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Çözümlü Örnekler ve Ödevler

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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)

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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

Düzlem Eğrilerinden Hatırlatmalar ve Uzay Eğrileri, İki Değişkenli ve İkinci Derece Fonksiyonlar ve Karşıt Gelen Yüzeyler

Uzayda yüzeyler: iki bağımsız ve tek bağımlı değişkenle tanımlanan sayısal fonksiyonlar. Yüzeylerin anlaşılması ve temel yüzeylerde çizimler: perspektif görünüm, eşit değer eğrileri ve kesitlerin çizimi. İki değişkenli ikinci derece kuvvet fonksiyonlarıyla verilen temel yüzeyler. Silindir yüzeyleri ve dönel yüzeyler. İki değişkenli özel bir yapı olarak karmaşık değerli fonksiyonlar. Mathematica, Mathlab, Ghostview… gibi yazılımlarla bilgisayarda çizimlerden örnekler.

Düzlemde Teğet ve Dik Vektörlerin Hatırlatılması, Çözümlü Örnekler ve Ödevler17:00

Uzay Eğrileri Yay Uzunluğu Cinsinden Eğrilik, Burulma, Teğet Vektör, Dik Vektör ve İkinci Dik Vektör, Fresnet Formülleri14:12

Uzay Eğrileri Vektör Fonksiyonun Parametresi Cinsinden Eğrilik, Burulma, Teğet Vektör, Dik Vektör ve İkinci Dik Vektör14:49

Çözümlü Örnekler ve Ödevler16:59

Enseigné par

Attila Aşkar
Prof. Dr.

Transcription

In the former part, a space

curve, the parameters defining the curve

According to t, the solution that we have achieved curvature

twisting, finding a unit vector We learned.

Now this is what we have learned to consolidate as well as

To view an interesting problem, Let practices.

Proceeding to the application, our formula Let's get together.

Such expressions of b t n,

Expression of the expression of curvature and torsion like this.

As provided herein, now calculation method goes like this.

The first step, because the curve of x x We define with.

We need to calculate the first three derivatives.

point x x2 x3 point point because, at t x point b

There are points x2, x2 points off there and in to, there x3 point.

Then, after the calculation of their these multiplications 2s

3-multiplying and multiplying these lengths account We have to this because they

formula is set in the third step in -putting them.

Now that's a spiral helix Let's examination,

ie before the construction of a helical Let parametric equations.

Maybe some of you have done, yapmadınızsa can be done easily.

As on a page,

draw a line of a circular shape that

I Bükel on a circle that is, a on a circular cylinder.

, On which the straight line drawn, us will give a helical spiral.

If we look a bit more digital, horizontal 2 pi times

a, b vertical length of 2 times pi If we choose, horizontal

we translate this into a circle is the radius of a circle wall.

This slope also seems to be a divided by b.

With these thoughts helix equation, x

A times the cosine t happen, as it on which we

The projection of this point moves falls on this circle.

On this circle is equal to theta If we think we're starting from scratch,

It is a times cosine theta x, y at a time is sine theta.

Parametric equation of a circle, but The difference between the circle, the circle always in the plane

Standing here on the way back in, b

in line with the rise of this the helix formed this happens.

Thus defining the helical vector here

Or seen as a cosine of theta t, We chose here as a parameter.

be a sine t and b t.

As we define the spiral

Let's just its derivatives.

immediately by the point x such that x from derivatives

t minus sine and cosine t come b.

When we take the second derivative of x points We take one more derivative.

I also point x3, x2 point again We know that with derivatives.

Now, in this formula, we point x and x2 noktanınvekt needed to multiplication.

The length of point x, the point x of the point x2 multiplying the length of a 3-in.

Now if we make them, with point x

vector product of the point x2 work determinants comes from.

i j k are writing.

Used as the first vector point x we put on line.

x2 point we put in the last row and We are opening.

I components of the first row and column tumbling these

zero, but the pros will become negative, the second

diagonal change of sign changes Due to changes in the EU once the sine t.

.mu.l we're looking at, minus the EU in a similar manner times the cosine cosine t involved.

K, .delta we're looking at here is an interesting There formation.

