Author: Tony Pay ★2017
Date: 2006-01-18 22:32
I wrote:
>> For the mathematically inclined -- but not if you're not -- this simple and, in retrospect, obvious realisation reminds me of the time a friend revisiting her school mathematics as a mature student asked me, why is it that the area of a triangle is a half length of base times height, whereas the volume of a tetrahedron is a third area of base times height? After thinking a moment, I was happy to be able to answer, "because the integral of x is 1/2 x squared, and the integral of x squared is 1/3 x cubed.">>
Someone complained that what I wrote was a little sketchy and therefore possibly confusing. Perhaps it helps to include the information that a further bit of her question was, is it anything to do with that the base of the triangle has two points, and the base of a tetrahedron three?
Now, that's a 'deep connection' type question of a sort that has to be encouraged -- even if in this case it's wrong -- so I was looking for the 'right' deep connection to offer her. And why I was pleased with my answer is that we don't often think of the 1/2 in the formula for the area of a triangle as arising from an integration, because we learn it in elementary geometry.
Yet clearly, if the triangle has height h and base b, we can think of the area A as given by an integration wrt x, from 0 to h, of the length of a horizontal line intersecting the sides of the triangle a vertical distance x from the vertex. By similar triangles, this length is proportional to x, with constant of proportionality k given by b = kh, so the area A is the integral:
- h
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| (b/h)xdx = (b/h)(1/2)h^2 = (1/2)bh
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- 0
And this argument goes over exactly to the case of the volume of the tetrahedron:
If the tetrahedron has height h and base area A, we can think of the volume V as given by an integration wrt x, from 0 to h, of the area of a horizontal slice of the tetrahedron a vertical distance x from the vertex. Again by similar triangles, this area is proportional to x^2, with constant of proportionality k given by A = kh^2, so the volume V is the integral:
- h
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| (A/h^2)x^2dx = A/h^2(1/3)h^3 = (1/3)Ah
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- 0
An interesting postscript to all this is that my friend, in her forties at the time, and having previously been fixer for the London Sinfonietta and the Orchestra of the Age of Enlightenment, passed her 'school' mathematics exams with flying colours, and then went on to do a degree in mathematics. (She's now researching and teaching at Sussex University, well beyond my level:-)
Tony
Post Edited (2006-01-18 22:36)
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