Wednesday, 9 December 2009

Fun With Parabolas in Calculus

Ok, two quick mental problems. Some folks will see one as somewhat easier than the other, but ...well heck, read on...

Problem One. A secant cuts the parabola y= x2 at x= -1 and x=1. Find the area of the region between the secant and the parabola..

Problem Two. A secant cuts the parabola y= x2 at x= 1+sqrt(5) and x=3+sqrt(5) . Find the area of the region between the secant and the parabola..

Ok, if two seems to hard, try this middle of the road alternate:

A secant cuts the parabola y= x2 at x= 0 and x=2. Find the area of the region between the secant and the parabola..

I still find it fascinating each year as I introduce my calculus class to this idea that the answers are all the same. I lead them eventually to conclude that the area between the secant and x2 depends only on the difference between the two x-values.

I was recently reminded of this little parabola property when the very kind Dave Renfro put me on to an article in the December Monthly from the MAA by Bettina and Tom Richmond entitled "How to Recognize a Parabola." Among a dozen really nice properties of parabolas, the first one was this little area relation. They included several that I had used recently in my blog about tangents to a parabola and circle. They also had another theorem closly related to these problems in their eleventh example which says (in my less rigorous language) that the area of the region between the secant and a parabola is 2/3 the area of the parallelogram that inscribes that region. The two ideas together mean that the perpendicualr distance between the secant and the tangent parallel to it when multiplied by the length of the secant is a constant depending only on the difference in the x values also. ... In the region cut off by the horizontal secant the distance from the secant to the vertex is 1 and the length of the secant is 2. Two-thirds their product gives the area of 4/3 sq units. In the region cut off by the secant from x=0 to x=2 the secant has a length of sqrt(20). The tangent parallel to the secant will be at x=1 (that would be Theorem 3 in the Richmonds' paper, "the average of the slopes at the endpoints is the slope at the midpoint" ). The perpendicular distance from the tangent point to the secant line is 1/sqrt(5) , and the product of these two distances is again 2.

The relation between the two above has started me wondering about their tenth theorem, which I had not observed before, and a possible generalization of the theorem. The theorem said that if you took the section formed by a horizontal secant and the parabola, and revolved it around the y-axis, the volume inside the bowl and the volume remaining from a cylinder when the bowl is cut out are the same. Ok... but driving to work I got to wondering, If we took and oblique secant, and rotated the area between the secant and the parabola about the line perpendicular to the secant passing through the point where the tangent would be parallel, would that also have the same volume? (I keep thinking it would, will leave the exercise prove or disprove to the reader).

By the time I get around to writing this, I'm thinking it is NOT the same as the horizontal case... but I keep thinking when I get to it there will be something pretty hiding here...
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