# If $\gamma$ is spherical, then the equation $\frac{\tau}{\kappa}=\frac{d}{ds}(\frac{\dot{\kappa}}{\tau \kappa^2})$ holds.

Question:

Let $\gamma (t)$ be a unit-speed curve with $\kappa(t)\gt0$ and $\tau(t)\neq0$ for all $t$. Show that, if $\gamma$ is spherical, i.e., if it lies on the surface of a sphere, then $$\frac{\tau}{\kappa}=\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau \kappa^2}\right).\tag{2.22}$$

Let $s$ be the arc length along the curve $\vec{\gamma}(s)$. Recall the definition of tangent vector and the Frenet-Serret formulas $$\vec{t} = \frac{d}{ds}\vec{\gamma}\quad(*1) \quad\text{ and }\quad \left\{\begin{array}{rrrrl} \dot{\vec{t}} =& &\kappa \vec{n}& &(*2a)\\ \dot{\vec{n}} =& -\kappa \vec{t}& &+ \tau\vec{b}&(*2b)\\ \dot{\vec{b}} =& &-\tau\vec{n} & &(*2c) \end{array}\right.$$ We have $$\begin{array}{rrll} & (\vec{\gamma} - \vec{\alpha})\cdot(\vec{\gamma} - \vec{\alpha}) & = r^2\\ \text{diff. once, (*1)} \implies & \vec{t} \cdot(\vec{\gamma} - \vec{\alpha}) & = 0 &(*3a)\\ \text{diff. again, (*2a, *1)} \implies & \kappa\vec{n} \cdot(\vec{\gamma} - \vec{\alpha}) + \vec{t}\cdot\vec{t}& = 0\\ \vec{t}\cdot\vec{t} = 1\implies & \vec{n}\cdot( \vec{\gamma} - \vec{a} )& = -\frac{1}{\kappa} &(*3b)\\ \text{diff. again, (*2b,*1)} \implies & (-\kappa\,\vec{t} + \tau\,\vec{b})\cdot(\vec{\gamma}-\vec{a}) + \vec{n}\cdot\vec{t} & = -\frac{d}{ds}\frac{1}{\kappa} = \frac{\dot{\kappa}}{\kappa^2}\\ \vec{t}\cdot\vec{n} = 0\text{ and (*3a)} \implies & \tau\,\vec{b}\cdot(\vec{\gamma}-\vec{\alpha}) & = \frac{\dot{\kappa}}{\kappa^2}\\ \iff & \vec{b}\cdot(\vec{\gamma}-\vec{\alpha}) & = \frac{\dot{\kappa}}{\tau\kappa^2}\\ \text{diff. again, (*2c,*1)} \implies & -\tau\,\vec{n}\cdot(\vec{\gamma}-\vec{\alpha}) + \vec{b}\cdot\vec{t} &= \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\\ \vec{b}\cdot\vec{t} = 0\text{ and (*3b)} \implies & \frac{\tau}{\kappa} & = \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right) \end{array}$$