When I saw the title at first I thought you found a bug on the _Microsoft Office_ ( *_Ms_* office), in one of its windows, and you are justifying why you didn't post a video earlier. 😮 In a hurry to help you out, I just read the 2015 year, so I thought _Microsoft Office _*_2015_* ..., I don't know this version!?!?! 🤔 Then a climax song, a camera approaching a paper on a window... _"Oh, God, he was evicted from his own office"_ 😲 A paper with two circles... _"Oh no, he is being threatened by a Symbology Math Secret Society (SMSS) after his 'ta-ta-ta' proposal"_ 😬 And then "Which one is a circle?"... _"Nah, it is just an Ophthalmic exam"_ . I can't see any difference in a 5-inch Android display. 🔵🔴🔠👓👀
I recall one of my undergraduate physics professors mentioning that humans are extraordinarily good at identifying perfect circles by eye, though I can't find a reference for it at the moment. Like many others, to me it was immediately clear that the blue shape was not a perfect circle. The red shape did not show any obvious defects just by examination. The math here is interesting in its own right, but I find the ability of the human brain to naturally identify perfect circles to be far more fascinating. Perfect circles are pretty rare in nature, so it seems strange that we should have developed the ability to instinctively identify them. What evolutionary purpose could that serve? Is it a second-order effect of how our visual system works, meaning that we're just really good at visual pattern recognition and most of us at this point have seen countless perfect circles? This to me is much more interesting than proving using coordinates that one of these shapes is a circle.
Our eyes have nerve cells that always try to switch off their neighbours in order to produce contrast. Everytime you see a perfekt circle exactly the same cells light up, and the neural pathway gets stronger, simply because it is being used. When you now see the wrong circle, a good part of those neurons light up, but some dont as they usually would -> that makes you feel that there's something "off" about the blue circle. This of course is really simplifying the whole process :)
@@sachadossantos7972 i will try my best. What do you want me to elaborate on? The nerve cells in the eye, how using neurons makes the pathways stronger or anything else?
I originally guessed red circle just by looks. Once the graph was added: 3 points define a circle. The three contact points with the axes are (0,0) (6,0) and (0,2) Because of the first two, the x coordinate of the middle must be 3. likewise the y coordinate must be 1 because of (0,0) and (0,2) [it's easier to see this visually] This places the center at (3,1) and each point was 3 units left/right tand 1 unit up/down, meaning they all have the same length thanks to pythagoras. Since this is the only thing you can truly read off the graph, and it's consistent with a circle, you can conclude the red one is the correct answer
@@blackpenredpen Interesting, the reasoning you use does not require advanced undergraduate Maths (lucky for me), competence in lower division Maths is all that's required.
@@blackpenredpen But it's so obvious, is it not? I mean the red one is so perfectly round, but the blue one is clearly squished, as if someone had accidentally stepped on it or something. Maybe the edge of the video is helping.
I just knew it wasnt the blue one since it looked kinda weird around the lower right arc (4th Quadrant). The red one looked like a perfect circle, so it's what I chose.
I used that one theorem that states two chords AB and CD of the same circle intersecting at E will give AE × EB = CE × ED. Assuming the parts of the axes as chords of the blue circle and the origin as the point of intersection, we get 4×2=3×3 or 8=9 which is absurd. Therefore, the blue one is definitely not a circle.
David Flores People are stupid, or hypocritical, perhaps both. You do not solve geometry problems based on diagrams, which are almost never to scale. You solve them based on information you are verbally given.
By similarity, In blue circle, the point of intersection divides each one in 2 segments, there multiplication should be same But that is not the case. As 8 is not 9. For the red circle, yes it is a circle, in fact the circumcircle of a 2 , 6 and 2√10 triangle with radius √10. Is it right though???? Hey BPRP, it would make my day if I get a heart.
Though always fun to watch, much of the Maths presented by Black Pen Red Pen are past my limited understanding. Very happy to have followed this one very well.
