Hyperbolas are great, they're used to describe gravity assists, the shape power lines or chains make when hanging off two points is a Catenary.

This post is about the hyperbolic functions $\cosh(t), \sinh(t)$ that describe hyperbolics parametrically. In particular I show how you might have discovered them yourself!

This is the unit hyperbola, described by $x^2 - y^2 = 1$. We want to find $x(t), y(t)$ that describe this curve. Since they are always on the curve they must satisfy

Knowing how $x(t),y(t)$ change will help us find $x(t)$ and $y(t)$, so let's differentiate using implicit differentiation

Hey! That means $x' = Cy$ and $y' = Cx$ for some constant $C$ (since rational numbers are unique up to a constant multiple) This reminds me of $e^{Ct}$, except there are two functions in a cycle instead of one.

We can use this to find all the derivatives in terms of $x$ and $y$

Hey! If differentiating $x$ gives us $Cy$ and vise versa, then differentiating $x(t) + y(t)$ should stay the same up to a multiple of $C$, Let's do this

If we denote $s(t)$ to be the sum $x(t) + y(t)$ then we have $s'(t) = Cs(t)$.

This reminds me of $e^{Ct}$ since it's derivative is $C$ times itself, so I'm going to write $s(t)$ as a Maclaurin Series and try to find $s(t)$ in terms of $e^{Ct}$, but first we need to decide what $s(0)$ is, let's look at the graph again

Where should our parametric equations start? All points work fine, but we'll pick $(1, 0)$ for two reasons

1. It's a natural seeming starting point
2. Having $x(0) = 1, y(0) = 0$ will drastically simplify our maclaurin series

Now that we have our starting point, we can find the maclaurin series of $s(t)$

You might recognize this series, it's the series for $e^{Ct}$!, therefor $s(t) = e^{Ct}$

Now I want to find $x(t)$ and $y(t)$ individually, going back to the original equation I notice it's a difference of squares, meaning we can factor it as

Adding the sum and difference we can recover the individual equations!

Let's check our work, our functions should satisfy $x(t)^2 - y(t)^2 = 1$

Great! our $x(t)$ and $y(t)$ work, and they are $\cosh(t)$ and $\sinh(t)$ when $C=1$.

Arc length reparameterization

Picking $C=1$ seems like an arbitrary choice, let's try to reparameterize in terms of arc length to give $C$ geometric meaning

You can try to evaluate this integral, but you can't! this is a Nonelementary integral (see wolframalpha).

Relation to other conics

If you noticed the similarity between $\cosh, \sinh$ (for describing hyperbolas) and $\cos, \sin$ (for describing circles and ellipses) You might be wondering about expressing other conics parametrically.

The only other conic is a parabola (as circles are a special case of an ellipse). parabolas are described by $y = ax^2$, which is already a function!

Visually parabolas (3) look different from the other conics, they just barely hug the cone

I like to think of this in terms of their Eccentricity

• Ellipses have eccentricity less then one
• Parabolas have eccentricity equal to one
• Hyperbolas have eccentriciy greater then one

Parabolas really are the special child in the family! They are the only conic with a single eccentricity. Here's a desmos graph showing this off

I plan to write about parabolas some time, since they show up everywhere in algebra I want to become more acquainted with their amazing properties