Pi for parabolas -- the universal parabolic constant.

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[@demenion3521]

for sqrt(1+u²) i would always go with hyperbolic substitution with u=sinh(t). the double angle formulas for hyperbolic functions make the calculation very straight forward and lead to the equivalent result sqrt(2)+arsinh(1) which is arguably even nicer.

[@danielmilyutin9914]

Same thoughts

[@arcuscotangens]

Agreed.

[@thechosentwins6994]

well i remmebber quit a lot of fromulae as iit-jee student int(sqrt(a^2+x^2))=xsqrt(a^2+x^2)/2+a^2*ln(x+sqrt(a^2+x^2))/2

[@danielmilyutin9914]

@@thechosentwins6994 Remembering is okay, helps to save some time. But one must be able to derive what he remembers or it may be shallow knowledge.

[@thechosentwins6994]

@@danielmilyutin9914 bro ezz use by parts as 1 and the functionand use substitution

[@minamagdy4126]

Fun fact, an alternate definition of an ellipse, and even a circle, based on that of a parabola, is as such. The construction is a focus at point (0,1), and a directrix circle passing through the origin with center "due north" of the focus. The definition, similarly to a parabola, is that the ellipse is the locus of all points equidistant from the circular directeix (measured at closest point) and the focus. This is exactly the same as the regular definition of an ellipse as that equality can be extended to be the sum-of-lengths constant between the focus and the directrix center (also called the alternate focus) equalling the radius of the directrix circle. It nicely extends to a parabola as said radius goes to infinity, a sort of 1/0 infinity where the other side produces hyperbolas for "negative" radii. The eccentricity is the distance between the two foci divided by the directeix radius, which indeed equals zero for a circle and approaches one as the shape approaches a parabola.

[@yunoewig3095]

That would be called Pa…

[@CasualMitosisCollective]

I'd rather call it Pi-rabola.

[@woody442]

For the most part my brain just thought "pow pew pa", so I was right all along!

[@keonscorner516]

Parabox

[@nikolakosanovic9931]

Why Pi is not ci

[@Tanvir_Ahmed_Earth]

"πm" is pronounced peyam So, π should be (p((i+a)/2))) pe thus a parabola should be perabola

[@ddognine]

This is a nice refresher of what analytic geometry is all about which I always found was an odd addition to the title of my calculus texts even though none of my professors ever went out of their way to point out what exactly in our calculus text qualifies as analytic geometry.

[@frankjohnson123]

I made it an embarrassingly long time in my math education without knowing what "analytic geometry" actually is. I conflated the term in my head with differential geometry, but when I looked it up one day it made a lot of sense.

[@MyOneFiftiethOfADollar]

Pi constant for all circles not so surprising, but would not expect the ratio of the two quantities you initially wrote down to be a constant for all parabolas! Pretty amazing. Don’t remember seeing this in analytical geometry.

[@PunmasterSTP]

This video was parabolamazing, and the high quality of Michael’s videos is always constant! 👍

[@JayTemple]

Did the Dean from Community help you with that word? lol

[@PunmasterSTP]

@@JayTemple He actually did not, but if you want to hear a pun about a specific topic, just let me know!

[@whilewecan]

Wonderful. I did not know there exists some constant for parabola like the case of a circle. Thank you.

[@HAL-oj4jb]

It's pretty obvious that all parabolas are similar when you think about eccentricity. Ellipses have 0

[@manucitomx]

Great video, professor. It rang some bells from Calc II. As always, thank you.

[@jonathanbeeson8614]

I always like to get a feel for the value of these constants. In this case P = 2.295587...

[@ericbischoff9444]

Not that different from pi. Lower than pi, because pi applies to a "rounder" shape, therefore longer.

[@egillandersson1780]

About 3π/4 😉

[@freddupont3597]

About 3, just like pi, if you are an engineer! LOL

[@QuantumHistorian]

It feels like the "natural" parallel of pi ought to be the ratio between the arc length and the latus rectum that bounds it. That just _looks_ more like a diameter and circumference. Bringing in the distance between the focal point and the directrix is just weird to me IMO.

[@SzanyiAtti]

I agree that looks more natural, but I think the focal length is more fundemental to the parabola. EDIT: After thinking about it, I realized that the length of the latus rectum is 4f, so using that would just simply cut the constant in half. It is analogous to tau and pi, where it is natural to choose either the diameter or the radius.

