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 Post subject: Bertrand curves
PostPosted: Sat Nov 21, 2015 6:12 am 
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We call Bertrand curves, two curves $\overrightarrow{r_1},\overrightarrow{r_2}:I\subset\mathbb{R}\longrightarrow \mathbb{R}^3$ which have the property: For every $t\in I$ the normal lines at $\overrightarrow{r_1}(t),\;\overrightarrow{r_2}(t)$ are identical. Prove that:
(i) the distance between the points $\overrightarrow{r_1}(t)$ and $\overrightarrow{r_2}(t)$ is constant, for every $t\in I$.
(ii) the angle between the tangent lines at $\overrightarrow{r_1}(t),\;\overrightarrow{r_2}(t)$ is constant, for every $t\in I$.

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 Post subject: Re: Bertrand curves
PostPosted: Mon Jul 18, 2016 9:58 am 
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We give a solution:

WLOG we consider a natural parameter $s$ for the two curves. The normal line of $\overrightarrow{r_1}$ at $\overrightarrow{r_1}(s)$ has equation $\varepsilon_1:\overrightarrow{y}=\overrightarrow{r_1}(s)+\lambda(s)\,\overrightarrow{n_1}(s)\,,\; \lambda\in \mathbb{R}$ and the normal line of $\overrightarrow{r_2}$ at $\overrightarrow{r_2}(s)$ has equation $\varepsilon_2:\overrightarrow{y}=\overrightarrow{r_2}(s)+\mu(s)\,\overrightarrow{n_1}(s)\,,\; \mu\in \mathbb{R}$.
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  1. Because the two lines are identical, we have that $\overrightarrow{r_1}(s)\in\varepsilon_2\,,\; \overrightarrow{r_2}(s)\in\varepsilon_1$, $\overrightarrow{n_2}(s)=\pm\overrightarrow{n_1}(s)$ and \begin{align*}
    \overrightarrow{r_2}(s)-\overrightarrow{r_1}(s)=\lambda(s)\,\overrightarrow{n_1}(s)\quad&\Longrightarrow\quad\dot{\overrightarrow{r_2}}-\dot{\overrightarrow{r_1}}=\dot{\lambda}\,\overrightarrow{n_1}+\lambda\,\dot{\overrightarrow{n_1}}\\
    &\Longrightarrow\quad\overrightarrow{t_2}-\overrightarrow{t_1}=\dot{\lambda}\,\overrightarrow{n_1}+\lambda\,\big(-\kappa_1\,\overrightarrow{t_1}+\tau_1\,\overrightarrow{b_1}\big)\\
    &\Longrightarrow\quad\overrightarrow{t_2}-\overrightarrow{t_1}=\dot{\lambda}\,\overrightarrow{n_1}-\lambda\,\kappa_1\,\overrightarrow{t_1}+\lambda\,\tau_1\,\overrightarrow{b_1}\\
    &\Longrightarrow\quad\big(\,\overrightarrow{t_2}-\overrightarrow{t_1}\big)\cdot\overrightarrow{n_1}=\big(\dot{\lambda}\,\overrightarrow{n_1}-\lambda\,\kappa_1\,\overrightarrow{t_1}+\lambda\,\tau_1\,\overrightarrow{b_1}\big)\cdot\overrightarrow{n_1} \\
    &\stackrel{\overrightarrow{n_2}=\pm\overrightarrow{n_1}}{=\!=\!=\!\Longrightarrow}\quad\pm\overrightarrow{t_2}\cdot\overrightarrow{n_2}-\cancelto{0}{\overrightarrow{t_1}\cdot\overrightarrow{n_1}}=\dot{\lambda}\,\cancelto{1}{\overrightarrow{n_1}\cdot\overrightarrow{n_1}\big.}-\lambda\,\kappa_1\,\cancelto{0}{\overrightarrow{t_1}\cdot\overrightarrow{n_2}}+\lambda\,\tau_1\,\cancelto{0}{\overrightarrow{b_1}\cdot\overrightarrow{n_1}} \\
    &\Longrightarrow\quad\cancelto{0}{\overrightarrow{t_2}\cdot\overrightarrow{n_2}}=\dot{\lambda}\\
    &\Longrightarrow\quad\lambda(s)=c\\
    &\Longrightarrow\quad\big|\,\overrightarrow{r_2}(s)-\overrightarrow{r_1}(s)\big|=c\,\cancelto{1}{\big|\,\overrightarrow{n_1}(s)\big|}\,.
    \end{align*} So the distance between the points $\overrightarrow{r_1}(s)$ and $\overrightarrow{r_2}(s)$ is constant, for every $s\in I$.

  2. For the tangent vectors of $\overrightarrow{r_1}$ at $\overrightarrow{r_1}(s)$ and $\overrightarrow{r_2}$ at $\overrightarrow{r_2}(s)$, we have \begin{align*}\frac{d}{ds}\big(\,\overrightarrow{t_1}\cdot\overrightarrow{t_2}\big)&=\dot{\overrightarrow{t_1}}\cdot\overrightarrow{t_2}+\overrightarrow{t_1}\cdot\dot{\overrightarrow{t_2}}\\
    &=\kappa_1\,\overrightarrow{n_1}\cdot\overrightarrow{t_2}+\overrightarrow{t_1}\cdot\big(\kappa_2\,\overrightarrow{n_2}\big)\\
    &\stackrel{\overrightarrow{n_2}=\pm\overrightarrow{n_1}}{=\!=\!=\!=\!=}\pm\kappa_1\,\cancelto{0}{\overrightarrow{n_2}\cdot\overrightarrow{t_2}}\pm\kappa_2\,\cancelto{0}{\overrightarrow{t_1}\cdot\overrightarrow{n_1}}=0\quad\Rightarrow\\
    \overrightarrow{t_1}\cdot\overrightarrow{t_2}&=\cos\alpha\,.
    \end{align*} So, the angle between the tangent lines at $\overrightarrow{r_1}(s),\;\overrightarrow{r_2}(s)$ is constant, for every $s\in I$.

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