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Homework Help: Differential Geometry Question

  1. Feb 10, 2008 #1
    1. The problem statement, all variables and given/known data
    Assume that [tex]\tau(s) \neq 0[/tex] and [tex]k'(x) \neq 0[/tex] for all [tex]s \in I[/tex]. Show that a necessary and sufficient condition for [tex]\alpha(I)[/tex] to lie on a sphere is that [tex]R^2 + (R')^2T^2 = const[/tex] where [tex]R = 1/k[/tex], [tex]T = 1/\tau[/tex], and [tex]R' = \frac{dr}{ds}[/tex]

    2. Relevant equations
    [tex]\alpha(s)[/tex] is a curve in R3 parametrized by arc length
    [tex]k = curvature = |\alpha''|[/tex]
    [tex]\tau = torsion = -\frac{\alpha' \times \alpha'' \cdot \alpha'''}{k^2}[/tex] (note sign; this is opposite of some conventions)

    3. The attempt at a solution
    I've approached this from 2 directions, but I haven't gotten them to meet. First, a necessary and sufficient condition is that [tex]|\alpha - P|[/tex] is constant, where P is the center of the circle. Alternatively, [tex](\alpha - P) \cdot \alpha' = 0[/tex].
    And I've expanded out [tex]R^2 + (R')^2T^2 = const[/tex] to get
    [tex]\frac{(\alpha' \times \alpha'' \cdot \alpha''')^2 + (\alpha'' \cdot \alpha''')^2}{(\alpha'' \cdot \alpha'')(\alpha' \times \alpha'' \cdot \alpha''')^2} = const[/tex]
    Also, I'm going to guess that the const on the right hand side is some function of the radius of the sphere, maybe the square of it (which would be [tex](\alpha - P) \cdot (\alpha - P)[/tex]), because what else is constant in a sphere?

    But I don't know where to go from here. I'm just looking for a hint at an avenue of approach, please nothing specific.
  2. jcsd
  3. Feb 10, 2008 #2
    So if I assume that the "const" in question is the square of the radius, that could make R and (R')T the lengths of respective legs of a right triangle whose hypotenuse is the radius. I know that R is the radius of curvature, so drawing a diagram I'm guessing that the center P is
    [tex]P = Rn - (R')Tb[/tex]
    (where n is the normal vector [tex]\alpha''/|\alpha''|[/tex] and b = [tex] t \times n [/tex] is the binormal vector).
    So then the sphere radius I want to test, [tex](\alpha-P) \cdot (\alpha-P)[/tex], becomes [tex](\alpha - Rn + (R')Tb) \cdot (\alpha - Rn + (R')Tb)[/tex]
    [tex]= \alpha \cdot \alpha - 2 R \alpha \cdot n + 2 R' T \alpha \cdot b + R^2 + (R')^2T^2[/tex]
    Now if I assume that [tex]R^2 + (R')^2T^2 [/tex] is constant I need to show that [tex] \alpha \cdot \alpha - 2 R \alpha \cdot n + 2 R' T \alpha \cdot b [/tex] is constant to show the sphere radius is constant. Am I on the right track?
  4. Feb 10, 2008 #3
    Revised guess for center P:
    [tex]P = \alpha + Rn - (R')Tb[/tex]
    so now the constant radius follows immediately and I simply have to show that P itself is constant.
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