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PROFESSOR: Today we're going
to keep on going
with related rates.
And you may recall that last
time we were in the middle of
a problem with this geometry.
There was a right triangle.
There was a road.
Which was going this way,
from right to left.
And the police were up here,
monitoring the situation.
30 feet from the road.
And you're here.
And you're heading this way.
Maybe it's a two lane highway,
but anyway it's only going
this direction.
And this distance was 50 feet.
So, because you're moving, this
distance is varying and
so we gave it a letter.
And, similarly, your distance
to the foot of the
perpendicular with the
road is also varying.
At this instant it's 40, because
this is a 3, 4, 5
right triangle.
So this was the situation that
we were in last time.
And we're going to pick
up where we left off.
The question is, are you
speeding if the rate of change
of D with respect to t is
80 feet per second.
Now, technically that would be
- 80, because you're going
towards the policemen.
Alright, so D is shrinking
at a rate of
- 80 feet per second.
And I remind you that 95 feet
per second is approximately
the speed limit.
Which is 65 miles per hour.
So, again, this is where
we were last time.
And, got a little question
mark there.
And so let's solve
this problem.
So, this is the setup.
There's a right triangle.
So there's a relationship
between these lengths.
And the relationship is that
x ^2 + 30 ^2 = D ^2.
So that's the first relationship
that we have. And
the second relationship
that we have, we've
already written down.
Which is dx / dt - oops, sorry.
dD / dt = minus 80.
Now, the idea here is relatively
straightforward.
We just want to use
differentiation.
Now, you could solve for x.
Alright, x is the square root
of D^2 - 30 squared.
That's one possibility.
But this is basically
a waste of time.
It's a waste of your time.
So it's easier, or easiest, to
follow this method of implicit
differentiation, which
I want to encourage
you to get used to.
Namely, we just differentiate
this equation
with respect to time.
Now, when you do that, you have
to remember that you are
not allowed to plug
in a constant.
Namely 40, for t.
You have to keep in mind what's
really going on in this
problem which is that x is
moving, it's changing.
And D is also changing.
So you have to differentiate
first before
you plug in the values.
So the easiest thing is to use,
in this case, implicit
differentiation.
And if I do that, I get 2x dx /
dt is equal to 2D (dD / dt).
No more DDT left.
We hope.
Except in this blackboard.
So there's our situation.
Now, if I just plug in,
now I can plug in.
In values. so this is after
taking the derivative.
And, indeed, we have here 2
times the value for x which is
40 at this instant.
And then we have dx / dt.
And that's equal to 2
(D), which is 50.
And then dD / dt is - 80.
So the 80's cancel and we
see that dv / dt = -
100 feet per second.
And so the answer to the
question is yes.
Although you probably wouldn't
be pulled over for this much
of a violation.
So that's, right, it's more than
65 miles an hour, by a
little bit.
So that's the end of
this question.
And usually in these rate of
change or related rates
questions, this is considered
to be the answer to the
question, if you like.
So that's one example.
I'm going to give one more
example before we go on to
some other applications implicit
differentiation.
So my second example is going
to be, you have a it is a
conical tank.
With top of radius 4
feet, let's say.
And depth 10 feet.
So that's our situation.
And then it's being
filled with water.
So, is being filled
with water.
At 2 cubic feet per minute.
So there is our situation.
And then the question is, how
fast is the water rising when
it is at depth 5 feet?
So if this thing is half-full
in the sense, well not
half-full in terms
of total volume.
But half-full in terms of
height, what's the speed at
which the water is rising.
So, how do we set up
problems like this?
Well, we talked about
this last time.
The first step is to set up
a diagram and variables.
And I'm just going to
draw the picture.
And I'm actually going to
draw the picture twice.
So here's the conical tank.
And we have this radius,
which is 4.
And we have this height,
which is 10.
So that's just to allow me to
think about this problem.
