The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Hi mind your decisions I have a problem for you: In a trapeze abcd (ab||dc) The ac and bd diagrams are cut at point e the ratio between the bases is 4:9 and the area of the triangular bec is 108 What is the area of the whole trapeze?
-"the first method is simple" me: "alright, this might be the first time where i can solve a hard Math problem on my ow..." -"shows space rocket schematic" me: "okay, next video"
@@schwartzseymour357 Just watch the first method, the other ones are waaaay too abstract and you're needlessly complicating things. With some knowledge of geometry and testing your 3D logical insight it's fairly doable. If you lack these concepts, yeah this video is as good as impossible for any newbie.
@@janjan55555 I disagree. If you know integral calculus, the third method is VERY straight forward, and is definitely the way I would have went before he even mentioned the geometric solutions.
@@janjan55555 To me the integral method feels brute force and it’s the way I went about solving it before watching the explanation. I was expecting some clever/creative geometry trick and it turned out to be really simple. I just couldn’t see it.
this may sound funny but it's how they did it before calculus/geometry was common, in fact no so long ago, math became more common in 18th century or so
@Aryan Sharma you think from perspective of today. Note that only few in only few nations knew this in ancient times. Rest of nation was not able to read or write. What more geometry was expressed primarily graphically, there wasn't kind of todays' equations.
For those who are wondering where π came for θ, it is the radian system or circular system for angles. 180 degree = π radian. And 1 radian = 57 degree (approx). *And 1 radian can also be represented by the angle formed at the centre of a circle by 2 radii separated by an arc of 1 unit length (on the circumference of the circle)*
I'm proud of myself for this one. It is the first time I solved one of your math puzzles. It took me like an hour but I got the right result of 9.77 I'm only 14 years old and I've never had anything like this in school before.
@@commonlogicart1604 I found some method to calculating the different circle pieces area, and subtracting them. But it's almost a year ago, so I don't remember exactly what I did.
you know i've been watching a couple of your vids and there's usually a common theme in finding the correct answer. Either you use coordinates or geometry. When using geometry there's always one rule... find triangle and you'll find the answer. Just throwing that out. It's amazing to how much extent you can use this simple concepts to solve something like this
Option 4 - Depending on how many sig figs you need. Step 1: Find a sheet of graphing paper. Step 2: Sse a compass and rulers to draw out the figure. Step 3: Spend the last 3 hours of the exam to count the number of little squares.
This is essentially how some computer programs compute areas. They just make the squares small enough so that they get the desired precision and of course they can count the squares much faster :)
I work at a railway testing lab. There's a test that involves pressing a rubber pad in between a mechanical actuator and a cast iron mold with small high relief squares carved on it, over which a powder is spread. You place the rubber pad, press it using a specific pressure, and leave it there for a while. Afterwards you remove the pressure, shoots a photo of the rubber side with the square marks the powder made, transfer that photo to a computer, and have it calculate the ratio powdered area vs. non-powered area by literally counting how many pixels in the photo got this or that color and dividing one number by the other.
Next video: Put many circles inside a square; make most of them pass from intersections of other circles. Find the most absurd looking piece of area. And then solve for it.
I paused the video and solved the problem on my own before watching the full video. I approached it geometrically. I did it in my head without even using paper and pencil. I thought it was quite simple. I continued to watched the full video just to make sure I got the same answer. I was surprised that all the 3 methods in the videos are so long. Below is the outline of my method. Using the same labelling as 00:34 of the video, Area of shape DPQ = Area of shape DPQC - Area of shape DQC. It is easy to see that the shape DPQC repeated 4 time will fill the square. So the area of DPQC is just 25. Next step is to solve for area of shape DQC. Let midpoint of DC be E. Area of DQC = Area of sector QEC + Area of quadrilateral DEQA - Area of sector DAQ. It can be easily checked that Angle QEC = 2 * Angle QDC = Angle QAD = 2 * tan^-1(0.5). Therefore, Area of shape DQC = pi * 5^2 * (2 * tan^-1(0.5)/2pi) + 2 * 1/2 * 5 * 10 - pi * 10^2 * (2 * tan^-1(0.5)/2pi) =50 - 75*tan^-1(0.5) Finally, Area of DPQ = Area of shape DPQC - Area of shape DQC = 25 - (50 - 75*tan^-1(0.5)) = 75*tan^-1(0.5) -25 This is the same answer. I didn't finish reading all the 1.5k comments. But so far among the comments I've read, I haven't seen a similar method. This is the first time I see this problem. I'm not sure whether or not this method above has been posted before elsewhere on the internet.
I got the area of the yellow area a differrent way: Area of the semicircle = 12.5π (Solve with πr^2 × 1/2) Multiple it with four 12.5π × 4 = 50π It is like you draw four semicircles inside the square. Or imagine as mirroring the two given semicircles. Now you can see it covers the whole square and also the yellow area 4 times overlaying. Which means 50π = area of the circle + 4 × yellow areas 50π = 100 + 4y (y=yellow area) y = (50π - 100)/4 y = 25/2(π-2) y = 12.5π - 25
As an alternative, one might use symmetry and scale to simplify the solution. Steps: 1) 'Symmetrize' the picture by drawing another quarter circle radius 10, centered at the TOP, RIGHT vertex of the square. This divides the square into 8 regions, of which 3 are pairwise identical, thereby resulting in 5 distinct shapes ( with areas x1 thru x5). Our target region (shown in red in video) is one of the paired regions (say x2). 2) Express the areas of the small semicircle, the large quarter circle, and the square as linear equations in the variables x1 thru x5. 3) Thus far we have 3 linear equations, (and 5 unknowns) so we need 2 more. These are a) The area of the small 'lens' region formed by the intersection of the two semicircles, which can be easily computed as described in the video, or otherwise. and b) The area of the larger 'lens' formed by the intersection of the two large quarter circle. Since the larger lens has the same shape and twice the linear dimensions as the smaller lens, its area is 4 times as large. 4) Now we have a complete set of equations, from which x2 can be easily obtained.
Thought that would be the "best" solution as well when trying the problem on my own, but sadly the equation system is underdetermined and you can't solve for x2 (the red "wanted" region) with these equations alone. You somehow need to calculate the area of one of the "paired" regions on their own and for that you always need the right most corner of the red region (or its symmetrical counterpart). And at that point you can just use the calculus solution from the get go...
It included the most important step I remember from my calculus class - "annnd... skip some steps". Though in my class this was the "and from there... it's obvious" step :P
I dont think so, its easier to error with calculus, and more work demanding in my opnion. If you can work with basic algebra or geometry and have a good trig perception you should use it. Calculus is too powerfull, like trying to catch a fly with a gun when you have a fly swatter.
Your first "little lens" can be calculated much simpler, namely as the 5x5 square without the two other shapes, so that's 25 - (25 - quarter circle) - (25 - quarter circle), so that is a half circle of radius 5, minus 25 all in all, making it 12.5pi - 25. That's a whole lot faster than your "lens" calculation.
