Fun fact, the letter π was chosen to describe the ratio of circumference/diameter because of the word 'περιφέρεια' (~periphery) that is the 'περίμετρος' (perimeter) of a circle.
I find it truly incredible how many words in English come from Greek yet the spoken Greek is so incomprehensible to native English speakers who have never studied Greek.
Just for fun, I pulled out my old Calculus book from college and integrated using the equation for a circle (after solving for y to get y= sqrt(r^2-x^2). I simply integrated from 0-1 (quadrant 1 of a circle with radius r with its center at (0,0)) and multiplied by 4. Sure enough...Area still equals (pi)r^2 !
The first I estimated PI, I was a sophomore in high school. I used nested octogons: 1 inscribed, 1 circumscribed. That initial estimate was 3.16. It wasn't too hard later to determine a formula for estimating PI as a lim(N->Inf) for nested N-gons.
There are many ways you can integrate it, if you do a basic single intergral by solving for y then you have to use trig substitution. However you could do it much easier and faster with double intergrals in polar coordinates.
@@oliviervancantfort5327 One problem. In order to determine the value of PI, you can't use PI. At least, those were always the rules I applied for myself when I'd pursue that value out of curiosity.
r² is the area of a square implied by two radii of a circle of radius r that are at 90° to each other. 4r² is therefore the area of the square in which the circle with radius r is inscribed. If πr² is the area of the inscribed circle, then π/4 is the fraction of a square with sides 2r that is taken up by its inscribed circle. Maths is a beautiful thing.
Thank you!! I've just done it!! Upon learning of William Jones 1706 work you referenced, I went to Wikipedia and found out that this work (with the first use of the Greek letter pi for representing the ratio between a circle's circumference and its diameter) is titled "Synopsis Palmariorum Matheseos." Then within mere minutes, I found myself buying a copy of it on Amazon!! Thank you so much, Presh, for enlightening me!!!!
Here is a simple method: Picture the concentric rings from the third approach. The innermost ring has a circumference of zero (it's basically a point). The outer ring has a circumference of 2πr. Now add up all the circumferences by integrating 2πr from 0 to r. It's a very simple integral that results in πr^2
If calculus and its full notation and explication had already been discovered, this would not be very hard. These proofs all anticipate calculus by a fair period of time, at least in the definition of a limit…
A little detail you have to handle carefully: the areas of the circumference lines that you propose to integrate are exactly zero. A reasoning based on rings with infinitesimal thickness, or equivalent, has to be established.
@@xnick_uy Isn't this the basis of finding areas using Calculus? It's the same logic as adding up rectangles with infinitesimal thickness. Or is it just a coincidence that it produced the right result?
@@BillionFires Yes, but the proper formal calculation needs a few steps. The area of a ring with inner and outer radii a and b, respectively, is pi*(b^2-a^2). But we can't use this expresion to prove what the area of the circle is (is the other way around). We have to show that if we set b = a + dr and let dr -> 0, the area of such a ring goes to 2*pi*dr (which is correct but we have to prove it without using the formula we are trying to demonstrate).
@@BillionFires exactly, integration is basically summing up the area under a curve/graph.These clever , well thought methods predate calculus by a long time period.
If you want the area of a non-circular ellipse, then you can use the formula πab, where a and b are the major and minor axis. For a circle, a and b would be the same.
It seems like the common insight to all these proofs is: 1) The circular curve can be approximated by a sequence of straight lines placed regularly in some pattern close to the curve (ie a regular piecewise linear approximation). 2) The more lines there are, the more closely the sequence approximates the curve with less distortion to the area. In my opinion, calculating π must have been much more difficult than finding the formula for the area given pi. If I recall correctly, I heard somewhere that Archimedes approximated π by finding the perimeter of an inscribed (inside the circle) regular polygon as the video showed, but also a circumscribed (outside the circle) regular polygon. Then he could measure or calculate the perimeter of the outer (circumscribed) and inner (inscribed) regular polygons and use this to figure out some sort of upper and lower bound on π (I might be citing this story incorrectly).
Yeah I hear that too, I also heard that Archimedes as an Ancient Greek was not comfortable with the idea of limits, so he presented it by saying π always lies between two specific numbers given by inscribed and circumscribed regular polygons, where the polygons may have as many sides as desired.
Yes that is true. It was one of the oldest methods for calculating pi, albeit very inefficient. Archamides stopped at the 96-gon and still only managed to get that pi was between 3.1408 and 3.1429, only guranteeing him 2 correct decimal places
From first principles, we define A = 2πr. Now consider the infinitesimal increase in area δA arising from an infinitesimal increase in radius δr. That is very close to 2πr.δr. So δA ≈ 2πrδr which leads to δA/δr ≈ 2πr. In the limit as δr tends to 0, the approximations becomes an equality and we get dA/dr = 2πr. That can be solved for A by taking the integral from 0 to r and we immediately find A = πr^2. It's not particularly rigorous, but once you can see that the rate of increase of the area of a circle wrt its radius is 2πr, it should be obvious what the area is.
Love this history of math stuff. I know middle school and high school students don't really care about where math comes from (I didn't), but I do think it's vital to teach it before college. You never know whose interest could pique.
The answer to this question is quite easy, but a little in involved. You are give a straight line on a flat plane of a length, doesn’t matter how long but the whatever the length is you subdivided the line into 2 equal parts, each part represents a unit scale (for simplicities sake). We establish the center as the origin and one reference point at one end of the segment. The length of the line is 2, but we should imagine an esoteric line traveling back to the reference point so in actuality the line is length 4. What is the enclosed area. To have an area you need some sort of enclosure in s second dimension, but the two lines superposition so there is zero. Segments =2 Cum. Len = 4 Area = 0 So at the origin we bisect segment and stretch it so that now we have a square joining points equal distance on two orthogonal axis. The 4 points form 4 chords. We pretend we don’t know the length or any thing about sines and cosines. So if the line going from the reference point and back is a flat loop and we stretch it out the center we are adding length into to each half a segment. 2/2 = 1. The length the line is traveling towards is 1 unit from the circle and since its point was at the origin the distance traversed of the new point is 1. Thus new point it how far from old point, let’s just do one. The reference point is 1 unit in original dimension say a, in the new dimension created the new point is b So length is SQRT((a(new) - a(ref))^2 + (b(new) - b(ref))^2) = SQRT ((0-1)^2 + (1-0)^2) = SQRT(1+1) = SQRT (2). So the tilted square is composed of 4 chords, each chords on an imaginary ⭕️. What is the area under the chords? To arrive at the area we need to project a new radii that bisects each chord. At the intersect of each chord we have a bisector that junctions with 2 equal half chords in this case it’s the SQRT(2)/2=SQRT(1/2). This bisector length is SQRT(1- SQRT(1/2)^2) = SQRT(1/2). Thus treating the chord as the base the area = halfchord * bisector which is SQRT(1/2)^2 = 1/2, the combined area under the chords is 4*1/2 = 2 # c p A/c A 2 2.0 4 0 0 4 1.4 5.7 0.5 2 The answer is right here we just cannot see it yet. I will point it out where I hid it. When I said we will use the chord as the base this means we used the bisector as the height, so I did a change of basis. But the I calculated based on the halfchord and bisector, which is correct, but as we will see that in calculating the area we’re removing piecemeal half the perimeter. I will do one more. So we are going to create 8 new chords. Again we start at our reference point (1,0). So we have defined the length of the halfchord or bisector as SQRT(1/2). The new chord is SQRT((bisector-1)^2 + (halfchord-0)^2)^2. The segment outside of the bisector is (1- SQRT(1/2))^2 = 1.5 - SQRT(2) the square of the halfchord is SQRT(1/2)^2 Chord “⭕️/8” = SQRT((1.5 - SQRT(2))+(1/2))=0.7653, its bisector = 0.923 and the area/ea = 0.3535 and total area is 2.828 If we repeat this process about 23 more times eventually we get to a point where the bisector is indistinguishable from 1 on most devices and does not change, each new chord is equal to the previous halfchord. But unchanging is the per chord calculation of area, area is 1/2 base x height. So the question is why this is the case. The ⭕️ is an abstraction created by humans, it’s actually points on a line connected by chords. A rectangle is another abstraction created by humans. We defined this as 4 rectilinear line segments and using this system we define simple area. But the area swept by each line segment,chord on circle, is not rectilinear. When we bisect the chord we generate 2 right triangles, when the diagonal sides are abutted to each other they are rectilinear and we have a calculable area, but in doing that, in 2-D space the displacement of the chord in simple area with one dimension now superpose half upon itself. So there is one detail here we need to make this work. In the beginning of the problem I set the parameter of the argument. Whatever the length is I am assigning a new length of 2 new units, I then mystically stretched the line by 2 units so that I could fill in an empty area. These are artificial processes, they are not real, they begin in the imagination. In doing this I forced the radius to be 1, in the third step I forced out another dimension. So let’s argue the original line length is 200, I set its length to 2, I stretched its length to 400 (4) so it would return to its reference point, then I processvely stretched it in another dimension to 628.32. In essence I forced most of the line into 2 dimensional space by creating a curvature. The force is 100 outward in every direction. And so to make this work in need to multiply the area I made by the area scale factor which is r^2. Where is this, let’s look at the first second dimensional stretch. I forced out in the second dimension 100 units. I then defined a line going from the second dimension back to the first. I then defined the area as area between the line and the origin as defined by to radii. So now I have an area centered around a direction vector (45°), so that area is 0 units/height on one end and 141.42 units per height one the other end of each radii. But the radii point along different vectors, the bisector splits the difference. In doing this we create a new basis z for each radii which displaces along z a fraction of the displacement along x (for the reference point). As a consequence that displacement is the cosine of the angle which centers the bisector between to radii that with the chord form the area. The halfchord just happens to be the sine of half the angle of the chord. Thus in this case area along each bisector is (r * sine = halfchord)(r*cosine = bisector) = r^2 sin*cos where angle is 1/2 the chord. As perimeter goes to 2pi, cosine goes to r and area goes to the r^2 * sine of 1/2 the chords angle. In this case To der
Regarding the last example, I was thinking that you can integrate the different circumferences for the radius going from 0 to r, which is \int_{0}^r 2xpi dx= r^2pi.
