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How to find the perimeter of an ellipse???

Note by Anirudha Nayak
3 years, 6 months ago

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Let the ellipse be : \(\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1\)

Parametric coordinates : \(\displaystyle (x,y) = (a\cos\theta,b\sin\theta)\)

\[ds^2 = dx^2+dy^2\]

\[ds^2 = \left((\frac{dx}{d\theta})^2 + (\frac{dy}{d\theta})^2\right)d\theta^2\]

\[ds^2 = (a^2\sin^2\theta+ b^2\cos^2\theta)d\theta^2\]

\[ds = \sqrt{(a^2\sin^2\theta+ b^2\cos^2\theta)}d\theta\]

Integrate it from \(\displaystyle 0\) to \(\displaystyle 2\pi\), and you will get the result.

Anish Puthuraya - 3 years, 6 months ago

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Good luck integrating that monster.! (it hasn't been solved yet)

Anish Puthuraya - 3 years, 6 months ago

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i am surprised This question came in VIT couldnt solve it maths was at a good level indeed when is urs or have u already given it

Anirudha Nayak - 3 years, 6 months ago

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@Anirudha Nayak It was yesterday. What was the exact question?

Anish Puthuraya - 3 years, 6 months ago

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@Anish Puthuraya i have posted it

Anirudha Nayak - 3 years, 6 months ago

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@Anirudha Nayak how was d exam wht r u expecting

Anirudha Nayak - 3 years, 6 months ago

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@Anirudha Nayak atleast 100 to aana chahiye.

Anish Puthuraya - 3 years, 6 months ago

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@Anish Puthuraya me tooo simple paper though

Anirudha Nayak - 3 years, 6 months ago

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@Anish Puthuraya ans for d question

Anirudha Nayak - 3 years, 6 months ago

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Were there any options? I am asking because as Anish said that the value of the integral has no clear form. Maybe they intended to ask something different??

Sudeep Salgia - 3 years, 6 months ago

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\(\pi(2^{1/2}),2\pi(2^{1/2}),\pi(2+2^{1/2})\) and one other i dont remember them clearly

Anirudha Nayak - 3 years, 6 months ago

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There is no exact formula but use the famous Indian mathematician Ramanujan came up with this better approximation:search Ramanujan circumference of ellipse formula

Mardokay Mosazghi - 3 years, 6 months ago

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