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Hi there.

We have seen before that gradient,

gradient, but in Cartesian coordinates We saw that in terms of x and y variables,

After that x, y, z three We also generalize variable.

But they do not need to tell We've talked about this before now,

circular coordinates are also very important In nature, a coordinate system,

technology in many systems of thought, profession cycle

In the event a circular coordinates also of paramount importance

because the problem in terms of these coordinates expression can be a lot easier.

In addition, the example of infinite space an infinite plane waves

Think of the spread, than the circular coordinates

It is understandable because the semi- circle diameter tending to infinity

You can think of it as for him, and phrases are easier than extraordinary.

Now after giving this rationale here We'll start with a gradient, but we'll see

In the Cartesian coordinate previously As the most basic size gradient.

Thereafter, divergence, curl and NEW readily able to obtain Laplacians.

We do these operations, gradient We know in Cartesian coordinates,

components and a digital function x y according to the size of the partial derivatives.

Gradient of a single processor DDX variable derivatives

Two of receiving processors variable generalization,

Because these two variables vector processor to be qualified.

Now here we gradient of f circular coordinates will account.

Let's consider what elements have them.

There are times where i j is a clearly written,

their anti circular e r e theta coordinates.

Circular coordinates We hope that you remember.

A point on the plane x, y As we will show coordinates

made with a center distance r and the x-axis can be expressed in terms of theta angle.

Cartesian coordinates of this circular coordinates where the relationship between

projecting a point on the X axis will be times when it is clear that the cosine of theta.

on the y-axis or its projection of parallel lines

Get it by r times the sine of theta is We know we have seen in many places,

we do not know the learning a good opportunity.

This way I drew it again here.

Of course, we say r is constant When we get a circle.

When we say theta is equal in fixed

we obtain an accurate All points because these

We combine on the right to the center When the angle theta of the composition.

This means that this right the locus of points.

Here i e r and e theta and j with drew.

Now that we know will point to

Combining central direction e for the theta-theta is

When we change the the unit vector is obtained,

Another view perpendicular to this e r in terms of the tangent to the circle.

Now our goal is to find them.

Where x is the Cartesian coordinates, but the circular

variable is written here in terms of here, as we know from the simple geometry.

E r is this, that of the vector x

The function of the r and theta If we keep a constant theta

When we take the derivative with respect to r theta means that we maintain a constant partial

derivative means taking a univariate function is transformed into the vector function

r is then only a function of x, say these derivatives will give a tangent vector.

As can be seen by taking derivative by r cosine theta and sine theta involved.

Similarly, if we keep r constant theta function of x only,

In the third part of the initial te As we have seen in a space,

shows a curve on the plane in this case.

As you know, this means keeping a constant r

on the circle means to advance If we show here on the circle.

Theta is changing on the circle.

Theta obtained varies a tangent vector magnitude.

A tangent theta derivative vector, but it gives

here because it can not be a unit vector Judging will be based at r.

Get derivative minus sine of theta is the cosine of theta-theta, but there will be.

Therefore, if we divide the unit vector r we obtain that regard, we divide r.

Now we i, r and e

i and j have achieved in terms of theta.

However, we need the opposite because we want to find a gradient,

There are also good j in gradient, it is very easy There are two equations with two unknowns here.

i and j are unknown, although Even now, as the vector of unknown

it can be calculated in three ways In fact, using vector attributes

two equations with two unknowns Not so here is j

in the first equation to kosinÃ¼sl you multiply the second equation sinus

See if you can hit here sine cosine wherein the sine and cosine occurred occurred.

By subtracting expenses j'l terms.

Here we hit kosinÃ¼sl occurred for the cosine square,

here we stand for sinus sinus to square that has occurred interests

sine squared plus cosine square I have found was right here.

First we were hit kosinÃ¼sl,

We had a second hit and had brought the sinus Did you look like what you see, and e, where e is

Tetala terms like that.

Other kinds of things in a simple geometry You can find it here, do not want to

This vector e r i and e theta Get on the projection, see

Get out of here by the projection of E i direction cosine theta is coming.

E theta projection on See if you get in the opposite direction

Considering located in an e theta theta e in the opposite direction by turns,

I will say that you get a minus Thank interior can also be achieved with the cross.

I'm showing them, especially because these are pretty practical algebraic operations,

This can be useful elsewhere.

No size thereof when a vector from the opposite direction to find components

because the inverse transformation for many sought transformation is happening on the ground.

I take it, see if you hit i'yl This term is orthogonal to the j

Because the length of a fall i i i times a cosine theta be a data here.

So i e r on component cosine theta.

The second equation is also used If you hit the i'yl vector

j'l terms will fall again I will still give you a good time.

Where i i theta,

E theta we find the components on See it turns out negative sine.

This may also be done like J, of course, multiplying the

what can be achieved if the road Whether we find this solution.

Now because we have made ??significant progress the following statement, we express the gradient

they will translate in terms of theta and e e r i and j have found a little before we found him.

