The production quality of these videos never ceases to amaze me: The fluiditiy of the animations seamlessly demonstrating the ideas as they are being narrated. The impeccable pacing in the script that dives into the real unexpectedness and wonder of math. The passion and care that are woven through it all. You really are today's Feynman to me! Thank you!
An absolutely brilliant video! I don't understand how the visualisations get better and better through each video, simply superb. I particularly enjoyed the alternating flashing technique to emphasise complicated parts of the video. You are the Shakespeare of Maths please never stop I am sure you will inspire some great minds in the future.
I'm already inspired - kind of addicted to math now The quaternions and projection series explaind the connection between s -doamain and z domain in DSP with out mentioning either. this guy is a genius at teaching. Maybe the next Plato or something like that
@Matthew Burson - Thank you. I didn't know he had a github repo. This guy is very nice and smart. Indeed, may he inspires a new generation which will have better tools, better understanding, and solid foundation to continue the "science work".
@@wilfroberts637 not quite - he uses latex for notation, but the animations are made with his own library called manim (which I assume comes from math anim).
I'm watching this video after a year and it makes much more sense to me than ever. I remember when I first learnt about the surface area of sphere, it itched me and I searched UA-cam for that but it didn't made that much sense to me but now, I'm satisfied. Thanks 3Blue1Brown!
You should hear old scientists complain. Basically, everything that took 5-10 years and/or tons of money can be done in a week to a year for a fraction of the relative cost.
Evariste Galois not everyone can be like him. Also, sadly, there is a ton of stuff they are required to cover so that they have barely anytime to do stuff like this
He's recommended a book before called Measurement by Paul Lockhart. It's a really good book and I would also recommend it. That said, this guy should definitely write a book of his own
In general, it is 4^k/(nCk). It seems that the integer multiple case appears only for k=1. Assume there exist an integer p such that 4^k/(nCk)=p. Since 4^k/(nCk) is monotonically increasing as k increases, we consider p>4 cases. Then, k*Log(4/p) = Log(nCk). In addition, LHS is negative for p>4 while RHS is positive, for positive integers k. Hence, only for p=4, LHS=RHS=0. This interesting case seems like following the law of small numbers.
I am considering only the odd dimensional cases. But it is also suprising that the transcendental number Pi does not appear for odd dimensional cases, while it is not for even dimensional cases.
Actually, there is another way to find surface area of circle, I actually noticed it when I was in my high school, if you differentiate the volume of sphere w.r.t. radius, then you get total surface area of sphere... The same case applies for Circumference of circle and area of circle, the circumference of circle is derivative of area of circle w.r.t. radius. I don't know if this is just a coincidence or there is actually some relation. You can also apply this rule to total surface area of cube and volume of cube etc.
@@NTdredd Yeah, that follows from the definition of the derivative. When you know why things work the way they do, that is precisely when math starts getting interesting.
I am from India. The IIT JEE is considered the toughest exam here, and probably in the world for 17 year old students. And I don't think that out of thousands of students who crack it with amazing grades, actually know anything with this precision.
@@youtubeshorts2911 The standard method Grant used was new to me, but to be honest I solved this problem using the method he has devised on his own, way back in class XI. Do not underestimate anyone. And Grant is not a Ph.D. He is a graduate from Stanford University. Ph.D doesn't make you knowledgable, hunger and patience do. Peace.
@@youtubeshorts2911 shivansh joshi , i agree with u what u think bout sarthak bro I am in 10 preparing for jee While I was reading about traingle I got the angle bisector theorem that it divides the opposite side by the same ratio of the two other sides First i proved it pratically by with the help of goemetry(meansurment ) and the i tried to prove it theoretically by using properties Although i wasn't able to prove it theoretically but then i saw it and i found it, i was very happy that i proved it pratically. Just telling because sarthak bro thinks that students dont understand concepts in depth. Bless me for jee 🙏 I want to secure AIR
Thanks for this video,i remember when i was in 10th class i ask my teacher about surface area and volume of sphere , he said no need to know that just learned the formula , so thanks for this .And one more thing that can make a video of volume of a sphere
The channel Think Twice has the best explanation for the volume of a sphere that I’ve ever seen. It uses Cavalieri’s principle (which it explains), and the face that the area of a pyramid with height h and base area A has volume hA/3 (which it does not explain, but is clear with some elementary calculus, and has some cute visual proofs). Go check it out!
I love 3b1b so much. 18 months out from the last time I sat in a maths classroom, I happened to see a picture of a problem on a whiteboard in the background of a photo shoot that was solving for the area of a sphere. I got curious and decided to look up the maths (because I thought it was wrong, yikes), and came upon this video. Whereas another video easily could have delved straight into calculus that I definitely no longer remember, this video ended up not only answering my question in detail, but left me saying, out loud, "How cool is that?". You are such a fantastic communicator and orator, and I'm so excited to see where you go as a new subscriber.
I've always thought geometry is the best way to introduce many mathematical concepts. And why haven't I watched 3blue1brown before? This is very much my kind of explanation. However, as tired as I am, I might have to skip getting the paper out. I'll just have to watch this one again some time. :) Edit: Optical illusion at 12:08 -- when separated, the rings appear to shrink latteraly.
1. Circumference: 2pi*sin(th)*R; Area of the ring: 2pi*sin(th)*R^2*dth 2. Area of the shadow: 2pi*sin(th)*cos(th)*R^2*dth = pi*sin(2th)*R^2*dth 3. For them to differ by a factor of 1/2, sin(th) must be equal to sin(2a) (where a is the other angle). So th=2*a. 4. Mapping area of each ring on the top of the sphere on to the shadow (halving the angle for each) we get a circle of shadows, whose radius is R/2, and whose area is 1/4*pi*R^2. After doing the same for the bottom portion, the total area is 1/2*pi*R^2. It is exactly 2 times less the area of the rings, so the area of half a sphere is pi*R^2 and so the area of the whole sphere is 2*pi*R^2, which means there's a hole in my argument but the general idea is correct I guess. * Correction * Actually, when I said that the radius of the mapping is equal to half the radius of the sphere I was wrong as it must be equal to √2/2 since the angle is 45° and cos(45°)=√2/2. And so the area of the portion of the shadow that we get after the mapping is equal to π*(√2/2*R)^2 = π/2*R^2. This way we get the right answer if we proceed with my steps sketched above, in the main part of the comment.
@@TheFlue2000 the circumference 2pi*sin(th)*R is the same and the projection of thickness, 2pi*R*dth, is 2pi*Rcos(th)*dth. multiply them together for the result.
That's because you weren’t counting only on only the top ring, but every even (or odd) ring in the whole sphere. So in the end, the area of the odd rings happen to be 2 times that of the shadow (a circle), so getting both the sum will be (2+2) times the area of the shadow, hence Asphere=4*pi*R^2
@@TheFlue2000 For #2, you can subtract the areas of the circles around the inner and outer edges of the shadow rings, one of which has radius R sin θ, and the other has radius R sin θ + R cos θ dθ. (That + might be a - depending on which exact triangle you use, but it works out the same.) Remember that area of a circle is πR², and (dθ)²=0.
Okay this just changed my perspective of circular areas I always thought it was just multiplying two 2piR which are perpendicular to each other This however makes WAY more sense
If education is required to improve, then I will vote for this channel. The animation is superbly great which properly matches the movement of the eye, a great way to learn even with beginners and non-mathematicians. Moreover, to create videos like this, it takes a trench-level of understanding of the topic. What this channel is teaching probably isn't being taught in some schools and universities. It dives into the most fundamental concepts/roots and answers the derivation of formulas we learned in schools. You cant call math a beautiful subject instantly, but in this way, you can see that it is indeed extremely beautiful and interesting. Kudos to this channel and I am thankful that I am born in this era of technology.
I feel you should sell more mathematical stuff on the 3b1b store, things like mechanical calculators, harmonic analyzer, some mathy visualization tools, some new version of chess u invented etc etc etc..... That represents you better than clothing with math printed on it...
