the area of a lune -- its harder than you may think!

We calculate the area of a lune using integration via trigonometric substitution.
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[@xxxx015]

The fact that he proofs the formulas every single time i just love that some teachers just tell students to memorize or note it some where

[@TechToppers]

Yeah! Typical 3b1b style. But, I won't say that copying.

[@anushrao882]

This is one of the instances where grammar does matter lol.

[@TechToppers]

@@anushrao882 *Grammar*

[@shademits173]

Still no one solve the following geometry problem kzbin.info/www/bejne/pHK2gqirl7B-m7M kzbin.info/www/bejne/poq8f6iniK9qraM

[@tymoteuszzdunek8358]

You can still solve this question using high school level methods, area of the semicircle - area of a segment.

[@chrispizz1327]

I think it's interesting to be able to solve this with minimal and casual tool

[@arieljorgerossi5046]

How would u do it?

[@robertmauck4975]

If you connect the two intersection points, they form a diameter of the smaller circle. The area of the lune is then the area of half of the small circle minus the overlapping segment of the large circle. You can calculate the segment's area as the area of the sector between the radii of the large circle to the intersection minus the triangle made by the radii and the diameter of the small circle.

[@mokkapatisiddharth5793]

@@arieljorgerossi5046 The diameter of the smaller circle is a chord to the larger circle. So, the area of the dune is basically half the area of the smaller circle minus the area of the minor segment of the aforementioned chord. So now the problem boils down to finding the area of the minor segment, which can be easily determined using basic trigonometry and geometry.

[@ZannaZabriskie]

You are right. Easier way, you can do the math on your head.

[@cranil]

I guess he didn't stop

[@AmooBaktash]

The geometrical solution is also easy. It would be the area of half circle with radius 3 ( = 9 * pi/2), minus the area of the sector with angle 2 * arctan(3/5) radius 5 ( = 25 * arctan(3/5) ), plus the area of the triangle with base 6 and height 4 ( = 12).

[@richardbloemenkamp8532]

Indeed I did this and to me it is faster, easier, more intuitive and you don't have to know any anti-derivative formula's. It was actually easier than I thought (as opposed to the video-title.)

[@josephmartos]

Integration is more fun xD

[@AmooBaktash]

@@josephmartos The fun is in they eyes of the beholder :)

[@goodplacetostop2973]

Well no good place to stop so a HOMEWORK instead... A6 - Putnam 1971 : α is a real number such that 1^α, 2^α, 3^α, ... are all integers. Show that α ≥ 0 and that α is an integer.

[@eduard9929]

It is not that hard I think. If a is negative ,then for a large base b we have b^a is a positive subunitar number, so it isn't an integer. If a is a positive rational (but not natural) number, then a = x/y , with x and y natural numbers. Thus b^a = b^(x/y)=c => b^x = c^ y . For b= 2 we have 2^x = c^y => c is a power of 2 => y divides x => x/y is natural. And now if somehow a is irrational, we know the Gelfond-Schneider theorem which says that if x is natural and y is irrational algebric number, then x^y is transcendental , so it can't be an integer ! In conclusion, a is an integer >=0 .

[@liab-qc5sk]

sol is let a

[@goodplacetostart9099]

But always a Good Place To Start 0:01

[@goodplacetostart9099]

How do you be so early every time man ?

[@goodplacetostop2973]

@@goodplacetostart9099 Because I know at what time Michael uploads. It’s straightforward 😛

[@tomhapke3942]

"Look, the moon is shining so bright tonight! What a romantic evening... " Michael Penn or any other mathematician: " Well, let me tell u something... "

[@sweetcornwhiskey]

This solution necessarily requires that the intersection points of the smaller circle with the larger circle be collinear with the center of the smaller circle, which is not stated. Good problem, though!

[@deepjyoti5610]

Thanks

[@gioelechristille4650]

I think they are always collinear. You could probably prove this by solving the system of equations

[@trelligan42]

@@gioelechristille4650 Nah, this placement is special because it's using a semicircle (upper semicircle) of the smaller circle. It seems to be commonly used in textbooks, but it's possible to slide the small circle down so (for example) it intersects the origin of the larger circle. This would result in a smaller lune, and invalidate the initial triangle used for finding the points of intersection.

[@tobycooper4639]

thats what i was thinking, and it seems to me like the centre of the small circle can't be colinear with the intersections. If the centre is colinear, then we have the smaller circle centred at (0, 4), and it has a radius of 3, but somehow it still goes into the negative y axis?