@@pelinalwhitestrake3367 You have the right to tell others not to do this, not to do that. But you at least have to tell them *why* because they have the right to choose whether to follow you or not.
It’s amazing how much school doesn’t teach you. From the area of a circle to the quadratic formula, they teach a lot of hows, but not a lot of whys. It makes sense for brevity’s sake, but it’s almost like they’re trying to make you hate math by making it a memory game rather than a problem solving game.
It really depends on your school(s) and your teacher(s). Most of my math teachers in high school were pretty good at explaining both why and how. For example, we derived the quadratic formula, so we knew why it worked.
@@pvanukoff In a good Grammar School in England with a Maths Teacher who had been Football Goalie for England, really, we were shown the principles at O Level but then at A Level we were expected to be able to reproduce the proofs if asked. There was and still is a lot wrong with the Maths Curriculum. So much time wasted on Cos (A+B), Sin (A-B) etc etc. But this was good. So Wikipedia experts where did I do to school ?
the definition of π is the ratio of a circle's perimeter over its diameter. I think for modern people the infinitessimal pizza slices is the most intuitive formula for the surface. if you take a pizza's worth of slices and arrange them with touching sides but alternating the side the crust is on, you get a paralellogram. cut the slice on tthe extreme end in half and you have a rectangle with some crust sticking out. The more slices, the less sticks out. till the point the shape becomes indistinguisable. what surface is this rectangle? it's the length of a pizza slice in one direction, and half the pizza's circumference in the other, because we've been alternating the crust. This brings us to dimensions of r by πr, proving the formula
I wondered about this very thing years ago. With the relatively modern benefit of calculus you can prove the relation easily enough, but fascinating to see the ancient arguments which somehow I missed out on. Thanks!!
The circumference is 2*pi*r. The area of a triangle is base * height/2. The base gets arithmetically smaller and smaller towards the top across the height. With the same reasoning, the area of a circle must be 2*pi*r * r/2 = pi * r^2.
That first proof (using the regular n-gon) becomes even nicer if you use τ (tau) as the circle constant (ratio of radius to circumference ≒ 6.28) rather than π. C = τr A = ½Cr = ½τrr = ½τr² Which maintains the ½pa form from the n-gons.
BUT MY DOUBT IS STILL NOT CLEARED . I always had the doubt about the discovery of pi . Like you have told , old mathematicians found out that every circles circumference to diameter ratio is a constant , which was later given the notation pi . But we tell pi is an irrational number , and from what i know an irrational number cannot be a ratio . If the old guys were trying to find out a ratio then how did we end up in an irrational number?
An irrational number is a real number that cannot be expressed as the ratio of two integers. π is defined as the ratio of circumference to diameter, and these quantities cannot both be integers at the same time.
Oh rational numbers were ratios of integers , sorry my bad So for ever circle whose radius is a whole number , the circumference would be an irrational number, which is a multiple of pi How did the old guys measured the circumference of circles soo accurately I think today we know pi's value upto tenthousandth place or something. This value must have come from measuring circles right ? Then how were they so precise in doing such measurements Or does this have to do something with the equation of circle
@@nothing-jw2ns Archimedes proved that 3 10/71 < π < 3 1/7. The reasoning went something like this, but without trig. First, note that the value of π is exactly equal to the area of the unit circle, A = π(1)^2 = π. Let p_n be the regular n-gon inscribed in the unit circle, and let P_n be the regular n-gon circumscribed about the unit circle. Thus, the area of the unit circle is bounded between the area of p_n and the area of P_n, or A(p_n) < π < A(P_n) Let 2𝜽 = 360º/n be the common central angle of p_n and P_n. Orient p_n and P_n so that the positive x-axis bisects one of the n isosceles triangles that make up the polygons. By trigonometry, the area of p_n is given by the combined areas of the n isosceles triangles, or A(p_n) = n*(Area of isosceles triangle) = n*1/2*(base)*(height) = n*1/2 * (cos(𝜽))*(2sin(𝜽)) = n/2 * 2cos(𝜽)sin(𝜽) = n/2 * sin(2𝜽) (by double-angle identity for sine) = n/2*sin(360º/n) (since 2𝜽 = 360º/n) Similarly, for area of P_n: A(P_n) = n*(Area of isosceles triangle) = n*1/2*(base)*(height) = n*1/2 * (1)*(2tan(𝜽)) = n*tan(𝜽) = n*tan(180º/n) (again, since 2𝜽 = 360º/n, 𝜽 = 180º/n) This gives the following bounds for π: A(p_n) < π < A(P_n) n/2*sin(360º/n) < π < n*tan(180º/n) for all n = 1, 2, 3, 4, .... Choosing n = 100 will give A(p_n) ≈ 3 10/71 and A(P_n) ≈ 3 1/7.
The original definition of π in Greek mathematics was not as the circumference constant (C/d) but as the area constant (A/r²). That A = π × r² was essentially proven in proposition XII.2 of Euclid's Elements. But Euclid never mentioned a circumference constant. That, as the video describies, is due to Archimedes.
Hey, I'm a brazilian fan of your channel, and I loved it last year when you solve our Pinocheo math problem from OBMEP. This year we again had an interesting problem about some flowers which, if you want to, I can translate to you.
This is the second video of yours that I have watched related to Pi. In both videos you only go back in history to Archimedes. The approximation for Pi goes back much farther than that. There is an ancient Egyptian papyrus dating back some 3,500 years describing an estimation for Pi. I believe this papyrus is held in the British museum. Back then they used fractions to denote the value as decimals hadn't been invented yet. The value for Pi they came up with 3,500 years ago was approximately 3.16. Or, 4(8/9)^2. Close enough for Egyptian government work I guess, LOL.
@@andoletube -- What's interesting is that the 3,500 year old papyrus showed how to figure out angles, areas of a triangle, etc. It was basically a math lesson for students.
One more explanation is the one you can call the ratio of a circle to a square with equal radius and apothem using the percentage of the circle's π (constant of 3.14159) and the square's perimeter-apothem ratio (constant of 4) which is approximately 80%: (3.14159÷4)×area of the square(diameter or 2×radius)^2 or (π÷4)×D^2=(π÷4)×(2r)^2=(π÷4)×(4r^2)=πr^2
Find the function F(n) = the area of a regular polygon of n sides inscribed in a circle of radius r. Then take the limit of F(n) as n tends to infinity. I did it as an undergraduate taking calculus 2. I found it convincing.
r square is the area of the square of the same dimension of a circle. pie is the golden ration that shaves off the excess area in a square to form its corresponding circle. i.e., pie defines the ratio difference between the area of a square and a circle of the same dimension.
While target shooting at 300 yards I found something strange. 1 MOA (1/60 of a degree) at 300 yards is pi, the cord dimension of 1MOA included angle in inches. I checked this on my calculator it rounds to 10 digits so I used EXCEL. (300 yds) 10800" x 1/120° sin x 2 = π and it is correct to 98 digits.
