why the area under the tangent line of 1/x is special

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2 жыл бұрын

Visit 👉 brilliant.org/blackpenredpen/ to learn more calculus (20% off with this link)!
Learn why the area under the tangent line of y=1/x is very special. We need to know how to use the power rule to take the derivative of 1/x and how to write the equation of a tangent line. Then we just need the area of a triangle formula, which is just 1/2*base*height. This is a very classic calculus problem and it's great for Calculus 1 or AP Calculus AB students. Subscribe for more math for fun videos 👉 ‪@blackpenredpen‬
#calculus #blackpenredpen #apcalculus #chanllenge
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Пікірлер: 420

@blackpenredpen 2 жыл бұрын

Learn calculus in a Brilliant interactive way 👉 brilliant.org/blackpenredpen/ You can help this channel a lot by simply clicking the link and checking them out. Thank you 😃

@Mnemonic-X 2 жыл бұрын

Do you know why Einstein was really stupid? I know why.

@CONEJOI666 2 жыл бұрын

Thanks to teach this exersice, my teacher of calculus when appear this exersice, explained so horrible and i cannot understand nothing, but seeing ur video i have clear the method

@ahmadtim2537 2 жыл бұрын

For the function f(x)=m/x^n, the area under the tangent line when x=a is (m×(n+1)^2)/(2n×a^(n-1)) so it that's why the area under the tangent line of f(x)=1/x is special. Edit: thanks for being the best math guy on youtube.

@ryanb1874 2 жыл бұрын

your serious, all angles of tangent will produce same area?

@ewancox7017 2 жыл бұрын

I'm jealous of people who have him as his teacher

@blackpenredpen 2 жыл бұрын

My students might think otherwise 😆

@ewancox7017 2 жыл бұрын

@@blackpenredpen haha, I guess I can't really complain when I go to a Maths based College in England

@combcomclrlsr 2 жыл бұрын

You mean envious. Jealousy is fear of losing someone to another.

@JP-lz3vk 2 жыл бұрын

@@blackpenredpen I assume the students want to learn from a passionate teacher like yourself and are not forced to be there. But the best teachers demand the best from the their students and since you have a KZfaq channel, no-one can say they were not warned....

@RS-jq4oc 2 жыл бұрын

@@JP-lz3vk well said.

@leonardsmith9870 2 жыл бұрын

I nearly failed math in high school but you've really re-ignited my passion for math at the ripe age of 31. I've recently upgraded my high school math subjects with high 90s in order to get into university! Thank you for what you do and getting me back into it!

@Roescoe 2 жыл бұрын

This man is subbed to Animenz. Good taste :D

@grgfrg7 2 жыл бұрын

Thanks for posting my friend I am also 31 and thinking about going back to university to study a different field. Good to see others in my situation making it happen

@Smallpriest 2 жыл бұрын

@@Roescoe He's also subbed to Pan Piano, what a man of culture

@gaetanramos7903 2 жыл бұрын

Good luck with university man, that's awesome.

@StichyWichy21 2 жыл бұрын

General formula for area for a function f(a): A(a) = af - ½a²f' - f²/2f' From this it can be found the only functions for which area is constant are of the form f(x) = Ax + B (which is trivial), and f(x) = A/x

@davidseed2939 2 жыл бұрын

thank you very much. usually my comments are solutions. In this case, my comments were questions. It makes it more interesting (and less tiring!)

@peterclark5244 2 жыл бұрын

I was going to try and puzzle this out, thanks!

@saif_saleh_ 2 жыл бұрын

What is this called? I'd like to learn more about its derivation, but I can't find anything online.

@mahmoudfathy2074 2 жыл бұрын

@@saif_saleh_ just do as shown in this video. But generalizing f(x), it's not rigorous i guess but it will end up with this formula. Start: m = f'(a) p = (a, f(a))

@virusweb7080 2 жыл бұрын

@@mahmoudfathy2074 It is rigorous actually. af is the rectangle which I'm sure you can visualize. There are 2 triangles left to add. One of which will have its base as a. Since the slope is f', the other leg has to be of length a*f' and same reasoning for the other triangle.

