Intuitive Guide to Hyperbolic Functions – BetterExplained

If the exponential function $e^x$ is water, the hyperbolic functions ($\cosh$ and $\sinh$) are hydrogen and oxygen. They're the technical,rarely-discussedparts that combine into a famous whole.

Admittedly, the hyperbolic functions were tucked into a dark part of my attic. They were defined with strained motivations ("Need yet another way to build a hyperbola?") then crammed intotables of integrals, soon to be forgotten. I couldn't think with them.

After much struggle, I found their purpose:

What are the hyperbolic functions ($\cosh$ and $\sinh$)?The even/odd parts of the exponential function ($e^x$) that, funny enough, can build a hyperbola.
Why are parts of the exponential called hyperbolic?That's the modern name. These functions areso darn good在制作双曲线的时候，他们被定型为那个角色。(Similarly, sine isn'tjust about circles, and we shouldn't name it "circular sine"!)
Why are hyperbolic functions useful?一个更好的框架是:为什么$e^x$的部分是有用的?我们现在有了“迷你对数”和“迷你指数”，它们是$e$著名性质的部分版本。
I can handle it: how do hyperbolas connect to exponentials?Hyperbolas come from inversions ($xy = 1$ or $y = \frac{1}{x}$). The area under an inversion grows logarithmically, and the corresponding coordinates grow exponentially. If we rotate the hyperbola, we rotate the formula to $(x-y)(x+y) = x^2 - y^2 = 1$. The area/coordinates now follow modified logarithms/exponentials: the hyperbolic functions.
Actually, I couldn't handle it.That's ok. We'll build up to it. These functions took many years to be discovered, and their behavior is hardly obvious.

这篇文章相当技术化:我们研究的是氢，而不是水。如果在课堂上出现了双曲函数，你没有太多的选择，最好有一个直觉。如果你是为了乐趣而学习，不要纠结于细节，这才是学微积分的学生应该做的。

As a prerequisite, have these insights in mind:

$e^x$is the process of continuous, 100% growth
2022世界杯预选赛 is the time for $e^x$ to grow to a given value

Let's dive in.

Contents

Part 1: Exponential Viewpoint
第二部分:双曲函数是什么样的?
Part 3: Applications of the Exponential Interpretation
Part 4: Geometric Viewpoint
Part 5: Applications of the Geometric Interpretation
Summary
References
Other Posts In This Series

Part 1: Exponential Viewpoint

Background: Odd/Even Functions

Chemical compounds can be separated into constituent atoms; math objects are similar.

The number 13 can be split into an even part (12) and odd part (1). They combine to the whole: 13 = 12 + 1.

This even/odd split works for functions, too:

$\displaystyle{f(x) = f_\text{even}(x) + f_\text{odd}(x)}$

Functions are tricky to separate because they have multiple values. The separation we look for is between the future values ($x > 0$) and the past ones ($x < 0$).

To see what the future and past have in common, take their average:

$\displaystyle{f_\text{even}(x) = \text{avg}[\text{future} + \text{past}] = \frac{f(x) + f(-x)}{2}}$

To see how the future and past differ, average their gap:

$\displaystyle{f_\text{odd}(x) = \text{avg}[\text{future} - \text{past}] = \frac{f(x) - f(-x)}{2}}$

Can we combine these parts to get the original?

$f_\text{even}(x) + f_\text{odd}(x) = \frac{f(x) + f(-x)}{2} + \frac{f(x) - f(-x)}{2} = \frac{f(x) + f(-x) + f(x) - f(-x)}{2} = \frac{2f(x)}{2} = f(x)$

Neat trick. We can split any pattern into its even and odd parts.

Taylor series

TheTaylor Series (Math DNA)expresses a function as a polynomial:

$\displaystyle{f(x) = c_0 + c_1 x + c_2 x^2 + c_3x^3 + \cdots}$

The even exponents ($x^0, x^2, ...$) are symmetric in the past and future (for example: $x^2 = (-x)^2$), and the odd exponents are anti-symmetric ($x^3 = - (-x)^3$).

