Let $F:M\times[0,T)\to(N,\bar{g})$ be a one-parameter family of embedded hypersurfaces that solves a geometric flow $$\frac{\partial}{\partial t}F=\lambda n,$$ where $\lambda$ is a function to be designated as the speed of the flow and $n$ is the unit normal vector. Now I would like to prove the evolution of the induced metric: $$\frac{\partial}{\partial t}g_{ij}=2\lambda h_{ij}$$ Here is an excerpt from a proof by Huisken and Polden:
I was wondering how the second displayed equation comes and how it leads to the third displayed equation.
I suppose what the authors do is simply differentiate $g_{ij}$ w.r.t. $t$, so I get $$\frac{\partial}{\partial t}g_{ij}=\bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^i},\frac{\partial F}{\partial x^j}\right)+\bar{g}\left(\frac{\partial F}{\partial x^i},\frac{\partial}{\partial t}\frac{\partial F}{\partial x^j}\right).\tag{1}$$ But how do I combine these two terms to find $$\frac{\partial}{\partial t}g_{ij}=2\bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^i},\frac{\partial F}{\partial x^j}\right)?$$ $\bar{g}$ is a symmetric $2$-tensor field, so I do have $\bar{g}(\partial_i,\partial_j)=\bar{g}(\partial_j,\partial_i)$, but does this principle apply equally well to the RHS of (1)?
The next question is how I can manage to expand the product derivative $\frac{\partial}{\partial x^i}(\lambda n)$. Honestly, I do not know what this derivative is supposed to mean.
Could someone please provide me with some details? Thank you.
Edit. Why can we switch the roles of $i$ and $j$: $$\begin{align} \bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^i},\frac{\partial F}{\partial x^j}\right)+\bar{g}\left(\frac{\partial F}{\partial x^i},\frac{\partial}{\partial t}\frac{\partial F}{\partial x^j}\right)&=\bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^i},\frac{\partial F}{\partial x^j}\right)+\bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^{\color{red}j}},\frac{\partial F}{\partial x^{\color{red}i}}\right)\\ &=\bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^i},\frac{\partial F}{\partial x^j}\right)+\bar{g}\left(\frac{\partial}{\partial t}\frac{\partial F}{\partial x^{\color{red}i}},\frac{\partial F}{\partial x^{\color{red}j}}\right)? \end{align}$$
$\newcommand\gb{\bar{g}}$\begin{align*} \partial_tg_{ij} &= \partial_t(\gb(\partial_iF,\partial_jF))\\ &= \gb(\partial_t(\partial_iF),\partial_jF)+ \gb(\partial_iF,\partial_t\partial_jF)\\ &= \gb(\nabla_i(\partial_tF),\partial_jF)+ \gb(\partial_iF,\nabla_j(\partial_tF))\\ &= \gb(\nabla_i(\lambda n),\partial_jF) + \gb(\partial_iF,\nabla_j(\lambda n))\\ &=\gb(\partial_i\lambda n + \lambda\nabla_in,\partial_jF) + \gb(\partial_iF,\partial_j\lambda n + \lambda \nabla_jn)\\ &= \lambda(\gb(\nabla_in,\partial_jF) + \gb(\partial_iF,\nabla_jn)). \end{align*} So set $$ 2\lambda h_{ij} = \lambda(\gb(\nabla_in,\partial_jF) + \gb(\partial_iF,\nabla_jn)), $$ i.e., $$ h_{ij} = \frac{1}{2}(\gb(\nabla_in,\partial_jF) + \gb(\partial_iF,\nabla_jn)), $$
Finally,recall that $\nabla n$ is the Weingarten map $$\gb(\nabla_in,\partial_jF) = \gb(\partial_iF,\nabla_jn) $$ is the second fundamental form. Therefore, the last equation in the excerpt holds.