Here's a cute problem (from Robert M. Young, Excursions in Calculus, p. 244): "What is the average straight line distance between two points on a sphere of radius 1?"
(Answer to follow.)
If any of my students are reading this: no, this should not be interpreted as a hint to what will be on the final exam.
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14 comments:
Fun problem, especially since I'm relearning my calculus and am looking for reasons to use it. I got 64 / 15\pi, but did so much calculation along the way that I assume 1) I did something wrong somewhere and 2) there must be a simpler way to do it. At least the size of the answer is plausible.
(And indeed, I just found the actual solution and I was wrong. Now to find my mistake...)
4/pi ?
4/3 only one integral to compute! XD
Maybe I'm slow, but I don't get what "average" means in this case. How are the different "rings" weighted? If a particular line of latitude has twice the circumference of another, does it have twice as many infinite points from the given points? The same number?
I don't know, but 4/3*(pi^2) is a good number. :)
good but not correct in my opinion! its only 4/3 !
It's unbelievable: This simple problem got 5 different answers (1 deleted) by professional mathematicians! We're talking about the bad effects of pocket calculators here!
Oh yes, I am slow. Now I get what "average" means.
+1 for Chris Lusto's 4/pi
i don't think the solution depends on pi.
as i cut pi in the nominator & denominator and in my integral there are only 2 sini (where pi) has an non decimal value!
The answer in general is (4/3)*R. In the given case R = 1, hence it is 4/3.
its quite clearly 4/3
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