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
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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
Türev Uygulamalarından Seçme Konular
İki değişkenli fonksiyonlarda Koordinat dönüşümleri ve Jakobiyan. Diverjans, Rotasyonel ve Laplasyen. Dairesel koordinatlarda gradyan. Doğanın dört temel kısmi türevli denklemi: dalga, sızma, Laplace denklemleriyle Schrödinger denkleminin tanıtılması. İki değişkenli sonuçların üç değişkene ve mümkün olan durumlarda “n” değişkene genellenmesi. Karmaşık değerli fonksiyonların özel yapısıyla kısmi türevler ve tam türev.
Meet the Instructors
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hi there.
So far extremes We saw three types of problems.
The first is that the local extreme value local minimum and maximum values??.
Near a point here forever We're talking about small areas.
Secondly, that the absolute extreme values the smallest maximum values??.
Wherein a closed finite We're talking about regions,
that the boundaries of the region, including which is formed in the interior, including
largest and smallest We learned to find value.
Lastly particular restriction a function under the smallest and most
finding large outlier values we learned to do it in three different
see that the path has undergone a most suitable for applications optimum
method of calculation of the LaGrange We saw that with the multiplier.
Now we are entering a new field Up to now we have not touched it.
Here we want to find the integral of a that the largest and smallest values.
Important unknowns here different from the other cases.
In other cases, our unknown algebraic values, that is, a
point's coordinates.
However, a function wherein We will try to find out.
It's our job to find the function but that this function
equations to find the right college Many of onset
In the field of civil engineering and foundation used in many fields of science.
Let's start with an example.
This is an example from physics,
In physics, we know that the kinetic energy of a mass times velocity squared divided by two times.
There are also potential energy, these two a move is going to be one,
of Newton's laws work in this We know that.
Force equals mass times acceleration, force negative derivative of this potential energy.
Now that there's an equivalent structure.
We get the job done hesaplasak kinetic work is being done and
hence potential energy a job is done.
Potential energy minus the work done by Coming to the mark because the potential energy
How much of an existing power If you have done so you can consume is
you consume minus sign for You must put the resulting surplus
potential energy to get the work done.
If you are not related to physics It does not matter much,
Imagine that such a function.
This is because the function is not t is known as a function of
x V given as a function of x, but do not know where the equivalent.
For now, we leave it just short then equal to the time that Newton's laws
We will see that, but nature rules Newton's law that for a
The equivalent forms is I want to remind you.
This in many cases there is only Newton but not in the laws of nature,
in operations research, and then
There are such problems in economics.
Our problem is generally as follows: An unknown function y
The function of the x and y have a given function depends on us that x
and Y is bonded to Y in the derivative This is such an integral connected.
If we recall the previous example The task was doing x t y
tasks that are connected to the variable x x Instead of T-dependent variable y
There was also because we were getting and derivatives There kinetic energy derivative.
Let's assume for now that a water function get fixed ends.
We do not know where such a y y
We are looking to these integral Let the critical values??.
This problem of an outlier, but an upper In previous classes, because it is a
functions that were given the value of algebraic
that the extreme values ??of algebraic unknowns wanted.
Here, x is a function of So we're looking for a y a
our knowledge, our function.
When critical values We have the following in place of x
Let x plus delta x change function is changing and how
it's also the largest or smallest We find that we can remember.
Now we good here, it s Let's take a change in y.
Of course I will change it I change it the delta
Let's show this to y'yl of function derivatives
similar to a delta delta x in y.
Here an algebraic number x in exchange A change in the function here
Of course, the base year is changed, the change in y Let's just say it changes the base of the delta y.
As we have done:we We are holding steady at two points,
y is the function of these two points This function passes a
We give change for any x y change is happening at the point of the delta.
Of course, because it is hard tips this mutant functions
passes through two points.
Now let's see we have the following major F
There was function of xy and y base x is a function of a change
because they do not do x, but there is a number of we're making changes in the function
plus delta y y y y base was the base year is the base year plus the delta.
This function F plus delta F
Let's say that a function F in the delta exchange shows.
The same two variables As in functions
detecting the change in these functions We want to find.
We can show that the Taylor series.
This function Taylor When we opened with a series of
function in y and y base values ??will be.
Will be the first derivative of y y'yl delta by the second
y as y base derivative Delta will be the base year.
This, we calculate the delta F. From this expansion for F.
We will remove Fe each other and this will take you back to stay.
Now continue this series We can make, but also in function
As we have seen this value before 'll have to reset.
How the requirement as a condition of take the extreme values ??of a function
of the first term of Taylor As should be zero.
This plus the delta while I I I because the change in y
I will change that when we give I will say that this delta.
These two integral combine sugar, so if i remove it from the heat plus the Delta II
of course, the following will be and y is y'l increased üssül
The term which is negative this increase and this integral over x.
This initiative saw the first-order
I have seen the term here The first term in f
And the second term in f changes in them.
Now, of course, changed when y Delta also changed the base year to year as
from each other, but these two changes not independent, ie, changes in derivatives
such as the definition of derivatives is equal to the derivative of the change of change.
As there geometry thereof You can see from the figure that
y is equal to the delta derivatives.
This delta delta y base y is equal to the derivative
from the continuity of functions We have to assume that it comes in.
Now, if there is such a term See dy df base here
If we look at the base of the delta y df dy had a base.
This year the base of the delta DDX here We use the delta as y
but here there is almost vari base able to get a partial integration.
