Verify by integration that the perimeter of a circle is 2 (pi) r

[@basqye9]

Great video my friend. Just wanted to say you have the best penmanship of any math youtuber I've seen. It's an undervalued skill but a critical one.

[@kingbeauregard]

Cole from Cole's World of Mathematics has terrific penmanship too, but she's using paper, not a chalkboard. I think our man Prime is operating at a disadvantage.

[@ThenSaidHeUntoThem]

​@@kingbeauregardThat's so kind of you ❤

[@JeanMarieGalliot]

Too bad there is no demonstration of de general formula for an arc length, which is the most interesting part

[@thorhilda]

@PrimeNewtons I am not disputing that this exercise is valid for practicing integration techniques. However, it appears to me that it does not adequately demonstrate the circumference of a circle. The issue lies in the subtle introduction of the circumference’s definition through trigonometric substitution. Trigonometric functions inherently relate to the ratio between the radius and circumference. This relationship is fundamental to defining the number π (pi). While showing that the circumference is a multiple of the radius using their ratio, it does not constitute a substantial proof. It’s akin to trivially asserting that the ratio is proven by the ratio itself. In other words, the trig substitution, at 12:09, implies already what we are trying to demonstrate. 2πr is already included as an axiom. In a sneaky convoluted way, we are saying, "If we assume C=2πr, lets show that C=2πr, in twelve easy steps that includes injecting the ratio C/r = 2π." I think you need to use the fondamental definition of the integral, the limit, instead.

[@ingiford175]

You can use Euler substitution instead.... Never tried it but it should work Edit: By the time you get to arc length formula, derivatives of sin/cos are well defined and should not be an issue.

[@roihemed5632]

Yeah it seemed to me like circular reasoning as well

[@thorhilda]

@@roihemed5632 LOL

[@thorhilda]

@@roihemed5632 😄

[@KahlieNiven]

totally agreed.

[@steveb.8491]

I never thought Calculus could be so soothing. You make it look so elegant and fantastic! Thank you! Just subscribed!

[@_Lucian_V]

Salutations, Prime Newtons! I've been engrossed in your captivating content for a mere couple of days, and I must confess, your intellectual prowess never fails to leave me astounded. Consider me an ardent disciple, for your channel has undoubtedly earned itself a devoted subscriber.

[@PrimeNewtons]

Wow, thank you!

[@tonyscott1658]

I like this guy! Elegant! He's a good teacher!

[@sameermansour1659]

You are great man ! your explanation is very simple , but so pure!

[@beeswax4909]

Brilliant Explanation ... No words ... just on the point ... Thank you ...

[@alocin110]

You remind me of my great teacher Mr. Salahudeen. He left our college in Karachi, Pakistan, for Nigeria to start a teaching profession there. The way you are explaning things here makes me believe you are his student. Anyway, your method of teaching is what is the true purpose of education: "learning, not just solving". Thank you for sharing your knowledge and teaching the universal language of the universe. I subscribed to your channel: just because you are a true image of my inspirer.

[@jadenredd]

your channel is rapidly growing unc! glad to say i was here supporting you early 🤣🤣

[@blackholebusterisgood6057]

I really didn't understand the calculus part but I knew about the variables, the perimeter, and the arc length. This video was great to watch, keep it up!

[@johnroberts7529]

Another tour de force on the Maths front ...thanks again. I have let some of your recent videos slip by; this is a nice one to break me back in again. 😊

[@samkim5576]

congratulations Newton. You are a great mathematics teacher. I have learnt a lot from you even though my undergraduate degree was in Mathematics and Chemistry. I taught mathematic at higher school level for three years. Well done my brother. Please visit Nairobi, Kenya

[@ingiford175]

This is near the heart of Spivak's Calc. He defines pi as the area of a circle, uses that to define an area, and uses that area to define cosine and then sin, and through a cleaver use of inverse functions he can get the derivative of those functions without the limit of sin(x)/x = 1 as x goes to 0. Then later he proves that the pi he defined is the same as the pi that is normal by using this.

[@mishradevansh308]

😊 thanks for enhancing our knowledge in calculus. Although these questions and topics are not currently in my academic course. But I still watches your vdo to get extra knowledge .the reason might be I can't sleep without solving maths questions. If I do so then I wouldn't be able to meet the sleep . Thes days. Whenever I get free time I used to watch ur vdos nd it helps alot ❤

[@richardawoyade8019]

Best explanation I've seen in years

[@steflabbe]

J'adore cette démonstration. Celle du volume de la boule est sympa aussi 😅 Merci pour tes supers vidéos 👍

[@KahlieNiven]

sauf que ce n'est pas une démonstration, juste un raisonnement circulaire.... sin et cos sont de facto definis par rapport à pi qui est défini comme rapport entre périmètre et diamètre.

