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 $$\frac{\tau}{\kappa}=\frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau \kappa^2}\right).\tag{2.22}$$
Please help me doing this question. 
In fact, there is its solution as I posted below. But I don't understand the answer. Please explain this more clearly. Thank you for helping. 
Solution:

 A: It is really a chasing of definition. I have supplied the info needed to derive each step below. I hope that is clear enough.
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
$$\begin{array}{rrll}
& (\vec{\gamma} - \vec{\alpha})\cdot(\vec{\gamma} - \vec{\alpha}) & = r^2\\
\text{diff. once, (*1)} \implies & \vec{t} \cdot(\vec{\gamma} - \vec{\alpha}) & = 0 &(*3a)\\
\text{diff. again, (*2a, *1)} \implies & \kappa\vec{n} \cdot(\vec{\gamma} - \vec{\alpha}) + \vec{t}\cdot\vec{t}& = 0\\
\vec{t}\cdot\vec{t} = 1\implies & \vec{n}\cdot( \vec{\gamma} - \vec{a} )& = -\frac{1}{\kappa} &(*3b)\\
\text{diff. again, (*2b,*1)} \implies & (-\kappa\,\vec{t} + \tau\,\vec{b})\cdot(\vec{\gamma}-\vec{a}) + \vec{n}\cdot\vec{t} & = -\frac{d}{ds}\frac{1}{\kappa} = \frac{\dot{\kappa}}{\kappa^2}\\
\vec{t}\cdot\vec{n} = 0\text{ and (*3a)} \implies & \tau\,\vec{b}\cdot(\vec{\gamma}-\vec{\alpha}) & = \frac{\dot{\kappa}}{\kappa^2}\\
\iff & \vec{b}\cdot(\vec{\gamma}-\vec{\alpha}) & = \frac{\dot{\kappa}}{\tau\kappa^2}\\
\text{diff. again, (*2c,*1)} \implies & -\tau\,\vec{n}\cdot(\vec{\gamma}-\vec{\alpha}) + \vec{b}\cdot\vec{t} &= \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)\\
\vec{b}\cdot\vec{t} = 0\text{ and (*3b)} \implies & \frac{\tau}{\kappa} & = \frac{d}{ds}\left(\frac{\dot{\kappa}}{\tau\kappa^2}\right)
\end{array}$$
