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So what exactly is an osculating plane? r'(t) X r (t)? Am I right?

  1. Sep 27, 2010 #1
    So what exactly is an osculating plane? r'(t) X r"(t)? Am I right?

    From what I know it is a plane that kisses a point on the curve at point t of curve f(t) and it is a plane normal to r'(t) X r"(t)? Am I right?

    So to look for this osculating plane I simply have to find the cross product of r'(t) and r"(t)?

    What does it look like on a xyz graph?

    EDIT: By the way per definition of the equation of a plane: a(x-x0) + b(y-y0 + c(z-z0) I need to find

    a) a point on the plane (x0,y0,z0)
    b) a vector normal to the plane <a,b,c>

    Hence if I am looking for the osculating plane I would need:

    a) a point on the osculating plane (x0,y0,z0)
    b) B(t) which happens to be the vector normal to my osculating plane B(t) = T(t) X N(t), from this I find my desired <a,b,c>

    I put it all together and I get my osculating plane of form: a(x-x0) + b(y-y0 + c(z-z0). (Right or wrong?)
    So what exactly is the cross product r'(t) X r"(t) for? Cutting down time so I dont have to go through the tedious process of finding T(t) and N(t)?
    Last edited: Sep 27, 2010
  2. jcsd
  3. Sep 28, 2010 #2
    Re: So what exactly is an osculating plane? r'(t) X r"(t)? Am I right?

    You are right. Since the osculating plane passes through the curve at some point, you can take x0, y0, z0 as a point on the curve.
    You don't have to find T and N (by the way, what do you mean by N?) to find a, b and c, but just r' and r''. What's so tedious on differentiating two times the curve?
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