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PROF.
JERISON: We're starting
a new unit today.
And, so this is Unit 2, and
it's called Applications
of Differentiation.
OK.
So, the first application, and
we're going to do two today,
is what are known as
linear approximations.
Whoops, that should
have two p's in it.
Approximations.
So, that can be summarized
with one formula,
but it's going to take
us at least half an hour
to explain how this
formula is used.
So here's the formula.
It's f(x) is approximately equal
to its value at a base point
plus the derivative
times x - x_0.
Right?
So this is the main formula.
For right now.
Put it in a box.
And let me just describe
what it means, first.
And then I'll describe
what it means again,
and several other times.
So, first of all,
what it means is
that if you have a
curve, which is y = f(x),
it's approximately the
same as its tangent line.
So this other side is the
equation of the tangent line.
So let's give an example.
I'm going to take the
function f(x), which is ln x,
and then its derivative is 1/x.
And, so let's take the
base point x_0 = 1.
That's pretty much
the only place where
we know the logarithm for sure.
And so, what we plug in
here now, are the values.
So f(1) is the log of 0.
Or, sorry, the log
of 1, which is 0.
And f'(1), well,
that's 1/1, which is 1.
So now we have an
approximation formula which,
if I copy down
what's right up here,
it's going to be ln x is
approximately, so f(0)
is 0, right?
Plus 1 times (x - 1).
So I plugged in here,
for x_0, three places.
I evaluated the coefficients and
this is the dependent variable.
So, all told, if you
like, what I have here
is that the logarithm of
x is approximately x - 1.
And let me draw a
picture of this.
So here's the graph of ln x.
And then, I'll draw in the
tangent line at the place
that we're considering,
which is x = 1.
So here's the tangent line.
And I've separated a
little bit, but really
I probably should have drawn
it a little closer there,
to show you.
The whole point is that
these two are nearby.
But they're not
nearby everywhere.
So this is the line y = x - 1.
Right, that's the tangent line.
They're nearby only
when x is near 1.
So say in this
little realm here.
So when x is approximately
1, this is true.
Once you get a
little farther away,
this straight line,
this straight green line
will separate from the graph.
But near this place
they're close together.
So the idea, again, is that
the curve, the curved line,
is approximately
the tangent line.
And this is one example of it.
All right, so I want to
explain this in one more way.
And then we want to
discuss it systematically.
So the second way that
I want to describe this
requires me to remind
you what the definition
of the derivative is.
So, the definition
of a derivative
is that it's the limit, as delta
x goes to 0, of delta f / delta
x, that's one way of
writing it, all right?
And this is the
way we defined it.
And one of the things that
we did in the first unit
was we looked at this backwards.
We used the derivative knowing
the derivatives of functions
to evaluate some limits.
So you were supposed
to do that on your.
In our test, there were
some examples there,
at least one example,
where that was the easiest
way to do the problem.
So in other words, you can
read this equation both ways.
This is really, of course, the
same equation written twice.
Now, what's new about
what we're going to do now
is that we're going to take
this expression here, delta f
/ delta x, and
we're going to say
well, when delta x
is fairly near 0,
this expression is
going to be fairly
close to the limiting value.
So this is
approximately f'(x_0).
So that, I claim, is the same
as what's in the box in pink
that I have over here.
So this approximation formula
here is the same as this one.
This is an average
rate of change,
and this is an infinitesimal
rate of change.
And they're nearly the same.
That's the claim.
So you'll have various exercises
in which this approximation is
the useful one to use.
And I will, as I said, I'll be
illustrating this a little bit
today.
Now, let me just explain why
those two formulas in the boxes
are the same.
So let's just start over
here and explain that.
So the smaller box is the same
thing if I multiply through
by delta x, as delta f is
approximately f'(x_0) delta x.
And now if I just
write out what this
is, it's f(x),
right, minus f(x_0),
I'm going to write it this way.
Which is approximately
f'(x_0), and this is x - x_0.
So here I'm using the
notations delta x is x - x0.
And so this is the
change in f, this
is just rewriting
what delta x is.
And now the last step is
just to put the constant
on the other side.
So f(x) is approximately
f(x_0) + f'(x_0)(x - x_0).
So this is exactly what I had
just to begin with, right?
So these two are
just algebraically
the same statement.
That's one another
way of looking at it.