Sejam A=(0, 4), B=(3, 0), C=(0, -2), D=(-3, 0) e P=(0, 0). Pelo teorema das cordas, A, B, C e D estão sobre uma mesma circunferência se, e só se, PA•PC=PB•PD. PA=4, PC=2, PB=PD=3 4•2≠3•3 então a curva azul não é um círculo
Much simpler, use geometry! In a right triangle, the center of the circumscribed circle is in the middle of the hypotenuse. So, the middle of between two points (0,2) and (6,0) ... Easy: Center = (3,1) For the radius, we use the Pythagorean theorem: R² = 3² + 1² => R = (10) ^ (1/2) Thanks, for all of your videos. They are great. =-=-=-==--=-=-=-=-=-=-=-=-=-= Beaucoup plus simple, utiliser la géométrie ! Dans un triangle rectangle, le centre du cercle circonscrit se trouve au milieu de l’hypoténuse. Donc ici, c'est le milieu entre les deux points (0,2) et (6,0)... Facile : Centre = (3,1) Pour le rayon, on utilise le théorème de Pythagore : R² = 3²+1² => R=(10)^(1/2) Merci, pour toutes vos vidéos. Elles sont formidables. Amicalement, Us.
You can visually see that the blue one isn't the circle even without the lines if you compare the top of the blue to the bottom in relation to it's curvature
I'm guessing it's when theta is zero, sin theta is zero but cos theta is 1, but we see the point (6,0), so we know we have 6cos theta in the equation. And when theta is pi/2, sin theta is 1 but cos theta is zero and we see the point (2,0), so we know the equation features 2sin theta. But why do we not see the point (-6,0) when theta = pi?
@@nathanielb3510 Theta is not defined for the curve at pi. In this example theta is somewhere between (-pi/4)ish and (pi/2)ish. I'm approximating the domain of theta based on the diagram, but the point is the same. Edit: Think of d(lnx)/dx at x=-2. We don't see it when we plot the derivative because x
@@mrfreezy7457 Is it just for this diagram that pi is not defined, or in general? I put that equation in desmos and it shows the same graph as this video. Is there some rule for when theta is defined and not?
@@nathanielb3510 I just done the same thing with sliders. theta=pi is in the domain when the coefficient of cos(theta) is less than zero, but theta=0 is no longer in the domain. I don't have an explanation yet as to how to find the domain without the graph. But, I'll keep looking into it.
Well I thought that yeah the y axis cuts the blue circle in a symmetric manner but still i didn't consider as the diameter I rather went on to assume it as a chord Now we have two chords of same length each 6 units long one parallel to the x axis and another parallel to the y axis ,,,on the top of that the centre (if it's a circle) needs to be equidistant from the chords! Now since they are x and y axes, so for a point to be equidistant from the axes the point must lie on y=x or y=-x thus our center will either be (a, a) or (a,-a) then the center comes out to be either the origin or somewhere outside the circle given! So if we consider the case of the origin this means that the chord parallel to x axis is a diameter and so should be the chord parallel to the y axis and for the chord parallel to y axis to be one of the diameters the center must lie exactly in between the chord which is not the case!
The Blue was not a circle, I used the *power of point theorem* that states if 2 chords intersect then the product of the length of the 2 line segment formed from the 1st chord equals the product of the length of the length of the line segments formed from the second chord. Using this theorem, it the blue was circle then *9=8* which is not true, hence *red is a circle* 😀
visually if you look at the blue circle and tilt your head 90 degrees it becomes very clear that it isnt a perfect circle. the red circle doesnt change when you do the same thing also i think what makes it a little difficult to tell is the positioning of the blue circle being at a bottom corner i think if we were to swap the red and blue position it would be much easier to tell because the circular aspect of the blue circle is a adjacent to the red circle but the bottom which is clearly distorted doesnt really have anything to compare it to.