I've had a similar reasoning to do that part (square - 2 whole circles and then divide by 4). I wish there is a close way (substract areas) to solve the second part... i'm pretty sure there is one but i can't find it
His first “little lens” could be calculated in the way you showed, but what about the second “little lens”? You probably will need a different method to solve it, on the other hand, his method of the first “little lens” was applicable to the second lens, thus, saving him time and making the overall solution easier, at least in my opinion. How would you calculate the second “little lens”?
@@--Za did the exact same thing and the second part is bothering, you need to get the area of about 6.47 by using the circles but I cant seem to figure out how
I'm proud to report I sat through the entire video without skipping. That alone is an accomplishment given how little I understand anything that was said.
Calculus is mostly used to program computers and calculators which uses the same definite way to solve one type of questions. Intelligent ways are much more diverse and allows shortcuts that cannot be discovered by machines. It also varies from one person to another, signifies creativity. Unlike calculus, geometry defines intelligences that a machine will never have. (Unless somebody invented artificial intelligence that will eventually cause global extinction.)
I just recently discovered your channel. I did the problem with both methods, the area subtraction using geometry (since that stood out to me) and the calculus way (because that just made sense integrals for the win). I set up the problems in my head and didn't particularly solve for an exact solution. Once that was done I finished watching the video. I m glad my physics/math brain is always on the ball great problem. Math is just awesome!
9:00 instead of using trig functions, we can also apply the similar triangles rule. Find the ratio of the kite triangle and smaller triangle hypotenuse, then their areas have ratio to the square
I once solved this really hard Q in class which no one could solve and then the teacher told us how to do it and his method was completely different than mine but we had the same answers.
Area of intersection of two semi circles can be calculated by drawing all four semi circles inside a square. Suppose area of semi circle is A, area of intersection of two semi circles is B and area of the square is C than 4*A - 4*B = C. Because when you add areas of four semi circles together, each intersection will be counted twice so you have to subtract four times of the area of intersection in order to make it equal to the area of squire. For a square of side 10, the value of B is comes out to be 14.26. It is first half of the solution.
Don't feel bad, unfortunately this problem calculus wise is very reliant on formulas from table of integrals. The solving was largely done by the reduction formulas, which I don't think any reasonable calc teacher would expect u to have memorized because they don't pop up that often, outside of situations like this where the curves happen to be parts of circles.
@@kameronrogers4000 no it would if you have enough practice of indefinite integration during 12th std, or even if you face such difficulty later you could always look on internet, these are basic reduction
I don't know of a way to do it without calculus. This isn't one of those "clever" problems that has a trick solution once you see it. Instead, it's a lot of grinding calculation.
Wow! You really had a lot of work with this one! Great job! I used a mix of methods. For the 5x5 lens, I used geometry b/c its very easy to calculate: Area of 1/2 circle with radius 5 - 1/4 Square area. For the 5x10 lens I used calculus, but I turned the original pic upside down. So then it became integral of smaller circle from x=0 to 8 minus the integral of the larger circle from x=0 to 8. To solve the "sqrt's" integrals I had to look it up in a table of common indefinite integrals. PS: I also did it in the geometrical way, but my final solution uses arcCos function b/c I used the cosine law to determine the angle of the circumference arcs.
I barely passed Diff-EQs (5th straight Sem of Math getting a BSME) 45 years ago; and have been retired 25 years now; and I have to say WOW, I barely remember doing these practically in my head, and I'm dizzy right now. Whee, this is fun!
okay, at the end of the video, you asked if i got this question and what method I used, but I did Not in fact get an answer, but I had the thought process of what methods I would have used which were all correct. I haven't been in math for two going onto three years, but I knew and followed everything you were explaining. Made me smile
I ended up a lot closer than I expected. I've been challenging myself recently to not use a calculator while solving problems, so when I tried using the geometric solution mentioned here, I ran into a major snag when having to calculate the inverse tangent. I ended up taking a best guess on it based on what looked right for the problem and ended up with a final answer of "9.9 or a little less". I'm surprised that I ended up using the correct method though. Usually geometry is my Achilles Heel for mathematics. I didn't even consider using calculus to solve for the answer, which surprises me given how much I tend to default to it.
As an observation, the formula found for the area of a lens in the video is for the special case of a lens where the two circles meet at right angles - in general, you won't have right-angled triangles, and end up with a longer general formula (the standard version is given in terms of r and R, the two radii, and d, the distance between the two centers)
@@goyonman9655 It is in two dimensions so the integration should be straightforward. I haven't checked the proof. The narrator seemed to be skipping bits from the part I listened to and his calculus explanation didn't seem clear. The key point is equating two of the curves to determine the x,y limit points through which you need to integrate. When people skip bits of an explanation most usually it's because they are so clever they expect everyone else sees the missing bits as obvious or they don't fully understand the proof. If he comes back and corrects me then fine but at least i hope he comes back with more information.
Same with me. As soon as I saw the picture I thought this is to complicated to think about. Let's just integrate and get it over with. But I knew for sure there must be a geometric solution.
🤯 I get the second solution best out of all of them but I wouldn't have got there in a million years and I wouldn't be able to do it even now without watching this again!
I've graduated high school 2 years ago and I almost forgot those math elements.. Now I remember that I was genius in math when I was a student and now I realized that my brain is rotten since graduation... Better study again. I've been wasting my time since graduation playing games and internet surfing... Thank you so much.
I literally just finished calc 3 this morning so it was still very fresh in my head. I was thinking "oh, you could do polar coordinates and use single or double integrals (whatever you fancy), broken up by the intersection point, with the origin being in the top-right corner" I actually hate that these solutions were significantly easier and used a much more basic version of math
Yea exactly, I was counting it by transforming into polar coordinates and in my head it was much easier with a double integral, than three basic integrals in cartesian
I literally wrote a python code to do this computationally: start with an nxn grid. Imagine a circle at point (0,0) of radius n. (circle 1) Imagine a circle at point (0,n/2) of radius n/2. (circle 2) Imagine a circle at point (n/2,0) of radius n/2. (circle 3) Loop through literally every point in the nxn grid. Add to variable A if the point is: inside circle 1 AND outside circle 2 AND inside circle 3. The ratio A/n^2 is the percentage of the square containing the area we want. Multiply by 100 square units and you have the answer to the question. I blew n up to 5000 and waited about 45 seconds for my code to run and I got 9.77, so there is a "Red Green" solution lol.
Something tells me we may be able to solve this with a system of equations. I named each section with constants a --> f (with b being the objective), and got to b + e = pi*(10)^2/8. Only worked on it for a couple of minutes but I'm sure I can represent e (which would the the lower right section of the quarter circle) in terms of the total area of the square then I'm good! Thanks for sharing this Presh!
I did the same and got to a matrix 6x6. I’ve tried solve it using Cramer’s rule but didn’t work. So I had to appel to the integral method and found 11,0743. I still didn’t watch the video.