[@jotazuma]

Sure, a very special case, not stated. Can be solved trivially by 8th graders.

[@nilsastrup8907]

Easier solution with only geometry and trigonometry: Notice that the area is equal to half the area of the small sircle minus the (sircelsector-the triangle) between (3,4) and (-3,4)(from his coordinate system). Therefore we only need to 1) find the area of the small half sircle and 2) the sircle sector: 1) A=0.5pi*9=4.5pi 2)A=25*arcsin(5/3)-6*4*0.5 =25arcsin(5/3)-12 Wanted area=4.5pi-25arcsin(5/3)+12

[@shivansh668]

Always 🙂

[@jesusalej1]

There is no sircle!! There are two circles!

[@z4zuse]

Small semi circle - (large segment - 2 x 345triangle) Large segment = (pi/asin(3/5)) x 25pi = 25 asin(3/5)

[@vinc17fr]

I haven't looked at this 13-minutes video, but the area is: small semi-circle − (big circular sector − associated triangle[3,4,5]), which gives (9/2)π−25·arcsin(3/5)+12 without calculations!

[@anonymous_4276]

What method did you use? The one he used (and I as well) was to integrate.

[@vinc17fr]

@@anonymous_4276 No need to integrate or to do any non-trivial calculations. Denote P and Q the intersection points, and O the center of the big circle. The area of the lune is the area A1 of the small semi-circle above (PQ) minus the area of the part of the big disk above (PQ). And the area of the part of the big disk above (PQ) is the area A2 of the circular sector OPQ minus the area A3 of the triangle OPQ. Thus the result is A1 − A2 + A3, where A1, A2 and A3 are very easy to determine: A1 is the area of the disk πr² divided by 2, i.e. A1 = (9/2)π. Let's do A3 before A2. Denote M the midpoint of [PQ], and consider the triangle OMP. You have OP = 5 and MP = 3, and a right angle at M. Thus OM = 4 (this is the well-known right triangle with side lengths 3, 4, 5), and A3 = 3×4 = 12. A2 is the angle of the sector multiplied by R²/2 (if the angle is 2π, you get the full disk, whose area is πR²), or the angle ∠MOP multiplied by R²; due to the right angle at M, ∠MOP = arcsin(MP/OP) = arcsin(3/5), so that A2 = 25·arcsin(3/5).

[@anonymous_4276]

@@vinc17fr thanks a lot! I really appreciate the time you took to write this out.

[@jackhandma1011]

The figure is kinda misleading. The center of the larger circle should be outside the smaller one. Still a good video though.

[@elijahflynt3217]

woah

[@angelradness5542]

I’m really glad to read this comment, the image was totally bugging me too, and I thought that was why.

[@il_caos_deterministico]

That's the first thing I noticed!

[@elijahflynt3217]

@@timshort9787 it is because the top is oblong

[@jotazuma]

@@elijahflynt3217 Huahahahua!

[@kriswillems5661]

It took me a while to realise that the small circle and the bigger one cross at a the point with the same y-coordinate as the center of the small circle. This is not obvious and should be clearly stated in the question.

[@Maazin5]

Yes I had the same confusion. I was wondering from where he got the distance between the two centers

[@flatisland]

the miracle of substitution - it amazes me time and time again :-)

[@leickrobinson5186]

... “and that’s a good place to stop“. Say it. Say it!! :-D

[@Bayerwaldler]

The picture on the board looks misleading. If the little circle has radius 3 and is centred at (0,4) then it cannot intersect the y-axis below the x-axis.

[@Invincible2203]

Nice but does that even matter

[@shivansh668]

Ohh yaa 👍

[@psioniC_MS]

This is simple: 1/2 * area of small circle - circular segment in bigger circle (chord length 6) = (π*3²)/2 - (arcsin(3/5)*5² - (6*4)/2) = 9π/2 - 25*arcsin(3/5) + 12.

[@fulla1]

I did the same thing and got the same result with only two lines and a little picture.

[@esinbass6570]

I am always searching for online tutorials to help me with my university study and I always struggle to find half decent ones. I appreciate your work, you made it so clear and very easy to follow! Thank you!

[@amiramaz]

I think a given is missing for the intersection of the circles. One could shift the upper circle up or down to change the area of the sector.

[@ricardocavalcanti3343]

I think he realized that at 00:35.