We can quickly extract the even/odd parts by separating the function's Taylor series into even/odd exponents:

$\begin{align*} f(x) &= f_\text{even}(x) + f_\text{odd}(x) \\ f_\text{even}(x) &= c_0 + c_2 x^2 + c_4x^4 + \cdots \\ f_\text{odd}(x) &= c_1 x + c_3x^3 + c_5x^5 \cdots \end{align*}$

Even and Odd Exponentials

Ok, we have our trick. Why not try to split up the famous exponential function?

$\begin{align*} e^x_{\text{even}} &= \text{avg}[e^x, e^{-x}] = \frac{e^x + e^{-x}}{2} = \cosh(x) \\ e^x_\text{odd} &= \text{avg}[e^x, -e^{-x}] = \frac{e^x - e^{-x}}{2} = \sinh(x) \\ e^x &= e^x_{\text{even}} + e^x_{\text{odd}} = \cosh(x) + \sinh(x) \end{align*}$

Instead of the awkward $e^x_{\text{even}}$ and $e^x_{\text{odd}}$, we call the even part $\cosh$ (hyperbolic cosine) and the odd part $\sinh$ (hyperbolic sine). (The pronunciationvaries.)

Now, why the adjective "hyperbolic"? Euler used the quantity $(e^{x} + e^{-x})$ without giving it a special name. Lambert later called them "transcendental logarithmic functions", and even later the ability to build hyperbolas was seen. That use case has stuck.

Call me old-fashioned, but parts of $e^x$ are interesting for that reason alone. I don't need a hyperbola to justify their utility.

第二部分:双曲函数是什么样的?

Let's graph $\cosh$ and $\sinh$ with their parent:

$e^x$ is our standard exponential, with Taylor series: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$
$\cosh(x)$ is a gentle bowl. It's roughly parabolic, then expands exponentially, with Taylor series: $\cosh(x) = 1 + \frac{x^2}{2!} + \frac{x^4}{4!} + ...$
$\sinh(x)$ looks linear at first, then grows exponentially also: $\sinh(x) = x + \frac{x^3}{3!}...$

Now for a usually-confusing question:What does the parameter $x$ in $\cosh(x)$ really mean?

没什么特别的!它是相同的$x$，我们给任何指数，通常增长的时间:$e^x = e^{rt} = e^{100\% t}$。

We can see the hyperbolic trig functions as:

$\cosh(x)$: What is theeven part$e^x在$x后面的$ ?
$\sinh(x)$: What is theodd part$e^x在$x后面的$ ?

A few gut checks:

What's $\cosh(0)$? At $x=0$ (aka $\text{time} = 0$), we haven't moved from 1.0 (the exponential starting point). The average is 1, and there's no separation, so $\sinh(0) = 0$.
As time goes on, the future grows and the past ($e^{-x}$) vanishes. The average value becomes $\frac{e^x + 0}{2}$, and the average difference becomes $\frac{e^x - 0}{2}$. For large $x$, we'd expect $\cosh(x) \sim \sinh(x) \sim 0.5 e^x$.

Now, how about theinverse双曲函数如$\text{acosh}$(也称为$\text{arccosh}$或$\text{arcosh}$)?"Inverse hyperbolic cosine" sounds scary, but think of it like this:

$\ln(a)$: How long until $e^x$ reaches value $a$?
$\text{acosh(a)}$: How long until theeven$e^x$的一部分达到$a$?This must take longer than $\ln(x)$, since we're only using theevenpowers in our exponential growth Taylor series.
$\text{asinh(a)}$: How long until theodd$e^x$的一部分达到$a$?同样，这必须需要更多的时间$\ln(x)$。

In the graphs above, $\text{acosh}$ and $\text{asinh}$ require more time (i.e., are above) the natural log.