Hi Presh, many thanks for this gem. It's a lot more simple though. Focus on area PDA (via inner arc). I call it the battle axe. It's area is obviously 25. You instantly also know the area of PD, the double-sided symmetrical konvex lens which I call the fish. (done via computing semi-circle minus 25). Now work out area BCD (via point Q), I call it the "kite", like a hangglider (done by the "square" minus "quartercircle"). Thereafter you work out area QBC using year 10 high school maths (drawing a line from Q orthogonal to line BC) followed by integrating separately on both sides of this line "under" (here to the right of) the two arcs of the respective circles. Take QBC away from the "kite" and you get QCD, known as the cleaver. Take away cleaver and fish from already known semi-circle and yield PQD being 9.7735... Woody
@@ShittingStar07 . QBC looks like a tail fin of a fish or a double-sided ramp. I come across this shape quite often . Even though calculus feels like cheating, the simplicity to work out the area is too quick and easy. I draw a line from Q orthogonal to BC and integrate both sides and add them together. It's merely 2 simple circles, easy. Two things to watch out for. We integrate above the circle not under it. Second thing, since there are two different radii, 3 component's change when writing down the integral, not two.
This is solvable using geometry and algebra only. Algebra can find the coordinates of the right-most circular intersection using the general equation for circles and the known centers. Knowing that, a sufficient system of triangles can be constructed such that the desired area is obtained purely by addition / subtraction of triangular areas and arc segment areas. I will confess it took 2 hrs to come to this conclusion.
I used the 2nd method with some modifications. When calculating the blue shape I first calculated both of the sectors, and then I took the total "kite" area (= 50) and substracted one of the sectors from it to get the part of the other sector that is outside the blue lens. And then I simply subtracted this area from that other sector to calculate the blue shape. Seems to be easier.
What his teacher hears: So if 3 - 2 = 1 it means that 1 is the answer What I hear: So if the power of xy squared divided by 5.73x10 to the power of -82 you can calculate the coefficients of angles bdgx2 and x over y to the power of 68 in an a standard form and that’s how you calculate the efficiency of booster 2 in a 2500 litre rocket...
When you showed that the length of the side of square is 10, my first guess was "the area is somewhere around 10". And then you showed that it is around 9.77. :D
I remember a few years ago I wasn't able to solve this. I was able to solve it today having learned calculus. The geometric way requires less calculations, but the critical thinking involved in that method would have taken me longer to be honest. So the calculus way was the fastest way to solve it for me.
1. 3D print the shape but give every edge the same height wall 2. Measure how much water it hold and how much red area holds 3. Red water divide by all water X100 4. Area of Square X Step 3 = Area of red
I did integrals but I’m finding a common theme. Whenever looking for an area of a circular shape, find a right angle triangle that splits the needed area into a right triangle and a proportion of the circle
My ignorant younger self took AP Stats instead of AP Cal..so I did this the long geometric way, and essentially just nodded my head as you narrated the on-screen calc operations, like, "Yes, the audio appears to be in sync with the video here," so, long story short... 🙏 I appreciate your videos, which make me wish I'd taken more maths courses at university, because I always enjoyed the subject and I ought to have continued to be good at it (Well, that wasn't very short, was it?) ✌️
You can divide the square in to 4 smaller squares each with a area of 25 and assume the circles radius to be 5 and 10 by completing them, the result is similar and can be calculated by normal math.
For the calculus solution, I used polar coordinates. Rotated the figure and shifted the origin to the point where the left, top point (0,5) is now located at. Doing so, all the curves pass through the origin and r-theta functions are easy to set up. Used A= 0.5 x integral of r^2 d theta. All the cartesian equations of circles can be converted easily to polar x = r* cos(theta) etc. This is a shorter and less painful method than using cartesian system shown here. All the angular limits are easy to find using regular trig/geometric formulae.
After watching the solutions, I realize that I'm simply mathematically undereducated, I didn't have the tools to solve it with what little trigonometry I remember from school. I guess, I'll have to start from scratch with basic maths if I want to be able to tackle problems like these... Oh well, it was fun trying to derive what I could with what little I could remember.
The second method seemed to me to just be the first method worked out twice with specific values for a and b. In both cases you find all the triangles etc.
The way I did it was drawing the diagram and putting A, B, C, D, E and F into each of six parts of the jigsaw puzzle, with B being the shaded bit. If we take half of one side of the square as r, the area of the square is 4r2 then express B = 4r2 - A - C - D - E - F and you can find other relations where two of the letters add together to make half an area of a circle etc etc and eventually all other variables cancel out to give the answer.
I used method 1, except: first, I was too lazy to do the math for the larger lens, so used an online calculator for that (area=24.043), and second, I found the area of the smaller lens by a simpler method. Not having the graphics available, I describe it thus: the smaller lens is found by calculating 1/2(area of left-side semicircle) - 1/2( 1/2 the box area - area of upper semicircle), or 1/2(39.270) - 1/2(50-39.270) = 14.270. So large lens less small lens is 24.043 - 14.270 = 9.773.
I looked at that and thought: It is pretty straightforward with calculus, though it would take a while with multiple integrals. With geometry, that's tough. EDIT: I wrote that before watching. That makes me feel better.
You posted before my comment appeared...The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Could you work out the area of the semi circles and the quarter circle, then add them up, then take away the area of the square, then since you have two layers of the required area then dive it by 2?
You posted before my comment appeared...The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Here is another way of showing the geometrical method: For the semi-circles, we have the circle area formula of πr², so: half circles area = 25π/2 = 12.5π quarter circle area = 100π/4 = 25π Considering that the small lens goes between two identical circles, we can claulcat it as: smalllens = semi_circle1 + semi_circle2 - kite smalllens = 12.5π + 12.5π - 25 smalllens = 12.5 x (π-2) or smalllens = (π-2) x side² / 8 And for the biglens we need the lens formula: biglens = side² x (6 x arctan(0,5) - 4 + pi) / 8 f = side² - 12.5π - 25π + biglens Then we can solve: Answer = biglens - smalllens
I paused the video for about 45-60 days and after multiple attempts and probably about 5-8 hours I did solve it as in second solution :D #kingoftheworldfeeling :)
@@shyamsharma1746 About 1+ months ago, I found the problem on MSE and there was a comment by Presh pointing to this vid. The vid already exist then because I indeed watched it and commented. I don't know why the published date now was changed to Jan 10, 2019. There must be a mess-up by YT or some other reasons I don't know. Today is the 11th, it is unlikely that video of this nature could receive so many comments in just one day.
I did't solve it but just watched the solutions. Very interesting, and of the three I think the second solution is simpler and better. The Calculus solution is very drawn out and would need a person who is good at it - we all can be good at Calculus, just keep practicing and memorize first pricipals. Thanks for posting and really enjoy your maths videos.
I feel like if I just studied, revisited and rediscovered the maths I learned in polytechnic I'd be able to solve using both trigo & calculus. I really need a jumpstart course on that.