There is an explanation for this. It involves trigonometry, including measuring angles in radians. For convenience, let τ (tau) represent the number of radians in a full turn: 2π, or roughly 6.28. So, τ radians is the same as 360 degrees. 1 MOA is 1/60 of a degree, so it is 1/21,600 of a full turn. Because there are τ radians in a full turn, this means that 1 MOA is τ/21,600 radians. Now, we want to calculate what 1 MOA is at 300 yards, or 10,800 inches. Let's call this length x. Imagine you are target shooting, standing at the point A. Draw a line segment from A to the point of the target's center, T. This has a length of 10,800 inches. Now at the target point, go at a right angle from the previous line segment, and draw a line segment of length x. We'll call the endpoint B (for bullet). Now draw a final line segment from A to B, and all the line segments together form a triangle. Specifically, it is a right triangle, with its right angle at the target point T. Now for the trigonometry. We know that the tangent of an angle is the ratio of the opposite side to the adjacent side: tan(A) = (opp) / (adj) Here, the angle A is τ/21,600 radians, as mentioned before. The opposite is x, which is the length we want to find, and the adjacent side is the 10,800-inch line segment from you to the target. Let's plug that in: tan(τ/21,600) = x / 10,800 Doing a little algebra, we multiply both sides by 10,800 to solve for x: 10,800 * tan(τ/21,600) = x Or, if you want the x on the left: x = 10,800 * tan(τ/21,600) So how do we find an approximation for this value? Well, the tangent of an angle is the same as the sine of that angle, divided by the cosine of the same angle: x = 10,800 * sin(τ/21,600) / cos(τ/21,600) Let's think about the unit circle definition of the trig functions, where the unit circle is the circle of radius 1. When you go a certain number of radians around, that's the same as traveling that distance along the unit circle, counterclockwise starting from the right side. Now, note that τ/21,600 is a very small number. For very small numbers, the sine of that number of radians is about the same as that number itself. Meanwhile, the cosine of a very small angle is about 1. So now let's return to our equation and put those approximations in. We'll use the approximately equals sign (≈) to signify that it's only approximate. x = 10,800 * sin(τ/21,600) / cos(τ/21,600) x ≈ 10,800 * (τ/21,600) / 1 And now it's just arithmetic: x ≈ 10,800 * (τ/21,600) x ≈ 10,800τ / 21,600 x ≈ τ / 2 Remember, τ is double of π, so π is half of τ: x ≈ π So, x is approximately π, meaning the length we were looking for is approximately π inches. And we are done.
My previous reply said, I went to some online calculators to find the actual decimal representation of 10,800 * tan(τ/21,600), and the number of matching digits is nowhere near 98. For reference, I will put the true value of π followed by what I got: 3.14159265... 3.14159274... I don't know why there's a discrepancy between our results.
@@isavenewspapers8890 a calculator rounds the numbers past 9 sometimes internal to 14. I set excel to use 100 digits pi and sin and the difference was 0.0.....all the way to 98 and that is still rounded. Oh and you should use sin not tan (Isosceles triangle, included angle) 2 right triangles. the side C is 10800 (hypotenuse) small side is π. My hand held TI-30Xa says the difference is 0.000000011
@@ronaldmontgomery8446 Let me see if I understand correctly. Your calculation is 10800 * sin((1/120) deg) * 2, and this results in a number extremely close to π, right? I ran this past multiple calculators: Desmos, WolframAlpha, Google, and my own Casio fx-300ES PLUS from high school. All seemed to report the same value for this expression: 3.141592642... Indeed, taking the difference between π and this, we get: 0.000000011... Or, in scientific notation, this is approximately: 1.1 * 10^(-8) This matches the result from your TI-30Xa. It still seems strange that Excel doesn't agree, though.
@@ronaldmontgomery8446 Desmos, WolframAlpha, the Google calculator, and my Casio fx-30ES PLUS all report a value of 0.000000011 for π - (10800 * sin((1/120) deg) * 2), the same as your TI-30Xa. I don't know why Excel doesn't agree.
Here's how to derive the volume of a sphere. Set up the equation, y = sqrt(R^2 - x^2), as the outline of its cross section. This comes from the equation of a circle, x^2 + y^2 = R^2, solved for y. Consider an infinitesimal disk at each x-position along this shape, to form the body of revolution. It has a radius of y, and an infinitesimal thickness of dx. Each disk is a cylinder, so its volume is dx times the area of the circular face. Thus each volume element, dV, will be given by: dV = pi*y^2 dx Put it together with our equation for y: dV = pi*(R^2 - x^2) dx And integrate from x = -R to x = +R, to "add up" all these infinitesimal disks. V = integral dV V = pi*integral (R^2 - x^2) dx from -R to +R Simple application of the power rule. Boost each exponent by 1, and have the new exponent join the coefficient in the denominator: V = pi*[R^2*x - 1/3*x^3] evaluated from x = -R to +R V = pi*[(R^3 - 1/3*R^3) - (-R^3 + 1/3*R^3)] Combine and simplify the fraction, and we get: V = 4/3*pi*R^3
The last one resembles the shell method of integration (concentric circles). The formula for the area of a circle is the integral of the formula of the circle's circumference.
These are genius, they're so intuitive. I did it myself by summing the areas of n isosceles triangles with apex angle (2π/n) and equal side lengths r, using the fact that sin α = α for small α, and tending n to infinity; but that's hugely complicated compared to any of the wonderful proofs here.
The third example is almost the radial integral: Let A(x) be the area of a circle of radius x. Then A(x+h) - A(x) is the area of a radial slice of the circle, and it must be limited by h*2pi*x and h*2pi*(x+h), when you try to convert the slice into a rectangle. By the ordinary "sandwich"-method, you get A'(x)=2pi*x, and thus A(x)=pi*x^2, which ends the proof 😊. Note that the derivative of the area is the circumference, which makes perfectly sense if you think about it a few seconds 😎.
I remember at one time thinking π = 22/7 exactly. I missed the point being that it's just a very good approximation that works for a lot of everyday calculations.
I still Remember i studied about the pie chapter at the Finals of my grade 7 it was fun i was statiesd that i could solve some questions Perimeter=πr^2 and the second Area= 2πr those are the 2 formulas i like it very much.
Draw a infinite number of radii by rotating a single radius around the centre of the circle an infinite small angle theta and then add them all up. Then the area is the double integral from 0 to 2pi of the integral from 0 to r of r dr d-theta.
Area = 4 ∫ √(r²-x²)dx between 0 and r let x = rcos(x) then dx = -rsin(t) dt Area = -4r² ∫ sin²(t) dt between π/2 and 0 Since cos²(t) + sin²(t) = 1 and cos²(t) - sin²(t) = cos(2t) we get sin²(t) = ½ (1 - cos(2t)) So... Area = 2r² ∫ 1 - cos(2t) dt between 0 and π/2 = 2r² [ t - 2sin(2t) ] between t=0 and t=π/2 = πr²
At school we were just told the πr² formula. One afternoon I pondered upon the derivation of this formula and came up with the pizza and the onion methods. This shows that it does not take a genius to figure it out. One evening when learning about relativity I wondered how much time would slow down depending upon your velocity, so I drew myself a simple spacetime diagram and came up with an equation, which a few years later found in a physics book to be the Lorentz transformation. My point is that things are often far more simple than you might imagine. To begin with a problem might appear somewhat daunting, yet on reflection having solved it you see how simple it is.
During our school days , we did read about derivation of area of a circle . It was there in famous Geometry book by HALL and STEVENS . Both 1st and 2nd methods were mentioned .
Imagine you have a circle. You also have a bigger circle which means that the C and d for that circle are X times larger compared to the smaller one. So the the bigger circle has circumference of X*C and diameter of X*d. The ratio between the the two is then (X*C)/(X*d) which is equal to just C/d.
@@jimi02468I believe the comment was asking how we know that the two measurements get scaled up by the same amount in the first place, which this reply does not justify.
@@isavenewspapers8890 Well if you think about the fact that two circles of different sizes are otherwise identical. What does it mean to be identical in that way? It means that the ratio between any two lengths is the same for both shapes. I mean, imagine that you have two maps or something but they are different sizes. One of them fits to a page of a book and the other fills up the whole table. If one of the maps shows that the distance from your home to a grocery store is, for example, 1/4 of the distance between your home and a football stadium, that's obviously going to be the same ratio of 1/4 regardless of which map you use, or even in real life. Just like any map will have the same ratio between any two distances, any circle will have the same ratio C/d.
Empirically by measuring many circumferences and dividing by its diameter thus achiving an estimate. They did it as accurately as possible to get an aproximate value. Hear he explaining it from 0:30 to 1:00
@@desrtsku I guess there is many ways to do this. The most simple method and the one that I know of is to simply surrond a perfectly round object (it could even be a draw of a circle made with compass) with a thin cord of some kind that doesnt stretch. Surround it inch by inch accurately . Then simply straight up the cord and mesure it with a ruler. It just happens that even in Archimedes time they already had compasses. Quite rudimentary indeed. But draw a big enough circle and you will easily get it right at least 2 decimals of the pi number
@@desrtsku Archimedes determined pi, by calculating the perimeter of a 96-gon, and bounding it by its cross-flats "diameter" and its "cross-points" diameter. People continued with this method of bisecting polygons to calculate pi, until Newton discovered a method that was much more computationally efficient.
Too long since I did A Level double maths but it will be on Wikipedia without a doubt. There are about half a dozen different sums of series which give pi. That is the subtlty of this topic. This video shows the principals about distance and area. Calculating pi itself is quite different, but just as cool if not more so.