[@amiramaz]

@@ricardocavalcanti3343 thank you didn't notice

[@damianbla4469]

During my calculations I noticed something nice about arcsin: Area(Lune) = Area(half of the small cirlce) - Area(sector of the large circle) + Area(trangle with sides 3,4,5) "Area(sector of the large circle)" can be calculated in two ways: 1st way: Area(sector of the large circle) = (1/2) * (2*alpha) * R^2 = alpha * R^2 = arcsin(3/5) * 5^2 = arcsin(3/5) * 25 = 25 * arcsin(3/5) 2nd way: sin(2*alpha) = 2 * sin(alpha) * cos(alpha) = 2 * (3/5) * (4/5) = (24/25) sin(2*alpha) = (24/25) so 2*alpha = arcsin(24/25) Area(sector of the large circle) = (1/2) * (2*alpha) * R^2 = = (1/2) * arcsin(24/25) * 5^2 = = (1/2) * arcsin(24/25) * 25 = = (25/2) * arcsin(24/25) Now we compare both formulas: Area(sector of the large circle) in the 2nd way = Area(sector of the large circle) in the 1st way (25/2) * arcsin(24/25) = 25 * arcsin(3/5) || :25 (1/2) * arcsin(24/25) = arcsin(3/5) || *2 arcsin(24/25) = 2 * arcsin(3/5) And now I think we could give someone a task to do: "Prove that arcsin(24/25) = 2 * arcsin(3/5)"

[@MichaelRothwell1]

It's interesting that you can solve Michael's general indefinite integral using the area of a segment (as well the target definite integral as many have pointed out).

[@Sam_on_YouTube]

You could also do this generally, with big radius a and small radius b. The integral would be 2int(sqrt(a^2-b^2)+sqrt(b^2-x^2)-sqrt(a^2-x^2)dx) from 0 to b 2(xsqrt(a^2-b^2)+b^2/2×arcsin(x/b)+x/2×sqrt(b^2-x^2)-a^2/2×arcsin(x/a)-x/2×sqrt(a^2-x^2)) from 0 to b 2xsqrt(a^2-b^2)+b^2×arcsin(x/b)+x×sqrt(b^2-x^2)-a^2×arcsin(x/a)-x×sqrt(a^2-x^2)) from 0 to b [2bsqrt(a^2-b^2)+b^2×arcsin(b/b)+b×sqrt(b^2-b^2)-a^2×arcsin(b/a)-b×sqrt(a^2-b^2))] - [2(0)sqrt(a^2-b^2)+b^2×arcsin(0/b)+0×sqrt(b^2-0^2)-a^2×arcsin(0/a)-0×sqrt(a^2-x^2))] 2bsqrt(a^2-b^2)-b×sqrt(a^2-b^2))+pi×b^2/2-a^2×arcsin(b/a) bsqrt(a^2-b^2)+pi×b^2/2-a^2×arcsin(b/a)

[@Kamyak]

If y-axis is the line joining the two centres then how can we prove that the line joining center of smaller circle to one of the point of intersections is parallel to x-axis. The distance between two centers is not necessarily 4. If we try to construct it then we can take it to be any number.

[@Dysan72]

0:35 he forgot to mention how exactly the circles were related. Due to the fact that the smaller circle is raidus 3 and he states that distance to the incept of the circles from the y-axis is 3 we can infer that the intecepts are on the diameter. and from that that the origine is 4 above the origin.

[@BecozPro]

Suggestion: there's a grass field with unit radius and you tie a goat to the fence along the perimeter with a piece of rope. The goat will eat any grass it can reach, based on the length of the rope. How long does the rope need to be in order for the goat to eat exactly half the grass in the field?

[@alexandertownsend3291]

That sounds like a high school geometry problem to me. I won't give away the answer, but here is how you solve it. Think of the formula for the area of a circle. Call the radius of the bigger circle r and the radius of the inner circle s. Use algebra and solve for s.

[@Gab92260]

1.1587...

[@alexandertownsend3291]

@@Gab92260 Not quite right. Keep in mind that I said "inner" circle.

[@CliffLeeHippo]

I'm expecting "This is good place to stop" to ensure I won't miss something important.

[@Dionisi0]

I think you are wrong When you are assuming the distance fron y axis to the intersection if the circles is 3, since 3 is the radius of the smaller circle that means that the point in the y axis is the center of that circle, but you got a distance bigger that the actual the radius inside the smaller circle

[@kriswillems5661]

Exactly the question was missing something information. The center of the small circle and the intersections have the same y-coordinate.

