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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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 Part 1 Comment 9 of 9 |
By symmetry, we can consider just x<=0, y<=0.  Then there are two cases: x<=1 and x<1.

With x<=1 the closest point of the segment is (x,0) and the furthest point of the segment is (-1,0).  The radii of the two circles are y and sqrt(y^2 + (x+1)^2).  The area between these is pi*(x+1)^2.

With >1 the closest point of the segment is (1,0) and the furthest point of the segment is (-1,0).  The radii of the two circles are sqrt(y^2 + (x-1)^2) and sqrt(y^2 + (x+1)^2).  The area between these is pi*(4x).

The two area formulas are equal for the boundary value x=1.  Furthermore, the line A=pi*(4x) is tangent to the parabola A=pi*(x+1)^2 at the point (1,4*pi).

 Posted by Brian Smith on 2016-02-29 12:58:41

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