to calculate $\frac{\d}{\d t}\operatorname{Vol}(F(N,t))|_{t=0}$. Given local coordinates $x_i$ on $N$ we know that $$\operatorname{Vol}(F(N,t))=\int_N\sqrt{\det\left(g\left(\frac{\partial F}{\partial x_i},\frac{\partial F}{\partial x_j}\right)\right)}\operatorname{vol}_N.$$ Let $p\in N$ and choose our coordinates $x_i$ to be normal coordinates at $p$: i.e. so that $\frac{\partial F}{\partial x_i}(p,t)=e_i(t)$ satisfy $g(e_i(0),e_j(0))=\delta_{ij}$. If $g_{ij}(t)=g(e_i(t),e_j(t))$ and $(g^{ij}(t))$ denotes the inverse of the matrix $(g_{ij}(t))$ then we know that $$\frac{\d}{\d t}\sqrt{\det(g_{ij}(t))}|_{t=0}=\frac{1}{2}\frac{\sum_{i,j}g^{ij}(t)g_{ij}'(t)}{\sqrt{\det(g_{ij}(t))}}|_{t=0}=\frac{1}{2}\sum_ig_{ii}'(0).$$

Now, if we let $\nabla$ denote the Levi-Civita connection of $g$, then $$\begin{aligned}
\frac{1}{2}\sum_ig_{ii}'(0)&=\frac{1}{2}\sum_i \frac{\d}{\d t}g\left(\frac{\partial F}{\partial x_i},\frac{\partial F}{\partial x_i}\right)|_{t=0}\\
&=\sum_ig(\nabla_Xe_i,e_i)\\
&=\sum_ig(\nabla_{e_i}X,e_i)=\operatorname{div}_N(X)\end{aligned}$$ since $[X,e_i]=0$ (i.e. the $t$ and $x_i$ derivatives commute). Moreover, we see that $$\operatorname{div}_N(X)=\sum_ig(\nabla_{e_i}X,e_i)=\operatorname{div}_N(X^{\rm T})-\sum_ig(X^{\perp},\nabla_{e_i}e_i)=\operatorname{div}_N(X^{\rm T})-g(X,H)$$ (since $\nabla_{e_i}\big(g(X^{\perp},e_i)\big)=0$) where ${}^{\rm T}$ and ${}^\perp$ denote the tangential and normal parts and $$H=\sum_{i}\nabla^{\perp}_{e_i}e_i$$ is the *mean curvature vector*. Overall we have the following.

The first variation formula is $$\frac{\d}{\d t}\operatorname{Vol}(F(N,t))|_{t=0}=\int_N \operatorname{div}_N(X)\operatorname{vol}_N=-\int_N g(X,H)\operatorname{vol}_N.$$

The $\operatorname{div}_N(X^{\rm T})$ term does not appear in the first variation formula because its integral vanishes by the divergence theorem as $N$ is compact without boundary. In general, it will still vanish since we assume for our variations that there exists a compact submanifold of $N$ with boundary which contains the support of $X^{\rm T}$ and so that $X^{\rm T}$ vanishes on the boundary.

We deduce the following.

$N$ is a 