JOEL LEWIS: Hi.
Welcome back to recitation.
Today I wanted to talk
about something that's
mentioned in the notes but
wasn't covered in lecture
because of the exam review.
So this is, the subject is
hyperbolic trig functions.
So first I just wanted to define
them for you and graph them
so we can get a little
bit of a feeling for what
these functions are
like, and then I'm
going to explain to you why
they have the words hyperbolic
and trig in their names.
So these are some
interesting functions.
They're not, they don't--
aren't quite as important
as your usual, sort of
circular trig functions.
But yeah, so let me introduce
them and let me jump
in just with their definition.
So there are two
most important ones.
Just like a regular
trigonometric functions
there's the sine and
the cosine and then
you can write the other
four trigonometric functions
in terms of them.
So for hyperbolic
trig functions we
have the hyperbolic cosine
and the hyperbolic sine.
So the notation here,
we write c o s h.
So the h for hyperbolic.
So hyperbolic cosine.
And usually we
pronounce this "cosh."
And similarly, for
the hyperbolic sine
we write s i n h,
for hyperbolic sine,
except in the reverse order.
And we usually pronounce this
"sinch," so in American English
as if there were an
extra c in there.
Sinch.
OK.
So these functions have
fairly simple definitions
in terms of the exponential
function, e to the x.
So cosh of x is defined to be e
to the x plus e to the minus x
divided by 2.
And sinh of x is defined to be
e to x minus e to the minus x
divided by 2.
So if you remember what
your graph of e to the x
looks like, and your
graph of e to the minus x,
it's not hard to see that the
graphs of cosh x and sinh x
should look sort of like this.
So for cosh x, so we see as
x gets big, so e to the minus
x is going to 0.
It's not very important.
So it mostly is driven
by this e to the x part.
And as x gets negative and
big, then this is going to 0
and this is getting
larger and larger.
So we got something
that looks like this.
So it looks a little
bit, in this picture
it looks a little
bit like a parabola,
but the growth here is
exponential at both sides.
So in fact, this is growing
much, much, much faster
than, say, 1 plus x squared.
So it's a much steeper curve.
OK.
And it reaches its
minimum here at x
equals 0-- it has the
value 1 plus 1 over 2.
So its minimum there
is at x equals 0,
it has its minimum value 1.
For sinh, OK, so we're taking
the difference of them.
So it's similar when x
is positive and large,
e to the x is big,
and e to the minus
x is pretty small,
almost negligible.
So we got exponential
growth off that side.
When x becomes
negative and large,
e to the x is going to 0, e to
the minus x is becoming large,
but it's-- we've got
a minus sign here.
So as x goes to minus
infinity, this curve
goes also to minus infinity.
And again, the growth here
is exponential in both cases.
And if you were curious, say
about what the slope there
at the origin is, you could
quickly take a derivative
and check that that's passing
through the origin with slope 1
there.
OK.
So this is a sort
of basic picture
of what these curves look like.
They have some nice properties,
and let me talk about them.
So for example,
one nice thing you
might notice about
these functions
is that it's easy to
compute their derivatives.
Right?
So if we look at d/dx of cosh
x, in order to compute that,
well, just look at the
definition of cosh.
So it's really just a sum of
two exponential functions.
Exponential functions are
easy to take the derivatives.
Take the derivative of e to
the x, you get e to the x.
Take the derivative of e to
the minus x, well, OK, so
it's a little chain rule, so
you get a minus 1 in front.
So the derivative of cosh x is
e to the x minus e to the minus
x over 2.
But we have a name for this.
This is actually just sinh x.
So the derivative
of cosh is sinh,
and the derivative
of sinh, well, OK.
You look at the same
thing, take this formula,
take its derivative.
Well, e to the x, take its
derivative, you get e to the x.
e to the minus x,
take its derivative,
you get minus e to the minus
x, so those two minus signs
cancel out and become a plus.
So this is e to the x plus
e to the minus x over 2,
which is cosh x.
So here you have
some behavior that's
a little bit reminiscent of
the behavior of trig functions.
Right?
For trig functions, if you
take the derivative of sine
you get cosine.
And if you take the
derivative of cosine
you almost get back sine,
but you get minus sine.
So here you don't have
that extra negative sign
floating around.
Right?
So you, when you take
the derivative of cosh
you get sinh on the nose.
No minus sign needed.
So that's interesting.
But the real reason
that these have
the words trig in their name is
actually a little bit deeper.
So let me come over here
and draw a couple pictures.
So the normal trig
functions-- what sometimes we
call the circular trig functions
if we want to distinguish them
from the hyperbolic trig
functions-- they're closely--
so circular trig
functions, they're
closely related to
the unit circle.
So the unit circle
has equation x squared
plus y squared equals 1.
It's a circle.
Well, close enough, right?