All right, so now,
I want to go through
some systematic
discussion here of
several linear approximations,
which you're going
to be wanting to memorize.
And rather than it's being
hard to memorize these,
it's supposed to remind you.
So that you'll have a lot
of extra reinforcement
in remembering
derivatives of all kinds.
So, when we carry out these
systematic discussions,
we want to make
things absolutely as
simple as possible.
And so one of the
things that we do
is we always use the
base point to be x_0.
So I'm always going
to have x_0 = 0
in this standard list of
formulas that I'm going to use.
And if I put x_0 =
0, then this formula
becomes f(x), a little
bit simpler to read.
It becomes f(x)
is f(0) + f'(0) x.
So this is probably
the form that you'll
want to remember most.
That's again, just the
linear approximation.
But one always has
to remember, and this
is a very important thing, this
one only worked near x is 1.
This approximation here really
only works when x is near x_0.
So that's a little addition
that you need to throw in.
So this one works
when x is near 0.
You can't expect it
to be true far away.
The curve can go
anywhere it wants,
when it's far away from
the point of tangency.
So, OK, so let's work this out.
Let's do it for
the sine function,
for the cosine function,
and for e^x, to begin with.
Yeah.
Question.
STUDENT: [INAUDIBLE]
PROF.
JERISON: Yeah.
When does this one work.
Well, so the question was,
when does this one work.
Again, this is when x
is approximately x_0.
Because it's actually the
same as this one over here.
OK.
And indeed, that's
what's going on when
we take this limiting value.
Delta x going to 0 is the same.
Delta x small.
So another way of saying it
is, the delta x is small.
Now, exactly what we mean by
small will also be explained.
But it is a matter to
some extent of intuition
as to how much, how good it is.
In practical cases,
people will really
care about how small it is
before the approximation is
useful.
And that's a serious issue.
All right, so let me carry out
these approximations for x.
Again, this is
always for x near 0.
So all of these are
going to be for x near 0.
So in order to make
this computation,
I have to evaluate the function.
I need to plug in
two numbers here.
In order to get this expression.
I need to know what f(0) is and
I need to know what f'(0) is.
If this is the function f(x),
then I'm going to make a little
table over to the right here
with f' and then I'm going
to evaluate f(0), and
then I'm going to evaluate
f'(0), and then read off
what the answers are.
Right, so first of all if
the function is sine x,
the derivative is cosine x.
The value of f(0),
that's sine of 0, is 0.
The derivative is cosine.
Cosine of 0 is 1.
So there we go.
So now we have the
coefficients 0 and 1.
So this number is 0.
And this number is 1.
So what we get here is 0 + 1x,
so this is approximately x.
There's the linear
approximation to sin x.
Similarly, so now this
is a routine matter
to just read this
off for this table.
We'll do it for the
cosine function.
If you differentiate the
cosine, what you get is -sin x.
The value at 0 is 1, so
that's cosine of 0 at 1.
The value of this
minus sine at 0 is 0.
So this is going back
over here, 1 + 0x,
so this is approximately 1.
This linear function
happens to be constant.
And finally, if I do need e^x,
its derivative is again e^x,
and its value at 0 is 1, the
value of the derivative at 0 is
also 1.
So both of the terms here,
f(0) and f'(0), they're both 1
and we get 1 + x.
So these are the
linear approximations.
You can memorize these.
You'll probably remember them
either this way or that way.
This collection of
information here
encodes the same
collection of information
as we have over here.
For the values of the
function and the values
of their derivatives at 0.
So let me just emphasize again
the geometric point of view
by drawing pictures
of these results.
So first of all, for the sine
function, here's the sine
- well, close enough.
So that's - boy, now that is
quite some sine, isn't it?
I should try to make the two
bumps be the same height,
roughly speaking.
Anyway the tangent line
we're talking about is here.
And this is y = x.
And this is the function sine x.
And near 0, those things
coincide pretty closely.
The cosine function, I'll
put that underneath, I guess.
I think I can fit it.
Make it a little smaller here.
So for the cosine
function, we're up here.
It's y = 1.
Well, no wonder the
tangent line is constant.
It's horizontal.
The tangent line is horizontal,
so the function corresponding
is constant.
So this is y = cos x.
And finally, if I draw y = e^x,
that's coming down like this.
And the tangent line is here.
And it's y = 1 + x.
The value is 1 and
the slope is 1.
