I was doing some research when I found out this vector field, and I couldn't find its equations nowhere. How could I get the equation that describes this vector field, using only vector calculus and linear algebra (if it is possible)?
Edit: The equation of an ellipse in polar coordinates is the following: $$ r(\theta)=\frac{\mu}{1 + \varepsilon \cos\theta} \rightarrow r(\theta) \;(1 + \varepsilon \cos\theta) = \mu \rightarrow r + r\varepsilon \cos\theta = \mu \\ $$ Transforming into cartesian coordinates: $$ \sqrt{x^2 + y^2} + \varepsilon x=\mu $$ And rearrange to do an implicit derivtion, arriving to: $$ \frac{dy}{dx} = -\frac{\mu \varepsilon + x (1 - \varepsilon^2)}{y} $$
So this is the slope each of the vectors in the field should have. What I should do next?
The picture suggests that the vector field consists of the vectors that are tangential to those lines where a function $f(x,y)$ defined on $\mathbb R^2$ is constant. We know that the gradient of $f$ is orthogonal to those lines. Therefore I think a way to calculate the vector field is to set it to $$ \begin{pmatrix}\displaystyle-\frac{\partial f}{\partial y}\\\displaystyle \frac{\partial f}{\partial x} \end{pmatrix}. $$