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Solving for a unit speed curve given curvature and torsion (diff. geo)

  1. Jan 21, 2007 #1
    1. The problem statement, all variables and given/known data
    Find the unit speed curve alpha(s) with k(s)=1/(1+s^2) and tau defined as 0.

    2. Relevant equations
    Use the Frenet-Serret equations

    K(s) is the curvature and tau is the torsion

    T= tangent vector field (1st derivative of alpha vector)
    N= Normal vector field (T'/k(s))
    B= Binormal vector field (T x N)

    K(s) is defined as the norm of T'
    3. The attempt at a solution

    Ok I wrote out the matrix for the Frenet-Serret and found the differential equations to solve:

    T' = T/(1+s^2) *which implies => (alpha(s))'' = (alpha(s))'/(1+s^2)
    N'= -N/(1+s^2) *which implies =>N' = -(alpha(s)''/(1+s^2)

    Which so far is really quite easy. But here is the kicker, when I try to solve the diff. eq's and go back and check my solutions, my k(s) value is not correct. I feel rather stupid for asking this, but I seem to have forgotten how to treat a systems of diff. eqs. Any thing to kick start my memory for solving this system would be great.

    Thanks for any help in advanced.
  2. jcsd
  3. Nov 27, 2010 #2
    The answer is a little tricky:

    Since the curve is torsionless, then the curve is planar, so that one can think that it lies completely on the xy-plane.

    So, you have to find a curve a(s)=(x(s),y(s)). The tangent vector then is a'(s)=(x'(s),y'(s)), and we have the condition that x'^2(s)+y'^2(s)=1, since it is unitary. Here we do the trick: Since the ways to chose x and y are infinite, we assume that they are the antiderivative of a cosine and a sine, with the same angle:


    So, the unitary condition on the tangent vector is fulfilled. We need now to find our function f.
    To do this, we differentiate the tangent vector and then we calculate the norm of the new vector. Using the chain rule, we obtain that f'(s)=k(s). So, f(s)=int(k(s))+c, and you can then go back to find the curve you're looking for.
  4. Dec 15, 2011 #3
    sorry for opening this old thread but i have to find a curve with given curvature and zero torsion. I read the answer but i couldn't understand the trick to solve the differential system. Is there anybody who could be more precise? Thanks
  5. Dec 15, 2011 #4

    You don't need to solve a differential sistem.

    The fact that your curve has zero torsion means that it is planar, so that you can study it as a curve on the xy-plane. This means that you can denote it as a(s)=(x(s),y(s)). But your curve has unit speed, which means that ||a'(s)||^2=x'(s)^2+y'(s)^2=1. A natural assumption then is to set x'(s)=cos(f(s)) and y'(s)=sin(f(s)). Integrating these expressions we obtain

    y(s)=int(sin(f(s)))+b. (1)

    Now, if we differentiate the tangent vector a'(s), we obtain a''(s)=f'(s)(-sin(f(s)),cos(f(s))) using the chain rule. We know that the curvature k(s) is defined as


    where n(s) is the normal vector. This yields f'(s)=k(s). Integrating again, we find that f(s)=int(k(s)). Once we obtain f(s), we substitute in (1) and hence obtain a(s).
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