PI is an irregular number, so the calculated area is only an estimate, but the actual area is a real number, hell an integer if you defined a new number system where PI is the unit value.
Literally 15 minutes ago I was trying to remember what the area of a circle was and after some thought came up with a variant of Abrahams method. The weirdest thing is i never heard of that one before and yet thats the one that occurred to me.
How about the other simple calculation for 3rd method? The area contain circles perimeter with r=0 to r=R so integrain of (2pi*r) and r from (0 to R) will be (2* pi * R^2 / 2) or simply (pi * R^2)
One thing if you try the second method on a sphere to get the surface area, you will get the wrong answer.... 3blue1brown did a video on why it breaks for a sphere.
r is raised to the power of the dimensions the radius is filling. A 2D circle is power 2. A 3D circle (sphere) is power 3. So a 1D ‘circle’ on a straight line is just r, and a 4D circle (Hyper sphere) is r raised to the power 4. Dunno what Pi does in these instances though…I mean it just stays Pi in circles and spheres, but what about higher and lower dimensions? Further brain stretching…what happens with -ve dimensions, ie. Less than a point.
nice music as the circle was unwrapped in the 3rd proof. Interesting how each proof used limits to get from a concrete figure towards infinity and a circle.
The animated is still wrong if the slice are not thin enough, because what you get are not stack of rectangles as illustrated, but rather stack of trapezoids.
Just curious, as I am old. Don't schools still have teachers write out this proof on the blackboard in ninth grade in Beginning Geometry? As I recall, much of the year was spent on proofs like this and how to solve complex questions combining various properties of curves and polygons.
That’s really wild. But it occurred to me that a guy invented a way to wheel. Then another guy decided to reinvent the wheel. Then some third dude again reinvented the wheel.
A circle, centered around the orgin with radius R can be defined by the equation: x^2+y^2=r^2. But, when does the limit as N -> inf. sided polygon inscribed inside a circle, centered around the Orgin approach the circle equation? Also, if it approaches the shape of a circle, then how come polygons corner points are connected with line segments, but 2 points on a circle's circumference aren't connected with a line segment, being shaped more like a curve between 2 points? Moreover, can it be proved that this area between the polygon and the curve of the circle converges, and approaches ZERO as the number of points on the infinite sided polygon approaches infinity?
That's why the dollars is going to drop when people ready to tell the true ( actually the diference of value of π is inversely proportional to the difference of radii of the circles. Meaning larger the radius smaller π
If the curvature remains constant at the perimeter the depth of the wedge will not equal the radius unless the wedges base is a single point ie zero width. Another words at that point the wedge can no longer be used to estimate area. Or given the inability to perform the required measurements approaching infinity the value of Pye can not be finitely determined?
Because pi is, by DEFINITION, the ratio of the circle's circumference to its diameter. So: C = pi*D But D = 2*r, so C = pi*(2*r) To get the area, we can integrate all the little annular rings, each of thickness dr, which just means integrating that circumference formula from 0 to R. The integral of 2*r*dr is r^2, evaluated between the limits: Area = pi*R^2 - pi*0^2 = pi*R^2 Q.E.D. Couldn't be simpler or more obvious.
Perhaps you could also expand on how these ancient civilizations came up with the approximation of 3.14###. before the decimal system we use today came into being. I understand that they must have had some way of doing this but nothing I've found gives a satisfactory answer.
Never was a fan of Method #2. Method #1 would be great to teach in schools, while method #3 (via numberphile and 1minutephysics) is my personal favorite
Its how much the radius line rotates thus it equals radius×circumfrence but since the raduis line has two ends so its rotation is half .therefore its =2pieR×R/2 =2pie R^2. Thats how i thought it
Method4:- Consider a thin ring inside a circle, just like method3, with radius gradient dX and circumference 2piX. Integrate 2piX with respect to X and we get the area.
None of the equations account for a circle's boundary. A boundary, to be a boundary, must have some measure to it. That's how we know that math is an abstraction of reality, not reality itself. There actually are no boundaries.
I think the question is how to establish a way to prove that this constant ratio of circumference to diameter that we call pi is equal to 3.14...etc . After that areas and volumes follow.
So Archimedes just happened to be walking around barefooted one day around 250 BC, and thought, hmmm just what is the area of that there circle. Why did he want to know that ?
Pi is also found in the Bible… 1 Kings 7:23 He made the Sea of cast metal, circular in shape, measuring ten cubits from rim to rim and five cubits high. It took a line of thirty cubits to measure around it.
Yes, except it's wrong: to measure around it should have taken a bit more than 31 cubits, not 30. Which has resulted in literally centuries of inerrancy apologists trying to find excuses for it.
Measurements for construction often do not need to be exact. Approximating π as 3, with an error of 4.5%, may have been "good enough" for their purposes. A cubit is not an exact unit of measurement either, since different people have different sized bodies. According to Wikipedia "The cubit is an ancient unit of length based on the distance from the elbow to the tip of the middle finger.", and "lengths typically ranged from 44.4 to 52.92 cm". (~ 16-19% error). Lastly, π is irrational, so any estimation is approximate. Edit: Maybe the cubit length could possibly have been standardized across the individual measurements, if they used some sort of measuring tool or if the same person's arm was used for all the measurements. I suppose this depends on the details of how the objects were constructed, which is now lost to history. Maybe they just constructed the circle of a certain size, and then measured the dimensions after, and the lengths were later rounded (ie maybe the writer rounded to 1 significant figure). The writers probably were not concerned with exact measurements, as this is not a math textbook. Maybe it was not a perfect circle. It seems that real life can be more complex and tricky than the Platonic idealized mathematical models we use to understand it.
Interesting that the ancients used arguments that took a limit to infinity, when the formalism to support the concepts of limits or of infinity wasn’t established.
Archimedes has formalized a version of it. He called it a method of exhaustion. Instead of just having an inscribed polygon he also used an outer polygon to bound the area of a circle between two values. Then increasing the number of sides of the polygon the lower bounding value and upper bounding value would approach a fixed value that would be the area of the circle.
Fun fact, the letter π was chosen to describe the ratio of circumference/diameter because of the word 'περιφέρεια' (~periphery) that is the 'περίμετρος' (perimeter) of a circle.
oh and the first letter looks like a pi
@@StrawPancake The first letter IS pi lol
periphereia
perimetros
I find it truly incredible how many words in English come from Greek yet the spoken Greek is so incomprehensible to native English speakers who have never studied Greek.
@@Tiqerboy It's all Grssk to me.
Just for fun, I pulled out my old Calculus book from college and integrated using the equation for a circle (after solving for y to get y= sqrt(r^2-x^2). I simply integrated from 0-1 (quadrant 1 of a circle with radius r with its center at (0,0)) and multiplied by 4. Sure enough...Area still equals (pi)r^2 !
The first I estimated PI, I was a sophomore in high school. I used nested octogons: 1 inscribed, 1 circumscribed. That initial estimate was 3.16. It wasn't too hard later to determine a formula for estimating PI as a lim(N->Inf) for nested N-gons.
There are many ways you can integrate it, if you do a basic single intergral by solving for y then you have to use trig substitution. However you could do it much easier and faster with double intergrals in polar coordinates.
@@jeffreyestahl Lim n -> oo [ 2^n * sqrt(2 - sqrt(2 + sqrt( 2 + ... ))) ] with n many square roots.
Much easier to integrate in polar coordinates. Just integrate r dr from 0 to 2pi, which is basically the calculus equivalent of first method. 😊
@@oliviervancantfort5327
One problem. In order to determine the value of PI, you can't use PI. At least, those were always the rules I applied for myself when I'd pursue that value out of curiosity.
I value much more this type of content and historical background investigation than solving nearly-impossible questions/puzzles. Amazing video!
r² is the area of a square implied by two radii of a circle of radius r that are at 90° to each other.
4r² is therefore the area of the square in which the circle with radius r is inscribed.
If πr² is the area of the inscribed circle, then π/4 is the fraction of a square with sides 2r that is taken up by its inscribed circle.
Maths is a beautiful thing.
Nicely done.
Thank you!! I've just done it!! Upon learning of William Jones 1706 work you referenced, I went to Wikipedia and found out that this work (with the first use of the Greek letter pi for representing the ratio between a circle's circumference and its diameter) is titled "Synopsis Palmariorum Matheseos." Then within mere minutes, I found myself buying a copy of it on Amazon!! Thank you so much, Presh, for enlightening me!!!!
congratulations 👏you are truly passionate,i love people like you❤❤
@@kelumo7981 Thank you!!! ❤❤
The animations in this video were very good, Presh!
I love how excited you get, Presh! Keep it up! 👏👏👏
archimede's method was so good. felt like a plot twist. thats why i love maths
Math history is amazing. This was excellent! Thank you.
