# Mean curvature of a level set

On page 256 http://zakuski.utsa.edu/~jagy/papers/Michigan_1991.pdf there is a formula for the mean curvature of a level set of a function. Im interested in the case n=3. How do you prove this formula? I can prove it for the case of a graph, so i thought i would use the implicit function theorem to make it a graph locally, but i dont see how I will get the original function in the final answer. Alternatively, I'm aware this equation is just the divergence of the normal vector, but how do you show that H can be defined this way? I seem to need to refer to a chart, but i can't seem to find one that gives a nice answer. Any help would be much appreciated

Tom

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As I recall, what I did originally was just to rotate everything. Put another way, if your level surface is a graph over the xy plane, $z=f(x,y),$ in such a way that $f(0,0)=0$ and $\nabla f(0,0) = (0,0),$ then the mean curvature at the origin is just the Laplacian $\Delta f(0,0),$ although I always liked to divide by 2, or $n$ in $\mathbb R^{n+1}.$ As soon as you need to rotate in order to get the level surface into that position, instead of just the Laplacian you get a mixture of products of first and second partials because of the rotation step. I think i say this in the paper, take the formula for a level set and just plug in $g(x,y,z) = f(x,y) - z$ and see what you get.