Intuitively, I see the various hyperbolic functions as modified exponentials and logarithms:

$\cosh$ and $\sinh$ are partial/delayed exponential curves (with new behavior near zero, where the past still has influence).
$\text{acosh}$ and $\text{asinh}$ are slower versions of the natural log. It takes $\ln(3) = 1.09$ units for $e^x$ to grow from 1 to 3, but it takes $\text{acosh}(3) = 1.76$ units for just the even part of $e^x$ to do the same.

Part 3: Applications of the Exponential Interpretation

Since $\cosh$ and $\sinh$ are mini exponentials (those little cherubs!), we'd guess they still have interesting properties.

Function equal to its own second derivative

$e^x$ is the function equal to its own derivative ($f' = f$).

$\displaystyle{e^x \xrightarrow{\text{d/dx}} e^x \xrightarrow{\text{d/dx}} e^x \dots }$

Ok. And Sine and cosine are functions equal to their ownfourthderivatives ($f'''' = f$).

$\displaystyle{\sin(x) \xrightarrow{\text{d/dx}} \cos(x) \xrightarrow{\text{d/dx}} -\sin(x) \xrightarrow{\text{d/dx}} -\cos(x) \xrightarrow{\text{d/dx}} \sin(x) }$

重复一次导数，重复四次导数……这是一个很大的差距。Anything in-between?

You bet. The hyperbolic functions equal their ownsecondderivative ($f'' = f$):

$\displaystyle{\frac{d}{dx} \sinh(x) = \frac{d}{dx} \frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x}}{2} = \cosh(x)}$

$\displaystyle{\frac{d}{dx} \cosh(x) = \frac{d}{dx} \frac{e^x + e^{-x}}{2} = \frac{e^x - e^{-x}}{2} = \sinh(x)}$

Unlike $\sin$ and $\cos$, there's no awkward negative signs in the cycle, just a toggle:

$\displaystyle{\sinh(x) \xrightarrow{\text{d/dx}} \cosh(x) \xrightarrow{\text{d/dx}} \sinh(x) \xrightarrow{\text{d/dx}} \cosh(x) \xrightarrow{\text{d/dx}} \dots }$

Neat. The hyperbolic functions are like "half exponentials" because it takes two derivatives to complete the cycle. This is why they're useful in calculus -- not because we care about the coordinates on a hyperbola!

Understanding Integral Tables

You'll see hyperbolic functions in tables of tricky integrals and derivatives:

Ignore the specifics. Let's see the general pattern without getting lost in the details.

We have our standard relationship:

$\displaystyle{\frac{d}{dx} \ln(x) = \frac{1}{x}}$

由于双曲函数是指数函数的变化，我们期望$\frac{d}{dx}\text{asinh}$类似于$\frac{1}{x}$。From the table:

$\displaystyle{\frac{d}{dx} \text{asinh}(x) = \frac{1}{\sqrt{x^2 + 1}}}$

When $x$ is large, the "$+ 1$" doesn't matter much, and the derivative becomes:

$\displaystyle{\frac{d}{dx} \text{asinh}(x) = \frac{1}{\sqrt{x^2 + 1}} \sim \frac{1}{\sqrt{x^2 }} \sim \frac{1}{x}}$

Ah! The function $\text{asinh}(x)$ behaves like $\ln(x)$, except for small $x$ where the $+1$ term still matters. Without this exponential perspective, I'd haveno cluewhat the derivative of $\text{asinh}$ should resemble. (You want the rate of change of the inverse function that determines the y-coordinate a in a hyperbola? What?)

New descriptions of Sine and Cosine

Remember, we can now split the exponential function whenever we want:

$\displaystyle{e^x = \cosh(x) + \sinh(x)}$

那么，如果我们代入$ix$呢?We'd get

$\displaystyle{e^{ix} = \cosh(ix) + \sinh(ix)}$

ButEuler's formulasays:

$\displaystyle{e^{ix} = \cos(x) + i\sin(x)}$

Whoa. The even/odds parts of each function must be the same, so:

$\displaystyle{\cosh(ix) = \cos(x)}$

$\displaystyle{\sinh(ix) = i \sin(x)}$

If we feed the imaginary axis to our everyday exponential function, we get the trig functions which live on a circle. The rotation is happening via the parameter $ix$, vs. in the function $e^{ix}$. (Instead of rotating the steering wheel, we're rotating the engine, so to speak.)

