# Why are hyperbolic functions defined by area?

I have successfully derived the hyperbolic functions in terms of exponentials from the graphical definition:

For area $$u/2$$ bound by the unit parabola ($$x^2 - y^2 = 1$$), a ray from the origin to a point $$(a,b)$$ on the hyperbola and the $$x$$-axis, $$\cosh u = a$$ and $$\sinh u = b$$.

With some calculus and algebra, this definition can yield: $$\sinh u = \frac{e^u-e^{-u}}{2}$$ $$\cosh u = \frac{e^u+e^{-u}}{2}$$

However, I am not very content with my derivation. I understand that the graphical definition is supposed to compare with the definition of trigonometric functions from the unit circle, as $$\sin u$$ and $$\cos u$$ can similarly be defined by bound area. My question is: why? Why not define hyperbolic functions from angle? From arc length? Why area specifically? Answers from other questions from this website and others states that "it is the only definition that carries over". If anyone can elaborate on this, it will be greatly appreciated, thanks!

Edit: It has been brought to my attention that you can define hyperbolic functions from arc length. Despite this, the definition in terms of area is still the most popular and most used. I maintain hopeful that an intuitive reason exists.

• It is not the only definition that carries over. You can also define hyperbolic functions using arc length; you just have to measure the arc length properly (in the Minkowski pseudometric rather than the Euclidean metric). See, e.g., this answer. – Micah Jun 1 '20 at 2:33
• Thanks! However, I would also like to know why this definition is the most "popular" one per se. – chematwork Jun 1 '20 at 2:37
• @WillJagy I know that definition exists, but I am looking at the graphical definition, not the one based on differential equations. – chematwork Jun 1 '20 at 2:44

The unit circle is the locus of points with unit distance from the origin in the Euclidean metric: that is, the metric corresponding to the standard dot product $$(x_1,y_1)\cdot (x_2,y_2)=x_1x_2+y_1y_2$$. You can define the standard trig functions as parameterizations of the unit circle in the Euclidean metric, using either arc length or area; both definitions will be equivalent.

Similarly, the unit hyperbola is the locus of points with unit distance from the origin in the Minkowski metric: that is, the metric corresponding to the nonstandard dot product $$(x_1,y_1) \cdot_M (x_2,y_2)=-x_1x_2+y_1y_2$$. Again, you can define the hyperbolic functions as parameterizations of the unit hyperbola in the Minkowski metric, using either arc length or area; both definitions will be equivalent.

So why does it appear that you can define the hyperbolic functions using area, but not using arc length? Because you're studying the hyperbola in the Euclidean metric! The Euclidean arc length of a curve $$\displaystyle \int_C \sqrt{dx^2+dy^2}$$ is usually different from the Minkowski arc length $$\displaystyle \int_C \sqrt{-dx^2+dy^2}$$ of that curve, so switching metrics leads to a different arc length parameterization for the hyperbola.

However, both the Euclidean and Minkowski definitions of area turn out to be equivalent. To see this, note that we can just show it for parallelograms, and then integrate to get the result for arbitrary shapes.

The signed area of the parallelogram spanned by the vectors $$v_1=\left$$ and $$v_2=\left$$ can be computed via the dot product as follows. First, let $$v_1^\perp=\left<-y_1,x_2\right>$$ be one of the two vectors perpendicular to $$v_1$$ (that is, with $$v_1^\perp \cdot v_1=0$$), and with the same magnitude as $$v_1$$. Then the signed area of the parallelogram spanned by $$v_1$$ and $$v_2$$ is $$x_1y_2-y_1x_2$$, which can be conveniently written in the form $$v_1^\perp \cdot v_2$$.

Working in the Minkowski metric, if $$v=\left$$, then the vector perpendicular to $$v$$ with the same magnitude is $$v^\perp=\left$$. So if $$v_1=\left$$ and $$v_2=\left$$, then $$v_1^\perp \cdot_M v_2=-y_1x_2+x_1y_2=x_1y_2-y_1x_2$$, which is coordinate-wise identical to the expression for the area in the Euclidean metric.

TL;DR: if we define the circle or hyperbola in the most natural metric for that particular curve, we can get the trig/hyperbolic functions using either arc length or area.

But most of the time, we stick to the Euclidean metric when doing coordinate geometry. The area definition of hyperbolic functions turns out to be the same in either metric, but the arc length definition doesn't, so it ends up looking like we need to use the area definition and not the arc length definition.

• Took me some time to understand the answer, that's why it took me long to accept it, sorry! P.S. Do you have any resources on Minkowski metrics? It's the first time I've heard about it. Any help is greatly appreciated! – chematwork Jun 1 '20 at 12:16
• I think most of this is stuff I learned incidentally while trying to understand other things (e.g., special relativity and hyperbolic geometry). The beginning part of this pdf looks like a good reference; they also recommend this book which may be a little more accessible. – Micah Jun 1 '20 at 14:55

I am inclined to think of the hyperbolic functions as functions that are defined to be similar to the trigonometric sine and cosine functions, and happen to have a geometric expression in terms of areas.

$$e^{ix} = \cos x + i\sin x\\ \cos x = \frac {e^{ix} + e^{-ix}}{2}\\ \sin x = \frac {e^{ix} - e^{-ix}}{2i}$$

Which are very similar to the definitions of the hyperbolic functions.

$$\cos ix = \cosh x\\ \cosh ix = \cos x\\ \sin ix = i\sinh x\\ \sinh ix = i\sin x$$

And the similarites between these functions continues.

The solution to the differential equation

$$y'' = - y$$ is $$A\sin x + B\cos x$$

and the solution to

$$y'' = y$$ is $$A\sinh x + B\cosh x$$

$$\frac {d}{dx} \sin x = \cos x\\ \frac {d}{dx} \sinh x = \cosh x\\ \frac {d}{dx} \cos x = -\sin x\\ \frac {d}{dx} \cosh x = \sinh x$$

etc.