I'm trying to find a formula for the average distance between a point,$\ P$ and a curve segment,$\ C$

When I tried to use the formule for the mean of a function with a distance function $$\frac{1}{a+b}\int_{a}^{b}(\|f(t),P\|)dt$$ where I defined $\ C$ in terms of $\ f:{\rm I\!R} \to {\rm I\!R}^2, C= \{f(i)|a \leq i \leq b\}$

and $\ P = (0,0), a=0$ and $\ b=1$, the average distance between $\ P$ and $f(t) = (t,t^2)$ should be the same as the distance between $\ P$ and $f(t) = (t,\sqrt{t})$, as the 2 curve segments are just mirror images. This is not the case when I use my formula.

Does anyone why my formula is wrong and which formula is the correct one?

  • $\begingroup$ Shouldn't you be parameterising $C$ by arc-length? $\endgroup$ – Lord Shark the Unknown Sep 1 '17 at 18:56
  • $\begingroup$ Averages need to be with respect to something, and when you change the parameterization, you change what the average is with respect to. If you parameterize the curve with respect to arc length and find the average distance with respect to that, you will get a different answer than if you parameterize it with respect to, say, the $x$ coordinate. $\endgroup$ – Steve Kass Sep 1 '17 at 18:57

The correct formula should be $$\frac{\int_a^b ||f(t),P||dl}{\int_a^b dl}$$

So for $$f(t)=(t,t^2)$$ you have $$\frac{\int_0^1 \sqrt{t^2+t^4}\sqrt{1+4t^2}dt}{\int_0^1\sqrt{1+4t^2}dt}$$

For $$f(t)=(t,\sqrt{t})$$ you have $$\frac{\int_0^1 \sqrt{t^2+t}\sqrt{1+\frac{1}{4t}}dt}{\int_0^1\sqrt{1+\frac{1}{4t}}dt}$$

Let $u=\sqrt{t}$, it equals $$\frac{\int_0^1 \sqrt{u^4+u^2}\sqrt{1+\frac{1}{4u^2}}2udu}{\int_0^1\sqrt{1+\frac{1}{4u^2}}2udu}=\frac{\int_0^1\sqrt{u^4+u^2}\sqrt{1+4u^2}du}{\int_0^1\sqrt{1+4u^2}du}$$

So the two integrals are clearly the same.


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