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Math Help - Calculcating the length of a curve using area integration of delta function

  1. #1
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    Calculcating the length of a curve using area integration of delta function

    Suppose we have some well behaved curve y=f(x) lying in a plane. The length of a curve is given by

    L = int( sqrt( 1+f'(x) ) )dx

    I want to express this integral using the delta fuction as an integral over some area

    A = int( delta( y-f(x) ) )dxdy.

    Can this be done? If so, how to prove it?

    Thanks.
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  2. #2
    A Plied Mathematician
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    Assuming you mean the Dirac delta function here. I think you should use different function labels in order to avoid confusion. Try this on for size. Let

    g(x)=\sqrt{1+(f'(x))^{2}}. Then

    \displaystyle L=\int_{a}^{b}g(x)\,dx.

    Let \epsilon>0. Then the integral

    \displaystyle \int_{a}^{b}\int_{g(x)-\epsilon}^{g(x)+\epsilon}y\,\delta(y-g(x))\,dy\,dx<br />
=\int_{a}^{b}g(x)\,dx.<br />

    Does that work for you?
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  3. #3
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    Where does the y in front of the delta function come from?
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  4. #4
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    It came from my head. My reasoning went like this: the Dirac delta function (technically, a distribution), under an integral sign, is a "replacer". That is, you actually define the Dirac delta distribution by its action under an integral sign:

    \displaystyle\int f(x)\delta(x-a)\,dx=f(a).

    So the Dirac delta distribution "picks out" one value of the domain, and substitutes it into the function f(x).

    Now, I wanted to end up with g(x), and I wanted to use the delta distribution \delta(y-g(x)). Hence, the function into which I wanted to replace would have been a simple y.

    Make sense?
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