[@iabervon]

How about taking the shortest segment from the focus to the curve, taking an arc through that closest point bounded by the line perpendicular to the segment, and choosing the constant to be the length of the arc divided by the length of the segment? That's a construction that works for both curves and gives pi for a circle (length of semicircle over radius).

[@9WEAVER9]

a new era has dawned

[@stanleydodds9]

I think this is very natural; think of pi as the ratio between a semicircular arc and the radius. The arc is analogous to a semicircular arc, the latus rectum is analogous to the diameter, and the distance between the focal point and the directrix (which is exactly half the latus rectum) is analogous to the radius (which is exactly half the diameter).

[@angel-ig]

That's just like arguing about pi vs tau

[@titush.3195]

Nice, TIL that asinh(1) = ln(1 + √2)

[@restcure]

A touch of math and two new (to me) bands that work well together What a beautiful day!

[@stanleydodds9]

I would just do a hyperbolic substitution, x = sinh(u). Then you immediately get the integral of cosh^2 from -arcsinh(1) to arcsinh(1). This is a standard integral, but you can do it by parts or by the double angle formula if you want. Also you get the answer in the neater form, sqrt(2)+arcsinh(1). You can always turn arcsinh into log if you want. I think that's easier and more straight forward than splitting up the integral in a non-obvious way, doing integration by parts, and then using the standard integral of sec.

[@mellevin6029]

Pretty mysterious, all these transcendental constants coming out of algebraic things.

[@2712animefreak]

There should be a similar constant for the unit hyperbola (eccentricity sqrt(2)). The line perpendicular to the symmetry axis at one of the foci cuts off a "cap" from one of the branches.

[@Jack_Callcott_AU]

This is something they never taught me at school, or at university. I am grateful for finding out about it. I checked it out on Wikipedia. Like π it is transcendental. Thanks.

[@Jack_Callcott_AU]

@@micahmeneyerji 👍

[@goodplacetostop2973]

15:23

[@minecraftmovieman1]

I wish i paused this video before you did it and tried for myself. Great question for someone in calc 2

[@fd_472]

Nice result. Love it

[@cernejr]

approx. 2.296 . The curve is approx. 1.15 times the length of the straight line (i.e. 15% longer). From the picture I would have guessed that the ratio is larger, like 1.3 .

[@IntelR]

So there is a formula for the length of a parabola in terms of this constant? Like the circumference?

[@isaacdeutsch2538]

It seems to me this is less of a parabola's "pi" and more of a parabola's "tau" as the focal length is the length of the semilatus rectum. It feels more natural to me to use the ratio between the arc and the latus rectum, like pi is for a circle... I wonder why that isn't used as the parabolic constant.

[@kutuklukoy9150]

"All parabolas are similar. And there is only one true parabola." Parabolati Confirmed?

[@GeoffryGifari]

satisfying. Is the parabolic constant in any way related to pi? can they be mapped into each other?

[@CTJ2619]

another great video - thanks

[@abrahammekonnen]

2:18 Didn't know that was the official definition of a parabola nice to know.

[@rmandra]

Thanks!

[@BadlyOrganisedGenius]

At 0:21 this briefly becomes a Bill Wurtz video

[@sonarbangla8711]

I thought Penn to be able to explain the magic of the q-series. I am waiting eagerly.

[@tomkerruish2982]

Another antiderivative for the secant function is the inverse gudermannian.

[@cicik57]

can you define this as function for any 2 order curve depending on excentricity?

[@skylardeslypere9909]

If you could agree on which ratio to take, I guess you could. What would you propose we take as the 'constant' for ellipses and hyperbolas?

[@sharpnova2]

@@skylardeslypere9909 they can't have a constant since they have two varying attributes. parabolas and circles only have one

[@skylardeslypere9909]

@@sharpnova2 that's why I put it in quotations. I assumed it'd be clear from context as Kreo already mentioned it would depend on the eccentricity.

[@jardozouille1677]

Very interesting. Are there also constants for ellipses and hyperbolas ?

[@CM63_France]

Hi, 8:14 : you can already take into account the parity of the function at this point, 9:45 : ok. May be we can define such a constant for hyperbolas as well, provide we only consider those with perpendicular asymptotes.

[@bioengboi137]

9:55 I normally just put an x in front of the derivative if I don't know the immediate answer then fill in the rest of the antiderivative. In this case sqrt(1+x^2) ~ x int[x dx] = 1/2 x^2: 1/2 x sqrt(1 + x^2) + ... => [x sqrt(1 + x^2) + arcsinh(x)]/2 + C

[@Deejaynerate]

I think it's pretty interesting that the silver ratio pops up here, let alone as a logarithm.