[@bilalabbad7954]

Thanks

[@caladbolg8666]

Nice. As others have mentioned, it's be simpler with geometry though. Lune = small semicircle - big sector + big section, and all of those are almost trivial.

[@kevinmorgan2317]

The solution is assuming that the two points of intersection of the circles are the end points of a diameter of the small circle. It doesn't have to be like that. Given the set up, the solution can be simplified. Great exposition though of the technique used. Thanks.

[@rafciopranks3570]

1:26 How is the distance between center and (0,0) equal to 4, when the distance to the lowest point of that circle (under x axis) is 3?

[@txikitofandango]

Oh wow, with the circle positioned correctly, the area of the lune is the area of (half the small circle) - (area of the sector) + 2(area of 3-4-5 triangle) Then it's obvious how you get 9pi/2 - 25arcsin(3/5) + 12 :)

[@txikitofandango]

Okay, and now to watch the video and learn some calculus

[@GaryFerrao]

WHAT. NO. 0:34 I really want to know how you positioned those circles in the plane. You drew the top circle going below the x-axis which is clearly more that 4 away from its centre, but the radius is 3. How is this possible? It can only be if the circle doesn't cross the x-axis.

[@m.fadhiilhaekal7693]

Amazing

[@ramziabbyad8816]

I think I once saw that you can solve this type of problem quite generally using Green's theorem and the center of mass integral.

[@txikitofandango]

I did it geometrically. The area of the lune is (small circle) - (sector of big circle bounded by small circle) - (sector of small circle bounded by big circle - central kite) That gave 9pi - 25arccos(5/6) - (9arccos(-7/18) - 5sqrt(11)/2). Which can be simplified, because arccos(-a) = pi - arccos(a): 9arccos(7/18) - 25arccos(5/6) + 5sqrt(11)/2

[@txikitofandango]

Ah, I had assumed that the small circle goes through the center of the large one. Time to do the problem over, positioned correctly :)

[@VerSalieri]

The segment formed by the point of intersection and the center of the smaller circle is perpendicular to the axis formed by joining the centers of the circles. But, what if it wasn’t? I think, we should probably find a general formula for it..but... the pen is all the way in the other room....

[@tomatrix7525]

Great video.

[@engjayah]

Another easy approach: Consider x-axis as the common chord between the circles. So that the equations of small and large circles will be x^2 + y^2 = 9 and x^2 + (y+4)^2 = 25 respectively. As a result the area of the Lune will be the difference between area of a Semi Circle of radius 3 and the Sector of the big circle above the x-axis. By symmetry and integration the Area of Sector = 25arcsin (3/5) -12 From which we arrive at the same answer in a few steps!

[@petek1365]

Its quicker to integrate y = cos^2(x) using the substitution y = (cos 2x+1)/2

[@smoggert]

Calculating the area of cartman's hat.

[@RunstarHomer]

0:36 "I should have mentioned how the two circles are positioned more carefully" Yeah, this problem is impossible without more information.

[@ABc-es4nv]

From the picture in the thumbnail i supposed that the small circle was passing through the center of the bigger one and because of that got a different solution.

[@tushroy81]

The position of the small circle may vary. Hence the area of the shaded person will aslo vary.

[@billydoherty4041]

Michael, you have assumed that the smaller circle meets the larger one at a point where the horizontal line goes through the centre of the small circle. In general that is not the case. The area depends on how "high up" you place the small circle on the y-axis. In other words you have looked at one particular case only. Love your channel though. Keep it up!

[@SubtleForces]

That lune was interestingly positioned with its full diameter exactly visible. I am wondering about two questions. Firstly, what would the general formula look like if the diameters were R and r and it is positioned as in the video? Secondly, is there a general formula as the lune rises? What I mean by rising is that it goes from exactly hidden where the tips are aligned to the point where the smaller circle has fully risen above the other (when the area is trivially the area of the small circle). In other words, is there a function for the area of the lune as it goes from nothing to a full circle?

[@thibaultpillon5725]

The integral of cos^2 is classicaly solved using the formula cos^2 x=1/2(1+cos 2x). His way was a little bit strange...

[@shanmugasundaram9688]

I have doubt about the line joining the intersecting points of the circles passes through the centre of the smaller circle.Also the picture is wrongly drawn.

[@MichaelRothwell1]

Michael forgot to state this at the beginning.

[@alexortiz9777]

I can't get past the figure, how does the smaller circle have a radius 3 and radius > 4?

