Hexagon, meet parabola.
Рет қаралды 30,456
3 жыл бұрынWe look at a nice problem involving calculating the area between a pair of parabolas and an inscribed regular hexagon.
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@mrzimcrnce1389 3 жыл бұрын
Might have been easier to just calculate the top right area of the hexagon via the formula, and just integrate the quadratic function from 0 to root(3/2), and then subtract
@HitoPrl 3 жыл бұрын
Exactly my thoughts, much easier.
@malignusvonbottershnike563 3 жыл бұрын
Plot twist; Michael's choosing a more complicated method so that he can get the video to 10 minutes and get more ad revenue.
@xevira 3 жыл бұрын
@@malignusvonbottershnike563 Or just he's showing a method on an easy example that can be used for something more complicated that may not necessarily lend itself to using easy to use formulas.
@Kokurorokuko 3 жыл бұрын
Exactly
@goodplacetostop2973 3 жыл бұрын
@@malignusvonbottershnike563 In that case, Michael would farm the hell out of the Overkill series 😂
@mooncowtube 3 жыл бұрын
When you got the formula for the top parabola, I just integrated that from 0 to sqrt(3/2) to get the area under the first quadrant and the subtracted one-and-a-half equilateral triangles (each equilateral triangle being sqrt(3)/4) and quadrupled the answer. It seemed quicker and simpler.
@johnwalthall4937 3 жыл бұрын
More proof that hexagon is the bestagon
@enolastraight577 3 жыл бұрын
That sounds SO Sheldon Cooper.
@johnwalthall4937 3 жыл бұрын
@@enolastraight577 true but it’s also cgp grey’s newest video.
@der.Schtefan Жыл бұрын
Somebody send this to Cgp grey
@Tiqerboy 3 жыл бұрын
Your diagram looks like two fish who try to eat each other, but it's a tie.
@MTd2 3 жыл бұрын
It actually looks like a white shark trying to eat me!
@Dexaan 3 жыл бұрын
There's always a fish of the same size
@asadsabir7718 3 жыл бұрын
HEXAGON IS THE BESTAGON!!!!
@drwhominer 3 жыл бұрын
knew id find this down here, bestagons everywhere
@yourmathtutorvids 3 жыл бұрын
LOL I swear video brainwashes its viewers. Every time i see i hexagon, I now repeat that in my head
@vii-ka 3 жыл бұрын
at first i thought this video was about how parabolas are better than hexagons
@ThaSingularity 3 жыл бұрын
Haha beat me to it
@tuffleader2448 3 жыл бұрын
kzfaq.info/get/bejne/qs5_nMmmrdmalY0.html for the uninitiated
@canaDavid1 3 жыл бұрын
When I saw the thumbnail and title, I thought it was a cgpgrey response... But then I read the username, and realized it was math.
@Bodyknock 2 жыл бұрын
From 6:15 on there’s no need to calculate where the parabola and hexagon intersect. Since we already have the equation of the top parabola, the total area between the top and bottom parabolas is just twice the area between the x axis and the top parabola by symmetry, and you can easily calculate that area by integrating the parabola. The area of a hexagon with side length 1 is also easy to calculate, so the combined area of the red regions is just 2x (area under top parabola - 1/2 area of hexagon).
@larswilms8275 3 жыл бұрын
10:31 now you have four time the area of the top parabola from 0 to sqrt(3/2) minus the area of the hexagon (which is 6 times the area of the equilateral triangle i.e. 1/2 *1 * 1/2sqrt(3)) just to simplify things.
@MatematicamentecomGRibeiro 3 жыл бұрын
excellent video! never stop! I hope one day to make videos as good as yours 🤙🏾❤️
@goodplacetostop2973 3 жыл бұрын
15:54 Hello Michael, hello everyone. Have a lovely Sunday. Don’t forget to take some time for yourself. Assume x and y are chosen at random from integers {1,...,n}. Let p_n be the probability that x+y is a perfect square. Compute lim(p_n • √n), as n→∞. Express the result in the form (a√b + c)/d, where a, b, c, d are integers.
@LukeCollins 3 жыл бұрын
Is repetition allowed? Can I pick the same number from {1,..,n} twice?