[@qdrtytre]

About 2.2955871 for the curious.

[@SuperYoonHo]

Nice sir

[@garydetlefs6095]

It is of interest to note that the decimal expansion of this constant is listed in the online encyclopedia of an integer sequences www.oeis as sequence # A103710 and also that P/6 is the average distance between two random points in a unit square. An article in Wikipedia on the universal parabolic constant develops the value in about five lines. It is also of interest to note that if we change the constant to sqrt(2) - ln(sqrt(2)+1) we obtain what is referred to as the universal equilateral hyperbolic constant, a ratio of areas rather than Arc lengths which is discussed in sequence number A222362. So glad that I ran across you on KZbin Dr Penn. You are an amazing mathematician, a wonderful teacher and just about the best math presence online one could find. Thank you for all your dedication and hard work

[@rhaq426]

nice animations at the start

[@JayTemple]

Fun fact: There's no closed-form expression for the circumference of an ellipse. Seeing what went into finding the parabolic constant, I no longer find this surprising!

[@leostein128]

The circumference of an ellipse is given in closed form in terms of a well-studied "complete elliptic integral". There's no sharp line that divides which functions are allowed for "closed form" expressions and which ones are not!

[@JayTemple]

@@leostein128 At my first meeting with the graduate advisor where I earned my Master's, I said I wanted to look for the antiderivative of e^(x^2). He told me it had already been proven that "There is no closed-form integral" for that function. We both understood what the term meant. I will grant you, however, that from a computational standpoint, there's no real difference between, say, 2(pi)r and an infinite series whose limit happens to be 2r times some other sequence/series.

[@Cjendjsidj]

What does a closed form mean? π * a * b isn't closed?

[@aniruddhvasishta8334]

@@Cjendjsidj I believe this is for the area, not circumference

[@donaldbustell]

Is this constant used, or perhaps of use, anywhere? Similar to the way pi shows up all over physics.

[@artsmith1347]

From wiki: The average distance from a point randomly selected in the unit square to its center is d_avg = P / 6

[@donaldbustell]

@@artsmith1347 Cool! Thanks. Now that leads to another analysis: "Where did '6' come from?"

[@artsmith1347]

@@donaldbustell Can you not see it in the wiki article?

[@donaldbustell]

@@artsmith1347 I guess my question was more rhetorical -- along the line of "oh great! Another rabbit hole to distract me", and I hadn't bothered to look up the article. But I have now and so the entry to that rabbit hole will bug me until I go figure it out. Thanks again.

[@artsmith1347]

Also found in tweet by @TamasGorbe via a blog by CarlRobitaille: "Plot the function y=exp(-x) over the positive x-axis and rotate the graph about the x-axis. The surface you get has area π × the parabolic constant." This is unlike Gabriel's horn (also called Torricelli's trumpet), for which the surface area is infinite.

[@garekbushnell3454]

Is there a similar constant, or pair of constants, for a 3rd order polynomial?

[@Your_choise]

the sqrt(1+u^2) for the parabola feels related to the equation of a half circle sqrt(1-x^2)

[@abd-7883]

9:53. √1+u^2 can be directly integrated by parts . No need of using trigonometric function . You used it and end up with integration by parts again 😅😅

[@spiderwings1421]

how? if you integrate 1 and differentiate sqrt(1+u^2) (thats the only combo you can do here), you get u ^2/sqrt(1+u^2) but i dont see how this makes it any better? we could do trig sub here let u = tan(t) but that just brings us back to sec^3...

[@robertveith6383]

What you wrote is equivalent to 1 + u^2. You need grouping symbols: sqrt(1 + u^2). Or, use that radical symbol or the format of ( . . . )^(1/2).

[@edwardlulofs444]

Fun. Thanks.

[@n8cantor]

Latus rectum? Damn near latus killed 'em!

[@alre9766]

Awesome, it never occurred to me that parabolas had a constant. P=2,205587149392638...

[@bernhardbauer5301]

Looks wrong to me.

[@bernhardbauer5301]

2.29558714939263807403

[@frankjohnson123]

2.2955871493926...

[@chrstfer2452]

Makes me wonder when this was discovered. I know Pi is an ancient discovery, but its proof doesnt require integration. Is there another proof of this that doesnt require 17th century math?

[@artsmith1347]

Likewise. I infer this constant was unknown to the ancient Greeks, who wrote books about conic sections. Logarithms were unknown until the 1600s. A quick search did not reveal that an ancient Greek did an approximation of P comparable to Archimedes' estimate of pi.

