Recently I am reading book on mean curvature flow by carlo mantegazza.There I found a problem on hypersurfaces stated below : Show that if the hypersurface $M \subset {R}^{n+1}$ is locally the graph of a function $f:R^n\rightarrow R$,then we have $$g_{ij}=\delta_{ij}+f_if_j, $$ $$\nu=-{(\nabla f,-1)\over \sqrt{1+|\nabla f |^2}},$$ $$h_{ij}={Hess_{ij}f\over\sqrt{1+|\nabla f |^2}},$$ $$H={\Delta f \over\sqrt{1+|\nabla f |^2}}-{Hess f(\nabla f,\nabla f)\over (\sqrt{1+|\nabla f |^2})^3}=div\left({\nabla f \over\sqrt{1+|\nabla f |^2}}\right) $$where $f_i=\partial_i f$ and $Hess f$ is the Hessian of the function $f$. I guess that it is a simple problem and I can feel the intuition behind it.But please anybody help me the explicit calculation associated with the problem.

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    $\begingroup$ Hint: a parameterization of the shape is given by $(x_1, \ldots, x_n) \mapsto (x_1, \ldots, x_n, f(x_1, \ldots, x_n))$. Call that parameterization $S$. Can you compute $X_i = \partial S / \partial x_i$? That's the $i$th basis vector for the tangent space, and $g_{ij}$ is just $X_i \cdot X_j$. (Incidentally, in the formulas above, $f_i$ denotes the $i$th partial derivative of $f$.) $\endgroup$ Jun 7, 2015 at 16:33


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