@goodplacetostop2973 3 жыл бұрын
@@LukeCollins The exercise doesn’t mention at all the repetition, so I assume you can pick any number you want, repeated or not.
@TheMaster-tq7cl 3 жыл бұрын
(4√2-4)/3
@LukeCollins 3 жыл бұрын
Here is my solution, this was a beautiful problem: Suppose x and k² - x are both in {1,...,n}. In particular, 1 + x ⩽ k² ⩽ n + x. For each given x, there are ⌊√(n+x)⌋ - (⌈√(1+x)⌉ - 1) ways to choose a square in 1 + x ⩽ k² ⩽ n + x (so that k² - x is also in {1,...,n}). Thus the total number of ways two integers in {1,...,n} add up to a square is the sum of ⌊√(n+x)⌋ - (⌈√(1+x)⌉ - 1) where x ranges from 1 to n. Since we want to work out a limit, we can approximate this general term to √(n+x) - √(1+x) + O(1). Using summation by parts, i got that the sum is n√(2n) - (4/3) n √(n+1) + O(n). Thus the limit is √2 - 4/3 = (3√2 - 4)/3, is this correct?
@goodplacetostop2973 3 жыл бұрын
@@LukeCollins I have (4√2 - 4)/3, or 4(√2 - 1)/3. I’ll wait a bit then post the suggested solution.
@DrJest 3 жыл бұрын
Guess I'm too lazy, I integrated the parabola, then calculated the Area of the exagon (perimeter * apothem / 2), divided by 4 and subtracted^^'
@Dysan72 3 жыл бұрын
Same, I'm just happy that I got the same answer as him. It's been a while since I've done calc.
@camrouxbg 3 жыл бұрын
Nothing wrong with that at all. I was thinking of that way, but I like his way too.
@ayubjikani5401 3 жыл бұрын
In first quadrant, after integrating parabola then simply calculating the area of two triangles(Equlitrial+Right) And Subtracting it from area under the parabola and its 4 times will give your answer. And that method would be easier. Isn't it? Edit : ❤😊
@Thecoffeefreak 3 жыл бұрын
don't even need to do two triangles. The quadrilateral formed has an even simpler calculation.
@wolo0048 3 жыл бұрын
Great Problem! I think this would be an interesting problem to solve without Calculus using the Archimedean formula for the area of a parabolic segment minus the area of a hexagon using the apothem. It would be cool to see the original geometric argument Archimedes used to prove his formula on your channel or perhaps just using modern Calculus to do so. Thanks for the great problem, I'll be using this with my Calculus students!
@jkid1134 3 жыл бұрын
It is almost disingenuous to call the Archimediean approach void of Calculus
@Jeathetius 3 жыл бұрын
I feel like everyone commenting about how the calculus is so much more complicated than necessary are blowing right by the fact that we used two trig identities instead of Pythagoras.
@vitalikpu 3 жыл бұрын
One of the first problems I actually solved on my own! Thank you for sharing
@timurpryadilin8830 3 жыл бұрын
Michael, please make some videos about the stereometrical problems. For example, finding cuts of polyhedrons, etc. That would definitely lead to very beautiful 3D images on the board!
@Lightning_Lance 3 жыл бұрын
o no, this maths channel isn't just squaring the circle. It's hexagoning the parabola. I've gone too deep.
@richardfarrer5616 3 жыл бұрын
Once you have the formulae for the parabolae, why not just note that the area between then is twice the area under the top one between the intersection points (i.e. integrate it between -sqrt(3/2) and sqrt(3/2)), and the area of the hexagon is 6 times the areas of an equilateral triangle of side 1. Then just take the difference. Super easy - barely an inconvenience.
@tomatrix7525 3 жыл бұрын
Great stuff as expected
@scoutswell 3 жыл бұрын
I love your videos!
@humester 3 жыл бұрын
A = 4( area under parabola in 1st quadrant ) - 12( area of triangle = (1/2)(1/2)( (root 3) / 2)). Also, while playing around with this, I noticed that the area of any equilateral triangle is (√3/4)(x^2), where x is the length of a side.
@lmattas 3 жыл бұрын
very nice and educational video, thank you!