Great point! The most honest answer here is that producing harmonic analyzers would be much harder, specifically in that getting a third party to handle logistics would be harder. I want to spend most of my time on videos, and for me, the store is a nice little way to promote expressions of a love of math while not pulling me away from the main pursuit too much. That said, I do agree with you, so will put in some more thought here... Anyone know a good harmonic analyzer supplier ;)
Dude, please take me as your apprentice. I'm binging on your videos desperate to know more driven by the need to destroy the "geniuses" by creating circumstances where intelligence borderlines knowledge. At a point I felt so dumb I wanted to paint the wall with my brains, but no that takes courage and I don't got that either. For all 🤘
The mapping from a slice to its shadow consists in replacing the factor 2 sin(- theta) by the factor sin(-2 theta). As far as I can tell, there is then no way of avoiding an integral computation, because this mapping does not give us a constant ratio between the slice and its shadow. Nor does it seem possible to exploit geometrically the substitution of 2 theta for theta. (Note: The angle from the pole being negative, I take its negative to get a positive slice surface; if instead we use an angle alpha from the equator, we will instead have a factor 2 cos(alpha) that gets replaced by sin(2 alpha), i.e. cos(pi/2 - 2 alpha).)
You have a big ego don't you, thinking such an arrogant thing. Did you ever think it maybe from people who are new to channel or don't understand the material.
Or just people who thought there would be a more direct relation between four circles and a sphere of the same radius. They were left 3blue1brown-balled.
15:25 is the question in the title of the video 15:38 has the answer “I’m sorry, but the princess is in another castle” So for 15+ minutes, all I could think about was how the distance of an object from its light source and projected surface change the size of the shadow despite its own consistency
I absolutely love your video series. They are so informative and the animation makes it all the more easier to relate and understand the concepts. Thank you. Re: the final exercise, after the step where we determine the area of the ring = \(2R^{2}\sin \left(x ight)\mathrm{d}x\) Why don't we just integrate it from 0 to pi: \(\int _{0}^{\pi }2R^{2}\sin \left(x ight)\mathrm{d}x\) ?
Awesome video! Could you also do a video on why the volume of a sphere is 4/3 pi r^3? Both by splitting it into tiny pyramids and by having the circular cross sections be equal to the annular cross sections made by subtracting a cone from the middle of a cylinder. Also maybe include how the surface area's "circumference", surface area and volume formulas are all derivatives/integrals of each other.
If the plane on which the shadow is cast is not perpendicular to the light source, what inclination would the plane have to be such that the elliptical shadow cast is equal to the surface area of the sphere?
I think this problem is exactly the reason why our world map today is not accurate but we dont have a better way to draw a world map that matches the scale
Wow! lots of technology to understand the area of a sphere! Very Clever! Wonder what Archimedes would have done if he had known all of these technologies...
Answer: area of a surface ring = 2πR^2*sin(θ)dθ area of a shadow = 2πRsin(θ)*cos(θ)Rdθ = πR^2*sin(2θ)dθ = 0.5*(2πR^2*sin(2θ)dθ) This is half the surface area of a surface ring with angle 2θ. Only half the surface area of the sphere will be covered by the rings as for every dθ covered on the projection 2dθ is covered on the sphere. So the xy disc has an area of half, half the surface area of the sphere. So the surface area of the sphere is 4 times the area of the xy disc i.e. the sphere’s shadow.
How lucky we feel about the day we found this channel...or probably it found us..!! They(UA-cam feeds) say ..it is indeed the best maths channel presently. ❤️
I really appreciate your efforts to show beautiful geometric proofs, without using calculus. I would like to ask you a question if you don't mind: is it possible to find out what are the geodesics on a sphere, without using calculus?
im particularly proud of myself for realising that the shadow will be the largest crossection of the circle given light rays are parallel, which is pir^2.
I have to say, trying to relate the area of a circle to its surface, via rings, was too unintuitive to only try to get us to figure it out on our own. I find that "intuitiveness" is not really about relating specific shapes directly or indirectly to other ones, but rather is a process in the brain that relates to speed. If you gather information in your brain fast enough to connect it all together, into *something*, anything, that is meaningful, then it will be understood as intuitive. Intuitiveness isn't about creating a (what I consider to be) convoluted way of finding your answer, and the only thing that connects your starting point and end result is a similar looking shape, but is about using simple enough terms, and concepts, to deduce the answer fast enough that your brain does not have to remind itself of some parameter that is precise, that you may have forgotten, or forgotten the significance of, by the time you reach the next step, or one of the next steps in the method. In that sense, explaining things quickly, like you did with the previous triangle example was much more intuitive than the final Ring Method. I'm not particularly good at math, so perhaps I could have done better, but I did just spend 4 hours in total trying to answer this, and getting frustrated over it. Now that I read a comment explaining it better, I understand the implications better, and perhaps that's all that was needed, a little learning. But to my taste, I would advise against using this ring method again, especially in an educational video, unless of course you're explaining it fast enough to make intuitive. I did like your animations though. Thanks for at least putting this out there.
I fear that there is (more than) a fault in my reasoning :-D The ratio between a ring area on the half sphere and its projection on the plane (at distance d from the origin) goes from 1 to 0, when we move from the pole to the equator (so that d goes from 0 to R). Let's call it r, where r=1-d/R. In the r-d plane, r draws the right triangle (0,1) (R,0) (0,0) which has R/2 area in the [0,R] domain. Let's consider the ratio between a ring area on the upper half sphere, and its projection on the bottom half sphere: we get the expression r=1 in the r-d plane, which yields a rectangle with R area, in the [0,R] domain. The sum of the areas of the projections is obviously πR² (the circle with radius R), and because this sum maps to the R/2 area in the r-d plane, the value R in r-d plane (that corresponds to the total area of the rings on the half sphere), has to map to the double of the circle area, 2πR². The area of the two halves is therefore 4πR². Anyway awesome video Grant, awesome.
Whonactually understood any of this video..its convoluted and impossible tonunderstand and turns me off from bring curious to understand things because itntakes so damn long and is so confusing tonunderstand one simple thing..sheesh..
@@leif1075 The point is to gain a general understanding of the principles and techniques at work. You may know the formula from another method of derivation, but Grant's videos emphasize the methods as well as the result they talk about. It helps to cultivate better mathematical thinking.
@@MohaMMaDiN55 this is mathematics not physicsm, but your logic calculus is invalid because in calculus you have to have continuous matter and there's NO continuous matter in real life. talking about you know the the normal derivative and such things
Because good teachers are extremely rare, even more so in maths. And the technology to do this is finally becoming nice enough to do this kind of thing. I hope we see a big push in the next 5-10 years with this kind of teaching.
After watching this you may have the general idea of how the maths behind this work, but have you tried doing it yourself ? You probably won't be able to do it. That's why mathematics aren't taught like this. You need to do it yourself, sit in front of a blank piece of paper and try the demonstration on your own.
7:53 "sometimes easy to forget why we're doing this".. its amazing how you pay attention to where one might loose track, after having mastered the topic.
I also do that when i use tutor my friends. i think in their own perspective and think of problems they might not be getting and i tutor them base on self-explanatory conditions and not those complicated concepts that no one can even logically answer like "how did the long division method got formulated", it should be self-explanatory.
you just helped me make the connection. The name of the channel is 3blue1brown, and in the animated classroom there are 3 blue pi student creatures and 1 brown pi teacher creature
Realmente es un excelente video, la claridad de la explicación, la nitidez y belleza de las imágenes, la sencillez del lenguaje, este video deberîa ser usado por muchos profesores en sus clases..
