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
Çözümlü Problemler I: Uzayda Çizgisel Entegraller
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En provenance du cours de Koç University
Çok değişkenli Fonksiyon II: Uygulamalar / Multivariable Calculus II: Applications
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Koç University
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À partir de la leçon
Düzlemde Green, Uzayda Stokes ve Green-Gauss Teoremleri
Vektör alanlarının tanıtılması; vektör alanlarıyla türev ve entegral. Düzlem eğrilerinde entegraller. Entegralin yörüngeye bağlı ve yörüngeden bağımsız olması. Düzlem eğrilerinde birinci ve ikinci Green teoremleri. Düzlemdeki Green teoremlerinin vektörler, rotasyonel ve diverjansla gösterimi.
Rencontrer les enseignants
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hi our previous session
in the plane of two variables function of the
integrally on the trajectory from the three We have seen how to move dimension.
Further wherein the three types We have identified the problem.
Theorem Stoks second type, the third type Green-
Was the implementation of Gauss's theorem.
We are now, of course, the first We start with the type of problems.
The first problem is a problem as follows, an x, y, z given function
that the integral over a curve We want the integral over ds,
As the function e to the minus z x squared plus y squared multiplied by the
DSi will go from A to B on the integral, How are we going, given the road.
Road along a helix.
He defined how the helix?
A circle with a radius Are you going to step on the b
See x squared plus y squared yi When you receive a frame involved,
Would you say that going on a circle x-y plane in the plane of the projection
but also in terms of the z direction Are you constantly rising down
you can see the top of this DNA in a biology
As one of the double helix,
proteins, such as helical or We saw so much easier than the current issues,
a minaret rising above when riding on a circle longitudinal
can rise, Where If you are asked where to go?
from the zero point zero.
T is zero point zero zero mean point is zero because the intermediate compound in sinus t
t is equal to zero means zero zero When we returned back to zero again in y's
See two pi times the rise b t really say that we're going on two pins,
providing two pins as t When a cosine two pins
sinus is a d are two pi because it is zero is zero.
So we compute this integral t
of zero to two pi We'll find much change.
Do this supremely simple,
where we saw each term Instead of going yereştir.
At first we calculate the DS one.
point x in the plane of square frames years has DSi points.
Dots derivatives with respect to t.
Here, too, there is a point in the z or of the position vector of the vector x to t
do this by taking the derivative We will account for the length of the value,
norms will account.
See derivative of x with respect to t comes from the negative sine cosine,
t is derived from the cosine-sine coming that is staying here for a just a little.
Means taking the length thereof, first component squared plus
The second component of the frame components plus the square of the third component.
See the first two components of the square of a t squared plus a squared cosine squared sine-squared
t is a frame only for because it comes sine cosine squared t
b is a positive frame for the frame t To facilitate the writing of a frame plus
See if b squared c squared square root of the c dt ds will be times that we receive.
We will take them into place, from A to B is equal to T
zero meant the opposite, b t is equal to two pie was coming the opposite.
Where z instead of Z We will have placed him bt,
there is a square cosine of x squared t have a square plus y square,
a squared sine-squared t, easy to get here, of course, he calculates
which is important because this kind of mixed not to fiddle with the accounts to understand the concept
sine squared plus cosine squared t here, as a t
comes a square a square hard Because you're getting out,
c dt ds have found here, too, he also c We take it out to get that fixed,
bt just stayed inside to minus from zero to two pi dt stayed.
Of course, this easy integration; a square c here is the integral of e to the minus CT,
bt minus minus b to divide.
Less integral because it is the order of We're changing plus translation.
Means that t is equal to zero before We'll find t is equal to zero would be in
because it is above zero.
E pide two pi over two times b As you can see very easily found
too much of a bivariate vector No difficulty of this integral in space.
The second integral given the following structure, slightly different structure.
In a previous DSi directly had here in the integral, t
Once the product becomes u.
The first vector component E z times over the year,
The second component to the minus x minus x, the xy third component.
This is where are we going?
Here you need to define the curve.
We will go from A to B, but what ways We need to know that we're going.
This is called double parabola with this function,
components of the position of the x component t y component of the vector
t squared divided by two, z component of the cube at t divided by three.
Here, double parabola, The reason is called double parabola
a parabola in the xy plane There xz projection as well as
There is a parabola percent even There are also a parabola.
Tr y to see that if you say that x y will be x squared divided by two,
a parabola.