Now, it turns out because we
have a varying depth and so
on, and this is just the top.
That I'd better depict also the
level at which the water
actually is at present.
And furthermore, it's better to
do this schematically, as
you'll see.
So the key point here is to
draw this triangle here.
Which shows me the 10 and shows
me the 4, over here.
And then imagine that I'm
filling it partway.
So maybe we'll put that
water level in in
another color here.
So here's the water level.
And the water level, I'm going
to depict that horizontal
distance here, as r.
That's going to be
my variable.
That's the radius of the
top of the water.
So I'm taking a cross-section
of this, because that
geometrically the only thing
I have to keep track of.
At least initially.
So this is our water level.
And it's really rotated
around.
But I'm depicting just this one
half-slice of the picture.
And then similarly,
I have the height.
Which is this dimension there.
Or, if you like, the
depth of the water.
So the water has filled us up,
up to this point here.
So I set it up this way so that
it's clear that we have
two triangles here.
And that one piece of
information we can get from
the geometry is the similar
triangles information.
Namely, that r / h = 4 / 10.
So this is by far the most
typical geometric fact that
you'll have to glean
from a picture.
So that's one piece of
information that we get from
this picture.
Now, the second thing we have
to do is set up formulas for
the volume of water,
and then figure out
what's going on here.
So the volume of water
is, of a cone.
So again, you have to know
something about geometry to do
many of these problems. So you
have to know that the volume
of a cone is 1/3 base
times height.
Now, this one is upside down.
The base is on the top and it's
going down but it works
the same way.
That doesn't affect volume.
So it's 1/3 and the base is
pi r^2, that's the base.
And times h, which
is the height.
So this is the setup
for this problem.
And now, having our
relationship, we have one
relationship left that
we have to remember.
Because we have one
more piece of
information in this problem.
Namely, how fast the
volume is changing.
It's going at 2 cubic
feet per minute.
It's increasing, so that
means that dv / dt = 2.
So I've now gotten rid of all
the words and I have only
formulas left.
I started here with the words,
I drew some pictures, and I
derived some formulas.
Actually, there's one
thing missing.
What's missing?
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: Exactly.
What you want to find.
What I left out is
the question.
So the question is, what
is dh / dt when h = 5?
So that's the question here.
Now, we've got the whole problem
and we never have to
look at it again if you like.
We just have to pay attention
to this piece here.
So let's carry it out.
So what happens here.
So look, you could do this by
implicit differentiation.
But it's so easy to express r
as a function of h that that
seems kind of foolish.
So let's write r as 2/5 h.
That's coming from this
first equation here.
And then let's substitute
that in.
That means that v = 1/3
pi ( 2/5 h^2) h.
And now I have to differentiate
that.
So now I will use implicit
differentiation.
It's very foolish at this point
to take cube roots to
solve for h.
You just get yourself into
a bunch of junk.
So there is a stage at which
we're still using implicit
differentiation here.
I'm not going to try to solve
for h as a function of v.
Instead I'm just differentiating
straight out
from this formula, which
is relatively easy to
differentiate.
So this is dv / dt, which
of course = 2 and if I
differentiate it I just get this
constant, pi / 3, this
other constant, (2/5)^2.
And then I have to differentiate
h^3.
(h^2)h.
So that's 3h^2 ( dh / dt).
That's the chain rule.
So now let's plug in
all of our numbers.
Again, the other theme is, you
don't plug in numbers, fixed
numbers, until everything
has stopped moving.
At this point, we've
already calculated
our rates of change.
So now I can put
in the numbers.
So, 2 = pi 3 ( 2/5)^2( 3), and
then h was 5, so this is 5^2.
And then I have dh / dt.
There's only one unknown thing
left in this problem,
which is dh / dt.
Everything else is a number.
And if you do all the
cancellations, you see that
this cancels this.
One of the 2's cancels - well,
this cancels this.
And then one of the
2's cancels that.
