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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.

F.A.
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1190
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    An old differential geometry exercise whose solution had appeared on the Internet multiple times, e.g. here: http://math.stackexchange.com/questions/198119/prove-that-a-curve-is-spherical-iff-it-satisfies-the-relation – Francis Oct 20 '13 at 09:38
  • The question you sent its link and my question are not same:( @francis – 1190 Oct 20 '13 at 09:42
  • Sorry, I dont understand the question you sent. It is to complecated. :( @Francis – 1190 Oct 20 '13 at 09:50
  • Using notations in another answer, you have: $$\begin{array}{rl} \vec{t}\cdot(\vec{\gamma} - \vec{\alpha}) & = 0\ \vec{n}\cdot(\vec{\gamma} - \vec{\alpha}) & = -\frac{1}{\kappa}\ \vec{b}\cdot(\vec{\gamma} - \vec{\alpha}) & = \frac{\dot{\kappa}}{\tau\kappa^2} \end{array}$$ – achille hui Oct 20 '13 at 09:52
  • Two conditions are actually equivalent, you should see that by taking derivative on both sides of your equation using chain rule. – Francis Oct 20 '13 at 10:05
  • I am confused too much. So sorry. I am grateful of you if someone Will solve this. By using both question, it is impossible to solve this question. Because I am a newlearner @achillehui – 1190 Oct 20 '13 at 10:13

1 Answers1

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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.

Francis
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  • Hmm, I guess, from here, I will continue to solve in the link begining from the line "Now, we go the other direction, so assume". Isnt it? – 1190 Oct 20 '13 at 10:38
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    @B11b: Not really, see my updated answer. – Francis Oct 20 '13 at 21:02