*3.1. Extrinsic Curvature Using Differential Geometry*

The curvature of a circle at every point on the circumference is equal to the inverse of the radius of the circle. Thus, the curvature of a circle is a constant. A circle with a smaller radius bends more sharply and, therefore, has a higher curvature.

We assume that the first and second derivatives of the nonlinear smooth scalar function *h* exist. The curvature of the curve *y* = *h*(*x*) at a point *x* is equal to the curvature of the osculating circle at that point. The extrinsic curvature at the point *x* is defined by [36–38],

$$\kappa(\mathbf{x}) := \frac{\left| \frac{d^2 h(\mathbf{x})}{d\mathbf{x}^2} \right|}{[1 + (\frac{dh(\mathbf{x})}{d\mathbf{x}})^2]^{3/2}} = \frac{|h(\mathbf{x})|}{[1 + \dot{h}(\mathbf{x})^2]^{3/2}}. \tag{36}$$

The first derivative of *h* at a point *x* is given in (26). The second derivative of *h* with respect to *x* is given by

$$h(\mathbf{x}) = \frac{d^2 h(\mathbf{x})}{d\mathbf{x}^2} = an(n-1)\mathbf{x}^{n-2}, \quad n = 2, 3, \dots \tag{37}$$

Thus, using ˙ *h*(*x*) and ¨ *h*(*x*) in (36), we can calculate the extrinsic curvature *κ*(*x*) at any point *x* by

$$\kappa(\mathbf{x}) = \frac{an(n-1)\mathbf{x}^{n-2}}{[1+(an)^2\mathbf{x}^{2(n-1)}]^{3/2}}.\tag{38}$$