Here is a simple method: Picture the concentric rings from the third approach. The innermost ring has a circumference of zero (it's basically a point). The outer ring has a circumference of 2πr. Now add up all the circumferences by integrating 2πr from 0 to r. It's a very simple integral that results in πr^2
If calculus and its full notation and explication had already been discovered, this would not be very hard. These proofs all anticipate calculus by a fair period of time, at least in the definition of a limit…
A little detail you have to handle carefully: the areas of the circumference lines that you propose to integrate are exactly zero. A reasoning based on rings with infinitesimal thickness, or equivalent, has to be established.
@@xnick_uy Isn't this the basis of finding areas using Calculus? It's the same logic as adding up rectangles with infinitesimal thickness. Or is it just a coincidence that it produced the right result?
@@BillionFires Yes, but the proper formal calculation needs a few steps. The area of a ring with inner and outer radii a and b, respectively, is pi*(b^2-a^2). But we can't use this expresion to prove what the area of the circle is (is the other way around).
We have to show that if we set b = a + dr and let dr -> 0, the area of such a ring goes to 2*pi*dr (which is correct but we have to prove it without using the formula we are trying to demonstrate).
@@BillionFires exactly, integration is basically summing up the area under a curve/graph.These clever , well thought methods predate calculus by a long time period.
Wow, thank you so much for enlightening on very fundamentals of formula for a Circle's area.
If you want the area of a non-circular ellipse, then you can use the formula πab, where a and b are the major and minor axis. For a circle, a and b would be the same.
You mean the semi-major and semi-minor axes.
I like this format as an addition to the usual problems.
Keep on posting these videos 🙏
It seems like the common insight to all these proofs is:
1) The circular curve can be approximated by a sequence of straight lines placed regularly in some pattern close to the curve (ie a regular piecewise linear approximation).
2) The more lines there are, the more closely the sequence approximates the curve with less distortion to the area.
In my opinion, calculating π must have been much more difficult than finding the formula for the area given pi. If I recall correctly, I heard somewhere that Archimedes approximated π by finding the perimeter of an inscribed (inside the circle) regular polygon as the video showed, but also a circumscribed (outside the circle) regular polygon. Then he could measure or calculate the perimeter of the outer (circumscribed) and inner (inscribed) regular polygons and use this to figure out some sort of upper and lower bound on π (I might be citing this story incorrectly).
Yeah I hear that too, I also heard that Archimedes as an Ancient Greek was not comfortable with the idea of limits, so he presented it by saying π always lies between two specific numbers given by inscribed and circumscribed regular polygons, where the polygons may have as many sides as desired.
Yes that is true. It was one of the oldest methods for calculating pi, albeit very inefficient.
Archamides stopped at the 96-gon and still only managed to get that pi was between 3.1408 and 3.1429, only guranteeing him 2 correct decimal places
From first principles, we define A = 2πr. Now consider the infinitesimal increase in area δA arising from an infinitesimal increase in radius δr. That is very close to 2πr.δr.
So δA ≈ 2πrδr which leads to δA/δr ≈ 2πr.
In the limit as δr tends to 0, the approximations becomes an equality and we get dA/dr = 2πr.
That can be solved for A by taking the integral from 0 to r and we immediately find A = πr^2.
It's not particularly rigorous, but once you can see that the rate of increase of the area of a circle wrt its radius is 2πr, it should be obvious what the area is.
Love this history of math stuff. I know middle school and high school students don't really care about where math comes from (I didn't), but I do think it's vital to teach it before college. You never know whose interest could pique.
The answer to this question is quite easy, but a little in involved.
You are give a straight line on a flat plane of a length, doesn’t matter how long but the whatever the length is you subdivided the line into 2 equal parts, each part represents a unit scale (for simplicities sake). We establish the center as the origin and one reference point at one end of the segment.
The length of the line is 2, but we should imagine an esoteric line traveling back to the reference point so in actuality the line is length 4. What is the enclosed area. To have an area you need some sort of enclosure in s second dimension, but the two lines superposition so there is zero.
Segments =2 Cum. Len = 4 Area = 0
So at the origin we bisect segment and stretch it so that now we have a square joining points equal distance on two orthogonal axis. The 4 points form 4 chords. We pretend we don’t know the length or any thing about sines and cosines.
So if the line going from the reference point and back is a flat loop and we stretch it out the center we are adding length into to each half a segment. 2/2 = 1. The length the line is traveling towards is 1 unit from the circle and since its point was at the origin the distance traversed of the new point is 1.
Thus new point it how far from old point, let’s just do one. The reference point is 1 unit in original dimension say a, in the new dimension created the new point is b
So length is SQRT((a(new) - a(ref))^2 + (b(new) - b(ref))^2)
= SQRT ((0-1)^2 + (1-0)^2) = SQRT(1+1) = SQRT (2). So the tilted square is composed of 4 chords, each chords on an imaginary ⭕️.
What is the area under the chords? To arrive at the area we need to project a new radii that bisects each chord. At the intersect of each chord we have a bisector that junctions with 2 equal half chords in this case it’s the SQRT(2)/2=SQRT(1/2). This bisector length is SQRT(1- SQRT(1/2)^2) = SQRT(1/2). Thus treating the chord as the base the area = halfchord * bisector which is SQRT(1/2)^2 = 1/2, the combined area under the chords is 4*1/2 = 2
# c p A/c A
2 2.0 4 0 0
4 1.4 5.7 0.5 2
The answer is right here we just cannot see it yet. I will point it out where I hid it.
When I said we will use the chord as the base this means we used the bisector as the height, so I did a change of basis. But the I calculated based on the halfchord and bisector, which is correct, but as we will see that in calculating the area we’re removing piecemeal half the perimeter.
I will do one more.
So we are going to create 8 new chords. Again we start at our reference point (1,0).
So we have defined the length of the halfchord or bisector as SQRT(1/2). The new chord is SQRT((bisector-1)^2 + (halfchord-0)^2)^2. The segment outside of the bisector is (1- SQRT(1/2))^2 = 1.5 - SQRT(2) the square of the halfchord is SQRT(1/2)^2
Chord “⭕️/8” = SQRT((1.5 - SQRT(2))+(1/2))=0.7653, its bisector = 0.923 and the area/ea = 0.3535 and total area is 2.828
If we repeat this process about 23 more times eventually we get to a point where the bisector is indistinguishable from 1 on most devices and does not change, each new chord is equal to the previous halfchord. But unchanging is the per chord calculation of area, area is 1/2 base x height.
So the question is why this is the case.
The ⭕️ is an abstraction created by humans, it’s actually points on a line connected by chords. A rectangle is another abstraction created by humans. We defined this as 4 rectilinear line segments and using this system we define simple area. But the area swept by each line segment,chord on circle, is not rectilinear. When we bisect the chord we generate 2 right triangles, when the diagonal sides are abutted to each other they are rectilinear and we have a calculable area, but in doing that, in 2-D space the displacement of the chord in simple area with one dimension now superpose half upon itself.
So there is one detail here we need to make this work. In the beginning of the problem I set the parameter of the argument. Whatever the length is I am assigning a new length of 2 new units, I then mystically stretched the line by 2 units so that I could fill in an empty area. These are artificial processes, they are not real, they begin in the imagination. In doing this I forced the radius to be 1, in the third step I forced out another dimension. So let’s argue the original line length is 200, I set its length to 2, I stretched its length to 400 (4) so it would return to its reference point, then I processvely stretched it in another dimension to 628.32. In essence I forced most of the line into 2 dimensional space by creating a curvature. The force is 100 outward in every direction. And so to make this work in need to multiply the area I made by the area scale factor which is r^2. Where is this, let’s look at the first second dimensional stretch. I forced out in the second dimension 100 units. I then defined a line going from the second dimension back to the first. I then defined the area as area between the line and the origin as defined by to radii. So now I have an area centered around a direction vector (45°), so that area is 0 units/height on one end and 141.42 units per height one the other end of each radii. But the radii point along different vectors, the bisector splits the difference. In doing this we create a new basis z for each radii which displaces along z a fraction of the displacement along x (for the reference point). As a consequence that displacement is the cosine of the angle which centers the bisector between to radii that with the chord form the area. The halfchord just happens to be the sine of half the angle of the chord.
Thus in this case area along each bisector is (r * sine = halfchord)(r*cosine = bisector) = r^2 sin*cos where angle is 1/2 the chord. As perimeter goes to 2pi, cosine goes to r and area goes to the r^2 * sine of 1/2 the chords angle.
In this case
To der
Regarding the last example, I was thinking that you can integrate the different circumferences for the radius going from 0 to r, which is \int_{0}^r 2xpi dx= r^2pi.
idk i just imagine having 3 squares of side r, and then there's a little bit left to fill once you like cut the corners
Exactly how I imagine it, to remember it more easily.
You are me Bruh😂
See my separate comment about thinking about FOUR squares of side r, which is the square in which the circle is inscribed...
