PROFESSOR: OK, this lecture
is about the slopes, the
derivatives, of two of the
great functions of
mathematics: sine
x and cosine x.
Why do I say great functions?
What sort of motion do we
see sines and cosines?
Well, I guess I'm thinking
of oscillations.
Things go back and forth.
They go up and down.
They go round in a circle.
Your heart beats and
beats and beats.
Your lungs go in and out.
The earth goes around the sun.
So many motions are repeating
motions, and that's when sines
and cosines show up.
The opposite is growing
motions.
That's where we have powers of
x, x cubed, x to the n-th.
Or if we really want
the motion to get
going, e to the x.
Or decaying would be
e to the minus x.
So there are two kinds here.
We're talking about the ones
that repeat and stay level,
and they all involve
sines and cosines.
And to make that point,
I'm going to have to--
you know what sines and cosines
are for triangles from
trigonometry.
But I have to make those
triangles move.
So I'm going to put the triangle
in a circle, with one
corner at the center, and
another corner on the circle,
and I'm going to move
that point.
So it's going to be
circular motion.
It's going to be the
motion that--
the perfect model of repeating
motion, around
and around the circle.
And then the answer we're going
to get is just great.
The derivative of sine x turns
out to be cosine x.
And the derivative of cosine x
turns out to be minus sine x.
You couldn't ask for more.
So my interest is always to
explain those, but then I want
to really--
we're seeing this limit stuff
in taking a derivative, and
here's a chance for me
to find a limit.
This turns out to be the crucial
quantity: the sine of
an angle divided by the angle,
when the angle goes to 0.
Of course, when it's at 0,
the sine of 0 is 0, so
we have 0 over 0.
This is the big problem
of calculus.
You can't be at the limit,
because it's 0
over 0 at that point.
But you can be close to it.
And then if we drew a graph, had
a calculator, whatever we
do, we would see that that ratio
is very close to 1, but
today we're going to actually
prove it from the meaning of
sine theta.
Now remember what
that meaning is.
So back to the start
of the world.
Actually back to Pythagoras,
way, way back.
The key fact is what you
remember about right
triangles, a squared plus b
squared equals c squared.
That's where everything starts
for a right triangle.
I don't know if Pythagoras
knew how to prove it.
I think his friends
helped him.
A lot of people have
suggested proofs.
Einstein gave a proof.
Some US president even
gave a proof.
So it's a fundamental fact, and
I'm going to divide by c
squared, because I'd like the
right hand side to be 1.
So if I divide by c squared, I
just have a squared over c
squared plus b squared
over c squared is 1.
And I'm going to make that
hypotenuse in my picture 1.
So then this will be the a over
c, and that ratio of the
near side to the hypotenuse
is the cosine.
So what I have here is
cosine theta squared.
Let me put theta in there.
Theta is that angle
at the center.
And what's b?
So this is a over c.
That's cosine theta.
B over c is this point,
and that's sine theta.
And they add to 1.
So that's Pythagoras using
sines and cosines.
So this is the cosine.
And this vertical distance
is sine theta.
OK, so that's the
triangle I like.
That's the triangle that's
going to move.
As this point goes around the
circle at a steady speed, this
triangle is going to move.
The base will go left and
right, left and right.
The height will go up and down,
up and down, following
cosine and sine.
And we want to know things
about the speed.
OK, so that's circular motion.
Now I've introduced
this word radians.
And let me remind you what they
are and why we need them.
Why don't we just use 360
degrees for the full circle?
360 degrees.
Well, that's a nice
number, 360.
Somebody must have thought it
was really nice, and chose it
for measuring angles
around the world.
It's nice, but it's
not natural.
Somebody thought of it,
so it's not good.
What we need is the natural
way to measure the angle.
If we don't use the natural way,
then this is the sine--
if I measure this x
in degrees, that
formula won't be right.
There will be a miserable factor
that I want to be 1.
