area and arc length

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[@MagnusSkiptonLLC]

0:24 I like that mathematical proof.

[@tofu8676]

logic 101

[@MrQwefty]

If you take the negative root of the equation you just get cosh(-x+c) which is the same! It's great because this is exactly the use of the properties of the hyperbolic functions :D

[@blackpenredpen]

MrQwefty yes : )

[@angelmendez-rivera351]

There is a general solution which accounts for this as well as for y = 1. More specifically, y(x) = cosh([cosh^-1][y(0)] + [sgn(dy/dx)]*x). sgn(dy/dx) = -1 when choosing the negative square root, but using the sgn function is a more rigorous method to choose the root. You can also choose the positive rot for sgn(dy/dx) = 1. Finally, this even covers the situation of y = c for when dy/dx = 0. This is more general than y = +/- 1. You can solve for the allowed sgn(dy/dx) or y(0) by substituting into the original y = sqrt[1 + (dy/dx)^2]. You find that cosh([cosh^-1][y(0)] + [sgn(dy/dx)]*x = sqrt(1 + [sgn(dy/dx)]^2*sinh([cosh]^-1[y(0)] + [sgn(dy/dx)]*x)^2). If sgn(dy/dx) = 0, then simply y(0) = 1 by force, and otherwise, you simply obtain the hyperbolic cosine-sine identity of squares. This also requires the definite integral method rather than using the indefinite integral.

[@paulangus6298]

I love the dramatic sound whenever hyperbolic functions are mentioned. Was never able to memorise what they're all about.

[@MichaelRothwell1]

They are the even and odd parts of e^x.

[@azzteke]

Memorise? where's the problem? Just a simple combination of e-functions.

[@JivanPal]

The constant function y = 1 is a trivial solution; You can see this from graphical inspection, or by realising that the equations shown in the video ( that is, y = √( 1+ (dy/dx)^2 ) and the equivalent equation dy/dx = √(y^2 - 1) ) have a valid solution when dy/dx = 0; namely, y = 1.

[@TheKranzi70]

Of course y = 1 works also.

[@OrigamiPie]

That was my first thought as well

[@reIONEre]

That's not a function of x. In fact, it's a function of zero variables.

[@raghur561]

Ione Manuel yes it is, y= 1+0x, it's integral can still be taken with respect to x

[@mirkorokyta9694]

y=1 works, and it is a function of x, if you define it to be. (I mean, everything is a question of a definition, so defining y=f(x), where f(x) =1 fir all x gets you there). In fact BPRP is still taking this solution into account until he divides by sqrt(y^2-1) not discussing what happens when this equals 0. Well, it happens that you lose this so to speak trivial solution y=1.

[@xwtek3505]

@@reIONEre Constant function is still a function.

[@AndDiracisHisProphet]

of course. you can always moce the function up or down to achieve that

[@blackpenredpen]

AndDiracisHisProphet that's the secret

[@AndDiracisHisProphet]

oh no, i spoiled it!

[@AndDiracisHisProphet]

*evilish my friday evening cocktail sip*

[@its_w4yne]

Yes, for a specific a and b. But with some function it works for any a and b. Like y=cosh(x+C) or y=1.

[@AndDiracisHisProphet]

no, it always works for any a and b. of course the shift changes if you change a and b

[@qmzp2]

I love these dramatic sound effects

[@williamadams137]

Next Question : Find a function g(x) such that the perimeter of the region bounded by the x-axis, x = a , x = b and g(x) equals the area bounded by the perimeter.

[@JimmyXOR]

Cosh and Sinh are just sums of exponential functions, cosh(x)=(e^x+e^-x)/2 and sinh(x)=(e^x-e^-x)/2. He could easily write the answer of the last integral as (e^5-e^-5-e^2+e^-2)/2 instead of 70.576. Also, he divided away the trivial solution where y is a constant (y=1).

[@quahntasy]

Dirac Delta function has Area= 1. But Arc length nearly Equal to 0. This is an interesting video. Loved it.

[@aldoushuxley5953]

Why is the Arc length nearly Zero? When you approximate dirac Delta by other functions (dirac Delta is just an distribution) yes, it is true, that the function is Zero almost everywhwre, but the Peak around Zero is still "infinitely" high... I am Not certain if you are correct about the Arc length, but IT doesnt seam that trivial too me... Greatings, Jonathan

[@yannicstoll157]

„function“

[@MA-qp4zx]

Arc length of practical dirac delta isn't zero...it's 1/a where a is the duration of the signal..... What you are saying is actually an ideality over concept where the width a is taken as limit a->(tending to)0......and hence the magnitude jumps towards infinity which is shown by an arrow.... Practically,we have a width for delta function....and we use it extensively in communication where we sample the analog signals to digital form.... Ideal sampling uses concept of ideal delta function but the practical situation is different.