@davidgillies620 3 жыл бұрын
A1 + A2 is just the area under the parabola minus a quarter of the area of a unit regular hexagon, which is in turn 3/2 of the area of a unit equilateral triangle, or 3 sqrt(3)/8. So putting it all together, 4 times x - 2x^3/9 at sqrt(3/2) (parabola), - 3 sqrt(3)/2 (hexagon) for the answer (which can be written as (8 sqrt(2) - 9)/(2 sqrt(3)).
@tomatrix7525 3 жыл бұрын
10:10, if people are confused by this quickly implemented ‘trick’ it’s really actually no trick but obvious stuff, if you notice. ‘Disjoined’ domains as he says, means we are integrating over a x axis lenght that is continuous but disjoined, so notice where the upper bound of the first integral end is where the second starts, so he can combine these bounds and integrate the function -2/3x^2 over that new longer bound, to yield the equivalent, yet easier result
@CM63_France 3 жыл бұрын
Hi, Yes, I agree, the area of the (quarter of the) hexagon is easy to calculate, and then substract from the area between the curve and the x axis. For fun: 1 "I'll go ahead and", 1 "I'll may be go ahead and", 1 "so let's may be go ahead and do that", 1 "now what I want to do", 1 "and so on and so forth".
@fredfrancium 3 жыл бұрын
I want more video like this AMAZING
@udic01 3 жыл бұрын
7:53 since the question is geometry based, why not calculate A1+A2 by integrating the parabola in the 1st quadrant and subtracting the area if the trapezoid under it? [Integral(1-2/3x^2) from 0 to sqrt(3/2)] - [sqrt(3)/2 * (1+1/2)/2]
@leif1075 3 жыл бұрын
Would you know how to calculate that if you didnt know cooridnates of points on the hexagon by knowing that thing abiutb equilateral triangles that i didnt know..i don't think you could..
@udic01 3 жыл бұрын
@@leif1075 i am sorry but i didnt understand your intention. If you could explain again please. I think you didnt understand my intention. I am writing about calculating the integral after you know the points. (That's why i wrote 7:53 at the beginning of my original reply) We have to use cartesian geometry because we are talking about parabola. And we have to use geometry because we are talking about hexagon and its angles. So i was suggesting to make the calcualtions easier by calculating the trapezoid's area (which is a known basic formula in geometry) and the integral of parabola (which is a known basic easy integral.)
@tharatcheat6988 3 жыл бұрын
Thank for your brilliant and convenient solution 😘😘
@laurensiusfabianussteven6518 3 жыл бұрын
Problem solving aside, i love your drawing on the blackboard plus.. Hexagon is the bestagon yeay
@artsmith1347 3 жыл бұрын
Agreed. Well drawn and well planned because the solution didn't involve messing up the diagram. The problem solving was good, too.
@Christian_Martel 2 жыл бұрын
I took a different route and got the same answer! My calculus courses are 25 years in the past, this is a really good practice to keep my integrals in shape. 😂
@nanamacapagal8342 Жыл бұрын
By 5:48 you can take a quick shortcut: just integrate the eye shape enclosed by the two parabolas, and then subtract the area of the hexagon. The lines very clearly don't exceed the parabola so we can cut some corners here A(regular hexagon) = 6 * basal triangle = 6 * 1/2 * 1 * sin(pi/3) = 3 * sqrt(3)/2 A(eyeball) = integral[-sqrt(3/2), sqrt(3/2)]((1 - (2/3)x^2) - ((2/3)x^2 - 1) dx) = integral[-sqrt(3/2), sqrt(3/2)](2 - (4/3)x^2 dx) = 2 * integral[0, sqrt(3/2)](2 - (4/3)x^2 dx) //because it's even = 2 * (2x - (4/9)x^3)](sqrt(3/2), 0) = 2 * (2 * sqrt(3/2) - (2/3) * sqrt(3/2)) - 0 = 2 * sqrt(3/2) * (4/3) = 8 * sqrt(3) / (3 * sqrt(2)) = 4 * sqrt(6) / 3.