Hi, I had a question concerning square integers and their square, it's a bit trivial but maybe you might have an explanation So if you've ever looked at the list of square numbers from 1 - 100, they sort of work like complementary integers; kind of like co-functions where the sin(A) = cos(B), given A + B = 90°. So for example lets take 72 and 28 and their square 5184 and 784 respectively. Notice that the last two digits match up. But along with that, if you take the remaining digits - 51 and 7 and subtract them you get 44. And 72 - 28 is also 44. If anyone knows why this is the case please let me know, unless it's a complete coincidence (as far as I know this also works with any other "perfect sum" pair, where the sum of two perfect square integers is 100)
I got you. We're starting with two integers, x and y, which add to 100. So y = 100-x. Now, if we're taking the difference of the squares, we get x^2 - (100-x)^2. Multiplying out the right side gets us to x^2 - (x^2 - 200x + 10000). The squares cancel out leaving just 200x - 10000. Since we're just using integers, this means the last two digits of the squares must be the same, since the difference is a multiple of 100, meaning they have to subtract to zero. And as for the first two digits, the initial difference of the two numbers is x - (100-x), which simplifies to 2x - 100. This is the same as the difference of squares divided by 100, meaning it will show up in the third and fourth digits. Anything else?
It's a consequence of our numbering system using base 10. If you were in base 7, doing two numbers which add up to 49, which is 100 in b7, would give much the same result.
The blue shape has radius 3 vertically and radius sqrt(10) on the diagonals where it meets the x axis. Since the radius is not constant, it is not a circle.
I used the inscribed angle theorem for both circles. Came to the conclusion that the blue circle can not form a right triangle while the red can. for the red one, I looked at the pints (0,2) (0,0) and (6,0) which forms a right triangle with the hypotenuse of sqrt(2^2+6^2)=sqrt40=2sqrt10 r is therefore sqrt10
Hey Dada ... I think the gap between any 2 numbers in the x and that of in the y axes are not same .... For that reason , w. r.t. this perticular coordinate system it's not a circle.... Am I right?
I found out minutes ago that you and some others in the youtube math community know each other, and that was exciting, but then I found out that you're not on speaking terms (referring to Flammable Maths) and that made me sad. I love your content ("you" being the whole math community) and it is unfortunate that you had such a negative outcome in the end. Oh well. Unless you're nutella, you can't make everyone happy. Keep up the great content :)
We could also prove that blue curve is not a circle by intersecting chords theorem. Because 4*2 ≠ 3*3 but yes i didn't know how to prove that Red one is indeed a circle so that was neat!
The red solution seems to take an unfounded leap. Aren't there an unlimited number of ellipses centred at (3,1) that pass through (0,0),(0,2) and (6,0)?
Before watching, based on the thumb nail I saw a red and blue one. It looks like the blue one has a flat tire. So now I'll watch and see how this goes.
I figured it out because the red circle is defined using 3 points (the minimum required) and the blue one is defined with 4 which could over define it and he stated only **one** of the circles was right, which had to be red.
I don’t know exactly how but I’m assuming you multiply both sides by r, which makes r^2, and because of Pythagorean theorem, you can substitute r^2 with x^2+y^2. Then move all the x and y onto one side of the equation, and make it all equal to 0. Then just complete the square
Rajat Kaushik This is extremely easy. In the interval 0 < x < π, -1 < cos(x) < 1. Since [x] < x for non-integral x, [cos(x)] = 0 in the interval 0 < x < π/2, and [cos(x)] = -1 in the interval π/2 < x < π. Therefore, the integral simplifies to the integration of dx from π/2 to π, which is simply π/2.
Someone already gave an answer, but in the future you use wolfram alpha for this kind of homework help. Or for simple ones like this just google the indefinite integral of the function.
Never mind. You made a typo (greatest integral instead of greatest integer). I somehow just read/interpreted this as just “integral.” So I thought you were just asking for a simple integration. I’m actually not familiar with the greatest integer function (taken a fair amount of math, but not a major). Edit: Not a typo I think, just a different name for it. “Integral” as an adjective form of “integer” isn’t a use I was familiar with.
The blue circle is considered an oblete spheriod, the red circle is an actual circle. The blue circle's sides stick out more and the top and bottom, the earth is an example of an oblete spheriod because the poles suck in and the sides stick put more.
lol. I thought the question was "which one is a circle" without the axis, I paused the video, took a careful look and decided that red one is a circle.
Okay for the blue one you use the argument that the vertical line is a diameter, that is incorrect. It may look like a diameter but could be a chord. Thus what you are doing is comparing two chords which intersect at the origin. Use the Intersecting Chords theorem to conclude it can't be a circle. In one case it's 2*4 while in the other it's 3*3 not equal this not a circle.
