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Math Help - Find the Curvature

  1. #1
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    Find the Curvature

    Find the curvature of the space curve.
    r(t) = -4i + (4+ 2t)j + (t^2 +2)k

    * I was just wondering if my solution looks good so far????

    r ' (t)= 0i +2j + 2tk

    ll r ' (t)ll = sqrt(4 +4t^2) = 2 + 2t^2

    = T(t) = r ' (t) / ll r ' (t) ll = 2j + 2tk / 2j + 2t^2K

    * After this I use the quationat rule??? 99% sure???
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  2. #2
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by sderosa518 View Post
    Find the curvature of the space curve.
    r(t) = -4i + (4+ 2t)j + (t^2 +2)k

    * I was just wondering if my solution looks good so far????

    r ' (t)= 0i +2j + 2tk

    ll r ' (t)ll = sqrt(4 +4t^2) = 2 + 2t^2

    = T(t) = r ' (t) / ll r ' (t) ll = 2j + 2tk / 2j + 2t^2K

    * After this I use the quationat rule??? 99% sure???

    \sqrt{4+4t^2}=\sqrt{4(1+t^2)}=2\sqrt{1+t^2}
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  3. #3
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    Also, you do NOT divide a vector by a vector. The denominator is the number 2\sqrt{1+ t^2}, not a vector.
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  4. #4
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    here is my solution to T ' (t).

    T(t) = r '(t)/ llr '(t)ll = 2j + 2tk / 2(1+t^2)^ (1/2)

    T '(t) = 2(1+t^2)^ (1/2) (2K) - (-1/2) (1+t^2)^(-1/2) (2t) (2j +2tk) / (2(1+t^2)
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  5. #5
    MHF Contributor alexmahone's Avatar
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    Quote Originally Posted by sderosa518 View Post
    here is my solution to T ' (t).

    T(t) = r '(t)/ llr '(t)ll = 2j + 2tk / 2(1+t^2)^ (1/2)

    T '(t) = 2(1+t^2)^ (1/2) (2K) - (-1/2) (1+t^2)^(-1/2) (2t) (2j +2tk) / (2(1+t^2)
    Please learn to use LaTeX.

    In any case, your computation of T'(t) seems to be incorrect. Remember that the formula for curvature is \displaystyle \kappa=\frac{1}{v}|\frac{dT}{dt}|.
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  6. #6
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    [Math]
    T '(t) = 2(1+t^2)^ (1/2) (2K) - (-1/2) (1+t^2)^(-1/2) (2t) (2j +2tk) / (2(1+t^2) [Math]
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