@LorxusIsAFox 3 жыл бұрын
I decided to bash this one open with coordinate geometry and calculus, so I was pleasantly surprised to see that that's how he starts anyway, including solving for the equation of the parabola. After that, though, I directly calculated the area of the hexagon and subtracted from the relevant integral for the area between the parabolas.
@atirkahn 3 жыл бұрын
For me it was a lot easier to just find the whole area of one of the parabolas between the two points on the x axis, then times that by 2 for the other parabola as well and then minus the whole thing from the area of the hexagon, a lot easier to grasp instead of sectioning the red area into pieces...
@beanhwak 3 жыл бұрын
use the area of the trapezium in the irst quadrant
@Taterzz 3 жыл бұрын
i would imagine using the formula for a trapezoid's area would have made this easier, but i think he did it this way on purpose so as to give a general solution method for situations like this that don't always have nice shapes.
@fellipeparreiras4435 3 жыл бұрын
Relaxing!
@bryanbischof4351 3 жыл бұрын
That hexagon looks perfect, holy crap.
@Walczyk 3 жыл бұрын
why do we need the equation of a line? once we have the parabola and the end points can't we integrate, double it, and subtract the area of a hexagon?
@islandfireballkill 3 жыл бұрын
There are many ways to get the answer. Do you happen to know off the top of your head what the area of a unit sided hexagon is?
@nilton61 3 жыл бұрын
Why not integrerate the paranola from 0 to sqrt(3)/2 and subtact the area of THE parallell Trapeze?
@MrRyanroberson1 3 жыл бұрын
also weird how the area is super close to 2/3. i wonder if any other shapes hold such properties?
@senhueichen3062 3 жыл бұрын
After 6:00, once we have the equation for the parabola, then compute the area under the parabola, subtract half of the area of the hexagon...that is it....I feel not necessary to find equation of the line connecting two adjacent vertices of the hexagon.
@JalebJay 3 жыл бұрын
Would the exam accept the common knowledge that a unit hexagon has an area of 3\sqrt(3)/2 then just do A = 4*\int_0^{\sqrt(3/2)} (1-2x^2/3) dx - 3\sqrt(3)/2 instead of the calculus you used?
@jesusthroughmary 3 жыл бұрын
this is what I did
@jesusthroughmary 3 жыл бұрын
Even if you didn't know the area of a unit hexagon, it is trivial given the fact that it is equivalent to six equilateral triangles.
@CamAlert2 3 жыл бұрын
CGP grey started a new cult.
@KungFuPanda1223 3 жыл бұрын
hexagons are bestagons
@notafeesh4138 3 жыл бұрын
Wait... a HEXAGON?!
@Thecoffeefreak 3 жыл бұрын
after deriving the parabolic formula why not just subtract the area of the quadrilateral in quadrant one from the area of the integral of the parabola from (0,1) to (sqrt(3),0)) then multiplying the difference by 4? Much easier that way.
@Sam_on_YouTube 3 жыл бұрын
The stuff starting at 5:48 was a waste. Take your integral and then use basic trig for the area of the quarter hexagon. The 30/60/90 triangle has an area of 1/2*1/2*root(3)/2, or root(3)/8. The quarter hexagon is 3 times that, or 3*root(3)/8. Just subtract that from your single integral that covers the curve from the y intercept all the way to the x intercept.
@DaszekGD 3 жыл бұрын
Will it not be easier to calculate area under parabola from (-sqrt(3/2);0) to (sqrt(3/2);0) (i mean integral and then subtract area of 3 equilateral triangles and then double this area to get final result
@ruathak1106 3 жыл бұрын
Integrate from 0 to sqrt(3/2) and then subtract 3*sqrt(3)/8.
@giuseppebassi7406 3 жыл бұрын
Was it possible to calculate without calculus?
