# A Variational derivative of mean curvature

#### Hologram0110

Hello Physics Forums,

I have a simple parametric surface in R3 <x,y,z(x,y)>. I've calculated the the usual mean curvature:

H= ((1+hx^2)hyy-2hxhyhxy+(1+hy^2)hxx)/(1+hx^2+hy^2)^3/2

I needed to take the variational derivative of this expression. Since it has second order spatial derivatives the equation is:

d/dh - d/dx d/dhx - d/dy d/dhy + d^2/dx^2 d/dhxx + d^2/dxdy d/hxy +d^2/dyy d/hyy

To simplify the math I grouped the terms by d/dx and d/dy and I found that :
-dH/dhx + d/dx dH/dhxx + 1/2 d/dy dH/dhxy = 0
and
-dH/dhy + d/dy dH/dhyy + 1/2 d/dx dH/dhxy = 0

I've encountered this in my work by brute forcing it with Mathematica. However, it seems like there might be a more fundamental or elegant way of getting the result. Is there some well known reason for this? I'm an engineer working in material science. I've never taken a course on variational calculus or differntial geometry so please forgie me if this is obvious.

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