@OCTAGRAM 2 жыл бұрын

Students are often not taught properly hyperbolic functions, hyperbolic rotation and pseudo-Euclidian geometry. Circle stays intact after rotation by any angle. Similarly hyperbola stays intact after hyperbolic rotation by any angle. I can apply hyperbolic rotation to transform any tangent triangle into a single well-known one, for which I already know the area. A nice property of hyperbolic rotation is that from the Euclidian interpretation area stays the same. Because Euclidian and pseudo-Euclidian areas are the same. Distances are not the same, areas are. Thus, hyperbolically rotated triangle have the same area as well-known one. Hyperbolic rotation can be viewed as area-preserving affine transformation. Yet another reason for constant area. There is a similar thing in higher dimension. Take an infinite cone and a hyperboloid inside with asymptotes matching infinite cone walls. Now you can draw tangent plane to any point of hyperboloid, and for the same reasons the volume will be the same. Using cone axis rotation and hyperbolic rotation we can move any hyperboloid point to a single well-known point with well-known volume. In terms of special relativity, Lorentz transformation keep the light cone intact (although lightwave length changes, but it has no impact on our task). Lorentz transformations also keep intact the spacelike surface of all events equally distant from an origin. This surface is hyperboloid and a light cone is a degenerated case of hyperboloid. Back to 2D, there is one more thing you can do. Take two hyperbolas with same asymptotes. Take tangent line to one hyperbola and intersect with another hyperbola. You'll enclose a constant area. It works very similar to two concentric circles and a chord that is tangent to the inner circle. In Euclidian space we just cannot degenerate concentric circles into something as remarkable as a light cone. Yet another idea for 2D: find an ellipse that has axes same as hyperbola asymptotes, and is tangent to hyperbola in a chosen point. This time we are relying on affine transformation properties.

@snacke4500 2 жыл бұрын

Got this on a test not to long ago! I happened to solve the question and the answer blew my mind! One of my favourite questions!

@jakobr_ 2 жыл бұрын

This probably has some connection to the fact that the rectangle with one corner at the origin and the opposite corner on y=1/x has the same area=1 no matter what point on the line you choose. The connection to the “area under tangent line = 2” is easy to see geometrically once we’ve shown the tangent’s intercepts are twice the coordinates of the point on the curve. Can anyone think of a non-calculus way of getting to that point?

@jeremydavis3631 2 жыл бұрын

Here's one (it helps to draw the figure as you read about it): Consider the rectangle OBAC O(0,0), A(a,1/a), B(a, 0) and C(0, 1/a). Clearly, the area of this rectangle is 1. Now split the rectangle along its diagonal OA, forming two right triangles: OAB and OAC. Reflect triangle OAB across line AB and segment OAC across segment AC. This forms a new triangle ODE with D(2a,0) and E(0,2/a) and with exactly twice the original area, i.e. 2. To prove that segment DE actually passes through A: Angle BAC is a right angle, and the reflections preserved angles. Therefore, angles DAB and EAC are complementary, and the the three angles altogether are supplementary. So lines AD and AE are parallel. Since they both contain A, they are the same line. To prove that line DE is indeed tangent to y=1/x: We already know that A is on the curve, and given the shape of the curve and the negative slope of line DE, in order for the line and curve not to be tangent, there must be a second intersection point, Z, such that Z≠A. Suppose we find such a point at (z,1/z). Then line EZ must contain point D(2a,0). The equation of line EZ is y=(2/a-1/z)/(-z)x+2/a. Plugging in y=0 leads to x=2z/(a(2/a-1/z)). By our supposition, x=2a. So, solving for z: 2a=2z/(a(2/a-1/z)) a=z/(2-a/z) 2a-a^2/z=z 2az-a^2=z^2 z^2-2az-a^2=0 z=(2a±√(4a^2-4a^2))/2 z=2a/2 z=a But if z=a, then Z=A, which is a contradiction. Therefore, A is the only intersection of line DE and the curve defined by y=1/x, which are therefore tangent. Therefore, the initial construction was valid, and the area of the triangle must always be 2. Q.E.D. (That was a fun challenge. 😊)

@GopaiCheems 2 жыл бұрын

Wow, wonderful observation! Indeed, that area of rectangle being 1 would come to use. We draw the tangent at A(a,1/a). Then we draw the rectangle with O(0,0) and (a,1/a) as opposite vertices. We know this area is 1. Let the other two vertices be B,C. Now we find the tangent equation. The slope is -1/a^2 hence the equation is (y-1/a) = -(x-a)/a^2. Clearly, the x-intercept (y=0) = 2a Hence done, as you pointed out how the intercepts being twice of the coordinates of the point are enough. But for completeness, we draw AB, OA and AC to split the triangle in 4 regions. Two of these are equal to each other, as are the other two. Hence area of rectangle = 1/2(Area of triangle) = 1. For a non-calculus derivation of the slope equation, we write y = mx + c xy = 1 Hence x(mx + c) = 1 => mx^2 + cx -1 = 0 has only one solution. So discriminant must be zero. Hence c^2 = -4m. Now substituting x=a, y=1/a finishes.