@user-mn9lj6to4i 3 жыл бұрын
I think, the square of area between parabolas is an intergral, the square of a hexagon is just 6* 1*1*sin(pi/3)
@MrRyanroberson1 3 жыл бұрын
let's see... so hexagons uniquely break down into six equilateral triangles, therefore the area in the hexagon is six times the unit triangle (with area sqrt(3)/4) and therefore has area sqrt(3) * 3/2. Next: the quadratic functions are vertical reflections, and therefore all area above the x axis contained by the down-opening one is half of the area bounded between the two quadratics. The down-opening quadratic has a vertex at the apex, which is at (0,1). Now the final ingredient is that other vertex on the hexagon. Notice it is a vertex of the equilateral triangle with points (0,0) and (0,1). Therefore the third point has y=1/2 and x=sqrt(3)/2. Now we have a quadratic: y = ax^2 + 1 (already considering the point (0,1) and symmetry) with the constraint 1/2 = a(3/4) + 1, therefore a = -2/3. With a function for the quadratic, we just integrate to get its general area cubic: -2x^3 /9 + x, and we can determine where it intersects the x axis with the difference of squares: -2/3 (x^2 - 3/2) = -2/3 (x - sqrt(3/2))(x + sqrt(3/2)); the quadratic has points (±sqrt(3/2), 0), so we want to integrate from one to the other. with area(x) = -2x^3 /9 + x; area(sqrt(3/2)) - area(-sqrt(3/2)) = 2 sqrt(2/3). The final area is the difference of the quadratic-bounded area (twice the previous number) and the hexagon area (sqrt(3) * 3/2). We get the answer: 4 sqrt(2/3) - 1.5 sqrt(3) ~ 2/3
@KazeShiniSK 3 жыл бұрын
2:09 if it bisects the angle it bisects the side length as well, 1 divided by 2 is 0.5 no need for Sin 30 for that
@shriramsahoo1274 3 жыл бұрын
That's not true , an angle bisector does not coincide with the median. This is only because it is isosceles that u caan say that in this case only
@KazeShiniSK 3 жыл бұрын
there is only one case in this video
@shriramsahoo1274 3 жыл бұрын
@@KazeShiniSK The derivation of your intuition as well is completely based on trigonometry which is a very fundamental concept that can always be used and it is of no use simply adding such superfluous comments when there is actually no better holistic efficiency or accuracy
@Grundini91 3 жыл бұрын
SInce the parabola is an even function couldn't you have just taken twice the integral from -sqrt(3/2) to sqrt(3/2) to get the area between the parabolas. Then taken the fact that we were dealing with a unit hexagon and that a hexagon is made up of 6 equilateral trianges to just find the area of one of the triangles and multiply that by 6 to get the area of the hexagon. Then just subtract the area of the hexagon from the area between the two parabolas to get the area of the red region?
@mathsamtube2741 3 жыл бұрын
Great
@tobyfitzpatrick3914 3 жыл бұрын
If you look up the word "overkill" in the dictionary, there's a link to this video.
@pierreabbat6157 3 жыл бұрын
Saying "square root of three over two" for two different numbers is confusing. I'd say "square root of three halves" or "square root of one and a half" for √(3/2).
@phasm42 3 жыл бұрын
Brutal
@leif1075 3 жыл бұрын
Is it really that commonly known that you can make 6 equilateral trisngles out of a hexagon? I've never heard that? Did anyone else..and if you didnt know that how would you figure out the coordinates of those points..i got the height is 1 minus the base squared fpr the small triangles you can make on the side of a hexagon but then how to solve for that base?? I don't see how..you'd eitherbhave to know that fact about hexagons and six equilateral triangles or you are stuck there..
@maddyf2063 3 жыл бұрын
Wouldn’t it be easier to do the integral of the parabola in the first quadrant then take away the area of the trapezium?
@Smevv 3 жыл бұрын
Exactly what I was thinking since it’s symmetric you can just find the area under the parabola and then subtract the area of the trapezium and then just multiply by 4
@lordvenusianbroon 3 жыл бұрын
I agree that the specific problem could be done in an easier manner as you describe. But from a teaching perspective, doing it a different slightly more complex way helps illustrate the sort of methodolgy that would be useful if you have a shape or line that you can't readily calculate the area under the lower curve purely from geometric reasoning that using a regular hexagon gives you. Also makes it easy to check both ways, which is good for those just coming to calculus.
@Smevv 3 жыл бұрын
@@lordvenusianbroon I suppose you’re right gotta view it from everyone’s perspective, if you get the right answer you get the right answer hence we can’t complain, a good part about maths is revisiting your work and seeing how you can improve on it too
@maddyf2063 3 жыл бұрын
@@Smevv Efficiency is good but so is universality
@solarisone1082 3 жыл бұрын
@@maddyf2063: Yeah. I thought the whole "without loss of generality" thing was big with mathematicians.
