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Given a curve $\gamma$ with curvature and torsion $\kappa$ and $\tau$ respectively, if I scale the curve as $\tilde{\gamma} := \dfrac{\gamma}{c}$, then I get $\tilde{\kappa} = c\kappa$ and $\tilde{\tau} = c\tau$, where $c$ is a positive real constant.

But if I start with two curvature and torsion functions $\kappa, \tau$ determining a curve $\gamma$ and then modify them using $c$ as $c\kappa, c\tau$, will the corresponding new curve be related to $gamma$ the same way as in the earlier case??

I dont think so, but then again I am confused as to what is happening?

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    $\begingroup$ Think about what it means for curvature and torsion to determine a curve. $\endgroup$ – Anthony Carapetis Apr 11 '16 at 8:21
  • $\begingroup$ @AnthonyCarapetis well from what I understand about the fundamental theorem of space curves, any set of curvature torsion functions uniquely determines a space curve, but I am not able to gather how that helps me establish the relationship between the curves??Can you add something more?? $\endgroup$ – Vishesh Apr 11 '16 at 8:26
  • $\begingroup$ @AnthonyCarapetis, On second thoughts, I think I get what you said, the curvature and tosion functions and space curves are in a 1-1 correspondence with each other, so both the ways yield the same curve. Is that right?? $\endgroup$ – Vishesh Apr 11 '16 at 8:31
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The converse is also true. As the fundamental theorem of space curves show, the curvature and torsion uniquely define the curve itself. So given $\kappa$ and $\tau$ at every point, it is possible to draw out the curve uniquely (up to rigid transformation).

You have shown that for a new curve $\tilde{\gamma} =\frac{\gamma}{c}$ , then you would immediately know that it is just scaling the curve by a ratio. The curvature and the torsion is just $c$ times the original's. By the same logic, if you have $c$ times the original curvature and torsion, they uniquely define the scaled curve.

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  • $\begingroup$ Thanks a lot. The question seems really silly once you look bacak, in fact right after Anthony Carapetis' comments. $\endgroup$ – Vishesh Apr 11 '16 at 9:18

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