A simpler expression for arcsin(x) in terms of arctan(x) can easily be derived from a right triangle in which sin(y) = x (take opposite site to have length x and hypothenuse to have length 1, e.g.). Then the adjacent side has length sqrt(1 - x*x). Now the angle can be expressed as: - y = arcsin(x) - y = arctan(x / sqrt(1 - x*x) Since both expressions must yield the same angle, we deduce that arcsin(x) = arctan(x / sqrt(1 - x*x)). (this also holds in case x < 0, since both arcsin and arctan then flip signs; but we have to take x -1 and x 1 - although 'limit-wise' it still holds)
@18:30: ...and build confidence for people....nope...unfortunately my confidence has flown out of the window...I noticed the circles/quarter circles, square, left overs from deduction of the above, and all related it to AB, pi, etc. I thought I was there, started watching the video...and then, my confidence flew right out of the window, it became worse when the integrals came in for what I thought wasn’t that difficult, especially when in some of your video’s you talk about 5th graders in china... But, let’s stick to the inspire part and try your next video with an open mind. Thx! And keep posting these video’s.👍🏻
All it takes is to draw up the limits of integration in the Cartesian coordinate system(Set point A as (0;0)), go to the polar coordinate system and take the double integral. We had this on the test last semester.
The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus
Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions
And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Hi Mind your decisions, isn't the solution just a specially sheared version of a slice of the upper semicircle limited by points P and Q?
Hey, I solved it in a different way that does not require as much knowledge of trigonometry!
Now noone has an excuse for saying "first"
Hi mind your decisions I have a problem for you:
In a trapeze abcd (ab||dc)
The ac and bd diagrams are cut at point e the ratio between the bases is 4:9 and the area of the triangular bec is 108
What is the area of the whole trapeze?
Matheus Mendes yeah I also did it , well I used basic geometry and one equation system.
Just check the resolution of picture and count the red pixels. It will be surprisingly accurate.
Yeah this is big brain time.
That is actually a legit answer. The method is Monte Carlo method.
Tell me this
Sfhnxdkmvddghcssasfhkkmnjggcghjhy cnbvyul iyvvjygvbjuydfvjjnnb hgjjbffhjgtynoolmnbc hhyrszxvnklopplkkjhgh
That is the basis for calculus.
@@technosiam4480 I see what you mean.
-"the first method is simple"
me: "alright, this might be the first time where i can solve a hard Math problem on my ow..."
-"shows space rocket schematic"
me: "okay, next video"
I'm just watching it to see what I DON'T understand.
@@schwartzseymour357 Just watch the first method, the other ones are waaaay too abstract and you're needlessly complicating things. With some knowledge of geometry and testing your 3D logical insight it's fairly doable.
If you lack these concepts, yeah this video is as good as impossible for any newbie.
@@janjan55555 I disagree. If you know integral calculus, the third method is VERY straight forward, and is definitely the way I would have went before he even mentioned the geometric solutions.
@@williamhexberg6748 I do and I still disagree with your method
@@janjan55555 To me the integral method feels brute force and it’s the way I went about solving it before watching the explanation. I was expecting some clever/creative geometry trick and it turned out to be really simple. I just couldn’t see it.
Another solution: print it on a paper, cut out the red part, weight it, divide by the total weight.
this may sound funny but it's how they did it before calculus/geometry was common, in fact no so long ago, math became more common in 18th century or so
@ then print a modified picture of it without the red part colored in
Lame
@Aryan Sharma you think from perspective of today. Note that only few in only few nations knew this in ancient times. Rest of nation was not able to read or write. What more geometry was expressed primarily graphically, there wasn't kind of todays' equations.
Engineers vs mathematicians
For those who are wondering where π came for θ, it is the radian system or circular system for angles. 180 degree = π radian. And 1 radian = 57 degree (approx). *And 1 radian can also be represented by the angle formed at the centre of a circle by 2 radii separated by an arc of 1 unit length (on the circumference of the circle)*
this is a 7th grade problem
@@Mike-rw1jw akshully I could do it at 1st grade 🤡
I'm proud of myself for this one. It is the first time I solved one of your math puzzles. It took me like an hour but I got the right result of 9.77
I'm only 14 years old and I've never had anything like this in school before.
Ayo good job!!
@@Soefae lol, you answer on my 11 month old comment. But anyways, thank you
@@TheDreamingDriftersStories out of curiosity, if you didn’t learn it in school then where did you learn it? And which method did you use?
@@commonlogicart1604 I found some method to calculating the different circle pieces area, and subtracting them.
But it's almost a year ago, so I don't remember exactly what I did.
I'm 14 too and I get 21.4
you know i've been watching a couple of your vids and there's usually a common theme in finding the correct answer. Either you use coordinates or geometry. When using geometry there's always one rule... find triangle and you'll find the answer.
Just throwing that out. It's amazing to how much extent you can use this simple concepts to solve something like this
Solve this problem guys. This is a challenge for you. Problem given in the link below.
ua-cam.com/video/pgWx7CipFlo/v-deo.html
Yup. This is exactly why trigonometry is taught as a standalone course and not as the subset of geometry that it is.
Option 4 - Depending on how many sig figs you need.
Step 1: Find a sheet of graphing paper.
Step 2: Sse a compass and rulers to draw out the figure.
Step 3: Spend the last 3 hours of the exam to count the number of little squares.
This is essentially how some computer programs compute areas. They just make the squares small enough so that they get the desired precision and of course they can count the squares much faster :)
마인드 유어 디시젼
@@최정운-t5h dak chua!!
I work at a railway testing lab. There's a test that involves pressing a rubber pad in between a mechanical actuator and a cast iron mold with small high relief squares carved on it, over which a powder is spread. You place the rubber pad, press it using a specific pressure, and leave it there for a while. Afterwards you remove the pressure, shoots a photo of the rubber side with the square marks the powder made, transfer that photo to a computer, and have it calculate the ratio powdered area vs. non-powered area by literally counting how many pixels in the photo got this or that color and dividing one number by the other.
That's what I would do in exam..
Next video: Put many circles inside a square; make most of them pass from intersections of other circles. Find the most absurd looking piece of area. And then solve for it.
@Tokisaki Kurumi using only a ouija board
And then realise a 7 year old Korean kid solved it quicker than you did.
U don't have any other work or what?🤔😑
haha, this one is so complicated i don't want to watch through it.
AhmedHan that actually sounds fun
I paused the video and solved the problem on my own before watching the full video. I approached it geometrically. I did it in my head without even using paper and pencil. I thought it was quite simple. I continued to watched the full video just to make sure I got the same answer. I was surprised that all the 3 methods in the videos are so long. Below is the outline of my method. Using the same labelling as 00:34 of the video,
Area of shape DPQ = Area of shape DPQC - Area of shape DQC.
It is easy to see that the shape DPQC repeated 4 time will fill the square. So the area of DPQC is just 25.
Next step is to solve for area of shape DQC. Let midpoint of DC be E.
