When is the derivative of area the perimeter?

[@transcendental477pi8]

Interesting vid! Enjoy your work. For the questions at the end, we should expect an n-sided regular polygon to approach its inscribed circle in the limit as n->∞, so the area and perimeter of the regular n-gon should approach the usual formula for area and circumference of a circle. For the following, all limits are as n->∞: Perimeter (L'Hopital method): lim Pn(r) = lim 2*n*r*tan(pi/n) = lim (2*r*tan(pi/n)) / (1/n) = (L'Hopital) lim (2*r*sec^2(pi/n)*pi*d/dn(1/n)) / (d/dn(1/n)) = lim 2*r*(1^2)*pi = 2*pi*r, the circumference of its inscribed circle. Area (Series method): lim An(r) = lim n*r^2*tan(pi/n) = lim n*r^2*sin(pi/n)/cos(pi/n) = lim n*r^2*sin(pi/n) = lim n*r^2*(pi/n - (pi/n)^3/3! + (pi/n)^5/5! - (pi/n)^7/7! + .....) = lim r^2(pi - pi^3/(n^2*3!) + pi^5/(n^4*5!) - pi^7/(n^6*7!)+....) = lim r^2*pi = pi*r^2, the area of its inscribed circle. Either L'Hopital or Maclaurin series could be used for the perimeter or area limit calculations...whichever is used is just preference!

[@avi8365]

can't we just say n=1/t and then the limit is just trivial

[@EpicMathTime]

@@avi8365 No, I don't see how that does anything at all.

[@avi8365]

@@EpicMathTime If you put n=1/t then limit changes to lim t to 0 r^2 tan(πt)/t multiply and divide by π lim t to 0 πr^2×tan(πt)/πt now πt goes to 0 as t goes to 0 which means the limit t to 0 tan(πt)/πt=1(standard result) hence total limit is πr^2 the only reason this is a nice variable change is cause we can arrive at a standard result or if you don't remember it the differentiation is pretty easy if one wanna use L'Hopital

[@xriccardo1831]

@@avi8365 we could just use asymptotic equivalences

[@MrRyanroberson1]

the point being that both forms exploit tan(x->0)/x ~ 1

[@crashplague3608]

There is not a single person on earth that would correctly guess your sub count after watching your vids.

[@iwantblueplate4288]

So true. I thought he had at least a million subs.

[@aaronsabela7921]

After reading this comment, subscribing was never a hesitation

[@stumbling]

94,358. How did I do?

[@ultrio325]

overestimated by 20x

[@willnewman9783]

Nice video. Note that this works for a very large class of shapes if you modify it in the correct way: Let R be a subset of the plane. Define A(r)=Area({x| the distance between x and a is less than r, for some point a in R}). Then A'(0)=length of the boundary of R, assuming R is nice enough. And this generalise to higher dimensions too, in the obvious way

[@EpicMathTime]

I gave a talk on this to the undergraduate math majors at my university a few weeks ago, and one of the professors that attended brought up the very same generalization.

[@scotthall5256]

As n->inf, tan(pi/n) -> pi/n (which can be shown by Maclaurin series). Substitute into original equation and solve. So, as n->inf, P(r)=2pi*r and A(r) = pi*r^2 as expected.

[@JamalAhmadMalik]

AH... so right!

[@stumbling]

Math: That only works for circles. Also Math: What if we make everything circles?

[@bobus_mogus]

Woah. Seriously quality content!

[@EpicMathTime]

:)

[@jeffreyh.1436]

An alternate solution: We know that A=rs=rp/2 and also that p and r are both directly proportional to side length. Thus, plugging in p=kr for some constant k yields dA/dr=d(kr^2/2)/dr=kr=p and thus derivative of area is indeed the perimeter. This proof generalizes to all polygons with inscribed circles (i.e. the result holds for polygons like rhombuses as well).

[@bartholomewhalliburton9854]

I can't believe this comment doesn't have more likes! Good observation

[@jonathanratcliff8632]

My man with the quality vid! 10/10

[@RunstarHomer]

Incredible, what a beautiful result. I'm so glad I was introduced to your channel thru Flammable Maths. It's quickly become one of my favorite channels on KZbin. Please, keep up the good work.