@GopaiCheems 2 жыл бұрын

@@jeremydavis3631 ah, identical I wonder if we can completely dispose with even coordinates and solve a purely geometric problem in terms of hyperbolas

@oioficina3043 2 жыл бұрын

youtuuu.tokyo/Dv1PhhYxWho

@bachirblackers7299 2 жыл бұрын

Jacob Roggy you may use the same point (a,1/a) to be the intersecting point of major and minor axes of some Ellipse, same ellipse is kissing both x axes and you axes yes he must have area of Pi ,

@euva209 2 жыл бұрын

Simple but thought-provoking. It can also be shown that for any function y=C/x, the area of the triangle will be equal to 2C. If you start with y = Cx^n, it can be shown by using the same approach that the only way the area is independent of the tangent point (x1,y1) is if n= -1.

@sysyphenf8ewtfr603 2 жыл бұрын

4:12 - 4:19 it seems you forgot to cut this out in editing lol

@tomasgalambos3115 2 жыл бұрын

Is anyone else as impressed as I am over how quickly and smoothly he switches markers? O.O

@NotBroihon 2 жыл бұрын

1/x is the only function that works. I did the math for f(x) = 1/x^n (with n > 0) and the resulting area A is: A = (1/2)(n + 2 + 1/n) * a(1 - n) where a is the x coordinate of the tangent point. As you can see only for n = 1 (aka 1/x) this result is independent of a. For 1/x^2 A is 9/(4a), 1/x^3 A is 8/(3a^2) etc. I got the formular for A by integrating the tangent function t(x) from 0 to the root of t(x). The generalized tangent function is: t(x) = -n / (a^(n + 1)) * x + (1 + n)/a^n And the root is a/n + a. t(x) can be derived from f(a) and f'(a). Edit: it's just a triangle, no need to integrate the tangent function lmao Result is the same though.

@inexistente8622 2 жыл бұрын

I always love to see math extended to a "n-limit" lul. Its so nice to see a formula that surfices a value you can choose. (If i said anything off or didnt understand something properly, pardon my english lmao.)

@decare696 2 жыл бұрын

I did the general case for any function (amounted to solving a differential equation) and the only solutions are f(x) = ax+b and f(x) = c/x. The former isn't very interesting (kind of self-evident), so 1/x and scaled versions are really the only functions with this property.

@inexistente8622 2 жыл бұрын

@@decare696 Im really curious to know what you meant by "any function". Im still learning Diferential Equations, so its not like i could grasp whatever method you used X). Still, sounds marvelous!

@pedrosso0 2 жыл бұрын

@@decare696 I did general case and got the equation (y/y')^2-x^2 y''/2=0 But I don't know how to solve for y

@_Blazing_Inferno_ 2 жыл бұрын

It looks like you’re missing a ^, where it would be a^(1-n) rather than a(1-n), unless I’m mistaken. I was confused after reading the equation the first few times.

@ArjunKumar-king06 2 жыл бұрын

I totally love your videos.....They are the best and they are vivid to understand!

@josueramirez7247 2 жыл бұрын

I really enjoyed this demonstration. These days, I don’t get to do a lot of math exercises so I really appreciate the challenge at the end.

@bingcheng8419 2 жыл бұрын

1/x is the only function. Here is proof: Assume 'a' as parameter, and points are (2a,0) and (0,2/a), and area of triangle 2. The line function is y=-1/a^2x+2/a . If x is any increasing function of a: x(a), so y=-1/a^2*x(a)+2/a. This parametric curve should have a slop of -1/a^2. Solving the differential equation for x(a), get x(a)=a

@CoDMunichTV 2 жыл бұрын

well, what about 2/x or 3/x. they all have the same size triangle formed by any tangent.

@jafetguzman5012 2 жыл бұрын

@@CoDMunichTV 2/x and 3/x are all 1/x. Essentually, you are taking the parent funciton (1/x) and multiplying it by a constant (2 or 3).

@martyguild 2 жыл бұрын

@@jafetguzman5012 i mean yeah, but that means 1/x is not the only function: a/x for all a in the reals would account for every function satisfying this property.

@jafetguzman5012 2 жыл бұрын

@@martyguild I don't disagree. I assumed that's what 1/x meant, but in less mathy words.

@gaetanbouthors 2 жыл бұрын

@@jafetguzman5012 there are other fucntions that could work, this is only proof for triangles with an area of 2. 2/x and 3/x do work but with different areas, however it isn't what was implied by the original comment, and isn't obvious either

@lavy9740 2 жыл бұрын

Hey man, I just wanna say that you are my favorite math guy on the internet. You're energy and love for mathematics is infectious and makes me proud to be studying the magic of arithmetic

@volodymyrgandzhuk361 2 жыл бұрын

To generalize, if we consider the hyperbola y=(ax+b)/(cx+d) (of course, c≠0 and ad-bc≠0, otherwise it would be a line and not a hyperbola), then the area of the triangle formed by its asymptotes and any tangent is (2|ad-bc|)/(c²).

