Though the title includes "(geometric) evolution equations", my question is really more of how to use normal coordinates to help us prove an identity involving components of a tensor. My sole experience with these coordinates is the proof of the second Bianchi identity $$\nabla\mathrm{Rm}(X,Y,Z,V,W)+\nabla\mathrm{Rm}(X,Y,V,W,Z)+\nabla\mathrm{Rm}(X,Y,W,Z,V)=0.$$ To prove this identity, we show the LHS vanishes at an arbitrary point $p$. The justification can be tedious, but if we choose normal coordinates centered at $p$, then large amounts of effort can be dispensed. The whole process is fine with me, but some other techniques seem to be needed if I want to prove instead that $$\frac{\partial}{\partial t}g_{ij}=2Hh_{ij},\tag{1}$$ which describes how the induced metric $g:=F^*k$ on the immersion $F:M\to(N,k)$ evolves under the mean curvature flow $$\partial_t F_t=H\nu.$$ Those $h_{ij}$ are the components of the second fundamental form. Now fix $p\in M$. If I succeed in justifying (1) at $p$ with normal coordinates $(x^i)$ centered at $p$, then can I conclude that $$\frac{\partial}{\partial t}\widetilde{g}_{\alpha\beta}=2H\widetilde{h}_{\alpha\beta}\tag{2}$$ for any (NOT necessarily normal) coordinates $(\widetilde{x}^\alpha)$ around $p$?
This question may sound stupid, but I need someone experienced to check if my idea works:
At $p$, the rule for a change of coordinates tells us $$\widetilde{g}_{\alpha\beta}=g_{ij}\frac{\partial x^i}{\partial\widetilde{x}^\alpha}\frac{\partial x^j}{\partial\widetilde{x}^\beta}.$$ Now, at $p$, $$\begin{align} \frac{\partial}{\partial t}\widetilde{g}_{\alpha\beta}&=\frac{\partial x^i}{\partial\widetilde{x}^\alpha}\frac{\partial x^j}{\partial\widetilde{x}^\beta}\frac{\partial}{\partial t}g_{ij}\\ &=\frac{\partial x^i}{\partial\widetilde{x}^\alpha}\frac{\partial x^j}{\partial\widetilde{x}^\beta}(2Hh_{ij})\\ &=2H\left(\frac{\partial x^i}{\partial\widetilde{x}^\alpha}\frac{\partial x^j}{\partial\widetilde{x}^\beta}h_{ij}\right)\\ &=2H\widetilde{h}_{\alpha\beta}. \end{align}$$ Did I get it? Thank you.