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Homework Help: Curvature question

  1. Feb 25, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the unit tangent, normal and binormal vectors T,N,B , and the curvature of the curve
    x=−4t y=−t2 z=−2t3 at t=1.

    2. Relevant equations

    3. The attempt at a solution

    I found T=(-4/sqrt(56),-2/sqrt(56),-6/sqrt(56)) which is correct. But I keep getting N wrong? PLease help me, thank you!
  2. jcsd
  3. Feb 25, 2010 #2


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    Looks good as far as it goes. Hard to say what you are doing wrong unless you show us what you are doing to get N.
  4. Feb 25, 2010 #3
    N = T'/|T'|
    so the first term should be 0; but the webwork keeps denying that answer.
  5. Feb 26, 2010 #4
    N = (0,-2/sqrt(148),-12/sqrt(148)) and that gives a wrong answer

    Attached Files:

    Last edited: Feb 26, 2010
  6. Feb 26, 2010 #5


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    Not 100% sure of this (im at work), but you get the train of thought:
    Try the answers in bold
    r(t) = (-4t)i + (-t^2)j + (-2*t^3)k

    v(t) = r'(t) = (-4)i -(2*t)j - (6*t^2)k
    ||v(t)|| = sqrt(2*(4+t^2+9*t^4))

    T(t) = v(t)/||v(t)|| =((-4)i + (-2*t)j + (-6t^2))/sqrt(2*(4+t^2+9*t^4))

    Evaluate at t=1

    T(1) = ((-4)i + (-2)j + (-6)k)/(6503/869) = 869*(-4i-2j-6k)/6503

    N(t) = T'(t)/||T'(t)|| = diff(T,t)/length(diff(T,t))

    N(t) = 0i - (2/3) j - (4*t)k
    evaluate at t=1

    N(1) = 0i - 2/3j -4k

    B=T x N = [ 4*t^2, -16*t, 8/3]
    at t=1

    B(1) = 4i - 16j + 8/3k
  7. Feb 26, 2010 #6
    I believe your answers are incorrect, as I have all T,B, curvature correct in the attached image above. Your T, however, has the same i value as mine, which is 0 and incorrect.
  8. Feb 26, 2010 #7


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    Maple rules.
  9. Feb 26, 2010 #8
    Okay Maple rules do give correct answers. Thanks. But I don't understand why they would give different answers to the "conventional" methods. I'm confused about when to use Maple rules and when not?
  10. Feb 26, 2010 #9


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    I made a mistake earlier while using the conventional method. The correct way is:

    N(t) = T'(t)/||T'(t)|| = (r'(t) x (r''(t) x r'(t)))/(||r'(t)||*||r''(t) x r'(t)||) = <304/(8*sqrt(157)*sqrt(14)), 40/(8*sqrt(157)*sqrt(14)), -216/(8*sqrt(157)*sqrt(14))>

    DUH. It all checks out.
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