In sine's case, we have an awkward imaginary term to shuffle around:

$\displaystyle{\sin(x) = \frac{\sinh(ix)}{i} = -i \sinh(ix)}$

These connections are more useful in complex analysis (still learning...), normally you'd prefer to pass real parameters into functions.

New Trig Identities

A few new identities doesn't hurt. With some quick algebra, we can turn regular trig identities into their hyperbolic versions. Starting with:

$\displaystyle{\cos^2(x) + \sin^2(x) = 1}$

We swap in $\cos(x) = \cosh(ix)$ and $\sin(x) = -i \sinh(ix)$ to get:

$\begin{align*} \cosh^2(ix) + [-i\sinh(ix)]^2 &= 1 \\ \cosh^2(ix) + i^2\sinh^2(ix) &= 1 \\ \cosh^2(ix) -\sinh^2(ix) &= 1 \end{align*}$

Now, the term $ix$ is just a parameter. To simplify things, just say $z = ix$ and write:

$\displaystyle{\cosh(z)^2 -\sinh(z)^2 = 1}$

We can leave out the specific parameter and write:

$\displaystyle{\cosh^2 -\sinh^2 = 1}$

这种模式通常有效。To convert a regular trig formula to its hyperbolic equivalent (Osborne's Rule), swap:

$\cos^2 \Rightarrow \cosh^2$
$\sin^2 \Rightarrow -\sinh^2$ (due to the $i$ when converting $\sin$ into $\sinh$)

Machine Learning Functions

While looking for applications of $\cosh$, I ran across this function, used inMachine Learning:

$\displaystyle{\text{logcosh(x)} = \ln(\cosh(x))}$

一个直接的解释是令人困惑的:“取双曲线中x坐标的自然对数”。嗯?这代表什么?

The exponential perspective makes it simpler:

We don't want to take the natural log of the regular exponential, since $\ln(e^x) = x$. A function like $x$ (or rather, the absolute value $|x|$) is too pointy, and doesn't have a clean derivative at zero.
However, $\cosh(x)$ onlyresemblesthe regular exponential function. Its natural log will onlyresemble$f(x) = x$. The resulting function $\ln(\cosh(x))$ looks parabolic near zero, and linear as we grow:

这是怎么回事?For small $x$:

$\cosh(x) \approx 1 + \frac{x^2}{2}$. These are the first few terms of its Taylor series.
$\ln(1 + x) \approx x$. The time to grow from 1.0 to 1 + .01 at 100% interest is only .01 units (not enough time for compounding)
结合这些近似，我们得到:$\ln(\cosh(x)) \ prox \ln(1 + \frac{x^2}{2}) \ prox \frac{x^2}{2}$。

As $x$ increases, we approach an offset line:

$\displaystyle{\text{logcosh(x)} = \ln(\cosh(x)) \approx \ln(\frac{e^x + 0}{2}) = \ln(e^x) - \ln(2) = x - \ln(2)}$

Cool! We have a parabola-line hybrid. No need for hyperbolas, we're dealing with variations of the exponential function.

Bonus: Lining Things Up

Maybe we can get $\text{logcosh}$ to turn into the line $y = x$, instead of the offset line $y = x - \ln(2)$. We can add a term to undo the $-\ln(2)$:

$\displaystyle{y\ =\ln\left(\cosh\left(x\right)\right)+\ln\left(2\right)\left(1-e^{-0.1x^{2}}\right)}$

Notes:

Our new function (purple) still looks like a parabola for small $x$.
The $x^2$ in the in $e^{-0.1x^2}$ makes the function symmetric.
Adjusting the constant $0.1$ changes how fast we approach the line $y=x$.

