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Lenght in a circle

  1. Jan 10, 2005 #1

    quasar987

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    Hi,

    I'd like to know the name for the lenght in a circle that I will describe:

    Consider 2 arbitrary points lying on the perimeter of a circle. I'm talking about the shortest distance distance between these those points. (i.e. the straight line joining them.)

    Thx
     
  2. jcsd
  3. Jan 10, 2005 #2

    jamesrc

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    It's called a chord.
     
  4. Jan 10, 2005 #3

    HallsofIvy

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    Yep, a chord.
     
  5. Jan 10, 2005 #4

    quasar987

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    Ok, I have a confusion involving a cord. In physics, we often make the transformation from an infinitesimal cord to an infinitesimal arc: d(cord) = d(arc). And I think it's used in some math proofs too.

    For exemple, imagine a position vector [itex]\vec{r}(t)[/itex] of fixed norm, rotating around the z axis (i.e. [itex]\vec{\omega} = \omega \hat{z}[/itex]). Let's say the radius of the circle described by the motion of its tip/head/arrow is R. After a time [itex]\Delta t[/itex], it has rotated an angle [itex]\omega \Delta t[/itex] and then they (the manuals) say that it can be seen that when [itex]\Delta t[/itex] is small, [itex]||\Delta \vec{r}||[/itex] (which is a cord), is very near the lenght of the arc [itex]\Delta s[/itex] subtended by the angle [itex]\omega \Delta t[/itex], and thus, poof, [itex]||d\vec{r}|| = ds[/itex].

    And while this seems to be true intuitively, I have never seen a proof of this statement. And when I try to do it, here's what I get:

    I start from a cercle of radius R and a cord [itex]\delta[/itex] subtended by an angle [itex]\theta[/itex]. I find that the lenght of the cord is given by

    [tex]\delta = 2Rsin\left(\frac{\theta}{2}\right)[/tex]

    Therefor,

    [tex]d\delta=Rcos\left(\frac{\theta}{2}\right)d\theta[/tex]

    While

    [tex]ds=rd\theta[/tex]

    So

    [tex]d\delta=cos\left(\frac{\theta}{2}\right)ds[/tex]

    A result indicating that even the differential version is just an approximation because only true for a principal angle [itex]\theta=0[/itex].
     
    Last edited: Jan 10, 2005
  6. Jan 10, 2005 #5

    StatusX

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    it follows from the identity

    [tex] \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 [/tex]

    Also, the mistake in your derivation is that theta is a function of the chord length.
     
    Last edited: Jan 10, 2005
  7. Jan 11, 2005 #6

    quasar987

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    Thanks for your reply StatusX

    At which point does that identity fit in?


    Could you elaborate?
     
  8. Jan 11, 2005 #7

    dextercioby

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    Sure,if "y" is a function of "x",then "x" is a function of "y",right??So he basically didin't say anything new...You were right,though...Your calculations were corrrect.I guess u knew that,but u have my confirmation...

    Daniel. :smile:
     
  9. Jan 11, 2005 #8

    StatusX

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    A chord of length 2 R sin(theta) is subtended by an angle of 2 theta (in radians) and so has an arclength of 2 R theta. The identity means d(chordlength)/d(arclength) = 1.

    I'm sorry, I just glanced at your last step and assumed you had forgotten that if arclength is 0, then theta must be as well. You did forget that, but for a different reason. What you derived is how an chord of finite length changes with an infinitessimal change in arclength. What you wanted was the ratio of a differential chord to the differential arclength it subtends. To get the differential you're looking for, just take theta=0.
     
    Last edited: Jan 11, 2005
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