So all-told what we have here
is that dh / dt = 1 / 2 pi.
And so that happens to
be feet per second.
This is the whole story.
Questions, way back there.
STUDENT: [INAUDIBLE]
PROFESSOR: Where did I get --
the question is where did I
get r = 2/5 h from.
The answer was, it came from
similar triangles,
which is over here.
I did this similar
triangle thing.
And I got this relationship
here. r / h = 4/10, but then I
canceled it, got 2/5 and
brought the h over.
Another question, over here.
STUDENT: [INAUDIBLE]
PROFESSOR: The question
was the following.
Suppose you're at this
stage, can you write
from here dv / dh.
So, suppose you're here.
And work out what this is.
It's going to end up being
some constant times h^2.
And then also use dv / dt
= dv / dh ( dh / dt).
Which the chain rule.
And the answer is yes.
We can do that, and indeed
that is what my
next sentence is.
That's exactly what
I'm saying.
So when I said this
- sorry, when you
said this, I did that.
That's exactly what I did.
This chunk is exactly dv / dh.
That's just what I'm doing.
OK.
So, definitely, that's
what I had in mind.
Yeah, another question.
STUDENT: [INAUDIBLE]
PROFESSOR: You're asking
me whether my
arithmetic is right or not?
STUDENT: [INAUDIBLE]
PROFESSOR: Pi - per second.
Oh.
This should - no, OK, right.
I guess it's per minute.
Since the other one
is per minute.
Thank you.
Yes.
Was there another question?
Probably also fixing my
seconds to minutes.
Way back there.
STUDENT: I don't understand,
why did you
have to do all that.
Isn't the speed 2 cubic
feet per minute?
PROFESSOR: The speed at which
you're filling it is 2 cubic
feet, but the water level is
rising at a different rate,
depending on whether you're
low down or high up.
It depends on how wide the
pond, the surface, is.
So in fact it's not.
In fact, the answer wasn't
2 cubic - it wasn't.
There's a rate there.
That is, that's how much
volume is being added.
But then there's another number
that we're keeping
track of, which is the height.
Or, if you like, the
depth of the water.
OK.
So this is the whole point
about related rates.
Is you have one variable,
which is v, which
is volume of something.
You have another variable,
which is h, which is the
height of the cone
of water there.
And you're keeping track
of one variable in
terms of the other.
And the relationship will always
be a chain rule type of
relationship.
So, therefore, you'll be able to
- if you know one you'll be
able to figure out the other.
Provided you get all of
the related rates.
These are what are called
related rates.
This is a rate of something
with respect to
something, etc. etc.
So it's really all about
the chain rule.
And just fitting it to word
problems. So there's a couple
of examples.
And I'll let you work
out some more.
So now, the next thing that I
want to do is to give you one
more max-min problem.
Which has to do with this
device, which I
brought with me.
So this is a ring.
Happens to be a napkin ring,
and this is some parachute
string, which I use when
I go backpacking.
And the question is if you have
a - you think of this is
a weight, it's flexible.
It's allowed to move
along here.
And the question is, if I fix
these two ends where my two
hands are, where my right hand
is here and where my left hand
is over there.
And the question is, where does
this ring settle down.
Now, it's obvious, or should be
maybe obvious, is that if
my two hands are at equal
heights, it should settle in
the middle.
The question that we're trying
to resolve is what if one hand
is a little higher
than the other.
What happens, or if
the other way.
Where does it settle down?
So in order to depict this
problem properly, I need two
volunteers to help me out.
Can I have some help?
OK.
So I need one of you to hold the
right side, and one of you
to hold the left side.
OK.
And just hold it against
the blackboard.
We're going to trace.
So stick it about here, in
the middle somewhere.
And now we want to make sure
that this one is a little
higher, alright?
So we'll have to - yeah, let's
get a little higher up.
That's probably good enough.
So the experiment
has been done.