@hvok99 2 жыл бұрын

This is so cool! I love how clean this problem is. Thanks for sharing

@yagof6365 2 жыл бұрын

Great explanation, professor. Thanks a lot!!!

@slinkyslider 2 жыл бұрын

4:01 when I am presenting in front of the class

@AdoNir 2 жыл бұрын

Stopped at 0:23, took 5 minutes for myself to solve it, got the answer 2 for any tangent point, got amazed, still watched till the end because you deserve it!

@blackpenredpen 2 жыл бұрын

Thanks!

@SuperMtheory 2 жыл бұрын

Your video inspired me to write a Desmos demonstration for my students. Thanks!

@mikeschieffer7899 2 жыл бұрын

At 5:26 when y=1/x^2 the area of the triangle is not constant. The area is 9/(4a) where a is the x coordinate of the tangent point.

@seroujghazarian6343 2 жыл бұрын

the general formula of the area of the triangle under the tangents to y=1/x^n at (a,1/a^n) is ((n+1)^2)/(2n×a^(n-1))

@AJ-et3vf 2 жыл бұрын

Awesome video! Thank you!

@larrysherk 2 жыл бұрын

Very good. Great example clearly explained.

@felixmarquezy6772 2 жыл бұрын

I love your videos thanks for helping me to progress faster in math !

@blackpenredpen 2 жыл бұрын

Happy to help!

@amanshah5233 2 жыл бұрын

Brilliant sir, you always make my doubt so clear.

@koukanayoub2381 2 жыл бұрын

Great video as always, keep going ❤️❤️

@jessiemeanwell 2 жыл бұрын

This is so cool!! I might make my own video on this problem with some pretty animations...

@MJ-np8kg 2 жыл бұрын

4:06 , him considering re-recording the video is just epic at this moment 😅

@Ch1pp007 2 жыл бұрын

That was really interesting. Thank you.

@yamilbaret403 2 жыл бұрын

Excelente! Maravilloso aporte! Lo voy a agregar a mi cuaderno. Muchas gracias! Saludos de Argentina!

@agenolmedina9159 2 жыл бұрын

Very cool!!! I love this video!

@lazarjovanovic3642 2 жыл бұрын

We had similar question on the admission exam for the college back in 2015. The function was exp(-x), and the question was: "If the tangent line intersects the axes in points A & B, and the coordinate start is denoted with 0, what is the maximum area of the triangle ABO?"

@jacklardner8229 2 жыл бұрын

One of my favorite questions from Calc 1

@tanavposwal Жыл бұрын

Keep posting these special obsevation it helps a lots in exams

@Jha-s-kitchen 2 жыл бұрын

Learning school maths as KZfaq videos and facts is really awesome... I learnt Basic calculus from 3b1b and some adv. calculus and integrations from blackpenredpen, just calculus or just, by him... you are great.

@jeremiasbollati9859 2 жыл бұрын

Fantastic result !

@talkgb 2 жыл бұрын

to find similar functions, we look for a general length and width of the triangle which is nicely just the respective x and y intercepts of the tangent. therefore, for length, we solve f’(a) (x - a) + f(a) = 0 for x and get x = a - f(a)/f’(a). for width, we solve y = f’(a)(0 - a) + f(a) and get y = f(a) - af’(a). therefore the area of the triangle is A = (a - f(a)/f’(a))(f(a) - af’(a))/2. expanding this, we get A = [2af(a) - f(a)^2/f’(a) - a^2f’(a)]/2. turning into quadratic for specific case A = 2 gives a^2f’(a)^2 + (4 - 2af(a))f’(a) + f(a)^2 = 0 isolating f’(a), we get f’(a) = [af(a) - 2 +/- 2 sqrt(1 - af(a))]/a^2 which looks better as y’ = (xy - 2 +/- 2 sqrt(1 - xy))/x^2 sadly we can’t solve this besides test functions but we can confirm that only polynomials that are degree -1 monomials could work for Y (or technically 0, but that can be trivially ruled out) because otherwise the square root will remain, which cannot be in the derivative of a polynomial.

@SteveAcomb 2 жыл бұрын

The real fun bit is when you form a solid by rotating this curve around the x axis and get a cone with finite internal volume but infinite surface area; an endless funnel you could fill with paint but never paint

@plazmyx5998 2 жыл бұрын

ooh that's nice

@technowey 2 жыл бұрын

That is super cool! Thank you for making videos!

@blackpenredpen 2 жыл бұрын

😃

@technowey 2 жыл бұрын

This is cool! Thank you for another excellent video.

