Why is the gradient vector zero inside a circular level curve I've been trying to find an awnser to my question but can't seem to find it.
Say we have a level curve graph with a circular level curve. Then, inside there is a critical point. I understand in the point of view that in every direction we either increase or decrease but I dont get the intuition of why the gradient is 0 at that point. I mean, If we go in the y direction and the x direction the function either increase or decrease so it shouldnt be zero. But at the same time the gradient should be pointing in all direction or in none.. and all the gradient when approching the critical point are either pointing in all direction or pointing in the same direction.
Anyways, my question is, why is the gradient the null vector inside a circular level curve?
Thank you!
 A: If I'm interpreting yo correctly you are given a function $f:\>{\mathbb R}^2\to{\mathbb R}$, and know that for a certain $c\in{\mathbb R}$ it has a simply closed level curve $\gamma_c: \ f(x,y)=c$. By Jordan's curve theorem this curve encloses an open interior $\Omega$ whose closure $\bar\Omega=\Omega\cup\gamma_c$ is a compact set. If $f$ is not constant on $\Omega$ it takes, e.g., some values $>c$ on $\Omega$, hence assumes a global maximum $M>c$ at a point  $p\in \bar\Omega$. This point $p$ is necessarily in the interior $\Omega$, hence has to be a zero of $\nabla f$.
A: Are you familiar with the idea of the gradient vector being the projection (onto the $xy$-plane, in this example) of the lower normal to a small planar patch tangent to the function?  Then at a local minimum or maximum, the tangent patch is horizontal, so its normal projects to the zero vector.
You can do the same thing in single variable calculus : everywhere the derivative is nonzero, the sign of the derivative tells you which way to go to increase the function, and the magnitude of the derivative tells you how rapidly the function is increasing in that direction.  Of course, the projection of the lower normal to the tangent line of the function at a point onto the $x$-axis tells you the same thing.  When the derivative is zero, the tangent line is horizontal and the normal points straight down, so its projection onto the $x$-axis is zero.
Comment on "lower normal":  A line has two normal vectors.  If we disallow vertical lines (which we do for tangent lines/planes/etc. whose slope is given by the derivative), then there is a normal that points "up" and one that points "down", where these directions correspond to the component of the vector in the dependent variable direction pointing "in the direction of increasing dependent variable" and "in the direction of decreasing dependent variable", respectively.  The "lower normal" makes the consistent choice to always pick the normal pointing down (so that the projection onto the independent variables gives the direction of increasing dependent variable).