We now see where
this thing is.
But, so now hold on tight.
This thing stretches.
So we want to get it stretched
so that we can
see what it is properly.
So this thing isn't heavy enough
for this demonstration.
I should've had a ten-ton
brick attached there.
But if I did that, than I would
tax my, right, I would
tax your abilities to.
Right, so we're going to try
to trace out what the
possibilities are here.
So this is, roughly speaking,
where this thing - and so now
where does it settle.
Well, it settles about here.
So we're going to put that
as x marks the spot.
OK, thank you very much.
Got to remember where
those dots --
OK, all right.
Sorry, I forgot to
mark the spots.
OK, so here's the situation.
We have a problem.
And we've hung a string.
And it went down like this and
then it went like that.
And we discovered where
it settled.
Physically.
So we want to explain this
mathematically, and see what's
going on with this problem.
So this is a real-life
problem.
It honestly is the problem you
have to solve if you want to
build a bridge.
Like, a suspension bridge.
This, among many problems.
It's a very serious and
important problem.
And this is the simplest
one of this type.
So we've got our shape here.
This should be a straight
line, maybe not quite as
angled as that.
The first, we've already drawn
the diagram and we've more or
less visualized what's
going on here.
But the first step after the
diagram is to give letters.
Is to label things
appropriately.
And I do not expect you
to be able to do
this, at this stage.
Because it requires a
lot of experience.
But I'm going to
do it for you.
We're going to just do that.
So the simplest thing
to do is to use the
coordinates of the plane.
And if you do that, it's also
easiest to use the origin.
My favorite number is 0 and
it should be yours, too.
So we're going to make
this point be (0, 0).
Now, there's another fixed
point in this problem.
And it's this point over here.
And we don't know what
its coordinates are.
So we're just going to give
them letters, a and b.
But those letters are
going to be fixed
numbers in this problem.
And we want to solve it for
all possible a's and b's.
Now, the interesting thing,
remember, is what happens when
b Is not 0.
If b = 0, we already have a clue
that the point should be
the center point.
It should be exactly that x,
the middle point, which I'm
going to label in a second,
is halfway in between.
So now the variable point that
I'm going to use is down here.
I'm going to call this
point (x, y).
So here's my setup.
I've now given labels
to all the things on
the diagram so far.
Most of the things
on the diagram.
So now, what else
do I have to do?
Well, I have to explain
to you that this is a
minimization problem.
What happens, actually,
physically is that the weight
settles to the lowest point.
That's the thing that has the
lowest potential energy.
So we're minimizing
a function.
And it's this curve here.
The constraint is that we're
restricted to this curve.
So this is a constraint curve.
And we want the lowest
point of this curve.
So now, we need a little bit
more language in order to
describe what it is
that we've got.
And the constraint curve, we
got it in a particular way.
Namely, we strung some string
from here to there.
And what happens at all these
points is that the total
length of the string
is the same.
So one way of expressing the
constraint is that the length
of the string is constant.
And so in order to figure out
what the constraint is, what
this curve is, I have to
describe that analytically.
And I'm going to do that by
drawing in some helping lines.
Namely, some right triangles
to figure out
what this length is.
And what the other length is.
So this length is pretty
easy if I draw a
right triangle here.
Because we go over x
and we go down y.
So this length is the square
root of x^2 + y^2.
That's the Pythagorean
theorem.
Similarly, over here, I'm going
to get another length.
Which is a little
bit of a mess.
It's the vertical.
So I'm just going to label one
half of it, so that you see.
So this horizontal
distance is x.
And this horizontal distance
from this top point with this
right angle, over there.
It starts at x and ends at a.
The right-most point is
a in the x coordinate.
So the whole distance
is a - x.
So that's this leg of
this right triangle.
And, similarly, the vertical
distance will be b - y.
And so, the formula here, which
is a little complicated
for this length, is the square
root of (a - x)^2 + (b - y)^2.
