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Hopefully the results of the previous session bi

Have you reviewed these results and there

We were there:a equals f x y z

Requirements for the extreme value function

condition of the first order derivatives to zero qualification is

If the condition of the critical point nature gives us.

Top half, great value, the minimum value or whether it is a saddle point.

Here, too, the determinant of the second order

the determinant obtained from derivatives important.

If this mark increased the maximum value I might.

May be the smallest value.

The way it can differentiate with respect to x look at the second order derivatives.

If you increase again the opposite of our intuition will be the smallest value is bi,

because if the count minus the minimum value It feels like I need to be.

However, the opposite of surplus value.

This is also the largest value minus the value see that.

If this determinant is now less than zero f

because we do not look for the sign of x x It does not matter.

This is the only type of plus or minus sign not be able to maintain.

Then becomes a saddle point.

Now we see the results of this second We will get through to prove.

The idea here is different.

Y is zero, x is zero a Taylor series We used to be around that.

Around the x minus x is zero when the judges

The term of the term before, but it multiplier is zero We have seen that it should be.

Then squared term, the second We are looking at very term.

Second-order terms, because the judges most active terms.

For example, the term cube larger than x minus x is zero when it is small.

There approaches the main logic x is zero y series is open around zero.

Here, the chain will use derivatives.

Now we say f x y z equals

the largest and smallest points on Let's take.

Suppose now that on the surface b You are at the point where

all curves passing through infinitely many There are curves.

Any one of these typical bi so y is equal to x on y

complete with a chain rule for derivatives Let derivative account.

Now you see it does not do x to y If you put it in terms of function

Effective basically univariate turned to the function would be.

Here we are taking advantage of this.

This two-variable function in any chain derivatives along the curve

a function of one variable, while the full Coming to calculate derivatives business.

Because the function of full-derived single We multivariate download.

Now here's the chain derivative rule us from x to complete the partial derivatives of these derivatives

According to y or x and y are partial derivatives of shows that the derivative.

Of functions of one variable that we We know that there must be zero.

So the problem now univariate We take the function indirgere.

That any of this to be zero along a curve and thus

can be provided for all curves The only cure for this partial with respect to x

y is derived based on the partial derivatives zero is.

It is no coincidence that as the years Base bi You can select the

it without every single one of them is zero You can reset.

But even if all the curves, that for There is the possibility of providing one.

He zeros of the first order derivatives be.

As you can see Taylor series obtained by The results of the functions only

variable bi course artificially bringing it to function any

along the curve, any curve surface but we can account for

All curves mean any curve means for achieving it.

Here we find this first condition.

In the second condition the same thing.

Gene functions of one variable in the second If higher order derivatives

the minimum value, if the axis of greatest did you know that.

The result is equal to zero in the b could not shake.

We keep our sake he also Think.

Means to calculate the second order derivative of full first order

complete the full derivative of the derivative one more time say you get.

E, to take full derivative know what it was.

BI units will take partial derivatives with respect to x.

This size is the same for the full derivative d * Accounts that have this.

Plus the same size alone see here

partial derivatives of the same size according to y will take.

We will then multiply this number by the base year.

As you can see here, the second derivative f x X, we have calculated the term.

x has y'l terms.

Because here we opened the second with respect to x future derivatives.

Get out of here by derivatives with respect to x one more time x y'l next term.

See this.

From here, the derivative with respect to x to y is one more time

When you take the derivative with respect to a more x y'l next term.

These two going for him here.

Here you see the second derivative of y by bi y

base got here, b y base have here.

y are the square of the base year in terms y'l.

See also here Bi derivative of y

means to get by here, where a product There are, finally, y

accordingly receive derivative derivative

accordingly take, but by the derivative of y

to get the derivative thereof with respect to x get it to x

One more by the derivative of y with respect to x

from there take the derivative of y two üssül The term is coming.

Now we want it to be a plus.

Look for it to be plus all of these The term is supposed to be the pros.

Now here again to complete the square

'll get out of it by means of f x x.

Thus one, stays here.

We split it.

We split it.

two year term üssül fell.

Fell due to the following:Because here requirements in the critical condition

In none of these terms is zero at the point x is zero y.

So this year fell two üssül terms.

Therefore, only had these three terms backwards.

As you can see here b y üssül with customers is this

If we take the first karen b here again since there.

One plus of this term, the coefficient half Get y

While this frame with a base hit the first two The term produce.

Wherein a plus b is a square.

B twice this term.

We also produce one extra term, but bi.

This square of this term.

Here again is not correct to say coincidence.

Structure as a result of base frames b y is there.

And that there are frame c.

Coefficient in the first.

It also would remove.

The common denominator here when we bought the frame is there.

Therefore, these coefficients are multiplied.

Yet as you can see the full second order derivatives plus

To be valuable, this coefficient be zero,

is greater than zero and the f x x of gene must be greater than zero.

This we found earlier, Taylor as a series of

As a result, we found there should be exactly the same.

The maximum value of one variable gene We know from the function.

The second order derivatives must be zero.

Sorry, must be less than zero.