When you come here a divided our When the three x cube cubic prune
will be a parabola, geometry, such but just that as the account
where x is also not so important to know, y, z will yereştir what we see,
There dx means that x t would be here by dx dt.
T square by one-half years that the dye, dt dx becomes.
t be dt times, z a t divided by three cubes by that t would be dt dz square too.
DX will put them to take place, dy, dz t.
Easy to them via e z z,
that one-third of our cube by t here you via e y to z, we put
with a higher z to y multiplied by the DX.
Y is a square divided into two t but there are also dt dt
the common factor in all that outside as we hoarded.
Second term e to the minus x, According to the minus x xt that
e to the minus t dy, dy dt is the time we find the t.
So here goes t.
Here again is a common factor dt x times
There are times dz y, x times y t one-half cubic
dz is square and t is t d in five t that's finally an integral,
ha to know the limits We're going to need from A to B.
Look here not given t but the integral over t.
See a zero b initially zero,
a is the y component of the x component zero in the z-component.
Following means that t is zero, a point.
B above us at the point of two thirds given two to eight,
Let's look on this curve?
If This-
that if we do not provide the point b a t not on the curve, but really to t
If you see two data here two t squared divided by two to four or two,
eight divided by three in the z-frame, means that b point of this curve gives the equation,
therefore, say that on the curve the integral from zero to two,
If you want to deal with now that the integral This important aspect of the problem is
but if you say u t cube cube t, d Deliver Please buluyor t is squared.
So here's this integral to do.
Where e is there now minus t times t wherein the partial done by integrals,
If you say that you have the base of this request,
there are times minus any base,
have to be given to a base above t minus the integral means
This conventional fractional integration is done through.
T in five of the integral t easy six divided by six
place them here by mistake have not done this you will find the number.
What is important to establish this integration, After him an account of this integral.
That's a good thing to do it, but this problem is that the basis weight
integral can be edited.
The third integral as follows: A given vector,
Calculation of the integral dx u once asked, but that,
integrally connected to the orbit 'll determine whether,
u is independent from orbit potential are saying.
Here the two variables.
Do you know the fit criteria were negative vx.
The third component is really here was not out of here,
but when the third component rotational mean, you're writing ijk'y,
radians of the components dx, dy, dz are writing.
After that u We are writing to bring components.
The components shown in zxy, zxy.
Now i see here To find components
we are taking the first line, We take the first column.
Y is a derivative of y minus derivative of x by z,
x is derived by z zero So one is staying here.
When we arrived we start with j axis,
plus or minus signs plus As for going.
Gene first row and j We take the column.
See here x by y by derivative minus derivative of z z.
But in the jar so that the negative staying here for a.
Already shows the following criteria, If zero rotational
independent of the orbit, If nonzero dependent orbit.
Already one from 0 enough to be different.
But just to finish third problem here we're calculate component.
Any number already nonzero clear that although the zero vector.
So it's our basic criteria.
Necessity and sufficiency Question
the condition is not provided.
Therefore, this integral depends on the orbit.
Hence there is no potential of the.
Therefore, this problem We do not calculate too.
Is connected to its orbit not given to orbit.
Therefore the problem, integral is undefined, it can not calculate.
From A to B in this example to a gene given, but not put in orbit.
The problem says that the integral of the value orbit
Determine whether it is connected says.
Then you are independent from orbit There also needs to provide a trajectory,
You can find potential potential After finding was independent from orbit
b, value of the potential You can enjoy a the value.
Now we'll move again.
Gene competence, even necessity rotational condition is zero.
Again here, the rotational ijk'y are writing for.
DDX, FDI, ddz'y are writing.
Vector given first, second,
The third component of the third as rows are calculated.
I still see component If we consider the first column and
We're taking the first line, This remains backwards.
See here, by XY himself from x to y are derivatives,
x've written here, xz'n from the derivative with respect to z has an x,
x minus x, it was zero, that means a would be the potential promises.
When we look at component j,
XY himself x to see here Y has one derivatives.
But we start with j axis for that.
The last time to the latter, According derivative of AI z
here comes a year too, The minus sign here, but see,
j thus have a negative sign, The plus sign is going to say.
As you can see, this is zero.
At the same Calculate the third If you make it zero.
So the rotational zero.
Here are just now realize that, There is a potential of the vector field,
If a gradient because a
The curl of the gradient is zero almost identical to those provided.
If the potential of this integral is independent of the orbit.
Now let's find them.
Given vectors are writing here again.
