Ders çok değişkenli fonksiyonlardaki ikili dizinin birincisidir. Burada çok değişkenli fonksiyonlardaki temel türev ve entegral kavramlarını geliştirmek ve bu konulardaki problemleri çözmekteki temel yöntemleri sunmaktadır. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler Bölüm 1: Genel Konular ve Düzlemdeki Vektörler Bölüm 2: Uzayda Vektörler, Doğrular ve Düzlemler; Vektör Fonksiyonları Bölüm 3: Düzlem Eğrilerinden Hatırlatmalar ve Uzay Eğrileri, İki Değişkenli ve İkinci Derece Fonksiyonlar ve Karşıt Gelen Yüzeyler Bölüm 4: Özel Yapıdaki İki Değişkenli Olarak Karmaşık Fonksiyonlar, İki Değişkenli Fonksiyonlarda Kısmi Türev ve İki Katlı Entegralin Temel Tanımları; Limit Kavramının Gerekliliği ve Anlatımı Bölüm 5: Türev Hesaplama Yöntemleri Bölüm 6: Türev Uygulamaları Bölüm 7: İki Katlı Entegraller ve Uygulamaları ----------- The course is the first of the sequence of calculus of multivariable functions. It develops the fundamental concepts of derivatives and integrals of functions of several variables, and the basic tools for doing the relevant calculations. The course is designed with a “content-based” approach, i. e. by solving examples, as many as possible from real life situations. Chapters Chapters 1: General Topics and Vectors in the Plane Chapters 2: Vectors in Space, Lines and Planes; Vector Functions Chapters 3: Reminders of Plane Curves and Space Curves, Quadratic Functions and Variables, Surfaces Chapters 4: Special Two Variables Complex Functions, the Basic Definition of Partial Derivatives and Two Storey Integrals in Two Unknown Functions ; Necessity and Details of Limits Chapters 5: Methods of Derivative Calculations Chapters 6: Application of Derivatives Chapters 7: Two Storey Integrals and Applications ----------- 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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Из курса от партнера Koç University
Çok değişkenli Fonksiyon I: Kavramlar / Multivariable Calculus I: Concepts
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Türev Hesaplama Yöntemleri
İki değişkenli sayısal açık fonksiyonlarla tanımlanan yüzeyde teğet düzlem ve diferansiyel. Zincirleme türev yöntemi ve tam türev. Yöne göre türev. Gradyan. Koordinat dönüşümü ve Jakobiyan. Taylor serileri. Kritik noktalar, en büyük ve en küçük değerler. Türev hesaplamalarının üç ve “n” değişkenli fonksiyonlara genellenmesi.
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Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hello.
So far we have done with a gradient accounts.
Gave meaning from time to time.
Here tidy gradient as I want to open a little more meaning.
A very important concept because the gradient.
Two variables partial derivatives gradient processor consists of.
This first component to the second component We can find the way.
That when affecting a function f on f with respect to x, and
y components of f are obtained by We are.
As for the three dimensions x and y as well as the bi
We are adding these three variables but also our influence on function.
I put dot here.
Although the addition of a fourth four will work b.
50 Although many variables, that 50 units to be variable, it will.
Then of course x y z such that x is not an x two x I need to go as three x 50.
Conceptually, therefore, a great No difference
Size 50 to go to two or three dimensions account between making.
A significant difference in two dimensions, of course, two
Our variable as a function of time have the possibility of drawing.
Therefore, this is a premonition to improve our contributes.
Even if the three variables need to move four-dimensional space is needed.
Because the three independent variables.
Bi fourth-dependent.
Therefore, we are losing the ability to boot.
But it's not too serious bi losses.
Because the account that prevents us from doing things b does not exist.
Already a bit of the power of mathematics
in short, to be able to go beyond our intuition provide.
I want to start with two arguments.
Two variables from three different perspectives gradient I want to give a detailed review.
First, in terms of the tangent plane.
Now when working with a gradient
the equation of the tangent plane we saw before.
We buluyo tangent plane equation as follows: z
is equal to the function y is zero, x is zero value.
Function of the partial derivatives, with respect to x multiplying the partial derivatives of x minus x reset.
Similarly, the partial derivative to y but that x is zero y
This calculated value is zero at zero it shows y minus y be nullified.
We know it from the equation of the plane plane perpendicular vector
f x, f y, and will be minus one.
We have seen this in several ways.
We rather minus the old Artillerie 've seen.
But it is important to strike the
Artillerie getting minus the use of Not much difference between b.
We two E vector, the vector My ayrıştılal.
BI them only for the part of the are found.
In the section for non Öbürkü.
Just as you can see for which of
indeed part of the gradient components forms.
What this means is this:in the space of tangent plane When we receive the projection, where clearly
I wrote in the e, perpendicular to the surface
When we receive the projection of the vector We find a gradient.
Because here, bringing zeros
the projection of the third component have received We're going.
If it shows a large gradient with N f this is going on.
Here again in a second review
We have already encountered this N. direct review, to interpret.
These components of the surface normal vector I see that process.
A little more detail, all of them something which is actually intertwined with each other
but we look like f x y z equals though but If b is constant and z.
z mean to hard surfaces, where z is equal to f
x and y surface of a curved surface such Suppose that,
z is equal to the cross-section when taken z'yl a fixed horizontal plane.
This means a cross section of the two we find.
This would be such a line.
Wherein on the curve, this curve we take the projection of x and y for a fixed
curve, wherein the projection of the curve, steep gradient vector that is happening here.
