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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: part 1 not finished. | Comment 6 of 9 |
(In reply to part 1 not finished. by Jer)

Jer,

May I point out that you asked for the 'annulus area'? If the origin is selected then don't the ends of the segment  sweep over each other giving an annular area of zero?

Having said that I note that your post and Charlie's comment nevertheless treat the swept out circular area as being relevant for cases where at least one of x,y is less than or equals 1. But It seems to me that that is no longer an 'annulus area'?

On further reflection, a more formidable difficulty might be that the '4+4' formula doesn't accurately describe what happens when one of x,y is very near the origin (smaller than 1) and the other is greater than 1, because then the closest point on the swept 'rectangle' to x,y is not actually at a vertex but a point on one of its sides. I have corrected my earlier answer to handle this.

Edited on December 21, 2010, 2:13 pm
 Posted by broll on 2010-12-21 11:52:12

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