This is just amazing. Getting to know this now after all these years
@1:20 Actually, Liu Hui (劉徽) found the fraction approximation 3927/1250, not the decimal representation.
Don't say this name to any Russian person.
@@pelinalwhitestrake3367
You have the right to tell others not to do this, not to do that. But you at least have to tell them *why* because they have the right to choose whether to follow you or not.
The symbol pi, has been around THOUSANDS of years. It just was used for the phoneme p.
presh checkout the 2024 JEE advanced paper, all the questions are good and u can get to the solution in one step
You should Email him on his address given at the end of his videos
What do you use to make your animations? They're superb!
Great video, Presh. Love your videos, but this one is one of my favorites ... really well explained, thank you!
It’s amazing how much school doesn’t teach you. From the area of a circle to the quadratic formula, they teach a lot of hows, but not a lot of whys. It makes sense for brevity’s sake, but it’s almost like they’re trying to make you hate math by making it a memory game rather than a problem solving game.
It really depends on your school(s) and your teacher(s). Most of my math teachers in high school were pretty good at explaining both why and how. For example, we derived the quadratic formula, so we knew why it worked.
@@pvanukoff In a good Grammar School in England with a Maths Teacher who had been Football Goalie for England, really, we were shown the principles at O Level but then at A Level we were expected to be able to reproduce the proofs if asked.
There was and still is a lot wrong with the Maths Curriculum. So much time wasted on Cos (A+B), Sin (A-B) etc etc. But this was good.
So Wikipedia experts where did I do to school ?
Great summary video!
the definition of π is the ratio of a circle's perimeter over its diameter. I think for modern people the infinitessimal pizza slices is the most intuitive formula for the surface. if you take a pizza's worth of slices and arrange them with touching sides but alternating the side the crust is on, you get a paralellogram. cut the slice on tthe extreme end in half and you have a rectangle with some crust sticking out. The more slices, the less sticks out. till the point the shape becomes indistinguisable. what surface is this rectangle? it's the length of a pizza slice in one direction, and half the pizza's circumference in the other, because we've been alternating the crust. This brings us to dimensions of r by πr, proving the formula
I wondered about this very thing years ago. With the relatively modern benefit of calculus you can prove the relation easily enough, but fascinating to see the ancient arguments which somehow I missed out on. Thanks!!
The circumference is 2*pi*r.
The area of a triangle is base * height/2. The base gets arithmetically smaller and smaller towards the top across the height. With the same reasoning, the area of a circle must be 2*pi*r * r/2 = pi * r^2.
That first proof (using the regular n-gon) becomes even nicer if you use τ (tau) as the circle constant (ratio of radius to circumference ≒ 6.28) rather than π.
C = τr
A = ½Cr = ½τrr = ½τr²
Which maintains the ½pa form from the n-gons.
BUT MY DOUBT IS STILL NOT CLEARED . I always had the doubt about the discovery of pi . Like you have told , old mathematicians found out that every circles circumference to diameter ratio is a constant , which was later given the notation pi . But we tell pi is an irrational number , and from what i know an irrational number cannot be a ratio . If the old guys were trying to find out a ratio then how did we end up in an irrational number?
An irrational number is a real number that cannot be expressed as the ratio of two integers. π is defined as the ratio of circumference to diameter, and these quantities cannot both be integers at the same time.
Of course an irrational number can be a ratio...
Oh rational numbers were ratios of integers , sorry my bad
So for ever circle whose radius is a whole number , the circumference would be an irrational number, which is a multiple of pi
How did the old guys measured the circumference of circles soo accurately
I think today we know pi's value upto tenthousandth place or something. This value must have come from measuring circles right ? Then how were they so precise in doing such measurements
Or does this have to do something with the equation of circle
@@nothing-jw2ns Archimedes proved that 3 10/71 < π < 3 1/7. The reasoning went something like this, but without trig. First, note that the value of π is exactly equal to the area of the unit circle, A = π(1)^2 = π. Let p_n be the regular n-gon inscribed in the unit circle, and let P_n be the regular n-gon circumscribed about the unit circle. Thus, the area of the unit circle is bounded between the area of p_n and the area of P_n, or
A(p_n) < π < A(P_n)
Let 2𝜽 = 360º/n be the common central angle of p_n and P_n. Orient p_n and P_n so that the positive x-axis bisects one of the n isosceles triangles that make up the polygons. By trigonometry, the area of p_n is given by the combined areas of the n isosceles triangles, or
A(p_n) = n*(Area of isosceles triangle)
= n*1/2*(base)*(height)
= n*1/2 * (cos(𝜽))*(2sin(𝜽))
= n/2 * 2cos(𝜽)sin(𝜽)
= n/2 * sin(2𝜽) (by double-angle identity for sine)
= n/2*sin(360º/n) (since 2𝜽 = 360º/n)
Similarly, for area of P_n:
A(P_n) = n*(Area of isosceles triangle)
= n*1/2*(base)*(height)
= n*1/2 * (1)*(2tan(𝜽))
= n*tan(𝜽)
= n*tan(180º/n) (again, since 2𝜽 = 360º/n, 𝜽 = 180º/n)
This gives the following bounds for π:
A(p_n) < π < A(P_n)
n/2*sin(360º/n) < π < n*tan(180º/n) for all n = 1, 2, 3, 4, ....
Choosing n = 100 will give A(p_n) ≈ 3 10/71 and A(P_n) ≈ 3 1/7.
Loved it ❤ my fav proof is the 2nd one
The original definition of π in Greek mathematics was not as the circumference constant (C/d) but as the area constant (A/r²). That A = π × r² was essentially proven in proposition XII.2 of Euclid's Elements. But Euclid never mentioned a circumference constant. That, as the video describies, is due to Archimedes.
Hey, I'm a brazilian fan of your channel, and I loved it last year when you solve our Pinocheo math problem from OBMEP. This year we again had an interesting problem about some flowers which, if you want to, I can translate to you.
This is the second video of yours that I have watched related to Pi. In both videos you only go back in history to Archimedes. The approximation for Pi goes back much farther than that. There is an ancient Egyptian papyrus dating back some 3,500 years describing an estimation for Pi. I believe this papyrus is held in the British museum. Back then they used fractions to denote the value as decimals hadn't been invented yet. The value for Pi they came up with 3,500 years ago was approximately 3.16. Or, 4(8/9)^2. Close enough for Egyptian government work I guess, LOL.
3.16 is pretty bad though - bad enough to suggest they missed something important otherwise they would have got closer.
@@andoletube -- Not bad for 3,500 years ago though. Apparently it worked good enough for them considering the construction they did.
@@ironcladranchandforge7292 Or that the construction they did neatly sidestepped anything circular...Pyramids, after all...
@@andoletube -- What's interesting is that the 3,500 year old papyrus showed how to figure out angles, areas of a triangle, etc. It was basically a math lesson for students.
One more explanation is the one you can call the ratio of a circle to a square with equal radius and apothem using the percentage of the circle's π (constant of 3.14159) and the square's perimeter-apothem ratio (constant of 4) which is approximately 80%:
(3.14159÷4)×area of the square(diameter or 2×radius)^2
or
(π÷4)×D^2=(π÷4)×(2r)^2=(π÷4)×(4r^2)=πr^2
Find the function F(n) = the area of a regular polygon of n sides inscribed in a circle of radius r.
Then take the limit of F(n) as n tends to infinity. I did it as an undergraduate taking calculus 2. I found it convincing.
Great video! It's fascinating to see that they came to such an understanding 2,000 years ago.
r square is the area of the square of the same dimension of a circle. pie is the golden ration that shaves off the excess area in a square to form its corresponding circle. i.e., pie defines the ratio difference between the area of a square and a circle of the same dimension.
While target shooting at 300 yards I found something strange. 1 MOA (1/60 of a degree) at 300 yards is pi, the cord dimension of 1MOA included angle in inches. I checked this on my calculator it rounds to 10 digits so I used EXCEL. (300 yds) 10800" x 1/120° sin x 2 = π and it is correct to 98 digits.
There is an explanation for this. It involves trigonometry, including measuring angles in radians.
For convenience, let τ (tau) represent the number of radians in a full turn: 2π, or roughly 6.28. So, τ radians is the same as 360 degrees.
1 MOA is 1/60 of a degree, so it is 1/21,600 of a full turn. Because there are τ radians in a full turn, this means that 1 MOA is τ/21,600 radians.
Now, we want to calculate what 1 MOA is at 300 yards, or 10,800 inches. Let's call this length x.
Imagine you are target shooting, standing at the point A. Draw a line segment from A to the point of the target's center, T. This has a length of 10,800 inches. Now at the target point, go at a right angle from the previous line segment, and draw a line segment of length x. We'll call the endpoint B (for bullet). Now draw a final line segment from A to B, and all the line segments together form a triangle. Specifically, it is a right triangle, with its right angle at the target point T.
