Show that for a curve lying on a sphere of radius r with nowhere vanishing torsion, the following equation is satisfied:
$$\left(\frac{1}{\kappa}\right)^2+\left(\frac{\dot{\kappa}}{\kappa^2\tau}\right)^2=r^2$$
Please help me doing this. Honestly, I could not do anything. So I cannot show my efforts. Thank you.
This is to illustrate the relationship between two question: write $\rho=1/\kappa$ and $\sigma=1/\tau$. We have $$\rho^\prime=\left(\frac{1}{\kappa}\right)^\prime=-\frac{k^\prime}{k^2}.$$ Your equation thus becomes $$\rho^2+(\rho^\prime\sigma)^2=r^2.$$ Take derivative on both side, we get $$2\rho\rho^\prime+2(\rho^\prime\sigma)(\rho^{\prime\prime}\sigma+\rho^\prime\sigma^\prime)=0,$$ manipulate the terms few time we have $$\frac{\rho}{\sigma}+(\rho^{\prime\prime}\sigma+\rho^\prime\sigma^\prime)=\frac{\rho}{\sigma}+(\rho^\prime\sigma)^\prime=0,$$ which is equivalent to the equation in the link.
Remark:
Judging by your question you actually want to go from a spherical curve to its condition. Here is a hint on how to start: consider a unit-speed curve $\alpha$ lies on a sphere with center $\mathbf{c}$ and radius $r$. (since an arbitrary-speed curve involves only reparameterization, the result here can be extended.) We can thus write $$(\alpha-\mathbf{c})\cdot(\alpha-\mathbf{c})=r^2.$$ Take derivative on both side, $$2(\alpha-\mathbf{c})^\prime\cdot(\alpha-\mathbf{c})=0.$$ Recall that $T=(\alpha-\mathbf{c})^\prime$, so $$T\cdot(\alpha-\mathbf{c})=0.$$ Keep on this track and utilize different Frenet formula, you will eventually get to the desired result.