Ders çok değişkenli fonksiyonlardaki iki derslik dizinin ikincisidir. Birinci ders türev ve entegral kavramlarını geliştirmekte ve bu konulardaki problemleri temel çözme yöntemlerini sunmaktadır. Bu ders, birinci derste geliştirilen temeller üzerine daha ileri konuları işlemekte ve daha kapsamlı uygulamalar ve çözümlü örnekler sunmaktadır. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler Bölüm 1: Multivar 1'in Özeti, Dairesel Koordinatlarda Entegraller Bölüm 2: Türev Uygulamalarından Seçme Konular Bölüm 3: Çok Değişkenle Zincirleme Türev ve Jakobiyan Bölüm 4: Uzayda Yüzey ve Hacım Entegralleri Bölüm 5: Düzlemde Akı Entegralleri Bölüm 6: Düzlemde Green, Uzayda Stokes ve Green-Gauss Teoremleri Bölüm 7: Stokes ve Green-Gauss Teoremleri ve Doğanın Korunum Yasaları ----------- The course is the second of the two course sequence of calculus of multivariable functions. The first course develops the concepts of derivatives and integrals of functions of several variables, and the basic tools for doing the relevant calculations. This course builds on the foundations of the first course and introduces more advanced topics along with more advanced applications and solved problems. The course is designed with a “content-based” approach, i. e. by solving examples, as many as possible from real life situations. Bölümler Bölüm 1: Summary of Multivar I, Integral in Circular Coordinates Bölüm 2: Topics of Derivative Applications Bölüm 3: Chain Derivatives with Multi Variables and Jacobian Bölüm 4: Surface and Volume Integrals in Space Bölüm 5: Flux Integrals in the Plane Bölüm 6: Green in Plane, Stokes in Space and Green-Gauss Theorems Bölüm 7: Stokes and Green-Gauss Theorem and Nature Conservation Laws ----------- Kaynak: Attila Aşkar, “Çok değişkenli fonksiyonlarda türev ve entegral”. Bu kitap dört ciltlik dizinin ikinci cildidir. Dizinin diğer kitapları Cilt 1 “Tek değişkenli fonksiyonlarda türev ve entegral”, Cilt 3: “Doğrusal cebir” ve Cilt 4: “Diferansiyel denklemler” dir. Source: Attila Aşkar, Calculus of Multivariable Functions, Volume 2 of the set of Vol1: Calculus of Single Variable Functions, Volume 3: Linear Algebra and Volume 4: Differential Equations. All available online starting on January 6, 2014
Green Teoremleriyle Çözümlü Problemler
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From the course by Koç University
Çok değişkenli Fonksiyon II: Uygulamalar / Multivariable Calculus II: Applications
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Koç University
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From the lesson
Düzlemde Akı Entegralleri
Uzayda kapalı yüzeylerle tanımlanan hacımlar; sonsuz küçük hacımların birleştirilmiş yaklaşımla elde edilmesi. Jakobiyan ve sonsuz küçük hacım. Kartezyen, silindir ve küresel koordinatlarla uzaydaki yüzey ve cisimlerde üç katlı entegral hesaplamaları. Uzayda küre, koni, paraboloitler gibi temel yüzeylerle tanımlanan hacımlardan örnekler ve bunları içeren hacım hesapları.
Meet the Instructors
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hi there.
About the preceding session We prove two theorems.
One of them is quite We have seen in detail.
This projection of the t integral along a closed curve.
You can see the expansion of this integral D x plus y as any other.
The two-storey integral to the have proven to be equal.
Similarly, on the projection of the n gene expression also appears here
and the gene thereof in a two storey We have seen that the integral is equal to.
Mathematically, a work cycle integral, one cycle integral,
that it can now.
Less if you say u to v, If u and v in the first turn to.
Similarly here Remove the identical two-storey
integral, but in the back of their physical meaning is different.
This is one of the on the tangent vector
i.e. projection on walls this integral accumulation.
By this a second wall One measure of a side passage.
Because it is the vector perpendicular at each point of the surface
Upon projection of the unit vector 're getting and we collect them.
This is the u n d s point gives the integral.
Open their software.
Now we want to see a few examples.
