calculuscurvaturedifferential-geometrymultivariable-calculus
The curvature equation for implicit functions, level sets is usually given in two forms: one is the divergence of the gradient of the unit normal:
$\kappa = \bigtriangledown \cdot \frac{\bigtriangledown \phi}{|\bigtriangledown \phi|}$
and the other is
$\kappa = \frac{\phi_{xx}\phi_y^2 – 2\phi_x\phi_y\phi_{xy} + \phi_{yy}\phi_x^2}{(\phi_x^2+\phi_y^2)^{3/2}}$
How do we derive the second equation from the first?
There are a number of characterizations of curvature. The "most intuitive" and geometric one, in my opinion, is the inverse of the radius of curvature. That is, intuitively, the radius of the circle that lies tangent to the curve in a neighborhood of the point (more formally, the osculating circle.
You may think of it as the limit of the distance to the intersection pont of two normals of arbitrarily close points on the curve.
Your lecturer's definition coincides nicely with this one.
I'll first explain where the dot product comes from. If $$\vec v = \alpha \vec w$$ Then taking the definition of $\vec v \cdot \vec w$ as $|v||w|\cos\theta$ where $\theta$ is the angle between them. We have that since $|w| = \alpha|v|$, and $\theta = 0$ because they are scalar multiples, and therefore in the same direction,$$\vec v \cdot \vec w = \alpha|v|^2$$ and since $|v|$ is 0 for a unit vector (as in your case): $$T'\cdot U = \kappa |\gamma'|$$ And division gives your second formula.
From $\| \gamma'(s) \| = 1$, it follows that $$0 = \frac{d}{ds} \langle \gamma'(s),\gamma'(s) \rangle = 2\langle \gamma''(s),\gamma'(s)\rangle.$$
So $\gamma'$ and $\gamma''$ are perpendicular. Assume that $\gamma''(s) \neq 0$. Thus $N(s) := \gamma''(s)/\|\gamma''(s)\|$ locally defines a unit normal field to $\gamma$. Since $\gamma$ parametrizes a level set of $\phi$ and $\nabla \phi$ is orthogonal to its level sets it follows that $\nabla \phi / \|\nabla \phi\|(\gamma(s)) = \pm N(s)$. Consequently
$$\kappa(s) = \|\gamma''(s)\| = |\langle \gamma''(s),N(s)\rangle| = |\frac{d}{ds} \langle \gamma'(s),N(s)\rangle - \langle \gamma'(s),\frac{d}{ds}N(s)\rangle| = |\langle \gamma'(s),\frac{d}{ds} N(s)\rangle|.$$
Since $\left\|N(s)\right\| = 1$ it follows as for $\gamma'$ that $\langle N,\frac{dN}{ds} \rangle = 0$, thus $\frac{dN}{ds}$ is a multiple of $\gamma'$ and $\left\|\frac{dN}{ds}\right\| = |\langle \gamma'(s),\frac{dN}{ds}\rangle|$. Thus
$$\kappa(s) = \left\|\frac{dN}{ds}(s)\right\| = \left\|\frac{d}{ds} \frac{\nabla \phi}{\|\nabla \phi\|}\right\|.$$
Thus it remains to see that $|\nabla \cdot N| = \left\|\frac{dN}{ds}\right\|$: For this one needs the following fact: Let $\{b_i\}$ be an orthonormal basis of $\mathbb R^n$. Then $$\nabla \cdot X = \sum_{i = 1}^n \langle DX(b_i),b_i\rangle$$
for any vectorfield $X : \mathbb R^n \to \mathbb R^n$ with differential $DX$.
Unfortunately (via google) I did not find a reference for this Edit: Using Riemannian geometry this follows since divergence is defined as the trace of the Levi Civita connection $\nabla$, which on $\mathbb R^n$ is given by $\nabla_XY = DY(X)$ for vector fields $X,Y$. From this we see
$$\left\|\frac{dN}{ds}(s)\right\| = \langle \gamma'(s),\frac{dN}{ds}(s)\rangle = \langle \gamma'(s),DN(\gamma'(s))\rangle + \langle N(s),DN(N(s))\rangle = \nabla \cdot N(s).$$
By $DN$ we mean the differential of $N$ (note that strictly speaking this does not exists, as $N$ is only defined along $\gamma$. But we can extend $N$ to a vectorfield defined on an open subset using $N = \pm \nabla \phi/||\nabla \phi||$ as above). The second equality holds since $\langle N,DN(X))\rangle = 1/2\partial_X\langle N,N \rangle = 0$ for any vector field $X$ ($\partial_X$ denotes the directional derivative in direction $X$).
Maybe these calculations are more complicated than they could be, but I don't see a simplification right away.
Best Answer
Just expand in coordinates: $$\begin{align}\kappa &= \nabla \cdot \frac{\nabla \phi}{|\nabla \phi|} = \nabla \cdot \frac{(\phi_x,\phi_y)}{\sqrt{\phi_x^2+\phi_y^2}}\\ &=\left(\partial_x \frac{\phi_x}{\sqrt{\phi_x^2+\phi_y^2}}\right)+ \left(\partial_y \frac{\phi_y}{\sqrt{\phi_x^2+\phi_y^2}}\right) \\ &= \frac{\phi_{xx}}{\sqrt{\phi_x^2+\phi_y^2}} - \frac{\phi_x (\phi_x\phi_{xx}+\phi_y\phi_{xy})} {(\phi_x^2+\phi_y^2)^{3/2}} + \frac{\phi_{yy}}{\sqrt{\phi_x^2+\phi_y^2}} - \frac{\phi_y(\phi_x\phi_{xy}+\phi_y\phi_{yy})} {(\phi_x^2+\phi_y^2)^{3/2}} \\ &= \frac{\phi_{xx}(\phi_x^2+\phi_y^2) - \phi_x (\phi_x\phi_{xx}+\phi_y\phi_{xy}) +\phi_{yy}(\phi_x^2+\phi_y^2) - \phi_y(\phi_x\phi_{xy}+\phi_y\phi_{yy})}{(\phi_x^2+\phi_y^2)^{3/2}}\\ &= \frac{\phi_{xx}\phi_y^2 - 2\phi_x\phi_y\phi_{xy} + \phi_{yy}\phi_x^2}{(\phi_x^2+\phi_y^2)^{3/2}} \end{align}$$