Now for the trigonometry. We know that the tangent of an angle is the ratio of the opposite side to the adjacent side:
tan(A) = (opp) / (adj)
Here, the angle A is τ/21,600 radians, as mentioned before. The opposite is x, which is the length we want to find, and the adjacent side is the 10,800-inch line segment from you to the target. Let's plug that in:
tan(τ/21,600) = x / 10,800
Doing a little algebra, we multiply both sides by 10,800 to solve for x:
10,800 * tan(τ/21,600) = x
Or, if you want the x on the left:
x = 10,800 * tan(τ/21,600)
So how do we find an approximation for this value? Well, the tangent of an angle is the same as the sine of that angle, divided by the cosine of the same angle:
x = 10,800 * sin(τ/21,600) / cos(τ/21,600)
Let's think about the unit circle definition of the trig functions, where the unit circle is the circle of radius 1. When you go a certain number of radians around, that's the same as traveling that distance along the unit circle, counterclockwise starting from the right side.
Now, note that τ/21,600 is a very small number. For very small numbers, the sine of that number of radians is about the same as that number itself. Meanwhile, the cosine of a very small angle is about 1.
So now let's return to our equation and put those approximations in. We'll use the approximately equals sign (≈) to signify that it's only approximate.
x = 10,800 * sin(τ/21,600) / cos(τ/21,600)
x ≈ 10,800 * (τ/21,600) / 1
And now it's just arithmetic:
x ≈ 10,800 * (τ/21,600)
x ≈ 10,800τ / 21,600
x ≈ τ / 2
Remember, τ is double of π, so π is half of τ:
x ≈ π
So, x is approximately π, meaning the length we were looking for is approximately π inches. And we are done.
My previous reply said, I went to some online calculators to find the actual decimal representation of 10,800 * tan(τ/21,600), and the number of matching digits is nowhere near 98. For reference, I will put the true value of π followed by what I got:
3.14159265...
3.14159274...
I don't know why there's a discrepancy between our results.
@@isavenewspapers8890 a calculator rounds the numbers past 9 sometimes internal to 14. I set excel to use 100 digits pi and sin and the difference was 0.0.....all the way to 98 and that is still rounded. Oh and you should use sin not tan (Isosceles triangle, included angle) 2 right triangles. the side C is 10800 (hypotenuse) small side is π. My hand held TI-30Xa says the difference is 0.000000011
@@ronaldmontgomery8446 Let me see if I understand correctly. Your calculation is 10800 * sin((1/120) deg) * 2, and this results in a number extremely close to π, right?
I ran this past multiple calculators: Desmos, WolframAlpha, Google, and my own Casio fx-300ES PLUS from high school. All seemed to report the same value for this expression:
3.141592642...
Indeed, taking the difference between π and this, we get:
0.000000011...
Or, in scientific notation, this is approximately:
1.1 * 10^(-8)
This matches the result from your TI-30Xa. It still seems strange that Excel doesn't agree, though.
@@ronaldmontgomery8446 Desmos, WolframAlpha, the Google calculator, and my Casio fx-30ES PLUS all report a value of 0.000000011 for π - (10800 * sin((1/120) deg) * 2), the same as your TI-30Xa. I don't know why Excel doesn't agree.
Could you explain similarly how/why volume formulas work? I know that's kind of broad, but this was so well done I think you could explain it simply
Here's how to derive the volume of a sphere.
Set up the equation, y = sqrt(R^2 - x^2), as the outline of its cross section. This comes from the equation of a circle, x^2 + y^2 = R^2, solved for y.
Consider an infinitesimal disk at each x-position along this shape, to form the body of revolution. It has a radius of y, and an infinitesimal thickness of dx. Each disk is a cylinder, so its volume is dx times the area of the circular face. Thus each volume element, dV, will be given by:
dV = pi*y^2 dx
Put it together with our equation for y:
dV = pi*(R^2 - x^2) dx
And integrate from x = -R to x = +R, to "add up" all these infinitesimal disks.
V = integral dV
V = pi*integral (R^2 - x^2) dx from -R to +R
Simple application of the power rule. Boost each exponent by 1, and have the new exponent join the coefficient in the denominator:
V = pi*[R^2*x - 1/3*x^3] evaluated from x = -R to +R
V = pi*[(R^3 - 1/3*R^3) - (-R^3 + 1/3*R^3)]
Combine and simplify the fraction, and we get:
V = 4/3*pi*R^3
Pie are not square. Pie are round.
BINGO !!! We've been told a lie at school. Thanks for clearing that up.
A Pop Tart would be pi squared.
@@potemkineconomy1769, no, because Pop Tarts are rectangular.
Location of pie is in the sky...whether square or round....focus on that which is relevant...😜🤣😎
The last one resembles the shell method of integration (concentric circles). The formula for the area of a circle is the integral of the formula of the circle's circumference.
These are genius, they're so intuitive. I did it myself by summing the areas of n isosceles triangles with apex angle (2π/n) and equal side lengths r, using the fact that sin α = α for small α, and tending n to infinity; but that's hugely complicated compared to any of the wonderful proofs here.
Years and years of math education and I don't recall ever being taught any of these derivations. I was just told how to calculate it.
The third example is almost the radial integral:
Let A(x) be the area of a circle of radius x. Then A(x+h) - A(x) is the area of a radial slice of the circle, and it must be limited by h*2pi*x and h*2pi*(x+h), when you try to convert the slice into a rectangle. By the ordinary "sandwich"-method, you get A'(x)=2pi*x, and thus A(x)=pi*x^2, which ends the proof 😊.
Note that the derivative of the area is the circumference, which makes perfectly sense if you think about it a few seconds 😎.
I really like methods 2 and 3. This was a great video!
You can also do it by integrating the circumference by radius
I remember at one time thinking π = 22/7 exactly. I missed the point being that it's just a very good approximation that works for a lot of everyday calculations.
I still Remember i studied about the pie chapter at the Finals of my grade 7 it was fun i was statiesd that i could solve some questions Perimeter=πr^2 and the second Area= 2πr those are the 2 formulas i like it very much.
Draw a infinite number of radii by rotating a single radius around the centre of the circle an infinite small angle theta and then add them all up.
Then the area is the double integral from 0 to 2pi of the integral from 0 to r of r dr d-theta.
Area = 4 ∫ √(r²-x²)dx between 0 and r
let x = rcos(x)
then dx = -rsin(t) dt
Area = -4r² ∫ sin²(t) dt between π/2 and 0
Since cos²(t) + sin²(t) = 1
and cos²(t) - sin²(t) = cos(2t)
we get sin²(t) = ½ (1 - cos(2t))
So...
Area = 2r² ∫ 1 - cos(2t) dt between 0 and π/2
= 2r² [ t - 2sin(2t) ] between t=0 and t=π/2
= πr²
Great video. Thanks
You might want to re-read Archimedes' "Measurement of a Circle".
At school we were just told the πr² formula. One afternoon I pondered upon the derivation of this formula and came up with the pizza and the onion methods. This shows that it does not take a genius to figure it out.
One evening when learning about relativity I wondered how much time would slow down depending upon your velocity, so I drew myself a simple spacetime diagram and came up with an equation, which a few years later found in a physics book to be the Lorentz transformation.
My point is that things are often far more simple than you might imagine. To begin with a problem might appear somewhat daunting, yet on reflection having solved it you see how simple it is.
yeah or maybe it shows you are unusually smart
During our school days , we did read about derivation of area of a circle . It was there in famous Geometry book by HALL and STEVENS . Both 1st and 2nd methods were mentioned .
Thank you for an amazing video!
But how do you prove C/d for any circle is a constant (pi)?
you can use a similar argument to similar triangles having the same ratio between sides
Imagine you have a circle. You also have a bigger circle which means that the C and d for that circle are X times larger compared to the smaller one. So the the bigger circle has circumference of X*C and diameter of X*d. The ratio between the the two is then (X*C)/(X*d) which is equal to just C/d.
@@jimi02468I believe the comment was asking how we know that the two measurements get scaled up by the same amount in the first place, which this reply does not justify.
@@isavenewspapers8890 Well if you think about the fact that two circles of different sizes are otherwise identical. What does it mean to be identical in that way? It means that the ratio between any two lengths is the same for both shapes. I mean, imagine that you have two maps or something but they are different sizes. One of them fits to a page of a book and the other fills up the whole table. If one of the maps shows that the distance from your home to a grocery store is, for example, 1/4 of the distance between your home and a football stadium, that's obviously going to be the same ratio of 1/4 regardless of which map you use, or even in real life. Just like any map will have the same ratio between any two distances, any circle will have the same ratio C/d.
Question is how to prove that pi=3.14.....etc (other than using ropes to measure circumference and radius)
Okay. But how did they even calculate π?
