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
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.
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.
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
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 ?
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.
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
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.
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.
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 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.
@@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.
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.
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
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.
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
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.
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.
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.
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.
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.
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.
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.
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 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 😎.
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.
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.
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.
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.
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.
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 .
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.
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.
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
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.
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)
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
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 ?
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.
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.
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.
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.
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
Method 1 and 2 actually do not give us prove about the value of pi at all. All they gave is that pi is a ratio of C/d. All we got from these 2 methods is that A=Cr/2, which lead us back to square 1, because we need to know the value of C before we can calculate A and for that we need to have a way to calculate pi. Do you see the circular problem? The only method that gives us the answer is the third.
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?
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.
Why is there ambiguity going on recently ? Sum of angles inside a circle = 360 OR ♾️ , 0.99999 = 1 or ≠ 1 ( by law of scientific significant figures ) ,
Because "angles inside a circle" isn't a precisely defined mathematical term. If you mean interior angles, this is a specific term for polygons, where the interior angles always add to (n - 2)*180 degrees, where n is the number of polygon sides. In the limit as n goes to infinity for a regular polygon, the polygon approaches a circle. So plugging in n=infinity, shows us that the number of "interior angles" of a circle should add up to infinity. But it's difficult to say exactly what you mean by "interior angles", since they aren't angles between straight lines. If you just mean how many degrees are there in a full rotation, then the answer is 360 degrees. This is a different question entirely than the sum of the interior angles inside the shape. If the angles in question add to 360 degrees for a circle, then they add to 360 degrees for all shapes. It is the exterior angles of the circle that add up to 360 degrees; not the "interior angles".
![Tee Grizzley - Blow for Blow (feat. J. Cole) [Official Video]](http://i.ytimg.com/vi/wNLN5xsx5dc/mqdefault.jpg)
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.
I value much more this type of content and historical background investigation than solving nearly-impossible questions/puzzles. Amazing video!
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.
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!
archimede's method was so good. felt like a plot twist. thats why i love maths
I love how excited you get, Presh! Keep it up! 👏👏👏
Math history is amazing. This was excellent! Thank you.
Wow, thank you so much for enlightening on very fundamentals of formula for a Circle's area.
I like this format as an addition to the usual problems.
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.
You mean the semi-major and semi-minor axes.
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
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.
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.
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
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...
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 video, Presh. Love your videos, but this one is one of my favorites ... really well explained, thank you!
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.
This is just amazing. Getting to know this now after all these years
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
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
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.
Great summary video!
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.
What do you use to make your animations? They're superb!
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.
The symbol pi, has been around THOUSANDS of years. It just was used for the phoneme p.
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 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.
@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.
Great video! It's fascinating to see that they came to such an understanding 2,000 years ago.
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.
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
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...😜🤣😎
Thank you for an amazing video!
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.
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
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.
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.
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.
I really like methods 2 and 3. This was a great video!
You can also do it by integrating the circumference by radius
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.
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.
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
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.
Great video. Thanks
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.
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
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 😎.
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.
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.
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.
You might want to re-read Archimedes' "Measurement of a Circle".
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.
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.
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.
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 .
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.
Nice proofs. Very fascinating.
Loved this, thanks!
wow... very interesting... haven't thought about how the formula was derived thus far...
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
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.
Very nice - though sadly "rectangle" has been spelt as "rectange" on a few screens.
The ancients really had the ability of reasoning and mental extrapolation !
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.
Nothing explains it better than the pizza illustration.
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
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.
Amazing proofs
Ancient People having better imagination than us...
What is so delicious about πr² is that it's describing an infinite number of pieces of cake (or pizza).
This was amazing thanks
That was very enjoyable! 🤓
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)
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
It is sad that back in grade school eons ago we were made to memorize the formulas for basic shapes w/o any form of explanation of WHY.
This is a fine video !!
"exactly the approximation"? That ranks up there with some of the silliest expressions I've ever heard - along with the ridiculous "very unique"!
That was cool. Those old guys were pretty smart.
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 ?
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.
Circumference used to hate pi but it’s coming around around…
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.
What a great channel.
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.
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.
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
The last section says at the bottom "A Mechanical Derivation of the Area of the Sphere". Should it be "Circle"?
Method 1 and 2 actually do not give us prove about the value of pi at all. All they gave is that pi is a ratio of C/d. All we got from these 2 methods is that A=Cr/2, which lead us back to square 1, because we need to know the value of C before we can calculate A and for that we need to have a way to calculate pi. Do you see the circular problem? The only method that gives us the answer is the third.
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?
Was it also found empirically at that time that pi is a constant, that the ratio is same for all circles?
I had never seen the third method before, it's pretty nice (but was probably a pain to formulate and enunciate clearly back then).
It was derived by iteration of small slices
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.
Why is there ambiguity going on recently ?
Sum of angles inside a circle = 360 OR ♾️ , 0.99999 = 1 or ≠ 1 ( by law of scientific significant figures ) ,
Because "angles inside a circle" isn't a precisely defined mathematical term. If you mean interior angles, this is a specific term for polygons, where the interior angles always add to (n - 2)*180 degrees, where n is the number of polygon sides.
In the limit as n goes to infinity for a regular polygon, the polygon approaches a circle. So plugging in n=infinity, shows us that the number of "interior angles" of a circle should add up to infinity. But it's difficult to say exactly what you mean by "interior angles", since they aren't angles between straight lines.
If you just mean how many degrees are there in a full rotation, then the answer is 360 degrees. This is a different question entirely than the sum of the interior angles inside the shape. If the angles in question add to 360 degrees for a circle, then they add to 360 degrees for all shapes. It is the exterior angles of the circle that add up to 360 degrees; not the "interior angles".