![CAMPปลิ้น | EP.83[1/2] เปิดแคมป์หลอนต้อนรับ 2 ตัวพ่อแห่งวงการผี!](http://i.ytimg.com/vi/2uPuHYHgwiI/mqdefault.jpg)
This problem has been on my window since 2015.
Maybe *Dr. Peyam* pasted it there 😂
And you are solving it now. It's bad for a Mathematician 😁
67-3=64
When I saw the title at first I thought you found a bug on the _Microsoft Office_ ( *_Ms_* office), in one of its windows, and you are justifying why you didn't post a video earlier. 😮
In a hurry to help you out, I just read the 2015 year, so I thought _Microsoft Office _*_2015_* ..., I don't know this version!?!?! 🤔
Then a climax song, a camera approaching a paper on a window... _"Oh, God, he was evicted from his own office"_ 😲
A paper with two circles... _"Oh no, he is being threatened by a Symbology Math Secret Society (SMSS) after his 'ta-ta-ta' proposal"_ 😬
And then "Which one is a circle?"... _"Nah, it is just an Ophthalmic exam"_ . I can't see any difference in a 5-inch Android display.
🔵🔴🔠👓👀
@@qiwas4665 Nah, they were synthesized way earlier. They are just given the official IUPAC naming in 2015 November. Get your facts correct bro.
I recall one of my undergraduate physics professors mentioning that humans are extraordinarily good at identifying perfect circles by eye, though I can't find a reference for it at the moment. Like many others, to me it was immediately clear that the blue shape was not a perfect circle. The red shape did not show any obvious defects just by examination.
The math here is interesting in its own right, but I find the ability of the human brain to naturally identify perfect circles to be far more fascinating. Perfect circles are pretty rare in nature, so it seems strange that we should have developed the ability to instinctively identify them. What evolutionary purpose could that serve? Is it a second-order effect of how our visual system works, meaning that we're just really good at visual pattern recognition and most of us at this point have seen countless perfect circles? This to me is much more interesting than proving using coordinates that one of these shapes is a circle.
Our eyes have nerve cells that always try to switch off their neighbours in order to produce contrast. Everytime you see a perfekt circle exactly the same cells light up, and the neural pathway gets stronger, simply because it is being used. When you now see the wrong circle, a good part of those neurons light up, but some dont as they usually would -> that makes you feel that there's something "off" about the blue circle. This of course is really simplifying the whole process :)
@@matzehero3371 That's fascinating!
@@matzehero3371 can you please give an more in depth explanation? I think I'm not the only one interrested .
@@sachadossantos7972 i will try my best. What do you want me to elaborate on? The nerve cells in the eye, how using neurons makes the pathways stronger or anything else?
@@matzehero3371 the the nerve cells in the eye, I know how neurons send message to each other but don't know much more
I thought blue wasnt a circle because of its flatter bottom
How unmathematical!
I measured it with a screen ruler and it didn't had the same height and width, so no circle.
@@NeedsEvidence it's not unmathematical. A circle should have constant curvature
wasn't*
Me too
I originally guessed red circle just by looks.
Once the graph was added: 3 points define a circle. The three contact points with the axes are (0,0) (6,0) and (0,2)
Because of the first two, the x coordinate of the middle must be 3. likewise the y coordinate must be 1 because of (0,0) and (0,2) [it's easier to see this visually]
This places the center at (3,1) and each point was 3 units left/right tand 1 unit up/down, meaning they all have the same length thanks to pythagoras. Since this is the only thing you can truly read off the graph, and it's consistent with a circle, you can conclude the red one is the correct answer
But what is with the option that neither is a circle?
Exactly what I fought, after which I thought that red citcle is touching (3, - 2) and (3,4) but those are sqrt(10) ~3,16 and 1-sqrt(10) ~ -2,16.
SPOILER ALERT :)
I could see the blue figure wasn't a circle by looking at it, but that's bad mathematics because a diagram is not always to scale :)
Same the blue one looks a little flat on the bottom.
@hawkturkey Seen from the side, the Earth is this shape. The Southern Hemisphere is slightly bigger than the Northern.
The blue curve is a limaçon 🐌. The polar equation is r=3+sin(θ)
they did the math
@@OrangeC7 they did the monster math
@@inigo8740 It caught on in a flath!
