why the area under the tangent line of 1/x is special
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- Опубликовано: 30 сен 2024
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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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Do you know why Einstein was really stupid? I know why.
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
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.
your serious, all angles of tangent will produce same area?
I'm jealous of people who have him as his teacher
My students might think otherwise 😆
@@blackpenredpen haha, I guess I can't really complain when I go to a Maths based College in England
You mean envious. Jealousy is fear of losing someone to another.
@@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 RUclips channel, no-one can say they were not warned....
@@JP-lz3vk well said.
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
thank you very much. usually my comments are solutions. In this case, my comments were questions. It makes it more interesting (and less tiring!)
I was going to try and puzzle this out, thanks!
What is this called? I'd like to learn more about its derivation, but I can't find anything online.
@@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))
@@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.
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!
This man is subbed to Animenz. Good taste :D
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
@@Roescoe He's also subbed to Pan Piano, what a man of culture
Good luck with university man, that's awesome.
4:12 - 4:19 it seems you forgot to cut this out in editing lol
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?
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. 😊)
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.
@@jeremydavis3631 ah, identical
I wonder if we can completely dispose with even coordinates and solve a purely geometric problem in terms of hyperbolas
youtuuu.tokyo/Dv1PhhYxWho
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 ,
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.
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!
4:01 when I am presenting in front of the class
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
well, what about 2/x or 3/x. they all have the same size triangle formed by any tangent.
@@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).
@@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.
@@martyguild I don't disagree. I assumed that's what 1/x meant, but in less mathy words.
@@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
Area of this triangle is 2 "ahaa".
😆
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.
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.)
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.
@@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!
@@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
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.
This is definitely a property of rectangular hyperbolas, but would it work for hyperbolas in general?
Could you link to a source? I don't know what rectangular hyperbolas are
@@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.
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.
@@OCTAGRAM ah, wonderful!
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.
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.
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))
Wonderfully proven
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).
Since I lived in Singapore I like your singaporish English (don’t know where you live but you sound like a Singaporean
Sir you teach enjoying math🥳🥳
how'd he get three different colours of marker out of two markers ? The real question....
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
ooh that's nice
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
Why did you shave? 😢
If he did shave then what happened with you by him ?
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.
Is anyone else as impressed as I am over how quickly and smoothly he switches markers? O.O
Did it with y(x)=1/x^2 but doesn't work, area can be described as A(a)=9/(4*a), "a" being the tangent point.
Just saw a guy who did the general formula f(x)=x^n; wow.
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)
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)
hypotenUses!!
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.
youtuuu.tokyo/mE0pkxJNojS
I love your videos thanks for helping me to progress faster in math !
Happy to help!
OK THIS VID MAKES Me enjoy math
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?"
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.
The Y-intercept really got to him
😆
Where did the old kung fu master whith long beard go?
And who is this guy?
So confused rn
Interesting, does this have any engineering applications?
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.
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!
Thanks!
nice property! 😉
That pause at the y-intercept, I know you wanted to say, "Its right there" so badly lol
My intuition told me from the first moment, that the AREA should be constant just by the fact that x*y = const
Wow this was so cool!!
Thanks!
4:10 When you have too many tabs open so your computer starts lagging
All the functions that fulfill the constant area restriction must satisfy that:
-[y-y'*x]^2/(2*y')=A
where A is a constant (the area)
This comes from
A=x*Y/2
In the tangent line equation, you have that
Y-y = y'(x-a)
So,
Y=y'*x+y-y'*a
Y=m*x+b. Therefore m=y'; b=y-y'*a.
Then we have to find x and Y
Y=0=m*x+b
x=-b/m
And
Y=m*0+b
Y=b.
Replacing:
A=(-b/m)(b)/2=-b^2/(2*m)
A=-[y-y'*x]^2/(2*y')
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
How about when x approaches 0 or infinity? then the area wont be 2
General formula for f(x) = x^n:
A(a) = (n + 1)²/(2na^(n - 1))
😮!!
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.
Solve without calculus.
Coordinates of tangent point ( a,1a)
Area of rectangle from thst point to origin =1. Cut the rectangle diagonally down from yaxis to xaxis.
Slide one copy up to sitting on top of rectamgle and one copy to the right of the rectzngle.
The two small triangles each with area 1/2 b=a, h=1/a.
NOTE.. incomplete see discussion
But how can you prove that the copies are same as the rectangles obtained by tangent?
