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ContentsOn the approximation of: THE AREA OF A REGULAR POLYGONOn the approximation of: THE CIRCUMFERENCE OF AN ELLIPSETHE BINODAL LEMNISCATEA STRAIGHT LINEOn the approximation of: THE AREA OF A REGULAR POLYGONThe Journal of Recreational Mathematics, Vol.31(2) 81-82,2001 has an article on the approximation of the area of a regular polygon. There is sometimes a need to use the approximation when making an estimate, even though there may be an exact formula which uses trigonometry. It can almost be done in your head. The approximation is not too good for the equilateral triangle. It is quite close for the square. But, for n=5, the pentagon and upwards the error is slight. Let e be the length of one edge and n be the number
of edges, or sides. π = 3.1415926535897932384626433 approximately The approximation is intended for the ordinary citizen and Middle School, or Junior High School student who might like to find the area or for anyone who wants an estimate. The approximation is given by: A(n,e) = (n2 π)e2/4π .(1) No special trigonometry tables are required, Using 22/7, or a better approximation of π, makes little difference. You will still end up with an error of about 1% on the high side. Since we know the error is always on the high side by about 1%, a second estimate would be one percent lower, which is easy to calculate EXAMPLE Brian the gardener wanted to estimate the area of a pentagon flower bed he had in mind. It would be 2 meters along each edge and a regular pentagon. Using π=22/7, n=5, e=2 meters in Equation 1, we have:
A(n,e) = (n2 π)e2/4π
less .069545454 which is one percent =6.88 square meters which is the same as the exact formula by trigonometry.
On the approximation of the circumference of an ellipseAn approximation to the circumference of an ellipse was shown in the newsletter of the Statistical Association of Manitoba, Vol. 14, No. 5, January 1991. The approximation is closer than the error induced by using 22/7 for the value of p. The equation is:
S = 2(p
- (p
-2).e3).a
THE BINODAL LEMNISCATEIt can be shown that Cassinis Lemniscate is derived out of the equations of two circles: (x-a)2 + y2 = k12 (1) (x+a)2 + y2 = k22 (2) which when multiplied together produce: {(x-a)2+y2}{(x+a)2+y2)}=k12k22=b4 (3) and, depending upon the value of b become ovals, lemniscate, or ellipse. Now, in order to obtain another loop, or node, bring in another circle: x2 + y2 = k32 (4) and obtain, {(x-a)2+y2}(x2+y2){(x+a)2+y2)}=k (5) where the new k is to be determined. Set y=0 allows us to obtain and obtain where the curve intercepts the x-axis. (x-a)2x2(x+a)2 = x2(x2-a2)2 = k (6) or, x6 2a2x4 + a4x2 k = 0 (7) In order to have the intercepts at x=±A, and x=±B one also requires: (x-B)(x-A)2(x+A)2(x+B) = 0 (8) Both Equations 7 and 8 represent the curve from different points of view. The coefficients must be equal, so it requires solving the simultaneous equations: 2a2 = 2A2 + B2 (11) a4 = A4+2A2B2 (12) k = A4B2 (13) With effort, you will find: B2 = 4a2/3 (14) a2=3A2 (15) k=4a6/27 (16)
The nodes appear at: (±2a/√3, 0) and the x-intercepts are found at (±2a/√3, 0).
A STRAIGHT LINEExcerpted from: The article starts off by stating: It then discusses;
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