不久前，我还不知道如何让抛物线慢慢过渡成直线。但是把$\cosh$看作“抛物线短期，指数长期”给了我们一个线索:用自然对数来撤消“指数长期”的行为，给了我们“抛物线短期，线性长期”。

Hyperbolic Tangent

The hyperbolic tangent is used as a machine learning activation function:

$\displaystyle{\tanh(x) = \frac{\sinh(x)}{\cosh(x)} = \frac{\text{odd part of } e^x}{\text{even part of } e^x}}$

What's its meaning? It's the ratio between the odd and even parts of the exponential function after a given amount of time.

At $x=0$, the even part dominates (full symmetry between past and future)
As $x$ grows, the odd part catches up, and the ratio approaches 1 (equal parts symmetry and anti-symmetry)

The function $\tanh$ is nicely centered at 0 and smoothly varies between -1 and 1. As $x$ increases, $\cosh(x) \sim \sinh(x) \sim 0.5e^x$ and $\frac{0.5e^x}{0.5e^x} = 1$.

Part 4: Geometric Viewpoint

Ok. We've gone pretty far without talking about hyperbolas. Why?

Well, just look at it. They aren't an everyday shape. What's their purpose? Catenaries, orbital mechanics? That's anapplicationbut not the reason they exist.

Time to build an intuition.

The Inverse Function Hyperbola (xy=1)

Hyperbolas are built from inverse functions, like $y = \frac{1}{x}$. It's a simple, useful relationship: what's the inverse of $x=2$? $y=1/2$. Multiply them and we undo all scaling: $xy=1$.

A burning math question might be: sure, we have this inverse relationship, but what's the area underneath?

In calculus terms, this is:

$\displaystyle{\text{area} = \int_{1}^{x} \frac{1}{x} dx}$

Where

$\int_1^x$ specifies the boundaries. We start at $x=1$ and go to some upper value for $x$. (We can't start trapping area at $x=0$, since it shoots up to infinity.)
$\frac{1}{x} dx$ is the area we collect at each x-coordinate as we march along

This integral is hard. Thankfully, one definition of the natural log is:

$\displaystyle{\ln(x) = \int \frac{1}{x}}$

so we have a ready-made solution. Starting from $x=1$, what upper x-coordinate will trap 5 units of area? We want to solve:

$\displaystyle{5 = \int_{1}^{x} \frac{1}{x} dx = \ln(x)}$

Which means

$\begin{align*} 5 &= \ln(x) \\ e^5 &= e^{\ln(x)} \\ 148.41 &= x \end{align*}$

Wow! That's a large coordinate to trap 5 measly units of area.

In general, we have

$\displaystyle{e^{\text{area}} = \text{x-coordinate}}$

该位置的y坐标为$y = \frac{1}{x} = \frac{1}{e^{\text{area}}} = e^{-\text{area}}$。

The Rotated Hyperbola

做双曲曲线有很多种方法。If we rotate 45 degrees, we get something like this:

如何旋转方程$xy=1$?The standard way is with arotation matrix, but let's do the rotation withcomplex numbers.

Let's treat points as complex numbers: $(x, y) \rightarrow x + yi$. A sample point $(a + bi)$ is on therotatedhyperbola if, after undoing the rotation, we see our original $xy = 1$ relationship.

Candidate point: ($a + bi$)
45-degree counterclockwise rotation: $\frac{(1 + i)}{\sqrt{2}}$

$\displaystyle{\text{original point} = \text{candidiate} \cdot \text{rotation} }$

$\displaystyle{= (a + bi)\frac{(1 + i)}{\sqrt{2}} = \frac{1}{\sqrt{2}}[(a + bi^2) + (ai + bi)] = \frac{(a - b)}{\sqrt{2}} + \frac{(a + b)}{\sqrt{2}}i}$

Ok, we turned our candidate back to its original (pre-rotation) point, which is at

$\displaystyle{(x, y) = \left( \frac{(a - b)}{\sqrt{2}}, \frac{(a + b)}{\sqrt{2}} \right) }$

When does it have our $xy=1$ relationship?

