Arc Length of Polar Curves

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This calculus 2 video tutorial explains how to find the arc length of a polar curve.

Area of Polar Curves:

Area Between Polar Curves:

Arc Length of Polar Curves:

Surface Area of Polar Curves:

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Writing Equations of Ellipses:

Eccentricity of an Ellipse:

Area of an Ellipse:

Circumference of an Ellipse:

Hyperbolas - Conic Sections:

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Parabolas - Conic Sections:

Polar Equations of Conic Sections:

Calculus 3 - Intro to Vectors:

Final Exams and Video Playlists:

Full-Length Videos and Worksheets:

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How tf does this man become a master of every single scholarly concept available at a college. He really is a changing the game of how school is. These videos are better than my professor's lectures and videos, like, if The Organic Chemistry Tutor literally presented me with the same syllabus that my current school gave me, and provided me his videos as we go along the semester, that would be quality schooling right there.

Chrisymcmb

Just when I needed this.
Absolutely helpful.
Thanks a lot, sir.

DiutoAjoku

You literally saved my life!!!! i would not have passed calculus without these videos

littleWormGirl

The only issue I have is, we are not allowed to use calculators in the exam.
Please, can u solve the second example without a calculator? Thanks

DiutoAjoku

Thank you for all your content, however with this subject there should have been full calculations for those who aren't allowed to use calculators during classes or exams. Hard to see the in between of integrating from the final answer.

Sefidicus

Professor Organic Chemistry Tutor, thank you for an excellent video/lecture on the Arc Length of Polar Curves in Calculus Two. The examples/practice problems are simple to follow and understand from start to finish, however the integration of Arc Length of Polar Curves can be problematic. This is an error free video/lecture on YouTube TV with the Organic Chemistry Tutor.

georgesadler

solution of question no 2.(allow me let delta as x)

sqrt(sin from 1+sin2x=[cos x +sin x ]^2

then integration you will get 2sqrt(2)[sin(x/2)-cos(x/2)] from 0 to 2pi

However, if will get 0, and we know that left part and right part are symmetry line.

(i take left part because it is more easier for us to understanding)

so i change it into from pi/2 to 3/2pi and due it just one part, so don't forget multiply with 2

*final*
4sqrt(2)[sin(x/2)-cos(x/2)] from pi/2 to 3pi/2=8

limkenny

Thank you kind sir this will help me a lot since my exam is in just 2h

upsidedownChad

5:20 why you input beta and alpha angle as 2pi and 0, when I try to input, it gives me an answer of 0. But when I apply Pi and 0 as an angle, it results to 4 which times 2 is 8. It also confused me a bit because on other videos, they demonstrated as finding the half part of the cardoid and theb times two.

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