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What's a fast way to find the normal and binormal vectors?

  1. Jul 15, 2016 #1
    1. The problem statement, all variables and given/known data
    ##r(t)=(t^2)i+(1+\frac{1}{3}t^3)j+(t-\frac{1}{3}t^3)k##
    Find the tangential, normal, and binormal vectors for this TNB frame.

    2. Relevant equations
    ##T(t)=\frac{v(t)}{|v(t)|}##
    ##N(t)=\frac{T`(t)}{|T`(t)|}##
    ##B(t)=T(t)×N(t)##

    3. The attempt at a solution
    The problem isn't that I don't know how to solve this, it's just that I don't know how to solve this fast enough. I had about forty minutes to figure out this problem, to calculate the normal and tangent components of acceleration, and to find the velocity and acceleration of an unrelated polar equation. In the end, I didn't have enough time to finish all of the quiz. Here's what I did...

    ##v(t)=(2t)i+(t^2)j+(1-t^2)k##
    ##|v(t)|=\sqrt{4t^2+t^4+1-2t^2+t^4}=\sqrt{2t^4+2t^2+1}##
    ##T(t)=\frac{2t}{\sqrt{2t^4+2t^2+1}}i+\frac{t^2}{\sqrt{2t^4+2t^2+1}}j+\frac{(1-t^2)}{\sqrt{2t^4+2t^2+1}}k##

    And here's the time-consuming part that took up much of the allotted time.

    ##T_x`(t)=\frac{2(\sqrt{2t^4+2t^2+1})-2t(\frac{1}{2})(8t^3+4t)(2t^4+2t^2+1)^{-\frac{1}{2}}}{2t^4+2t^2+1}##
    ##T_x`(t)=\frac{4t^4+4t^2+2-8t^4-4t^2}{(2t^4+2t^2+1)^{\frac{3}{2}}}=\frac{2-4t^4}{(2t^4+2t^2+1)^{\frac{3}{2}}}##

    Then I had to do this for the y- and z-components, then the magnitude, and it just drained away so much time. I need a faster way to compute N(t), and in turn, B(t), in other words.
     
  2. jcsd
  3. Jul 15, 2016 #2

    Mark44

    Staff: Mentor

    I'm not aware of any faster way than the one you employed. These kinds of problems tend to be time-consuming, especially in calculating the N and B vectors.
     
  4. Jul 15, 2016 #3
    I see. Maybe I should e-mail the teacher about it, that it's too much to be giving in just one quiz. I just had a stroke of hindsight, by the way, for calculating the normal component of the acceleration. I could have used:

    ##a_N=\sqrt{|a|^2-|a_T|^2}=\sqrt{4+8t^2-2t^4-2t^2-1}=\sqrt{-2t^4+6t^2+3}##

    I should have known this, but I suppose that losing points on this question will help me remember it better in the future. Thanks for your input, by the way.
     
    Last edited: Jul 15, 2016
  5. Jul 16, 2016 #4

    micromass

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    Staff Emeritus
    Science Advisor
    Education Advisor
    2016 Award

    You could apply the non-arclength formulas. The advantage of this is that you don't need to differentiate a fraction: https://en.wikipedia.org/wiki/Frenet–Serret_formulas#Other_expressions_of_the_frame

    For example:
    [tex]\mathbf{r}(t) = (t^2, 1 + t^3/3, t - t^3/3)[/tex]
    [tex]\mathbf{r}'(t) = (2t, t^2, 1 - t^2)[/tex]
    [tex]\mathbf{r}''(t) = (2, 2t, -2t)[/tex]

    So

    [tex]\mathbf{r}'(t)\times \mathbf{r}''(t) = (-2t^3 - 2t(1-t^2), -(-4t^2 - 2(1-t^2)), 4t^2 - 2t^2) = (-2t, 2t^2 +2, 2t^2)[/tex]
    So
    [tex]\mathbf{B}(t) = \frac{(-2t, 2t^2 + 2, 2t^2)}{\|(-2t, 2t^2 + 2, 2t^2\|} = \frac{(-2t, 2t^2 + 2, 2t^2)}{\sqrt{8t^4 + 12t^2+4}}[/tex]
     
  6. Jul 16, 2016 #5
    Thank you for that link. I think I'll find those formulas useful next time I must calculate TNB or curvature.
     
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