Re: Average Distance to Circumference

On Feb 27, 9:25 am, Robert Israel
<isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote:
HWa...@xxxxxxxxx writes:
Hi,

First, I'd like to begin by saying this isn't a homework problem. I'm
a high school student, and this isn't exactly the kind of problem they
teach in geometry. It's something I came up with recently. I also
apologize if this is the wrong group.

So, assume we have a circle with a radius of ten meters. An insect is
placed on a random point in the circle and immediately begins crawling
in a random direction. On average, how far will it have to travel to
get to the edge of the circle? Assume it follows a straight line.
Also, what field of mathematics is this covered in?

It's a problem in calculus of several variables.

Suppose the insect is placed at a point at distance r from the centre
and crawls at an angle theta from the direction to the centre.
The distance it must crawl is d, where by the Law of Cosines
100 = d^2 + r^2 - 2 r d cos(theta). Solving this quadratic equation,
d = r cos(theta) + sqrt(100 - r^2 sin(theta)^2).

Unfortunately, if you're in high school the rest may not make
much sense to you yet. The joint probability density for r and
theta is f(r,theta) = r/(100 pi) for 0 <= r <= 10, 0 <= theta <= 2 pi,
so the expected distance is given by a double integral

int_0^{2 pi} int_0^{10} f(r,theta) (r cos(theta) + sqrt(100 - r^2
sin(theta)^2)) dr dtheta

Probably the ambiguous third side d(angle not between given sides)
admitting negative sign before radical -sqrt(100 - r^2.... would also
give rise to the same result by symmetry? I did not check it.

which works out to 80/(3 pi) or approximately 8.488263630 metres.
--

Robert Israel isr...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada

If we consider infinitesmal triangles area rho^2/2 dtheta, and weight
it with radius, then average radius after integrations would be 2 r/3.
Why is such an approach wrong?

Regards,
Narasimham






.

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