$\begin{align*} \frac{(a-b)}{\sqrt{2}}\frac{(a+b)}{\sqrt{2}} &= 1 \\ (a - b)(a + b) &= 2 \\ a^2 - b^2 &= 2 \end{align*}$

好的!我们的候选是在旋转的双曲线上，如果:$a^2 - b^2 = 2$或$\text{实量级}^2 - \text{虚量级}^2 = 2$。We can express this requirement in $(x, y)$ notation as:

$\displaystyle{x^2 - y^2 = 2}$

Almost there. Our original hyperbola ($xy = 1$) contains the point $(1, 1)$, which is a distance of $\sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2}$ from the origin. The constraint equation is really $x^2 - y^2 = r^2$.

If we set the radius to 1, we get a formula for the unit hyperbola:

$\displaystyle{x^2 - y^2 = 1}$

Tada! A nice, clean equation, but I also see it as $(x - y)(x + y) = 1$, which hints at the inverse relationship.

Ok. Time for the scary diagram you'll see in most textbooks:

我们有旋转过的双曲线，想要用红色表示a/2的面积单位。(完整的$a$单位包括x轴下的面积。什么函数决定了这个区域的坐标?

The solution turns out to be the even and odd parts of the exponential, $\cosh$ and $\sinh$. There's a 1950's pamphlet "Hyperbolic Functions" by V. G. Shervatot, thatgoes through the derivation. The key intuition is realizing that hyperbolas (generally speaking) trap area logarithmically, so the necessary coordinates grow exponentially:

$xy = 1$ hyperbola (trap area under curve)
- $\text{x-coordinate} = e^{\text{area}}$
- $\text{y-coordinate} = \frac{1}{e^{\text{area}}} =e^{-\text{area}} $
$x^2 - y^2 = 1$ hyperbola (trap $\frac{\text{area}}{2}$ per diagram)
- $\text{x-coordinate} = \frac{e^{\text{area}/2} + e^{-\text{area}/2}}{2} = \cosh(x)$
- $\text{y-coordinate} = \frac{e^{\text{area}/2} - e^{-\text{area}/2}}{2} = \sinh(x)$

In case it needs to be said: it'snot obviousthat the even/odd parts of the exponential function determine the coordinates that trap area in a rotated hyperbola.

(Aside: Hyperbolas can be defined in terms of distance to fixed points or a conic section, but this gives no intuition for why exponentials are involved.)

The Secant/Tangent hyperbola

One more confusion is why we neednewfunctions to parameterize the hyperbola, when existing trig functions do the trick:

Start with a circle
For any angle $x$, we have coordinates $(x, y)$ = $(\cos(x), \sin(x))$
If we invert the x coordinate (hey, it's what hyperbolas do), we get $x = \frac{1}{\cos(x)} = \sec(x)$
If we scale the y coordinate by thatsame inversion, we get $y = \sin(x) \cdot \frac{1}{\cos(x)} = \tan(x)$

Does this really make a hyperbola? It meets the requirements:

$x^2 - y^2 = 1$ (relationship needed for rotated hyperbola)
$\sec^2(t) - \tan^2(t) = 1$ (rearrangement of the famous $\sec^2(t) = 1 + \tan^2(t)$)

This video shows how the various parameterizations behave (open the calculator):

Our familiar trig functions ($\sec(t), \tan(t))$ trace thesame hyperbolaas the fancy new $(\cosh(t), \sinh(t))$. They just go a different speed. And the parameter $t$ is just an everyday angle we plug into trig functions.

虽然双曲线有多种参数化，但$\cosh$和$\sinh$是用指数定义的，是欧拉公式中$\sin$和$\cos$的类比。他们可以戴上官方双曲参数化皇冠。

Revisiting Inverse Functions

反双曲函数从坐标到面积。假设我在x坐标为5的单位双曲线上。我困住了多少区域?${作用}\文本(5)= 2.29美元。

The inverse functions are sometimes called $\text{arcosh}$ ("area hyperbolic cosine"). This forces us to think about the coordinate-to-area conversion. I prefer to think about exponentials, and use $\text{acosh}$ ("inverse hyperbolic cosine"). Area is one interpretation, don't force me into it.