Fred
Yay I like your term "circle-looking circle”! Btw are you posting it on the window intentionally for students' curosity (for 4 years)?
Yes. And if they don’t watch my video, they will keep guessing. lol
@@blackpenredpen Interesting, the reasoning you use does not require advanced undergraduate Maths (lucky for me), competence in lower division Maths is all that's required.
Ey vibing math
I like your enthusiasm!
@@blackpenredpen But it's so obvious, is it not? I mean the red one is so perfectly round, but the blue one is clearly squished, as if someone had accidentally stepped on it or something. Maybe the edge of the video is helping.
I just knew it wasnt the blue one since it looked kinda weird around the lower right arc (4th Quadrant). The red one looked like a perfect circle, so it's what I chose.
I used that one theorem that states two chords AB and CD of the same circle intersecting at E will give AE × EB = CE × ED.
Assuming the parts of the axes as chords of the blue circle and the origin as the point of intersection, we get 4×2=3×3 or 8=9 which is absurd. Therefore, the blue one is definitely not a circle.
I used same
I used my eyes only lol
Yes 8=9
They are both not a circle, because you can not have one on a screen with a finite amount of square pixels that are in a finite grid.
Exactly! What I thought when I first saw the thumbnail
Ok "Which of these two curves is the closest approximation to a circle assuming they're both in a flat Euclidean space."
But every pixel has a infinite amount of points ¯\_(ツ)_/¯
@@silverbladeii But all the points in a pixel turn on/off at the same time. Magnify enough, and the "circle" has a jagged edge.
David Flores People are stupid, or hypocritical, perhaps both. You do not solve geometry problems based on diagrams, which are almost never to scale. You solve them based on information you are verbally given.
By similarity,
In blue circle, the point of intersection divides each one in 2 segments, there multiplication should be same
But that is not the case. As 8 is not 9.
For the red circle, yes it is a circle, in fact the circumcircle of a 2 , 6 and 2√10 triangle with radius √10.
Is it right though????
Hey BPRP, it would make my day if I get a heart.
What is the name of your used theorem? I think I already see this.
@@MrRenanwill Use similarity of triangles.
Though the result is also called as power of a point theorem.
Me too I thought "3×3!=2×4". But I could tell before he put the coordinate axes that the blue curve is not a circle.
It's Euclid's Element 3:35.
2:19 notation made me cry...
Though always fun to watch, much of the Maths presented by Black Pen Red Pen are past my limited understanding. Very happy to have followed this one very well.
Sejam A=(0, 4), B=(3, 0), C=(0, -2), D=(-3, 0) e P=(0, 0). Pelo teorema das cordas, A, B, C e D estão sobre uma mesma circunferência se, e só se, PA•PC=PB•PD.
PA=4, PC=2, PB=PD=3
4•2≠3•3 então a curva azul não é um círculo
smart man you are!!!
Much simpler, use geometry!
In a right triangle, the center of the circumscribed circle is in the middle of the hypotenuse.
So, the middle of between two points (0,2) and (6,0) ... Easy: Center = (3,1)
For the radius, we use the Pythagorean theorem: R² = 3² + 1² => R = (10) ^ (1/2)
Thanks, for all of your videos. They are great.
=-=-=-==--=-=-=-=-=-=-=-=-=-=
Beaucoup plus simple, utiliser la géométrie !
Dans un triangle rectangle, le centre du cercle circonscrit se trouve au milieu de l’hypoténuse.
Donc ici, c'est le milieu entre les deux points (0,2) et (6,0)... Facile : Centre = (3,1)
Pour le rayon, on utilise le théorème de Pythagore : R² = 3²+1² => R=(10)^(1/2)
Merci, pour toutes vos vidéos. Elles sont formidables.
Amicalement,
Us.
You can visually see that the blue one isn't the circle even without the lines if you compare the top of the blue to the bottom in relation to it's curvature
I just used the intersecting chords theorem to determine whether the red or blue shapes were circles
he broke down all the algebra in my head with a single statement, "if you focus on the x-axis"
After 2 years, u probably, now, know it
How do we know the polar equation for a circle right away? Is it some method like "go 6 squares right, 2 up" or was it given from the beginning?
