PreliminariesThe circumference C of an ellipse must be computed using calculus. To do this, we set up a Cartesian coordinate system. We put the origin at the center of the ellipse, the xaxis along the major axis, whose length is 2a, and the yaxis along the minor axis, whose length is 2b. The eccentricity e is defined byThen the equation of the ellipse is0 <= e = sqrt(a^{2}b^{2})/a < 1. Now the formula for computing the arc length of any curve given by the parametric equations x = f(t), y = g(t), over the range c <= t <= d isx^{2}/a^{2} + y^{2}/b^{2} = 1, a >= b > 0. d s = INTEGRAL sqrt[(dx/dt)^{2}+(dy/dt)^{2}] dt. c
DerivationFor the above ellipse, we can use the parametric equationsThenx = a sin(t), y = b cos(t), 0 <= t <= 2, C = s, dx/dt = a cos(t), dy/dt = b sin(t). We can use the 4fold symmetry of the ellipse to rewrite this as2 C = INTEGRAL sqrt(a^{2}cos^{2}[t]+b^{2}sin^{2}[t]) dt. 0 This can be further rewritten, using cos^{2}(t) = 1  sin^{2}(t), and the definition of the eccentricity e, as/2 C = 4 INTEGRAL sqrt(a^{2}cos^{2}[t]+b^{2}sin^{2}[t]) dt. 0 Now this integral is a famous one. It is called a "complete elliptic integral of the second kind." It is one of those integrals that cannot be expressed in closed form in terms of the familiar functions of calculus, except if e = 0, when we have a circle. That means that this is the simplest formula possible for the circumference of a general ellipse./2 C = 4a INTEGRAL sqrt(1e^{2}sin^{2}[t]) dt. 0 Another way to write this is C = 4aE(/2,e).
EvaluationThis integral can be evaluated numerically, of course. Another way to compute its value is using an infinite series. SetThenx = (ab)/(a+b). This series converges pretty rapidly, especially when x is small, that is, when a and b are close together, that is, when e is small.infinity C = (a+b)(1 + SUM [(2n2)!/(n![n1]!2^{2n1})]^{2}x^{2n}), n=1 C = (a+b)(1 + x^{2}/4 + x^{4}/64 + x^{6}/256 + 25x^{8}/16384 + ...).
ExampleIf a = 15, b = 6, e = sqrt(21)/5, and x = 3/7, you findapproximately. With six terms, we get 7 significant figures of accuracy for this value of x (the correct answer being 69.03933778699452855...), even though a and b are not close together (and e = 0.916515... is not very small).C = (15+6)(1 + 9/196 + 81/153664 + 729/30118144 + 164025/94450499548 + 2893401/18512297918464 + ...), = 21(1 + 0.045918367 + 0.000527124 + 0.000024205 + 0.000001737 + 0.000000156 + ...), = 21(3.1415926536)(1.046471589), = 69.039336580,
ApproximationThe approximate formula can be found in Mathematical Tables from the Handbook of Chemistry and Physics, 10th ed. (1954), p. 315.C = 2 sqrt((a^{2}+b^{2})/2) The idea seems to be to use 2r, but for r use the rootmeansquare of the semimajor and semiminor axes. Then a > r > b, so one shouldn't be too far off. In fact, so the first order term is right, and the second order term is double what it should be, so this is not too awful an approximation, but not very good, either. It's about as good as using r = (a+b)/2.2 sqrt[(a^{2}+b^{2})/2] = (a+b)(1 + x^{2}/2  x^{4}/8 + ...), Better would be to use which agrees in both first and second order terms. If x^{4}/16 is negligibly small, then this gives right answers, whereas the original approximation does only if x^{2}/4 is negligible.C ~~ (sqrt[(a^{2}+b^{2})/2] + [a+b]/2), = (a+b)(1 + x^{2}/4  x^{4}/16 + ...), A short search of the Dr. Math archives turned up the following approximation due to Ramanujan: C ~~ (3a + 3b  sqrt[(a+3b)(b+3a)]) This is even better, because which agrees even in the thirdorder term!(3a + 3b  sqrt[(a+3b)(b+3a)]) = (a+b)(1 + x^{2}/4 + x^{4}/64 + x^{6}/512 + ...),
However, in the same paper
[Ramanujan, S., "Modular Equations and Approximations to ,"
which has a relative error of about (3/2^{17})x^{10} for small values of x, since this function has series expansionC ~~ (a+b)(1+3x^{2}/[10+sqrt(43x^{2})]) agreeing with the actual series for C through terms of the fifth order, and even having sixth order term close to right!(a+b)(1 + x^{2}/4 + x^{4}/64 + x^{6}/256 + 25x^{8}/16384 + 95x^{10}/131072 + ...), For a relatively compact formula, this is clearly the winner, although it's rarely clear how Ramanujan derived his formulas. Anyone interested enough to read this far would probably greatly enjoy taking a detour to read about his fascinating life and contributions to mathematics in the MacTutor History of Mathematics archive.

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