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
Yöne Göre Türev Çözümlü Örnekler
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En provenance du cours de Koç University
Çok değişkenli Fonksiyon I: Kavramlar / Multivariable Calculus I: Concepts
17 notes
À partir de la leçon
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.
Rencontrer les enseignants
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Directional derivative in the previous section I have seen the concept of
and its largest and smallest values account of how we have learned that we will.
Now, as usual, our approach these problems resolved
To reinforce the concepts and calculations We want to increase our ability.
As the first example.
Yet it because you're getting a simple surface to see
easy, or on accounts
review the main concepts stuck We can miss.
This is a paraboloid.
x and y, z is equal to zero, nine, means a paraboloid in the top nine.
Hence the downward pointing minus signs a paraboloid.
When we gave the paraboloid z zero xy-plane x squared plus y
the radius of which is equal to the square of nine three The circumferential cuts.
Here in the xy plane view.
As a perspective view where the We see the circle.
We stayed here for one or two points in the xy plane we choose.
One or two points here, in the xy plane.
Also in perspective view in the xy plane the point is here, but in the third dimension
We know that we come up here When we find the point on z.
To this point, the height x of a two year a and
Five out of nine to four that I remove We find means of four
From this point the height of the horizontal plane In the four.
We first question, one by one
The derivative in accordance with the direction finding we want.
However, if we look at the plane that this is a vector
a point in one or two places an in the direction of a vector.
Considering again a plane in three dimensions
In one direction, it does not pass through the center This vector.
This is when we rise above such base a three-dimensionally
vector such that it is projected that such that vector.
This surface and also with the point
find the largest and the smallest derivatives we want.
This foreboding can easily reply question.
Our paraboloid such a paraboloid our.
Now up on the paraboloid Heading largest slope occurs.
On the way down to the smallest slope occurs.
See, when this maximum slope direction
If we take this inward projection an outgoing vector.
Thus, the components will be minus minus.
However, the smallest slope minus sign will be going down.
In this direction at this point and direction vector
away from its center toward the projection will be a vector.
Foreboding them easily we can find.
Now let's move on to the calculations, the calculations easy.
We have such a simple formula, gradient We will calculate.
We'll take the inner product of the vectors given but u independence from the neck of the vector
to ensure that we divide the length of the where the unit vector so that it occurs.
Very easy gradient, where the f function
derivatives with respect to x and y by will take.
Minus two x minus two years.
Gradient in a two-point value When we look at the
will be minus two and minus four.
We're putting a minus instead of x, y instead We put four minus two.
And that's just the foreboding in the same direction.
Easy going.
The highest slope with my fingers in it
I showed up towards the gradient.
This will counter the vector
The x's will be too inward minus sign comes from there.
it
When the value of the account
dividing this by the length of the inner product will account.
This is a very easy thing.
but the units one by one to find the length of the vector We divides the square root.
Gradient for this one minus two minus four We're putting.
From multiplying minus two minus four minus six as we find.
The largest and smallest slopes in just as I said, in accordance with our internal önsezgi.
The largest value and the smallest value of the gradient
the absolute value of one plus sign One of the minus sign.
The largest and the smallest value is formed negative aspects which formed the greatest value
The negative components, it's just that I showed will be this kind of as a vector.
This projection minus minus its components.
Vector that goes down to the smallest value thereof for the
a vector projection toward the outside, therefore, components plus plus is happening.
So these are pretty easy calculations.
However, if we consider the way again
plane with a two-coordinate here we are.
Here we say that we found inside Tmax towards
vector, so the maximum, the largest slope This content occurs towards the direction.
We have also found this foreboding.
This is just like I showed up at
a vector towards the gradient is brings.
The gradient direction of the vector is zero a vector which is perpendicular.
See here for two and if we take one minus Two times two is four, minus one
times four minus four, zero, this front There is the importance of plus or minus.
Now comes zero anyhow.
It also shows the direction of the line.
This occurs at the zero slope in the direction large
and the smallest pitch in the direction in which the formation of steep.
Says that the general theory.
The second example we take the same surface, paraboloid facing down.
And you're getting the same point, this on the plane a couple of paraboloid
Select the item above point we find.
But as there is a difference this time, we call
In a previous post u were directly We are moving on a parabola.
This is the equation of the parabola y equals two times the square root of x.
If we take the square of this year is equal to the square of X, a parabola to be x-axis.
However, the y-axis, which we normally We are used because there are more parabolas y
equals x squared function was taking but nothing has changed here.
On this parabola progresses, more precisely in space, while moving surface
We will stay on this parabola izdüşümü so when you scroll in a two-point
We want to find the direction of the slope.
Again, an easy calculation, the formula is the same.
Unit gradient vector multiply.
Where does this unit vector, this curve through.
Let's do an inspection before it's a wonder two points on this curve.
Indeed, a data y to x is equal to two we find.
So the right spot, this parabola We are working on a dot.
At this point we need to find T vector.
You can find a very simple way We know that because the slope of derivatives.
If we account derivative, derivative of the square root of x one-half times the square root of x is happening.
But one or two in the front there is this dilemma
When x is equal to one another in simplification We find a slope that.
Slope of course, but a number of us, a vector You have to have a tangent vector.
This tangent vector, this means horizontal
When we go up a slope above We're going.
So we're going above a, those of us immediately gives the vector.
If there was something more complicated, perhaps
in the plane we saw in the third section slope of the tangent directly in
vector, or even finding the unit tangent vectors We used formulas
but a much simpler way wherein able solutions.
We need the unit tangent vector.
His take on the length of the u We divide.
It is already in a previous problem We tried to strike.
Therefore, the result is always the same turns.
Here we can provide as a.
I wonder if this value is the value in line with the is the greatest and the smallest claim that
we are staying within the value I really
The maximum value of the length of the gradient We know that.
Gradient that is minus two minus four length of first plus second squared
components of the square and the square root of four plus one six, twenty, twenty square root.
Is approximately four and a half.
We have seen that before.
It is the smallest value of the minus sign.
Therefore, we do not need an account.
We can now test the following monitoring, we can provide.
Really in line with slope
derivatives, directional derivatives, these two values Does staying in between, it's true.
Because this value is smaller than the maximum value, is bigger than the smallest value.
Therefore, as a result, we found to internalize
We're going to have a checksum.
So far, we've worked with two arguments,
three variables to generalize in this section we want.
But now take a break, that you We saw a little internalization of topics.
Problems by yourself again maybe Earn thoroughly solving skills.
These three variables in three dimensions then we learned
chain derivatives, directional derivatives and We will generalize the concept of gradient.
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