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Proof: Curvature Zero -> Motion along a line

  1. Sep 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Proof that, if a particle moves along a space curve with curvature 0, then its motion is a along a line.


    2. Relevant equations
    [tex]K=\frac{||r'(t)\times r''(t)||}{(||r'(t)||)^3}[/tex]
    (curvature of a space curve)


    3. The attempt at a solution
    Assume the curve is smooth, so r'(t) cannot be the zero vector. The numerator must be 0. I evaluate the cross product (set it to 0), and get the following equations.

    [tex]g'(t)h''(t) = h'(t)g''(t)[/tex]
    [tex]f'(t)h''(t) = h'(t)f''(t)[/tex]
    [tex]f'(t)g''(t) = g'(t)f''(t)[/tex]

    Here I don't know what to do to get to the equation of a line.

    Thank you in advance.
     
    Last edited: Sep 29, 2008
  2. jcsd
  3. Sep 29, 2008 #2

    Dick

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    The easiest way to see this to take the curve parametrization so that |r'(t)|=1. If you differentiate r'(t).r'(t) you see that r'(t).r''(t)=0 (so r' and r'' are perpendicular). Curvature=0 tells you also that r'(t)xr''(t)=0 (so r' and r'' are parallel). What does that tell you about r''(t)?
     
  4. Sep 29, 2008 #3
    Thank you very much for your answer.

    That would mean that r''(t) is the 0-vector. Which means that there is no acceleration, no change of direction, motion should be straight.

    However, my professor told me that this is not formal enough. I tried to arrive at a similar conclusion by using ||a x b|| = ||a|| ||b|| sin(theta).

    Is there no way to formally arrive "back" at the parameterized equation of a line by using the result I already have?
     
  5. Sep 29, 2008 #4

    Dick

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    If r''(t) is zero then r'(t) is a constant. So r'(t)=r'(0) for all t. Integrating r'(t) to get r(t) then gives you r(t)=r(0)+r'(0)*t, right? That isn't formal enough? It looks messy to try to argue starting with what you have to the conclusion r(t) is a line.
     
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