Empirically by measuring many circumferences and dividing by its diameter thus achiving an estimate. They did it as accurately as possible to get an aproximate value. Hear he explaining it from 0:30 to 1:00
@@hugooliveira4440 I'll adjust the question, then. How did they even "measure the circumference"?
@@desrtsku I guess there is many ways to do this. The most simple method and the one that I know of is to simply surrond a perfectly round object (it could even be a draw of a circle made with compass) with a thin cord of some kind that doesnt stretch. Surround it inch by inch accurately . Then simply straight up the cord and mesure it with a ruler. It just happens that even in Archimedes time they already had compasses. Quite rudimentary indeed. But draw a big enough circle and you will easily get it right at least 2 decimals of the pi number
@@desrtsku Archimedes determined pi, by calculating the perimeter of a 96-gon, and bounding it by its cross-flats "diameter" and its "cross-points" diameter. People continued with this method of bisecting polygons to calculate pi, until Newton discovered a method that was much more computationally efficient.
Too long since I did A Level double maths but it will be on Wikipedia without a doubt. There are about half a dozen different sums of series which give pi. That is the subtlty of this topic. This video shows the principals about distance and area. Calculating pi itself is quite different, but just as cool if not more so.
PI is an irregular number, so the calculated area is only an estimate, but the actual area is a real number, hell an integer if you defined a new number system where PI is the unit value.
Nice proofs. Very fascinating.
Very nice - though sadly "rectangle" has been spelt as "rectange" on a few screens.
Literally 15 minutes ago I was trying to remember what the area of a circle was and after some thought came up with a variant of Abrahams method.
The weirdest thing is i never heard of that one before and yet thats the one that occurred to me.
They used pi to represent the value because there was no Greek letter for cake, which as we all know is the superior circular dessert.
How about the other simple calculation for 3rd method?
The area contain circles perimeter with r=0 to r=R so integrain of (2pi*r) and r from (0 to R) will be (2* pi * R^2 / 2) or simply (pi * R^2)
One thing if you try the second method on a sphere to get the surface area, you will get the wrong answer.... 3blue1brown did a video on why it breaks for a sphere.
the surface of a sphere "bulges". It's not on a plane. That's why it won't work. You have to include the additional area of the bulge.
r is raised to the power of the dimensions the radius is filling. A 2D circle is power 2. A 3D circle (sphere) is power 3. So a 1D ‘circle’ on a straight line is just r, and a 4D circle (Hyper sphere) is r raised to the power 4. Dunno what Pi does in these instances though…I mean it just stays Pi in circles and spheres, but what about higher and lower dimensions?
Further brain stretching…what happens with -ve dimensions, ie. Less than a point.
nice music as the circle was unwrapped in the 3rd proof.
Interesting how each proof used limits to get from a concrete figure towards infinity and a circle.
Loved this, thanks!
wow... very interesting... haven't thought about how the formula was derived thus far...
"exactly the approximation"? That ranks up there with some of the silliest expressions I've ever heard - along with the ridiculous "very unique"!
Stop the vid at 8:44 you will see that Rabi Abraham bar Hiyya's proof resembles a Menorah.
What an intriguing coincidence. 🙂
Maybe it's not a coincidence. Maybe that's where he got his inspiration for the proof?
The animated is still wrong if the slice are not thin enough, because what you get are not stack of rectangles as illustrated, but rather stack of trapezoids.
@@rashidisw Your comment is correct but it says more about your sense of humour than anything else.
Just curious, as I am old. Don't schools still have teachers write out this proof on the blackboard in ninth grade in Beginning Geometry? As I recall, much of the year was spent on proofs like this and how to solve complex questions combining various properties of curves and polygons.
That’s really wild. But it occurred to me that a guy invented a way to wheel. Then another guy decided to reinvent the wheel. Then some third dude again reinvented the wheel.
A circle, centered around the orgin with radius R can be defined by the equation: x^2+y^2=r^2.
But, when does the limit as N -> inf. sided polygon inscribed inside a circle, centered around the Orgin approach the circle equation?
Also, if it approaches the shape of a circle, then how come polygons corner points are connected with line segments, but 2 points on a circle's circumference aren't connected with a line segment, being shaped more like a curve between 2 points? Moreover, can it be proved that this area between the polygon and the curve of the circle converges, and approaches ZERO as the number of points on the infinite sided polygon approaches infinity?
That's why the dollars is going to drop when people ready to tell the true ( actually the diference of value of π is inversely proportional to the difference of radii of the circles. Meaning larger the radius smaller π
I was thinking about cutting the circle to its center and turn it into a strip of paper where the breadth is r and its length is pi r
If the curvature remains constant at the perimeter the depth of the wedge will not equal the radius unless the wedges base is a single point ie zero width. Another words at that point the wedge can no longer be used to estimate area. Or given the inability to perform the required measurements approaching infinity the value of Pye can not be finitely determined?
Because pi is, by DEFINITION, the ratio of the circle's circumference to its diameter. So:
C = pi*D
But D = 2*r, so
C = pi*(2*r)
To get the area, we can integrate all the little annular rings, each of thickness dr, which just means integrating that circumference formula from 0 to R. The integral of 2*r*dr is r^2, evaluated between the limits:
Area = pi*R^2 - pi*0^2 = pi*R^2
Q.E.D.
Couldn't be simpler or more obvious.
The last section says at the bottom "A Mechanical Derivation of the Area of the Sphere". Should it be "Circle"?
Pi has been around for thousands of years, its use as the ratio of the circumference to the diameter of a circle is only s few hundred years old.
Perhaps you could also expand on how these ancient civilizations came up with the approximation of 3.14###. before the decimal system we use today came into being. I understand that they must have had some way of doing this but nothing I've found gives a satisfactory answer.
Was it also found empirically at that time that pi is a constant, that the ratio is same for all circles?
8:42 - Ironic that in the Rabi Abraham's solution, it appears as a many-candled Menorah while unwrapping!
This was amazing thanks
What a great channel.
Never was a fan of Method #2.
Method #1 would be great to teach in schools,
while method #3 (via numberphile and 1minutephysics) is my personal favorite
Its how much the radius line rotates thus it equals radius×circumfrence but since the raduis line has two ends so its rotation is half .therefore its =2pieR×R/2 =2pie R^2. Thats how i thought it
Method4:- Consider a thin ring inside a circle, just like method3, with radius gradient dX and circumference 2piX.
Integrate 2piX with respect to X and we get the area.
None of the equations account for a circle's boundary. A boundary, to be a boundary, must have some measure to it. That's how we know that math is an abstraction of reality, not reality itself. There actually are no boundaries.
I think the question is how to establish a way to prove that this constant ratio of circumference to diameter that we call pi is equal to 3.14...etc . After that areas and volumes follow.
So Archimedes just happened to be walking around barefooted one day around 250 BC, and thought, hmmm just what is the area of that there circle. Why did he want to know that ?
How come i can't find a similar demonstration for surface area or volume of sphere?
Is there any deeper meaning in the idea that the derivative of the area, with respect to the radius is the circumference?
Pi is also found in the Bible…
1 Kings 7:23
He made the Sea of cast metal, circular in shape, measuring ten cubits from rim to rim and five cubits high. It took a line of thirty cubits to measure around it.
Yes, except it's wrong: to measure around it should have taken a bit more than 31 cubits, not 30. Which has resulted in literally centuries of inerrancy apologists trying to find excuses for it.
God obviously isn't infallible. Or rather the Bible is doo-doo.
Measurements for construction often do not need to be exact. Approximating π as 3, with an error of 4.5%, may have been "good enough" for their purposes. A cubit is not an exact unit of measurement either, since different people have different sized bodies. According to Wikipedia "The cubit is an ancient unit of length based on the distance from the elbow to the tip of the middle finger.", and "lengths typically ranged from 44.4 to 52.92 cm". (~ 16-19% error). Lastly, π is irrational, so any estimation is approximate.
Edit: Maybe the cubit length could possibly have been standardized across the individual measurements, if they used some sort of measuring tool or if the same person's arm was used for all the measurements. I suppose this depends on the details of how the objects were constructed, which is now lost to history.
Maybe they just constructed the circle of a certain size, and then measured the dimensions after, and the lengths were later rounded (ie maybe the writer rounded to 1 significant figure). The writers probably were not concerned with exact measurements, as this is not a math textbook. Maybe it was not a perfect circle.
It seems that real life can be more complex and tricky than the Platonic idealized mathematical models we use to understand it.
@@davidlohmann5098 The variation in length is the variation between standards that were being used, not between each individual measurement.
Interesting that the ancients used arguments that took a limit to infinity, when the formalism to support the concepts of limits or of infinity wasn’t established.
Archimedes has formalized a version of it. He called it a method of exhaustion. Instead of just having an inscribed polygon he also used an outer polygon to bound the area of a circle between two values. Then increasing the number of sides of the polygon the lower bounding value and upper bounding value would approach a fixed value that would be the area of the circle.
Amazing proofs
That was cool. Those old guys were pretty smart.