Area of DQC = Area of sector QEC + Area of quadrilateral DEQA - Area of sector DAQ.
It can be easily checked that Angle QEC = 2 * Angle QDC = Angle QAD = 2 * tan^-1(0.5).
Therefore, Area of shape DQC = pi * 5^2 * (2 * tan^-1(0.5)/2pi) + 2 * 1/2 * 5 * 10 - pi * 10^2 * (2 * tan^-1(0.5)/2pi)
=50 - 75*tan^-1(0.5)
Finally, Area of DPQ = Area of shape DPQC - Area of shape DQC
= 25 - (50 - 75*tan^-1(0.5))
= 75*tan^-1(0.5) -25
This is the same answer.
I didn't finish reading all the 1.5k comments. But so far among the comments I've read, I haven't seen a similar method. This is the first time I see this problem. I'm not sure whether or not this method above has been posted before elsewhere on the internet.
Finally an elegant solution without huge calculations and variables. Bravo!
@@GGReaver Thanks for taking notice of my solution. I thought no-one would bother reading it 😆
Any mathematics enthusiast will be blessed by finding this channel...Hats off to the creator❤️🎉🙏👍
I used the 4 option.... I watched the video
Me too 😅
Sven Minoptra , 😂😂😂😂 I did the same
If I only had 5min to solve, I would answer "definitely less than 100" & walk away
exactly
"definitely less than 50" to be more accurate
@TheMutedGamer I would bet it was less than 18. (haven't watched yet)
Well definetely more than zero
@Yash Hriddha I bet it's 9.77
i have enough problems in my life
😂😂😂😂😂😂😂
This stuff is precisely to distract you from problems in life. XD
😂😂😂😂... Swears
LMAO relatable😂😂😂
Maths book: "Cute"
I got the area of the yellow area a differrent way:
Area of the semicircle = 12.5π (Solve with πr^2 × 1/2)
Multiple it with four
12.5π × 4 = 50π
It is like you draw four semicircles inside the square. Or imagine as mirroring the two given semicircles.
Now you can see it covers the whole square and also the yellow area 4 times overlaying.
Which means 50π = area of the circle + 4 × yellow areas
50π = 100 + 4y (y=yellow area)
y = (50π - 100)/4
y = 25/2(π-2)
y = 12.5π - 25
As an alternative, one might use symmetry and scale to simplify the solution.
Steps:
1) 'Symmetrize' the picture by drawing another quarter circle radius 10, centered at the TOP, RIGHT vertex of the square.
This divides the square into 8 regions, of which 3 are pairwise identical, thereby resulting in 5 distinct shapes ( with areas x1 thru x5). Our target region (shown in red in video) is one of the paired regions (say x2).
2) Express the areas of the small semicircle, the large quarter circle, and the square as linear equations in the variables x1 thru x5.
3) Thus far we have 3 linear equations, (and 5 unknowns) so we need 2 more.
These are a) The area of the small 'lens' region formed by the intersection of the two semicircles, which can be easily computed as described in the video, or otherwise.
and b) The area of the larger 'lens' formed by the intersection of the two large quarter circle.
Since the larger lens has the same shape and twice the linear dimensions as the smaller lens, its area is 4 times as large.
4) Now we have a complete set of equations, from which x2 can be easily obtained.
Thought that would be the "best" solution as well when trying the problem on my own, but sadly the equation system is underdetermined and you can't solve for x2 (the red "wanted" region) with these equations alone. You somehow need to calculate the area of one of the "paired" regions on their own and for that you always need the right most corner of the red region (or its symmetrical counterpart). And at that point you can just use the calculus solution from the get go...
I’m majoring in Mechanical Engineering and the calculus method seemed to be the easiest method. 😂
It included the most important step I remember from my calculus class - "annnd... skip some steps".
Though in my class this was the "and from there... it's obvious" step :P
I dont think so, its easier to error with calculus, and more work demanding in my opnion. If you can work with basic algebra or geometry and have a good trig perception you should use it. Calculus is too powerfull, like trying to catch a fly with a gun when you have a fly swatter.
1! +2!+ 3! +...........25! divided by 13 then what will be the remainder left?
@@theunique140 Using python.
x = 0
for i in range(1,26):
x += factorial(i)
x%13
= 9
@@theunique140
If you're still looking for an answer, you can do it like that
1! + 2! + 3! + ... + 12! + 0 (x!%13 = 0 for x>=13)
1(1 + 2(1 + 3(1 + 4(1+5(1+6(1+7(1+8(1+9(1+10(1+11(1+12))))))))))) % 13
11*(12+1) = 11*0
10*1
9*11 = 99, 99%13 = 8
8*9 = 72, 72%13 = 7
7*8 = 56, 56%13 = 4
6*5 = 30, 30%13 = 4
5*5 = 25, 25%13 = 12
4*0 = 0
3*1 = 3
2*4 = 8
1*9 = 9
Your first "little lens" can be calculated much simpler, namely as the 5x5 square without the two other shapes, so that's 25 - (25 - quarter circle) - (25 - quarter circle), so that is a half circle of radius 5, minus 25 all in all, making it 12.5pi - 25. That's a whole lot faster than your "lens" calculation.
I've had a similar reasoning to do that part (square - 2 whole circles and then divide by 4).
I wish there is a close way (substract areas) to solve the second part... i'm pretty sure there is one but i can't find it
Isn't this basically a way he went in the second option?
His first “little lens” could be calculated in the way you showed, but what about the second “little lens”? You probably will need a different method to solve it, on the other hand, his method of the first “little lens” was applicable to the second lens, thus, saving him time and making the overall solution easier, at least in my opinion. How would you calculate the second “little lens”?
I did the same.
@@--Za did the exact same thing and the second part is bothering, you need to get the area of about 6.47 by using the circles but I cant seem to figure out how
One of the best math channels of UA-cam, greetings from Brazil!
yes really
Papa Flamme ficou triste agora
Super exatas>>>
Xande god
Yeah
This and 3Blue1Brown
I'm proud to report I sat through the entire video without skipping. That alone is an accomplishment given how little I understand anything that was said.
The amount of work that goes into your videos is truly phenomenal 🙌
It's funny how calculus is 10 time easier than geometriy
that's because the calculus way is the robot way, the geometric way is the intelligent way.
avhuf why is it robot
No it's not
@@SticklyDaKing because u don't need to use a lot of ur brain , even a 10 th grade student can solve it that way given hr know all the formulas
Calculus is mostly used to program computers and calculators which uses the same definite way to solve one type of questions. Intelligent ways are much more diverse and allows shortcuts that cannot be discovered by machines. It also varies from one person to another, signifies creativity. Unlike calculus, geometry defines intelligences that a machine will never have. (Unless somebody invented artificial intelligence that will eventually cause global extinction.)