[@chamuvmg8650]

Nice problem, But is it really necessary to use integration? It seems it can be solved by using area of a segment...

[@YoutubeModeratorsSuckMyBalls]

There is strange thing on plot. You can see that distance between centers is 4, but distance from center of smaller circle to its circumfrence in direction to center of larger circle is 3, in which 4< 3

[@trelligan42]

"Drawing is not to scale."

[@jotazuma]

Huahauhauhashashah! Exactly! Moreover, that is a very special case of the problem, one in which the angle between the y-axis and the intersection of the circles is right.

[@diegofigueroa8307]

Yeah I was looking for a comment that point that out

[@ronaldjensen2948]

In the drawing on the board, the radius of the smaller circle is approximately 3.8 and the point is near (3.8, 3.2)

[@tcoren1]

He made an additional assumption that the center of the small circle is on the same y value as the intersections, an assumption he used liberally but never stated and didn't match his drawing, hence the discrepancy you saw

[@egillandersson1780]

I get the same formula, but I was too lazy to compute 😉. So I used a geometrical (and trigonometric) approach : find the area of the sector in the big circle, substract the triangle, then, in the small circle, substract the previous result of the semi-circle. And ... ta ta ta ... the result is the same !

[@ramziabbyad8816]

Try to put all the info in the beginning because I like to solve them first. In this case the small circle looks like it passes through the center of the larger circle thus I got a different answer.

[@ramziabbyad8816]

Actually, I guess you can solve for arbitrary shift a nm.

[@ramziabbyad8816]

I think ur pic is wrong, the little circle shouldn't go below the center of big circle.

[@oremilak]

Why not use the double angle formula?

[@antoniussugianto7973]

At beginning you didn't mention how far they overlapping?? You only mentioned their radius?

[@mikegallegos7]

dang that satisfies chaos !! lol

[@c-bass9968]

As subs as he busted out the Cartesian plain I just went ∫[-3, 3](sqrt(9-x^2)+4)dx-∫[-3, 3](sqrt(25-x^2))dx

[@robertapsimon3171]

The sketch of the circles is wrong as the original cannot be within the smaller circle if the y offset of the centre is 4 and the radius of the circle is 3. This threw me a bit when seeing the maths.

[@trelligan42]

"Drawing not to scale."

[@hannesstark5024]

At 3:25 why is the root equal to a*cos(theta) ?

[@Qermaq]

Why not note that the lune area is 1/2 the area of the circle r=3 minus the area of the chord section of the circle r=5 and theta = 2atan(4/3)?

[@advaykumar9726]

No good place to stop?

[@mondherbouazizi4433]

Hey you're missing something!!!! You forgot the "And that's a good place to stop." Unforgivable!

[@CraigNull]

Is a lune defined so that its width is the diameter of the smaller circle?

[@trelligan42]

Not according to Wikipedia - but this presentation is common in textbooks, using just the upper semicircle of the smaller circle.

[@shivansh668]

Hey, plz help me out . Should i take Thomas' Calculus or Stewart's Calculus As I'm starting learning Calculus after 2 months :)

[@trelligan42]

Stewart's with Early Transcendentals well thought of in general, but see if you like a particular instructor - and use their book.

[@tcoren1]

Why did you assume that the y value of the center of the smaller circle is the same as the y value of the intersection? That's certainly not true in general. Without it the triangle is no longer a right triangle / one of the sides is no longer of length 3

[@manucitomx]

What? No good place to stop? How do I know I can move on? This was a great problem.

[@SafetyBoater]

Is the location of the small circle defined only by the x coordinate of the point of intersection of the circles?

[@TedHopp]

This solution assumes that the intersection points of the two circles are on a diameter of the smaller circle. That is not a valid assumption and it is also not given in the problem statement.

[@ikarienator]

Well, Lune = Semicircle (9pi/2) - Slice (10atan(3/4)) + Triangle(12)

[@ZainAlAazizi]

4.5*pi*(1-cos(arctan(3/5))), just after I realized the angle value 😁

[@GaryFerrao]

OK i'm happy that it was not necessary to consider the intersections on the x-axis. Because he clearly drew a circle of radius greater than 4, but called that 3. Maybe that's what tripped him and he couldn't stop…

[@chrispizz1327]

Very good as you use to be !

[@atomicgeneral]

Nit: Smaller circle will not intersect x-axis . didnt matter for area calculation.