@technowey 2 жыл бұрын

I missed that I already posted below two months ago. I did remember seeing this before. I enjoyed watching it again. I'm an electrical engineer, and this is simple mathematics - and yet I never learned this in school. I'm sure there are many other cool mathematics-facts that I don't know.

@JohnVKaravitis 2 жыл бұрын

MIND BLOWN!

@gabinhamel4654 2 жыл бұрын

Very interesting and easy to understand 👌🏻

@silksongreactions 2 жыл бұрын

The Y-intercept really got to him

@blackpenredpen 2 жыл бұрын

😆

@andrewbeck9303 2 жыл бұрын

Wow this was so cool!!

@blackpenredpen 2 жыл бұрын

Thanks!

@baptiste5216 2 жыл бұрын

Amazing ! I wonder if there are more functions wich have this nice proprety of a constant area under the slope.

@fwosti3548 2 жыл бұрын

Fun math video. Fun proof. TY :)

@koeielul112 2 жыл бұрын

Things drawing attention: 1) Plush Pokemon icon 2) board marker juggling 3) Actual math stuff :D

@blackpenredpen 2 жыл бұрын

😆

@pwmiles56 2 жыл бұрын

That's an interesting question at the end. The answer is no, only y=1/x solves this particular problem i.e. A=2. A way to see this is to make lots of triangles, which will have their hypotenauses between points (2x,0) and (0,2/x). The hypotenauses border a curve called the envelope. So the envelope is y=1/x, that's the only solution. (You could put A=2*a^2 but that would just give y=a^2/x)

@Green_Eclipse 2 жыл бұрын

Yeah I found the same thing. The area being constant (rather than specifically 2) has 3 cases: y''=0 (line) y/y'-x=0 (subset of case 1) y/y'+x=0 (y=c/x) Which is either a line or proportional to 1/x. (I assumed that the tangent line just went to the x and y axis there could be some weird curve with self intersections but IDK how to compute that generically)

@azzteke 2 жыл бұрын

hypotenUses!!

@jakobr_ 2 жыл бұрын

Straight lines work as well. All tangent lines of a straight line are the same line, and therefore have the same area underneath. With an exception given to perfectly horizontal or vertical lines.

@oioficina3043 2 жыл бұрын

youtuuu.tokyo/mE0pkxJNojS

@eustacenjeru7225 2 жыл бұрын

I love your teaching skills

@QuantumOverlord 2 жыл бұрын

Had a quick go at proving this before watching the video. This is actually a great question that requires a fairly rudimentary knowledge of calculus but will be quite challenging at the same time. Would be good as a stretch exercise.

@DarkBoo007 2 жыл бұрын

That pause at the y-intercept, I know you wanted to say, "Its right there" so badly lol

@olz6928 2 жыл бұрын

In my calculus class I also proved that the middle point of the tangent line (from y-axis to x-axis) is always on the functio.

@BigDBrian 2 жыл бұрын

if it's given that it's a constant, it's very easy to see it has to be 2, because if you just take the diagonal at (1,1) then it's the area of a 1x1 square and two 1x1 triangles

@CoDMunichTV 2 жыл бұрын

i had this excact problem as a project in university but i did it using the taylor-polynomial. and then i made an equation for finding any function with the same properties, so any function where it does not matter where you put the tangent line and the area of the trianlge will always be the same. it was a very interesting problem thinking back now

@erdo4321 2 жыл бұрын

best math channel

@75blackviking 2 жыл бұрын

I love this guy's channel. He is so good at his profession.

@blackpenredpen 2 жыл бұрын

Thanks!

@gregthomson1064 2 жыл бұрын

It’s pretty impressive how flawlessly he switches marker colors.

@andrelucassen9229 2 жыл бұрын

Nice! It reminds me a bit of the subtangent as desribed in chapter 1 of "New horizons in geometry" by Apostol and Mamikon.

@vijaykulhari_IITB 2 жыл бұрын

Sir you teach enjoying math🥳🥳

@TheCrower77 2 жыл бұрын

im amazed at his ability to seamlessly switch between markers

@robertprobst3836 2 жыл бұрын

This problem was part of the section with random standalone problems in one of the 12th grade mathematics Abitur exams of East Germany in the 1960s.

@GopaiCheems 2 жыл бұрын

This is definitely a property of rectangular hyperbolas, but would it work for hyperbolas in general?

@RS-jq4oc 2 жыл бұрын

Could you link to a source? I don't know what rectangular hyperbolas are

@GopaiCheems 2 жыл бұрын

@@RS-jq4oc do you know that there are a pair of lines which hyperbolas never meet, but come really close to? A rectangular hyperbola is one for which these two lines (called asymptotes) are perpendicular to each other. In this case (y = 1/x), the X axis and Y axis are the two asymptotes.