Same question in my brain
I'm guessing it's when theta is zero, sin theta is zero but cos theta is 1, but we see the point (6,0), so we know we have 6cos theta in the equation. And when theta is pi/2, sin theta is 1 but cos theta is zero and we see the point (2,0), so we know the equation features 2sin theta.
But why do we not see the point (-6,0) when theta = pi?
@@nathanielb3510 Theta is not defined for the curve at pi. In this example theta is somewhere between (-pi/4)ish and (pi/2)ish. I'm approximating the domain of theta based on the diagram, but the point is the same.
Edit: Think of d(lnx)/dx at x=-2. We don't see it when we plot the derivative because x
@@mrfreezy7457 Is it just for this diagram that pi is not defined, or in general? I put that equation in desmos and it shows the same graph as this video. Is there some rule for when theta is defined and not?
@@nathanielb3510 I just done the same thing with sliders. theta=pi is in the domain when the coefficient of cos(theta) is less than zero, but theta=0 is no longer in the domain.
I don't have an explanation yet as to how to find the domain without the graph. But, I'll keep looking into it.
Well I thought that yeah the y axis cuts the blue circle in a symmetric manner but still i didn't consider as the diameter I rather went on to assume it as a chord
Now we have two chords of same length each 6 units long one parallel to the x axis and another parallel to the y axis ,,,on the top of that the centre (if it's a circle) needs to be equidistant from the chords! Now since they are x and y axes, so for a point to be equidistant from the axes the point must lie on y=x or y=-x thus our center will either be (a, a) or (a,-a) then the center comes out to be either the origin or somewhere outside the circle given! So if we consider the case of the origin this means that the chord parallel to x axis is a diameter and so should be the chord parallel to the y axis and for the chord parallel to y axis to be one of the diameters the center must lie exactly in between the chord which is not the case!
The Blue was not a circle, I used the *power of point theorem* that states if 2 chords intersect then the product of the length of the 2 line segment formed from the 1st chord equals the product of the length of the length of the line segments formed from the second chord.
Using this theorem, it the blue was circle then *9=8* which is not true, hence *red is a circle* 😀
Simple typing mistake.Hence Red* is a circle
Gotta confess... I'm getting kinda addicted to this channel and Dr. Peyam's.
Best and most worthy clickbait that has ever existed.
visually if you look at the blue circle and tilt your head 90 degrees it becomes very clear that it isnt a perfect circle. the red circle doesnt change when you do the same thing
also i think what makes it a little difficult to tell is the positioning of the blue circle being at a bottom corner i think if we were to swap the red and blue position it would be much easier to tell because the circular aspect of the blue circle is a adjacent to the red circle but the bottom which is clearly distorted doesnt really have anything to compare it to.
Hi, I had a question concerning square integers and their square, it's a bit trivial but maybe you might have an explanation
So if you've ever looked at the list of square numbers from 1 - 100, they sort of work like complementary integers; kind of like co-functions where the sin(A) = cos(B), given A + B = 90°. So for example lets take 72 and 28 and their square 5184 and 784 respectively. Notice that the last two digits match up. But along with that, if you take the remaining digits - 51 and 7 and subtract them you get 44. And 72 - 28 is also 44. If anyone knows why this is the case please let me know, unless it's a complete coincidence (as far as I know this also works with any other "perfect sum" pair, where the sum of two perfect square integers is 100)
I got you. We're starting with two integers, x and y, which add to 100. So y = 100-x.
Now, if we're taking the difference of the squares, we get x^2 - (100-x)^2.
Multiplying out the right side gets us to x^2 - (x^2 - 200x + 10000).
The squares cancel out leaving just 200x - 10000.
Since we're just using integers, this means the last two digits of the squares must be the same, since the difference is a multiple of 100, meaning they have to subtract to zero.
And as for the first two digits, the initial difference of the two numbers is x - (100-x), which simplifies to 2x - 100. This is the same as the difference of squares divided by 100, meaning it will show up in the third and fourth digits.
Anything else?
UltimateGamer16 I see where you are going with this, but this is nothing like a pair of cofunctions.