I put this in AutoCAD
😎😎😎
Ans - 9.7736
you are THE MAN
9.773570675
@@ahmedmonjid5696 Thanks man
@@mayanktomar5022 exactly
How did you measure the area of the shape? (I created the same things as shown in the video but I can't measure it)
I just recently discovered your channel. I did the problem with both methods, the area subtraction using geometry (since that stood out to me) and the calculus way (because that just made sense integrals for the win). I set up the problems in my head and didn't particularly solve for an exact solution. Once that was done I finished watching the video. I m glad my physics/math brain is always on the ball great problem. Math is just awesome!
9:00 instead of using trig functions, we can also apply the similar triangles rule. Find the ratio of the kite triangle and smaller triangle hypotenuse, then their areas have ratio to the square
I Just loooooove the feeling when you can use so many different solutions to find the same answers. Keep up the good work!
I once solved this really hard Q in class which no one could solve and then the teacher told us how to do it and his method was completely different than mine but we had the same answers.
I literally looked at the thumbnail and tried to solve it for 10 mins. My friend said “are you looking for something on the screen?”
1! +2!+ 3! +...........25! divided by 13 then what will be the remainder left?
I graphically solved it without paper after looking for hour and a half HAHAHAHA ocd
Thanks, dude you made me realized I need to retake Calculus and Integration.
Solve this problem guys. This is a challenge for you. Problem given in my channel.
I solved this problem using calculus, it was the fastest and easiest way, in my opinion. I really liked the other two geometric forms of solving it!
Area of intersection of two semi circles can be calculated by drawing all four semi circles inside a square. Suppose area of semi circle is A, area of intersection of two semi circles is B and area of the square is C than 4*A - 4*B = C. Because when you add areas of four semi circles together, each intersection will be counted twice so you have to subtract four times of the area of intersection in order to make it equal to the area of squire. For a square of side 10, the value of B is comes out to be 14.26. It is first half of the solution.
Me: I know calculus well
Me after watching this: I know nothing about calculus
Facts. I haven't even begun learning integral calculus, only differential, so I found the geometry a million times easier to understand.
This was basic calculus actually, you do have some problems
@@DanTheManTerritorial lmao, you have the reduction formulas memorized?
Don't feel bad, unfortunately this problem calculus wise is very reliant on formulas from table of integrals. The solving was largely done by the reduction formulas, which I don't think any reasonable calc teacher would expect u to have memorized because they don't pop up that often, outside of situations like this where the curves happen to be parts of circles.
@@kameronrogers4000 no it would if you have enough practice of indefinite integration during 12th std, or even if you face such difficulty later you could always look on internet, these are basic reduction
I am still struggling with this problem so I haven‘t watched the solution- it is driving me nuts! I am not yet ready to give up. Wish me luck. 🤔🙄😳😠😩
u done yet?
Affinizationz lmaoo 1 year later
I bet he stills solving it
Updates ?
I don't know of a way to do it without calculus. This isn't one of those "clever" problems that has a trick solution once you see it. Instead, it's a lot of grinding calculation.
And i was there thinking i can do it in my head with my calc 🤣
Solve this problem guys. This is a challenge for you. Problem given in my channel.
Wow! You really had a lot of work with this one! Great job!
I used a mix of methods. For the 5x5 lens, I used geometry b/c its very easy to calculate: Area of 1/2 circle with radius 5 - 1/4 Square area.
For the 5x10 lens I used calculus, but I turned the original pic upside down. So then it became integral of smaller circle from x=0 to 8 minus the integral of the larger circle from x=0 to 8. To solve the "sqrt's" integrals I had to look it up in a table of common indefinite integrals.
PS: I also did it in the geometrical way, but my final solution uses arcCos function b/c I used the cosine law to determine the angle of the circumference arcs.
I barely passed Diff-EQs (5th straight Sem of Math getting a BSME) 45 years ago; and have been retired 25 years now; and I have to say WOW, I barely remember doing these practically in my head, and I'm dizzy right now. Whee, this is fun!
okay, at the end of the video, you asked if i got this question and what method I used, but I did Not in fact get an answer, but I had the thought process of what methods I would have used which were all correct. I haven't been in math for two going onto three years, but I knew and followed everything you were explaining. Made me smile
2:02 Solution using geometry and trigonometry
6:49 Longer solution using geometry and trigonometry
12:06 Solution using calculus (integrals)
I ended up a lot closer than I expected. I've been challenging myself recently to not use a calculator while solving problems, so when I tried using the geometric solution mentioned here, I ran into a major snag when having to calculate the inverse tangent. I ended up taking a best guess on it based on what looked right for the problem and ended up with a final answer of "9.9 or a little less". I'm surprised that I ended up using the correct method though. Usually geometry is my Achilles Heel for mathematics. I didn't even consider using calculus to solve for the answer, which surprises me given how much I tend to default to it.
As an observation, the formula found for the area of a lens in the video is for the special case of a lens where the two circles meet at right angles - in general, you won't have right-angled triangles, and end up with a longer general formula (the standard version is given in terms of r and R, the two radii, and d, the distance between the two centers)
Can you state the longer one?
Absolutely. It's only going to be a right angle if you have a tangent which, as you say, is only in one special case. I'd stick with the calculus !
@@we-are-electric1445
But being bent at an angle other than 90 degrees would complicate the Cartesian plane needed for the integration
@@goyonman9655 It is in two dimensions so the integration should be straightforward. I haven't checked the proof. The narrator seemed to be skipping bits from the part I listened to and his calculus explanation didn't seem clear. The key point is equating two of the curves to determine the x,y limit points through which you need to integrate. When people skip bits of an explanation most usually it's because they are so clever they expect everyone else sees the missing bits as obvious or they don't fully understand the proof. If he comes back and corrects me then fine but at least i hope he comes back with more information.
@@we-are-electric1445
Remember the dimensions of the square gave him values of points on the axes and how to pinpoint their values on the plane
Same with me.
As soon as I saw the picture I thought this is to complicated to think about. Let's just integrate and get it over with. But I knew for sure there must be a geometric solution.
🤯 I get the second solution best out of all of them but I wouldn't have got there in a million years and I wouldn't be able to do it even now without watching this again!
I did not calculate it, I guesstimated 10%, being a building contractor has it's perks ! ;)
Lol
"its" used as a genitive gets no apostrophe.
@@cubeeater6458 dude, get a life.
whoa, that's great!
@@cubeeater6458 Ha ha. I couldn't do the maths problem, but like you I can see all the grammatical errors in the comments :-)
LETS GOOOO I ACTUALLY GOT ONE OF YOUR VIDEOS RIIGHTT ✊🏼😤 Damn I've never been so proud of myself lmao
I liked it at the end when he said "did you figured out this problem ?" XD
Please have mercy
I've graduated high school 2 years ago and I almost forgot those math elements..
Now I remember that I was genius in math when I was a student and now I realized that my brain is rotten since graduation...
Better study again. I've been wasting my time since graduation playing games and internet surfing...
Thank you so much.