So here are the two formulas
that are going to allow me to
set up my problem now.
So, my goal is to set
it up the way I did
here, just with formulas.
And not with diagrams
and not with names.
So here's what I'd like to do.
I claim that what's constrained,
if I'm along that
curve, is that the total
length is constant.
So that's this statement here.
The square root of x^2 + y^2
+ this other square root.
These are the two lengths
of string.
Is equal to some number,
L, which is constant.
And this, as I said, is what
I'm calling my constraint.
Yeah.
STUDENT: [INAUDIBLE]
PROFESSOR: So the question is,
shouldn't it be b plus y.
No, and the reason is that
y is a negative number.
It's below 0.
So it's actually the sum, -
y is a positive number.
Alright, so here's
the formula.
And then, we want to find the
minimum of something.
So what is it that we're
finding the minimum of?
This is actually the hardest
part of the problem,
conceptually.
I tried to prepare it,
but it's very hard
to figure this out.
We're finding the least what?
It's y.
We got a name for that.
So we want to find
the lowest y.
Now, the reason why it seems
a little weird is you might
think of y as just
being a variable.
But really, y is a function.
It's really y = y ( x ) is
defined implicitly by the
constraint equation.
That's what that curve is.
And notice the bottom point is
exactly the place where the
tangent line will
be horizontal.
Which is just what we want.
So from the diagram, the bottom
point is where y' = 0.
So this is the critical point.
Yeah?
STUDENT: [INAUDIBLE]
PROFESSOR: Exactly.
So I'm deriving for you, so the
question is, could I have
just tried to find y'
= 0 to begin with.
The answer is yes, absolutely.
And in fact I'm leading
in that direction.
I'm just showing you, so I'm
trying to make the following,
very subtle, point.
Which is in maximum-minimum
problems, we always have to
keep track of two things.
Often the interesting point
is the critical point.
And that indeed turns out
to be the case here.
But we always have to
check the ends.
And so there are several ways
of checking the ends.
One is, we did this
physical problem.
We can see that it's
coming up here.
We can see that it's
coming up here.
Therefore the bottom must be
at this critical point.
So that's OK, so that's one
way of checking it.
Another way of checking it is
the reasoning that I just
gave. But it's really
the same reasoning.
I'm pointing to this thing and
I'm showing you that the
bottom is somewhere
in the middle.
So, therefore, it is a place
of horizontal tangency.
That's the reasoning
that I'm using.
So, again, this is to avoid
having to evaluate a limit of
an end or to use the second
derivative test, which is a
total catastrophe in this case.
OK, now.
There's one other thing that you
might know about this if
you've seen this geometric
construction before.
With a string and chalk.
Which is that this curve
is an eclipse.
It turns out, this is a
piece of an eclipse.
It's a huge ellipse.
These two points turn
out to be the
so-called foci of the ellipse.
However, that geometric fact is
totally useless for solving
this problem.
Completely useless.
If you actually write out the
formulas for the ellipse,
you'll discover that you
have a much harder
problem on your hands.
And it will take you twice or
ten times as long, so, it's
true that it's an ellipse,
but it doesn't help.
OK, so what we're going to do
instead is much simpler.
We're going to leave this
expression along and we're
just going to differentiate
implicitly.
So again, we use implicit
differentiation on the
constraint equation.
So that's the equation which
is directly above me
there, at the top.
And I have to differentiate
it with respect to x.
So that's a little ugly,
but we've done
this a few times before.
When you differentiate a square
root, the square root
goes into the denominator.
And there's a factor of 1/2, so
there's a 2x which cancels.
So I claim it's this.
Now, because y depends on x,
there's also a y y' here.
So technically speaking, it's
twice this with a half.
2 / 2 times that.
So that's the differentiation
of the first
piece of this guy.
Now I'm going to do this second
one, and it's also the
chain rule.
But you're just going to have
to let me do it for you.
