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Arc length or a smooth curve

  1. May 3, 2007 #1
    Guys, I need your kind assistance. I am studying arcs length. Suppose a vectorial function with domain [a, b] (interval in R) and range in RxR. This range is a curve in the RxR plane.

    Take a partition P of [a, b]: a= t0, t1, t2,...., tn = b.

    We have a straight line which goes from F(t0) to F(t1), another straight line that goes from F(t1) to F(t2), etcetera. Thus we build a "polygonal". By definition, the supremum of set A = { lengths of the polygonals corresponding to any partition} is the arc length.

    But all books (in their drawings) assume that, for example, F(t2) can not be located in the curve that was "cut off" by the straight line which goes from F(t0) to F(t1). In other words: they assume that F(t2) can not go backwards and settle between F(t0) and F(t1).

    I simply do not understand where is their basis for such assumption.

    In two books they say that both component functions of F (lets call them f and g) have continuous derivatives and that there is not "t" in [a,b] such that (f'(t), g'(t)) = (0,0). I know that this implies that, for example, if we take F'(t1), at least one of the original component functions (f or g) is monotone in a interval that contains t1, but I fail to see how this conect with my question of the previous paragraph.

    Can you help me? Thanks.
     
    Last edited: May 3, 2007
  2. jcsd
  3. May 4, 2007 #2
    Guys, it is too obvious or too boring for droping some words to explain this?
     
  4. May 4, 2007 #3

    HallsofIvy

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    We may be having difficulty understanding what you are asking. The points used to approximate the curve by a "broken line" are chosen so that the line does not "go back on itself"! We don't have assume such a thing. We are free to choose the points and order them as we wish.
     
  5. May 4, 2007 #4

    Office_Shredder

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    No..... it's much easier

    Suppose the curve crosses back on itself

    Then F(t0)=F(t2) (say)

    But then F'(t0) = F'(t2).

    This sounds reasonable. Looking at the curve, however, it's clearly not so

    This is a contradiction, and hence the curve can't cross over itself

    EDIT: This isn't properly right now that I think about it, but I remember an argument similiar to it working
     
    Last edited: May 4, 2007
  6. May 4, 2007 #5
    In the books they begin as this: "Take a partition of [a, b], say t0, t1,..., tn, and make a polygonal line drawing straight lines from F(t0) to F(t1), from F(t1) to F(t2), etcetera". Seemingly the points of the partition are chose at random. I have never read that we chose (say) t2 so as to avoid that F(t2) may fall in the curve's section that is between F(t0) and F(t1).


    Nothing to coment about this paragraph(is the last one of my first post)?:

    In two books they say that both component functions of F (lets call them f and g) have continuous derivatives and that there is not "t" in [a,b] such that (f'(t), g'(t)) = (0,0). I know that this implies that, for example, if we take F'(t1), at least one of the original component functions (f or g) is monotone in a interval that contains t1, but I fail to see how this conect with my question of the previous paragraph.



    Shedder, I am not saying that the curve can not cut itself. It can. What I am asking is how can we know that the vertices of the polygonal aproximative line are always forward in the curve (regarding the previous vertex).
     
    Last edited: May 4, 2007
  7. May 5, 2007 #6

    Zen

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    Is this what you mean?
    [​IMG]
    or this:
    [​IMG]
    In the first case, it's clearly not a problem, since this will not be the partition resulting in the longest length.

    In the second case I think this will conflict with the continiuty of the function. (remember that (in my book at least) t0<t1<t2<t3<...<tn)

    [ img ] tag not supported?!
     
    Last edited: May 5, 2007
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