So we also we do on account of the cycle,
we do on the field at the same We need to find a result.
Our first example of this theorem is a useful, show n.
Such a linear integral given account.
As you can see quite mixed terms here.
Coordinates of the center of this team in that in the center and edges
the circumference of a square of two says on the accounts.
This is not something that can not be done.
But on four separate correctly 'll do the math.
There are quite complicated terms.
Now will you do it with respect to x, it, is not so easy a job.
x cube times sine squared x partial You'll make them with integral stuff.
This is quite a complicated job.
But we are now using this theorem, We prove theorems of Green,
instead of calculating it with the same result We can return to an area integral
and perhaps this area of integration We would be easy to find.
Now first, following by Green's theorem.
Integral given to us like that.
Green's theorem there is x plus y in D there.
So the first term where u x cube term.
The second term this sinus The term also has hiperbolikl.
Theorem tells us to make this account Is there xi instead of ALSA, that u
y If you leave the field integral on the even if you can find the same result.
See, there comes a time that we get x.
when u get y a minus involved.
But in theory a minus sign because it is a minus minus one.
Both times the integral over the area, that area twice.
The two sides of We know the area of ??the square.
Four square.
There are also a couple of factors here.
So eight square just as involved.
Of course, this sample a little of this highlight the usefulness of Theorem
designed for one example.
But in many cases this can be difficult.
So the integral cycle can be difficult.
Using Theorem space We can do the integral over.
Or area of integration It can be difficult to calculate.
Area integral was asked to do.
Integral to this cycle turning to easily account
brings in the possibility.
But this is not always going to be a very useful I'm guessing that there is something.
You can see.
That the direct integration A very long to make,
a lot of mistakes is an example to be made.
A second Green theorem we can do well with.
Because the second Green's theorem us says the following:Negative v times d *
plus u d y times, he says.
Means that the term is now y'l as we will describe.
y'l term is now here:So x, Our hiperbolikl term sinus is happening.
As there is in the negative x We will determine the multiplier.
Therefore it it has the minus sign will be.
As you can see here has changed In case you have entered the sign of a difference.
But the result will be the same, because here will take different derivatives.
These two variations of each other take the results will be the same.
We take the derivative of u with respect to x.
A derivative of u with respect to x.
Plus, you're getting v is the derivative with respect to y.
Refer to the outside where there is a minus, minus the inside.
This is the first term contains only x Get them to the falls derivatives.
Similarly, the terms herein y'l When we take the derivative with respect to x had fallen.
Here comes a a.
Again taking twice the area eight frames is achieved.
There is also an application like this, a generic application.
We uu minus one-half of x, y and and we take it first vector comprising Green
we apply Theorem, see D x plus v d y, v x minus y will be equal to u.
So once d x plus x minus y
d will be equal to y times two-storey integral to the right.
v where x.
there is a future of x.
Less than u y u got us here.
from Y minus future.
Here is a minus will be given for two.
So a divided area twice both times.
A means of finding a ground Find the general formula would have.
Let's make an app like this:Corner a, team that is at the center of coordinates.
A corner x axis on a wide.
One corner of the y axis on the b-high
triangle area Green Let's do a checksum with the theorem.
Because such a triangle We know what that area.
One-half times the length of these edges,
Because the height b, b is one-half times a is open.
But we of Green's theorem As an application,
Our hands are used to doing it.
Now we've found the following formula:Area A minus two times x plus x divided by y d d y.
If you get on the wall We find its area.
For example, if you receive map of Turkey,
if you make such a point of corners with may be subject to thousands of points.
If you want to calculate this area, This formula gives it to you.
Hes wall, When this integral on the account,
And when you take little pieces You can zoom in with them correctly.
This quantitative analysis is entering quantitative analysis, but also
their origins in basic information.
This too is a formula that can be used.
Now let's do it.
We're going from A to B.
Its width, base a.
Then we go from b to c.
E the right of the equation can be written easily.
where y equals we've already given.
y is equal to b divides a.
Because of the slope, b is the height of a base, according to
b divided missing, but because descending anything.
For him, have a negative x.
when x is equal to y is equal to zero.
Right.
when x is equal to y is equal to zero.