My hearty thanks:from India
I literally just finished calc 3 this morning so it was still very fresh in my head. I was thinking "oh, you could do polar coordinates and use single or double integrals (whatever you fancy), broken up by the intersection point, with the origin being in the top-right corner"
I actually hate that these solutions were significantly easier and used a much more basic version of math
Yea exactly, I was counting it by transforming into polar coordinates and in my head it was much easier with a double integral, than three basic integrals in cartesian
I literally wrote a python code to do this computationally:
start with an nxn grid.
Imagine a circle at point (0,0) of radius n. (circle 1)
Imagine a circle at point (0,n/2) of radius n/2. (circle 2)
Imagine a circle at point (n/2,0) of radius n/2. (circle 3)
Loop through literally every point in the nxn grid.
Add to variable A if the point is:
inside circle 1 AND outside circle 2 AND inside circle 3.
The ratio A/n^2 is the percentage of the square containing the area we want.
Multiply by 100 square units and you have the answer to the question.
I blew n up to 5000 and waited about 45 seconds for my code to run and I got 9.77,
so there is a "Red Green" solution lol.
share it please
What you did is known as monte carlo integration
GG
nice
that seems unnecessarily brute force...
Something tells me we may be able to solve this with a system of equations. I named each section with constants a --> f (with b being the objective), and got to b + e = pi*(10)^2/8. Only worked on it for a couple of minutes but I'm sure I can represent e (which would the the lower right section of the quarter circle) in terms of the total area of the square then I'm good! Thanks for sharing this Presh!
Did you manage to do it that way, I'm really curious because I have been trying that myself but no results yet
I did the same and got to a matrix 6x6. I’ve tried solve it using Cramer’s rule but didn’t work. So I had to appel to the integral method and found 11,0743. I still didn’t watch the video.
I used hyperoperated calculus with stable equality with a trig failsafe, also including operation successorship. This method is unbeatable.
The method by the integrales is great !!! And it's a good mean to understand how powerful are the integral concept and tool !!! Very nice video !!!
As soon as I saw this picture, the word "calculus " came to my mind, even though I had no idea how exactly...
Solve this problem guys. This is a challenge for you. Problem given in my channel.
Hi Presh,
many thanks for this gem. It's a lot more simple though. Focus on area PDA (via inner arc). I call it the battle axe. It's area is obviously 25. You instantly also know the area of PD, the double-sided symmetrical konvex lens which I call the fish. (done via computing semi-circle minus 25). Now work out area BCD (via point Q), I call it the "kite", like a hangglider (done by the "square" minus "quartercircle"). Thereafter you work out area QBC using year 10 high school maths (drawing a line from Q orthogonal to line BC) followed by integrating separately on both sides of this line "under" (here to the right of) the two arcs of the respective circles. Take QBC away from the "kite" and you get QCD, known as the cleaver. Take away cleaver and fish from already known semi-circle and yield PQD being 9.7735... Woody
wha...
Exactly how I did but got stuck at the QBC part..
@@ShittingStar07 . QBC looks like a tail fin of a fish or a double-sided ramp. I come across this shape quite often . Even though calculus feels like cheating, the simplicity to work out the area is too quick and easy. I draw a line from Q orthogonal to BC and integrate both sides and add them together. It's merely 2 simple circles, easy. Two things to watch out for. We integrate above the circle not under it. Second thing, since there are two different radii, 3 component's change when writing down the integral, not two.
@@WoodyC-fv9hz can it not be done without using calculus or trigonometry?
This is solvable using geometry and algebra only. Algebra can find the coordinates of the right-most circular intersection using the general equation for circles and the known centers. Knowing that, a sufficient system of triangles can be constructed such that the desired area is obtained purely by addition / subtraction of triangular areas and arc segment areas. I will confess it took 2 hrs to come to this conclusion.
Exactly, i did it. And i got 9,78%
It took me 3 hours using the same method!
I am skipping this question in my exam.......btw thanks a lot for coming up with such awesome questions and solutions.
Ngl its an emotion
After 2.5 hrs of constant doing math i did itttt
Yess i did it
I used the 2nd method with some modifications. When calculating the blue shape I first calculated both of the sectors, and then I took the total "kite" area (= 50) and substracted one of the sectors from it to get the part of the other sector that is outside the blue lens. And then I simply subtracted this area from that other sector to calculate the blue shape. Seems to be easier.
I’m just gonna pretend I understand what he said
hahahahahaha so so good to know its not just me
i could solve the yellow one with an easier formula
but i will have to pretend to know about the blue one
What his teacher hears: So if 3 - 2 = 1 it means that 1 is the answer
What I hear:
So if the power of xy squared divided by 5.73x10 to the power of -82 you can calculate the coefficients of angles bdgx2 and x over y to the power of 68 in an a standard form and that’s how you calculate the efficiency of booster 2 in a 2500 litre rocket...
When you showed that the length of the side of square is 10, my first guess was "the area is somewhere around 10". And then you showed that it is around 9.77. :D
I remember a few years ago I wasn't able to solve this. I was able to solve it today having learned calculus. The geometric way requires less calculations, but the critical thinking involved in that method would have taken me longer to be honest. So the calculus way was the fastest way to solve it for me.
1. 3D print the shape but give every edge the same height wall
2. Measure how much water it hold and how much red area holds
3. Red water divide by all water X100
4. Area of Square X Step 3 = Area of red
i just drew this up in Fusion360 and had it measure the area for me.
I did integrals but I’m finding a common theme. Whenever looking for an area of a circular shape, find a right angle triangle that splits the needed area into a right triangle and a proportion of the circle
My ignorant younger self took AP Stats instead of AP Cal..so I did this the long geometric way, and essentially just nodded my head as you narrated the on-screen calc operations, like, "Yes, the audio appears to be in sync with the video here," so, long story short... 🙏
I appreciate your videos, which make me wish I'd taken more maths courses at university, because I always enjoyed the subject and I ought to have continued to be good at it (Well, that wasn't very short, was it?) ✌️
You can divide the square in to 4 smaller squares each with a area of 25 and assume the circles radius to be 5 and 10 by completing them, the result is similar and can be calculated by normal math.
For the calculus solution, I used polar coordinates. Rotated the figure and shifted the origin to the point where the left, top point (0,5) is now located at. Doing so, all the curves pass through the origin and r-theta functions are easy to set up. Used A= 0.5 x integral of r^2 d theta. All the cartesian equations of circles can be converted easily to polar x = r* cos(theta) etc. This is a shorter and less painful method than using cartesian system shown here. All the angular limits are easy to find using regular trig/geometric formulae.
After watching the solutions, I realize that I'm simply mathematically undereducated, I didn't have the tools to solve it with what little trigonometry I remember from school. I guess, I'll have to start from scratch with basic maths if I want to be able to tackle problems like these... Oh well, it was fun trying to derive what I could with what little I could remember.
The second method seemed to me to just be the first method worked out twice with specific values for a and b. In both cases you find all the triangles etc.
Solve this problem guys. This is a challenge for you. Problem given in the link below.
ua-cam.com/video/pgWx7CipFlo/v-deo.html
Love your vids! Please don't stop!
