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About two functions of the previous session, We got two different functions.

The first function of a circle We took one out from A to B,

he's still with the function of a square edge We went out or a line from A to B.

We went out with the first function, and each We found a different value at a time.

In the second function the same again We got orbit, a circle

We went through a square edge We went out and went right,

so this time we found the same results In the first example the function was due to orbital

results in the second embodiment functions results independent of the output trajectory.

We left asking the question.

Not independent from orbit this

In the second example of a coincidence, also does it matter?

We said that it is important Why this course now

We will see that it is important therefore worth examining.

We asked slightly different questions easier to answer

this is a question we have made a significant Are coincidence of circumstances under which

a through a path from A to B is independent of the value of the integral.

Now I will try to answer.

The first job of a theorem immediately.

In what circumstances is independent from orbit What are the sufficient and necessary conditions for that.

This dx plus there once u times dy integral outcome of

to be independent from orbit vx necessary and sufficient condition is less fit,

the second component i.e. the first variable derived first by

junction of the second variable derivative will be equal to each other.

Comes to mind right now, if vx to uy'y a function F is equal to refer

derivative with respect to x if I picked a V F I choose derivative of the function y

second derivative here inasmuch as when I bought dv dx

there was one more derivative with respect to x to y derivative is taken for the second complex derivatives.

whereby the derivative with respect to y u F is the derivative with respect to x, y in an in

Get derivative of complex derivative again but in a different order mixed derivatives,

but we know that Clairaut's theorem We found, the function and its derivatives

If there is sufficient continuity two partial derivatives are equal.

So it's vx

is equal to a sufficient uy'y is parallel to an F because they

second complex function Because it was equal to derivatives.

Instead, see if we put Instead df dx v instead of u

dx dx times that we put dy'y df df df plus

dy dy once divided this chain on derivatives

I have seen this is the exact differential of df using partial derivatives.

Fully differential for F is equal to the integral thereof.

B and the value thereof to a value in need to find it and

As you can see from A to B. How we are going to an independent,

happens only due to extreme values??, intermediate orbit is going independent.

Parallel requirements also prove to be able to here anyway if certain

independent of the orbit In order to be plus udx

dy'n have to be an exact differential if you got it in full diferansiyels

here to go back the second derivative is equal

vx is equal to that given by using comply with the condition are able to provide.

Since this F

it is an important thing give a name to the value

Even though we have a name worth trivial, all of the people all of the important

Is the name is important to him, We give a name to everything.

We call this potential, that the vector

u and v are the components of the We call the vector potential.

The above ingredients In terms of vector x,

The x derivative of f The second component of the vector

For the second variable in this vector we see that the derivative is based on Fy.

If we combine them F is the gradient of

See see that the gradient is very has been encountered in a different context,

infrastructure have the same thing here.

Gene gradient significantly We see that we come across.

The following is an equivalent in the previous theorem Udx plus the integral of this integral Vdy

be independent from orbit sufficient and necessary condition for the

That v is Fx Fy forms a potential

means exist, it's also so potentially

independence of the existence of orbit We show that the equivalent.

Independence from orbit conditions We've got the idea when looking for a potential,

by reversing it If there is potential in orbit

we see that it is independent.

Now, if it brings a new concept

for a closed orbit, we If we account for udx vdy'y

If it is zero, and a shows that come from the gradient

and in reverse to a If it appears from the gradient of this

of the integral along the closed loop requires to be zero.

Supremely easy to prove,

we closed orbit Let's start from a to b.

b a'yl infinitesimal distance that they may be close to each other

We take a point where artificial We're cutting the curve here,

closed curves bring an open curve.

If this potential where the integral As a fully differential can write

and its value before the As we have seen in this b F

but here in a minus value to b'yl They're going to limit to one another,

The limit means that we bring to the b closed curve is a curve that is open to the

cycle turns, so If you have enough trait conditions

Will be equal to the Fa and Fb

this value is zero, and it has proven so if u would be a gradient from the

If a closed curve derived one cycle for this

udx integral plus integral Vdy more abstract it becomes zero,

If we write self as an internal is a product of the components u and v

The vector components dx and dy is the inner product of the vector.

This more compact as and this integral writes

Closed-loop zero see that.

