TITLE: Derivative of area gives mean curvature?
QUESTION [7 upvotes]: My lecturer made a comment today about how 'the derivative of the area gives you the mean curvature' but I'm not really sure what he meant?
I guess what I mean is that I don't understand this definition:
"A surface $M \in \mathbb R^3$ 
is minimal if and only if it is a critical point of the area functional for all compactly supported variations"
I understand that a critical point is a stationary point (ie $d/dt=0$) and that if $H=0$ a surface is minimal. That is all I currently understand in this definition.
If anyone could explain this simply (as possible) I would really appreciate it!

REPLY [6 votes]: Thank Yuri who stated the formula. Here comes a simple proof. Using Yuri's notation,
$$dS^t=\sqrt{\det(g_t)}du^1\wedge du^2$$
Where ${(g_t)}_{ij}=\langle f_i, f_j\rangle=\langle \frac{\partial}{\partial u^i}, \frac{\partial}{\partial u^j}\rangle$ is the first fundamental form. From wiki, we have $\frac{d}{dt}(\det(g_t))=\det(g_t)tr(g_t^{-1}\dot{g_t})$, hence
$$\frac{d}{dt}dS^t=\frac{d}{dt}\sqrt{\det(g_t)}du^1\wedge du^2$$
$$=\frac{1}{2\sqrt{\det(g_t)}}\det(g_t)tr(g_t^{-1}\dot{g_t})du^1\wedge du^2$$
$$=\frac{1}{2}tr(g_t^{-1}\dot{g_t})dS^t$$
Also from wiki, $H=\frac{1}{2}tr(g_0^{-1}\mathrm{I\!I})$, so if we can show that $\dot{g_0}=2\phi\mathrm{I\!I}$ we'll get the desired result: $$\frac{d}{dt}dS^t|_{t=0}=\frac{1}{2}tr(g_0^{-1}\dot{g_0})dS=\phi tr(g_0^{-1}\mathrm{I\!I})dS=2\phi H dS$$
To show that $\dot{g_0}=2\phi\mathrm{I\!I}$, it's a direct computation:
$$(\dot{g_t})_{ij}=\frac{d}{dt}\langle f_i,f_j\rangle=\langle \frac{d}{dt}f_i,f_j\rangle+\langle f_i,\frac{d}{dt}f_j\rangle$$
$$= \langle \frac{\partial}{\partial_{u^i}}f_t,f_j\rangle+\langle f_i,\frac{\partial}{\partial_{u^j}}f_t\rangle$$
$$= \langle \frac{\partial}{\partial_{u^i}}\phi N,f_j\rangle+\langle f_i,\frac{\partial}{\partial_{u^j}}\phi N\rangle$$
$$= \phi \langle \frac{\partial}{\partial_{u^i}} N,f_j\rangle+\phi \langle f_i,\frac{\partial}{\partial_{u^j}} N\rangle$$
$$= \phi \langle N_i,f_j\rangle+\phi \langle f_i,N_j\rangle=2\phi\mathrm{I\!I}_{ij}$$