Context: I have just been introduced to mean curvature and would like to verify an elementary computation to ensure that I am not kidding myself when learning this material.
Let us recall that a smooth family $(M_t)_{t \in I}$ of embedded hypersurfaces flows according to mean curvature if $$\partial_t x = {H}(x),$$ where $x \in M_t$ and $H(x)$ is the mean curvature at $x$.
Let $g_{ij}$ denote the metric tensor of $M_t$. I want to verify that $\partial_t g_{ij} = -2H A_{ij}$.
Q1: Is $A_{ij} = -(\partial_i \nu)(\partial_j X)$, where $\nu$ denotes the unit outer normal?
Let us proceed: \begin{eqnarray*} \partial_tg_{ij} &=& \partial_t( \partial_i X \cdot \partial_j X) \\ &=& \partial_t(\partial_i X) \cdot (\partial_j X) + (\partial_i X) \cdot \partial_t (\partial_j X) \\ & = & \partial_i(\partial_t X) \cdot (\partial_j X) + (\partial_i X) \cdot \partial_j(\partial_t X) \\ &=& \partial_i (H\nu) \cdot (\partial_j X) + (\partial_i X) \cdot \partial_j (H \nu) \\ &=& 2H (\partial_i \nu)(\partial_j X) \\ &=& -2H A_{ij}. \end{eqnarray*}
Thanks for your time.
Your calculation is correct. Indeed the same is true when the ambient space is not Euclidean. When the immersion is not of coidimension one, we'll have
$$\partial_t g_{ij} = -2 \langle A_{ij}, \vec H\rangle.$$