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A regular curve $C$ in $\mathbb R^3$ is called a Bertrand Curve, if there exists a diffeomorphism $f:C \to D$ from $C$ onto a different regular curve $D$ in $\mathbb R^3$ such that $N_xC=N_{f(x)}D$ for any $x\in C$. Here $N_xC$ denotes the principal normal line of the curve $C$ passing through $x$, and $T_xC$ will denote the tangent line of $C$ at $x$. Prove that: a) The distance $|x-f(x)|$ is constant for $x\in C$; and the angle made between the directions of the two tangent lines $T_xC$ and $T_{f(x)}D$ is also constant.

I can prove that $|x-f(x)|$ is constant,since we can write curve $C$ as $\alpha(s):[a,b]\to \mathbb R^3$ ($s$ is the arc of $C$), and curve $D$ as$f(\alpha(s))$. So we have $$\alpha(s)+r\alpha''(s)=f(\alpha(s))$$. Differentiating both sides of the equation, we can get $\overrightarrow{t(s)}-rk\overrightarrow{t(s)}-r\tau \overrightarrow{b}=df\cdot \alpha'(s)$ and $r$ is a constant.

Second, in order to prove the angle between two tangent lines are constant, I want to differentiate $\langle \alpha'(s),\frac{df\cdot \alpha'(s)}{|df\cdot \alpha'(s)|} \rangle$ to get zero, but I can't.

Assume $\overrightarrow{t},\overrightarrow{n},\overrightarrow{b}$ are respectively tangent vector, principal normal vector, second normal vector with arc $s$ parameter of curve C, and $k,\tau$ are respectively curvature and torsion of C. This is my calculation:

$$\frac{d}{ds} \langle \alpha'(s),\frac{df\cdot \alpha'(s)}{|df\cdot \alpha'(s)|} \rangle = \langle \alpha''(s),\frac{df\cdot \alpha'(s)}{|df\cdot \alpha'(s)|} \rangle + \langle \alpha'(s),\frac{d}{ds}(\frac{(1-rk)\overrightarrow{t}-r\tau \overrightarrow{b}}{\sqrt{(1-rk)^2+r^2\tau^2}}) \rangle = \langle \alpha'(s),\frac{-rk'\overrightarrow{t}(s)-(1-rk)k\overrightarrow{n}(s)-r\tau'\overrightarrow{b}(s)-r\tau\overrightarrow{b'}(s)-((1-rk)\overrightarrow{t}(s)-r\tau\overrightarrow{b}(s))\cdot \frac{1}{2}\frac{-2(1-rk)rk'+2r^2\tau\tau'}{\sqrt{(1-rk)^2+r^2t^2}}}{(1-rk)^2+r^2\tau^2} \rangle.$$ Only choose terms about $\overrightarrow{t}$ in the numerator and make the inner product with $\alpha'(s)$. And I get the formula about is equal to $$\frac{-rk'((1-rk)^2+r^2\tau^2)+(1-rk)(rk'-r^2kk'-r^2\tau\tau')}{((1-rk)^2+r^2\tau^2)^{3/2}}=\frac{-r^2k'\tau^2-r^2\tau\tau'+r^3k\tau\tau'}{((1-rk)^2+r^2\tau^2)^{3/2}}$$ which is not equal to zero :(

Could anybody give some advise and hints to me?thank you! :)

share|cite|improve this question
The derivative of $f(\alpha(s))$ wrt $s$ isn't $df.\alpha'(s)$ but $df(\alpha(s)).\alpha'(\alpha(s))=J_f(\alpha(s))\alpha'(s)$. So when you compute the derivative of $\langle\alpha'(s),...\rangle$ you need to compute the derivative of $df(\alpha(s)).\alpha'(s)$. – Mercy King Jun 22 '12 at 22:36

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