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I Integrating a curve of position vectors

  1. Mar 29, 2017 #1
    I'm looking at different ways to express the derivative a curve, like circular and tangent/normal components.
    Is there no such way that lets you express a vector integral in terms of information from the vector you want to integrate?
     
    Last edited: Mar 29, 2017
  2. jcsd
  3. Mar 29, 2017 #2

    jedishrfu

    Staff: Mentor

    Can you provide some explicit examples for your curve and how you use circular and tangent/normal components?

    My understanding is that you are referring to the Frenet Serret formula of a space curve is that right?

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

    Can you provide a context for the vector integral you're thinking about?
     
  4. Mar 29, 2017 #3
    No specific example at the moment, but I wanted some opinions if it would definitely be impossible or not.
    Yeah, the Frenet Serret frame has come up in my searches, though I haven't investigated it yet.

    I was able to create an expression for the derivative of a vector where I'm adding an angle value to the vector angle
    and dividing a magnitude value with the vector length. Both values depend on x(t)*y'(t) - y(t)*x'(t) and x(t)*x'(t) + y(t)*y'(t).
    Then I did the same for the integral of a vector, this time subtracting the same angle value and dividing the same magnitude
    value by the length of the vector to be integrated. The values depended on the same quantities, but this time I had to integrate
    the rectangular components: X(t)*y(t) - Y(t)*x(t) and X(t)*x(t) + Y(t)*y(t).

    Expressions of derivatives depend on information of what is derivated, but is it impossible to write expressions of integrals depending on information of what is integrated (and without a sum/integral operator)?
    I guess it boils down to that derivating is "lossy", so we can make chain, product rules for it, but not how to go the other way (in all cases)?
     
    Last edited: Mar 29, 2017
  5. Mar 29, 2017 #4
    This is what I got:

    derivative of (x(t), y(t)) = l(t)/sqrt(x(t)^2 + y(t)^2) * (cos( atan2(y(t),x(t))+a(t) ), sin( atan2(y(t),x(t))+a(t) ))
    where
    l(t) = sqrt( (x(t)*y'(t) - y(t)*x'(t))^2 + (x(t)*x'(t) + y(t)*y'(t))^2 )
    a(t) = atan2( x(t)*y'(t) - y(t)*x'(t), x(t)*x'(t) + y(t)*y'(t) )

    integral of (x(t), y(t)) = l(t)/sqrt(x(t)^2 + y(t)^2) * (cos( atan2(y(t),x(t))-a(t) ), sin( atan2(y(t),x(t))-a(t) ))
    where
    l(t) = sqrt( (X(t)*y(t) - Y(t)*x(t))^2 + (X(t)*x(t) + Y(t)*y(t))^2 )
    a(t) = atan2( X(t)*y(t) - Y(t)*x(t), X(t)*x(t) + Y(t)*y(t) )
     
  6. Mar 29, 2017 #5

    jedishrfu

    Staff: Mentor

    @fresh_42 may have some ideas here as he's published some excellent insight articles on the various forms of derivatives.
     
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