So this is how to remember
it graphically if you like.
This analytic picture
is extremely important
and will help you to
deal with sines, cosines
and exponentials.
Yes, question.
STUDENT: [INAUDIBLE]
PROF.
JERISON: The question
is what do you normally
use linear approximations for.
Good question.
We're getting there.
First, we're getting a
little library of them
and I'll give you
a few examples.
OK, so now, I need
to finish the catalog
with two more examples which
are just a little bit, slightly
more challenging.
And a little bit less obvious.
So, the next couple that we're
going to do are ln(1+x) and (1
+ x)^r.
OK, these are the last two
that we're going to write down.
And that you need
to think about.
Now, the procedure is
the same as over here.
Namely, I have to write down
f' and I have to write down
f'(0) and I have to
write down f'(0).
And then I'll have
the coefficients
to be able to fill in
what the approximation is.
So f' = 1 / (1+x), in the
case of the logarithm.
And f(0), if I plug in,
that's log of 1, which is 0.
And f' if I plug
in 0 here, I get 1.
And similarly if I do it for
this one, I get r(1+x)^(r-1).
And when I plug in f(0),
I get 1^r, which is 1.
And here I get r
(1)^(r-1), which is r.
So the corresponding statement
here is that ln(1+x) is
approximately x.
And (1+x)^r is
approximately 1 + rx.
That's 0 + 1x and
here we have 1 + rx.
And now, I do want
to make a connection,
explain to you what's going
on here and the connection
with the first example.
We already did the
logarithm once.
And let's just point out
that these two computations
are the same, or
practically the same.
Here I use the base point
1, but because of my,
sort of, convenient
form, which will end up,
I claim, being much
more convenient
for pretty much
every purpose, we
want to do these things
near x is approximately 0.
You cannot expand the logarithm
and understand a tangent line
for it at x equals 0, because
it goes down to minus infinity.
Similarly, if you
try to graph (1+x)^r,
x^r without the 1 here,
you'll discover that sometimes
the slope is infinite,
and so forth.
So this is a bad
choice of point.
1 is a much better choice
of a place to expand around.
And then we shift things so
that it looks like it's x = 0,
by shifting by the 1.
So the connection with the
previous example is that
the-- what we wrote before I
could write as ln u = u - 1.
Right, that's just recopying
what I have over here.
Except with the letter u
rather than the letter x.
And then I plug in, u = 1 + x.
And then that, if
I copy it down,
you see that I have
a u in place of 1+x,
that's the same as this.
And if I write out u-1,
if I subtract 1 from u,
that means that it's x.
So that's what's on the
right-hand side there.
So these are the
same computation,
I've just changed the variable.
So now I want to try to
address the question that was
asked about how this is used.
And what the importance is.
And what I'm going to do is
just give you one example here.
And then try to emphasize.
The first way in which
this is a useful idea.
So, or maybe this is
the second example.
If you like.
So we'll call this
Example 2, maybe.
So let's just take
the logarithm of 1.1.
Just a second.
Let's take the logarithm of 1.1.
So I claim that, according to
our rules, I can glance at this
and I can immediately see
that it's approximately 1/10.
So what did I use here?
I used that ln(1+x)
is approximately x,
and the value of x
that I used was 1/10.
Right?
So that is the
formula, so I should
put a box around these
two formulas too.
That's this formula here,
applied with x = 1/10.
And I'm claiming that 1/10 is
a sufficiently small number,
sufficiently close to 0,
that this is an OK statement.
So the first
question that I want
to ask you is,
which do you think
is a more complicated thing.
The left-hand side or
the right-hand side.
I claim that this is a
more complicated thing,
you'd have to go to a calculator
to punch out and figure out
what this thing is.
This is easy.
You know what a tenth is.
So the distinction
that I want to make
is that this half, this
part, this is hard.
And this is easy.
Now, that may look
contradictory,
but I want to just do
it right above as well.
This is hard.
And this is easy.
OK.
This looks uglier, but
actually this is the hard one.
And this is giving us
information about it.
Now, let me show
you why that's true.
Look down this column here.
These are the hard
ones, hard functions.
These are the easy functions.
What's easier than this?
Nothing.
OK.
Well, yeah, 0.
That's easier.
Over here it gets even worse.
These are the hard functions
and these are the easy ones.
