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Differential geometry : Tangent vector & reparameterization

  1. Dec 13, 2015 #1
    1. The problem statement, all variables and given/known data
    Problem statement uploaded as image.

    2. Relevant equations
    Arc-length function
    eq0014M.gif
    3. The attempt at a solution
    Tangent vector:
    r=-sinh(t), cosh(t), 3

    Now, I just need to reparameterize it using arclength and verify my work is unit-speed. Will someone give me a hint? Should I use the arc-length function to accomplish this.
     

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  3. Dec 13, 2015 #2

    Ray Vickson

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    What is preventing you from trying it for yourself?
     
  4. Dec 13, 2015 #3
    It's that I have no means of checking the solution, so before I invest in it, I would like to know if my method is correct (assuming that I integrate correctly).
     
  5. Dec 13, 2015 #4

    Mark44

    Staff: Mentor

    This isn't the tangent vector.
    In future posts, please show more of your work. What you have here just barely qualifies as a problem attempt.
     
  6. Dec 13, 2015 #5
    Tangent vector
    upload_2015-12-13_18-41-41.png

    Now, compute the norm of the tangent vector:
    upload_2015-12-13_18-55-58.png
    Using this, make the following substitution
    upload_2015-12-13_19-1-35.png
     
  7. Dec 13, 2015 #6

    Mark44

    Staff: Mentor

  8. Dec 13, 2015 #7
    How do I compute the arclength, without knowing the range? For example [0 <= t <= 2pi]

    Another shot at the arc length of the tangent vector

    [itex]\sqrt{(9+9*sinh(t)^2+16*cosh(t)^2)}dt =^*
    25 cosh^2(t) =
    25 sinh^2(t)+25
    [/itex]

    * Using a trigonometric identity

    PS: csgn is code used specifically by maple. It' not necessarily a mathematical function
     
    Last edited: Dec 13, 2015
  9. Dec 13, 2015 #8

    Mark44

    Staff: Mentor

    The arc length function in your relevant equations gives the arc length in terms of a parameter t.
    The last expression above is not helpful, but the one before it is helpful. What happened to the square root?
    Do you know what it means, though? I've never seen it, but I don't use Maple.
     
  10. Dec 13, 2015 #9
    Right, the square root should've been preserved, in the above equation. Here it is, in all of its glory:

    [itex]\sqrt{25cosh^2(t)}[/itex]
    So, is this equation the reparameterization of the tangent vector?

    csgn(x) is the sign function of real AND complex numbers; where csgn = complex signum.
     
  11. Dec 13, 2015 #10

    Ray Vickson

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  12. Dec 13, 2015 #11
    upload_2015-12-13_20-51-51.png
    Plot of the 25cosh^2(t) function; the norm of the tangent vector
     
    Last edited: Dec 13, 2015
  13. Dec 13, 2015 #12

    Ray Vickson

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    This is not relevant. The question is what cosh(t) looks like, not its square. You should not even need to do an actual plot; just picture it in your mind.
     
  14. Dec 13, 2015 #13
    upload_2015-12-13_21-9-33.png

    Plot of [itex]5 \sqrt{cosh(t)}[/itex]

    I can picture it in my mind. What am I supposed to "see"?
     
    Last edited: Dec 13, 2015
  15. Dec 13, 2015 #14
    I get this for the parameterization by arclength of the tangent vector
    [itex]int(5*sqrt(cosh(t)^2), t = 0 .. 1) = 5 sinh(1)[/itex]
     
  16. Dec 13, 2015 #15

    Mark44

    Staff: Mentor

    This is not a parameterization -- it's a number.

    As I said before...
    IOW, ##\int_0^t 5 \sqrt{\cosh^2(w)} dw##
     
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