[@Sanatan_saarthi_1729]

Euclid has left the chat

[@buxeessingh2571]

The colouring of the thumbnail looks like the beginning of an episode of "The Partridge Family."

[@AnkhArcRod]

Obviously, the integral method is an overkill! It is much easier to solve this problem using simple geometry concepts of area of sector and area of a triangle.

[@davidt939]

0:59 wrong, this is not a consequence but the missing input of your problem if you want to be able to solve it.

[@godfreypigott]

You didn't state in the setup of the problem than the diameter of the smaller circle is a chord of the larger circle.

[@bopaliyaharshal2399]

Op

[@firefly618]

Why do you always complicate things with integrals and stuff? This can be solved with basic trig! The lune is half of the 3-circle, plus the 5,5,6 triangle below it, minus the corresponding sector of the 5-circle (corresponding to the large angle in the triangle.) The large angle in the triangle is 2·sin⁻¹(3/5) Therefore the area of the lune comes out to: 3²π/2 + 3·4 - 5²π·2·sin⁻¹(3/5)/2π = 9π/2 + 12 - 25·sin⁻¹(3/5) ≈ 10.0496…

[@notananimenerd1333]

Lol I wasted like 1 hour at the starting by seeing the figure to prove that the secant of the intersecting points of the circle is perpendicular to the large circle’s radius! After that I gave up and went 20 seconds ahead just to see that he hadn’t mentioned that earlier and I got really mad afterwards! Though on knowing that info I was able to solve this problem using elementary geometry to get the answer 9.999 approx

[@seanmacfoy5326]

Was it not a good place to stop? Is the video still going?! D=

[@LopedeCrypto]

I usually try to solve all your problems after watching the first 15 seconds of video, and it always takes me longer than you. For some reason, this problem took me like 1 minute and you took 13. What happened here? It takes high school level math to solve it.

[@david203]

There are three lunes defined by two circles, in general. Just saying.

[@az0rs]

is arcsin same as the inverse sin or is it something else?

[@TechToppers]

Literally same.

[@az0rs]

@@TechToppers oh ok ty !

[@DontMockMySmock]

All that calculus seems like overkill for a problem that can be done much simpler with just trigonometry.

[@DeclanMBrennan]

There's another approach to the same answer without calculus here: kzbin.info/www/bejne/l4KtmK1jhdyDpbs

[@palmtoptigeri9797]

Let z be a complex number. Show that if |(z+1)(z^2+z+1)|

[@thebeerwaisnetwork8024]

If he doesn't say that's a good place to stop then the video ending is a figment of your imagination. Everything you see beyond that is an abstract illusion. In other words, you're still watching the video.

[@The1RandomFool]

It wasn't a good place to stop today :(

[@-UMT-VIGNESHKRISHNAS]

10.0496392212

[@-UMT-VIGNESHKRISHNAS]

Final answer

[@davidjames1684]

This guy needs to learn how to draw to scale. Also, a fairly simple solution would be to draw it to scale, scan it into a computer, have it do a flood fill in the area of interest, and count up the pixels. Then compare that to the # of pixels of either circle, compute the ratio, then compute the area. It is an approximation, but a very good one. Probably off by no more than 1% if done correctly.

[@thomasmay6215]

It doesn’t really matter that the figure is off. Once he specified the point of intersection the figure doesn’t really matter. Also what you described a well know Monte Carlo technique.

[@davidjames1684]

@@thomasmay6215 That is NOT Monte Carlo. It is pixel counting and using ratios. Also, why not draw it close to scale so as not to possibly confuse the viewers? What he did was just being lazy.

[@thomasmay6215]

@@davidjames1684 I amazed that you are so entitled that, while describing the person who provides free math content every day, you can even can even find the word lazy in your vocabulary. While constructive criticism is one thing, that, plus your original comment, is just rude.

[@davidjames1684]

@@thomasmay6215 Being accurate is not being rude. Correcting (or just pointing out) errors is also not rude. What am I supposed to do, stroke him for drawing not to scale?

[@jesusalej1]

Arctan(1)=pi/4, not pi/2.... 'why make it easier if can be harder'

[@ricardocavalcanti3343]

It's arcsin(1), not arctan(1).

[@jesusalej1]

@@ricardocavalcanti3343 I know. I was just sayng.

[@goodplacetostart9099]

Good Place To Start at 0:01 HOMEWORK:Suppose 'a' and 'b' are positive real Numbers such that a+b=1 find extremum value of (a^a)(b^b)+(a^b)(b^a)