What are Hyperbolic Trig Functions Functions of? Circular trig functions take in an angle and spit out a ratio. What do hyperbolic  functions take in (I know it's a number, but what geometrically does it represent)? 
I've seen images that suggest they're a function of area, and others describe $(\cosh(t),\sinh(t))$ as a parametric, which makes me think that t is arc length covered as you move along the unit hyperbola, just like how with a circular parametric trig function t can be interpreted as arc length. It seems like it should be the function of an angle, but it isn't.  
 A: The connecting factor amongst the hyperbolic and circular functions and the unit circle and hyperbola is not the angle subtended by the curve but rather the area bounded by it. Here are two images I pulled from the internet:


Consider the point $\mathrm H(\cosh w, \sinh w)$ on the unit hyperbola and the point $\mathrm C(\cos z, \sin z)$ on the unit circle, both plotted on the $xy$-plane. Then


*

*$w$ equals twice the area bounded by $\rm OH$, the hyperbola, and the $x$-axis,


and since the area of the unit circle is $\pi(1)^2=\pi$, then


*

*$z$ equals twice the area bounded by $\rm OC$, the circle, and the $x$-axis.



So to summarize, the relationship amongst the argument $t$ of any of the aforementioned functions and the area $a$ bounded by their characteristic locus is $t=2a$.


You mention “arc length covered as you move along the unit hyperbola.” Let’s consider the integral formula for arc length
$$L = \int_\alpha^\beta\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2}\,dt$$
and compare the arc length $L(\mathrm H(T))$ traced along the unit hyperbola from point $(1,0)$ to $(\cosh T, \sinh T)$ and the arc length $L(\mathrm C(T))$ traced along the unit circle from point $(1,0)$ to $(\cos T, \sin T)$:
$$\begin{align}
\mathrm H 
&: \begin{cases} x=\cosh t \\ y=\sinh t\end{cases} \\[2ex]
&: \begin{cases} {dx}/{dt} = \sinh t \\ {dy}/{dt} = \cosh t\end{cases}
\\[4ex]
\mathrm C 
&: \begin{cases} x=\cos t \\ y=\sin t\end{cases} \\[2ex]
&: \begin{cases} {dx}/{dt} = -\sin t \\ {dy}/{dt} = \cos t\end{cases}
\end{align}$$
So
$$L(\mathrm H(T)) = \int_0^T \sqrt{ \cosh^2 t + \sinh^2 t } \, dt$$
$$L(\mathrm C(T)) = \int_0^T \sqrt{ \cos^2 t + \sin^2 t } \, dt$$
Now invoke the identities $\sin^2t+\cos^2t = 1$ but $\sinh^2t+\cosh^2t = \exp2t-\sinh(t)\cosh(t)$ and you will see that
$$L(\mathrm C(T)) = \int_0^T \sqrt{ 1 } \, dt = t\bigr|_0^T = T$$
$$\begin{align}
L(\mathrm H(T)) &= \int_0^T \sqrt{ e^{2x} - {e^x-e^{-x}\over2} \cdot {e^x+e^{-x}\over2} } \, dt \\[2ex]
&= \int_0^T \sqrt{ e^{2x} - {e^{2x}-1-1-e^{-2x}\over4} } \, dt \\[2ex]
&= \int_0^T \sqrt{ {4e^{2x} - e^{2x}+2+e^{-2x}\over4} } \, dt \quad \cdots
\end{align}$$
Does that help you see why the nice relationship seen between $t$ and $L(\mathrm C(T))$ does not emerge with $t$ and $L(\mathrm H(T))$?
