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Unit tangent, unit normal, unit binormal, curvature

  1. Feb 23, 2012 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Question:
    "Find the unit tangent, normal and binormal vectors T, N, B, and the curvature of the curve
    x = 4t, y = -3t^2, z = -4t^3 at t = 1."

    Answer:
    T = 0.285714285714286 i - 0.428571428571429 j - 0.857142857142857 k
    N = -0.75644794981871 i + 0.448265451744421 - 0.476282042478447 k
    B = 0.588348405414552 i + 0.784464540552736 j - 0.196116135138184
    ϰ = 0.0445978383113072


    2. Relevant equations
    N = dT/dt / |dT/dt|


    3. The attempt at a solution
    I tried to use the equation from the "Relevant equations" part above. I know there are alternative ways but I want to figure out what I am doing wrong for this method.

    I (successfully) get the unit tangent vector to be:
    T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4)
    T = 2/7 i - 3/7 * t j - 6/7 * t^2 k

    I (unsuccessfully) get the unit normal vector to be:
    N = (3/7 i - 12/7*t k)/sqrt( (3/7)^2 + (12/7)^2 * t^2)

    What am I doing wrong?

    Any input would be greatly appreciated!
    Thanks in advance!
     
  2. jcsd
  3. Feb 23, 2012 #2

    Mark44

    Staff: Mentor

    The above is really T(t), the tangent vector for an arbitrary value of the parameter t.

    The problem asks for the unit tangent vector and unit normal vector at t = 1. IOW, it's looking for T(1), N(1), and B(1).
    See above. I didn't check your work, so if you have errors, I wasn't looking for them. Again, you want T(1), N(1), and B(1) - the unit vectors at a particular value of t.
     
  4. Feb 23, 2012 #3

    Dick

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    Science Advisor
    Homework Helper

    When you went from T = (4 i - 6t j - 12t^2 k)/sqrt(4^2 + 6^2 * t^2 + 12^2 * t^4) to T = 2/7 i - 3/7 * t j - 6/7 * t^2 k you put t=1 in the denominator. That's means you can't use the second expression to find dT/dt. You eliminated some of the t dependence. You need to use the first and use the quotient rule. BTW this is quite a messy problem.
     
  5. Feb 23, 2012 #4

    s3a

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    Sorry, I was doing a lot of stuff in my head so the t = 1 went on and off.

    I get |dT/dt| to be sqrt(144t^4 + 36t^2 + 16) and it doesn't seem that I can get rid of the square root.

    So, I am assuming it's hard to do it this way and that I shouldn't do it this way assuming it is possible. Is it possible though? (I am not asking for it computed that way but I would just like to know if it is possible to get past that step without using a computer or something of the sort.)
     
  6. Feb 23, 2012 #5

    Dick

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    Homework Helper

    I get something a LOT messier for |dT/dt|. I'm using a computer for this one and would feel sorry for someone who wasn't. It's pretty bad.
     
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