@OCTAGRAM 2 жыл бұрын

Yes, it will. You can apply affine transformation to general hyperbola to obtain rectangular one, for which you already know everything. Affine transformation scales area by constant factor, so equal areas remain equal. Tangent lines remain tangent.

@GopaiCheems 2 жыл бұрын

@@OCTAGRAM ah, wonderful!

@durianboi4565 2 жыл бұрын

Damn, the y-intercept got him really good lmao. Video was both interesting and entertaining, Thank you!!

@blackpenredpen 2 жыл бұрын

hahaha

@dushyanthabandarapalipana5492 2 жыл бұрын

Thanks!

@felinozaz87 2 жыл бұрын

Very good! :)

@cmilkau 2 жыл бұрын

The union of all such triangles with fixed area A covers exactly the area under A/(2x) and the hypotenuse midpoint traces that function.

@prgrmr8666 2 жыл бұрын

I got this question last semester I solved it this way: 1) you name the point where the tangent line crosses the x,y axis, lets call them a,b 2) suppose the tangent point at (x1, y1) Now we have three points (x1,y1) (0,a) (b,0) Now we calculate the slope for the line between the first two points, and the slope between the first and third point using delta(y)/delta(x) Now guess what, they're equal to each other and they= the derivative at x1 Now you have three equations to solve to get the value of a, b which means you have the dimensions of the triangle Sorry if anything wasn't clear

@zhelyo_physics 2 жыл бұрын

This is really interesting! 😀

@blackpenredpen 2 жыл бұрын

😆 glad you like it!

@yolanda6392 2 жыл бұрын

4:10 When you have too many tabs open so your computer starts lagging

@videomaths 2 жыл бұрын

très bonne explication !

@popcorn101cheese5 2 жыл бұрын

OK THIS VID MAKES Me enjoy math

@michaelmounts1269 2 жыл бұрын

really liked this…very elegant. you flashed 3 diagrams showing differing tangent points…kind of wished you displayed for s few more seconds

@marsvandeplaneet7786 2 жыл бұрын

I always did it aproach as rectangle from (0,0) tot the (x,y) of the dot, with surface 1

@alexufa2 2 жыл бұрын

блин. это офигенно. сразу видно разницу между "знаю формулу" и "понимаю теорию вероятностей"

@rahulbangre9152 2 жыл бұрын

oh i’ve seen this in one of my high school calculus textbooks before. I thought it was super neat :)

@southern_merican 2 жыл бұрын

Im impressed!

@matthewgiannotti3355 2 жыл бұрын

The triangle below the x-axis for the tangent line to the function y=ln(x) is another function that has this property with (a,f(a)) as intercepts

@charlescharles1270 2 жыл бұрын

super cool, man

@lunstee 2 жыл бұрын

I'd approach this in a manner following your early hint that it doesn't matter where on the curve you take a tangent. Note that the curve y=1/x is unchanged if you do a coordinate transform of (x,y) -> (x*A,y/A) : this is a horizontal scaling, together with a vertical scaling such that the two cancel each other and put the resulting curve right on top of the original curve. Not only is the curve invariant under this transform, so is the area under the tangent triangle. We're stretching the triangle by a factor of A in one axis, and shrinking it by the same factor in the other axis to no net effect on area. Since this transform doesn't affect the area of the triangle, we can transform the tangent point to (1,1). By symmetry of x and y, this is half of a 2x2 square, with area 2. The other approach to this is to note that the slope of the tangent, the derivative of 1/x is -1/x^2, while the slope of the line from the origin to (x,y) is y/x=(1/x) /x = 1/x^2, or the negative of the tangent line slope. This means the tangent line is a horizontal or vertical reflection of the line from origin to (x,y). If we draw vertical and horizontal lines from (x,y) to the axis, we partition the triangle into four smaller congruent triangles, two of which form a box of size x by y, with area 1. The total of the four small triangles is thus area 2. You can also avoid invoking calculus to get to this conclusion of the two slopes being negative of each other by following the initial invariance argument instead.

@davidgillies620 2 жыл бұрын

If f is a function that satisfies the necessary concavity/smoothness requirements etc. then A(f), the area bounded by the tangent line through the point {a, f(a)} and the axes, is a functional -(f(a) - a f'(a))^2/(2 f'(a)). That can only be a constant for f' ~ f^2.

@andrewporter1868 2 жыл бұрын

It would seem there may be a way to generally compute closed forms of \int_a^b f(t) dt using f'(x) by recursively constructing triangles and subtracting from the sum of the previous and current triangles' areas the conjunction their areas by computing an explicit formula. That would be really convenient.