[@silverhammer7779]

I wish I had you as an instructor for freshman calculus in engineering school. You don't just present it, you bring it to life.

[@JeanMarieGalliot]

Brillant.just wondering if it's not a kind of circular definition because arcsin already encompass PI in its definition.

[@LeonKayombo-yk4tc]

Great my sensei Aksanti. Et du courage!

[@CAG2]

Sorry to be pedantic, but isn't the circumference 2pi by definition? pi is defined as the ratio of the circumference of a circle to its diameter. I think this video instead shows that calculus is consistent, in that our definitions of arc length defined using integrals and derivatives coincide with what we previously knew, which is beautiful in itself.

[@Krmpfpks]

Yes, but it would have been helpful if the teacher had stated the objective first.

[@cmilkau]

Every "proof" of the circumference of a circle is necessarily circular (lovely pun haha), because the circumference of a circle is the *definition* of π.

[@kingbeauregard]

Each little chunk of the curve can be viewed as the hypotenuse of a little right triangle, where the horizontal side is dx, the vertical side is dy, and the hypotenuse is sqrt((dx)^2 + (dy)^2). That's just Pythagorean theorem. You can rewrite that as dx*sqrt(1 + (dy/dx)^2), and that's where we get the formula.

[@hishamhamed5033]

How about using cylindrical coordinates, where the element of length at radius r normal to the radius would be just r dθ. Integrating from 0 to 2π gets you there immediately. Cheers.

[@avantgarde04]

But we prove L= rd(theta) after we know that perimeter of a circle is 2πr By the unitary method we can say if Circle subtends an arc of 2πr length in 2π radians then for theta radians it will subtend (theta)×(2πr/2π) or r(theta)

[@hishamhamed5033]

@@avantgarde04 Interesting. Must have missed that. Thanks.

[@sobolzeev]

If you are ready to play vicious cycle, you may instantly parametrise circle as x=r cosθ, y = r sinθ. Then the arc length element ds = √[(x')² + (y')² ]dθ = rdθ. And you should only apply the data that a unit circle is 2π rad to finish the "proof".

[@philippecanepa4509]

Sorry, but can’t find the video you mentioned earlier regarding the perimeter.

[@MohammedAhmed-ws3ho]

How do you establish the following equations related to Distance, Velocity,, and Time using Calculus ? • a) S = ut + 1/2*at^2 where S is the Distance, u the Initial velocity, a is acceleration due to gravity and t is the time • b) V^2 - U^2 = 2aS where U is the initial velocity, and V the final velocity where

[@h1a8]

The proof is not valid (it contains circular reasoning). Arcsin(1) = pi/2 needs to be proven As using it creates a circular argument of why the arc length of a quarter of a unit circle is pi/2. Remember the definition of the radian. A radian is an angle whose arc is equal in length as the radius of the circle it sweeps out. Basically the proof directly follows from assumption that arcsin(1) = pi/2 without any Calculus.

[@Krmpfpks]

I don‘t really understand the reason to do it that way, if it‘s just for coming up with 2*pi*r then there’s way simpler versions to do this. If it‘s for learning this particular method or some of the techniques in this video it should have been stated at the beginning of the video?

[@Marina81505]

Amazing solution!❤👌

[@Dr_piFrog]

Why don't you use a polar coordinate system?

[@wes9627]

Nice. How about the perimeter of an ellipse represented by (x/a)^2+(y/b)^2=1 where x=a*cosθ and y=b*sinθ? Unfortunately there is no closed-form solution except when a=b, and the ellipse becomes a circle.

[@sebtek9158]

That's very interesting! But something bugs me: the formulas for sin(x), cos(x) etc. are based on the fact the perimeter of a circle is Pi times diameter. So, proving the perimeter is that, using sin or cos, is circular reasonning, no?

[@BleuSquid]

I'm here for the unintentional pun - "circular" reasoning 🤣

[@AdityaPawar803]

I exactly thought that, I am not even a mathematician but I understand that the value of pi by definition comes from the perimeter divided by diameter so any other way to prove this will be a circular way, though I think the point is to have fun with integration.

[@SuperDeadparrot]

ds = r * d theta where ds is the length of an infinitesimal chunk of arc length and d theta is the infinitesimal change of angle. Now integrate over the entire circle meaning integrate [ theta, 0, 2pi ]. We get C = 2pi * r as needed.