@@LilyKazami Nah it seems like ya hit the nail on the head, what I'm confused about is what is this supposed to mean?
@@angelmendez-rivera351 Yea Ik it isnt exactly but I was trying to explain it in a similar fashion so it didnt sound confusing
It's a consequence of our numbering system using base 10. If you were in base 7, doing two numbers which add up to 49, which is 100 in b7, would give much the same result.
i thought (x-a)^2+(y-b)^2=r^2 was the equation of a circle ( (a,b) is the centre and r is the radius)
The blue shape has radius 3 vertically and radius sqrt(10) on the diagonals where it meets the x axis. Since the radius is not constant, it is not a circle.
I used the inscribed angle theorem for both circles. Came to the conclusion that the blue circle can not form a right triangle while the red can. for the red one, I looked at the pints (0,2) (0,0) and (6,0) which forms a right triangle with the hypotenuse of sqrt(2^2+6^2)=sqrt40=2sqrt10 r is therefore sqrt10
I figured it out because I did a system of equations (given the intersection points) for the blue one and I saw there doesnt exist a solution.
U could have just used power of point...
just use pythagorean theorem of the triangle of sides 3 and 1. radius is root of: 3 square plus 1 square= root of 10
please, tell me the Function for that Circle-looking Shape!!...
Please....
(Blue one)
r=3+sinθ
Hey Dada ... I think the gap between any 2 numbers in the x and that of in the y axes are not same .... For that reason , w. r.t. this perticular coordinate system it's not a circle.... Am I right?
Red one. You can see that blue is vertically asymmetric
1:46 the issue is that they both need to be centered
Same forbth red one
I used the B.M.E. principle to solve it.
By My Eyes 😆
The blue one looked more deflated so I thought the red was the circle
integration of x ^x dx
Imposible.
Why
Is that supposed to say x^x? Regardless, the antiderivative of x^x is non-elementary, and can only be expressed as a double infinite summation.
How finde quadratic formula x^6-x^5+x^4-x^3+x^2-x+1?pls
Yay I could tell!
I haven’t learnt this yet but he makes it so you can understand it easily despite knowing nothing
i looked at them and knew blue wasn't a circle without hints, my eyes were able to see it bottom part wasn't the same es the upper part.
Can the circle blue the Pascal snail?
The magic number of completing squares is wonderful.
Very good video. Bye
Good day bprp! Can you do this?
If f(f(x)) = x^2 -1, find f(x).
I found out minutes ago that you and some others in the youtube math community know each other, and that was exciting, but then I found out that you're not on speaking terms (referring to Flammable Maths) and that made me sad.
I love your content ("you" being the whole math community) and it is unfortunate that you had such a negative outcome in the end.
Oh well. Unless you're nutella, you can't make everyone happy. Keep up the great content :)
Yes, the red one! The blue one look flattened at the bottom.
Red one
Can someone please explain the 2sin(theta) + 6cos (theta) to me? We've never been shown this in class....
Any chance you will be bringing back the Best Friend or the Me vs Calculus shirts for Christmas?
"which one is the circle?"
Me: "are they really drawn on the board?"
Hi!! Please solve this problem... I am not able to solve it!! Integration : (x³ cos(x/2) + 1/2)*√(4-x²) dx. Limit is from -2 to +2..
I got it right! Red is the circle
We could also prove that blue curve is not a circle by intersecting chords theorem. Because 4*2 ≠ 3*3 but yes i didn't know how to prove that Red one is indeed a circle so that was neat!
i got it right round one
Hi BPRP!
Hi Matt!
0:05 I think it's red, the blue one is squished at the bottom
I feel like a nerd for enjoying this
Well that was curious I spotted it immediately from the thumbnail
The red solution seems to take an unfounded leap. Aren't there an unlimited number of ellipses centred at (3,1) that pass through (0,0),(0,2) and (6,0)?
I figured this out using my advanced method of just looking at the damn thin.
The blue one is clearly less circular
2:17 Why do you want to multiply us by r? :-)
red
what a coincidence, I was just about to practice the piano piece on the side that can be heard in the background (jazz me blues)
Before watching, based on the thumb nail I saw a red and blue one. It looks like the blue one has a flat tire. So now I'll watch and see how this goes.