You are assuming that the other two rectangles form a tangent to the curve.
@@bobtivnan not two rectangles but two triangles and obviously they are similar since they are right angle triangles with a hypotenuse of the same slope. what is particular about this equation xy=1 or i think any equation xy=c, is that the two similar triangles are equal. I think that can be proved by the fact that the equation is symmetric in x and y, although i have not done that.
Yes, I meant triangles. They are obviously congruent. My point is that you have made an assumption that a tangent segment is formed. It may be possible to show that with a proof by contradiction, but it's a hole in your reasoning as it stands.
@@bobtivnan
|\
| \
~~.
|__|_\
in that image above… the period is the tangent point. The origin at the bottom left
the rectangle between them a single
tangent lines runs between the intercepts on the axes. formed from \\\
for any curve the triangles must be similar, because the corresponding angles are equal.
but in the case y=1/x, the two triangles are equal. and equal to half the rectangle.
in another case, eg y=1/x^2 then the two triangles would be similar but not equal.
For that assertion to be a proof, I need to show that the slope of the tangent to the function, is the slope of the diagonal of the rectangle. I hope to do that by virtue of the xy symmetry of the curve.
But I haven’t done that. does it work for (x-r)^2 +(y-r)^2 =r^2 ?
Things drawing attention: 1) Plush Pokemon icon 2) board marker juggling 3) Actual math stuff :D
😆
Hey bprp. Can you give a definitive proof that the equation (x+y)! (n+1) + y! ((x+y-n)!/(y-n-1)!) = (x+y)! (x+y-n) holds for all real numbers x, y and natural numbers n? I need that proof for my university-level calculus equation.
Either you have a typo, or the claim is false. Plug in some numbers to see that it doesn't work. Or note that for positive x and y, the LHS is greater than (x+y)! but the RHS is less than (x+y)^2.
@@turtlellamacow oh yeah sorry it's a typo rhs should include (x+y)! not (x+y)
I think this doesn’t hold for (x,y,n)=(0,2,1)?
I get the impression that you need to proof that (x+y)!( (n+1)-(x+y-n) + (y!/y!)*P(t=1 to n+1 of y-t)/P(t=1 to n of x+y-t) ) = 0
wherein P(t=a to b of f(t) ) means the product of all factors f(t) for t equal to all the integers a up to and including b, or, in other words: P(t=1 to n of x-t) = x!/(x-n)!
@@jeffreyh.1436 Yes... 6=8 is not true
This is really interesting! 😀
😆 glad you like it!
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.
Damn, the y-intercept got him really good lmao. Video was both interesting and entertaining, Thank you!!
hahaha
Great explanation, professor. Thanks a lot!!!
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|.
How come A/(2a) doesn't satisfy general solution at any C?
@@ВасилийТёркин-к8х 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)
The function f(x) has at the point (a|f(a)) the tangent t(x) = f'(a) * (x-a) + f(a). The zero point of the tangent is xN = -f(a)/f'(a) + a. The area that this encloses with the axes is calculated here as follows:
A(a) = (t(0) * xN)/2
= (-f'(a) * a + f(a)) * (a * f'(a) - f(a)) / (2 * f'(a))
= (-a^2 * f'(a)^2 + 2 * a * f'(a) * f(a) - f(a)^2) / (2 * f'(a))
= - (a * f'(a) - f(a))^2 / (2 * f'(a))
We now want the area A(a) to be constant regardless of the choice of a. Finally, using the notation y = f(x), we can formulate the following differential equation:
(x * y' - y)^2 + 2 * A * y' = 0,
where A is a real number corresponding to the desired constant area.
If one chooses the approach y = c/x^b, then follows:
c^2 * (b+1)^2 / x^(2*b) - 2 * A * c * b/x^(b+1) = 0
Obviously, 2*b = b + 1 must hold, so b = 1. Thus it follows: c = A/2.
If you choose a polynomial of the nth degree as approach, you can easily show that they cannot satisfy the differential equation (with the exception of the trivial linear function).
For a Taylor series (infinitely long polynomial), you find conditions for coefficients, but these ultimately lead only to the trivial solution.
Means: f(x) = A/(2*x) is the only function which solves the differential equation. Therefore, there is no other function with this property.
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
I totally love your videos.....They are the best and they are vivid to understand!
