Hyperbolic Trig Functions and the Unit Hyperbola

Takes the hypothesized hyperbolic trig functions, demonstrates a unit hyperbola and shows that the hypothesized functions really do match the unit hyperbola.

We know that if we take all of the points

in the *X, Y plane where x^2 + y^2 = 1*,

we get ourselves the unit circle.

Let me draw the unit circle.

That's my y-axis; this is my x-axis.

And the unit circle has the circle with the radius one.

So that's x=1, that's x=-1, that's y=1, that's y=-1

the unit circle looks something... let me draw it...

something like this, I think you get the point

Let's see if I can fill it in a little bit better.

So you realize that it's not a dotted circle.

There.

That's my best attempt at drawing the unit circle.

And we also know that the traditional trig functions,

or maybe we should call them the circular trig functions

are actually defined so that if you parameterize

so if you were to take *x=cos t*

and *y=sin t*

and you pick any t, right over here

and by definition it's going to sit on the unit circle

by definition, *x^2 + y^2 = 1*

so if you pick any t it's going to sit some place

on this unit circle.

Or another way to think of it is

if you vary t it's going to start tracing out this circle

And we know that t corresponds to the angle

with a positive x-axis, in this case,

that right over there is t.

Now wouldn't it be neat if there were a similar

analogy for, not the unit circle, but

something we could call the unit hyperbola?

So that's our little review of trigonometry right there;

our traditional trigonometry, now let's think about

the unit hyperbola.

Well, *x^2 + y^2 = 1* is a unit circle, I'll say that

*x^2 - y^2 = 1, I'm going to call this my unit hyperbola*.

Or a unit rectangular hyperbola. Hyperbola.

This is just a little bit of review from *Conic Sections*,

but it would look something like this:

It would look... something... that's my y-axis,

this is my x-axis, and then we can say,

well if y is 0, x can be ±1,

so you can think of that as the unit part

where it intersects the x-axis; that's +1, that's -1

and it has asymptotes, y=x and y=-x

We go through the intuition on that in the

Conic Section videos, y=x is that dotted line,

y=-x is that dotted line, right over there,

and then this thing is going to look like this.

It's going to have a right half that does something like this,

and does something like this, all a review of

*Conic Sections*, it gets closer to its asymptotes.

To y=x or y=-x

and the same thing on the left-hand side.

It's going to do something like that.

Wouldn't it be neat if we could parameterize

x and y with analogous functions so that

we get a similar type of property?

And you might guess what those functions are,

but let's actually try to verify it.

What would happen if x is equal to our hyperbolic

*cosine of t*, which is the same thing as

*e^t + e^(-t)*, all of that over 2

and y were to be equal to

*hyperbolic sine* of t, which is equal to

*e^t - e^(-t)* over 2. Wouldn't it be neat if there

were an analogy here; over here you pick any t

based on our circular trig functions, you ended up

with a point on the unit circle. Wouldn't it be amazing

if for any point t you ended up with a point on our,

what we're calling our *unit hyperbola*?

Well, in order for that to be true, with this parameterization

*x^2 - y^2* would need to be equal to one.

Let's see if that is the case!

So *x^2 - y^2* is equal to, well let's square this business

it's equal to e^(2t) plus

two times the product of these two things

*2e^t • e^(-t)*, this is e^0 here which is 1.

Plus e^(-2t), e^(-t)^2, all of that over 4

And then from that we will subtract y^2.

Minus, so the numerator's going to be

*e^(2t) - 2e^t • e^(-t) + e^(-2t)*, all of that over 4

So, immediately, a couple of simplifications here.

*e^t • e^(-t)*, that's just e^(t-t)

which is equal to e^0, which is equal to 1

This is going to be one, that's going to be one,

so we're going to have a 2 in either of those cases

and if we were to simplify it, all of this stuff over here

I'll do a numerator, so this is going to be equal to

over our [denominator] of 4

*e^(2t) + 2 + e^(-2t) - e^(2t)*

just distributing the negative sign

Plus two, and then minus e^(-2t)

Well this is convenient!

(Oh, I was writing it in black, a hard color to see)

This cancels with this,

This and this also add up to zero and you're left with

two plus two over four

which is indeed equal to one!

So this is a pretty good reason to call these two functions

hyperbolic trig functions.

These are the circular trig functions,

you give me a t on these parameterizations

we end up on the unit circle!

You vary t, you trace out the unit circle.

Here, for any real t, we're going to assume we're

dealing with real numbers,

for any real t we're going to end up on

the unit hyperbola right over here

and in particular we're going to end up on the right

so it's not exactly... over here pretty much any of these

points could be parameterized right here

over here we're going to end up

on a point on the right side of the unit hyperbola.

The reason why it's the right side

is... you go straight to the definition of

*cosh t*, this thing can only be positive

This thing can only be positive.

e^t can only be positive, e^-t can only be positive

so this is only positive.

But you give any t you will end up on this hyperbola!

Specifically the right side, if you want points on

the left hand side, you'd have to take the

*-cosh t and the sinh t*

to end up right over there.

But it's a pretty neat analogy.

We're looking at Euler's identity and we

kind of said, "oh, let's just start playing with these things!"

There seems to be a similarity here if we were to

remove the i 's and, all of a sudden, we've discovered

another thing! That there is this relationship

here there is this relationship between these trig functions

and the unit circle, here between our newly defined

hyperbolic trig functions and the unit hyperbola.

And you'd also find if you were to vary t it's going

to trace out... just as if you were to vary t here it

traces out the unit circle... if you trace t here it will trace

out the right-hand side, the right-hand side

of the unit hyperbola.

For this parameterization right here.