I figured it out because the red circle is defined using 3 points (the minimum required) and the blue one is defined with 4 which could over define it and he stated only **one** of the circles was right, which had to be red.
This is not a valid solution since the blue circle could very well be placed such that it's defined with 3 points
So, what's the answer?
Some say blue, some say red, but deep down we all know that....
The one on your T-shirt is the perfect Circle
"two circle-looking circles"
that must mean both are circles!
Neither is a circle because screens can never show perfect circles. If you zoom in close enough you notice pixel stairs.
0:15
The orange one is the real circle because the blue circle is flat on the bottom.
From the polar equation, how do we find the radius?
I don’t know exactly how but I’m assuming you multiply both sides by r, which makes r^2, and because of Pythagorean theorem, you can substitute r^2 with x^2+y^2. Then move all the x and y onto one side of the equation, and make it all equal to 0. Then just complete the square
He showed it in the video?
I suppose I have to be more explicit: how do we find the radius from the polar equation without changing to Cartesian first?
Red?
The red one!
My guess was the red one when I saw the video thumbnail and I was right
First time my brain decided to properly solve an image problem
I'm looking forward for '' 100 Partial Derivatives ''
blackpenredpen,
integrate [cosx]dx from limits 0 to pi
where [.] symbolizes greatest integral function.
Rajat Kaushik This is extremely easy. In the interval 0 < x < π, -1 < cos(x) < 1. Since [x] < x for non-integral x, [cos(x)] = 0 in the interval 0 < x < π/2, and [cos(x)] = -1 in the interval π/2 < x < π. Therefore, the integral simplifies to the integration of dx from π/2 to π, which is simply π/2.
Someone already gave an answer, but in the future you use wolfram alpha for this kind of homework help. Or for simple ones like this just google the indefinite integral of the function.
@@hewhogoesbymanynames its not simple lol
@@hewhogoesbymanynames can u tell me how to do
Never mind. You made a typo (greatest integral instead of greatest integer). I somehow just read/interpreted this as just “integral.” So I thought you were just asking for a simple integration. I’m actually not familiar with the greatest integer function (taken a fair amount of math, but not a major).
Edit: Not a typo I think, just a different name for it. “Integral” as an adjective form of “integer” isn’t a use I was familiar with.
The blue circle is considered an oblete spheriod, the red circle is an actual circle.
The blue circle's sides stick out more and the top and bottom, the earth is an example of an oblete spheriod because the poles suck in and the sides stick put more.
woah i finally learned how to do completing the square fast
lol. I thought the question was "which one is a circle" without the axis, I paused the video, took a careful look and decided that red one is a circle.
I can tell the blue drawing is an ellipse just by eyeballing it...
Just by looking at them i can easily tell you it's the red one
I could tell by eye
I’ve seen enough circles in my days to know the red one is a circle
When you substitute r^2 = x^2 + y^2, isn't that assuming it is a circle?
My mistake. It's just Pythagoras', and it's true for all functions.
Not gonna lie, the red one just immediately looks like a circle upon first glance
4th
Can someone help me to justify r =2sinx +6cosx
Really guessed red circle by looking thumbnail because blue circle bottom was not proper
I have not seen the video yet, but red feels like a circle, and blue feels squished
That thumbnail picture, lmao!
The blue one looks kinda oval
Okay for the blue one you use the argument that the vertical line is a diameter, that is incorrect. It may look like a diameter but could be a chord. Thus what you are doing is comparing two chords which intersect at the origin. Use the Intersecting Chords theorem to conclude it can't be a circle. In one case it's 2*4 while in the other it's 3*3 not equal this not a circle.
-1^2
The white one
When they went up I saw that the blue circle was pushed down and was a not a circle but red was not pushed in any direction and was a circle
The real problem outside my window: emo kids 😶
What is this sorcery
Anyone else managed to figure it out just by looking at them?
I could tell it was the red one just from looking at the thumbnail. I thought it was kind of obvious actually
The blue one didnt obey the chords theorem ( idk if that is its english name) however i didnt know how to prove that the red one was indeed a circle
holy shit the character arc of your beard to your new videos
looking at your title card i think that your shirt has the circle if not then the red one lol