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Homework Help: Differential Geometry: the Osculating Circle

  1. Sep 28, 2010 #1
    1. The problem statement, all variables and given/known data

    let f: J --> R^2 be a unit speed curve curve and define it's tangentially equidistant campanion by g(u) = f(t) + r*f'(t) for a fixed r>0. Show that the centre of the osculating circle of f at some u in J is the intersection of the line normal to f'(u) through f(u) and the normal line to g'(u) through g(u).

    2. Relevant equations

    T(u) = f'(u) as f is unit speed (ie. parametrized by arclength)

    N(u) = vector normal to f'(u) = R(T(u)) where R is a rotation in the plane by pi/2 in the counterclockwise direction.

    The centre of the osculating circle is f(u) + [1/k]*(N(u)), where k is the signed curvature of f.

    k := f''(u) dot product with R(f'(u)) for a unit speed curve.

    The Frenet-Serret equations also tell us that: N'(u) = -k*T(u)

    3. The attempt at a solution

    My attempts have thusfar consisted of applying the defition to find the normal line to f'(u) and g'(u) and setting them equal to eachother then trying to reduce with seriously no luck in proving the claim no matter how I try to manipulate them. I've tried realizing that:

    g'(u) = f'(u) + r*(f''(u)) ==> R(g'(u)) = vector pointing in direction normal to g'(u) = R(f'(u)) + r*R(f''(u)) = N(u) + r*(R(f'(u))' = N(u) + r*N'(u) = N(u) + r*(-k*f'(u))

    The last few steps can be justified by the fact that R is linear ==> its derivative is itself so that we can manipulate the term R(f''(u)) as shown (I hope) which at least brings the curvature into the fold and intuitevly makes sense in terms of how it is expressed.

    No matter what I've tried to do (even going so far as to seemingly uselessly express the desired point in terms of the orthonormal basis (T, N)) has got me no where. Any ideas I'm missing that could get me on the right track?
    Last edited: Sep 28, 2010
  2. jcsd
  3. Sep 29, 2010 #2
    First of all thanks to everyone who took time to read and consider this problem in order to help me out, I appreciate it.

    So I have a method to prove the claim. I'll sketch the ideas that lead to the proof quickly right now for this who are interested and post something more detailed when I have time for those who need a little more help understanding what to do.

    You start as I did above by finding vectors in the direction of the lines desired. It is key to discover that r*R(f'') = -r*k*T, which arises when you are finding a vector perpendicular to g', and understand why this is true.

    Then you simply express the line through f(s) perpendicular to f'(s) in the usual way: f(s) + c*N(s), where c is an arbitrary, real number parameter. Similarily, the line through g(s) which is perpendicular to g'(s) is g(s) + d*(N(s) - r*k*T).

    Then you express a point on each of these lines in term of the orthonormal basis, {T, N}, set the T component of line 1 equal to line 2 and the N component of line 1 equal to line 2. Simply solve the N component first, plug that information into the T component equality, send what you get into the equations you defined for the lines and the claim follows.
    Last edited: Sep 29, 2010
  4. Sep 29, 2010 #3
    Hi, thank you for taking the time to post this. I do have a question though - how did you go about expressing points on those vectors in terms of the orthonormal basis? This is the only part I'm stuck on, although I realize it's probably quite silly.

  5. Sep 29, 2010 #4
    Hi guys.
    This is a nice problem. But I think to prove it it's not so simple.
    I've already dedicated to it some time, but Ihave not found any noticeable results.
    With some time in my hands I will work on it.

    I see in the equations you have list, there is not the formula to compute the curvature by the derivatives of the curve. I guess you are aware of this curve because you cite other advanced formulas likeFrenet-Serret, so I think you already know it.
    Why you don't use it ? Maybe you've seen it not relevant.
    In any case I will you it to get some result. I already tried but I got nowhere.

    See you soon.
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