I am trying to figure out how to interpret the gradient of my scalar field at the position $$x_0$$.

The gradient of the function $$f: \mathbb{R}^n \supset \ U \longrightarrow \mathbb {R}$$ at the point $$x_0$$ indicates the direction in which the function value of $$f$$ increases the most, starting from the point $$x_0$$.

Considering my function: To determine the gradient at the point $$f(x, y) = x^3 - 3xy^2$$ we must first calculate the partial derivatives with respect to $$x$$ and $$y$$ and write them into a column vector.

The partiel derivate of $$x$$ at the position $$x,y$$ is: $$\frac{\partial f }{\partial x}(x,y) = 3x^2-3y^2$$ The partiel derivate of $$y$$ at the position $$x,y$$ is: $$\frac{\partial f }{\partial y}(x,y) = -6xy$$

Therefore I get: $$\nabla f(x,y) = \begin{pmatrix} \frac{\partial f }{\partial x}\\ \frac{\partial f }{\partial y} \end{pmatrix} = \begin{pmatrix} 3x^2-3y^2 \\ -6xy \end{pmatrix}$$

Plotting the scalar field I get the following:

The point $$(0,0)$$ is a critical point and a saddle. What does the gradient at $$x_0$$ tells me exactly when there is no increase?

When the gradient of $$f(x,y)$$ at $$x_0$$ is $$\nabla f =\vec{0}$$ it's telling you that your function is approximately constant near $$x_0$$ with respect to all directions. In the case of your question it means that there exists a neighbourhood of $$x_0$$, however close to it, such that the function is similar to an horizontal plane, i.e. there is not a direction with respect to which the function increases the most, starting from $$x_0$$.
• The function is definite in $(0,0)$, it's continuous in $(0,0)$ and both its partial derivatives are continuous there. So it's continuous and differentiable with a critique point saddle type.