@@Danilego I got this as the area of the shadow 2π*R^2*deltaø*cos(ø)*sin(ø). It's a complete mess, i'm not even sûre about the cos thing. I tried replacing the whole 2sin*cos thing by sin(2ø) but I didn't found any way out
@@jakub.kubicek stop blaming school. Even after watching this video, there are still people who doesnt even understand what is going on. People hate school becuz require real serious test. The test result determ if u are qualify for next grade or not up until the final grade which then u take college entrance exam. School need to prepare u for college entrance exam, it is more intense which is why it is more stressful. But on a youtube video, u are free to not understand anything u just watch but at the same time , u arent gonna pass any entrance exam anyime soon. These kind of video are good for inspiring new ppl to love math or for math entusiastic to learn more about math that they alr learn at school.
The most amazing thing is that Archimedes found the surface area of a sphere over a thousand years before calculus was even invented. While he didn't prove it to modern rigor, he can hardly be blamed for that.
Actually, his proof was bullet proof even by modern standards. He proved one miraculous lemma which underlay not only his proof that the area of a sphere is 2/3 the area of its circumscribed cylinder (lateral area + top & bottom) but also that it's volume is 2/3 the volume of this same cylinder! Legend has it that the sphere and its circumscribing cylinder were inscribed on his tombstone.
It helps a lot if you remember that Oscar Had A Hold On Arthur. Answers below the fold: Q1: Let's call the radius of the ring d. We have a right triangle with an angle of theta and a hypotenuse of R. In this case, d is opposite from theta. Using the above mnemonic, we can remember that O/H = sin(theta), ergo d = R sin(theta). The circumference of a circle is 2 pi R, so the inner circumference of the ring is *2 pi R sin(theta).* Thus the ring's area is approximately *2 pi R^2 sin(theta) d(theta).* Q2: The good news is that our inner radius d is the same as it was for the ring on the sphere, ergo the inner circumference will also be the same: 2 pi R sin(theta). What we need to figure out is the thickness of the ring's shadow. By drawing another right triangle where the hypotenuse is the thickness of the ring, R d(theta), we can see that the thickness of the shadow is adjacent to theta in our new triangle. Using the mnemonic above, we can see that A/H = cos(theta), ergo the thickness of the shadow = R cos(theta) d(theta). To finish off, we multiply these two to get an area of *2 pi R^2 sin(theta) cos(theta) d(theta).* Q3: Using the identity that 2 sin(theta) cos(theta) = sin(2*theta) reveals that we can rewrite the area of the shadow as *pi R^2 sin(2*theta) d(theta).* This is the same as the area of the ring except that we've dropped the 2 from in front, signifying that we've cut the value in half, but we've also doubled the value of theta. This means that the shadow at a given value of theta has half the area of the ring at double that theta value. For example, the shadow at theta = 30 degrees has half the area of the ring at 60 degrees. Thus, as we go to the next shadow, we skip past one of the rings and jump two rings ahead instead of one. Q4: Partially answered above, but as we compare each shadow to a ring on the sphere, we have to skip every other ring, jumping two rings ahead for each shadow. The other half of this puzzle is to remember that we only generated shadows from one hemisphere rather than the entire sphere. Since we skip one ring for each shadow, that means we need to use all of the rings from the entire sphere (except for the ones we jump over), instead of just using the rings from one hemisphere. An easy way to see this is to think about the last shadow, at theta = 90 degrees, which corresponds to half the area of the ring at 180 degrees, which is the last ring on the sphere. Q5: The area of the shadows sums up to the area of a circle of radius R. However, each shadow is only _half_ the area of one of the rings, and only half of the rings have been accounted for. A half of a half is one quarter. Ergo, a circle of radius R only has half the area of one hemisphere of the sphere, which in turn only has half the area of the whole sphere, and thus the area of the circle is one quarter that of the entire sphere.
Edit: Other comments explained my mistake. Thanks! That agrees with what I got, except that my area formulas has R instead of R^2. I've seen a few other commenters using R^2, so it is probably correct. I got the area as circumference*width where width is dø for the ring and cos(ø)dø for the shadow. To me this also makes sense that I multiply two distances to get area, rather than 3 (R, R, dø). How did you get the other R?
@@AnonymousAnonymous-ht4cm the width is linearly proportional to R. For example, at a fixed central angle, the arc length of an arc will increase by a factor of R as you increase the radius. Same reason why circumference is 2πR and not just 2π
Bravo sir. I'm actually glad that my intuition was leading me along a similar path. If half of the rings from one hemisphere gets us a circle of equal radius to the sphere, then the other half of the rings gets us two circles, and if we do the same on the other hemisphere we end up with four circles in total.
@@AnonymousAnonymous-ht4cm You are close. The widths are actually R*dø and R*cos(ø)*dø for the ring and shadow, respectively. That is where the extra R arises that you are missing. This should make intuitive sense because a sphere with a larger R will have wider strips for a given dø. Similarly, smaller spheres will have thinner strips for a given dø. You can also roughly think about this in terms of units. dø and cos(ø)*dø are essentially unitless. One needs to multiply these quantities by R to get a unit of length.
Congrats on such an amazing video, omg it's 3AM but here's how I did the exercise proof: 1. The circumference length of each ring is 2*pi*R*sin(theta), since the distance from the ring to the axis is R* sin(theta) (trigonometry). Hence, the area of a ring is 2*pi*R*sin(theta)*R*d_theta = 2*pi*R^2*sin(theta)*d_theta (1) 2. To calculate the area of a ring's shadow, I used some trig relations as well. In this case, the thickness of the ring shadow is R*d_theta*cos(theta). Therefore, the area of the ring shadow is 2*pi*R^2*sin(theta)*cos(theta)*d_theta (2) 3. Multiplying by both sides expression (2), we get 2*pi*R^2*2*sin(theta)*cos(theta)*d_theta I've put the number 2 right next to sin(theta)*cos(theta) to explicit the trigonometrical relation: 2*sin(theta)*cos(theta) = sin(2*theta) That specific angle is 2*theta. 4. So, this means that the area of the shadow of some ring with a corresponding angle of theta is equal to the area of a ring which has double of this angle. 5. Notice that, when theta ranges from zero to pi/2, we get to form all of the rings related to their shadows in the superior hemisphere. The total area of the shadow is the sum of all the thin ring shadows. Thus, that is equal to half the area of the hemisphere. That is, the area of the shadow (pi*R^2) is 1/4 of the surface area of the sphere. A(sphere) = 4*pi*R^2 Q.E.D.
I'm confused... When we sum up the areas of the each of the shadows of the rings, for all the rings the top hemisphere, we get the shadow of the sphere itself. That's equal to (pi*r^2). I get this step The shadow of a ring at an angle of theta has the same area as a ring at an angle of 2theta itself. Therefore, summing over all the shadows in the top hemisphere is the same thing as summing over the area of every other ring in the top hemisphere...this step I don't get at all. Why is this? Why does the fact that for any ring A, there's a ring B at double the angle of A that has the same area as the ring A's shadow, imply that when we sum over all the shadows if the rings in the top hemisphere, we get half the area of the top hemisphere...? Assuming that we do, we need to multiply this area by 4 to get the surface area of the sphere...so this step I get again. But I'm lost in the in-between step. Thanks!
Thank you for this, I was able to get most of the math right, but mistaking the cosine for a sine messed everything up, and figuring out that the result of sin(2theta)=sin(theta) was talking about two different angles was super helpful
A couple years ago, I was trying to estimate the amount of paint needed for an airliner. I based it on some simplifying assumption with regard to shapes and was surprised to find an equivalence between a hemisphere capped cylinder and open ended cylinder of the same overall length. Made me double check my math for the generalized case and indeed, (2 pi r)(l + 2r) describes both cases. So for my estimate paint usage, I just used the open ended cylinder with the length and radius for the fuselage and engines, plus a rough combination of triangles for the tail. (Wings, too, but their numbers stay separate since they use different products to prevent adhesion of ice)
@@h-Films Let: 1. number of questions done = f(t); unit = no of questions 2. time = T; unit = minutes The function f(t) does this: 1. when t = 0, x = 0 2. when t = 4, x = 4 The history teacher concludes that the gradient, f'(t) or df(t)/dt, is a constant value, which is 1 question per minute, obtained by this formula: f'(t) = [f(4)-f(0)]/(4-0) While it is valid if f(t) is a straight line, it is usually inaccurate otherwise since f'(t) often changes based on t (e.g. quadratic f(t)). This method is in a way similar to trapezium and newton-raphson approximations. Or, he's just saying that difficulty variations of calculus Q is very high, so you can't just say that all questions take the same time (1 minute) to complete.