Reversed problem: area is given, find the function. Area could be any number. (If the area is 4, the function is y = 2/x)
The ball he’s holding used to be the microphone right? Now it’s just branding?
It’s still a mic
Figured it out when I saw the thumbnail. :)
Still tapped here to leave a Like tho.
Thanks.
Your video inspired me to write a Desmos demonstration for my students. Thanks!
Sadly the general area for this triangle for 1/x^n is 1/2*(n+1)^2/n * a^(1-n), so sadly it does depend on a
Awesome video! Thank you!
Well isn't there a much simple solution of xy=1 and the area of the triangle is 2xy which is 2?
Also from what I know the rectangle area between a hyperbolia and its asymptote is also a contstant
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
Unexpected and delightful. Since we don't care about simplifying the line equation, why not substitute x=0 and y=0 back where it's still in the form x-x1 and y-y1 and save a few lines?
best math channel
y=2-x has the same property, because every tangent line is the same :3
lol
Another way to see it: Because the equation y=1/x is symmetric in x and y (which also means its graph is symmetric about the line y=x), the tangents at the points (a, 1/a) and (1/a, a) would enclose the same area with the axes, so it can't be proportional to a
I’m convinced the only difference between math students and teachers is that teacher can actually write legibly and draw accurate, realistic drawings
If f(x)=1/x^n, we get that the area of the triangle is equal to ((n+1)^2)/(2na^(n-1))
An interesting property of hyperbolas is that area of the triangle formed by tangent and asymptotes of a general hyperbola [(x²/a²)-(y²/b²)=1] is equal to 2ab
Here, if we have a=b, we get a rectangular hyperbola
A special case of the rectangular hyperbola is of the form (xy=c²)
Our equation, y=1/x is of this form.
Thus the area of that triangle becomes 2c², which when c=1, is 2.
Just another approach though...
This is cool! Thank you for another excellent video.
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.
I love it when cool things happen in math. I think only people who enjoy math would think it is cool. I told my wife about it and her response was basically, "so what.". I am still freaked that e^(i*pi) = -1 and other things like that. And how does pi even come out of a series, freaky.
If my math is right, it turns out a general formula for area under the tangent line of 1/(x^n) in the first quadrant is:
(((n+1)^2)(2n-1))/(2n(a^(n-1)))
At n=1, like in this video, it simplifies down to 2/(a^(1-1)), which is obviously just 2.
Cool little result!
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.
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
Just for the sake of variety, here is a method that doesn't use calculus.
We take the general straight line function y = mx + c and equate it to y = 1/x for its points of intersection with the curve:
1/x = mx + c
⇒ mx² + cx - 1 = 0 ... ①
Now isolate the tangent lines as the ones where both points of intersection coincide, i.e. where the discriminant of the quadratic equation ① is zero:
c² + 4m = 0
⇒ m = -c²/4
Hence the tangents to the curve have the form
y = c - c²x/4 ... ②
The y-intercept is c, and the x-intercept is given by:
c - c²x/4 = 0
⇒ x = 4/c
The area of the triangle formed by any such tangent line with the axes is half the product of the intercepts:
A = ½.c.4/c = 2 as required, independent of c.
If preferred we can integrate ② between 0 and 4/c to get the area:
A = ∫₀ ⁴/ᶜ (c - c²x/4)dx = [cx - ⅛c²x²]₀ ⁴/ᶜ = 4 - (c²/8)(16/c²) - 0 = 2.
So someone else showed that this property doesn't work for x^-n n!=1, however this trivially works for any linear equation that doesn't go through (0,0). I'd be wondering if this works for y = a/(x-b)+c or even a/(x-b) + cx + d.
However, this probably is limited to certain values of those constants. Take for example 1/(x+1)+1. At x=0 y = 2 and the area is 2. As x tends to infinity, the slope tends to 0 meaning the y intercept of the tangent approaches 1 as the x intercept approaches infinity.
Very interesting video. I have a question.
Maybe it's a little late, but could you or someone probe using differential equations that C/x is the only function that achieves this property?
I tried but I ended up with a non linear ODE (ordinary differential equation), and I don't know how to solve it. The equation is:
2A*y' = -(y - x*y')^2
If you substitute 1/x you can see it is a solution, but I would like to solve the equation with no prior information.
Thank you very much.
Pretty sure it only works for 1/x because working backwards locks us in to that formula.
