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Arc Length Confusion

  1. Jan 18, 2010 #1

    Char. Limit

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    Why is arc length of a function [tex]f(x)[/tex] from a to b defined as [tex]\int_a^b \sqrt{1+(f'(x))^2} dx[/tex]?

    Where they get the idea of squaring the derivative, adding 1, taking the square root, and then integrating it is beyond me.
     
  2. jcsd
  3. Jan 18, 2010 #2

    tiny-tim

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    They got the idea from Pythagoras :smile:

    √(1 + (dy/dx)2) dx = √((dx)2 + (dy)2) = ds :wink:
     
  4. Jan 18, 2010 #3

    Char. Limit

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    But aren't dx, dy, and ds all infinitesimals with no real meaning here?
     
  5. Jan 18, 2010 #4

    tiny-tim

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    They're all infinitesimals, but they are the limit of very small increments ∆x ∆y and ∆s with ∆s2 = ∆x2 + ∆y2 :smile:

    how would you define arc-length? :wink:
     
  6. Jan 18, 2010 #5

    HallsofIvy

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    If you don't want to use differentials, which you say are "infinitesimals with no real meaning here" then you will have to use the standard "Riemann sum" definition. Given a curve y= f(x) with x from a to b, divide the x-axis from a to b into n intervals, each of length [itex]\Delta x[/itex]. We can approximate the curve from [itex](x_i, f(x_i))[/itex] to [itex](x_i+ \Delta x, f(x_i+ \Delta x))= (x_i+ \Delta x, f(x_i)+ \Delta y)[/itex] where I have taken [itex]\Delta y= f(x_i+ \Delta y)- f(x_i)[/itex], by the straight line between those points. It's length, by the Pythagorean theorem, is [itex]\sqrt{(\Delta x)^2+ (\Delta y)^2}[/itex]. Now, factor [itex]\Delta x[/itex] out of that:
    [tex]\sqrt{(\Delta x)^2+ (\Delta y)^2}= \sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x[/tex]

    So the Riemann sum is
    [tex]\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x[/tex]
    which, in the limit, becomes
    [tex]\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx[/tex]

    I would be surprised if your text book didn't give all of that.
     
  7. Jan 18, 2010 #6

    tiny-tim

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    awww, you gave him the answer! :redface:
     
  8. Jan 18, 2010 #7

    Char. Limit

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    Well, since this isn't a homework problem, I think giving the answer is all right...

    I honestly have never seen the derivation before.

    Thanks for the help, both of you.
     
  9. Jan 24, 2010 #8
    Certainly in the limit [tex]\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x[/tex] does not become [tex]\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx[/tex] according to the definition of the Rieman integral.

    BUT ,by the mean value theorem we have that:

    Δy = f'(ξ)Δx ,where ξ is [tex] x\leq\xi\leq x+\Delta x[/tex] and now ,

    [tex]\sum\sqrt{1+ \left(\frac{\Delta y}{\Delta x}\right)^2}\Delta x[/tex] becomes :[tex]\sum\sqrt{1+ \left(f'(\xi)\right)^2}\Delta x[/tex] .This a Rieman sum which in the limit it becomes:


    [tex]\int_a^b \sqrt{1+ \left(\frac{dy}{dx}\right)^2} dx[/tex]
     
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