2006-10-27 21:24:59 · 4 answers · asked by kishore_sai_kishore 1 in Science & Mathematics ➔ Mathematics
Perimeter = 2 * pi * a {1 - (1 / 2)^2 * k^2 / 1 - [(1 * 3) / (2 * 4)]^2 * k^4 / 3 - [(1 *3 * 5) / (2 * 4 * 6)]^2 * k^6 / 5 - ...}
where k (the eccentricity) = sqrt(a^2 - b^2) / a
That should be enough so you can see the pattern of this infinite sum.
An approximation of this is usually given as :
Perimeter = pi * sqrt[2(a^2 + b^2)]
2006-10-27 23:11:57 · answer #1 · answered by falzoon 7 · 1⤊ 0⤋
Second answerer is correct, but it isn't straightforward.
A quarter of the perimeter -- i.e. say, the length of arc in the first quadrant -- is
integral of (1 + (dy/dx)^2) dx from 0 to a.
That integrand is
1 + bx/sqrt(a^2 - x^2)
and there isn't any primitive for it.
In fact it belongs to a class of integrals called (surprise! surprise!) "elliptic integrals".
Of course, for given values of a and b, you can use numerical integration to get as accurate an approximation as you need.
2006-10-27 22:27:27 · answer #2 · answered by Hy 7 · 1⤊ 0⤋
TAAITBOTB. INLOTSOTPSNTY
the answers are in the back of the book. if not, look over the shoulder of the person sitting next to you.
2006-10-27 21:27:51 · answer #3 · answered by hot.turkey 5 · 0⤊ 1⤋
use line integration
2006-10-27 21:28:56 · answer #4 · answered by The Potter Boy 3 · 0⤊ 0⤋