# Determining whether a particle is in stable/unstable equilibrium

A force defined by $$\vec F = (y^2\hat i + 2 x^2\hat j)$$ is exerted on a particle which is initially at the origin of the coordinate system. The particle is placed at rest right at the origin. Is this a stable situation?

If the problem one dimensional, I would have found $$\frac{-dF}{dx}$$ (because $$F=-\frac{dU}{dx}$$)and if this value was negative, unstable equilibrium and if positive, stable equilibrium.

I'm not sure how I do this for 2d/3d. Do I take $$-\nabla F$$ and proceed in the same way? If this is correct, what do I do if the value turns out to be $$0$$?

• Question for the student: should a saddle-point be considered stable or unstable? Why? Commented Nov 23, 2019 at 16:09
• @dmckee Ah, something I was wondering about but forgot to mention in the question. I think that it would be neither stable nor unstable since the direction of displacement would determine whether the particle tends to go back to its original position or not. Commented Nov 23, 2019 at 16:14
• The usual answer is "unstable" because you are worried about what unspecified perturbations will do to the system. Because you can't asumme the direction of "unspecified" will be exactly as needed, you expect them to send the system away from the equilibrium. Commented Nov 23, 2019 at 16:16
• Commented Nov 24, 2019 at 4:27