So I have to measure the angles
the right way, and
here's the idea of radians.
The measure of that
angle is this
distance around the circle.
That distance I'm going to call
theta, and I'm going to
say this angle is theta
radians when
that distance is theta.
So that now, what's
a full circle?
A full circle would mean the
angle went all the way round.
I get the whole circumference,
which is 2 pi.
So 360 degrees is
2 pi radians.
So the natural number
here is 2 pi.
This can't be helped, it's
the right one to use.
Radians are the right way
to measure an angle.
So now I'm ready to do the job
of finding this derivative.
OK.
Let me start at the
key point 0.
If we get this one, we get
all the rest easily.
So I'm looking at the graph
of the sine curve.
I'm starting at 0.
We know what sine theta looks
like, and I'm interested in
the slope, the derivative.
That's what this subject
is about, calculus,
differentiating.
So I want to know the
slope at that point.
And it's 1.
And how do we show
that it's 1?
So now I'm coming to the point
where I'm going to give a
proof that is 1.
And the proof isn't just for the
sake of formality or rigor
or something.
You really have to understand
the sine function, the cosine
function, and this is
the heart of it.
OK.
So we want to show
that slope is 1.
How am I going to do that?
That's the slope, right?
If I go a tiny amount theta,
then I go up sine theta.
So in this average slope, if
I take a finite step--
I could have called it delta
theta, but I don't want to
write deltas all the time.
So I just go out a little
distance theta and up to the
sine curve.
I stopped at the sine
curve by the way.
The straight line is a little
above the sine curve here.
And that ratio, up divided by
across, that's the delta sine
divided by delta theta.
And because it started at 0,
it's just sine theta is the
distance up, and theta is
the distance across.
So this is the average slope.
And of course you remember
what calculus is doing.
There's always this limiting
process where you push things
closer and closer to the point,
and you find the slope
at that point, sometimes called
the instantaneous
velocity or slope
or derivative.
Now here's the way it's
going to work.
I'm going to show that sine
theta over theta is
always below 1.
So two facts I want to prove.
I want to show that sine theta
over theta is less--
sorry, sine theta over theta--
well, let me get this right.
I might as well put
it the neat way.
I want to show that sine
theta is below theta.
This is for theta
greater than 0.
That's what I'm doing.
OK.
So that tells me that this
curve stays under that
straight line, that 45 degree
line, which I'm claiming is
the tangent line.
And it tells me when I divide
it by the theta, it tells me
that sine theta over
theta is below 1.
But now how much below 1?
Right now if I only know this,
I haven't ruled out the
possibility that the slope
could be much smaller.
So I need something below it.
And fantastically, the
cosine is below it.
So the other thing that I want
to prove is that the cosine--
and I'll let me do
it this way.
I'm going to show the tangent of
theta is bigger than theta.
Again, some range of thetas.
Positive thetas up
to somewhere.
I don't know, I think
maybe pi over 2.
But the main point is near
0, that's the main point.
So can I just rewrite--
do you remember what
the tangent is?
Of course, sine theta
over cosine theta.
So this is sine theta
over cosine
theta, bigger than theta.
We still have to prove this.
And now I want to bring the
theta down and move the cosine
up, and that will tell me that
sine theta, when I divide by
the theta and multiply
by the cosine theta.
So that was the same as that,
was the same as that.
And that's what I want.
That tells that this ratio is
above the cosine curve.
Do you see that if I can
convince you, and convince
myself that these are both true,
that this picture is
right, then--
I haven't gone into gory
detail about limits.
If you really insist,
I'll do it later.
But whatever.
You can see this has just got to
be true, that if this curve
is squeezed between the cosine
curve and the 1, then as theta
gets smaller and smaller,
it's squeezed to
equal 1 in the limit.
Allow me to say that that's
pretty darn clear.
OK.
Whatever limit meets.
So these are the main facts
that I need to show.
And I need to show those
using trig, right?