And what is the
nice relationship
between this circle
and the trig functions?
Well, if you choose any
point on this circle,
then there exists
some value of t
such that this point
has coordinates
cosine t comma sine t.
Now it happens
that the value of t
is actually the angle
that that radius makes
with the positive axis.
But not going to worry
about that right now.
It's not the key idea of import.
So as t varies through
the real numbers,
the point cosine t,
sine t, that varies
and it just goes
around this curve.
So it traces out
this circle exactly.
So the hyperbolic
trig functions show up
in a very similar situation.
But instead of
looking at the unit
circle, what we
want to look at is
the unit rectangular hyperbola.
So what do I mean by that?
Well, so instead of
taking the equation
x squared plus y squared
equals 1, which gives a circle,
I'm going to look at a
very similar equation that
gives a hyperbola.
So this is the equation
x squared minus y squared
equals 1.
So if you if you graph this
equation, what you'll see
is that, well, it passes
through the point (1, 0).
And then we've got
one branch here,
we've got a little
asymptote there.
So it's got a right
branch like that,
and also it's symmetric
across the y-axis.
So there's a symmetric
left branch here.
So this is the graph
of the equation
x squared minus y
squared equals 1.
So it's this hyperbola.
Now what I claim is
that cosh and sinh have
the same relationship to this
hyperbola as cosine and sine
have to the circle.
Well, so I'm fudging
a little bit.
So it turns out it's only the
right half of the hyperbola.
So what do I mean by that?
Well, here's what
I'd like to do.
Set x equals-- so we're going
to introduce a new variable,
u-- I'm going to set x equal
cosh u and y equals sinh u.
And I'm going to
look at the quantity
x squared minus y squared.
So x squared minus y squared.
So this is, so we use
most of the same notations
for hyperbolic trig
functions that we
do for regular trig functions.
So this is cosh squared
u minus sinh squared u.
And now we can plug in the
formulas for cosh and sinh
that we have.
So this is equal to e to the u
plus e to the minus u over 2,
quantity squared, minus e to
the u minus e to the minus
u over 2, quantity squared.
And now we can expand
out both of these factors
and-- both of these squares,
rather, and put them together.
So over 2 squared is
over 4 and we square this
and we get e to the 2u.
OK, so then we get 2 times e to
the u times e to the minus u.
But e to the u times e to the
minus u is just 1, so plus 2.
Plus e to the minus 2u minus
e to the 2u minus 2 plus e
to the minus 2u-- so same
thing over here-- over 4.
OK, so the e to the 2u's cancel
and the e to the minus 2u's
cancel and we're left
with 2 minus minus 2.
That's 4.
So this is 4 over 4,
so this is equal to 1.
OK.
So if x is equal to cosh u
and y is equal to sinh u,
then x squared minus y
squared is equal to 1.
So if we choose a point
(cosh u, sinh u) for some u,
that point lies
on this hyperbola.
That's what this says.
That this point-- OK, so
the point (cosh u, sinh u)
is somewhere on this hyperbola.
And what's also true is the
sort of reverse statement.
If you look at all such points,
if you let u vary and look--
through the real
numbers and you ask
what happens to this
point (cosh u, sinh u),
the answers is that it
traces out the right half
of this hyperbola.
If you go back to the
graph of y equals cosh x,
you'll see that the hyperbolic
cosine function is always
positive.
So we can't-- over here, we
can't trace out this left
branch where x is negative.
Although it's easy
enough to say what does
trace out this left branch.
Since it's just
the mirror image,
this is traced out by
minus cosh u comma sinh u.
So there's a-- so the
hyperbolic trig functions have
the same relationship to
this branch of this hyperbola
that the regular trig
functions have to the circle.
So there's where the words
hyperbolic and trig functions
come from.
So let me say one
more thing about them,
which is that we
saw that they have
this analogy with
regular trig functions.
Right?
So instead of satisfying
cosine squared
plus sine squared equals 1, they
satisfy cosh squared minus sinh
squared equals 1.
And instead of satisfying
the derivative of sine
equals cosine and the derivative
of cosine equals minus sine,
they satisfy derivative
of cosh equals
sinh and derivative
of sinh equals cosh.
So similar relationships.
Not exactly the
same, but similar.
So this is true of a lot
of trig relationships,
that there's a corresponding
formula for the hyperbolic trig
functions.
So one example of such
a formula is your--
for example, your angle
addition formulas.
So I'm going to just leave
this is an exercise for you.
So let me, I guess
I'll just stick it
in this funny little
piece of board right here.
So, exercise.
Find sinh of x plus y and cosh
of x plus y in terms of sinh
x, cosh x, sinh y, and cosh y.
So in other words, find
the corresponding formula
to the angle addition
formula in that case
of the hyperbolic
trig functions.
So I'll leave you with that.