[@michaeljburt]

Your videos are amazing. Great proofs, lots of detail, but also concise. Been loving every single one (I started with the binomial theorem one, which absolutely blew my mind). You're an incredible communicator, and I just hope that more people can find your channel.

[@EpicMathTime]

Thank you so much, these comments always make my day.

[@uhbayhue]

This is starting to become one of my favorite channels.

[@albertocalderamorante3597]

I found this channel some days ago and i have to say that this is the most complete and infravalorated channel i have ever seen

[@landsgevaer]

The perimeter is the derivative of the area if the contour of the shape expands outward perpendicularly over a distance dr everywhere when the "size parameter" is increased by dr. That indeed holds when parametrizing a regular polygon by the radius of the inner circle (or all polygons for which all sides touch that circle).

[@joelabraham708]

do you have a way to answer this classic algebra question: "A rectangle’s base is 8 cm longer than its height. The rectangle’s area is 84 cm2. what are the rectangle’s dimensions?" using calculus and the fact that perimeter is the derivative of area? I'm wondering if there might be a slicker|easier way than just the algebra that might be informative about the fundamental theorem of calculus?

[@landsgevaer]

@@joelabraham708 You can derive the area 2r*(2r+8) from an integral of the perimeter INT(16+8r)dr, and then the calculus remains the same. Little insight gained though.

[@joelabraham708]

@@landsgevaer WOW - I can't believe how quickly you replied! I've been toying with 'every rectangle has a diagonal that is y=(h/b)x and the area would be 2*int[(h/b)x]dx ... but I can't get it to the point of "SEE?! the derivative _is_ the perimeter, cuz, you know?! that's what that MEANS!!! the area is increasing by a piece of rectangle that is so thin it's literally the perimeter!" kind of thing - aaah, I guess I can't explain it well enough

[@eliyasne9695]

Great video! I took a slightly different approach to the proof. I think that this approach is better and more beautiful because its much more direct and it can generalise very easily. Instead of defining a regular polygon i just defined a polygon that has an inscribed circle. Then, just like you, i divided it into many triangles (each composed of a vertex, a neerby point of contact with the inscribed circle and the circles center center). Now, instead of trying to write down the formulas for its area and its outer side length i just treated it like its an area under a graph of some function (were the X axis is the edg from the center to the point of contact and the function is the outer side length). Now i can just use the fundamental theorem of calculus to automatically arrive at the conclusion we conjectured. So eventually, we can see that this theorem is just a special case of the fundamental theorem of calculus. Using this method we can generalise it to 3d polyhedra that have an inscribed sphere, too(with a 3d version of the FTOC). In fact, we can generalise it to any hyperpolyhedron of any dimensionality as long as it has an inscribed hyper sphere. I think it might even be possible to generalise it to some shapes in curved spaces.

[@heatshield]

Hey man. 👊 Steve sent me. Looking forward to your work here.

[@viplavbhende]

This is crazily interesting .I personally loved it so much . Thanks for the good work .

[@dvir-ross]

Great video! I was asking myself about this a while ago. Thanks!

[@stydras3380]

A god that gives us almighty theorems to prove? Some called George Frederic Berny Riemann?

[@RunstarHomer]

Pierre de Fermat

[@pbj4184]

@@RunstarHomer Lol I was going to write that

[@Govgrav506]

@@pbj4184 but you ran out of room didn't you

[@theultimatereductionist7592]

Just heard of this channel on 20181217 by someone named Steve at the NonSequitur channel. Steve mentioned having a math PhD friend.

[@jkid1134]

I’ll say, just from the circle example at the beginning (you know to be honest, just from the video title, when I was trying to think about how to get the area and perimeter of all n-gons in terms of the same variable), I as thinking about using either of the radii of the polygons. I was pretty surprised when you started off very specifically, on squares, and on side lengths. But I suppose that’s how it happened.

[@EpicMathTime]

Check out some of my math talk (sequences and series) with Steve Mcrae and Dr. Landon Curt Noll kzbin.info/www/bejne/h6m5mJeobMigbZI

[@williamrutherford553]

The first thing I thought when I saw those general formulas was, what about when n goes to infinity? You did start out with a circle, so it would be important to show those formulas work for a circle!

[@LucasFerreira-hy4sn]

wow dude,your videos are AMAZING.You have nothing compared to the number of subs you deserve.Now I wanna see the relation between volume and surface area.