Because it's just a little bit
too long for you to pay
attention to.
It turns out there's a minus
sign that comes out, because
there's a - x and a - y there.
And then the numerator, the
denominator is the same
massive square root.
So it's (a - x)^2 ( b - y)^2.
And the numerator is a
- x, which is what
replaces the x over here.
And then another term,
which is (b - y)y'.
I claim that that's analogous
to what I did
in the first term.
And you'll just have to check
this on your own.
Because I did it too
fast for you to be
able to check yourself.
Now, that's going to
be equal to what on
the right-hand side.
What's the derivative of
L with respect to x?
It's 0.
It's not changing
in the problem.
Although my parachute
stuff stretches.
We tried to stretch it to
its fullest extent.
So that we kept it fixed,
that was the goal here.
So now, this looks like
a total mess.
But, it's not.
And let me show you why.
It simplifies a great deal.
And let me show you
exactly how.
So, first of all, the whole
point is we're looking for the
place where y' = 0.
So that means that these
terms go away. y' = 0.
So they're gone.
And now what we have is the
following equation.
It's x / square root of x^2 +
y^2 squared is equal to, if I
put it on the other side the
minus sign is changed to a
plus sign. a - x divided by this
other massive object, (a
- x)^2 + (b - y)^2.
So this is what it's
simplifies to.
Now again, that also might
look complicated to you.
But I claim that this is
something, this is a kind of
equilibrium equation that can be
interpreted geometrically,
in a way that is very meaningful
and important.
So first of all, let me observe
for you that this x is
something on our picture.
And the square root
of x^2 + y^2 is
something on our picture.
Namely, if I go over to the
picture, here was x and this
was a right triangle.
And this hypotenuse was square
root of x^2 + y^2.
So, if I call this angle
alpha, then this
is the sine of alpha.
Right?
It's a right triangle, that's
the opposite leg.
So this guy is the
sine of alpha.
Similarly, the other side has
an interpretation for the
other right triangle.
If this angle is beta, then the
opposite side is a - x,
and the hypotenuse is what was
in the denominator over there.
So this side is sine of beta.
And so what this condition
is telling us is
that alpha = beta.
Which is the hidden symmetry
in the problem.
I don't know if you can actually
see it when I show
you this thing.
But, no matter how I tilt it,
actually the two angles from
the horizontal are the same.
In the middle it's kind
of obvious that that
should be the case.
But on the sides it's
not obvious that
that's what's happening.
Now, this has even, so that's
a symmetry, if you like, of
the situation.
These two angles are equal.
But there's something
more to be said.
If you do a forced diagram for
this, what you'll discover is
that the tension on the
two lines is the same.
Which means that when you
build something which is
hanging like this, it will
involve the least stress.
If you hang something very
heavy, and one side carries
twice as much load as the other,
then you have twice as
much chance of its falling
and breaking.
If they each hold the same
strength, then you've
distributed the load in a
much more balanced way.
So this is a kind of a balance
condition, and it's very
typical of minimization
problems. And fortunately,
there are nice solutions which
distribute the weight
reasonably well.
That's certainly the principal
of suspension bridges.
So yeah, one more question.
STUDENT: [INAUDIBLE]
PROFESSOR: OK, so the question
where it hangs, that is, what
the formula for x is and what
the formula for y is.
And other things, like the
equation for the ellipse and
lots of other stuff like that.
Those are things that I will
answer for you in a set of
notes which I will hand out.
And they're just a mess.
You see they're, just as in that
other problem that we did
last time, there's some
illuminating things you can
say about the problem and then
there's some messy formulas.
You know, you want to try to
pick out the simple things
that you can say.
In fact, that's a property of
math, you want to focus on the
more comprehensible things.
On the other hand,
it can be done.
It just takes a little
bit of computation.
So I didn't answer the
question of what
the lowest y was.
But I'll do that for you.