If the area = some constant C, in this case 2. Working backwards we get that the area of a triangle would work out to be 1/2(CA)(C/A) the A’s cancel to get 1/2CC as the area. But we also know the area is C so C=1/2(C^2). That’s 2C=C^2 or C=2. So an area of 2 is the only constant we can get this way. Meaning a Y intercept of 2/A and an x intercept of 2A are the only way to fit the property.
As we plot the point at X=A of the line with Y intercept 2/A and X intercept of 2A for every value of A it will only trace out 1/X
So, the only one solution for the diferential equation Constant=(1/2)(y-xy')(x-y/y') is y=1/x ?, I put it in a Dif. Eq. Sol. And I got y=Cx +/- 2(-C)^(1/2)
f=1/x=x⁻¹ so f´=-x⁻²=-1/x² pick point A(p,1/p) the tangent will have y=ax+b with a=f´(p)=-1/p² so the angent is y=-x/p²+b through A so 1/p=-1/p+b multiply by p 1=-1+bp b=2/p so tangent y=-x/p²+2/p . the height of the triangle is y for x=0 and the base is x for y=0, so lets calculate. x=0 -> y=2/p .
y=0 -> -x/p²+2/p=0 -x+2p=0 x=2p so h=2/p and b=2p area of triangle is A=bh/2 so A=2
note its not true for p=0
y = -x/a^2 + 2/a
y + x / a^2 = 2/a
Multiply by a/2 = 1 / (2 / a)
y / (2 / a) + x / (2 * a) = 1
so we have interceptions
(2*a;0)
(0;2/a)
It is like an ellipse.
sorry guys i have a math problem that I can't solve, it is to find the limit as x approaches 0+ of
((√(1-x+3x²))-cos(√x))/(x*ln(x))
it probably has to do with the taylor series
4:10 bprp has stopped responding
No matter how much I try, I simply can't deviate my look towards that pokeball microphone. It's just so... brilliant and... I NEED IT
it seems that if you have an inverse power function f(x)=1/x^n the equation for the tangent area becomes A=1/2 *((n+1)^2)/(na^(n-1)). In the case of n=1, a is irrelevant but only in that case
The equivalent triangle under e^(-x) x>=0, has maximum area at x=a=1. All triangle area are always less than 1? Agh well nice little fact to know I suppose, _mcih_ (might come in handy one day) シ
Clicked on video for curiosity.... the reason I watched past 50% was because I was wondering if there was a reason for the Pokeball. :D That poke-ball is one brilliant distraction! :D
There might be a way to prove this in a more simple way.
I intuitively understood that because of the function's symmetry and the fact it is bounded inside the +/+ quadrant, the tangent point would always be exactly in the middle of the tangent line but I couldn't find a way to conclusively prove it without calculations that are more complicated than taking the slope of the tangent and solving the problem that way.
Anyway, if one can prove that the tangent point is always exactly halfway along the tangent line then the area is simply 2(xy) with x and y being the coordinates of the tangent point. And since y=1/x, 2(xy) is simply 2.
hey, the (line) integral of the function line int[0,1] range of f(x)=b*sqrt(1-x^2), yes, its a circle or ellipse, b=1 or b1
My man shaved and looks straight 25 years old again
😆
haven't seen the sol, let me try
let's say the line is mx+n
it would mean mx+n = 1/x has 1 sol
mx² + nx - 1 = 0 so n² = -4m
the line mx+n intersect y=0 at -n/m
also intersect x=0 at n
so the area of triangle is ½ n (- n/m)
= -½(n²/m) = -½ (-4) = 2
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.
I should be making homework, yet I'm here watching a mathematics video. My level of procrastination confuses me.
Ight, something weird is happening which i cant upload my picture of my working out with 1/x^2. so yea... sad life. thank you for the help tho
(ill try to send the link here tho)
I love all your videos, but for once I am ready to do a question which btw is going to be horribly explained, hopefully someone gets the point.
I understood the video and I understand that it doesnt matter what the value is for a, these values cancel out. But this seems surprising when you take a very, very far away from the Y axis a value. Like, the function itself 1/x never touches the X axis and I guess u would have to take limits and so on. And yet, if you have an a thats very far away, were the tangent line then is barely paralel to the X axis, the area is still 2. It sounds like a contradiction that if we talk about infinities, we just have 2 as answer straight away. Would this exercise still hold when we are taking in account such a far away a value?
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.
Was bored doing school math
Watching this video to get rid of boredom 🥲