Introduction to Hyperbolic Trig Functions 

Dark Mage, the son of Johnny Cage

Actually its shine

Quick Mafffs Several ways it can be pronounced. I say shine myself. But yah. *Shine, brother of mine* How was that one?

Tanh, the friend of Sam

@@abdurrahmanlabib916 SHINE OF X=so shiny i cant see anything

I didn't want to know, but now I know.

You assumed that e^t identity is true. What if you want to derive based solely on the analytical trig intuition and not the logarithmic intuition?

@@ChristAliveForevermore it works either way though, and it’s beautiful seeing it in action

You know how to write a understandable mathematical comment pretty much 🤩

That is your shortcoming not something to be proud of really

@@pranavsingla5902 what is wrong with you? How is he proud of it in any way shape or form

@@pranavsingla5902 such arrogance, damn

@@pranavsingla5902 he never said he is proud of it...keep your vulgar comment to yourself

@@pranavsingla5902 Wow ever heard of something called " being humble " ?

Lilanarus hahaha

only rapidity is based on hyperbolic trig. otherwise your lorentz transforms and fourvectors require only rudimentary algebra s a mathematical prerequisite

Isn't it?

I wonder how many times he said that

its a done deal.

@@Prxwler isnet?

@@Chaudharys1 don dio

If you say all of the cables on the suspension bridge have no mass but the bridge-deck does have, with a homogeneous density and is also -horizontal- straight, then you can easily derive that the curve of the main carrying cables is indeed approximated by a parabola.

You are exactly right. It is a very common mistake to assume the curved cables in a suspension bridge are catenaries (hyperbolic cosine curves). In fact, they are not and to a very good approximation they are indeed parabolas. This is true since the road is fairly nearly horizontal and the weight of the road being suspended is generally much greater than the weight of the cables.

This is quite interesting. Somehow I never covered this topic adequately. Is there a function that interpolates between the two(catenaries & parabolas) ? l suppose based on the weight ratios &/or the suspended platforms straightness(to horizontal). I'd guess it must assume an infinite # of vertical hangers?

realcygnus Google "suspension bridge catenary" and there are links to a few papers which do that. The Wikipedia page on "catenary" has a brief discussion under "Suspension bridge curve" with links to a couple of papers.

thanks

What's the relation between i and φ? Idk that one lol

i*i+sqrt(2)^2=phi-phi+1

@@mukkupretski ¦:| Bruh, Dat doesn't count, the i turns into -1 and the φ is canceled

Quahntasy - Animating Universe : )

Each good teacher needs to do that.

"Never seen cosh and sinh in my life until I was asked to integrate it last week for Calc 2."... What?!? Be like: "Never seen a girl until I was married"

@@zohar99100 those are very different, maybe he was never taught hyperbolic trig and then suddenly he saw a question maybe by a different teacher who assumed the class knows hyperbolic trig and take the derivative of it

Well we already have the most simple ellipse: the unit circle

Djdjcjcjcj Jfnfjfidnf Actually, hyperbolas are in a way stretched out circles, where a = 1 & b = i.

Djdjcjcjcj Jfnfjfidnf In fact, by allowing complex numbers, any equation for any of the conic sections can be written in the form of (x/a)^2 + (x/b)^2 = 1.

Angel Mendez-Rivera multiplying by i is NOT a ‘scale’. It is more of a rotation in an argand diagram.

Tom Graham Yes, technically, but if your scalar field of a vector space with a complex coordinates is the set of complex numbers, then that still counts as scaling.

Everybody gangsta till matmaticians invent sech, csch and coth

Can I please know what is

Glad you like it!!!

comme au Portugal ou en Auvergne :p

Cotanshent Arcshcoshine

I love this

"RIGHT???" "WRONG!!!"

I remember learning the same, Stephen. Nice to have someone else confirm what I recall. Kind regards from the Shoalhaven.

SiR PuFFaRiN was j

And because therefore the load on it is proportional to the x-length not the arc length

I thought the same thing as well but I’m not sure

Nope. 06:20 t=2.area So, area=t/2

Thank you!

Would the cables on power lines or telephone masts be a better example, since they hang freely?

Pretty cool but e^(iτ/2)+1=0 isn't really cool

yeah you know when you're so smart that you invent another class of trig functions :\

Great question! And yes, it is considered to be a relationship, but it can be parameterised as a function of time (which might be why he used t as the variable, not theta). When rewriting the unit circle with parameter t, it will be a function in R2. If that doesn't make sense, don't worry! The point is just that relations can be written as higher dimensional functions.

Yes they are! cosh(x) = cos(ix) and sinh(x) = isin(ix)

cosh(t) = cos(t). Euler's expression pretty much sums up that. BTW, bprp has made a video on it

@@astudent9206 cosh(t)=cos(it).

for cosh: suppose you wanted to calculate cos(i). start with the maclaurin series for cosine and plug in i. you will find that cos(i) is equal to the sum from 0 to infinity of 1/(2n)!, which I will call S for brevity. Recall the maclaurin series for e^x, which i will call exp(x). notice S looks similar to exp(1), but there are a bunch of extra 1/[odd factorial] terms in exp(1). we can get rid of these extra terms by adding exp(-1) to exp(1). this will cancel all of the 1/[odd factorial] terms, but we will be left with extra 1/[even factorial] terms. we can divide by 2 to get rid of these extra terms, and after all this, we see that S is equal to (exp(1)+exp(-1))/2, which means cos(i) is equal to (exp(1)+exp(-1))/2. this can be generalized by instead doing cos(ix) to find that it will be equal to (e^x + e^-x)/2 and define this to be cosh(x). we can then find cosh(ix) using this definition of cosh and euler's formula to see cosh(ix)=cos(x)