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I'm learning differential geometry, specifically the theory of curves, and need help with the following exercise:

Consider the regular curve $\Gamma : \vec x = \vec x(s) \in C^4$ of $E^3$. If $\rho_1$ and $\rho_2$ are the radii of curvature of $\Gamma_1 : \vec x(s) = \mathbf{T}(s)$ and $\Gamma_2 : \vec x(s) = \mathbf{B}(s)$ respectively, show that $\rho_1^2 + \rho_2^2 = 1$.

We use $\mathbf{T}(s)$ and $\mathbf{B}(s)$ to denote the tangent and binormal vectors, respectively.

I'm sorry for my lack of effort but unfortunely I wasn't able to do much. My first thought was to manipulate the Frenet–Serret formulas but I couldn't reach any useful result. Also, I don't understand the importance of the first sentence of the problem. $\Gamma, \Gamma_1$ and $\Gamma_2$ are three different curves, so why is it important to know that $\Gamma \in C^4$ of $E^3$? We are only interested in the relationship between $\Gamma_1$ and $\Gamma_2$.

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  • $\begingroup$ The problem should be $\rho_1^2+\rho_2^2=1$ $\endgroup$ Sep 6 '16 at 3:00
  • $\begingroup$ @ReneSchipperus Yes, absolutely. I corrected my original post. Thank you. $\endgroup$
    – glpsx
    Sep 6 '16 at 3:03
  • $\begingroup$ Is it easier now ? $\endgroup$ Sep 6 '16 at 3:05
  • $\begingroup$ @ReneSchipperus I was working all along with the correct formulation of the problem. This was an unfortunate typo. I'm still stuck. $\endgroup$
    – glpsx
    Sep 6 '16 at 3:15
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    $\begingroup$ Yeah s is arc length for the ORIGINAL curve, but its not arc length for T and B. $\endgroup$ Sep 6 '16 at 3:45
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Here is the complete solution I indicated in the comments earlier, although OP has probably found it on his own by now.

The Frenet equations are $$\dot{\mathbf{t}}=\kappa \mathbf{n}$$ $$\dot{\mathbf{n}}=-\kappa \mathbf{t}+\tau \mathbf{b}$$ $$\dot{\mathbf{b}}=-\tau\mathbf{n}.$$

Given a curve $\mathbf{r}(t)$ where $t$ need not be arc length,

$$\kappa_{\mathbf{r}}=\frac{\dot{|\mathbf{r}} \times \ddot{\mathbf{r}}|}{|\dot{\mathbf{r}}|^3}$$

Taking first $\mathbf{r}=\mathbf{ t}$ we have $$\dot{\mathbf{ t}}=\kappa \mathbf{n}$$ $$\ddot{\mathbf{ t}}=\dot{\kappa} \mathbf{n}+\kappa(-\kappa \mathbf{t}+\tau \mathbf{b})$$ $$=\dot{\kappa} \mathbf{n}-\kappa^2 \mathbf{t}+\kappa\tau \mathbf{b}$$

Now $$\dot{\mathbf{t}} \times \ddot{\mathbf{t}}=\kappa^3\mathbf{b}+\kappa^2\tau\mathbf{t}$$ so $$|\dot{\mathbf{t}} \times \ddot{\mathbf{t}}|=\sqrt{\kappa^6+\kappa^4\tau^2}$$ $$=\kappa^2\sqrt{\kappa^2+\tau^2}$$

And $$|\dot{\mathbf{t}}|^3=\kappa^3$$ So $$\kappa_{\mathbf{t}}=\frac{\sqrt{\kappa^2+\tau^2}}{\kappa}.$$

Similarly, $$\kappa_{\mathbf{b}}=\frac{\sqrt{\kappa^2+\tau^2}}{|\tau|}.$$

The result is now clear.

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  • $\begingroup$ Thank you for taking the time to write a complete solution to the problem. Following your previous explanation I was also able to find the path to the correct solution. The fact that $\mathbf T$ and $\mathbf N$ are not parameterized by arc length does still confuses me though. I don't know if there is an easy way to understand this. $\endgroup$
    – glpsx
    Sep 6 '16 at 14:28
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    $\begingroup$ I am glad. Yeah $s$ is the arc length parameter for the original curve, $T$ and $B$ will have there own arc length parameters, which may be hard to calculate. In general non arc length parameterizations are forced on us in diff geo. $\endgroup$ Sep 6 '16 at 14:36
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HINT: Note that the unit tangent vector for both the curves $\Gamma_1$ and $\Gamma_2$ is $\mathbf N(s)$. Now use the chain rule to compute the derivative of $\mathbf N$ with respect to the arclength of $\Gamma_1$ and with respect to the arclength of $\Gamma_2$. In particular, if $\sigma_i$ is the arclength of $\Gamma_i$, note that $$\frac{d\mathbf N}{d\sigma_i} = \frac{\frac{d\mathbf N}{ds}}{\frac{d\sigma_i}{ds}},$$ and recall that $\dfrac{d\sigma_i}{ds}$ is the speed of the parametrized curve $\Gamma_i$.

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