Parabolic Segment
The arc length of the parabolic segment
| [画像: y=h(1-(x^2)/(a^2)) ] |
(1)
|
illustrated above is given by
and the area is given by
(Kern and Bland 1948, p. 4). The weighted mean of y is
so the geometric centroid is then given by
The area of the cut-off parabolic segment contained between the curves
can be found by eliminating y,
| x^2-ax-b=0, |
(13)
|
so the points of intersection are
| x_+/-=1/2(a+/-sqrt(a^2+4b)), |
(14)
|
with corresponding y-coordinates y_+/-=x_+/-^2. The area is therefore given by
![画像:int_(a-sqrt(a^2+4b))^(a+sqrt(a^2+4b))[(ax+b)-x^2]dx](/index.cgi/contrast/https://mathworld.wolfram.com/images/equations/ParabolicSegment/Inline40.svg){kind=link}
The maximum area of a triangle inscribed in this segment will have two of its polygon vertices at the intersections (x_-,y_-) and (x_+,y_+), and the third at a point (x^*,y^*) to be determined. From the general equation for a triangle, the area of the inscribed triangle is given by the determinant equation
Plugging in and using y_*=x_*^2 gives
| A_Delta=1/2[b+(a-x^*)x^*]sqrt(a^2+4b). |
(19)
|
To find the maximum area, differentiable with respect to x^* and set to 0 to obtain
so
| x_*=1/2a. |
(21)
|
Plugging (21) into (19) then gives
| A=1/8(a^2+4b)^(3/2). |
(22)
|
This leads to the result known to Archimedes in the third century BC, namely
| [画像: A/(A_Delta)=(1/6)/(1/8)=4/3. ] |
(23)
|
See also
Circular Segment, Geometric Centroid, ParabolaExplore with Wolfram|Alpha
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References
Beyer, W. H. (Ed.). CRC Standard Mathematical Tables, 28th ed. Boca Raton, FL: CRC Press, p. 125, 1987.Kern, W. F. and Bland, J. R. Solid Mensuration with Proofs, 2nd ed. New York: Wiley, p. 4, 1948.Referenced on Wolfram|Alpha
Parabolic SegmentCite this as:
Weisstein, Eric W. "Parabolic Segment." From MathWorld--A Wolfram Resource. https://mathworld.wolfram.com/ParabolicSegment.html