What is the mean distance between two randomly chosen points within a sphere of radius 1?
(show how you work it out as well)
2007-09-08 08:27:56 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
why are there so little genius or almost none
2007-09-08 08:34:09 · update #1
harry m - sorry but that not how or what it is neither is your answer
2007-09-08 08:36:13 · update #2
El Chupacabra - thats not it as well
2007-09-08 08:36:53 · update #3
ironduke8159 and slick nick - its not the answer
2007-09-08 08:39:10 · update #4
paulmaxpayton - I agree but my lecturer said this problem is most easily solved by introducing spherical coordinates. And solved in almost only a single page of work
2007-09-08 08:43:36 · update #5
SO WHAT IS THE ANSWER !!!!!!!!!!!!!
2007-09-08 08:46:36 · update #6
OK - I just called my ex senior - who happen to share his experience with same question before - hence he knows the answer given by my lecturer but not the working or solution.
The answer for the mean distance between two randomly chosen points within the sphere is 36/35 = 1,02857....
2007-09-08 09:24:02 · update #7
Sciencenut - you were correct when you said calculus method were used.The strange thing of this problem is that I was told it's far easier to solve in three dimensions than in two. More dimensions almost always implies more problems.
2007-09-08 09:27:24 · update #8
I think you have to specify if the points are all on the surface or if it is any two points contained within the volume of the sphere. I believe you are referring to the latter problem, making the above references incorrect.
If you specify that one of the two points is at the exact center of the sphere, then the problem becomes easy and the correct answer to this (highly simplified) problem is 0.5, times 2, I believe. Perhaps you can then prove that as the first point moves from the center to the outer surface, the average stays the same. i.e. 0.5. I don't know how to solve it but I'm sure there is a calculus method for doing it.
P.S. If one of the points is at the exact center of the sphere, then the average distance to any other point is 0.5. This descibes a sphere of "average radii", which is 1/2 the diameter of the original sphere. Then as your first point moves away from the center, the "sphere of average radii" changes to an elipsoid with your point as one of the foci, and the other focus of the elipsoid being the center of your original sphere. As your point finally gets to the outer surface of your original sphere, your "sphere of average radii" will once again become a sphere with a radius of 0.5. At least that is how I might try to set this problem up.
2007-09-08 09:02:46 · answer #1 · answered by Sciencenut 7 · 0⤊ 0⤋
Max is 2 you simply say the middle is 0 take the 2 and divide it by the 2 points..so 1
2007-09-08 15:38:14 · answer #2 · answered by Prisoner of Grace 3 · 0⤊ 1⤋
The furthest the two points can be separated is 2. The least they can be separated is 0. Hence the average distance will be 1.
2007-09-08 15:38:11 · answer #3 · answered by ironduke8159 7 · 0⤊ 1⤋
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X=(d1+d2)/2
so the maximum distance is the diameter or d1 is 2 and the minimum distance is nothing or 0... then (2+0)/2=1
mean distance=1
2007-09-08 15:35:24 · answer #4 · answered by Grand Phuba 5 · 0⤊ 0⤋
answer is 0.5^(0.5) but I'm afraid that I did it by monte-carlo, no analysis. - rats! that's wrong - my distribution was all skew-wiff. uniform distribution through the whole sphere gives just a little more than 1 as the answer.
2007-09-08 16:16:28 · answer #5 · answered by Anonymous · 0⤊ 0⤋
max D = 2 & min D = 0
I guess
the mean distance = 1
2007-09-08 15:35:12 · answer #6 · answered by harry m 6 · 0⤊ 1⤋
It's been done. This problem is fairly well known in statistical geometry. See the references.
2007-09-08 15:40:11 · answer #7 · answered by PMP 5 · 0⤊ 0⤋