Maybe I'll just mention one more
thing about this problem.
I'm using problems
from a completely
different point of view.
If you sort of roll the ellipse
around, you get the
same phenomenon from
each place here.
So it doesn't matter
where a and 0 are.
You'll get the same
phenomenon.
That is, the tangent line.
So, this angle and that
angle will be equal.
So you can also read that as
being the angle over here and
the angle over here are equal.
That is, beta, that is
the complementary
angles are also equal.
And if you interpret this as a
ray of light, and this as a
mirror, then this would be
saying that if you start at
one focus, every ray of light
will bounce and go to the
other focus.
So that's a property that
an ellipse has.
More precisely, a property that
this kind of curve has.
And in fact, a few years ago
there was this great piece of
art at something called the
DeCordova Museum, which I
recommended very highly to you
go sometime to visit in your
four years here.
There was a collection of
miniature golf holes.
So they had a bunch of mini
golf pieces of art.
And every one was completely
different from the other.
And one of them was called
hole in one.
And the tee was at one
focus of the ellipse.
And the hole was at the other
focus of the ellipse.
So, no matter how you hit
the golf ball, it always
goes into the hole.
No matter where it bounces,
it just, one bounce
and it's in the hole.
So that's actually a consequence
of the computation
that we just did.
Time to go on.
We're going to now talk
about something else.
So our next topic is
Newton's method.
Which is one of the greatest
applications of calculus.
And I'm going to describe
it here for you.
And we'll illustrate it on an
example, which is solving the
equation x^2 = 5.
We're going to find the
square root of 5.
Now, you can actually solve
any equation this way.
Any equation that you
understand, you can solve this
way, essentially.
So even though I'm doing it
for the square root of 5,
which is something that you
can figure out on your
calculator, in fact this is
really at the heart of many of
the ways in which calculators
work.
So, the first thing is to
make this problem a
little bit more abstract.
We're going to set f
( x ) = x^2 - 5.
And then we're going
to solve f (x) = 0.
So this is the sort of
standard form for
solving such a.
So you take some either
complicated or simple function
of x, linear functions are
boring, quadratic functions
are already interesting.
And cubic functions, as I've
said a few times, you don't
even have formulas
for solving.
So this would be the only method
you have for solving
them numerically.
So here's how it works.
So the idea, I'll plot
this function.
Here's the function, it's
a parabola, y = x^2 - 5.
It dips below 0, sorry,
it should be
centered, but anyway.
And now the idea here is
to start with a guess.
And square root of 5 is pretty
close to the square root of 4,
which is 2.
So my first guess is going
to be 2, here.
So start with initial guess.
So that's our first guess.
And now, what we're going to do
is we're going to pretend
that the function is linear.
That's all we're going to do.
And then if the function were
linear, we're going to try to
find where the 0 is.
So if the function is linear,
what we'll use is we'll plot
the point where 2 is on, that
is, the point (2, f ( 2 )),
and then we're going to draw
the tangent line here.
And this is going to be
our new guess. x = x,
which I'll call x1.
So the idea here is that this
point may be somewhat far from
where it crosses, but this point
will be a little closer.
And now we're going to do this
over and over again.
And see how fast it gets to the
place we're aiming for.
So we have to work out what
the formulas are.
And that's the strategy.
So now, the first step here is,
we have our guess, we have
our tangent line.
Which has the formula
y - y0 = m(x - x0).
So that's the general form
for a tangent line.
And now, I have to tell
you what x1 is.
In terms of this tangent line.
x1 is the x intercept.
The tangent line, if you look
over here at the diagram, the
tangent line is the
orange line.
Where that crosses the axis,
that's where I want to put x1.
So that's the x intercept.
Now, how do you find
the x intercept?
You find it by setting y = 0.
That horizontal line is y = 0.
So I set y = 0, and I get
0 - y0 = m (x1 - x0).
So I changed two things
in this equation.