[@angelmendez-rivera351]

How can you tell the arc length is nearly zero? You must calculate the derivative of the function to know the arclength, but I’m fairly certain the resulting equation has no closed form solution.

[@chandankar5032]

Thank you so much sir... Please upload these type of concept clearing topics...❤❤❤from india

[@mokoufujiwara1281]

y=1 seems works, take a range from x=a to x=b, area under y=1 = line segment length = b-a But, be careful, line segment is different from arc. But, should line segment be a subset of arc? This is a good question to discuss.

[@duggydo]

Very nice video. Only problem is lighting glare on whiteboard in some spots.

[@mokoufujiwara1281]

Doing it negative: y = cosh(-x+C) and cosh(-x) = cosh(x) We can simply write y=cosh(x+C)

[@pipertripp]

That was a cool exercise. Thx for sharing.

[@John-qm8mw]

You can get Y=1/sqrt{1-e^{2x+c}} if you use trigonometrical substitution instead of the hypervololic cosine . Please do not ignore me :)

[@blackpenredpen]

Got u! I will remember you and will always respond to you!!!

[@John-qm8mw]

@@blackpenredpen ohh boy, thanks :,3

[@djgulston]

Judging from the music, this is a great video to watching on Halloween! 😂

[@BigDBrian]

Of course you can. Take a reasonable arc length(if it's too steep it might not work.), e.g. the shape in the video. Now slide it up and down. It won't change the arc length, but it will change the area. So just slide it down or up until the values are equal.

[@asadyamin4867]

I'll just put down y, and you'll see y (why) lol 1:20

[@emorgan0085]

4:29 and let me just put down + c, and you'll c this c

[@purushotamgarg8453]

... and know you c

[@flowerwithamachinegun2692]

I see that you've discovered sound effects :))

[@oscartroncoso2585]

Keep up the great work!

[@blackpenredpen]

Oscar Troncoso thanks

[@user-vj7uc9tj7c]

@@blackpenredpen You missed a solution (y=1). The reason for that is when you had dy/dx = sqrt(y^2-1), you divided by zero (sqrt(y^2-1) is 0 for y=1), and got dy/sqrt(y^2-1) = dx, at 3:33.

[@khbye2411]

At what point in calculus will you need/find it convenient to know and remember hyperbolic trig functions?

[@sarahconner9433]

Calculus 2 and 3

[@williamadams137]

Wow this is so cool!👍👍😺

[@rafaellisboa8493]

can you plz make a video proving the arc length equation?

[@quasar9411]

Damn, that was really cool. Best thing i've seen today.

[@juedingyoutiao]

Love 2:30 when you move the camera up

[@MA-qp4zx]

That background music....just wow....You are a mathegangster sir....awesome 😉👨‍🏫

[@silasrodrigues1446]

Really really cool!!!

[@jayjay4752]

Can you do a video on partial differential equations?

[@ZelForShort]

Specifically practice problems

[@zeldasama]

I love how you have up on saying cosine hyperbolic fungion and such after one go and said co-sh, and si-sh. Hahaha great video. Loveg the halloween theme, scary stuff jumping around screen.

[@VIVEK2991]

Where's the video about deriving arc length formula?

[@purushotamgarg8453]

I think I can make it work on any function (I am not sure, please point out if I am wrong). Just take the interval to be [a,a+dx] and both, the arc length and area, would be literally 0. Here dx has its usual meaning.

[@granhermon2]

Awesome!

[@jojojorisjhjosef]

nice, how about an area under the curve that = the arc length squared? so like a straight at y = 2 line from 0 to 2

[@islandcave8738]

To clear up a bit, the d from the c and d at the end should not be confused with the d in dx and dy. The d from the dx and dy is actually the lower case delta.

[@Kumar-oe9jm]

Damn the that dramatic music

[@naturelover9767]

Nice more on this concept

[@papajack2205]

cool audio effects

[@leoyang1.618]

Blackpenredpen please watch my most recent video, a math one.

[@kkn5523]

Physically, won't it be incorrect to equate two quantities with different dimensions? Can't wrap my fat head around it 😖

[@zerglingsking]

They're just measures of length and area, not length and area ! ;) So it's ok to compare them

[@vacant4]

numerically equal

[@angelmendez-rivera351]

They do not have different dimensions. Dimensionless quantities can be made from every physical quantity.

[@teraflonik]

This is math not physics

[@HeyKevinYT]

I can weigh 100 lbs. but can also be 100cm tall

[@jardelkaique2522]

I found the solution using another way.. my answer was: (y + sqrt(y²-1)) = e^x is it the same thing or another fuction that satisfies the condition?

[@BluePi3142]

I got scared at the end.

[@rochib21]

Is 2:10 a sufficient condition? Can there be other solutions where the functions are not equal?

