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WZ
Numerade Educator

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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. $

Answer

$L \approx 4 a\left[\frac{\pi}{2}-\frac{\pi}{8} e^{2}-\frac{3 \pi}{128} e^{4}+\frac{5 \pi}{512} e^{6}\right]$

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Video Transcript

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