I plugged in 0 here, for y.
And I said that the place where
that happens is going to
be where x is x1.
So now let's solve.
And what we get here is -
y divided by - sorry,
-y0 / m = x1 - x0.
And now I can get a
formula for x1.
So x1 = x0 - (y0 / m).
I now need to tell you what this
formula means, in terms
of the function f.
So first of all, x0 is
x0, whatever x0 is.
And y0, I claim, is f (x0).
And m is the slope at
that same place.
So it's f' (x0).
And this is the whole story.
This is the formula which will
enable us to calculate
basically any root.
I'm going to repeat this
formula, so I'm going to tell
you again what Newton's method
is, and put a little more
colorful box around it.
So Newton's method involves
the following formula.
In order to get the n + 1st
point, that's our better and
better guess, I'm going to take
the nth one and then I'm
going to plug in this formula.
So f ( xn ) / f' ( xn ).
So this is the basic formula,
and if you like, this is the
idea of just repeating
what I had before.
Now, we've gone from geometry,
from just pictures, to an
honest to goodness formula
which is completely
implementable and very easy
to implement in any case.
So let's see how it works in
the case that we've got.
Which is x0 = 2.
f(x) = x^2 - 5.
Let's see how to implement
this formula.
So first of all, I have to
calculate for you, f' (x).
That's 2x.
And so, x1 = x0 -, so
f' , sorry, f ( x )
would be x0^2 - 5.
That's what's in
the numerator.
And in the denominator
I have the
derivative, so that's 2 x0.
And so all told, well, let's
work it out in two steps here.
This is - 1/2 x0 for the first
term, and then + (5/2 / x0)
for the second term.
And these two terms combine, so
we have here 1/2 x0 + 5/2
with an x0 in the denominator.
So here's the formula for
x1. in this case.
Now I'd like to show you
how well this works.
So first of all, we have x1 =
1/2 ( 2), if I plug in x1, +
5/4, which is 9/4.
And x2, I have 1/2 (
9/4) + 5/2 ( 4/9).
That's the next one.
And if you work this
out, it's 161/72.
And then x3 is kind of long.
But I will just write down what
it is, so that you can
see it's 1/2 ( 161/72) +
5/2 and then I do the
reciprocal of that.
So 72/161.
So let's see how
good these are.
I carefully calculated
how far off they are.
Somewhere on my notes.
So I'll just take a look
and see what I said.
Oh yeah, I did do it.
So, what's the square root of 5
minus - so here's n, here's
the square root of 5
- xn, if you like.
Or the other way around.
The size of this.
You'll have to decide on your
homework whether it comes out
positive or negative,
to the right or to
the left of the answer.
But let's do this.
So when n = 0, the
guess was 2.
And we're off by about
2 ( 10 ^ - 1).
And if n = 1, so that's this
9/4, that's off by
about 10 ^ - 2.
And then n = 2, that's this
number here, right?
And that's off by about
4 ( 10 ^ - 5).
That's already as good an
approximation to the square
root of 5 as you'll ever
need in your life.
If you do 3, this number here
turns out to be accurate to 10
^ - 10 or so.
This goes right off
to the edge of my
calculator, this one here.
So already, with the third
iterate. you're of way past
the accuracy that you need
for most things.
Yep, question.
STUDENT: [INAUDIBLE]
PROFESSOR: How come
the x0 disappears?
STUDENT: [INAUDIBLE]
PROFESSOR: So, from here
to here [INAUDIBLE]
STUDENT: [INAUDIBLE]
PROFESSOR: Hang on, folks,
we have a question.
Let's just answer it.
So here we have an x0, and here
we have - 1/2, there's an
x0^2 and an x which cancel.
And here we have a minus,
and a - 5/2 x0.
So I combine the 1 - 1/2,
I got + 1/2, that's all.
OK?
Alright, thanks.
We'll have to ask other
questions after class.
