For the level set method, $\phi(\vec{x},t)$ is the level set function in 3D and the level set $\phi(\vec{x},t) = 0$ forms the interface. For evolving $\phi$ the derivation says to imagine a particle $\textbf{x}(t)$ on the surface, then we differentiate with respect to t
$\frac{d}{dt}(\phi(\vec{x}, t)=0)$
Using chain rule we get
$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot \frac{d\vec{x}}{dt} = 0$
Then this equation is turned into
$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot \vec{V} = 0$
I guess $\vec{V}$ is the speed, and its elements are to replace $dx$, $dy$, .. that are elements of $d{\vec{x}}$. Hence we can substitute our own displacement and receive a result.
What I do not understand is they continue like this:
Seperate $\vec{V}$ into normal and tangential components
$\frac{\partial \phi}{\partial t} + \triangledown \phi \cdot (\vec{V_N}\vec{N} + \vec{V_T}\vec{T}) = 0$
Then since $\vec{N} = \frac{\triangledown \phi}{|\triangledown \phi |}$
We get
$\frac{\partial \phi}{\partial t} + V_N \cdot |\triangledown \phi| = 0$
What happened to $V_T$? Also how can the dot product of $\triangledown \phi$ with $\frac{\triangledown \phi}{|\triangledown \phi |}$ result in $|\triangledown \phi|$ ?
Link: http://www.cs.au.dk/~bang/smokeandwater2006/Lecture9_IntroToWaterAndLS.ppt