In the second step dx df and df r dy'n and

expressed in terms of theta are going to find out.

a derivative of f with respect to x chain will be derived

derivative with respect to x r by r theta derivative but also for

theta derivative of f depends on the theta derivative with respect to x is going on.

Similarly, it also means that df dy'y When we do that we need is

d theta dx dx, dy, and d is the theta dyer need of these four size

We know we can have them here The R squared is x squared plus y squared

its derivative with respect to x If we remove the two items is dx.

Here, because the two x's interests x and y independently of one another.

Therefore, the zero derivative of y with respect to x it would be, so we know the definition.

As you can see here d r d x's right.

But x is divided involved.

But the cosine of x divided by r We find that theta immediately.

Because it was teta'y cosine is x.

the more simple is the cosine of theta.

Similar operations are made for y.

relationship of x to y teta'yl the case.

Tangent theta, because a correct is on the x and y.

See x divided by y the following form

Turning to the ratio of y to x here becomes the tangent of the angle.

Here also derivatives If we make operations

so this is a very ordinary algebraic operations but a little caution needs to be done.

As you can see here d r d x, d r d y, d theta d theta d x and d y we find.

As we found them instead of When we place d f d r and d f d

the remaining theta,

it only in terms of theta The terms we have obtained happening.

Tool, we are looking for this.

As you can see how easy it was.

We found i and j, in terms of theta and e r e.

Now where d f d x, d f d y We find also in terms of r and theta.

Let's place to take them places.

Here is a summary that happen.

Expressed in Cartesian coordinates.

Now i and j we have found.

I just c that is, a d for the prior page

d x and d f d y found.

We're here take places.

d f d x the following expressions:i in here.

Similarly, the d f d y the following statement: Prior to this, we found a page.

See where we found.

We found we put these expressions.

j also found.

We are also placing.

Now that's fine, There are eight terms, see here.

When we hit them enunciating a, two, three, four.

Four to eight here.

We make them so sweet Simplification is going studies.

Look, I'll show you one.

There is also another ÃbÃ¼rki you Thank sine squared plus cosine squared

Complementing one.

See here for the first term d f d r front.

Square cosine theta occurs.

multiplied by e rs.

Are you looking for here is still the same is r square in front of the sinus occurs.

with e rs and sine-squared e to a square where the cosine

d f d r r times remain.

See how much was simplified.

Here the minus sign Coming sine and cosine.

Here the plus sign Coming sine and cosine.

They simplifies each other.

Similarly, see here d f is the sine squared theta for coming.

Here the cosine square d f d theta for coming.

They're going to e teta'yl.

As you can see when you put them in From that very sweet eight terms

simplification of how this gradient we obtain the expression.

Since a,

Any numerical function gradient in the circular coordinate this.

That gradient processor as emerges.

Of course the most basic of gradient was the case in Cartesian.

Now that we have a vector, that another vector of this vector,

and the inner product of a vector field we can obtain the vector product.

He will give us the divergence and curl.

This is a beginning to take on a gradient, inasmuch as this is a vector,

here it is also a vector gradient with a gradient If we multiply the inner product of two vectors is happening.

Here comes the Laplacian.

I'll go over these accounts.

For those interested in watching here I have given some guidance.

I want to know what has been achieved but the last.

This is a very commonly used.

They are obviously not there to remove.

Certain gradient.

Any of the vector E R and E theta We know the circular coordinates.

You need to shock.

You can see it when it hits the we encounter something interesting.

and e e r r and theta theta derivatives are coming.

We can do this particular e r.

E theta clear.

Derivatives thereof zero by r for one thing, because there are.

Derivative with respect to theta is also very clear.

They obviously.

When we place them The results of this is happening.

There are two equivalent use it.

See here a ternary representation.

How is that for terms such as d *

Or has the second derivatives based on derivatives.

Wherein these two terms can be combined.

See if you open these r u r times this

R is R times ago You can find it with derivatives.

Plus from derivatives by r You can also find this term.

It's here.

Hence the two have equivalent representation.

Sometimes this structure is more useful.

Sometimes this structure is more useful.

You also do not expect you to memorize.

Know the existence of such a thing alone your life will be much easier.

When examining the electromagnetic field, When examining the flow of a liquid

When the conservation of mass in a heat

In examining these types of flow processes 'll encounter are always transactions.

Rotational before laplacian NEW easier to do.

Because grad divergence learned in the previous step.

Bi, bi divergence of the vector of the following We have seen that the two equivalent notation.

Our vector of our numerical Although the gradient of the function

and we take it to the divergence in Cartesian As in the're getting laplacian.

These transactions do not have to again.

U see is because certain.

R f are derived by r.

According to certain theta theta derivative of f.

There follows a well run.

When we look back to the previous page Showing here anyway.

When you put this process, laplacian NEW You can find it in circular coordinates.

Divergence of two equivalent representation As the laplacian of

two equivalent representation is happening.

See the second degree, There are second order derivatives.

Was the case in Cartesian.

But a little different.

Because of curvature of the circular course coordinates, such as Cartesian happens.

Already a second difference of x and y.

