Thread: Calculcating the length of a curve using area integration of delta function

1. 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.

2. 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

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

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

Let $\displaystyle \epsilon>0.$ Then the integral

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

Does that work for you?

3. Where does the y in front of the delta function come from?

4. 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:

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

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

Make sense?