The way I did it was drawing the diagram and putting A, B, C, D, E and F into each of six parts of the jigsaw puzzle, with B being the shaded bit. If we take half of one side of the square as r, the area of the square is 4r2 then express B = 4r2 - A - C - D - E - F and you can find other relations where two of the letters add together to make half an area of a circle etc etc and eventually all other variables cancel out to give the answer.
I used method 1, except: first, I was too lazy to do the math for the larger lens, so used an online calculator for that (area=24.043), and second, I found the area of the smaller lens by a simpler method. Not having the graphics available, I describe it thus: the smaller lens is found by calculating 1/2(area of left-side semicircle) - 1/2( 1/2 the box area - area of upper semicircle), or 1/2(39.270) - 1/2(50-39.270) = 14.270. So large lens less small lens is 24.043 - 14.270 = 9.773.
I looked at that and thought: It is pretty straightforward with calculus, though it would take a while with multiple integrals. With geometry, that's tough.
EDIT: I wrote that before watching. That makes me feel better.
3 definite integrals if i count correctly that is if you chose 0 for y at conviniant position, not even watching :D
Sam two double integrals, the region can be broken down into two type 1 or two type 2 regions.
1! +2!+ 3! +...........25! divided by 13 then what will be the remainder left?
@@theunique140 0. 13 is a factor.
@@theunique140 Slightly more interesting variant I just thought of:
Divide by 13^13.
Same answer.
“Uploaded January 10, 2019”
1 month comments
W A I T
You posted before my comment appeared...The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus
Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions
And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
And that my boii is how a tough problem looks like
@@MindYourDecisions
Could you pls gimme some information about Math stack exchange
I solved this when I was a fetus in India.
THe calculus method is very elegant. Congratulations. I def lack the skills but I get some of the geometrical solutions thou.
Hello! I am from Brazil and I love your channel. Thanks for everything
Could you work out the area of the semi circles and the quarter circle, then add them up, then take away the area of the square, then since you have two layers of the required area then dive it by 2?
Wait a second, this video was a month ago. Was it only for patreons or something?
You posted before my comment appeared...The first comment is from 1 month ago because I posted it to Math StackExchange where I received great help: math.stackexchange.com/questions/3016718/a-way-to-find-this-area-without-calculus
Early comments also come from Patreon, as I do my best to give early access to supporters on Patreon: www.patreon.com/mindyourdecisions
And finally, thanks for 1 million subscribers! We are building one of the best mathematical communities in the world. Thank you.
Solve this problem guys. This is a challenge for you. Problem given in my channel.
why this have 69 likes ?
Here is another way of showing the geometrical method:
For the semi-circles, we have the circle area formula of πr², so:
half circles area = 25π/2 = 12.5π
quarter circle area = 100π/4 = 25π
Considering that the small lens goes between two identical circles, we can claulcat it as:
smalllens = semi_circle1 + semi_circle2 - kite
smalllens = 12.5π + 12.5π - 25
smalllens = 12.5 x (π-2)
or
smalllens = (π-2) x side² / 8
And for the biglens we need the lens formula:
biglens = side² x (6 x arctan(0,5) - 4 + pi) / 8
f = side² - 12.5π - 25π + biglens
Then we can solve:
Answer = biglens - smalllens
- Draw it on CAD software
- select the boundary
- Find the area
I paused the video for about 45-60 days and after multiple attempts and probably about 5-8 hours I did solve it as in second solution :D #kingoftheworldfeeling :)
I approached it with calculus.
BUT the geometric solutions were elegant.
Did you get the right answer or you just tried to solve it?
Solve this problem guys. This is a challenge for you. Problem given in my channel.
Nice and clear presentation. Great Job!
How can your comment be one month old...🤔
@@shyamsharma1746 About 1+ months ago, I found the problem on MSE and there was a comment by Presh pointing to this vid. The vid already exist then because I indeed watched it and commented. I don't know why the published date now was changed to Jan 10, 2019. There must be a mess-up by YT or some other reasons I don't know. Today is the 11th, it is unlikely that video of this nature could receive so many comments in just one day.
*3 Ways to Get Depressions (Insanely Work)*
I have no idea about calculus but I still watched it. I'm proud of myself.
Solve this problem guys. This is a challenge for you. Problem given in my channel
I found the second method the most intuitive, but thank god for calculus for making things much easier
I used an calculator. Definitely the best method.
MCAT skill: “idk looks like around 10% or so”
Mathematician / physics skills
* 25 minutes of work later = 9.77 *
i got 856/75
"So let's simplify"
Me: Yes, please
*complex equation turns into another complex equation*
Me: ok
I did't solve it but just watched the solutions. Very interesting, and of the three I think the second solution is simpler and better. The Calculus solution is very drawn out and would need a person who is good at it - we all can be good at Calculus, just keep practicing and memorize first pricipals. Thanks for posting and really enjoy your maths videos.
This is a great problem for AP Calc students!
3:50 how did you know that the lower angle is right?
DifferentStuff it’s a kite, so the angle on the left equals the angle on the right
@noahkupinsky1418 How you know its kite
I feel like if I just studied, revisited and rediscovered the maths I learned in polytechnic I'd be able to solve using both trigo & calculus.
I really need a jumpstart course on that.
Solve this problem guys. This is a challenge for you. Problem given in my channel.
It’s been 20 years and I still remember soh cah toa.
Mah man!!
Solve this problem guys. This is a challenge for you. Problem given in the link below.
ua-cam.com/video/pgWx7CipFlo/v-deo.html
A simpler expression for arcsin(x) in terms of arctan(x) can easily be derived from a right triangle in which sin(y) = x (take opposite site to have length x and hypothenuse to have length 1, e.g.). Then the adjacent side has length sqrt(1 - x*x).
Now the angle can be expressed as:
- y = arcsin(x)
- y = arctan(x / sqrt(1 - x*x)
Since both expressions must yield the same angle, we deduce that arcsin(x) = arctan(x / sqrt(1 - x*x)).
(this also holds in case x < 0, since both arcsin and arctan then flip signs; but we have to take x -1 and x 1 - although 'limit-wise' it still holds)
@18:30: ...and build confidence for people....nope...unfortunately my confidence has flown out of the window...I noticed the circles/quarter circles, square, left overs from deduction of the above, and all related it to AB, pi, etc. I thought I was there, started watching the video...and then, my confidence flew right out of the window, it became worse when the integrals came in for what I thought wasn’t that difficult, especially when in some of your video’s you talk about 5th graders in china...
But, let’s stick to the inspire part and try your next video with an open mind.
Thx! And keep posting these video’s.👍🏻
Took me a full hour, now i finally can go to bed
*10*
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*1*
*0*
*and that ends the official countdown of my brain cells*
Solve this problem guys. This is a challenge for you. Problem given in my channel.
My head hurts now.....
All it takes is to draw up the limits of integration in the Cartesian coordinate system(Set point A as (0;0)), go to the polar coordinate system and take the double integral. We had this on the test last semester.