I have to draw some graph
that convinces you.
And this isn't quite good
enough, because I just
sketched a sine graph.
I have to say where does
sine theta come from?
OK.
So this will be number one, and
this will be number two,
and when we get those two things
convincing, then we
know that sine theta over theta
is squeezed between and
approaches 1.
And then we'll know the story
at the start, and you'll see
that it becomes easy to
find these formulas
all along the curve.
OK.
Ready for these two?
Number one and number two.
OK, number one.
Why is sine theta--
oh, I can probably see
it on this picture.
Yeah, I can prove number
one on this picture.
Look, that piece was
sine theta, right?
And I want to prove
that this length--
what am I trying to prove?
That sine theta is
below theta.
Let me write it again
what it is to show.
In math it's always a good
idea to keep reminding
yourself of what it
is you're doing.
Sine theta is below theta.
OK.
So why is it?
And you see it here.
That was sine theta, right?
And where was theta?
Well, because we measured theta
in radians, theta is
this curvy distance that's
clearly longer.
The shortest way from this to
the axis there is straight
down, and that's sine theta.
A slower way is go round and
end up at not the nearest
point, and that was theta.
Is that good?
I could sometimes just to be
even more convincing, you add
a second angle, and you say OK,
there's 2 sine thetas and
here is 2 thetas, and clearly we
all know that the shortest
way from there to there
is the straight way.
So I regard this as done
by that picture.
You see we didn't
just make it up.
It went back to the fundamental
idea of where sine
theta is in a picture.
Now I need another picture.
Yeah, I need another picture
for number two to show that
tan theta is bigger
than theta.
That was our other job.
So essentially, I need that
same picture again.
Whoops, let me draw
that triangle.
Yeah, and it's got a circle.
OK, that's not a bad circle.
It's got an angle theta.
And now I'm going to--
math has always got
some little trick.
So this is it.
Go all the way out, so now the
base is 1, and this is still
the angle theta.
And what else do I know
on that picture?
Now I've scaled the triangle
up from this little one, so
the base is 1.
So what's that height?
Well, the ratio of the opposite
side to the near
side, that's what tangent is.
Tangent is the ratio--
whatever size the triangle--
is the ratio of the opposite
side to the near side.
Sine to cosine, here it's
tan theta is that
distance, and to 1.
Good.
OK, but now how am I
going to see this?
I have to ask you--
and it's OK--
to think about area instead
of distance for a moment.
What about area?
So what do I see of area?
I see right away that the area
of this triangle is smaller
than the area--
sorry, I shouldn't have called
that a triangle.
That's a little piece of pie,
a little sector of a circle.
So the area of this shaded--
did I shade it OK--
is less than--
so this is the area
of the sector.
Can I just call it the pie,
piece of a pie, is less than
the area of the triangle.
But we know what the area
of the triangle is.
What's the area of a triangle?
We can do that.
It's the base half, right?
1/2 times the base
times the height.
So the area of the triangle
is 1/2 times the
base 1 times the height.
OK.
Notice we've got the sign
going the right way.
We want tan theta to be
bigger than something,
so what do I hope?
I hope that the area of this
shaded part, the area of the
circular sector, is
1/2 of theta.
Wouldn't that be wonderful?
If I look at those areas,
nobody's in any doubt that
this piece, this sector that's
inside the triangle, has an
area less than the area
of the triangle.
So now I just have to remember
why is the area of this
sector, half of theta.
You know, there's another reason
why areas come up right
when we use a radians, when we
measure theta with radians.
So remember, just think about
this piece of pie compared to
the whole pie.
What's the area of the
whole piece of pie?
So I'm explaining 1/2 theta.
The area of the whole pie--
I'm going to get some terrible
pun here on the word pie.
Unintended, forgive it.
The area of that whole circle,
the radius is 1, we all know
what the area of a circle
is pi r squared.
r is 1, so the area is pi.