So that's the main advantage
of linear approximation
is you get something much
simpler to deal with.
And if you've made a
valid approximation
you can make much
progress on problems.
OK, we'll be doing
some more examples,
but I saw some more questions
before I made that point.
Yeah.
STUDENT: [INAUDIBLE]
PROF.
JERISON: Is this
ln of 1.1 or what?
STUDENT: [INAUDIBLE]
PROF.
JERISON: This is a parens there.
It's ln of 1.1, it's the
digital number, right.
I guess I've never used that
before a decimal point, have I?
I don't know.
Other questions.
STUDENT: [INAUDIBLE]
PROF.
JERISON: OK.
So let's continue here.
Let me give you some more
examples, where it becomes
even more vivid if you like.
That this approximation
is giving us something
a little simpler to deal with.
So here's Example 3.
I want to, I'll find the linear
approximation near x = 0.
I also - when I write this
expression near x = 0,
that's the same thing as this.
That's the same thing as
saying x is approximately 0 -
of the function e^(-3x)
divided by square root 1+x.
So here's a function.
OK.
Now, what I claim I want
to use for the purposes
of this approximation,
are just the sum
of the approximation formulas
that we've already derived.
And just to combine
them algebraically.
So I'm not going
to do any calculus,
I'm just going to remember.
So with e^(-3x), it's pretty
clear that I should be using
this formula for e^x.
For the other one, it may be
slightly less obvious but we
have powers of 1+x over here.
So let's plug those in.
I'll put this up so that
you can remember it.
And we're going to carry
out this approximation.
So, first of all, I'm going
to write this so that it's
slightly more suggestive.
Namely, I'm going to
write it as a product.
And there you can
now see the exponent.
In this case, r = 1/2, eh
-1/2, that we're going to use.
OK.
So now I have
e^(-3x) (1+x)^(-1/2),
and that's going to
be approximately--
well I'm going to
use this formula.
I have to use it correctly. x is
replaced by -3x, so this is 1 -
3x.
And then over here,
I can just copy
verbatim the other approximation
formula with r = -1/2.
So this is times 1 - 1/2 x.
And now I'm going to carry
out the multiplication.
So this is 1 - 3x
- 1/2 x + 3/2 x^2.
So now, here's our formula.
So now this isn't
where things stop.
And indeed, in this
kind of arithmetic
that I'm describing
now, things are
easier than they are in
ordinary algebra, in arithmetic.
The reason is that there's
another step, which
I'm now going to perform.
Which is that I'm going to
throw away this term here.
I'm going to ignore it.
In fact, I didn't even
have to work it out.
Because I'm going
to throw it away.
So the reason is
that already, when
I passed from this
expression to this one,
that is from this type
of thing to this thing,
I was already throwing away
quadratic and higher-ordered
terms.
So this isn't the
only quadratic term.
There are tons of them.
I have to ignore
all of them if I'm
going to ignore some of them.
And in fact, I only want to
be left with the linear stuff.
Because that's all I'm really
getting a valid computation
for.
So, this is approximately
1 minus, so let's see.
It's a total of 7/2 x.
And this is the answer.
This is the linear part.
So the x^2 term is negligible.
So we drop x^2 term.
Terms, and higher.
All of those terms
should be lower-order.
If you imagine x is
1/10, or maybe 1/100,
then these terms will end
up being much smaller.
So we have a rather
crude approach.
And that's really
the simplicity,
and that's the savings.
So now, since this unit
is called Applications,
and these are indeed
applications to math,
I also wanted to give you
a real-life application.
Or a place where linear
approximations come up
in real life.
So maybe we'll call
this Example 4.
This is supposedly
a real-life example.
I'll try to persuade
you that it is.
So I like this example because
it's got a lot of math,
as well as physics in it.
So here I am, on the
surface of the earth.
And here is a satellite
going this way.
At some velocity, v.
And this satellite
has a clock on it because
this is a GPS satellite.
And it has a time, T, OK?
But I have a watch, in
fact it's right here.
And I have a time which I keep.
Which is T', And there's
an interesting relationship
between T and T', which
is called time dilation.
And this is from
special relativity.
And it's the following formula.
T' = T divided by the
square root of 1 - v^2/c^2,
where v is the velocity
of the satellite,
and c is the speed of light.
So now I'd like to get a
rough idea of how different
my watch is from the
clock on the satellite.
