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!
@avi83655 жыл бұрын
can't we just say n=1/t and then the limit is just trivial
@EpicMathTime5 жыл бұрын
@@avi8365 No, I don't see how that does anything at all.
@avi83655 жыл бұрын
@@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
@xriccardo18315 жыл бұрын
@@avi8365 we could just use asymptotic equivalences
@MrRyanroberson14 жыл бұрын
the point being that both forms exploit tan(x->0)/x ~ 1
@crashplague36085 жыл бұрын
There is not a single person on earth that would correctly guess your sub count after watching your vids.
@iwantblueplate42885 жыл бұрын
So true. I thought he had at least a million subs.
@aaronsabela79214 жыл бұрын
After reading this comment, subscribing was never a hesitation
@stumbling3 жыл бұрын
94,358. How did I do?
@ultrio3252 жыл бұрын
overestimated by 20x
@willnewman97835 жыл бұрын
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
@EpicMathTime5 жыл бұрын
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.
@scotthall52565 жыл бұрын
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.
@JamalAhmadMalik5 жыл бұрын
AH... so right!
@stumbling4 жыл бұрын
Math: That only works for circles. Also Math: What if we make everything circles?
@bobus_mogus5 жыл бұрын
Woah. Seriously quality content!
@EpicMathTime5 жыл бұрын
:)
@jeffreyh.14362 жыл бұрын
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 Жыл бұрын
I can't believe this comment doesn't have more likes! Good observation
@jonathanratcliff86325 жыл бұрын
My man with the quality vid! 10/10
@RunstarHomer4 жыл бұрын
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.
@michaeljburt3 жыл бұрын
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.
@EpicMathTime3 жыл бұрын
Thank you so much, these comments always make my day.
@uhbayhue3 жыл бұрын
This is starting to become one of my favorite channels.
@albertocalderamorante35975 жыл бұрын
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
@landsgevaer4 жыл бұрын
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).
@joelabraham7084 жыл бұрын
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?
@landsgevaer4 жыл бұрын
@@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.
@joelabraham7084 жыл бұрын
@@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
@eliyasne96954 жыл бұрын
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.
@heatshield5 жыл бұрын
Hey man. 👊 Steve sent me. Looking forward to your work here.
@viplavbhende3 жыл бұрын
This is crazily interesting .I personally loved it so much . Thanks for the good work .
@dvir-ross2 жыл бұрын
Great video! I was asking myself about this a while ago. Thanks!
@stydras33805 жыл бұрын
A god that gives us almighty theorems to prove? Some called George Frederic Berny Riemann?
@RunstarHomer4 жыл бұрын
Pierre de Fermat
@pbj41844 жыл бұрын
@@RunstarHomer Lol I was going to write that
@Govgrav5063 жыл бұрын
@@pbj4184 but you ran out of room didn't you
@theultimatereductionist75925 жыл бұрын
Just heard of this channel on 20181217 by someone named Steve at the NonSequitur channel. Steve mentioned having a math PhD friend.
@jkid11345 жыл бұрын
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.
@EpicMathTime5 жыл бұрын
Check out some of my math talk (sequences and series) with Steve Mcrae and Dr. Landon Curt Noll kzbin.info/www/bejne/h6m5mJeobMigbZI
@williamrutherford5532 жыл бұрын
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-hy4sn5 жыл бұрын
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-Nothing5 жыл бұрын
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.
@TechClive5 жыл бұрын
Absolute chad
@Nobody-Nowhere-Nothing5 жыл бұрын
Alpha Chad
@Jack_Callcott_AU3 жыл бұрын
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.
@marccowan35853 жыл бұрын
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.
@kikosilva965 жыл бұрын
Great video, you need to upload more, i'm running out of videos to watch here!
@EpicMathTime5 жыл бұрын
I will be soon bro!
@SundararajanSrinivasan5 жыл бұрын
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
@whiteyplaysmighty85035 жыл бұрын
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?
@whiteyplaysmighty85035 жыл бұрын
Or alter the inscribed shape, such as an elliptical shape in a rectangle?
@SundararajanSrinivasan5 жыл бұрын
@WhiteyPlaysMighty I meant when only one circle could be inscribed. I am not too sure what happens when figures other than circles are used
@whiteyplaysmighty85035 жыл бұрын
@@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.
@SundararajanSrinivasan5 жыл бұрын
@WhiteyPlaysMighty how did you take the derivative wrt the inradius for the trapezoid?
@desrepeerc206 Жыл бұрын
Amazing video!
@biddu2683 Жыл бұрын
Fabulous!
@malexmartinez40073 жыл бұрын
Awesome!
@its_robbietime13333 жыл бұрын
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
@joshuamariano16635 жыл бұрын
Do you write in reverse? Or do you write in a mirror then you flip it in post?
@jkid11345 жыл бұрын
My guess is that the entire clip is mirrored, and that he writes normally.