My God, I didn't expect that.
Now what about this?
What fraction is this sector?
Well, the whole angle would be
2 pi, and this part of it is
theta, so I have the sector is
theta over 2 pi, that's the
angle fraction, times the
pi, the whole area.
Do you see it?
This piece of pie, or
pizza, whatever--
yeah, if I'd said pizza, I
wouldn't have had that
terrible pun.
Forget it.
So the area of this piece of
pizza compared to the whole
one is theta over
the whole 2 pi.
Suppose it was a pizza cut
in the usual 6 pieces.
Then this would be a 60
degree angle, but
I don't want degrees.
What would be the angle of that
piece of pizza that's cut
when the whole pizza's
cut in 6?
It would be 1/6 of 360.
That's 60 degrees.
But I don't want degrees.
It's 1/6 of 2 pi.
And this one is theta of 2 pi.
Anyway, the pis cancel.
Theta over 2 is the right
answer, and now we can cancel
the 1/2, and we've
got what we want.
That's pretty nice when you
realize that we were facing
for the first time, more or
less, the sort of tough
problem of calculus when I can't
really divide theta into
sine theta.
Sine theta, I can't
just divide it in.
I have to keep them both
approaching 0 over 0, and see
what that ratio is doing.
And now I said to conclude,
I'll go back and prove the
slopes, find the slopes
at all points.
OK, so at all points--
now let's start with sine x.
So what am I doing now?
I'm looking at the sine curve.
You remember it went up like
this and down like this.
I'm taking any point x.
Suddenly I changed the angle
from theta to x, just because
I'm used to functions of x.
We're just talking
letters there.
X is good, and this is
a graph of sine x.
X is measured in radians still.
OK.
So now what am I doing to find
the derivative at some
particular point?
I look at the sine there.
I go a little distance
delta x.
I go up to here, and
I look to see--
I want to know the change
in sine x divided by
the change in x.
And of course, I'm going
to let the piece
get smaller and smaller.
That's what calculus does.
The main point is my x is
now here instead of
being at the start.
I've done it for the start, but
now I have to do it for
all the other x's.
So there's the x.
There's the x plus delta
x, a little bit long.
In other words, can I write this
in the familiar way, sine
of x plus delta minus sine
there divided by delta x?
OK.
So again, we can't simplify
totally by just dividing the
delta x in.
We've got to go back
to trigonometry.
Trig had a formula for
the sine of a plus b.
Two angles added, then there's
a neat formula for it.
So the sine--
can I remind you of
that formula?
It is the sine of the first
angle times the cosine of the
second minus the cosine of the
first angle times the sine of
the second.
OK?
You remember this,
right, from trig?
The sine of a plus b
is this neat thing.
Now I have to subtract sine x.
So now can I subtract
off sine x?
When I subtract off sine x,
then I need a minus 1.
And now I have to divide
by delta x.
So I divide this by delta x, and
I divide this by delta x.
OK.
This is an expression
I can work with.
That's why I had to remember
this trig formula to get this
expression that I
can work with.
Why do I say I can
work with it?
Because this is exactly what
I've already pinned down.
Delta x is now headed for 0.
This point is going to come
close to this one.
So actually, I've got two terms.
This sine delta x over
delta x, what does that do
as delta x goes to 0?
It goes to 1.
That was the point of that
whole right hand board.
So this thing goes to 1.
Wait a minute.
That's a plus sign.
Everybody watching is going
to think, OK, forgot trig.
The sine of the sum of an angle
is the sine times the
cosine plus the cosine
times the sine.
Sorry about that one too.
OK, so sine of delta x over
delta x goes to 0.
And now finally, this goes to 1,
and actually I need another
little piece.
I need to know this piece, and
I need to know that that
ratio goes to 0.
So I need to go back to that
board and look again at the
cosine curve at 0.
Because this will be a slope
of the cosine curve at 0.
And the slope comes out 0
for the cosine curve.