So I'm going to use
this same approximation,
we've already used it once.
I'm going to write t.
But now let me just remind you.
The situation here is, we
have something of the form
(1-u)^(-1/2).
That's what's happening when
I multiply through here.
So with u = v^2 / c^2.
So in real life, of
course, the expression
that you're going to use
the linear approximation on
isn't necessarily itself linear.
It can be any physical quantity.
So in this case it's v
squared over c squared.
And now the
approximation formula
says that if this is
approximately equal to, well
again it's the same rule.
There's an r and then x
is -u, so this is - - 1/2,
so it's 1 + 1/2 u.
So this is approximately,
by the same rule, this is T,
T' is approximately t
T(1 + 1/2 v^2/c^2) Now,
I promised you that this
would be a real-life problem.
So the question is when people
were designing these GPS
systems, they run clocks
in the satellites.
You're down there, you're
making your measurements,
you're talking to
the satellite by--
or you're receiving its
signals from its radio.
The question is, is this
going to cause problems
in the transmission.
And there are dozens
of such problems
that you have to check for.
So in this case, what
actually happened
is that v is about 4
kilometers per second.
That's how fast the GPS
satellites actually go.
In fact, they had to decide to
put them at a certain altitude
and they could've tweaked
this if they had put them
at different places.
Anyway, the speed of light is
3 * 10^5 kilometers per second.
So this number, v^2 / c^2
is approximately 10^(-10).
Now, if you actually keep
track of how much of an error
that would make in a GPS
location, what you would find
is maybe it's a millimeter
or something like that.
So in fact it doesn't matter.
So that's nice.
But in fact the
engineers who were
designing these systems actually
did use this very computation.
Exactly this.
And the way that
they used it was,
they decided that because
the clocks were different,
when the satellite broadcasts
its radio frequency,
that frequency would be shifted.
Would be offset.
And they decided that the
fidelity was so important
that they would send
the satellites off
with this kind of,
exactly this, offset.
To compensate for the
way the signal is.
So from the point of
view of good reception
on your little GPS device, they
changed the frequency at which
the transmitter
in the satellites,
according to exactly this rule.
And incidentally, the reason
why they didn't-- they ignored
higher-order terms, the
sort of quadratic terms,
is that if you take u^2
that's a size 10^(-20).
And that really is
totally negligible.
That doesn't matter to
any measurement at all.
That's on the order
of nanometers,
and it's not important for
any of the uses to which GPS
is put.
OK, so that's a real example of
a use of linear approximations.
So. let's take a
little pause here.
I'm going to switch
gears and talk
about quadratic approximations.
But before I do that, let's
have some more questions.
Yeah.
STUDENT: [INAUDIBLE]
PROF.
JERISON: OK, so the
question was asked,
suppose I did this
by different method.
Suppose I applied the
original formula here.
Namely, I define
the function f(x),
which was this function here.
And then I plugged
in its value at x = 0
and the value of its
derivative at x = 0.
So the answer is, yes, it's
also true that if I call this
function f f(x), then it
must be true that the linear
approximation is f(x_0) plus
f' of - I'm sorry, it's at 0,
so it's f(0), f'(0) times x.
So that should be true.
That's the formula
that we're using.
It's up there in the pink also.
So this is the formula.
So now, what about f(0)?
Well, if I plug in
0 here, I get 1 * 1.
So this thing is 1.
So that's no surprise.
And that's what I got.
If I computed f',
by the product rule
it would be an annoying,
somewhat long, computation.
And because of
what we just done,
we know what it has to be.
It has to be negative 7/2.
Because this is a
shortcut for doing it.
This is faster than doing that.
But of course, that's a
legal way of doing it.
When you get to
second derivatives,
you'll quickly discover that
this method that I've just
described is
complicated, but far
superior to differentiating
this expression twice.
STUDENT: [INAUDIBLE] PROF.
JERISON: Would you have to
throw away an x^2 term if you
differentiated?
No.
And in fact, we didn't
really have to do that here.
If you differentiate
and then plug in x = 0.
So if you differentiate
this and you plug in x = 0,
you get -7/2.
You differentiate this
and you plug in x = 0,
this term still drops
out because it's just
a 3x when you differentiate.
And then you plug in
x = 0, it's gone too.
And similarly, if you're
up here, it goes away
and similarly over
here it goes away.
