The usual parametric presentation of the ellipse (blue in the diagram )
x 2 a 2 + y 2 b 2 = 1 , superscript x 2 superscript a 2 superscript y 2 superscript b 2 1 \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}\;=\;1,
using as parameter the eccentric anomaly t t t
x = a cos t , y = b sin t ( 0 ≦ t < 2 π ) . fragments x a t normal-, y b t fragments normal-( 0 t 2 π normal-) normal-. \displaystyle x\;=\;a\cos{t},\quad y\;=\;b\sin{t}\quad(0\leqq t<2\pi). (1)
\psaxes [Dx=9,Dy=9]-¿(0,0)(-3.5,-3.2)(3.5,3.5)
x x x y y y a a a b b b a a a P P P Q Q Q t t t If, according to the diagram, the point Q Q Q circumscribed circle corresponds the point P P P t t t polar angle of Q Q Q
Substituting to the arc length formula
s = ∫ t 1 t 2 ( d x d t ) 2 + ( d y d t ) 2 d t s superscript subscript subscript t 1 subscript t 2 superscript d x d t 2 superscript d y d t 2 d t s\;=\;\int_{{t_{1}}}^{{t_{2}}}\sqrt{\left(\frac{dx}{dt}\right)^{2}\!+\left(%
\frac{dy}{dt}\right)^{2}}\,dt
the limits 0 0 0 2 π 2 π 2\pi derivatives
we can express the perimeter length of the ellipse as
p = ∫ 0 2 π a 2 sin 2 t + b 2 cos 2 t d t . p superscript subscript 0 2 π superscript a 2 superscript 2 t superscript b 2 superscript 2 t d t p\;=\;\int_{0}^{{2\pi}}\!\sqrt{a^{2}\sin^{2}{t}+b^{2}\cos^{2}{t}}\;dt.
The expression may be converted by taking the factor a a a sin 2 t = 1 - cos 2 t superscript 2 t 1 superscript 2 t \sin^{2}{t}=1-\cos^{2}{t}
p = a ∫ 0 2 π 1 - ε 2 cos 2 t d t p a superscript subscript 0 2 π 1 superscript ε 2 superscript 2 t d t p\;=\;a\int_{0}^{{2\pi}}\!\sqrt{1-\varepsilon^{2}\cos^{2}{t}}\;dt
Here we have used the eccentricity ε = a 2 - b 2 a ε superscript a 2 superscript b 2 a \varepsilon=\frac{\sqrt{a^{2}-b^{2}}}{a} 0 ≦ ε < 1 0 ε 1 0\leqq\varepsilon<1 symmetry of the ellipse in regard to the coordinate axes, we get
p = 4 a ∫ 0 π 2 1 - ε 2 cos 2 t d t , p 4 a superscript subscript 0 π 2 1 superscript ε 2 superscript 2 t d t p\;=\;4a\int_{0}^{{\frac{\pi}{2}}}\!\sqrt{1-\varepsilon^{2}\cos^{2}{t}}\;dt,
and since on the interval [ 0 , π 2 ] 0 π 2 [0,\,\frac{\pi}{2}] sine function attains the same values as the cosine function (but in reverse order ), the length of the perimeter of ellipse may be written also as
p = 4 a ∫ 0 π 2 1 - ε 2 sin 2 t d t . p 4 a superscript subscript 0 π 2 1 superscript ε 2 superscript 2 t d t \displaystyle p\;=\;4a\int_{0}^{{\frac{\pi}{2}}}\!\sqrt{1-\varepsilon^{2}\sin^%
{2}{t}}\;dt. (2)
This is a complete elliptic integral of second kind. The name “elliptic integral ” is due to this task to determine the length of the perimeter of an ellipse (0 < ε < 1 0 ε 1 0<\varepsilon<1
The integral (2) is not expressible in closed form with elementary functions , but can be expanded e.g. to the series
p = 2 π a [ 1 - ( 1 2 ) 2 ε 2 1 - ( 1 ⋅ 3 2 ⋅ 4 ) 2 ε 4 3 - ( 1 ⋅ 3 ⋅ 5 2 ⋅ 4 ⋅ 6 ) 2 ε 6 5 - … ] . p 2 π a 1 superscript 1 2 2 superscript ε 2 1 superscript normal-⋅ 1 3 normal-⋅ 2 4 2 superscript ε 4 3 superscript normal-⋅ 1 3 5 normal-⋅ 2 4 6 2 superscript ε 6 5 normal-… p\;=\;2\pi a\left[1-\left(\frac{1}{2}\right)^{2}\frac{\varepsilon^{2}}{1}-%
\left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}\frac{\varepsilon^{4}}{3}-\left(%
\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{2}\frac{\varepsilon^{6}}{5}-%
\ldots\right]\!.
One quite accurate approximative formula, due to S. Ramanujan , is
p ≈ π ( a + b ) ( 1 + 3 λ 2 10 + 4 - 3 λ 2 ) p π a b 1 3 superscript λ 2 10 4 3 superscript λ 2 p\;\approx\;\pi(a\!+\!b)\left(1+\frac{3\lambda^{2}}{10+\sqrt{4-3\lambda^{2}}}\right)
with λ = a - b a + b λ a b a b \displaystyle\lambda=\frac{a\!-\!b}{a\!+\!b} 3 ε 20 2 36 3 superscript ε 20 superscript 2 36 \displaystyle\frac{3\varepsilon^{{20}}}{2^{{36}}}
