**1. Motivation.** There are several different motivations, let me give them here.

*Motivation One.* We had a problem: find a parametric curve which describes the circle. This problem gave us the trigonometric functions as its solution.

Suppose we had a hyperbola

(1).

What parametric curve produces this hyperbola?

*Motivation Two.* Recall we defined the trigonometric functions using and (and their linear combinations). What happens if we switched these with and respectively?

**2. Parametrizing the Circle.** Lets carefully consider how we parametrize the circle. If we draw a unit circle (i.e., a circle with radius 1) in the plane, then consider a slice of it:

What’s the area of the shaded region? Well, the total area is , and the portion is . So the slice’s area would be

(2).

But look, the point on the circle which is drawn lies at . So in other words,

(2′).

This is how we specify the parameter. If the point is above the axis, then the area is positive (as in our doodle); if we picked a point on the circle below the axis, the area would be considered “negative”.

**3. Hyperbola.** So we consider the “unit” hyperbola which has its graph be

Then we consider the same song-and-dance: draw a straight line from the origin, find the point where it intersects our hyperbola, and consider the area between the line and the axis (and between the axis and the hyperbola) as doodled thus

The area (the shaded grey region) is . Following our parametrization of the circle, we say that is the “hyperbolic angle”.

If the intersection point is below the axis, we say the area is negative and simply change the sign.

We say that the hyperbola is the parametrized curve

(3), and

where “cosh” (pronounced *kosh*) is the hyperbolic cosine funcion, and “sinh” (pronounced *sin-sh* or *cinch*) is the hyperbolic sine function. Compare this to the usual trigonometric functions.

**4.** Lets describe this area using integrals. First we will consider the triangular region formed by the axis and the intersection point:

As an integral, we can express this area as

(4)

where is the slope of the hypoteneuse.

But this is too much, we need to subtract out the excess area which we doodle as the shaded region in this diagram:

How do we describe this using integration? The error area is

(5)

and thus the area we want to find is

(6)

Now *evaluating* the integral in Equation (5) is tricky!

**5.** We calculate Equation (5) and the trick requires a change of variables

(7) and

which gives us

(8).

Recall the identity

(9).

We plug this into (8) which gives us [changing the to ]

(10a)

and using (9) to change into produces

(10b).

What now?

We consider the integral

.

What to do? We integrate by parts, writing

(11a) thus

and

(11b) thus

We write our integral as

(11c).

Integrating by parts gives us

(12).

Does this seem familiar?

We plug (12) into the right hand side of (10b) producing

(13a).

We add to both sides

(13b).

Divide both sides by 2

(13c).

Now we just need to evaluate !

Integrating secant is a little tricky. First we write

(14a).

Then we consider a change of variables

(14b)

then we find

.

Thus we have

(14b).

We can evaluate this quickly as

(14c).

Now we unroll our substitution, and write this as

(14d).

Thus our integral (14a) is

(14a’).

Great, so we can recall (13c) is

(13c).

Plugging in (14a’) to the right hand side gives us

(15).

What now?

Well, we have to keep unrolling our substitution. We had used , so

.

Now what? Well, we use these and rewrite Equation (15) as

(16).

We keep on unrolling, and find the solution to the indefinite integral

(16).

My goodness, that was a lot of algebra!

Are we done yet? No! We still have to compute the error area, which is

(17).

After some fumbling arithmetic, we find

(18).

That gives us the error area.

**6.** So now we have the error area, and the triangle’s area recall is

(4)

where is the slope of the hypoteneuse. So we have

(19).

So the total area is

(20).

Recall that the point , so we simplify this as

(20′).

That’s the area!

Why were we looking for this? Well, the hyperbolic angle (which we were describing in terms of the area) is

(21).

Observe, using Equation (20′), we have the hyperbolic angle be

(22).

How, oh how, can we find the parametric form of the hyperbola with this puny angle?

The first step is to rewrite (22) with , giving us

(22′).

We introduce hyperbolic “trigonometric” functions to parametrize the hyperbola:

(23a)

and

(23b).

What are the properties of these functions? And what are they, anyways?

**Exercise 1:**prove the identity .

From this exercise, we may “guess” that the parametric curve describing a hyperbola is

(24) and .

We observe that, if

then

(25a)

and, after some algebraic manipulations,

(25b).

Thus we find

(26a)

and

(26b).

Thus we deduce

(27a)

and

(27b).

Just as we prophesized earlier on!

**Exercise 2.** Find the derivatives of and .

**Exercise 3.** Construct the hyperbolic tangent “in the obvious way” and find its derivative.

**Exercise 4.** Construct the hyperbolic secant and hyperbolic cosecant, find their derivatives.

**Exercise 5.** Compute and .

**Exercise 6.** Calculate .

**References.** I acknowledge using Isaac Greenspan’s Deriving the Hyperbolic Trigonometric Functions `[pdf]`.