So the higher-order terms never
influence this computation
here.
This just captures the linear
features of the function.
So now I want to go on to
quadratic approximation.
And now we're going to
elaborate on this formula.
So, linear approximation.
Well, that should have
been linear approximation.
Liner.
That's interesting.
OK, so that was wrong.
But now we're going to
change it to quadratic.
So, suppose we talk about a
quadratic approximation here.
Now, the quadratic
approximation is
going to be just an elaboration,
one more step of detail.
From the linear.
In other words,
it's an extension
of the linear approximation.
And so we're adding
one more term here.
And the extra term
turns out to be related
to the second derivative.
But there's a factor of 2.
So this is the formula for
the quadratic approximation.
And this chunk of it, of
course, is the linear part.
This time I'll spell
'linear' correctly.
So the linear part
is the first piece.
And the quadratic part
is the second piece.
I want to develop this
same catalog of functions
as I had before.
In other words, I want
to extend our formulas
to the higher-order terms.
And if you do that
for this example here,
maybe I'll even illustrate
with this example
before I go on, if you
do it with this example
here, just to give you a
flavor for what goes on,
what turns out to be the case.
So this is the linear version.
And now I'm going to compare
it to the quadratic version.
So the quadratic version
turns out to be this.
That's what turns out to be
the quadratic approximation.
And when I use
this example here,
so this is 1.1, which is the
same as ln of 1 + 1/10, right?
So that's approximately
1/10 - 1/2 (1/10)^2.
So 1/200.
So that turns out,
instead of being
1/10, that's point, what is it,
.095 or something like that.
It's a little bit less.
It's not .1, but
it's pretty close.
So if you like,
the correction is
lower in the decimal expansion.
Now let me actually
check a few of these.
I'll carry them out.
And what I'm going to
probably save for next time
is explaining to you, so this
is why this factor of 1/2,
and we're going
to do this later.
Do this next time.
You can certainly do well to
stick with this presentation
for one more lecture.
So we can see this reinforced.
So now I'm going to work
out these derivatives
of the higher-order terms.
And let me do it for the
x approximately 0 case.
So first of all, I want to
add in the extra term here.
Here's the extra term.
For the quadratic part.
And now in order to figure
out what's going on,
I'm going to need to compute,
also, second derivatives.
So here I need a
second derivative.
And I need to throw in the value
of that second derivative at 0.
So this is what I'm
going to need to compute.
So if I do it, for example,
for the sine function,
I already have the linear part.
I need this last bit.
So I differentiate the
sine function twice
and I get, I claim
minus the sine function.
The first derivative
is the cosine
and the cosine derivative
is minus the sine.
And when I evaluate it at
0, I get, lo and behold, 0.
Sine of 0 is 0.
So actually the quadratic
approximation is the same.
0x^2.
There's no x^2 term here.
So that's why this is such
a terrific approximation.
It's also the quadratic
approximation.
For the cosine function,
if you differentiate twice,
you get the derivative is
minus the sign and derivative
of that is minus the cosine.
So that's f''.
And now, if I evaluate
that at 0, I get -1.
And so the term that I
have to plug in here,
this -1 is the coefficient
that appears right here.
So I need a -1/2 x^2 extra.
And if you do it for
the e^x, you get an e^x,
and you got a 1 and so
you get 1/2 x^2 here.
I'm going to finish these
two in just a second,
but I first want
to tell you about
the geometric significance
of this quadratic term.
So here we go.
Geometric significance
(of the quadratic term).
So the geometric
significance is best
to describe just by
drawing a picture here.
And I'm going to draw the
picture of the cosine function.
And remember we already
had the tangent line.
So the tangent line was
this horizontal here.
And that was y = 1.
But you can see intuitively,
that doesn't even
tell you whether this function
is above or below 1 there.
Doesn't tell you much.
It's sort of begging for there
to be a little more information
to tell us what the
function is doing nearby.
And indeed, that's what this
second expression does for us.
It's some kind of
parabola underneath here.
So this is y = 1 - 1/2 x^2.
Which is a much better
fit to the curve
than the horizontal line.
And this is, if you like,
this is the best fit parabola.
So it's going to be the
closest parabola to the curve.
And that's more or
less the significance.
It's much, much closer.
All right, I want
to give you, well,
I think we'll save these other
derivations for next time
because I think we're
out of time now.
So we'll do these next time.