( 4/3 ♊R^2 )' = 4♊R .. Surface is the differentiel of the volume ( différentiel in french )... Circonference ...the differential of the surface : 2♊R = ( ♊R^2 ) ' ....
I remember trying to prove this in highschool. It seemed impossible given the knowledge of a child but I wonder if I had a teacher like you that time, it would have been an enlightened day of my lifetime. Thanks for these elegant proofs ❤️
@@maheshm5463 I'm older still. I find calculus astounding. In school I found math nearly useless and when asked, no one could tell me why I should lean it except "you'll need it if you go to college." If anyone had just told me "basic math and algebra are the language of straight lines and planes (not always true, I know) but calculus is the language of curves and therefore, its the mathematical language of the Universe." It would have changed my perspective and therefore, my life. Tragic, really. Just 1 hour of the concept of integral calculus would have sparked my interest and I would have gone from "Learning because I have to" to "Learning because a WANT to".
I'm 60 and. I wish I had a maths teacher like this in school, instead we had a loud mouthed aggressive bully that terrified everyone in class, and no one was good at maths in my class, I wonder why ? He died of a Hart attack the year after I left school aged 40 and thought it's a pity he didn't die the year before I started. Now I'm disabled and at home most of the time and due to UA-cam I've found a new love for maths, I would like to start all over again from scratch as a beginner. If any one out there knows of some good videos on UA-cam with the same inspiring content as this please post a link. Thanks in advance 😀
The production quality of these videos never ceases to amaze me:
The fluiditiy of the animations seamlessly demonstrating the ideas as they are being narrated.
The impeccable pacing in the script that dives into the real unexpectedness and wonder of math.
The passion and care that are woven through it all.
You really are today's Feynman to me! Thank you!
well said.
also he makes it understandable to people watching even if theyre like in middle sch who didnt learn advanced maths or something
The craziest thing about the animation is that it’s all computer generated from code
While there are tools to help visualise geometry, the amount of work to produce such material is still so enormous!
@@THEELEMENTKING fr, i find that more impressive than doing it by hand. he managed to create a script to make fluid and clean animations for him
An absolutely brilliant video! I don't understand how the visualisations get better and better through each video, simply superb. I particularly enjoyed the alternating flashing technique to emphasise complicated parts of the video. You are the Shakespeare of Maths please never stop I am sure you will inspire some great minds in the future.
I'm already inspired - kind of addicted to math now The quaternions and projection series explaind the connection between s -doamain and z domain in DSP with out mentioning either. this guy is a genius at teaching. Maybe the next Plato or something like that
github.com/3b1b/manim
@Matthew Burson - Thank you. I didn't know he had a github repo. This guy is very nice and smart.
Indeed, may he inspires a new generation which will have better tools, better understanding, and solid foundation to continue the "science work".
Dude, Y the name 3blue1brown?
@@danielsouza2129 His eye color
Can we just take a moment to admire the quality of all animations in these videos ? It's just insane
I'm pretty sure all the animations are written in their own coding language as a part of latex if I remember correctly
@@wilfroberts637 not quite - he uses latex for notation, but the animations are made with his own library called manim (which I assume comes from math anim).
Hum bhi bnane baithe the manim se gand fat gya mera
well he did make a Python library to animate these videos for him
IIT
This guy knows how to explain every detail and knows exactly what questions will be asked and immediately answers them. This guy is truly amazing!
I'm watching this video after a year and it makes much more sense to me than ever. I remember when I first learnt about the surface area of sphere, it itched me and I searched UA-cam for that but it didn't made that much sense to me but now, I'm satisfied. Thanks 3Blue1Brown!
I grew up in the "primitive era", learning math was murder. We've come a long way in the last 60+ years.
OK, Boomer!
@@cdfactory I really don't wanna hear their hour long story about how they got to school
You should hear old scientists complain. Basically, everything that took 5-10 years and/or tons of money can be done in a week to a year for a fraction of the relative cost.
@@bringonthevelocirapture Except when they can't. Try designing an aircraft. Still damn hard.
@@Perririri trianon
3Blue1Brown: *Explaining mathematical concepts better than school ever could*
Evariste Galois not everyone can be like him. Also, sadly, there is a ton of stuff they are required to cover so that they have barely anytime to do stuff like this
I'm majoring in math, so I might be a math teacher some day. I'm planning on also completing a computer science degree, so hopefully I won't be.
I value proofs a lot more than my peers
My cal 4 professor talked about this when we did double integrals to find surface area, so it's not entirely fair to say none of them do this
Good luck getting a high school diploma merely by watching UA-cam videos for a few hours.
Make a video on how 3 cones make a cylinder
Fill 3 cones with water, pour them all in a cylinder of the same height, thats how my geometry teacher taught it
@@xicad1533 and same radius.
Mr. Virtual 𝕟𝕠 𝕠𝕟𝕖 𝕤𝕒𝕚𝕕 𝕒𝕟𝕪𝕥𝕙𝕚𝕟𝕘 𝕒𝕓𝕠𝕦𝕥 𝕡𝕣𝕠𝕠𝕗𝕤
@@chrisding1976 *_how do you make that font teach me the ways master_*
Tʜᴇ ғᴏɴᴛ ᴀᴘᴘ
My Calculus teacher: Now prove it with integration 💀
*ADMIT* *IT*
The *_Beauty_* *_of_* *_mathematics_* is the most satisfying thing ever..
Well.. it's Mathematics and Physics
@@ssuriset yeah
Ya
@@ssuriset well, it's science in general
You havent tried extasy
Please write a book on the beauty of mathematics!!!
He's recommended a book before called Measurement by Paul Lockhart. It's a really good book and I would also recommend it.
That said, this guy should definitely write a book of his own
Lots of good books on math already. 3B1B's strength is making videos on the beauty of math, so I just hope he keeps doing this!
A flipbook
Yes.i would buy it
A movie.
Even better: a 3D movie.
Grant, your math, narration and animation come together seamlessly like a Swiss watch. These videos are the very definition of 'Excellence'.
Your videos are stunning in their simplicity, a perfect blend of math, computer, programming, and speech. Well done!
10pm: 1 last video and I will go to sleep!
3am:
I'm falling asleep from boredom just watching it.
Wow
6am:
@@joet840 thats probably because you dont get it
Opposite affect on me, but I came back and glad I did.
Bro UA-cam. its 3 in the morning. Im not ready for math
In general, it is 4^k/(nCk). It seems that the integer multiple case appears only for k=1.
Assume there exist an integer p such that 4^k/(nCk)=p.
Since 4^k/(nCk) is monotonically increasing as k increases, we consider p>4 cases.
Then, k*Log(4/p) = Log(nCk). In addition, LHS is negative for p>4 while RHS is positive, for positive integers k.
Hence, only for p=4, LHS=RHS=0.
This interesting case seems like following the law of small numbers.
I am considering only the odd dimensional cases. But it is also suprising that the transcendental number Pi does not appear for odd dimensional cases, while it is not for even dimensional cases.
Actually, there is another way to find surface area of circle, I actually noticed it when I was in my high school, if you differentiate the volume of sphere w.r.t. radius, then you get total surface area of sphere...
The same case applies for Circumference of circle and area of circle, the circumference of circle is derivative of area of circle w.r.t. radius.
I don't know if this is just a coincidence or there is actually some relation.
You can also apply this rule to total surface area of cube and volume of cube etc.
@@NTdredd woah that's actually cool, never knew this
@@NTdredd Actually, this is not a co incidence. Derivative is good way to find it.