You hit it with the term plus the second
the integral term minus the first the second time derivative of the term
V as the integral base terms of getting to the integral
As these partial calculation of integrals the known functions of one variable.
Now if we place them.
See two terms from the term occurred a
them at the borders values ??of these terms also.
When we put it df There was no gelation did not dye it.
Here you have the base and delta y dy df
DX beats both of them We are also taking on integration.
There is a term in this limit.
It may be zero for Delta possible for all these years
term in brackets itself be zero, and
the boundary conditions be derived therefrom.
Where y is the fixed ends hard if you remember the following form
y, then y is fixed, delta in the year-end value will be zero.
If we use it to zero at the ends
If this will be a value or a given,
value will be given in ICU, but There is a second option against it.
I.e. they are not zero If it ends fixed
time becomes zero at one end of the base df dy.
At the other end of the base df dy is zero.
As you can see here that we obtain an equation that
a differential equation, we I do not yet know the solution.
However, many applications of this equation It is important to obtain even.
By the way here for him When searching for an extreme value
that end value of the integral function Finding the way we are treated.
Now I kick start it with an example.
One of the square root of the squared plus y base
dx'l multiplying the spring DSi We know that length.
We go from A to B. springs curves We want to find the shortest between.
In terms of the previous notation fx y
We were at the base of x and y do not look further.
That means that our function.
Our formulation of Euler's formula was,
We need to base df dy, We need dy'y df.
df dy y is not zero.
df base dye it with respect to x is not y So we're taking the derivative with respect to the base
a year divided by the square of the base is a plus in a split second if he thought the force
two will come through it two years The simplification will base future dilemma,
plus years of it at the base of one of minus one split second force will be formed.
As you can see it's only y a function that depends on the base.
Here df is zero dy For this we say that g
function in terms of the Euler equations derivative with respect to x must be zero.
So our equation is reduced to this simple.
So the derivative of y with respect to x y is zero the fixed base
should not base this gyu I need to be constant zero
y to be all the In order to be zero for
If a fixed base but gyu, if gy is constant at base year is fixed at the base.
y is constant base here but hard times x
plus c is a constant, including two is provided in such a manner.
Thus the derivative of y Please take a c y is equal to the base
Remove the derivative constant x two of a C constant for the derivative is zero.
Here we find right now, means connecting points a and b
the shortest of all the curves are correct.
We know this already, but in a flat space such need not be the type of
We know what happened in the plane, there We're going on a sample that has been revived.
Let's do this again, at least the business equation.
Here the intensity of the work done
The kinetic energy is the energy minus potential energy.
This integration time on the We find if we take the total business.
I do not want to enter into the physics of it here as the total workload
We passed so that energy We call the kinetic energy plus the already
potential energy with a minus sign have been brought, was ever made.
Let us now account for it.
Now the function f
There also does not seem t t x and x points.
But here there is a constant mass point x physics physics to communicate
You can also get one if you do not care one function here.
Now we have the Euler equation When we remember the Euler equation
were as follows:df dx point and this we need to t
also a need derivative There derivative of f with respect to x.
Let them account x to F wherein V is derived based derivative.
Moreover plain derivative.
Here we see it.
F is based on point x that the derivative is constant
twice by point x Does it would be the point x.
If you want to mix physics We know that this momentum.
According to Euler's equation dF dx point will be equal to the derivative with respect to t.
m here because of two points x F x Did we find the derivative point by point x,
The next time derivative with respect to t revenue received by the second derivative of x.
F, we find here the minus df dx
df by dx dx dV becomes a minus with a difference.
This drawback here is also a minus This equation comes plus
as we find.
V. If we get to the right job Remove the Euler equation we know the mass
times acceleration equals force.
This potential forces the derivatives of the minus sign.
As you can see here too With a simple extension now
we have seen extreme value problems We concluded a very basic.
By repeating this subject here We did not talk about the boundary conditions,
boundary conditions can be like you x is zero value is given in
or f
point x is given by the derivative momentum is also given.
Here we limit the initial Terms of problems occur,
us in this lesson they Our subject, but the following
To make the process a lot to you Your lesson will benefit.
Here you have a homework assignments are as follows:A vehicle,
e.g. a satellite Let's assume that launched this
simplified here reduced to one dimension.
At a point x in the unit path length q times y produces a fuel.
We want to find the trajectory yx.
minimal consumption of a and b Find the equation of the orbit that.
Of course, this differential equation will be quite a mixed
differential equations involved, but The solution to this except our topic,
but such an equation can be reached to see a significant gain for us.
This in many areas, for example, where We see a logistical problem.
The problem is as follows:consumption unit length qy burning, e per unit length of arc length ds.
Arc length of a base plus y dx times the square root of the square.
As you can see here, y and y
There are a function of x to the base clearly not connected.
Have such a function.
These are the Euler equation putting in orbit, the
differential equations are able to achieve.
so here comes the second year of the two base order differential equation turns out,
The solution to many of us, but it I do not care at this stage,
Go to the problems that such There are also specific benefits.
We're still standing here today.
After that we started in the first part again but that is just the beginning
we saw a presentation on a topic more fundamentally, we will be taking over.
All of these partial derivatives of our work To get your own integrals to get
an end in itself.
But at the same time with the partial derivatives differential equations have emerged.
This information the differential equation is happening in the preparation of
and is one of the key goals rules because of the nature of the people
The evolution of this type of activity Showing equations.
Goodbye.
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