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A New Quadrature of the Parabola
Posted:
Jan 2, 2002 3:18 PM
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The following is a quadrature of the parabola using neither infinitesimals
nor limits. It requires only the algebra of polynomial functions and the
geometry of similar figures. I haven't been able to find any reference in the
literature for a calculation like this. I would appreciate it if you let me
know of any such reference that you come across.
Consider the rectangle bounded by the four lines x=0, y=0, x=b, and y=b^2. We
want to determine the area of a "parabolic right triangle" lying within this
rectangle. The parabolic right triangle, or just p-triangle, has two legs
lying on the lines y=0, x=b, and a hypotenuse given by the curve y=x^2. Call
this p-triangle PT1.
The picture we will work with has two p-triangles sitting within the
rectangle. The first p-triangle has its right angle vertex in the corner of
the rectangle at the point (b,0). This is just PT1. The second p-triangle,
of equal area to the first, has its right angle vertex located at the point
(0,b^2) with legs along the lines x=0 and y=b^2. Call this one PT2. The area
of the rectangle, b^3, is given by twice the area of PT1 plus the area of the
"eye." This eye corresponds to the gap between the two p-triangles. If we can
relate the area of the "eye" to the area of PT1 the quadrature of the
parabola is solved.
We first subtract the area of PT2 from the area of the rectangle. The
remaining figure, eye and lower p-triangle, has a boundary given by
b^2-(b-x)^2. The remaining area is the area under this curve. If we then
subtract the area of PT1 from this remaining area we obtain the area of the
gap between the two p-triangles. The bounding curve for this gap is g(x)
=b^2-(b-x)^2-x^2. The area of the gap is given as the area under the curve
for g(x) over the interval from 0 to b. (Our eye, viewed algebraically, is a
little hill.)
The hill is symmetric around x=b/2. So the area of the whole hill is twice
the area of half the hill, say from b/2 to b. In turn this means that we can
compute the area of the hill as half the area of the original rectangle minus
the area of a p-triangle with legs lying on the lines y=0, x=b/2 and with
hypotenuse given by y=4x^2. Call this last p-triangle PT3.
The p-triangle, PT3, has an area four times the area of the p-triangle with
legs lying on y=0, x=b/2 and hypotenuse given by y=x^2. Call this p-triangle
PT4. Now it is also the case that the area of PT4 is one-eighth the area of
PT1. (Why?) Thus the area of PT3 is equal to one half the area of PT1.
Since the area of the gap (or the hill) is one half the area of the original
rectangle minus the area of PT3, then half the area of the gap must be equal
to half the area of PT1. Or to say the same thing twice as well, the area of
the gap is equal to the area of PT1. Hence the area of the rectangle is three
times the area under the parabolic triangle. QED.
For the area under the cubic, the area of the gap again has a quadratic
boundary. (Why?) The determination of the area under the cubic reduces to the
same type of argument as just presented. Try it. In turn these two cases act
as a model for determining the area under the curve y=x^n using a proof by
induction.
I think this is more than just a nice calculation. I think it has interesting
implications for calculus reform.
Bill Crombie
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