Maybe you can find more on quora or just google it.
@@NTdredd Yeah, that follows from the definition of the derivative. When you know why things work the way they do, that is precisely when math starts getting interesting.
I am from India. The IIT JEE is considered the toughest exam here, and probably in the world for 17 year old students. And I don't think that out of thousands of students who crack it with amazing grades, actually know anything with this precision.
You Said the correct thing Man math and physics is not a headache it's amazing if thought in the right way
@@youtubeshorts2911
The standard method Grant used was new to me, but to be honest I solved this problem using the method he has devised on his own, way back in class XI. Do not underestimate anyone. And Grant is not a Ph.D. He is a graduate from Stanford University. Ph.D doesn't make you knowledgable, hunger and patience do.
Peace.
@@sarthakgirdhar2833 Graduate ka matlab google kar le bhai. All the best with your GRE preparation.
@@youtubeshorts2911 shivansh joshi , i agree with u what u think bout sarthak bro
I am in 10 preparing for jee
While I was reading about traingle
I got the angle bisector theorem that it divides the opposite side by the same ratio of the two other sides
First i proved it pratically by with the help of goemetry(meansurment ) and the i tried to prove it theoretically by using properties
Although i wasn't able to prove it theoretically but then i saw it and i found it, i was very happy that i proved it pratically.
Just telling because sarthak bro thinks that students dont understand concepts in depth.
Bless me for jee 🙏
I want to secure AIR
And sarthak is it not precise
Thanks for this video,i remember when i was in 10th class i ask my teacher about surface area and volume of sphere , he said no need to know that just learned the formula , so thanks for this .And one more thing that can make a video of volume of a sphere
shaikh mohd Hamza that's bad your teacher was only teaching you for marks not for true education
@@itolukibami725 He probably did not know the derivation. This video is superb!
The channel Think Twice has the best explanation for the volume of a sphere that I’ve ever seen. It uses Cavalieri’s principle (which it explains), and the face that the area of a pyramid with height h and base area A has volume hA/3 (which it does not explain, but is clear with some elementary calculus, and has some cute visual proofs). Go check it out!
I love 3b1b so much. 18 months out from the last time I sat in a maths classroom, I happened to see a picture of a problem on a whiteboard in the background of a photo shoot that was solving for the area of a sphere. I got curious and decided to look up the maths (because I thought it was wrong, yikes), and came upon this video. Whereas another video easily could have delved straight into calculus that I definitely no longer remember, this video ended up not only answering my question in detail, but left me saying, out loud, "How cool is that?". You are such a fantastic communicator and orator, and I'm so excited to see where you go as a new subscriber.
02:20
The transition from Circles to the Triangle was OP! 🤟😻
I've always thought geometry is the best way to introduce many mathematical concepts. And why haven't I watched 3blue1brown before? This is very much my kind of explanation. However, as tired as I am, I might have to skip getting the paper out. I'll just have to watch this one again some time. :)
Edit: Optical illusion at 12:08 -- when separated, the rings appear to shrink latteraly.
1. Circumference: 2pi*sin(th)*R; Area of the ring: 2pi*sin(th)*R^2*dth
2. Area of the shadow: 2pi*sin(th)*cos(th)*R^2*dth = pi*sin(2th)*R^2*dth
3. For them to differ by a factor of 1/2, sin(th) must be equal to sin(2a) (where a is the other angle). So th=2*a.
4. Mapping area of each ring on the top of the sphere on to the shadow (halving the angle for each) we get a circle of shadows, whose radius is R/2, and whose area is 1/4*pi*R^2. After doing the same for the bottom portion, the total area is 1/2*pi*R^2. It is exactly 2 times less the area of the rings, so the area of half a sphere is pi*R^2 and so the area of the whole sphere is 2*pi*R^2, which means there's a hole in my argument but the general idea is correct I guess.
* Correction *
Actually, when I said that the radius of the mapping is equal to half the radius of the sphere I was wrong as it must be equal to √2/2 since the angle is 45° and cos(45°)=√2/2. And so the area of the portion of the shadow that we get after the mapping is equal to π*(√2/2*R)^2 = π/2*R^2. This way we get the right answer if we proceed with my steps sketched above, in the main part of the comment.
Can you explain how you got to 2?
@@TheFlue2000 the circumference 2pi*sin(th)*R is the same and the projection of thickness, 2pi*R*dth, is 2pi*Rcos(th)*dth. multiply them together for the result.
That's because you weren’t counting only on only the top ring, but every even (or odd) ring in the whole sphere. So in the end, the area of the odd rings happen to be 2 times that of the shadow (a circle), so getting both the sum will be (2+2) times the area of the shadow, hence Asphere=4*pi*R^2
@@TheFlue2000 For #2, you can subtract the areas of the circles around the inner and outer edges of the shadow rings, one of which has radius R sin θ, and the other has radius R sin θ + R cos θ dθ. (That + might be a - depending on which exact triangle you use, but it works out the same.) Remember that area of a circle is πR², and (dθ)²=0.
@@bjornfidder at that point I'm counting the area of only half of the rings. And the total area at the end must turn out to be 4*pi*R^2.
you just save my life in many whays, this is fkn amazing
thank you
these animations are astonishing. congrats.
학교에서는 구의 겉면적을 이중적분 혹은 카발리에르의 원리로 증명을 하곤 했는데 구의 겉면적을 수학 애니메이션으로 증명하는 획기적인 방법을 이 영상에서 제시한 것 같습니다. 문제를 푸는데는 다양한 방법이 있다는 사실을 알려준 좋은 영상인 것 같습니다.
Okay this just changed my perspective of circular areas
I always thought it was just multiplying two 2piR which are perpendicular to each other
This however makes WAY more sense
this also makes sense thanks !
I always used to love maths in school. And now that I saw mathematics in such beauty as you present it, I really start to miss it
If education is required to improve, then I will vote for this channel. The animation is superbly great which properly matches the movement of the eye, a great way to learn even with beginners and non-mathematicians. Moreover, to create videos like this, it takes a trench-level of understanding of the topic. What this channel is teaching probably isn't being taught in some schools and universities. It dives into the most fundamental concepts/roots and answers the derivation of formulas we learned in schools. You cant call math a beautiful subject instantly, but in this way, you can see that it is indeed extremely beautiful and interesting. Kudos to this channel and I am thankful that I am born in this era of technology.
I've never seen such a powerful animation study in my life. Great job man.. keep going 💪
these videos are a perfect way to keep my brain warm mathematically. thanks a bunch
All the other Pi's must be so annoyed at the first Pi's questioning
Unless they're relieved that someone is asking questions they had themselves.
I finally understood clearly how the surface area of a sphere is 4pir^2
I thought this was just going to be an revolution integral from 0 to pi.
I feel you should sell more mathematical stuff on the 3b1b store, things like mechanical calculators, harmonic analyzer, some mathy visualization tools, some new version of chess u invented etc etc etc..... That represents you better than clothing with math printed on it...
Great point! The most honest answer here is that producing harmonic analyzers would be much harder, specifically in that getting a third party to handle logistics would be harder. I want to spend most of my time on videos, and for me, the store is a nice little way to promote expressions of a love of math while not pulling me away from the main pursuit too much. That said, I do agree with you, so will put in some more thought here... Anyone know a good harmonic analyzer supplier ;)
@@3blue1brown MathMo deal striking!
"but why?"
The part we dont always get in class
Dude, please take me as your apprentice. I'm binging on your videos desperate to know more driven by the need to destroy the "geniuses" by creating circumstances where intelligence borderlines knowledge. At a point I felt so dumb I wanted to paint the wall with my brains, but no that takes courage and I don't got that either. For all 🤘
The mapping from a slice to its shadow consists in replacing the factor 2 sin(- theta) by the factor sin(-2 theta).
As far as I can tell, there is then no way of avoiding an integral computation, because this mapping does not give us a constant ratio between the slice and its shadow. Nor does it seem possible to exploit geometrically the substitution of 2 theta for theta.
