Area and Perimeter: Isoperimetric QuotientDate: 12/05/2001 at 06:41:20 From: Ace Subject: Isoperimetric Quotients I am stuck on what the isoperimetric quotient of a two-dimensional shape actually is. Is it a measure of compression? Please explain in detail. Ace Date: 12/05/2001 at 16:08:18 From: Doctor Peterson Subject: Re: Isoperimetric Quotients Hi, Ace. The isoperimetric quotient is a measure of the ratio of "inside" to "outside" in a shape; "fat" to "skin," you might say. It starts with the "Isoperimetric Problem," the question "Which shape, for a given fixed perimeter, gives the greatest area?" In order to compare shapes that have DIFFERENT perimeters, we can scale each shape to have a perimeter of 1, by dividing its linear dimensions by the perimeter. Since the area is proportional to the square of the linear dimensions, this will divide the area by the square of the perimeter. Therefore, the ratio of the area to the square of the perimeter (A/P^2) represents the area the shape would have if it were scaled so that its perimeter is 1. Then, the shape that has the greatest value for this ratio is the answer to the IP, and the greater the ratio, the more "stuff" the shape fits into the same "skin." We would like to scale this ratio so that the largest possible value is 1, making it easy to tell when a shape is close to the maximum. Since the circle gives the greatest value for this ratio (just imagine "blowing up a balloon," fitting as much air as you can into a fixed covering), we therefore divide A/P^2 by the ratio for a circle, which is A/P^2 = (pi r^2)/(2 pi r)^2 = (pi r^2)/(4 pi^2 r^2) = 1/(4 pi) This makes our Isoperimetric Quotient equal to 4 pi times A/P^2: IQ = (4 pi A)/P^2 For a circle, this will be 1; for something with no inside at all, it is zero. For anything else, it measures how close to a circle it is: the closer to one, the "fuller" or "fatter" the shape; the closer to zero, the "flatter" or "thinner" it is in this sense. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/