@zeroplays9915 5 ай бұрын

xy=c^2 (rectangular hyperbola) and let point be (ct, c/t) and then slope of tangent is dy/dx= (dy/dt) / (dx/dt) = -1/t^2 giving us equation of tangent as x + t^2 y = 2ct. Then put c=1 and get all 3 coordinates (by putting x=0 and then y=0) then apply area of triangle matrix to get A=2

@Starcanum- 2 жыл бұрын

Another cool observation one can make is that (a,1/a) is the midpoint of the line segment between the axes intercepts. So a hyperbola could be seen as the set of midpoints of hypotenuses of right triangles of a given area, with their legs lying on two intersecting lines. If we try to change the angle between the lines, we don't need to separately prove that the resulting locus will be a hyperbola too as it's just one linear transformation away from being the y=1/x, and all the tangent lines and triangle measurements scale accordingly no matter where they are. But if one likes to indulge in overcalculation of obvious things like I do, we could.. Nevermind, I did it but not going to post it as I got more and more lazy with formatting and it was hideous to begin with. It did cancel out in the end, which was mildly satisfying but then I knew that was always going to happen.

@anon6514 2 жыл бұрын

straight: y = mx + c curvy: y = 1/x Equate for the intersect: mx + c - 1/x = 0 Multiply by x: mxx + cx - 1 = 0 Discriminant of quadratic = 0 (one intersection point) cc + 4m = 0 c = 2sqrt(-m) m is the 1st derivative of 1/x eval'd at some x (call it k) m = -1/kk c = 2/k [ this is the height of the triangle ] To find the base: mx + c = 0 Subs in k for m and c: -x/kk + 2/k = 0 x = 2k [ this is the base of the triangle ] Area = 2 Nice problem. Surprising results.

@simplemathguy 2 жыл бұрын

Brilliant!

@shortnr Жыл бұрын

I'm a year late, but I thought I'd mention that you had some serious Michael Penn energy when you fumbled your words and started your thought over again at about 4:15 :)

@sk8erJG95 2 жыл бұрын

In general, if we have a curve f(x), then tangent line at x=a is given by y = f'(a)(x - a) + f(a), so the intercepts are given by y = f(a) - af'(a) and x = a - f(a)/f'(a). So the area of this triangle is A(a) = (1/2)(f(a) - af'(a))(a - f(a)/f'(a)) = (1/2)(af(a) - a^2f'(a) - f^2(a) + af(a)) = (1/2)(a[2f(a) - af'(a)] - f^2(a)). For this to be independent of a, we would need 2f(a) - af'(a) = 0 for all a, meaning 2f(a) = af'(a). Writing this as a differential equation gives y' = (2/x)y which can be solved via separation of variables: (1/y)dy = (2/x)dx ln(y) = ln(2/x) + C y = K*(2/x). So this shows the only functions with this property are multiples of f(x) = 1/x.

@bonbondojoe1522 2 жыл бұрын

General formula for f(x) = x^n: A(a) = (n + 1)²/(2na^(n - 1))

@blackpenredpen 2 жыл бұрын

😮!!

@Chuaaaaaaaaaaaaa 2 жыл бұрын

Incredibly beautyful.

@HanginInSF 2 жыл бұрын

Dude is explaining calculus in his second language with perfect grammar. Also he's only using one hand to do it.

@pingpong3311 2 жыл бұрын

1/x² would be 9/4a. Deriving the function would give you -2/x³. Your coordinate would be (a, 1/a²) Your slope would be -2/a³ Using the formula y-y1=m(x-x1), you would get y-1/a²=-2/a³(x - a). Distributing will give you y-1/a²=-2x/a³+2/a². Isolating y, you would get y=-2x/a³+2/a²+1/a², or y=-2x/a³+3/a², the equation in y=mx+b form. The y-intercept (also the height) is 3/a². Now we must find the base. If you set -2x/a³+3/a²=0, isolating for x, you will get x=3a/2. This is also the base. Last but not least, we have to now use A=bh/2. Plugging in the values, you will get A=(3a/2)(3/a²)/2. A=(9/2a)/2 Multiplying top and bottom by 2a will get you 9/4a. You're welcome.

@zubiprime 2 жыл бұрын

The rabbit hole actually goes WAY deeper with this if you use powers other than -1. If you do use calculus, you can make a graph (in Desmos) where *x* represents what *a* is in this video and *y* represents the area under the tangent line according to *a.* If the power you use is *n,* then the resulting graph will look like some coefficient times *x^(n+1).* If that isn’t insane I don’t know what is. I haven’t figured out what the coefficients are though, and I have no clue about a pattern for functions other than *x^n.* Thanks for reading.