The slope for the sine
curve came out 1.
Do you see how it's working?
So this is gone because
of the 0.
This is the cosine x times 1.
All together I get
cosine of x.
Hooray.
That's the goal.
That's the predicted plan,
desired formula cos x for the
ratio of delta of sine x over
delta x as delta x goes to 0.
Do you see that?
So we used a trig formula,
and we got the sine
right a little late.
Well, of course the reason I--
one reason I goofed was that the
other example, the other
case we need for the
second formula does
have a minus sign.
And it survives in the end.
So I would do exactly the same
thing for the cosines that I
did for the sines.
If there's another board
underneath here,
I'm going to do it.
Yeah, there is.
Now I want to know the delta
of cosine x over delta x.
That's what we do, we have to
simplify that, then we have to
let delta x go to 0.
So what does this mean?
This means the cosine a little
bit along minus the cosine at
the point divided by delta x.
Again, we can't do that division
just right away, but
we can simplify this
by remembering the
formula that cosign--
now let me try to remember it.
It's a cosine times a cosine
for this guy plus a sine--
no, minus a sine
times the sine.
That's the formula that we all
remember and go to sleep with.
Now divide by delta x.
Oh, first subtract cosine x.
So there was a cosine x, so I
want to subtract one of them.
OK?
And now I have to divide
by the delta x.
So I do that there.
I do it here.
And you see that we're in
the same happy position.
We're in the happy position that
as delta x goes to 0, we
know what this does.
That goes to 1.
We know what this does,
or we soon will.
That goes to 0, just the
way they did on the
board that got raised.
So that term disappeared
just like before.
This term survives.
It's got a 1, it's got now a
sine x, and it's got now a
minus sign.
So that's the final result,
that the limit
is minus sine x.
That's the slope of
the cosine curve.
And you wouldn't want
it any other way.
You want that minus sign.
You'll see it with second
derivatives.
So it's just terrific that
those functions, the
derivative of the sine is the
cosine with a plus, the
derivative of the cosine is
the sine with a minus.
OK.
And we've almost proved it, we
just didn't quite pick up this
point yet, and let me do that.
That will finish this lecture.
Why does that ratio
approach zero?
What is that ratio?
That ratio is coming from
the cosine curve.
The cosine curve at 0, the way
this ratio came from the sine
curve at 0.
Here I'm taking--
this is delta cosine.
There's lots of ways I can do
this, but maybe I'll just do
it the way you see it.
What's the slope of
the cosine at 0?
Yeah, I think I can ask that
without doing limits, without
doing hard work.
I'll just add the rest of the
cosine curve, because we know
it's symmetric.
What's the slope
at that point?
This is actually the most
important application of
calculus, is to locate
the place where a
function has a maximum.
The cosine, its maximum
is right there.
Its maximum value is 1, and it
happens at theta equals 0.
So the slope at a maximum--
all right, I'm going
to put this--
I could get this result by these
pictures, but let me do
it short circuit.
The slope at the maximum is 0.
OK.
Your intuition tells you that.
If the slope was positive,
the function
would still be rising.
It wouldn't be a maximum, it
would be going higher.
If the slope was negative, the
function would be coming down,
and the maximum would
have been earlier.
But here the maximum
is right there.
The slope has to be
0 at that point.
And that's the quantity that we
were after, because this is
the cosine of delta x.
There is the cosine
of delta x.
Here is the 1, here is the delta
x, and this ratio is
height over slope.
It gets to height over slope as
we get closer and closer.
It's the derivative, and
it's 0 at a maximum.
And my notes give another way
to convince yourself that
that's 0 by using these facts
that we've already got.
OK.
End of the-- so let me just
recap one moment, which this
board will do.
We now know the derivatives
of two of the great
functions of calculus.
We already know the derivative
of x to the n-th, and in the
future is coming e to the
x and the logarithm.
Then you've got the big ones.
Thank you.
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