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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:

enter image description here

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?

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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 $x_0$.

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  • $\begingroup$ Is the function semidefinit at 0,0 or indefinit with a saddle? Cause from your explanation I think it has no saddle point at 0,0 $\endgroup$
    – Valentine
    Feb 22, 2021 at 14:03
  • $\begingroup$ 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. $\endgroup$
    – anna
    Feb 22, 2021 at 14:49

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