@AdamRabczuk 3 жыл бұрын
It would be so much easier to just calculate the area under the parabola and then subtract the area of the hexagon
@schweinmachtbree1013 3 жыл бұрын
*flaps hand* and that's a good place to stop
@cdrundles 3 жыл бұрын
Archimedes would have found the area bounded by the parabola using triangles via method of exhaustion, and subtracted area of hexagon. No calculus!
@SycamoreGlitch 3 жыл бұрын
Hexagons are the bestagons
@quantabot1165 3 жыл бұрын
89K
@hanswm 3 жыл бұрын
Area of triangle =(1/2)… area of parabola = (2/3) …
@zdrastvutye 3 жыл бұрын
the shape of the parabola depends on r somehow s2.imagebanana.com/file/201110/qVKg8fLn.jpg his drawings are always accurate because mr.penn has got a goldfinger
@Xayuap 3 жыл бұрын
please, integrate, double and substract
@iamrepairmanman 3 жыл бұрын
Why not just calculate the area of the the two parabolas minus the area of a regular hexagon? No need to calculate parts
@jq747 3 жыл бұрын
Me: *uses calculator 4 steps ago*
@byronwatkins2565 3 жыл бұрын
Why not just subtract the hexagon's area from the total area between the two parabolas? There is no need to find the equations for the hexagon's sides.
@pow3rofevil 3 жыл бұрын
naaaisuh = nice versión koreana aajaj 🤭
@hamidkh5488 3 жыл бұрын
why didn't you use the area of Trapezius!?! . it was an easier way
@barbietripping 3 жыл бұрын
Just a litttttttle more than 2/3? Haven’t watched yet, but got (8sqrt6-9sqrt3)/6 using coordinates and integration. Let’s see how wrong I am today.
@barbietripping 3 жыл бұрын
What a relief, I almost exactly the same technique.
@h3xhexagonvn211 3 жыл бұрын
wut
@jkid1134 3 жыл бұрын
This is all fine and dandy as a problem solution, but is certainty not a way to solve the problem that's going to do well in a timed setting. More boomer math on this channel ahahaha
@zombiekiller7101 3 жыл бұрын
Early AF
@moonlightcocktail 3 жыл бұрын
Huh, I'm early...
@ngc-fo5te 3 жыл бұрын
What a long winded way of doing this.
@WidithaSamarakoon 2 жыл бұрын
Why are you wasting so much time when you could have just calculated right quarter of hexagon using area of a trapezium and then done the integration of parabola from 0 to root3/2.
@Deegius 3 жыл бұрын
It's a good job you are not an engineer having to earn a living by your maths (like I do). Such a long winded way to get the result. There is a standard formula for area under a parabola and another for area of hexagon, Would take no more than half a dozen lines to solve.
@lunkel8108 3 жыл бұрын
I agree that he probably solved it in a way that was a bit more complicated than necessary, but you do have to keep in mind that this is recreational math. Just having the answer in a lookup table in front of you would be more efficient but also kind of defeat the purpose.
@rajballabhyadav5089 3 жыл бұрын
Can you share the formula for finding area and nder a parabola? i think there shouldn't be any.
@Deegius 3 жыл бұрын
@@rajballabhyadav5089 base x 2/3 height = area
@Deegius 3 жыл бұрын
Base = 2 x root(3/2): Height = 1: Total area = 2 x 2 x (2/3) x root(3/2) Hexagon = 6 triangles: area = 6 x (1/2) x 1 x Root(3) / 2 Red area = 8/3 x root(3/2) - 6/4 x root(3)/2 QED
@rajballabhyadav5089 3 жыл бұрын
@@Deegius but this formula holds only for the area under straight line parallel to the tangent at vertex of the parabola and the parabola. Its for one special case not a general result even for straight lines.
@ho-kagoteatime7786 3 жыл бұрын
Hexagons are the bestagons
Nikita Zdradovskiy
Рет қаралды 6 МЛН