[@dinhthuan-thien8048]

I think it’s wrong because integral from a to b of (y - sqrt(1 + (dy/dx)^2) can work with any function of y?

[@singh2702]

From a to b there is a corresponding height. However these heights have created a length which is longer than a-b , this is why volume is always smaller than surface area. Each height's length is invisible to volume unlike area 👍

[@VaradMahashabde]

Wait, does the arc length formula come from the Euler's method video? WHY DID SKIP THAT?

[@shiv8757]

Please work on zeta function....ramanujan's work, reimann hypothesis

[@dolevgo8535]

was expecting Ood at the end, haha

[@friedkeenan]

Wouldn't y=1 work?

[@keithallatt9488]

What about letting y = 1? therefore dy/dx = 0 so area equals integral from a to b of 1, which is (b - a), arclength is integral from a to b of sqrt(1 + dy/dx) which evaluates to integral from a to b of 1, again being (b-a). So another solution is letting y = 1

[@blackpenredpen]

Keith Allatt that works too!

[@blackpenredpen]

In fact that the missing solution from my general solution. You can find it by letting sqrt(y^2-1) be 0. Since that is in the denominator.

[@madmuffin2511]

Can you say something to Integrals that can not be solved analytically? ...though i guess it might not be that much fun :)... I stumbled across: Integral( exp(x) x^4 /(exp(x)-1)^2 dx) from 0 to some x_m and it says it cant be solved, but seeing all ur nice integration tricks and stuff, I was wondering how u can say, that here there will not be any such tricks, but instead its just not analytically solvable. (This Integral comes up when calculating the specific heat of a solid in the Debye-Model)

[@medchs]

does two integrals being equal mean that he integrands are equal?

[@lucashoffses9019]

Would y=1 also work?

[@chillingpaully4137]

Yes

[@connerp6878]

Can you please do a proof of De Moivre’s Theorem. I have a math competition in a few months at a university a few hours away and I’d love to do it on this. Hope you see this thanks!!

[@mokoufujiwara1281]

One way around, use Euler's formula! Let y = cosx+isinx, dy/dx = -sinx+icosx = i*y, you have a differential equation that is easy to solve: dy/y = idx, integrate and you get ln y = ix, y=e^ix Using this formula, the De Moivre's Theorem becomes trivial.

[@connerp6878]

Mokou Fujiwara I’ve heard of that proof but I wanna see one done proving it for some kth value if that makes sense and I’d love to see him try it out.

[@mokoufujiwara1281]

Okay...try it the hard way.

[@jazzjohn2]

Can't two definite integrals on the same interval be equal with the Integrands unequal?

[@nutinmyass]

Nice Supreme shirt

[@blackpenredpen]

Brother Banks thanks

[@jazzjohn2]

Y =sin(x) + c from 0 to 2 pi is another. Choose c so c*2 pi equals the arc length.

[@vpambs1pt]

awesome

[@edwardhudson815]

where's the plus C after integrating the 1/sqrt(x^2-1) though?

[@sainathreddyvarikuti4835]

Is there any other functions ,where arc length is equal to area under curve ?? Or only coshx

[@hamsterdam1942]

First order differential equation have only one solution

[@simohayha6031]

@@hamsterdam1942 sinh?

[@hamsterdam1942]

@@simohayha6031 d/dx arcsinh (x) = 1/sqrt (x^2+1) Nope

[@apv]

Is this solution related to the fact that e^x is its own derivative?

[@russellchido]

My intuition told be answer would be a parabola, so this is funny.

[@eltapa5282]

Nice! Only works for cosh?

[@duckymomo7935]

No, there is also y = 1 (singular solutions)

[@theelk801]

I’m wondering if this is related to how cosh models a hanging chain

[@sjoerdo6988]

Why do the integrands have to be the same? two integrals with the same bounds being equal doesn't imply the integrands being equal right? for example: integral{0,1}x dx=integral{0,1}1/2 dx=1/2, even though x=/=1/2

[@PsyKosh]

What about non smooth solutions, where y' alternates between being equal to sqrt(y^2 - 1) and -sqrt(y^2 - 1)? we can have a wibbly wobbly function that still fulfills the criteria. Well, okay, FINE!, I guess differential arc length won't be defined properly at the interface points. But still... :)

[@Archik4]

Negative version y= cosh(-x+c). This is the same function

[@LYNXzTwist]

considering the level curve f(x,y)=c where y=cosh(x+c) you find at c=-x you have that y=1 considering y=cosh(x+c)=(e^(x-x)+e^-(x-x))/2=2/2=1

[@Cloud88Skywalker]

dramaticfxpen yay!

Cool. Btw: Nah, new sound effect package has been dropped in to make the show more dramatic :D

[@NCGamer20]

Hello can someone tell me how he got arc length value formula of root one plus dy by dx whole square.