We can derive the formulas for the inverse functions bysolving$x = \cosh(y)$ and $x = \sinh(y)$:

$\displaystyle{\text{acosh(x)} = \ln(x + \sqrt{x^2 - 1})}$

$\displaystyle{\text{asinh(x)} = \ln(x + \sqrt{x^2 + 1})}$

正如所料，这些看起来像修改过的对数。当$x$增长时，它们接近$\ln(x + x) = \ln(2x) = \ln(x) + \ln(2)$，即有偏移的自然对数。

Part 5: Applications of the Geometric Interpretation

The Catenary

The main application of the geometric view is that the $\cosh(x)$ is the shape a rope takes when hanging between two fixed points. It's not quite a parabola, it's acatenary curve, with the St. Louis Arch as afamous example. Here's a few more curves (source) that follow $\cosh(x)$:

The process to build this curve is fairly subtle:

First, create a rotated hyperbola with $x^2 - y^2 = 1$
Instead of using the hyperbola, make a graph ofjust the x-coordinate.
This graph ofjust the x-coordinatemakes a new curve, which models how the rope hangs

这个复杂的过程并不是如何发现$\cosh$的。There's a differential equation that models the forces inside a hanging rope:

$\displaystyle{\frac{dy}{dx} = \frac{s}{a}}$

Tosolvethe differential equation, we need the convenient exponential properties of $\cosh$, and wind up with:

$\displaystyle{y = a \cosh(\frac{x}{a})}$

It's cute that $\cosh$ parameterizes a hyperbola, but that interpretation has nothing to do with why it's the solution. I think "the catenary follows the even part of the exponential function" not "the catenary follows the x-coordinate of the hyperbola".

Shape Where Arc Length = Area

The area under the exponential $e^x$ equals the current value (plus a constant). Consider the region from $x=0$ to $x=2$:

Area under curve: $\int_0^2 e^x = e^2 - e^0 = 7.389 - 1 = 6.389$
Current value: $e^2 = 7.389$
Pattern: $\text{current value} = \text{area under curve} - 1$

A pretty clean connection, right? (Don't forget that $+C$)

Now how about $\cosh$?

Current value: $\cosh(x)$
Area under curve: $\int \cosh(x) = \sinh(x)$
Arc lengthof curve: $\int \sqrt{1 + (\cosh'(x))^2} = \int \sqrt{1 + \sinh^2(x)} = \int \cosh(x) = \sinh(x)$
Pattern: $\text{area under curve} = \text{arc length} = \sqrt{\text{current value}^2 - 1}$

The current value of $\cosh$ can be swapped in using the identity $\sqrt{\cosh^2 - 1} = \sinh$.

For large $x$, the $-1$ is negligible and $\sqrt{\text{current value}^2 - 1} \sim \text{current value}$. So, for large $x$, we get equality between area, arc length, and current value (imagine the green rope hanging down and just touching the x-axis). It's more connected than regular $e^x$, not bad!

(Intuition for another day: Math deals with unitless quantities. $13 \ \text{cm}$ is not directly comparable with $13 \ \text{cm}^2$. Yet in math class, we can solve $x = 1 + x^2$ and nobody cares that constant, linear and squared terms are used in conjunction.)

Hyperbolic Geometry

Theshape of the universemay be a hyperbola, andhyperbolic geometryis used in special relativity (beyond my pay grade). If we do live in a giant hyperbola, I, uh, may be forced to recant my "exponentials first" stance.

Summary

The hyperbolic functions can be seen as exponential functions (relating time and growth) or geometric functions (relating area and coordinates). Hyperbolas, generally speaking, have logarithmic area and exponential coordinates.

这是一段漫长的旅程，但这些功能已经不在我的阁楼上了。

Happy math.