[@pdorism]

The Greeks knew how to integrate definite integrals, they called it "the method of exhaustion".

[@stanleydodds9]

what do you mean by proof? The greeks knew that pi was a constant, and similarly it's very easy to show that this number is a constant. You don't need calculus for that. You just need to know that all parabolas are similar, in the same way that all circles being similar shows that pi is a constant. If you mean a proof that it's value can be given by some closed form expression, then I'm confused by what you think this "proof without calculus" is for pi. All the useful methods that I know for computing pi rely on calculus for their proof of correctness, perhaps just limits at the very least. Similarly, just to show that pi is irrational, I'd use calculus.

[@artsmith1347]

@@stanleydodds9 How is it "very easy to show that this number is a constant?" Without calculus, how would one be sure that the arc length varies with constant proportion for all parabolas? The length of an arc on a circle is not related simply to the length of an arc on a corresponding ellipse, where a = r but b= r * sin(theta). Even with calculus, there isn't a simple, closed form expression for the length of arc on an ellipse. How would the Greeks have arrived at a general expression for the blue arc length on the parabola in Dr. Penn's graph and related it to the L_2 distance?

[@stanleydodds9]

@@artsmith1347 I'll try to dumb it down for you: All parabolas are the same shape, in the same way that all circles are the same shape. All lengths inside these fixed, rigid shapes increase proportionally to each other when you scale them up or down. This includes arcs, straight lines, and anything else. Ever noticed that the perimeter of a square is always 4 times its side length? No matter how big the square is? This is because all squares are the same shape. However, rectangles don't have this property. Not all rectangles are the same shape. Ellipses and hyperbolas are not all the same shape. But circles and parabolas are.

[@yanmich]

I wonder if there is a way to prove that this ratio is constant the way Ancient Greeks did for circles, i.e., by using the method of exhaustion considering 2 arbitrary parabolas p and p'

[@giorgiobarchiesi5003]

9:15 it’s the area of half a circle

[@jensknudsen4222]

Does anyone happen to know if this constant is trancendental?

[@joeyhardin5903]

Yes, because ln(x) for any algebraic x (apart from 1 or 0) is transcendental

[@jensknudsen4222]

@@joeyhardin5903 Thanks!

[@RocketsNRovers]

so root2 + ln(1+root 2 ) is some good number and should be seen around more

[@whilewecan]

And,.. there are also some constants to ellipse and hyperbola???

[@matthewmorgan4765]

Does this have a listing on OEIS?

[@sven3490]

That constant us certainly irrational. But is it also transcendental? I apologize for my ignorance.

[@VincentiusErvinSantoso]

I think it's transcendental because of the "ln" (related to e, which is transcendental).

[@sven3490]

@@VincentiusErvinSantoso I agree with your assumption. But then again, ln (1)=0 and ln(e) = 1. I wonder if we can ask Mr. Penn to provide a proof in an upcoming video?

[@VincentiusErvinSantoso]

en.m.wikipedia.org/wiki/Transcendental_number ln a is proven to be transcendental if a is algebraic and not equal to 0 or 1. 1 + √2 is algebraic x = 1 + √2 x - 1 = √2 x² - 2x + 1 = 2 x² - 2x - 1 = 0 1 + √2 is not equal to 0 or 1, therefore ln (1 + √2) is transcendental.

[@RabbidSloth]

4:18 why can we assume the directrix and F are equal distances from the origin?

[@IoT_]

By the definition, the distance between the focus and a point on a parabola is the same as the distance between the same point and the directrix.

[@RabbidSloth]

Aha. Of course. Thanks for the help 👍

[@reynanhenry612]

Well I am expecting Pi in the final answer

[@ojas3464]

👍

[@jkid1134]

Have you seen Matt Parker's "One True Parabola"? To be sought out if you haven't

[@kristianwichmann9996]

Can this be generalized to an arbitrary eccentricity?

[@cheeseburger118]

The point of the video is that all parabolas are similar, and therefore don't have eccentricity :)

[@xifize]

@@cheeseburger118 Parabolas have an eccentricity of 1

[@carultch]

@@cheeseburger118 I think the OP's question is, can it be generalized for all conic sections of ANY eccentricity?

[@Mebasically]

I found it easier to do a hyperbolic trig sub (x=2fsinh(Φ))

[@ghijklabcdef]

4.59117

[@ralfbodemann1542]

Nice proposal. How are we gonna call that constant? - Pa? (Pi and Pa, Pa like "Parabola) - P = Penn's constant/number ?

