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Problem 86 Hard Difficulty

In Exercise 10.2.53 it was shown that the length of the ellipse $ x = a \sin \theta, y = b \cos \theta, $ where $ a > b > 0, $ is

$ L = 4a \int^{\pi/2}_0 \sqrt {1 - e^2 \sin^2 \theta} $ $ d\theta $

where $ e = \sqrt {a^2 - b^2}/a $ is the eccentricity of the ellipse.

Expand the integrand as a binomial series and use the result of Exercise 7.1.50 to express $ L $ as a series in powers of the eccentricity up to the term in $ e^6. $


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Calculus 2 / BC

Calculus: Early Transcendentals

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Video Transcript

The problem is in exercise 10.253. It shown that the length of the ellipse x is equal to sine theta y is equal to b times, cosine theta, why a is greater than b is greater than 0 is l is equal to 4 a times integral from 0 from from 0 to pi over 2, integral is square root of 1 minus x square times sine t square, where a is equal to square root of a square. Minus b, squared and divided by a is accentricity of the ellipse, extend it integrant as a binomial series and use the result of exercise 7.1 .50 to express l as a series of powers of the eccentricity up to the term in e to 6. So first we have 1 plus x to the power of 1. Half is equal to some and from 0 to infinity on half in x to m, so we half the integrant square root of 1 minus a square sine theta square is equal to sum, and from 0 to infinity on half n times negative, a square sine square To the power of n, which is equal to some and from 0 to infinity, while half n times negative 1 to n times e to 2 n times sine theta to the power 2, then the half l is equal to 4 a times. So the first term is integral 0 to pi over 2 times 1, the tea and plus 1 half times negative 1 times a square times integral from 0 to pi over 2 sine theta square d, theta plus 1 half times negative 1 half over 2 factorial times E to 4 times integral from 0 to pi over 2 sine t to the power 4 d, theta and plus 1 half times negative 1 half times negative 3, half over 3 factorial times negative 1 times e to 6 and times integral from 0 to pi. Over 2 sine theta to the power 6 d, theta and plus the the and by using the results of exercise 7.1 .50, we have integral, from 0 to pi over 2 sine theta to the power of a power of 2 n d. Theta is equal to 1 times 3 times 5 times dated times: 2 n minus 1 over 2 times 4 times the times 2 n times pi over 2 point. So behalf. This is equal to 4 a times pi over 2 minus pi over 8 times a square minus 3 pi over 128 times e to 4 minus 5 pi over 512 times e to 6 plus. Do.

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