Hyperbolic trig functions and the unit hyperbola Why they are called hyperbolic trigonometric functions
Hyperbolic trig 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.
- 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.
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