[@kappasphere]

When removing the r² from the denominator, I think it's easier to first do the u-substitution x=r u, dx=r du, because this makes factoring out the r² easier. This way you get L=int_0^1 r/sqrt(1-u²) du as the result, and can do your trig substitution on that: sin(t)=u, cos(t)dt=du L= int_0^π/2 r cos(t)/sqrt(1-sin²(t)) dt = int_0^π/2 r cos(t)/sqrt(cos²(t)) dt = int_0^π/2 r dt = π r/2 Take that x4, and you get as the full circumference of the circle: 2π

[@MohGhanem126]

I demonstrate the circumference (perimeter) and area of a circle formulas in a most simple and different way, using the SUMS instead of integration.

[@giorgioripani8469]

Amazing video my friend. As much as all your other lessons. The only iddue I can find is that you are using arsin(1) = pi/2 (the actual definition of the pength of the CD ircle perimeter)which is exactly what you are tryin to prove 😅

[@surendrakverma555]

Very good. Thanks Sir 👍

[@jperez7893]

A good technique to drive the point is to show that you should be able to derive the circumference by using the first and second quadrants, or even better, arbitrarily derive it using the 3rd quadrant

[@JSSTyger]

I have a problem for you to try to solve/show. I have an N-sided polygon fully inscribed in a circle with all side lengths equal. Something amazing happens when the number of side lengths approaches infinity. Can you prove that π = limit[(N/2)sin(360°/N) as N-->∞)?

[@PrimeNewtons]

I like this question. I did it in my head already. I may do a video on it. Good for Calc 1 students too.

[@upalsengupta5878]

Thanks for the excellent video, but can we find the same using polar equation s=r(.theta) , then differentiate dr= d(theta), and then integrating from 0 to π/2

[@upalsengupta5878]

Sorry ds= r. d(theta) 😊😊

[@PrimeNewtons]

Yes, definitely

[@МаксимАндреев-щ7б]

x=r cos(t), y=r sin(t), 0

[@satyam-isical]

Mathematicians are very funny First they define pi as circumference/diameter Then proves this😂

[@chemsdinesidha5254]

Wow! Magnifique vraiment...

[@juditesouza]

Why you did not used polar coordinates?

[@PrimeNewtons]

Not there yet

[@vincentmudimeli4430]

Good day Mr Newton where can I study maths and be like you I'm from South africa and I'm your great fan and inspired how so explain the concepts and approach to any problem hpw could I be like you I know imust be the best version of myself

[@simondancer]

It's a great video but I can't help but wonder that we only know that arcsin(1) is Pi/2 because we assume that the circumference of a circle is 2Pir. It's a bit of a "circular" definition.

[@Sg190th]

Yup. Calculus is the math that opens your eyes on why you have formulas used as a kid. Have you done the one with volume of a sphere's derivative being the surface area?

[@maxime9636]

Thank U so much 👍❤🙏🌹🌹🌹

[@_HAKSOZ]

I used a similar method to find an integral for the circumference of an ellipse, but i cannot solve the integral and im not even sure if it's solvable

[@hishamhamed5033]

Apparently not. The accurate formula is an infinite series solution. I've been there I'm afraid.

[@clancyschannel5965]

Yeah, we cant really use this formula on ellipses since it creates an integral which does not have an elementary function. The problem is that there is no exact formula for a perimeter of an ellipse which is actually sad :(

[@Necrozene]

By definition. Or using polar coordinates. Your way is so very difficult.

[@MounirCPIBOURABA]

I think if we do the maths in the right order the definition of cos and sin of any angel became after defined pi from perimettre of cercel The definition of pi is the ratio of cercel on daimettre Do you agree ?

[@jeffp.6092]

Ah yes I remember doing this in calculus, this is a nice refresher.

[@wonghonkongjames4495]

Good Morning，Sir， But this can be derived from the definition of Pi : Circumference of a Circle ÷ (2r) = 丌

[@edwardhaynes8857]

Practical calculation of the approx. perimeter is the square root of 10 X diameter of the circle.

[@MarcoMate87]

Excellent job!

[@Stacee-jx1yz]

The Paradox of the Circular Plane Contradictory: In Euclidean Plane Geometry, defining a circle as the set of points equidistant from a center point is paradoxically circular: C = {(x,y) : sqrt((x-a)^2 + (y-b)^2) = r} (Circle of radius r) This defines C using the algebraic distance function invoking C itself. Non-Contradictory Possibility: Infinitesimal Pluritopic Homotopy Theory C = {p : ∃q ∈ S1, p =r q} (Circle as monadic group quotient) Tπ = ⨀p ⨂q Γp,q(r) (Winding homotopy over relations) Defining circles C topologically as quotients of the monadic group S1 by pluralistic infinitesimal monadic relations Γp,q avoids circularity using homotopic methods.

