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Find the curvature of x = e^(t)

  1. Mar 17, 2009 #1
    1. The problem statement, all variables and given/known data

    Find the curvature of x = e^(t) y = e^(-t) z = t t = 0

    2. Relevant equations

    I've used the equation of

    k(t) = |r'(t) x r''(t) |/ |r'(t)|^3

    3. The attempt at a solution

    k(t) = |r'(t) x r''(t) |/ |r'(t)|^3


    = |e^t i + -e^(-t)j + 1k| x |e^t i + e^(-t)j + 0k| / |e^t i + -e^(-t)j + 1k|^3

    = |-e^(-t)i + e^(t)j +2k| / |e^t i + -e^(-t)j + 1k|^3

    Using t = 0


    = |-e^(0)i + e^(0)j +2k| / |e^0 i + -e^(0)j + 1k|^3


    = 2/1

    = 2

    Is this right ?

    regards
    Brendan
     
  2. jcsd
  3. Mar 17, 2009 #2

    cristo

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    Re: curvature

    I agree with everything to your penultimate line. The modulus of (-1,1,2) is not 2.
     
  4. Mar 17, 2009 #3
    Re: curvature

    How about now?

    = |-e^(0)i + e^(0)j +2k| / |e^0 i + -e^(0)j + 1k|^3


    = sqrt(2)/3


    Brendan
     
  5. Mar 17, 2009 #4

    cristo

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    Re: curvature

    Looks good to me.
     
  6. Mar 17, 2009 #5
    Re: curvature

    Thanks mate!

    Brendan
     
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