@kreglfromworld5 жыл бұрын
He's left handed, I hope that answers your question
@joejo12455 жыл бұрын
He did a Q&A he records the video then just flips it when he edits.
@skeletonrowdie17685 жыл бұрын
awesome video man, super creative.
@dominguezander85235 жыл бұрын
Very original idea and very nice video. The answer to your last question, by the way, is pi*r^2.
@David-km2ie5 жыл бұрын
This an amazing such a cool result
@epalegmail5 жыл бұрын
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
@rorygo4565 жыл бұрын
epalegmail i think the camera is inverted
@ottothelord57554 жыл бұрын
This was a great video!
@pk8234565 жыл бұрын
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.
@pk8234565 жыл бұрын
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_law5 жыл бұрын
would it work with an inscribed ellipse in a rectangle? what about 3d shapes and relationship with volume and surface area?
@David-km2ie5 жыл бұрын
Sounds interesting
@EpicMathTime5 жыл бұрын
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.
@ayanbanerjee72592 жыл бұрын
Wow great
@Mahmood429785 жыл бұрын
7:43 ....is actually on one of the Math GRE practice exams.
@Sooboor4 жыл бұрын
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
@sujathasuresh87332 жыл бұрын
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?
@99selfmade215 жыл бұрын
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?
@EpicMathTime5 жыл бұрын
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.
@99selfmade215 жыл бұрын
@@EpicMathTime what is then the derivative of tan x? and do i even get them? xD
@99selfmade215 жыл бұрын
Nevermind got it myself 😇
@nadiyayasmeen39284 жыл бұрын
Area and perimeter of a circle are the answers to the question asked at the end
@juanfa983 жыл бұрын
Nice
@jaymiller1093 жыл бұрын
I published this result - and the 3-dimensional analog - in 1977 in the Two Year College Math Journal.
@EpicMathTime3 жыл бұрын
Got a link? I'd love to read it. This problem is still one of my best math memories.
@jaymiller1093 жыл бұрын
@@EpicMathTime Sorry, I have only reprints.
@unpythonic3 жыл бұрын
@@EpicMathTime Found it on JSTOR: www.jstor.org/stable/3026559
@bobmarley99053 жыл бұрын
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😂😂
@benjaminbrady23854 жыл бұрын
I thought that benevolent gods name was Euler?
@nikogruben95735 жыл бұрын
I really enjoyed the vid, quality content! Although I did wonder, isn’t r the side lengths of the isosceles triangle, not the height?
@caladbolg86665 жыл бұрын
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 :)
@nikogruben95735 жыл бұрын
Caladbolg ooh, thanks
@LovepreetSingh-jv4uj3 жыл бұрын
Hey brother..WHAT IF WE PUT n=1.5 What does it actually means?
@gnikola20134 жыл бұрын
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.
@hugocarlsson36285 жыл бұрын
Amazing video
@imravishankar863 жыл бұрын
what if there is two variable in Area like A= l*b
@hank15195 жыл бұрын
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.
@Smitology2 жыл бұрын
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
@aakashsingh52885 жыл бұрын
Limit goes to area and perimeter of circle of radius r
@BlueGiant692025 жыл бұрын
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
@alimghazzawi37005 жыл бұрын
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
@sujathasuresh87332 жыл бұрын
Sir, u have drawn circum circle of regular polygons instead of drawing inscribed circle to prove the last theorem... isn't it wrong...
@EpicMathTime2 жыл бұрын
Do you mean those three graphics that appear? Yes, they aren't correct.
@sujathasuresh87332 жыл бұрын
Ok...sir 👍
@meltossmedia5 жыл бұрын
Damn, now I want to be a Mathematician
@EpicMathTime5 жыл бұрын
hell yea brah, do it
@meltossmedia5 жыл бұрын
@@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.
@melrobles85875 жыл бұрын
at 9:40 you switch to circumscribed circles not inscribed circles Does the result still hold? @ThePKFactor also spotted this as well
@EpicMathTime5 жыл бұрын
I really just meant to demonstrate the triangulation there, the circles can essentially be ignored.
@melrobles85875 жыл бұрын
@@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
@billygarvey6335 жыл бұрын
Are you writing backwards or left handed? Or something else? I must know
@camdenwhite46573 жыл бұрын
apothem!
@aggelosgkekas31134 жыл бұрын
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
@dominicellis18674 жыл бұрын
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
@Taricus5 жыл бұрын
10:30 Where'd the 2 go? @_@
@jomanaehab42005 жыл бұрын
He divided the angle in order to have a right angled traingle so that we can get tan of it
@DiegoMathemagician4 жыл бұрын
How is this theorem called?? (give it a name)
@EpicMathTime4 жыл бұрын
The Broadsword Theorem
@DiegoMathemagician4 жыл бұрын
@@EpicMathTime why a broadsword what the hell xD PD: great video, honestly
@DiegoMathemagician4 жыл бұрын
You know, i found a theorem named "Magic wand theorem", close enough
@alimghazzawi37005 жыл бұрын
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