[@KahlieNiven]

I got a small issue with this problem .... by definition pi IS the ratio between circumference and diameter (2r). How to prove a definition ? yes I know can use infinitesimal calculations and integrals , but still based on a defined value of pi. (sin and cos are defined from pi as pi is set to be the flat angle in radian) in other words, this was a circular reasoning about circles. PS : the work on integrals is still worth to be watched

[@temptemp563]

Sorry I stopped listening when you looked at the camera and smiled ...

[@glorrin]

This is a great video and explains it perfectly. I wonder if we can switch coordinate from the start maybe by saying that a circle is a function that takes the angle theta and return r f(theta) = r and we want to trace this function from 0 to 2pi and just do integral of r dtheta with theta from 0 to 2pi 2pi:0S r*dtheta [writing integrals is hard I am sorry about that] and result is [r*theta]2pi:0 which is 2pi*r But I dont remember why (or even if) what I just did is correct

[@xanderlastname3281]

Dumb question, wouldnt the integral of a circle be 0 since half of it is above the x axis and halve of it is below (they cancel out)?

[@PrimeNewtons]

We are looking for length and not area. Even if area, we are multiplying a positive by 4. So no zero.

[@xanderlastname3281]

@@PrimeNewtons ah

[@Jmcc150]

Since this is the definition of pi, nothing is being proved. It is literally a circular argument!

[@kimba381]

Very well done! I have always been suspicious of this one, as the fact that arcsin(1) = π is derived from C =2 π R. It seems circular to me. OO look, accidental pun.

[@vitotozzi1972]

you're a damned machine!!!! Awesome!!!

[@sobolzeev]

A vicious cycle detected. How do you know arcsin(1) before knowing that the quarter of a circle equals π/2?

[@GeoRedtick]

Since pi is defined as the number of diameters needed to go around a circle and the diameter is 2r then the circumference is an obviously pi2r.

[@chrisengland5523]

No, very clever, but no. I followed the calculation carefully and must admit, I admired the way he handled the integration. But at 15:15 he says "Arcsin(1) = π/2" (in radians). This is true, but it follows from the definition of π, which equals the circle circumference divided by its diameter. So basically, he's using something that follows from the definition of π to prove the definition of π. It's a circular argument (no pun intended). More generally, you can't prove a definition.

[@saarike]

Absolut wonderful!!!!!!

[@stevo4212]

I think your proof is circular because the sign function was only understood knowing that the circumference of a circle is 2pir.

[@yessinegebssi162]

wow you are amazing 😍😍😍 everyone should subscribe to this channel

[@benchapple1583]

Very beautifully presented. Unfortunately radians are defined by the presumption that the perimeter of a circle is 2πr. Thus you have a circular argument (no pun intended) and it's invalid. This is something I bothered all my maths professors with when reading maths at the OU. None of them had a solution other than something similar to yours and all were invalid. In fairness to my teachers they admitted this when challenged, they simply hadn't realised the error. I scoured the internet in vain. Then I decided to prove that for a circle Perimeter P and Radius R-> P is proportional to R. After 3 months of driving myself insane I found a solution. It came to me in a dream- something that perhaps only a mathematician would believe. I have never seen a valid proof other than mine, although one surely exists somewhere.

[@alithemathguy]

Use parametrics, x=Rcos t, y=Rsin t. Line integral from 0 to 2pi is 2piR

[@PrimeNewtons]

When we get there

[@ravichanana3148]

Differentiate the area w.r.t. the radius.

[@knupug]

Damn! That was exciting. Not joking. I'm a math nerd. What can I say.

[@serae4060]

I thought you don't need to proof this as this is literally the definition of Pi Pi=circumference/diameter=circumference/(2r)

[@mwilamusonda9687]

Brilliant

[@nyonkavincenttafeli7002]

What if the centre of the circle is not at the origin?

[@Nickesponja]

Do a change of coordinates first so it does sit at the origin

[@nyonkavincenttafeli7002]

@@Nickesponja so the demonstration is for a particular case(the circle must be at the origin). I say so cause the general equation of a circle is not y^2 + x^2 = r^2 as mentioned in the video

[@ShanthiVeluswamy]

Why prove it using calculus when you can do it using basic algebra Let C,D,R denote the circumference, diameter and the radius of a circle respectively. π=C/D D•π=C (2R)π=C 2πR=C

[@jean-paulbertuol4390]

Great !

[@HadarSHA256]

what a nice circular proof...

[@jwvdvuurst]

Luckely you do not have to do it everytime you need to know the circumference of a circle...

[@keithrobinson2941]

Sweet!

[@mauriziobelluomini]

if x goes from" 0 "to "r" teta goes from "0 "to pi/2

[@muzzi8520]

Me 11 years old: Say what??????????

[@whenlifeisintoxicated3048]

I am sorry for the teacher here. It is stupidity of highest order. How one gets Arcsin(1)=pi/2!! Think about it. Circular reasoning validated.