So we discovered the existence of an f.
So have a fun,
How do we find that f, How do we find the potential of this fun?
For the gradient of potential concept gives u.
So here is the x component gives.
gives yz'y.
derivatives with respect to x, the partial derivative gives yz'y.
This integration will take it.
This is integral to this one percent doing the hard tasks temporarily.
It brings an x multiplied with x, but We need also an integral constant.
y and z can be a function, In contrast, because we look to provide,
we take the derivative with respect to x percent of it comes but here too, since there were x
acting as a constant in the integral function at its derivative with respect to x is zero.
So we're on the right track.
Now here we have a little further.
F is a portion of a structure found.
Here two unknown at this time a function of unknowns.
Initially, for the three unknowns has a function,
were reduced to two unknowns.
Derivative of f to y gives V.
Because the second component of the vector, derivative of f to y.
From here, you have given here.
When we put this V
where the derivative of the function g 're getting when we receive them.
See, we take the derivative with respect to y of f When x and z which serves as a fixed
x and z are only here to stay, the x z have already been given here.
So here left and right terms
simplifies each other, g is the derivative with respect to y must have been zero.
Means G to y is zero means that g is independent of y.
So y may depend only on z.
Because x is not already connected.
As you can see from the three unknowns We have two unknowns in the function.
Wherein a single unknown functions.
If you put it w component, ie, where the
The third component of the vector z is a derivative of f.
F is the derivative with respect to z are writing here.
H is derived by z, h is az derivative is equal to the given function.
Here again, as you can see the specific parts simplifies each other.
DH backward slashes in dz'n 0 appears to be.
So h is a needs to be fixed.
Just because something is a function of z,
If you say you're necessarily be zero be a constant, but,
means securing the potential We find that with xyz.
If you just put here the constant either way you will not notice a thing.
Because once in a and b plus c minus one for the future of a time
unable to contribute, thus beginning since it can reduce.
means that at a given point of XYZ 1 2 3,
As b 6 5 4, wherein in b 6 5 4
1 2 3 In the XYZ values that a place them according to the difference of x,
The difference between y, z of the difference, results are obtained as shown here.
Now this one had the potential problems, There was a potential for the given function.
Now the third problem something a little bit different.
But part of a given vector given, the x component given
y component, given The third component is unclear.
And we call this vector at the
u times the orbit integral DX If you select to be connected w.
W. belirleyince means function is independent of the trajectory
Because there is a potential, including potential call.
This is given as homework but very easy steps.
Commonly, the rotational be zero.
But here is a wine unknown such as that for the previous ijk,
dx, dy, dz This vector also Write components obtained by
As the rotational happening here.
This has to be equal to the zero vector.
This vector equal to zero is to be a wy'n,
a means that the wx.
If you keep them as a result What emerges is w.
What also is W identical to the previous problem
maintaining its potential steps which is what emerges.
The problem here is different from the others, vectors have not been fully
and independent of this orbit We see that may be required,
to ensure that the w We find what ought to be.
By W. found, the orbit independent when, of course, has the potential to f.
Yet there's an assignment.
So all of these problems always easy to work.
There are two kinds of problems; Thank to the exactly as given and are asked
Is dependent on orbit trajectory independent of Does a single rotational criteria is zero.
Curl is zero in the plane two variables, the two-component vector
görmüşlük fit and x minus zero generalization of having three dimensions.
In this problem the component given vector,
have given component, but not given to wine components.
We say again to get independent from orbit.
Means to get independent of rotational means is zero.
The solution here has shown you the way.
Here you will find w, you will get the answer
given here has been given this w After trying locate potential.
Potentially, the same two previous in the sample or previous
as in the example of an NF a gradient occurs.
Here we find the f.
After you find a through f b Go to the call.
From A to B when you get up here in b
removing the value from the value of ultimately need is here.
Now we had a problem, I said three types.
The first of such problems of the three species We have seen various examples.
As you do the homework I left for the problems.
There are two types of this already very simple also in the first kind, you
completely given vector, depending orbit Do you do regardless of rotational'll see.
A full photo If an unknown portion, and
necessarily independent of orbit If you want to get,
then it would be zero rotational is determined that the unknown part.
Thus these problems not too difficult to type,
not too different from bivariate.
The next problem Stokes theorems related applications will be.
Here you'll see a little more difficult, the
integrally on it's surface.
But realizing method calculating integral
which have substantially even Editing problem.
This together in a more We will see in the session.
Goodbye for now.
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