PDE for the Mean Curvature Flow applied to a graph of a function

Question

Given a graph of a function $\gamma = (z,\rho(z))$, we know that the curvature $\kappa$ at the point $(z,\rho(z))$ is given by $$\kappa(z) = \frac{\rho''(z)}{\left(1+|\rho'(z)|^2\right)^{\frac 32}}.$$ I found in many reference materials that if we apply a mean curvature flow on $\gamma$ (which now evolves with time $t$), the resulting PDE will read as $$\partial_t\rho = \frac{\partial_{zz} \rho}{1+|\partial_z \rho|^2}.$$ Since I do not fully understand the derivation of this PDE given in all of the references I have found so far, I would like derive it on my own. Unfortunately, so far I only have the following wrong calculation by myself: According to the definition of MCF, we should have $$\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \kappa\,{\bf N},$$ where ${\bf N} = \frac{(-\rho'(z),1)}{\sqrt{1+|\rho'(z)|^2}}$ is the unit normal vector and $\frac{\mathrm{d} \gamma}{\mathrm{d} t} = (0,\partial_t \rho)$. However, these only allow me to obtain $$\partial_t\rho = \frac{\partial_{zz} \rho}{\left(1+|\partial_z \rho|^2\right)^2},$$ which is NOT the correct PDE that we are aiming for. Can someone point out my error and derive the right PDE? Thanks for any help!
Edit: It seems that if we use the relation $\frac{\mathrm{d} \gamma}{\mathrm{d} t} \cdot {\bf N} = \kappa$ instead of the vector identity $\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \kappa\,{\bf N}$, we can get the correct PDE. But obviously $\frac{\mathrm{d} \gamma}{\mathrm{d} t} \cdot {\bf N} = \kappa$ should be equivalent to $\frac{\mathrm{d} \gamma}{\mathrm{d} t} = \kappa\,{\bf N}$. This causes me a huge confusion...