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 Around the ring (Posted on 2010-12-20)
Take the line segment whose endpoints are the points (1,0) and (-1,0) and a point of rotation (x,y).

If the segment is rotated all the way around the point it will trace out an annulus.

Find simplified formula for the area of this annulus in terms of x and y.

Redo this problem for a rectangle with corners at (1,1), (-1,1), (-1,-1), and (1,-1).

 No Solution Yet Submitted by Jer Rating: 3.0000 (1 votes)

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 re(3): part 1 not finished. | Comment 8 of 9 |
(In reply to re(2): part 1 not finished. by Jer)

Noted.

Talking just about the segment problem and just about occasions where {-1<x<1}, or {-1<y<1} i.e. where the inner ring of the annulus does NOT fall on an end point of a segment, I note that the larger circle must pass through at least one of (1,0) and (-1,0), whereas the small annulus is pi*y^2. So, denoting (abs)x as |x|, (1+|x|)^2+y^2 is the square of the radius and the 'annular area' is pi*(|x|+1)^2, with a minimum of pi at (0,0).

Clearly either formula works at e.g. (1,1) since pi (|x|+1)^2= 4*pi*|x| if |x|=1.

Edited on December 22, 2010, 4:31 am
 Posted by broll on 2010-12-22 04:14:05

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