(Note: The angle from the pole being negative, I take its negative to get a positive slice surface; if instead we use an angle alpha from the equator, we will instead have a factor 2 cos(alpha) that gets replaced by sin(2 alpha), i.e. cos(pi/2 - 2 alpha).)
the dislikes are math teachers who are jealous :-)
The dislikes are who believe that the Earth is plane.
You have a big ego don't you, thinking such an arrogant thing. Did you ever think it maybe from people who are new to channel or don't understand the material.
Or just people who thought there would be a more direct relation between four circles and a sphere of the same radius. They were left 3blue1brown-balled.
The dislikes are bots
@@btdpro752 Nah. It's just you too stupid to understand a joke.
15:25 is the question in the title of the video
15:38 has the answer “I’m sorry, but the princess is in another castle”
So for 15+ minutes, all I could think about was how the distance of an object from its light source and projected surface change the size of the shadow despite its own consistency
I absolutely love your video series. They are so informative and the animation makes it all the more easier to relate and understand the concepts. Thank you.
Re: the final exercise, after the step where we determine the area of the ring = \(2R^{2}\sin \left(x
ight)\mathrm{d}x\)
Why don't we just integrate it from 0 to pi:
\(\int _{0}^{\pi }2R^{2}\sin \left(x
ight)\mathrm{d}x\) ?
Wow. You've managed to convince me that an equal area projection is somehow a natural way to view a map.
Awesome video! Could you also do a video on why the volume of a sphere is 4/3 pi r^3? Both by splitting it into tiny pyramids and by having the circular cross sections be equal to the annular cross sections made by subtracting a cone from the middle of a cylinder. Also maybe include how the surface area's "circumference", surface area and volume formulas are all derivatives/integrals of each other.
Please do one on Curvature (intrinsic/extrinsic and all that) relating to GR!
I have never seen this explained in such a lucid and simple manner. Hats off to you sir.
Wonderful video and clever insights really... Thanks!
This is why I loved Loomis for calculus: it trained my intuition without ever leading me astray.
Can someone enlighten me...how are these amazing animations created
Not often shown in this manner but amazing when it is . .
If the plane on which the shadow is cast is not perpendicular to the light source, what inclination would the plane have to be such that the elliptical shadow cast is equal to the surface area of the sphere?
a different level of knowledge game is going here , this is something special
how'd you animate 0:31-0:50? it's so realistic and smooth. i honestly cant even imagine making that, let alone rendering it.
he uses a custom made python software to animate math
its in the description
@@sugaristhenewwhite it was a joke the timestamp was of him irl
I feel blessed to have access to great materials than the earlier generations.
I think this problem is exactly the reason why our world map today is not accurate but we dont have a better way to draw a world map that matches the scale
Wow! lots of technology to understand the area of a sphere! Very Clever! Wonder what Archimedes would have done if he had known all of these technologies...
The length is not the width. Clarity is better than confusion.
Hopefully, by next year, I can understand this video better than now,
very good explaination cleared my maths doubts. thanks
Answer: area of a surface ring = 2πR^2*sin(θ)dθ
area of a shadow = 2πRsin(θ)*cos(θ)Rdθ
= πR^2*sin(2θ)dθ
= 0.5*(2πR^2*sin(2θ)dθ)
This is half the surface area of a surface ring with angle 2θ. Only half the surface area of the sphere will be covered by the rings as for every dθ covered on the projection 2dθ is covered on the sphere. So the xy disc has an area of half, half the surface area of the sphere. So the surface area of the sphere is 4 times the area of the xy disc i.e. the sphere’s shadow.
0:33 rare apparition of irl footage in a 3b1b video
This animation is beautiful! 👍
How lucky we feel about the day we found this channel...or probably it found us..!!
They(UA-cam feeds) say ..it is indeed the best maths channel presently. ❤️
I really appreciate your efforts to show beautiful geometric proofs, without using calculus. I would like to ask you a question if you don't mind: is it possible to find out what are the geodesics on a sphere, without using calculus?
The cylinder is more elegant, but the use of one HALF of the rings being made into a circle (then into triangles) is more intuitive
Visual proofs can be rigorous. Grant's animated presentational tools can facilitate such proofs.
im particularly proud of myself for realising that the shadow will be the largest crossection of the circle given light rays are parallel, which is pir^2.
Just to be more precise, a shadow caused by a light source that is at infinite distance.
Educational videos like this should be automatically payd by YT.
ONE dislike?
Who are you, one disliker?
How are you doing?
Everything ok with your life?
Sit here, let's talk...
Newton and Leibnitz are pleased by this explanation. 👌
I have to say, trying to relate the area of a circle to its surface, via rings, was too unintuitive to only try to get us to figure it out on our own.
I find that "intuitiveness" is not really about relating specific shapes directly or indirectly to other ones, but rather is a process in the brain that relates to speed. If you gather information in your brain fast enough to connect it all together, into *something*, anything, that is meaningful, then it will be understood as intuitive.
Intuitiveness isn't about creating a (what I consider to be) convoluted way of finding your answer, and the only thing that connects your starting point and end result is a similar looking shape, but is about using simple enough terms, and concepts, to deduce the answer fast enough that your brain does not have to remind itself of some parameter that is precise, that you may have forgotten, or forgotten the significance of, by the time you reach the next step, or one of the next steps in the method.
In that sense, explaining things quickly, like you did with the previous triangle example was much more intuitive than the final Ring Method. I'm not particularly good at math, so perhaps I could have done better, but I did just spend 4 hours in total trying to answer this, and getting frustrated over it. Now that I read a comment explaining it better, I understand the implications better, and perhaps that's all that was needed, a little learning. But to my taste, I would advise against using this ring method again, especially in an educational video, unless of course you're explaining it fast enough to make intuitive.
I did like your animations though. Thanks for at least putting this out there.
I'm a simple man. I see spherical or cylindrical coordinates systems, I skip.
Excellent video thanks a lot i loved it. Nothing is above thisss
I fear that there is (more than) a fault in my reasoning :-D
The ratio between a ring area on the half sphere and its projection on the plane (at distance d from the origin) goes from 1 to 0, when we move from the pole to the equator (so that d goes from 0 to R). Let's call it r, where r=1-d/R.
In the r-d plane, r draws the right triangle (0,1) (R,0) (0,0) which has R/2 area in the [0,R] domain.
Let's consider the ratio between a ring area on the upper half sphere, and its projection on the bottom half sphere: we get the expression r=1 in the r-d plane, which yields a rectangle with R area, in the [0,R] domain.
The sum of the areas of the projections is obviously πR² (the circle with radius R), and because this sum maps to the R/2 area in the r-d plane, the value R in r-d plane (that corresponds to the total area of the rings on the half sphere), has to map to the double of the circle area, 2πR². The area of the two halves is therefore 4πR².
Anyway awesome video Grant, awesome.
Pretty: a simple enough perspective for our trivial human brain to relate and understand.
But you forgot to mention how well a contact lens fits on a round eye! So a circle can fit perfectly on a sphere.
This is my Favorite lecture out of all math lectures that ive listened to all my life
massive respectt. o7
Normal people : Why?
3Blue1Brown : But why?
3blue1One?
@@giantrunt Fixed
Whonactually understood any of this video..its convoluted and impossible tonunderstand and turns me off from bring curious to understand things because itntakes so damn long and is so confusing tonunderstand one simple thing..sheesh..
@@leif1075 The point is to gain a general understanding of the principles and techniques at work. You may know the formula from another method of derivation, but Grant's videos emphasize the methods as well as the result they talk about. It helps to cultivate better mathematical thinking.
@@aviralsood8141 hello aviral
Aryan here
him: unwraps a circle into a triangle
me: you CaN dO tHaT?
Kayle Needler you can unwrap a circle into a triangle by cutting a perfect spiral from the edge to the exact centre
Actually you can’t do that unless I think if the circle is made up of an elastic material like rubber.