@philgaudreau1294 2 жыл бұрын

We can generalize the above by the following: The tangent of a function f(x) at the point x=a is given by y = f'(a)*(x-a) + f(a). the tangent has a zero at the point x = a - f(a)/f'(a) In other words, we would like to find all functions f(x) such that integral of the tangent of f(x) at the point a from 0 to a - f(a)/f'(a) is constant. In other words, integral_{0}^{a - f(a)/f'(a)} f'(a)*(x-a) + f(a) dx = C for some constant C. Carrying out the integration, simplifying and completing the square we obtain the following: - ( a*f'(a) - f(a) )^2 / (2*f'(a)) = C At first, this looks like a non-linear first order differential equation, which could be hard to solve BUT we can solve it with a nice trick! Rearranging the equation by multiplying by -2*f'(a) on both sides, we obtain: (a*f'(a) - f(a))^2 = -2*C*f'(a) Taking the derivative with respect to a, we obtain: 2*(a*f'(a) - f(a))*(a*f''(a) +f'(a) - f'(a)) = -2*C*f''(a) Simplifying further, we obtain: 2*f''(a)*( a^2 *f'(a) -a*f(a) + C) = 0 This implies that either f''(a) = 0 or a^2 *f'(a) -a*f(a) + C =0 or both Solving these differential equations, we obtain the functions: f(a) = m*a +b for constants m, b. or f(a) = C/(2*a) + d*a for constants d. We still need to verify that they satisfy the constraint integral_{0}^{a - f(a)/f'(a)} f'(a)*(x-a) + f(a) dx = C for some constant C. Plugging these functions into the constraint, we find that m = - b^2 /(2*C) and b is free. and that the constant d = 0. Hence, the only functions that satisfy this problem are: f(x) = b - b^2 * x /(2*C) for any b in the real numbers and f(x) = C/(2*x) BOOM QED!!!

@chuckaviator6423 2 жыл бұрын

For the area under the line tangent to f(x)=1/x², one can apply the same method to get the answer. In this case, the area will be 2.25/a where a is the x coordinate of the point where the line is tangent to the function. I'll try to find an answer for f(x)=1/x^n as well and update. Update: For f(x)=x^-n the area under tangent can be found with the formula ((n+1)/(2a^n))×((a/n)+a) here a is the x coordinate of the point where the tangent meets I tried this out on desmos and it works nicely

@hexagon8899 2 жыл бұрын

i actually predicted this before clicking on the video as visualising it in my mind it looked like it should do that

@user-fv3lz4gh1y 2 жыл бұрын

Since I lived in Singapore I like your singaporish English (don’t know where you live but you sound like a Singaporean

@bombsgamer 2 жыл бұрын

That was neat

@DJTejasMusic28 2 жыл бұрын

Awesome :)

@jpgm11 2 жыл бұрын

With the point-slope equation we get: y=f'(a)*[x-a]+f(a) Whose x-intercept (y=0) is x=a-f(a)/f'(a) and y-intercept (x=0) is y=f(a)-a*f'(a) so the area under the triangle formed by the tangent line is: A=1/2*[a-f(a)/f'(a)]*[f(a)-a*f'(a)] =1/2*[2*a*f(a)-a²f'(a)-f²(a)/f'(a)] So to get a constant area, f must satisfy the differential equation. 2*a*f(a)-a²f'(a)-f²(a)/f'(a)=2*A Using Matlab (I tried to solve the ED myself but i couldn't do it) the general solution for f is: f(a)=±C/2-C²/(8*A)*a Where all solutions are lines that wrap around the (particular solution) curve: f(a)=A/(2a) Therefore, the only functions that satisfy that the area of the triangle formed by the x and y axes and the tangent line to any point of it is constant are the straight lines and the one with the form y=C/x, where C is a constant that generates an area of 2*|C|.

@user-ik2kd9mb5t 2 жыл бұрын

How come A/(2a) doesn't satisfy general solution at any C?

@jpgm11 2 жыл бұрын

@@user-ik2kd9mb5t f(a)=A/(2a) is the singular solution of the differential equation, so it can not be obtained from the general solution. To find this solution you have to find a C(x) value so that ∂[F(C,x,y)]/∂C=0, where F(C,x,y)=0 is the general solution. In this case we have: F(C,a,f)=f+C²/(8*A)*a±C/2=0 So: ∂F/∂C=C/(4*A)*a±1/2=0 C/(4*A)*a=±1/2 C(a)=±2*A/a And replacing in the general solution we obtain the singular solution: f(a)=±C/2-C²/(8*A)*a f(a)=(2*A/a)/2-(2*A/a)²/(8*A)*a f(a)=A/a-A/(2*a) f(a)=A/(2*a)