Both length.

In circular coordinates a length of r.

However, an angle theta.

That is a dimensionless quantity.

Its square is already here There comes a run here in the denominator.

There is an r square.

Here also directly We see r squared.

So the length of the denominator There are square.

Here, too, this is really the square the square of the length of the leads, brings.

Because the dimensionless theta.

Curl is made with a similar process.

Finally gradient processor is known.

And a vector e theta e rs in terms of components

we have shown that when job Multiplication is staying in the calculation.

This again when he calculated by multiplying rectors are derivatives of the unit.

But this is not r e r and e theta derivatives to zero by r.

But each by theta derivatives have both.

And position them as well as rotational 're getting in circular coordinates.

As I told electromagnetism these equations

You can not remove without features.

Similarly, for a rigid body, for a liquid,

in a continuum mechanics problems again this magnitude are always useful.

Of course, the Cartesian coordinates As used where

If you had a circular piece of equipment,

of course if you look at his vibrations Cartesian coordinates are not suitable.

I need to work with circular coordinates.

Or a tornado or a wind,

b if you think the water cycle, if you think of a gas cycle,

Of course there is also still a circular coordinates would be more appropriate.

Possible to calculate directly the Laplacian area.

Two-time by taking partial derivatives.

This is a good exercise.

Those who want to leave.

But it is clear that the same result is found.

That with a little guidance I believe you can.

Now this circular coordinates NEW we found laplacian.

We can use the wave equation.

For example, emitted from an antenna, plane spread in E, if you think the waves.

You threw a stone, Think of the waves it produces.

A machine part, an annular well,

Did you hit the middle, where the waves are.

Here thereof equations Showing that way.

Here I e, I give you an example.

A good exercise is bi.

Laplace equation and the heat conduction in write it down in the habit of bringing a hand.

There is no difficulty.

Instead, as Laplace take copies of anything else other than

but nothing beyond bi habits It is also helpful in this kind of work.

One example, and promise.

I will not dwell on them.

Because they are very formulas applying another

things simple non-derivative transactions.

Here in Cartesian coordinates ago I want to find this

The function Panini gradient and laplacian.

Of course, it can not be said of the divergence.

Only a function numeric gradient and laplacian NEW available.

However, the divergence of the vector magnitude and can be obtained by rotation.

That is why, You can take these derivatives.

Derivatives of the formula directly You can do this by using or

here you find the x and E y inserting here,

this gradient and divergence formulas applying them can be obtained.

There is something interesting.

Cartesian Laplacian involved in the coordinate zero.

Circular coordinates zero turns.

And this is an important feature of the laplacian.

Which coordinate you go If you remove all zero in zero out.

I also e,

no formula for how all this works you can try to do without.

But if you try to do so, I would not recommend.

Calculates how long a lot of You will see what to do.

How this using formulas to emphasize that make your job easier.

A second assignment.

Here again, given a fine.

E, as in Cartesian coordinates.

To find its gradient and laplacian Panini We want, in Cartesian coordinates.

It's very easy.

You will receive derivatives with respect to x, You will receive the derivative with respect to y.

Second, will gather to take derivatives.

So to do this process.

After you do this directly

also being asked to do with formulas.

First thing to do here,

the x and y coordinates to translate into a circular, ie x r cosine theta,

Place for y r sine theta this We find r in terms of theta functions.

Then right formula You can apply.

To check your work The results for are provided herein.

Yet there's an assignment.

Here's to the same road, erm, There is also guidance.

But it is an important difference.

Previously always scalar,

that of numerical functions We found the gradient and laplacian Panini.

Here you are given a vector.

This divergence of the vector and We want you to find the curl.

These and e, the results for the given checking account, you have the chance.

Gene homework.

This is a little more work aims to improve your habits.

Gene X and Y given in terms of a vector.

This vector is no longer one at that

gradient can be found by enforcing but then it turns out a matrix.

And it normally divergence We want you to find the curl.

Certain results.

There's an example again.

All of these things,

all samples also teaches a little something but I do not want to go into them.

Make them as an exercise.

There is a challenge and when the divergence, rotational income,

We see here, the vector should be.

What happens when the gradient and Laplacian, A numeric function should be.

Now here's circular coordinates we finish.

As if we look at a general Cartesian We have seen in the two-storey integral coordinates.

Just behind the circular coordinates We saw a two-storey integral.

With this operation, also in derivatives in Cartesian coordinates, we see

Configure the import process, this gradient The basis of it, there are partial derivatives.

In the same process, this time a circular We see coordinates.

So both in Cartesian two-storey integral and derivative are seeing.

Their equivalent circular We see coordinates.

We have already seen two storey integral.

Here e, but a circular bivariate

coordinates the derivative process we are treated.

I hope that this assignment and parsed You can see examples, you can see,

You can examine a wont win.

We stand here today in this session.

Now we will begin a new chapter.

It is also very variable derivative chain.

We usually only one variable derivatives made ??by the chain.

We found a total derivative.

Here too variable The process will do.

Please goodbye for now.