[@Nobody-Nowhere-Nothing]

Nice remake of your very first vid my man. I like that you strive to make the videos interactive and (hopefully) force us to think. Best way to learn math is to do math. Keep doing what you're doing and you'll be huge.

[@TechClive]

Absolute chad

[@Nobody-Nowhere-Nothing]

Alpha Chad

[@Jack_Callcott_AU]

Great video. Now we know that r (the radius of the inscribed circle) is an important number for any regular polygon. There is a little formula about incircirles ( i.e. inscribed circles ) for any triangle which I really like, that is Area =r*p/2 with p the perimeter. This little fact leads to a proof of Heron's Formula.

[@marccowan3585]

Trivial - we note that the conjecture is true for circles. Then we note that the inscribed circle has perimeter and area directly proportional to the perimeter and area of the polygon we are concerned with, and the result follows.

[@kikosilva96]

Great video, you need to upload more, i'm running out of videos to watch here!

[@EpicMathTime]

I will be soon bro!

[@SundararajanSrinivasan]

This works for more than just regular polygons. It works for any triangle, a few quadrilaterals like rhombi, trapeziums. I think it can be generalized for ANY polygon. If you could try to check that it out it would be really cool :) Edit: I just managed to prove it, it works for any polygon for which an incircle can be described. Can I now give this as a challenge to you? :D

[@whiteyplaysmighty8503]

But wouldn't the number of inscribed circles have to be altered for certain shapes? Such as for a trapezoid of known dimensions, say height 1/2x and rectangular length of x (shorter side), wouldn't you need two inscribed circles? The process of finding this new measurement r would be the same, but imply 2 inscribed circles, thus r=ax=2*(some number)*x?

[@whiteyplaysmighty8503]

Or alter the inscribed shape, such as an elliptical shape in a rectangle?

[@SundararajanSrinivasan]

@WhiteyPlaysMighty I meant when only one circle could be inscribed. I am not too sure what happens when figures other than circles are used

[@whiteyplaysmighty8503]

@@SundararajanSrinivasan Ok, I wasn't sure if I missed something. I worked it would for the aforementioned trapezoid, but I had to assume some dimensions so it was more of a mental practice than anything rigorous.

[@SundararajanSrinivasan]

@WhiteyPlaysMighty how did you take the derivative wrt the inradius for the trapezoid?

[@desrepeerc206]

Amazing video!

[@biddu2683]

Fabulous!

[@malexmartinez4007]

Awesome!

[@its_robbietime1333]

that statement in the start of this video is why i had such a problem with geometry 1 because my teacher really wouldnt take the time to actually teach us how to do the thing in question she just gave a theorem and said prove it like what i dont even know how and my school was pretty stupid like why do i need to take geometry and trig before im allowed to calc 1 like i wouldnt even need to bother with calc 1 i could just skip straight to calc 2 it makes no sense to me

[@joshuamariano1663]

Do you write in reverse? Or do you write in a mirror then you flip it in post?

[@jkid1134]

My guess is that the entire clip is mirrored, and that he writes normally.

[@kreglfromworld]

He's left handed, I hope that answers your question

[@joejo1245]

He did a Q&A he records the video then just flips it when he edits.

[@skeletonrowdie1768]

awesome video man, super creative.

[@dominguezander8523]

Very original idea and very nice video. The answer to your last question, by the way, is pi*r^2.

[@David-km2ie]

This an amazing such a cool result

[@epalegmail]

May I ask which equipment do you use to write on the video screen like that, and how do you write in everything straight even though you are behind the "chalk board"? I find that very dope

[@rorygo456]

epalegmail i think the camera is inverted

[@ottothelord5755]

This was a great video!

[@pk823456]

I know this is a piddly detail, but the pictures at 9:37 don't match up with what you're saying, as the circles are on the outside (circumscribed I guess, not positive on the word), where as with the basic examples and wording of the proof the circles were inscribed. Cool video nonetheless really crisp.

[@pk823456]

So using the circumscribed circles from the pictures you get P(r) = 2n*r*sin(pi/2) and A(r) = (1/2)*n*(r^2)*sin(2*pi/n).

[@cunningham.s_law]

would it work with an inscribed ellipse in a rectangle? what about 3d shapes and relationship with volume and surface area?