Do it now assume that E from A Are you going to b doing a job

What happens when I make up from A to B. zero means that the work you do

This means that if a force We called energy is protected.

That inspired here if u are a if you come from the gradient vector field

protected areas in English conservative It's called conservative in Turkish

is used, protected areas, conserved stretches said.

In this way the physics work source

When you multiply the force Do you get a job.

Roads, sorry forces on the road If you multiply the energy you gain speed.

See here strongly ways where the product of

dx dt with a strongly dt'yl divide and multiply,

according to the time derivative of the position where he If you rate occurred comments here.

So we hit rate of energy forces, in this

When integrating the work done on time, We find here the job done directly.

Here's to you from a closed b, this work going,

How to do business you go are independent or

as the equivalent of an A start Turn off again for a aa

If you are going when you get done zero work do you interpret the energy is preserved.

Here the term in mathematics this term from the physics

mathematics across the sheltered area conservative areas protected area

terms protected area, If you love which, being interpreted.

Now I want to make a few examples.

Let y be the first component of the vector,

the second component of x we get the following We want to calculate the integral udx

uu plus Vdy to dx'l We stand dy'yl V.

Now you have the following questions, Does this integral is dependent on orbit,

Is orbit dependent, Is it independent from orbit?

I.e. if a ui u This vector components

u and v of the vector potential Determine if you come this potential.

If you are in this potential a way from A to B.

coordinate points reported herein We want to find the integral of.

Now once it's integral Is dependent on orbit

Do you understand that regardless of the condition supremely simple.

We saw a little before the second theorem component by the first component,

derivative with respect to the first variable, ie vx

the first component of the second derivative suit based on variables.

We're looking at vx v is itself x its derivative with respect to x

We're looking at one of his y a derivative thereof according to y.

So this is enough and the condition efficacy and necessity condition is provided,

This means that the integral is independent of the orbit.

In a very simple way We can examine it.

So there is a potential is independent of orbit,

the x component of this potential first component,

y derivative equal to the second component Means that fx will be equal to y,

If we take this integral with respect to x integral with respect to x on y fixed x,

but also will the integration constant,

hard to say at this integral gyu may be only a function of y.

F to provide the contrast Let partial derivative with respect to x from the first term

income year, the second term can only depend on y And because it is the derivative thereof is zero.

So partly found F this game yet because we are more

an integral constant unclear.

Hopefully, this F see here We use the derivative with respect to x,

If we use the derivative with respect to y of FX it income from derivatives x to y, the y,

g is the derivative with respect to y this g income base year, it will be equal to x.

Right to left, we see that X, x counterbalance each other already

dengeleyemesek made ??a mistake We can conclude that.

Here g means that the zero base We will arrive at that judgment.

Of functions of one variable the derivative is a constant of zero,

means that g is a constant.

Because we can get this fixed constant potential on its own

not important to be important with derivatives As hard as if I put it to you

derivatives will be reduced, If you want to put so hard here

When you get gradient From the derivative with respect to x y,

where the derivative with respect to y from x, this 'll find us at the beginning of the vector.

To find the value of the integral inasmuch as there is a potential,

The expression here is an exact differential.

We are already seeing this YDX plus xdy XY himself is fully differential.

In addition, we have proved the following theorem the potential value of this integral

b minus the final point, initial point potential

XY final point was about seven or eight We got here by way of example

starting point coordinates of the two.

Here we put them seven or eight x

and in the second one and for y two of these results we obtain.

So here without integral to which is independent from orbit

How do we do already in orbit is connected to the orbit should give

Is independent from orbit directly able to obtain from the right potential.

A second example of this previously than we did in the previous session

The sample we studied in the previous session You will recall from an a

We were going to point b and its three different We had trajectory, along a circle,

along the sides of a square and going from A to B directly.

Now on three different orbits We found the same results, but these

proof does not mean that only these three trajectory that gives the same value.

We asked whether this coincidence, We can now immediately answer here.

If this integral only these three Not orbit for the entire trajectory

This qualification is independent from orbit condition should ensure that,

i.e., the first component derivative thereof according to y,

The second component minus two xy component in turn with respect to x,

The second component ie the first variable, first component

A second variable There are cross-derivative.

If we take the derivative with respect to y u minus

coming two years, If we take the derivative with respect to x v is still two

Here comes the year therefore immediately able to achieve these results.

