Results 1 to 4 of 4

Thread: Principal Normal and Bertrand Curves

  1. #1
    Dec 2007

    Principal Normal and Bertrand Curves

    I have a problem here, I need it by monday, I'll appreciate any help.

    I'll define first a BERTRAND CURVE, it is a curve $\displaystyle \alpha$in $\displaystyle R^{3}$ such that there exists another curve $\displaystyle \beta$, distinct from it and a bijection f between $\displaystyle \alpha$ and $\displaystyle \beta$ such that for each corresponding points in f, $\displaystyle \alpha$ and $\displaystyle \beta$ have the same principal normal.

    I need to show this things.
    1. Every plane curve is a Bertrand curve.

    I already know that If $\displaystyle \alpha(t)$ is the plane curve, I can use any of its involute of the form $\displaystyle \alpha(t)+ k*N(t)$. $\displaystyle N(t)$here is the principal normal of $\displaystyle \alpha$. The bijection is also obvious. But my problem is, we haven't discussed involutes in class, so I have to show that their principal normals are the same. I'm using the formulas but I can't prove it.

    2. A curve with a nonzero curvature $\displaystyle \kappa$ and a nonzero torsion $\displaystyle \tau$ is a Bertrand curve if and only if $\displaystyle a \kappa + b \tau = $ where a,b are some constants.
    Follow Math Help Forum on Facebook and Google+

  2. #2

    i also need it by monday

    Hi Diane!
    This is not much but I hope this can help. This is for the first part without using involutions. What I've done so far is the following:
    First represent the curve
    $\displaystyle \alpha(t) = \delta^1(t)u^1 + \delta2(t)v^1, \delta^1(t)u^2 + \delta^2(t)v^2 $
    where $\displaystyle u$ and $\displaystyle v$ span the plane containing the curve and $\displaystyle \delta(t) = (\delta^1(t), \delta^2(t), 0)$.
    It's like a change of basis from {$\displaystyle u$, $\displaystyle v$} to {(1, 0, 0), (0, 1, 0)}
    Then we can take the curve
    $\displaystyle \beta = (\delta^1(t) + at + c)u + (\delta^2(t) + bt +d)v $
    where $\displaystyle a$, $\displaystyle b$, $\displaystyle c$, and $\displaystyle d$ are nonzero constants (this is to ensure that we get a distinct curve). This curve has the same domain as $\displaystyle \alpha$ and observe that if we differentiate this, we get
    $\displaystyle \beta'' = \alpha'' $
    Hence, $\displaystyle N_\alpha = N_\beta $ as desired.
    If you consider the function
    $\displaystyle f((t, \alpha(t)) = (t, \beta(t)) $,
    It can be verified that this is a bijection since $\displaystyle \beta$ is well-defined and since $\displaystyle \beta$ has the same domain as $\displaystyle \alpha$.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Dec 2007
    Waaaahhhh!!! Gelo's here...(nakakahiya naman that I have to consult in his forum)
    Thanks for the reply, no one here seems to take interest in Advanced Geometry. Well, I posted this days ago, I already have an answer different from yours but it's a little complicated. I don' have anything else. Gosh, I'm really having a hard time in Math 146. Bahala na...
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Jan 2009

    bertrand curve

    Let α be a C3 curve with τ ≠ 0.
    Prove that α is a circular helix if and only if α has at least two Bertrand mates.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: Oct 20th 2011, 10:16 PM
  2. Product of non principal ideals is principal.
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: Jun 12th 2011, 01:17 PM
  3. Principal Unit Normal and Curvature
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 17th 2011, 11:39 AM
  4. Replies: 0
    Last Post: Sep 29th 2010, 04:11 PM
  5. Comparison of two normal curves
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: Apr 13th 2010, 08:14 AM

Search Tags

/mathhelpforum @mathhelpforum