Ensure that it is less than zero for

we can do it in square brackets there is nothing.

This should be valuable plus.

Because it minus sign not possible to guarantee.

This must be pros cons of de f x x that is enough.

This gives sufficient conditions.

Now of functions of one variable bi optimization job.

So finding the largest and smallest values often very useful, important.

Physical problems such as the energy

The minimum value of the smallest events developing.

Or some technology issues the smallest of things

We want to be or to be the biggest we want.

In addition also some surfaces B one variable in understanding the critical

points, end points of a curve was useful in understanding the nature of our business.

The genes in these two variable function critical

An important contribution of the surface points to figure offers.

We will see two examples in this regard.

Yet here I summarize these results.

Ded, the first order derivatives of the zero finds, wherein X

We will find y is zero zero two unknown From the two equations.

The second determinant're looking at doing this.

What you get is the determinant axis can be a maximum.

What may be the biggest what the smallest value.

Here you will saddle point.

B do not need to look into other things.

But the greatest value of this determinant is increased may be, may be the smallest value.

Our intuition to determine that it still In contrast to me, at least as it is,

minus the minimum value if the count

seems like there is always of a student well since.

But the accounts show exactly the opposite.

These coefficients are looking for.

Now let's see examples.

I do not want to see too many examples.

Very complex iii.

I once let's start from the basic surface.

See a paraboloid x squared plus y squared.

We saw it in te first lesson.

Paraboloid facing upward.

If you say y f x y z from zero.

x upwards of z'yl outgoing bi parabo, parabola.

If you say that x is zero vertical them with the cross-section plane is to take.

This is still facing upward parabola parabola.

Therefore, if you get what sections such such as bi Cup

cup mouth upward angle, facing as trophies.

See if you put a minus sign each time at

in each section of a parabola facing down.

So this creates a hill.

Plus minus plus or minus signs here While the paraboloid is happening.

We call them because elliptic paraboloid There is a circle here.

z is equal to zero if you choose not hard bi plus valuable

x squared plus y squared bi bi hard if you choose the circle is going on.

I.e. with the plane of constant z is When we cut hoops

're getting a special ellipse and circle state.

Elliptic Paraboloid them going.

This exceeds, one facing one up facing down.

When we look to see was plus or no plus or minus signs axis

geometry, math or something thinking two If you can mark both plus.

Both may be missing.

Here may be minus one plus other is

or as the first plus minus other I might.

These are the most basic quadratic us showing surface.

As second-degree surfaces obtained at b can be:

X, in a second degree of y multiplied function.

Or it may be less marked.

These surfaces are very basic surfaces.

How in the single-valued function before We learn a correct equation.

After that, we learn the parabola, We learn hyperbole.

We learn that functions like.

This in these two variables counterparts.

Now just tell me where.

Let's find the critical points of this function.

derivatives with respect to x two x, y-derivative two years.

X is zero and y is the reset them to zero.

It is understood to be zero.

Here, let the second order derivatives.

f x x, f y is going on two.

If you take the derivative of f x to y here Since there is no zero year.

If you take the derivative of f with respect to x of y where In the absence of the x zero.

So those on the second diagonal becomes zero.

This determinant when he calculated two times two four.

Here an additive to be zero, does not come.

Means that four surplus.

Now this may be the greatest value, the may be little value.

But for x wherein x is two, plus It is valuable because it is the smallest value.

Let's look at the geometry of it right now.

This is a clear upward this kind of work paraboloid.

Any particular segment thereof open up a parabola.

we take the horizontal plane z equals z

wherein the hoops for surplus will be released.

Or are they equal level lines.

Let vice versa.

Minus the ancient marked.

See again the first order derivatives of the

critical point is zero at zero I understand.

Because these unknowns x and y, both the equation.

This f x is equal to zero, f y is equal to zero.

We put x to take them to the determinant the second derivative will be minus two.

According to the second derivative of y will be minus two.

We put them on the diagonal.

derivatives with respect to x to y of the receipt of the derivatives will be zero.

Because it has no year is here.

Hence the definition of a partial derivative x doing the hard tasks.

X and y from one another, or another perspective independent variables.

F x y is zero for him.

zero for x in y.

According to theorem have already become equally well.

These are zero.

But on the plus plus had previous bi.

You turn the determinant was four.

Here is a minus to minus four determinants involved again.

So here you be the largest value of There you have the smallest value of b.

No saddle point.

f x x're looking at here are eksiç.

The biggest minus is worth.

In the previous problem, so this is f x x plus the two were increasing.

Was greater than zero.

Then came out the smallest value.

In contrast, the surfaces are as follows: Paraboloid facing down.

The vertical sections in this instance no x zero

If you say negative y z equals the square down facing parabola.

If you receive it in the horizontal plane sections such a

Think of it as a half watermelon thing again.

It has cut circles with a horizontal plane Remove according to your location.

Cut with the vertical plane of the parabola interests.

Here is a paraboloid, but it would also paraboloid facing down.

Now we have the job, plus it marks

may be plus or minus minus or plus or minus I might.