Theoretically
@@MohaMMaDiN55 this is mathematics not physicsm, but your logic calculus is invalid because in calculus you have to have continuous matter and there's NO continuous matter in real life.
talking about you know the the normal derivative and such things
abdullah almasri This actually has nothing to do with what I said. I wasn’t talking about either calculus nor continuous matter.
But why on earth maths isn't taught like this?
Because good teachers are extremely rare, even more so in maths. And the technology to do this is finally becoming nice enough to do this kind of thing. I hope we see a big push in the next 5-10 years with this kind of teaching.
After watching this you may have the general idea of how the maths behind this work, but have you tried doing it yourself ? You probably won't be able to do it. That's why mathematics aren't taught like this. You need to do it yourself, sit in front of a blank piece of paper and try the demonstration on your own.
Because you'd have to wait for somebody to animate everything 😓
you dont have 1000 mintues every class...
Cause there are hundreds of formulas and teacher cant spend 20 minutes for each one.
7:53 "sometimes easy to forget why we're doing this".. its amazing how you pay attention to where one might loose track, after having mastered the topic.
and your face makes me pay attention to your saying
Yeah I noticed how well placed that was!
A well planned video.
This is why this is one of my favorite channels.
I also do that when i use tutor my friends. i think in their own perspective and think of problems they might not be getting and i tutor them base on self-explanatory conditions and not those complicated concepts that no one can even logically answer like "how did the long division method got formulated", it should be self-explanatory.
Brown: *explains*
Blue: But why?
Brown: *angry noises*
you just helped me make the connection. The name of the channel is 3blue1brown, and in the animated classroom there are 3 blue pi student creatures and 1 brown pi teacher creature
@@dayzimlich Yeah you are welcome
Lol
as the 464th like that's hilarious
@@dayzimlichcorrect.
Your channel is just ABSOLUTELY AMAZING! I love your videos! Thank you for producing such nice Math content for the world. 👏🏻👏🏻👏🏻
Procopio por aqui hein????
Tudo pela matemática!
Concordo, esse canal é um dos melhores que já vi.
3Blue1Brown dinamita na matemática
Realmente es un excelente video, la claridad de la explicación, la nitidez y belleza de las imágenes, la sencillez del lenguaje, este video deberîa ser usado por muchos profesores en sus clases..
E aí Procopio!! Concordo, muito bom mesmo.
I'm so smart I did the exercice all by thought without a piece of paper
And I also got it all wrong
It's not that hard to not use a piece of paper to do this if you are careful.
The risk I took was calculated, but man am I bad at math
I took the time to actually get the paper and pencil and do it
But still got it wrong
@@Danilego I got this as the area of the shadow 2π*R^2*deltaø*cos(ø)*sin(ø).
It's a complete mess, i'm not even sûre about the cos thing. I tried replacing the whole 2sin*cos thing by sin(2ø) but I didn't found any way out
@@EidosGaming same dude I got stuck right there
See most people find maths stressful and anxiety inducing... but somehow, this man has made it relaxing and beautiful.
Compulsory schooling is to blame
I don't think it's the concepts people find stressful, if 3b1b started keeping quizzes and grades for his viewers, a lot would run away
still stressful to me. i can’t accept cylinder equal to sphere. i can accept it is close but not equal.
@@jakub.kubicek stop blaming school. Even after watching this video, there are still people who doesnt even understand what is going on. People hate school becuz require real serious test. The test result determ if u are qualify for next grade or not up until the final grade which then u take college entrance exam. School need to prepare u for college entrance exam, it is more intense which is why it is more stressful. But on a youtube video, u are free to not understand anything u just watch but at the same time , u arent gonna pass any entrance exam anyime soon. These kind of video are good for inspiring new ppl to love math or for math entusiastic to learn more about math that they alr learn at school.
@@fannyalbi9040 the concept is simple, compare an arc to its projection
The math class is extended because the 3rd pi is still asking “why”.
Whole class hates that pi.
@Sanchi Sharma but WHY
I am one of the pi material
But that pi will be the one which will end up learning the most.
I AM THAT PI!!
Sanchi Sharma that pi believes 9/11 never happened
The most amazing thing is that Archimedes found the surface area of a sphere over a thousand years before calculus was even invented. While he didn't prove it to modern rigor, he can hardly be blamed for that.
Yup
How did he work it out?
Actually, his proof was bullet proof even by modern standards. He proved one miraculous lemma which underlay not only his proof that the area of a sphere is 2/3 the area of its circumscribed cylinder (lateral area + top & bottom) but also that it's volume is 2/3 the volume of this same cylinder!
Legend has it that the sphere and its circumscribing cylinder were inscribed on his tombstone.
Absolutely!
He didnt
It helps a lot if you remember that Oscar Had A Hold On Arthur.
Answers below the fold:
Q1: Let's call the radius of the ring d. We have a right triangle with an angle of theta and a hypotenuse of R. In this case, d is opposite from theta. Using the above mnemonic, we can remember that O/H = sin(theta), ergo d = R sin(theta). The circumference of a circle is 2 pi R, so the inner circumference of the ring is *2 pi R sin(theta).*
Thus the ring's area is approximately *2 pi R^2 sin(theta) d(theta).*
Q2: The good news is that our inner radius d is the same as it was for the ring on the sphere, ergo the inner circumference will also be the same: 2 pi R sin(theta). What we need to figure out is the thickness of the ring's shadow. By drawing another right triangle where the hypotenuse is the thickness of the ring, R d(theta), we can see that the thickness of the shadow is adjacent to theta in our new triangle. Using the mnemonic above, we can see that A/H = cos(theta), ergo the thickness of the shadow = R cos(theta) d(theta). To finish off, we multiply these two to get an area of *2 pi R^2 sin(theta) cos(theta) d(theta).*
Q3: Using the identity that 2 sin(theta) cos(theta) = sin(2*theta) reveals that we can rewrite the area of the shadow as *pi R^2 sin(2*theta) d(theta).* This is the same as the area of the ring except that we've dropped the 2 from in front, signifying that we've cut the value in half, but we've also doubled the value of theta. This means that the shadow at a given value of theta has half the area of the ring at double that theta value. For example, the shadow at theta = 30 degrees has half the area of the ring at 60 degrees. Thus, as we go to the next shadow, we skip past one of the rings and jump two rings ahead instead of one.
Q4: Partially answered above, but as we compare each shadow to a ring on the sphere, we have to skip every other ring, jumping two rings ahead for each shadow. The other half of this puzzle is to remember that we only generated shadows from one hemisphere rather than the entire sphere. Since we skip one ring for each shadow, that means we need to use all of the rings from the entire sphere (except for the ones we jump over), instead of just using the rings from one hemisphere. An easy way to see this is to think about the last shadow, at theta = 90 degrees, which corresponds to half the area of the ring at 180 degrees, which is the last ring on the sphere.
Q5: The area of the shadows sums up to the area of a circle of radius R. However, each shadow is only _half_ the area of one of the rings, and only half of the rings have been accounted for. A half of a half is one quarter. Ergo, a circle of radius R only has half the area of one hemisphere of the sphere, which in turn only has half the area of the whole sphere, and thus the area of the circle is one quarter that of the entire sphere.
Dis is the best
Edit: Other comments explained my mistake. Thanks!
That agrees with what I got, except that my area formulas has R instead of R^2. I've seen a few other commenters using R^2, so it is probably correct.
I got the area as circumference*width where width is dø for the ring and cos(ø)dø for the shadow. To me this also makes sense that I multiply two distances to get area, rather than 3 (R, R, dø). How did you get the other R?
@@AnonymousAnonymous-ht4cm the width is linearly proportional to R. For example, at a fixed central angle, the arc length of an arc will increase by a factor of R as you increase the radius. Same reason why circumference is 2πR and not just 2π
Bravo sir. I'm actually glad that my intuition was leading me along a similar path. If half of the rings from one hemisphere gets us a circle of equal radius to the sphere, then the other half of the rings gets us two circles, and if we do the same on the other hemisphere we end up with four circles in total.
