Arc Length of Polar Curves
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- Опубликовано: 15 дек 2024
- This calculus 2 video tutorial explains how to find the arc length of a polar curve.
Area of Parametric Curves: • Area of Parametric Curves
Area of Polar Curves:
• Finding Area In Polar ...
Area Between Polar Curves:
• Finding Area Bounded B...
Arc Length of Polar Curves:
• Arc Length of Polar Cu...
Surface Area of Polar Curves:
• Surface Area of Revolu...
_____________________________
Writing Equations of Ellipses:
• Writing Equations of E...
Eccentricity of an Ellipse:
• Eccentricity of an Ell...
Area of an Ellipse:
• Area of an Ellipse
Circumference of an Ellipse:
• Circumference of an El...
Hyperbolas - Conic Sections:
• Hyperbolas - Conic Sec...
______________________________
Parabolas - Conic Sections:
• Finding The Focus and ...
Polar Equations of Conic Sections:
• Polar Equations of Con...
Calculus 3 - Intro to Vectors:
• Calculus 3 - Intro To ...
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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.
You literally saved my life!!!! i would not have passed calculus without these videos
4:31 270degree should be 3pi/2
Just when I needed this.
Absolutely helpful.
Thanks a lot, sir.
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
your teacher will give you problems that will not require a calculator for the final answer, basically you can leave it in a fraction form
Try to always work less with the calculator. You will be more efficient without it.
I also feel you. It isn't legit until TOCT covers it.
That integral was a beast, I wouldn't expect that on a test. Anything we're there is a radical and trig functions under it scares me.
You can solve the integral as follows:
so under the root you have (1+sinx)^2 + (cosx)^2 (I am writing x instead of theta)
when you expand that you get 1 + 1 + 2sinx (Using (a+b)^2 and (sinx)^2 + (cosx)^2 = 1)
so that's 2 + 2sinx under the root
Thats 2(1 +sinx) under the root
But we can write sinx as 2sin(x/2)cos(x/2), and we can replace 1 with sin^2(x/2) + cos^2(x/2).
So within the bracket we basically have a^2 + b^2 + 2ab where a is sinx/2 and b is cos x/2
SO you can rewrite this as (sin(x/2) + cos(x/2)^2
Then you can cancel the root and its a simple integral.
Hope that helps :P
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.
Did you pass?
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 RUclips TV with the Organic Chemistry Tutor.
Thank you kind sir this will help me a lot since my exam is in just 2h
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.
yh i got 0 aswell not sure how he got 8 tbf...
Same
how can he get 8 guys please help, I found 0 as well.
@@2late4coffee Your calculator needs to be in radians
solution of question no 2.(allow me let delta as x)
sqrt(sin x+1)=[cos(x/2)+sin(x/2)]^2-------->come 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
what is the identity of 2+sin(theta)? Can you please show the manipulation...thank you
Thanks so much. My teacher gave us a long assignment on this and didn't show a single example, how saddening.
can anyone explain why the answer is 8 at 5:34? I've plugged this into an integral calculator and it's showing the answer to be 4pi
what if there's no bounds given?
How do you decide the intervals if it's not given cuz i came across a question "r = 2acostheta" with no intervals given
you have to graph and and figure it out yourself ngl. for the most part it is going to be from 0 to 2pi, but if its something like an incomplete spiral, things will be different
Simply use desmos or your graphing calculator(or your own knowledge, if you're confident) and see over what pi values the graph exists
How do you know that 6 sin theta is sphere
It’s one of those things that you just have to know or memorize.
Yeah just remeber it
no learn it go on RUclips and books and learn how to graph it memorize it then as a shortcut but first understand it.
why you did not calculate the integeral -_-
Always finish the calculations
Thank you ❤
thank you ☺
Thank you!!
My guy
brooooo why don't u solve everything am I meant to type calculator during the exam or prove the way I got the answer
ikr, how do you integrate that last one?
thank u sm man
He makes the hardest math look so easy and simple he's a real Legend
thanks....
جوانە
Stop skipping steps at the end
Thanksss 🤍🌷 you are so helpful for me for my exams🌸🌸🤍...