We are at b.
y when x is equal to zero, the
where x is zero, see When this term is reduced.
to simplify each other and b y is equal to the remains back.
So we provide that too.
Already two points on a line When you have provided all of them can provide.
Now we still inside this integral one's cycle "trigonometric
In order to see the direction from A to B. from b to c c a need to come to.
Of course, we can start from wherever you want.
From the start b c b, c, also could make on a b.
But all the same thing anyway.
Would be changed as a collection.
Now let's see.
b on x is equal to zero on a b.
x is a constant for this differential,
dx is equal to zero, it becomes infinitely small.
So this is zero, not necessarily
Although this was any hard this showing changes things would be zero, icon.
Then we'll go to c.
now we're going from c.
Here I jumped queue, these respects, CA easy.
Because over the years where c equals constant, y is zero.
He also becomes zero when d y.
If you are on a b c y We found the equation.
particular year.
Because here in the x-y There was a need.
particular year.
Here we take derivative d
b y d x is equal to minus one-half strikes.
So minus b times a d x d in the income year.
Now let's place them instead.
See the way from A to B. where d x is zero, x is zero.
Therefore, d x hence zero.
This term is falling.
This term is zero for x falls.
So from a zero where a In the integral over b, d x.
The second one-half of the formula again comes to be had in this
The dilemma of a split.
We're going from b to c.
y going from B to C expression given here.
There is a minus sign.
A minus minus y times x divided by y b.
We put it.
d x. We do not have anything on it.
Plus x times dy.
We put x, d y d x minus b divided.
As you can see, all x became integral on.
Falls in the third integral.
He falls.
The third integral ie from c y is equal to zero was a while.
This was the first fall term.
y to be zero, this first The term is dropped, but also y
hence zero also fell in the second term.
Therefore, the third The term becomes zero.
Therefore, it remains the only one to remain contribution towards staying on.
Him when he calculated a There are two slashes here.
See here when we calculate where various studies will be simplified.
Here are minus x b a'yl hit, b divided by a'yl who is possessed of.
Here is the Artillerie.
Therefore, this X, The term took each other.
Only here back a minus b divided by a times a.
That is a simplification was also just had a minus b.
Less're putting in here, d x.
This is important because we will be missing When the integral of x from A
So we're doing is going to zero with a value greater than the following limits.
Change the order in which it When the minus sign is a plus.
Then here comes a b.
d x is the integral of x.
x is the value of a.
Zero at zero.
We know one-half times the base increased height formula is happening.
Now it's much disdain.
Because it can be very random way.
As an example you these irregular I told you to calculate the area of a pentagon.
They are suitable for computer programs may make it very practical.
I said just the area of ??Turkey If you wanted to find just such a
using the formula a little small You can make the border dividing into pieces.
What is the advantage?
The following advantages:Space-squares You can also find splitting.
But of a rectangle Although one on one side,
Although a thousand points on one side, on the other side, though one thousand,
a thousand times a thousand million smallest rectangle is formed.
This is on a million whether you will collect accounts.
However, over the wall When you go every thousand
four thousand because it is in the corner of the wall gathering and shock will do.
Therefore your account thousand times on the order of shrinking.
One million of the rectangle to calculate the area
four thousand instead of the correct 'll do the math on.
This quantitative analysis is very is a formula that is used.
Now we finish this section.
In this section, let us consider what we are doing.
We have two sizes of Green theorems we have seen.
We have two sizes of Green of the three dimensions will now theorem.
Because we live in the real three-dimensional physical environment.
These two dimensions to it without would be very difficult to pass.
Our approach is already all events Once you understand them to see in two dimensions
After going three-dimensional transition easier Our approach from the beginning that he
In the first part of the course of this It was always like that in the second part.
I think this is a useful approach.
Many people also like this approaches are already using.
We'll adjourn now.
Thereafter, the sixth week of classes We will now, and we found that there are two dimensions
The results will generalize to three dimensions and Stokes' theorem to this new and Green-
It's called the Gauss theorem.
Some of them only at the Gauss theorem says.
But Stokes and Gauss' theorem first in a plane
and the second of Green's theorem is nothing other than generalized.
Your Goodbye.
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