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Relation between unit tangent/normal vectors, curvature, and Lin. Alg.

  1. Jun 7, 2014 #1
    Hey there,
    This isn't a homework question, it's for deeper understanding. So I'm learning about unit normal/tangent vectors and the curvature of a curve. I have a few questions/points.

    1) So my book states that we can express acceleration as a linear combination of the acceleration in the same direction as T (the tangent vector of the curve), and the acceleration in the normal/perpendicular direction (direction of N, the normal vector).
    --- One thing I'm wondering is: Say we are dealing with a three-dimensional curve, r(t)..
    --- The tangent vector to the curve is perpendicular to the position vector; wouldn't the unit normal
    --- vector just be parallel to the position vector? Since the unit normal vector is perpendicular to the tangent vector.
    --- In that case, could we not just take the derivatives of the position vector and be able to construct the unit tangent and unit normal vectors from that by dividing by their magnitude? Or is that what my book is saying, just in a very convoluted way?

    2) I'm taking a Linear Algebra course right now, and I feel as though I can translate what I'm learning in Calculus to what I've learned in Lin. Alg., but I'm not quite there right now.
    --- An alternative formula for figuring out the curvature is given in terms of the magnitude of the cross product of v, the velocity vector, and acceleration, the derivative of the velocity vector, divided by the magnitude of the velocity cubed.
    --- That sounds like we want a vector perpendicular to v and a divided by some scalar. Is there another way to find the curvature, then?
    --- Like taking a vector, removing its projections onto v and a? That would get a vector orthogonal to both, I could then take its magnitude and divide it by something?

    I'm a bit lost in trying to integrate my knowledge of Calculus with my knowledge of Linear Algebra. If I take the second derivative of the position function, r(t), aren't the components of it orthogonal? If not, can I just make an orthogonal basis, and then work from there? My head is spinning a bit. :)

    Thanks,
    ModestyKing
     
  2. jcsd
  3. Jun 7, 2014 #2

    jedishrfu

    Staff: Mentor

    The frenet-serret formulas may help with understanding the first part:

    http://en.wikipedia.org/wiki/Frenet–Serret_formulas

    Also just because the tangent vector is perpendicular to the position doesn't mean the normal vector is parallel to the position vector. If it were a 2D curve that would be true but in 3D there is additional freedom to not be parallel.

    It might help to study the curve as a circle vs the curve as a helix. The wikipedia article shows an animated helix.
     
  4. Jun 8, 2014 #3
    Hi jedishrfu,
    I think I get what you mean. In R3, there are three basis vectors, and the tangent vector could be just one of them - a vector orthogonal to it could be one of two other basis vectors. Given that the tangent vector and position vector are orthogonal, then, could I not form the normal vector by taking any old vector and removing its projections onto the tangent and position vectors?

    Thank you so much for the reference! I doubt I would have found that on my own. This just impacted me, as it's along the lines of what I was looking for:
    "The Frenet–Serret formulas were generalized to higher-dimensional Euclidean spaces by Camille Jordan in 1874.

    Suppose that r(s) is a smooth curve in Rn, parametrized by arc length, and that the first n derivatives of r are linearly independent.[2] The vectors in the Frenet–Serret frame are an orthonormal basis constructed by applying the Gram-Schmidt process to the vectors (r′(s), r′′(s), ..., r(n)(s))."

    All of which I know how to do and/or verify. Another interesting one that is exactly along the lines of what I was looking for is:

    "An alternative way to arrive at the same expressions is to take the first three derivatives of the curve r′(t), r′′(t), r′′′(t), and to apply the Gram-Schmidt process. The resulting ordered orthonormal basis is precisely the TNB frame. This procedure also generalizes to produce Frenet frames in higher dimensions."

    This makes me very happy. Thanks so much!!
     
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