Proving that the sum of the squared radii of curvature of two curves is constant

Question

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 $\Gamma_2$.

Answer 1

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.

Answer 2

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 $\Gamma_i$.