That's right, I'm sorry for this cross-section curve a vector perpendicular.
Yet another thing that the vertical surface.
These constants that when we cut
This co-equal values, lines are producing is here.
Perpendicular to this vector.
If we draw this further in x-y plane that f x y is equal to that fixed curve
it will curve up and down b The view from the perspective of the interests that kind of thing.
x is zero at zero perpendicular vector y.
In a slightly different as both two and three Let those of size.
But it's all up and down each other supplemental information.
f x and y equals c curve.
I'm not talking here of a surface.
Just off the bi function given by a curve in the plane.
This eye-parametric equivalent, Consider the representation.
t is a function of x and y, t is again that is a function y.
We complete the derivative of f with respect to t Let's take.
Partial derivative with respect to x by chaining rules derivative derivative of x with respect to t.
partial derivatives with respect to t y to y derivative.
But it is hard for the right side of it derivative would be zero.
This short, fully derived from the partial derivatives consisting
gradient with component D x,
consisting of d h u vector multiplication, inner is the product.
We see that this inner product is zero.
This means that the gradient of this curve steep is a vector.
If we go for three dimensions x and y are equal again Let the hard surface.
Although these three variables Here restrictions
due to only two of the variables is constant, touch, is free.
This surface infinitely many x is equal to x in t is a function of,
y is a function of t, z is equal to The curves are a function of t.
We choose one of these bi Let's think.
This large total derivative of f with respect to t again the product of the partial derivatives times this x,
derivative of the product of x, y derivative product, the product of z derivatives.
This d x, d t, d y, d z, d t location
so that the derivative of the vector We know that the tangent of the curve.
A location in a space orbit parameter
derivative gives a tangent vector.
This one is not the length of the tangent vector.
Because he can be a random parameter.
But as a result of this surface
gradient of a curve with tangent vector by multiplying
This is not another b Siya, the inner product where b is not Siya another term.
The first component of the first component thereof,
wherein the second component of the gradient of the tangent
The second component of the third grade with component
that the product of the third component of the tangent giving.
If this is zero the two orthogonal means.
So in the current identity of the surface we define the random
of a curve, so all of the curves multiplied with a gradient will be zero.
Consequently, all of the curves, from that point The all curves tangent
gradient is steep gradient for the surface is steep.
And here we see the following common feature of these size independent.
Two boyutda a curve in the body.
This curve is steep gradient.
A surface in three dimensions.
This surface is perpendicular to the gradient.
Although the N-dimensional space of N dimensions N minus will be a one-dimensional gradient perpendicular to the surface.
Therefore, regardless of the size of major The geometric objects, fun
constants is equal to the geometric objects for gradient is steep.
We can do it as visually.
Where x and y are equal in two dimensions for a fixed or zero function curves to give this b.
Suppose you have that kind of curve.
Perpendicular to the gradient of the curve.
So perpendicular to the tangent.
When we received for our three-dimensional minus y equals zero surface.
Gradient perpendicular to the surface.
As a concept a bit further following we try to understand,
that on the surface on this curve curve taken
Although any curve perpendicular to the vertical What if we come for the gradient vector.
But this curve with the horizontal plane itself bi Suppose z is equal to z reset the cut.
Here you have a curve.
Gradient of this curve will be small for him.
If larger, x, y and z is zero in the z
If we say that only a function of two variables will be.
This curve.
Small gradient of the curve f.
Projection of each other at the two work We also see that.
But here is the most basic means gradient which object one.
Now if the curve curve, face perpendicular to the surface, then is a vector.
Here is an example're doing.
This we know from simple geometry bi situation.
Let a circle.
Here we take this circle.
That at any point of its vertical
is the line connecting the center point We know.
Gradient in this direction should be.
When we get a sphere of any
said vertical center point of the sphere passes.
Therefore, this should provide each other.
When we received the gradient from the center have a right to be.
Indeed, the equation of the circle.
Taking the gradient of its two x is zero, two y is zero, x
zero point zero x and y from the center of this is a direction passing.
There are two factors, but it is important
a gradient that goes from the center In line with that.
Similarly, in the sphere when we received x z y is zero zero zero.
But these two direction twice does not change.
This is really a gradient that x is zero y
connecting the centers of z is zero point zero Accordingly, we see that.
If we do not put it as z z z is zero if It is here that the equation of a circle.
The gradient of this circle on the
that is projected in space of the gradient we see.
Where x is zero, y is zero, because the same involved.
Now we have finished a major part.
If you think this part of things too much bi does not exist.
A lot depends on the tangent plane.
Tangent plane and the full chain derivatives derivatives have achieved.
It's a bit of a different direction based on interpretative derivatives have achieved.
According to this aspect of the derivative gradient us because that leads to directional derivatives
we calculate the unit with a gradient direction We have seen that vector multiplication.
Therefore, we have learned much much more No number of concepts.
But I need to be able to work on these accounts.
In understanding the concept of the accounts he not that hard.
Thereafter, the addition of derivative applications We will see some applications.
These Taylor series and extreme value.
In this section we have finished more than I have seen methods of calculation.
But I also saw a bit of practice.
Here, the calculation methods now
We know accepting applications will do.
These extreme values are already finding the best optimization problems, which we call type.
These are examined in the next session We'll start.
Until then, goodbye.
I hope you geçiyos out of these issues.
Solved examples also yourself efforts to
Are you at the same time and with the work Your bakıyos.
So also do the quizzes quizzes Do not forget.
Goodbye.
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