[@David-km2ie]

Sounds interesting

[@EpicMathTime]

The analogue will work in all dimensions. As for other shapes, I believe similar conclusions can be made, but I haven't thought about it in a while.

[@ayanbanerjee7259]

Wow great

[@Mahmood42978]

7:43 ....is actually on one of the Math GRE practice exams.

[@Sooboor]

I know it's late, but still. Fitrst, let's calculate the limit of n*tan(pi/n) when n --> infinity (Because There's no math editor in the comments section, I'll just right lim). lim n*tan(pi/n) = lim n*sin(pi/n)/cos(pi/n) By Cauchy's definition of the limit of a function of n when n tends to infinity, (The limit is L when n tends to infinity, if for every epsilon>0, exists a real number M such that for every x>M, |f(x)-L|0. That means that 1/n is defined. Which means that: lim n*tan(pi/n) = lim sin(pi/n)/(cos(pi/n)/n) = lim sin(pi/n)/(pi/n*cos(pi/n)/pi) = lim sin(pi/n)/(pi/n)*pi/cos(pi/n) lim pi/n = 0 (when n tends to infinity) So, because lim sinx/x = 1 when x tends to zero: lim sin(pi/n)/(pi/n) (when n tends to infinity) = lim sinx/x (when x tends to zero) = 1 cosx is continuous for every x, which means that it is continuous at x=0. So, lim cos(pi/n) = cos(0) = 1 So, lim pi/cos(pi/n) = pi/1 = pi And because of arithmetic with limits, lim sin(pi/n)/(pi/n) * lim pi/cos(pi/n) = lim sin(pi/n)/(pi/n)*1/(cos(pi/n)/pi) = lim sin(pi/n)/(pi/n*cos(pi/n)/pi) = lim n*tan(pi/n) So, lim n*tan(pi/n) = pi (when n tends to infinity) So, lim nr^2*tan(pi/n) = r^2*lim n*tan(pi/n) = r^2*pi = pi*r^2 = Area of a circle with radius = r And, lim 2nr*tan(pi/n) = 2r*lim n*tan(pi/n) = 2r*pi = 2pi*r = circumference of a circle with radius = r

[@sujathasuresh8733]

Sir, is it true for any polygon.... Can we prove that the derivative of area of any polygon equal to its perimeter if the area and perimeter are expressed in terms of the radius of its incircle?

[@99selfmade21]

Question! : the last one the limit n->inf tan(pi/n)*n = pi easyto show with l'hôptial, but what if i am using degree? so the limit n->inf tan(180/n)*n = pi. How can i prove that while staying in degree?

[@EpicMathTime]

If you are in degrees, the derivatives of the trigonometric functions are not the same anymore. Account for that, and you can stay in degrees and solve the problem.

[@99selfmade21]

@@EpicMathTime what is then the derivative of tan x? and do i even get them? xD

[@99selfmade21]

Nevermind got it myself 😇

[@nadiyayasmeen3928]

Area and perimeter of a circle are the answers to the question asked at the end

[@juanfa98]

Nice

[@jaymiller109]

I published this result - and the 3-dimensional analog - in 1977 in the Two Year College Math Journal.

[@EpicMathTime]

Got a link? I'd love to read it. This problem is still one of my best math memories.

[@jaymiller109]

@@EpicMathTime Sorry, I have only reprints.

[@unpythonic]

@@EpicMathTime Found it on JSTOR: www.jstor.org/stable/3026559

[@bobmarley9905]

When he said it’s obvious I literally got confused for sec... but then realize it’s obvious cuz it’s turning into a circle😂😂

[@benjaminbrady2385]

I thought that benevolent gods name was Euler?

[@nikogruben9573]

I really enjoyed the vid, quality content! Although I did wonder, isn’t r the side lengths of the isosceles triangle, not the height?

[@caladbolg8666]

Keep in mind that r is the radius of the _inscribed_ circle, so it'll be the height of the characteristic triangle. The side lengths of the triangles would be the radii of the _circumscribed_ circle :)

[@nikogruben9573]

Caladbolg ooh, thanks

[@LovepreetSingh-jv4uj]

Hey brother..WHAT IF WE PUT n=1.5 What does it actually means?

