Hyperbolic functions and the unit hyperbola

We know that if we take all of the points in the *X, Y plane<i>where</i>x^2 + y^2 = 1*, we get ourselves the unit circle. Let me draw the unit circle. That's my <i>y-axis</i>; this is my <i>x-axis</i>. And the unit circle has the circle with the radius one. So that's <i>x=1</i>, that's <i>x=-1</i>, that's <i>y=1</i>, that's <i>y=-1</i> 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 <i>circular</i> 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 <i>t</i>, 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 <i>t</i> it's going to sit some place on this unit circle. Or another way to think of it is if you vary <i>t</i> it's going to start tracing out this circle And we know that <i>t</i> corresponds to the angle with a positive <i>x-axis</i>, in this case, that right over there is <i>t</i>. 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 <i>hyperbola</i>. Well, *x^2 + y^2 = 1* is a unit circle, I'll say that *x^2 - y^2 = 1<i>, I'm going to call this my</i>unit hyperbola*. Or a unit rectangular hyperbola. <u>Hyperbola</u>. 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 <i>y-axis</i>, this is my <i>x-axis</i>, and then we can say, well if <i>y</i> is 0, <i>x</i> can be ±1, so you can think of that as the unit part where it intersects the <i>x-axis</i>; that's +1, that's -1 and it has asymptotes, <i>y=x</i> and <i>y=-x</i> We go through the intuition on that in the Conic Section videos, <i>y=x</i> is that dotted line, <i>y=-x</i> 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 <i>y=x</i> or <i>y=-x</i> 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 <i>x</i> and <i>y</i> 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 <i>x</i> is equal to our <i>hyperbolic</i> *cosine of t*, which is the same thing as *e^t + e^(-t)*, all of that over 2 and <i>y</i> were to be equal to *hyperbolic sine* of <i>t</i>, 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 <i>t</i> 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 <i>t</i> 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 <i>is</i> the case! So *x^2 - y^2* is equal to, well let's square this business it's equal to <i>e^(2t)</i> plus two times the product of these two things *2e^t • e^(-t)*, this is <i>e^0</i> here which is 1. Plus <i>e^(-2t)</i>, <i>e^(-t)^2</i>, all of that over 4 And then from that we will subtract <i>y^2</i>. 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 <i>e^(t-t)</i> which is equal to <i>e^0</i>, 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 <i>e^(-2t)</i> 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 <i>t</i> on these parameterizations we end up on the unit circle! You vary <i>t</i>, you trace out the unit circle. Here, for any real <i>t</i>, we're going to assume we're dealing with real numbers, for any real <i>t</i> 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 <i>any</i> of these points could be parameterized right here over here we're going to end up on a point on the <i>right</i> 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. <i>e^t</i> can only be positive, <i>e^-t</i> can only be positive so this is only positive. But you give any <i>t</i> 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<i>and the</i>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>i</i> 's and, all of a sudden, we've discovered another thing! That there is this relationship here there is this relationship between <i>these</i> trig functions and the unit circle, here between our <i>newly</i> defined hyperbolic trig functions and the unit <u>hyperbola</u>. And you'd also find if you were to vary <i>t</i> it's going to trace out... just as if you were to vary <i>t</i> here it traces out the unit circle... if you trace <i>t</i> here it will trace out the right-hand side, the right-hand side of the unit hyperbola. For this parameterization right here.