Inasmuch as the fundamental theorem is provided

provided that a qualifying condition There is potential, and this potential

derivative with respect to x is the first component data a second component to y gives derivative.

Thus, this potential can be calculated immediately,

df dx x squared minus y squared would say that.

That the integral with respect to x If we take x to x squared divided by the cube of three years

y is a constant square y with respect to x for x squared,

There are of course a minus sign at the beginning, but this is an important integration constant

This problem may also be of importance but also a function to y

it may be because the derivative with respect to x Let's look to gain blocking until.

Three derivatives of f with respect to x where x squared here for

remains only takes you three x squared.

Here y derivative with respect to x squared so that it acts as a constant

partial derivative of the derivative with respect to x only the y remain square with a minus sign.

HX by the x, Hy also with respect to x derivative is zero because x and y from one another

it is independent of the derivative with respect to x, we As you can see this beginning we provide.

Now that hyun uncertain F. take place it in there,

v is divided dy is dF We know there was given to us minus

wherein said two xy integral minus two xy

i.e. it to y to y F If we take the derivative of the first term is reduced because

between zero derivative of x to y is independent of x and y.

Here, the derivative with respect to y from the negative you can see it comes to two years minus two xy

As for the left and right minus two xy counterbalance each other, but on the right side

it also had a to y h he comes from the base of derivatives.

Here the term mutually xy'l To simplify stabilize the base of h

we see that zero, so derivative of one variable is zero

is a constant function h is from here We see that there must be a constant.

Now inasmuch as the potential The value of this integral has

The only potential trajectories of independent Depending on the value on both ends.

This time our potential at this c Let's place, using, where c is there too.

See one third x cube minus xy squared plus c,

it coordinates and in a zero, was zero in B and IIIA.

In this example you will remember If a trajectory going from A to B.

This component of x in place Put falls to zero the first term,

falls in the second term, remains only one car.

In the second place this in between zero to y The term X, the term is falling but remains.

He is also a cube divided by three, there is less than a one in that will bring.

Minus one-third of a cube.

There are also c.

You can see the car comes twice.

With a plus sign with a minus sign.

Cancel out.

Here's to him from the beginning that this

c can get zero because to take each other already.

And as you can see the potential here We find a negative divided by using a cube.

Let us remember this example.

This example illustrates the problem as follows.

Here x cube minus y cube minus two x

y the following integral over trajectories were calculated separately, were calculated.

We have seen that this integral It is integral in independent,

The structures of these functions and as a result of these three orbits

separately integrally on We found a cube minus divided by three.

Potential now than it now We found directly,

minus as a cube divided by three.

There is a counter example.

The following functions are provided.

Sine, hyperbolic cosine of x y, plus there once d y.

No there is not, as yet unknown.

Question.

This is a very typical problem.

This integral may be addicted to the orbit,

be independent How have to be chosen?

there is in this case from the orbital are independent, there is a potential.

When there is potential Do you find this potential.

Right now, we're looking at.

Necessary and sufficient condition.

It was u'y first term.

Thereof, with the y-derivative.

Now we find.

Derivative thereof in the y sinus x remains the same.

Hyperbolic cosine of y to y derivative of the sine hyperbolic y.

This has to be equal to x.

Now let's read the equation backwards.

So, v is the partial with respect to x derivative of sine x, y hyperbolic sine.

If we calculate this integral The term X, varies only.

Income but a cosine x income minus sign.

Because the derivative of cosine of x minus one for the sinus

Let's get the negative should start in the sinus.

Hyperbolic sine y de unchanged.

Yet an integration here may be constant partial

For derivatives with respect to x is may be a function of y.

Let's do provide.

Let v is the derivative with respect to x.

Revenue minus sine x cosine of x.

Thus is a plus.

g is the derivative with respect to x is zero just because this gun y-dependent.

Potential immediately easy to find.

the derivative of f with respect to x

since it is the reverse Let's forget that we can rotate ui,

so u have written here, was given.

Sine, hyperbolic cosine of x as y.

That the integral with respect to x If we like what you see

from the integral sine cosine of x x comes with a minus sign.

A cosine y.

Here comes a hi again.

That's a vague function also can go.

You can get zero.