[@schweinmachtbree1013]

I believe pi (greek letter p) stands for "perimeter", so maybe you could call it alpha standing for "arc"

[@Happy_Abe]

Why can we assume the directrix is at y=-f?

[@tracyh5751]

we can always translate a parabola to wherever we want because translating a curve does not change its lengths.

[@ingobojak5666]

Consider the situation at x =0, i.e, on the symmetry axis y. The distance from the focus F to the parabola is f (we have called it that). The parabola is defined as those points where the distance from the directrix is equal to the distance from the focus. Thus in this position, as the focus is f directly above the vertex of the parabola, the directrix must be another f further down under the vertex of the parabola. Then we find equal distances from the vertex to the focus directly above and to the directrix directly below. If the x-axis contains the vertex, this means the directrix must be the line parallel to the x-axis which passes through (0, -f).

[@Happy_Abe]

@@ingobojak5666 Great explanation thank you!

[@edmundwoolliams1240]

A physicist would say it’s equal to 2*sqrt(2)

[@jamessierpinski3617]

Amogus in the thumbnail!

[@danibarack552]

Kinda weird that you can express the parabolic constant using square roots and the logarithm, while there is no such simple expression for the circle constant, pi

[@taimao2]

No F's left

[@Boe1771]

Which is ~2.3 :D

[@Ensign_Cthulhu]

7:11 "And this part here is 2f." Proof?

[@cheeseburger118]

The parabola is defined as the set of points equidistant from the focus and the directrix. Since f is by definition the distance from the focus to the vertex, the distance from the vertex to the directrix is also f, so the total is 2f

[@schweinmachtbree1013]

set y=f in y=x^2/4f

[@matthewryan4844]

He had just determined the equation was y=x^2/4f and it has height f at that point, and (2f)^2/4f=f

[@robertveith6383]

@@schweinmachtbree1013 y = x^2/(4f)

[@robertveith6383]

@@matthewryan4844 y = x^2/(4f), etc.

[@joeeeee8738]

This has a ln in its value. Why do I feel pi can be written in terms of e somehow?

[@joeyhardin5903]

It can, in an analogous way: pi = -sqrt(-1) * ln(-1)

[@angeldude101]

Personally I'd probably be more likely to write it as ln(-1) / sqrt(-1) to make it clear that it's just cancelling a coefficient. You could also doing | ln(-1) | in case you can't decide which square root of -1 to use.

[@joeyhardin5903]

@@angeldude101 Actually I agree; it looks much neater but I wrote it the other way to make it a bit clearer in KZbin comment format. Although there's only one sqrt(-1) - negative i doesn't count

[@angeldude101]

@@joeyhardin5903 And why not? Complex numbers are not ordered, so you can't make the claim that i is greater than -i. This is what happens when you try to find an inverse for a function that isn't invertible. And of course that's without counting quaternions and other hypercomplex number systems that have an _infinite_ number of square roots of -1.

[@joeyhardin5903]

@@angeldude101 Oh yes I believe you're right. I had assumed that in order for sqrt(x) to be a function, we had arbitrarily chosen i to be the _only_ squareroot of -1, but I suppose that convention would break down when it comes to the rest of the complex plane, where as you said, there is no sense of greater or lesser numbers.

[@juliusking5126]

If we make u=(e^x-e^(-x))/2, it will be more interesting.

[@DOROnoDORO]

should be called ψ smh

[@stevencarr4002]

He said 'rectum'.... he,he,he

[@ericbischoff9444]

"Latus Rectum Of Parabola", it's the name

[@hkg9571]

And he drew the line from the latus rectum to the directrix (L2) in brown ;-)

[@hkg9571]

Prof. Penn has a mischievous mind. At 9m20s "No more fs (eff-s)" !

[@kendebusk2540]

It's been a while, but I believe rectum in Latin means straight.

[@ericbischoff9444]

@@kendebusk2540 According to the wiktionnary, adjective "rectus" is a participle of verb "rego", to keep straight. So, yes.

[@theartisticactuary]

Rectum 😂

[@bobstreet2491]

Annulus! (As my complex analysis lecturer used to say, rather too often.)

[@avaraportti1873]

>define circle using the radius >define circle constant using diameter ??????

[@schweinmachtbree1013]

𝜏 supremacy

[@michaelempeigne3519]

define the latus rectum

[@Frahamen]

hehe. Rectum.