Explore
For Enterprise
Join for Free
Log In

Arts et humanités
Business
Informatique
Science des données
Technologies de l'information
santé
Mathématiques et logique
Développement personnel
Sciences physiques et ingénierie
Sciences sociales
Apprentissage des langues
Diplômes
Certificats

Explorez l'intégralité de Coursera

Parcourir
Rechercher
For Enterprise
Connexion
Inscrivez-vous gratuitement

Yöne Göre Türev Çözümlü Örnekler

Loading...

Çok değişkenli Fonksiyon I: Kavramlar / Multivariable Calculus I: Concepts

Université Koç

4.7 (19 notes) | 6.6K étudiants inscrits

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

Visualiser le programme de cours

Avis

4.7 (19 notes)

5 stars
16 ratings
4 stars
1 ratings
3 stars
1 ratings
1 star
1 ratings

GZ

Jun 02, 2016

Aynı hocadan diff dersini de bekliyoruz. Hızı x1.5 e alınca gayet başarı. Teşekkürler Koç

BC

Nov 04, 2015

Türkiye için önemli bir başarı.\n\nEmeği geçenlere teşekkürler.

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.

Yöne Göre Türev ve Gradyan15:49

Yöne Göre Türev Çözümlü Örnekler12:09

Üç Değişkenle Zincirleme Türev, Yöne Göre Türev ve Gradyan13:41

Gradyanın Anlamı14:53

Enseigné par

Attila Aşkar
Prof. Dr.

Transcription

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.

Explorer notre catalogue

Rejoignez-nous gratuitement et obtenez des recommendations, des mises à jour et des offres personnalisées.

Coursera propose un accès universel à la meilleure formation au monde, en partenariat avec des universités et des organisations du plus haut niveau, pour proposer des cours en ligne.

© 2019 Coursera Inc. Tous droits réservés.

Télécharger dans l'App Store

Disponible sur Google Play

Coursera
À propos
Direction
Carrières
Catalogue
Certificats
Certificats MasterTrack™
Diplômes
For Enterprise
For Government

Communauté
Learners
Partners
Developers
Beta Testers
Translators

Se connecter
Blog
Facebook
LinkedIn
Twitter
Instagram
YouTube
Blog Tech

Plus
Conditions
Confidentialité
Aide
Accessibilité
Presse
Contact
Répertoire
Filiales