@@AnonymousAnonymous-ht4cm You are close. The widths are actually R*dø and R*cos(ø)*dø for the ring and shadow, respectively. That is where the extra R arises that you are missing. This should make intuitive sense because a sphere with a larger R will have wider strips for a given dø. Similarly, smaller spheres will have thinner strips for a given dø.
You can also roughly think about this in terms of units. dø and cos(ø)*dø are essentially unitless. One needs to multiply these quantities by R to get a unit of length.
Congrats on such an amazing video, omg it's 3AM but here's how I did the exercise proof:
1. The circumference length of each ring is 2*pi*R*sin(theta), since the distance from the ring to the axis is R* sin(theta) (trigonometry). Hence, the area of a ring is 2*pi*R*sin(theta)*R*d_theta = 2*pi*R^2*sin(theta)*d_theta (1)
2. To calculate the area of a ring's shadow, I used some trig relations as well. In this case, the thickness of the ring shadow is R*d_theta*cos(theta). Therefore, the area of the ring shadow is 2*pi*R^2*sin(theta)*cos(theta)*d_theta (2)
3. Multiplying by both sides expression (2), we get 2*pi*R^2*2*sin(theta)*cos(theta)*d_theta
I've put the number 2 right next to sin(theta)*cos(theta) to explicit the trigonometrical relation:
2*sin(theta)*cos(theta) = sin(2*theta)
That specific angle is 2*theta.
4. So, this means that the area of the shadow of some ring with a corresponding angle of theta is equal to the area of a ring which has double of this angle.
5. Notice that, when theta ranges from zero to pi/2, we get to form all of the rings related to their shadows in the superior hemisphere. The total area of the shadow is the sum of all the thin ring shadows. Thus, that is equal to half the area of the hemisphere. That is, the area of the shadow (pi*R^2) is 1/4 of the surface area of the sphere. A(sphere) = 4*pi*R^2 Q.E.D.
Why didn't you use the appropriate symbols (i.e. π )? >.>
I'm confused...
When we sum up the areas of the each of the shadows of the rings, for all the rings the top hemisphere, we get the shadow of the sphere itself. That's equal to (pi*r^2). I get this step
The shadow of a ring at an angle of theta has the same area as a ring at an angle of 2theta itself.
Therefore, summing over all the shadows in the top hemisphere is the same thing as summing over the area of every other ring in the top hemisphere...this step I don't get at all. Why is this? Why does the fact that for any ring A, there's a ring B at double the angle of A that has the same area as the ring A's shadow, imply that when we sum over all the shadows if the rings in the top hemisphere, we get half the area of the top hemisphere...?
Assuming that we do, we need to multiply this area by 4 to get the surface area of the sphere...so this step I get again.
But I'm lost in the in-between step. Thanks!
Thank you for this, I was able to get most of the math right, but mistaking the cosine for a sine messed everything up, and figuring out that the result of sin(2theta)=sin(theta) was talking about two different angles was super helpful
Great explanation for the process, it is hard to explain that good.
Well thats a mere calculus proof...
No offense and of course i respect the fact that you tried it out all by yourself , but yeah its too easy.
A couple years ago, I was trying to estimate the amount of paint needed for an airliner. I based it on some simplifying assumption with regard to shapes and was surprised to find an equivalence between a hemisphere capped cylinder and open ended cylinder of the same overall length. Made me double check my math for the generalized case and indeed, (2 pi r)(l + 2r) describes both cases. So for my estimate paint usage, I just used the open ended cylinder with the length and radius for the fuselage and engines, plus a rough combination of triangles for the tail. (Wings, too, but their numbers stay separate since they use different products to prevent adhesion of ice)
Sooo, how close were you to getting it right with these approximations?
@@matthewrigby6089 His paint usage was much lower than expected and he now has a striped truck
the guy from the textbook problems is real 0.0
Bro gets his name called out on the math text book and it isn't just coincidence
1+¢=€¢-1
I love 0:40 when he tried to cover the surface of the sphere with circles
And how he gets frustrated lol
Yes! It somewhat scratches an itch.
Geography teacher : doing 4 questions in 4 minutes is the same as doing 1 question in one minute
Calculus : *am i a joke to you?*
yes explain
@@h-Films
Let:
1. number of questions done = f(t); unit = no of questions
2. time = T; unit = minutes
The function f(t) does this:
1. when t = 0, x = 0
2. when t = 4, x = 4
The history teacher concludes that the gradient, f'(t) or df(t)/dt, is a constant value, which is 1 question per minute, obtained by this formula: f'(t) = [f(4)-f(0)]/(4-0)
While it is valid if f(t) is a straight line, it is usually inaccurate otherwise since f'(t) often changes based on t (e.g. quadratic f(t)). This method is in a way similar to trapezium and newton-raphson approximations.
Or, he's just saying that difficulty variations of calculus Q is very high, so you can't just say that all questions take the same time (1 minute) to complete.
Chopin?
Chopin I love u
@@alwaysseverus741 my homie's tryna flex out here man
I'm torn. Is the math more beautiful or is it the animations?
The animations describe the math and are created with math, so I'd argue that they are the same thing.
You need much more time to imagine all that without 3b1b. Years. Imagine a student learning the first time.
The math _is_ the animations.
Maybe the brain behind the animations
It's all Mathematics
17 minutes of heaven.
12 minutes of heaven, then he gives homework!
YOUR NOT MY REAL DA.. uhm.. TEACHER
Joske Tobben lol
15october91 17 min 1 sec
Ended No Nut November on a good foot.
If only the pen paper drawings were just as easy to play around with~
I think that particularly was the biggest bottleneck for me back then!
Alright, we can delete UA-cam now. This is the winner. It's over.
too real
( 4/3 ♊R^2 )' = 4♊R ..
Surface is the differentiel of the volume ( différentiel in french )...
Circonference ...the differential of the surface :
2♊R = ( ♊R^2 ) ' ....
@Fcrgazul
That's the photo
Hahaha
@@jean-claudepecqueur625 gemini?
I left PUBG to watch this.
Lol I just did this
Wtf same
Sampad Barik lol
No one asked
U r a noob
I remember trying to prove this in highschool. It seemed impossible given the knowledge of a child but I wonder if I had a teacher like you that time, it would have been an enlightened day of my lifetime.
Thanks for these elegant proofs ❤️
mmm yummy onions
I'm now in my 40s and I find math more interesting than how I felt during my high school learning.
G Dunken I am 55 and feel the same
@@maheshm5463 I'm older still. I find calculus astounding. In school I found math nearly useless and when asked, no one could tell me why I should lean it except "you'll need it if you go to college."
If anyone had just told me "basic math and algebra are the language of straight lines and planes (not always true, I know) but calculus is the language of curves and therefore, its the mathematical language of the Universe."
It would have changed my perspective and therefore, my life.
Tragic, really. Just 1 hour of the concept of integral calculus would have sparked my interest and I would have gone from "Learning because I have to" to "Learning because a WANT to".
Welcome to maths. Unlike what it may have seem to be, it's a wonderful universe !
Lol, read my comment above. I am 70.
I'm 60 and. I wish I had a maths teacher like this in school, instead we had a loud mouthed aggressive bully that terrified everyone in class, and no one was good at maths in my class, I wonder why ?
He died of a Hart attack the year after I left school aged 40 and thought it's a pity he didn't die the year before I started. Now I'm disabled and at home most of the time and due to UA-cam I've found a new love for maths, I would like to start all over again from scratch as a beginner. If any one out there knows of some good videos on UA-cam with the same inspiring content as this please post a link. Thanks in advance 😀
This is the best mathematics channel on UA-cam. There is literally no competition. I want these videos played at my funeral.
hopefully not very impatient...!
'At my funeral' #Atruemathlover
lol
Wouldn't an endless loop of one of his animations on a screen on your tombstone be yet cooler ;-)?