[@ulfhaller6818]

kzbin.info/www/bejne/boqQpKGgncp8b68

[@salimkaddouri5266]

what is the unit of arc length, and what is the unit of area. It's strange to put square meter equal meter ( m^2=?? m). Is it possible?

[@shivamprasadyadav9263]

Please can you tell me why am i not getting the correct answer in the following ques. •Find the derivative of Tan^-1[√(x+1)/(x-1)] for |x|>1 The solution is -1/2|x|√x^2-1 Method to use- I know one method where you put x=Sec2theta.. and solve it But when I solved it directly using derivative of Tan inverse then chain rule ,I didn't get the |x|, can you tell me why... please.. If the question is non. Understandable,please tell me I will paste a drive Link containing a photo of this question.

[@sugarfrosted2005]

Integral equations!

[@jacoboribilik3253]

I am in doubt about this part 2:03 . If two definite integrals are the same, does that necessarily imply both integrands are the same too? I mean, it could be a solution but why can't there be other solutions to that equation? Someone mentioned below another solution could be 1. How do we know there are not more function that satisfy the equality? My knowledge in differential equations is little.

[@6c15adamsconradwilliam3]

He just wants to show us a possible solution.

[@vpambs1pt]

One question, what if the area is negative, will the arc lenght be negative as well, so the value keeps being the same?

[@matthewhammans4365]

Nuno Mateus hyperbolic cosine is only defined for y > 1 for all real values of x, so the area couldn’t be negative

[@vpambs1pt]

Nice, thanks

[@duckymomo7935]

even if the area is 'negative' the idea is to only integrate with ∫|f| so that the function is measurable

[@mcmage5250]

The dum dum dumm part got me scared 😦😦😦

[@37metalgearsolid]

y(x)=1 works just as well! :)

[@alejrandom6592]

first thing that came to mind is f(x)=1 (trivial but it works) my question is, why doesn't it appear on the solution to the diff eq.?

[@johnfife4370]

What’s with all the creppy music at the beggening . BTW love your channel

[@emir2591]

Y=1?

[@sagoot]

Y= -1

[@mjz5853]

How about restricting the area to be the absolute value?

[@victor666]

What’s up with the sound effects

[@remlatzargonix1329]

It could have the same numeric value, but it can't have the same dimensions...length is in length units, area is in (length units) squared.

[@swimcoach3702]

Funny that practically no one noticed that feet and square feet are not the same measurement, although the idea was to concentrate on only the numbers. However, he should have disclosed this at the end since failure to match units has failed many a groundbreaking discovery!

[@lunaticluna9071]

oh my cosh!

[@crosisbh1451]

I had a feeling that an exponential function was going to show up. I was pleasantly surprised it was a hyperbolic function.

[@faith3174]

cosh is essentially e^x with a fancy name

[@GdFireLord]

cosh is equal to (e^x+e^(-x))/2 so it's actually just an exponential function in disguise.

[@pauloat]

spooky

[@maehmaehmaeh9560]

Obviously f(x) = 1 also is a solution to the problem, but the solution doesnt seem to allow other answers than cosh(x). What am i missing here?

[@andrewweirny]

At 4:00 - supposedly Gauss used that same integration technique.

[@jordanshemilt6065]

Forgot constant solution of y=+-1 since this makes dydx =0

[@SVP-uy9qb]

Before watching the video: Obviously yes

[@SVP-uy9qb]

Lol it's 0:28 and he already answered

[@danielhaselbauer2549]

How do u know the integral goes from 2 to 5??

[@danielhaselbauer2549]

@@HL-iw1du i thought that a and b have to be specific values so that the equation is satisfyed thats why i didnt get it. Thx alot man

[@6c15adamsconradwilliam3]

The values of a and b can be any real number.

[@johnny_eth]

You didn't deduct the most obvious case: y=1+0x. Area[2,5]=3 ArcL[2,5]=3

[@alexsanderbenatti3596]

1:14 "Let me put a "y" right here *and you will see **_y_* "

[@historybuff0393]

Early Halloween shenanigans?

[@xevira]

dun dun DUUUUUNNNNNNNNNN (yay!)

[@williamtachyon2630]

Bluepenredpengreenpen.

[@shlokgupta9353]

I understood everything as it was really well represented, but why use that music 😂😂

[@kaukokaipuuz498]

wait why is 1+(sinx)^2=cosx...

[@ankurrai8677]

🙌🙌🙌👍👍👍

[@hellgate25000]

is this the only non-trivial function that satisfies this property? If not, then how may one find other functions?

[@hellgate25000]

And if it is, then what would be the best way to prove it?

[@LYNXzTwist]

yeah this is the only way, the trivial solution of course would be y=0.