[@gnikola2013]

Ok, so something cracy happened to me. I didn't watch the whole video because I wanted to see if I could get to the result myself. And I could. Now, the surprising part is, that I came up with exactly the same notation for area and perimeter, An(r) and Pn(r), although I didn't use the bar. I find that a bit crazy.

[@hugocarlsson3628]

Amazing video

[@imravishankar86]

what if there is two variable in Area like A= l*b

[@hank1519]

Nice job! I am sure that you are aware that the derivative of formula for the volume of a sphere gives the formula for its surface area.

[@Smitology]

Yes, and this theorem generalises to regular polyhedra too. The "radius" of such a regular polyhedron is given by the perpendicular distance from the centre to one of its faces

[@aakashsingh5288]

Limit goes to area and perimeter of circle of radius r

[@BlueGiant69202]

Is this related to the definition of the vector derivative and the multi-dimensional generalization of the fundamental theorem of Calculus? The perimeter is a 1-dimensional boundary. The area is 2-dimensional. geocalc.clas.asu.edu/pdf/SIMP_CAL.pdf geocalc.clas.asu.edu/pdf/MultCalc.pdf geometry.mrao.cam.ac.uk/2015/11/ga2015-lecture-6/ www.mathopenref.com/polygonincircle.html

[@alimghazzawi3700]

I think if u take these formula with the length from a half of a side to a center of pentagon or hexagon .etc It might be true

[@sujathasuresh8733]

Sir, u have drawn circum circle of regular polygons instead of drawing inscribed circle to prove the last theorem... isn't it wrong...

[@EpicMathTime]

Do you mean those three graphics that appear? Yes, they aren't correct.

[@sujathasuresh8733]

Ok...sir 👍

[@meltossmedia]

Damn, now I want to be a Mathematician

[@EpicMathTime]

hell yea brah, do it

[@meltossmedia]

@@EpicMathTime Dotson already got to me fam, and I'm a year deep into Physics. I'm about to take a Proofs class though, so we'll see what happens then.

[@melrobles8587]

at 9:40 you switch to circumscribed circles not inscribed circles Does the result still hold? @ThePKFactor also spotted this as well

[@EpicMathTime]

I really just meant to demonstrate the triangulation there, the circles can essentially be ignored.

[@melrobles8587]

@@EpicMathTime ohhhhh actually that clears up my confusion now! Anyway the Real Analysis part of your 'if math subjects were people' vid had me rolling. That's what had me subscribed lmao

[@billygarvey633]

Are you writing backwards or left handed? Or something else? I must know

[@camdenwhite4657]

apothem!

[@aggelosgkekas3113]

We compute the limit as n→+∞ (which by the order of the flammy brotherhood I will denote as "L") of n*tan(π/n). We have: L[n*tan(π/n)] = L[n*sin(π/n)/cos(π/n)] = L[1/cos(π/n)]*L[n*sin(π/n)] = L[1/cos(π/n)]*L[π*sin(π/n)/(π/n)] = Κ t=π/n n→+∞: t→0 THUS: K = limt→0[cost]*limt→0[πsint/t] = 1*π*1 = π We plug that in the given expressions (the other terms are constant in terms of n) and we arrive at the formulas for the circle And it was much harder than I expected to type this in a normal keyboard

[@dominicellis1867]

I knew that the tangent line and tangent are deeply and intrinsically connected this proves that the circle the triangle dx dy and |dx + dy| encloses connects the the angle theta opposite of dy to the fundamental theorem of calculus bridging the gap between geometry and the absolute differential nature of the ever changing universe

[@Taricus]

10:30 Where'd the 2 go? @_@

[@jomanaehab4200]

He divided the angle in order to have a right angled traingle so that we can get tan of it

[@DiegoMathemagician]

How is this theorem called?? (give it a name)

[@EpicMathTime]

The Broadsword Theorem

[@DiegoMathemagician]

@@EpicMathTime why a broadsword what the hell xD PD: great video, honestly

[@DiegoMathemagician]

You know, i found a theorem named "Magic wand theorem", close enough

[@alimghazzawi3700]

Now you are writing in something but we are seeing the video from the opposite side so we are supposed to see the writings reversed so I think either u r reversing the sides of the video which is most likely or u r writing in reverse of sides So I think u r left handed (observation )😉☺️ About the video I think it’s a nice topic