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Compute the curvature

  1. Sep 10, 2014 #1
    1. Compute the curvature.

    [itex]α(t) = (cos^3t, sin^3t)[/itex]

    This is not a unit-speed curve. I want to use [itex]κ(t) = \frac{||T'(t)||}{||σ'(t)||}[/itex]

    When I find [itex]α'(t)[/itex] and then its norm, I run into an impasse. Am I supposed to use a trig identity?
     
  2. jcsd
  3. Sep 10, 2014 #2

    ehild

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    What was that impasse? Do you know how to get the norm of a vector?

    ehild
     
  4. Sep 10, 2014 #3

    haruspex

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    Please post your working, or we cannot tell where you are stuck.
     
  5. Sep 10, 2014 #4
    Sorry. It was late last night. Maybe I'll get a different answer today. Heh.

    [itex]α(t) = (cos^3t, sin^3t)[/itex]

    [itex]α'(t) = (-3cos^2(t) sin(t), 3sin^2(t)cost(t)[/itex]

    [itex]||α'(t)|| = \sqrt{(-3cos^2(t) sin(t))^2 + (3sin^2(t)cos(t))^2} [/itex]

    [itex]||α'(t)|| = \sqrt{9cos^4(t) sin^2(t) + 9sin^4(t)cos^2(t)} [/itex]
     
    Last edited: Sep 10, 2014
  6. Sep 10, 2014 #5

    haruspex

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    What common factor can you take outside the square root?
     
  7. Sep 10, 2014 #6
    :rolleyes:

    I have no excuse. Thanks. I see it. Let me try to work the rest of the problem out.

    [itex]κ(t) = \frac{1}{3cos(t)sin(t)}[/itex]
     
    Last edited: Sep 10, 2014
  8. Sep 10, 2014 #7
    3. Show that the curve is planar (i.e. contained in a plane). Hint: Show that its torsion is identically zero. Of course, I need to set the torsion [itex]τ[/itex] = 0.

    [itex]α(t) = (\frac{1+t^2}{t}, t+1, \frac{1-t}{t})[/itex]

    [itex]α'(t) = (1-t^{-2},1, -t^{-2})[/itex]

    [itex]||α'(t)|| = \sqrt{2(t^{-4}-t^{-2}+1})[/itex]
     
  9. Sep 10, 2014 #8

    haruspex

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    What's your equation for torsion? And how is [itex]||α'(t)||[/itex] useful?
     
  10. Sep 10, 2014 #9
    I see one entirely in terms of alpha up to its third derivative, a cross product, norm, etc.

    The other one involves B'(t), N(t), and alpha.

    I found alpha-prime because it's used to find T, T', which is then used for N, and then B, and so forth.
     
  11. Sep 11, 2014 #10

    haruspex

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    That's the one. But remember you are only interested in whether it is zero, which simplifies matters a little.
    Yes, you need α', but I asked why its norm was interesting here.
     
  12. Sep 11, 2014 #11
    Oh, indeed. I suppose that I can simply show that the numerator [itex]α^{'''}(t)\cdot (α^{'}(t) \times α^{''}(t))[/itex] is zero.
     
  13. Sep 11, 2014 #12

    haruspex

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    Yes.
     
  14. Sep 12, 2014 #13
    Thanks. I have one more problem, and the HW is due today at 5 PM.

    [itex]Let σ be a unit-speed plane curve. Define a new curve (called parallel curve of σ) β by

    β(t) = σ(t) + 2N(t),

    where N(t) is the principal normal at α(t). Show that, if 2κ(t) < 1 for all t, where κ(t) is the curvature of σ at σ(t), then β(t) is a regular curve and that its curvature is κ(t)/(1-2κ(t)).

    I know that σ being a unit-speed plane curve implies that ||σ'(t)|| = 1 and that τ(t) = 0. Also, if β(t) is a regular curve, then ||β(t)|| = 1.[/itex]
     
    Last edited: Sep 12, 2014
  15. Sep 12, 2014 #14

    haruspex

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    Sorry, only just saw this.
    No, it is only required that it is differentiable and the derivative never changes sign.
    Try to get an expression for the derivative in terms of that of the original curve.
     
  16. Sep 12, 2014 #15
    [itex]β(t) = α(t) + 2N(t)[/itex]
    [itex]β'(t) = α'(t) + 2N'(t)[/itex]
    [itex]β'(t) = α'(t) + 2(-κT + τB)[/itex]
    [itex]β'(t) = α'(t) + 2(-κT)[/itex]
    [itex]β'(t) = α'(t) - 2κα'(t)[/itex]
    [itex]β'(t) = α'(t)(1 - 2κ)[/itex]

    [itex]||α'(t)|| = 1 \implies α'(t) ≠ 0[/itex]
    If [itex]κ(t) < \frac{1}{2},\ β'(t) > 0, \implies β'(t) \ is \ regular.[/itex]
     
    Last edited: Sep 12, 2014
  17. Sep 13, 2014 #16

    haruspex

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    Looks right.
     
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