Here it is the third type of plus minus.

When we look at this f x two x's going on.

year for two years is going negative.

Look in front of you so negative.

X is zero to zero by them again eşitleyin equal y equals zero.

Zero.

Yet the critical point x is zero mean and At points where y is zero.

But again, we calculate the determinant of f x x, f y and wherein

f x x two, for y minus two mixed derivatives zero.

derivative with respect to x and y is not in them.

So when calculating the determinant here minus four.

What's the maximum of what can be here anymore may be minimal.

Here b is a saddle point.

As we look at this, it becomes bi evil: When we see that y y Collapse

zero, it means that we take the x-axis a section along.

z equals x squared involved.

z is equal to x squared upwards outgoing parabola.

Similarly, we say x is zero at this time y direction

involved a parabola facing down.

So what kind of bi If you cut in the previous

Domestic definite upward or downward pointing was parabola.

They are cutting it with the horizontal plane See here for some x and y, we fixed

If you say, x squared minus y squared of the b hyperbola equation.

If you see that x is a constant increase to cut.

Because let's say a plus.

x When you make zero one year can not provide because it is here minus y squared.

So it along the x axis plus with those cuts.

the negative values along the y axis cuts.

These are as hyperbolic saddle point shows the geometry around.

Conversely minus signs, plus we take we, bi had received in the previous plus or minus.

Here, too, is still missing as you can see f x will be.

Minus two x's.

f y will be the notion of residual.

Two years.

They again to zero when occurs at a critical point zero zero.

Take the second order derivatives f x x minus y plus two for two.

Snow, mixed derivatives is zero.

Because it is always the same event is repeated again.

When we take the derivative with respect to y that zero interests.

On top of this, we take the derivative with respect to x zero.

These determinants minus.

There are means in which a saddle point.

BI is exactly the same as before.

That is not entirely sure.

You turn on the other plus was that.

You turn the second was negative.

But all in all still negative determinants You turn was.

What we have here now zero, we take x to y parabola pointing downward direction.

See this bi parabola facing down.

As this is a valley.

If you like this aşıyos two sides of the valley valley is there.

Please aşıyos bi hill in the valley.

Bi bi valley, on the hill.

Yet it is equally hard for our horizontal ptane

fragments, which contour lines of equal We call value lines.

It will be like.

If you compare bi alone as before z

intersection with X is not in surplus value.

So herewith a, the intersection will be.

y'yl which is also a plus

values the bus, where this time intersection will be.

But here may appear simpler than y zero means to say

y is equal to zero, the section vertical plane a parabola facing down.

Nothing like this will be a parabola.

Conversely, x is equal to zero if you receive this x equals

an upwardly in the zero to y parabola.

This is contrary've received a saddle You put turn.

Its like he Siya b.

If it reverses the previous bi interests.

Or contour lines of equal value if it b is rotated 45 degrees to the previous.

Them a little thinking for yourself you can easily find what matters.

These are the main surfaces.

See you there as we take x y Now Siya b interesting turns.

x, f x y the time we get involved.

Because the y are doing the hard process.

When we take the derivative with respect to y of x fixed doing the task.

Behaving as hard.

We do not want change because x is a partial in derivatives.

Gene wherein x is zero, y is zero, there is The

but if you pay attention in öbürkü where x years had had here.

Although the results here conversely

critical points of the same gene involved terms.

But when we look at the determinant f x x zero.

Because no where x.

But when we look for an interest x to y.

One more because this f x to y of derivative When taken an interest.

As you can see the bridge, the first diagonal zero zero turns on.

The second one involved an on the diagonal.

When this determinant is zero minus accounts a.

A negative result turns out.

So, again, these determinants minus sign

A saddle point is obtained for We are.

This is how the saddle points that belong to

Judging by the previous 45-degree bi is returned.

That the lines, the lines of equal value the hyperbolas.

Because you say z is equal to the constant b y x is equal to b divided by a constant.

We know these hyperbolas will be.

wherein the first constant value of z for

and third quarters in a plane minus values will be here.

They are a bit of yourself into thinking you will find.

Rotated 45 degrees to be obtained.

When we look at minus sign now I do not want to repeat it.

Completely similar things.

Again a negative determinants involved.

Unlike in previous bi plus one plus There was a negative one, but the result still involved.

There are still bi saddle point.

Bi this shape it according to a preceding 90 degree of back.

So x and y are missing the times

the first and third plane curves on Do ediyos obtained.

Plus get here when it was Your ediyos.

For example, you say z y is equal to one minus a divide x.

We know this since high school.

z is equal to x minus one-half.

He gives a parabola line.

Here it is drawn as a double-stranded parabola gives the line.

Now that we have seen the basic surfaces.

Here and now I want to focus.

Because your eyes off these basic surfaces I need to know.

You wake up at three in the night, submitted z equals x squared

What is y squared minus the count of the saddle surface you also need.

So much so that these fundamental bi thing.

Now a little bit different then We will look at the surface.

But I want to call it b.

We are also revising them in a thoroughly I hope that you will internalize.

Goodbye.