See a squared sine-squared t, where a square For frame t is the cosine,

staying here is just a square, so this expressed very simply appeared.

If we multiply this by taking away the point x3 See here x3 point

We found point x x2 point.

It is also very interesting from the inner product simplification going on, see

EU sine-squared t, b times a here,

cosine squared t, then the two combination

b involved because a square sine-squared

t is 1 to t plus cosine square, a is multiplied by the square zero

For one thing does not come from there, see How easy was this triple product.

When we look at the length of point x, x

point was here, see the first one here frame, plus

second squared, sine-squared t plus for giving a squared cosine squared t 1,

third-squared plus b squared here, plus b squared is coming from Accra to say that,

If we define it as the definition of c frame, x is a fixed point of the square.

Gene vectors with point x x2 point

Taking the length of the product, see here again, sine-squared t,

kosinüzs frame t is happening here, ab, a square b squared is coming.

Here comes again than a square, the accounts here

You can easily see, the AC turns out, this Coming up with a simple structure.

When we put them to behold If we rewrite this

We found the size, t, by definition, or the formula

x at point x on the length of the point It was part of the x point

We found that a certain length c, divided by the it turns out.

because we need to point t n, t point t point

longitudinal division, t, t certain point now easier to find, Sinisa will be cosine,

will be minus two is sine cosine out to be negative for the negative

turns out to be constant b is divided by c to b its derivative zero.

So we did not find t point point their

divide here by dividing the length of the c going.

Already it is one of the neck

easily see, cosine squared plus sinus

square point x and x2 port vector b multiplication, we know that in line,

the unit vector divided by its length This work will be of this size here

We take the size of its longitudinal split time

As you can see a time it turns out to b.

Now here are a variety of delivery possible.

The inner product and the inner product of t

is zero because it is orthogonal need.

See here sine cosine t t here t sine cosine t, different signs

At the same coefficients, where the reset This is because the product is zero.

Again let's look at the product of n and b.

Sine cosine, sine and cosine again

different sign, and the third is zero, again output from the zero.

If you hit it in a similar way to zero in We see that go.

There were as curvature and torsion,

point x of the point multiplied by x2 length, it

We have calculated the point where AC x

c divided by the cube of the length of the point x respectively.

You can see from the division of c cubes As we come to this conclusion.

And finally

iii of this triple product of these two product longitudinal

are divided here b divided by c square is çıky.

Helical structure to emphasize the privileged because I got

output constant curvature, torsion also constant output,

How the curvature at each point on the circle

If that is going hard at it on the helix on spiral

at every point of the curvature and torsion constant involved.

that does not change from point to point.

If we reverse the problem, curvature and found torsion is constant curve

one has to say a huge curve in the universe, both curvature and torsion is constant.

Now comes to mind right now, this spiral, 've read the double helix

Perhaps you've heard at least, consists of a helix structure of the DNA

I wonder where I wonder if there is a relationship molecules that together form

Seems curvature and torsion

A constant travels around the subject our work

Or something out of nothing in excess, but The interesting thing is certain.

Now I want to make a second example.

Route and leave it a little.

It's called double parabola, because at that If you look at the third

size is not taken instead of t x, y de 1 divided by 2 x squared

Or y not be here again

t is difficult to say at x x cube would be divided by 3.

So you have a parabola in two directions.

Someone one second-degree parabola tertiary

double parabola parabola it from that angle curve is called.

T on the curve equals 1, and in general, for t

curvature and torsion unit vectors Please account.

As the values t is 1, it I want to give the solution immediately.

So I do not want to give the interim, the same as in the previous problems.

We will take the derivative of x, the second derivative We will take, we'll take the third derivative,

we see their point x of the point x2 vector

multiplication, will account for the triple product and t is equal to 1 will receive the value.

We make them the following conclusions emerge.

Search for you to do it accounts I'm waiting, I'm giving it as homework.

As in the general account; all of a sudden t equals

Let's not one, that's still several variants take them

When we put this t function as we find them.

Issues to be considered here, Unlike spiral at any point,

both torsion and curvature of the tangent, We see that changed.

Where t is equal to 1 if we put this

generally results from a previous custom We're reaching consequences.

Now here are a few more papers.

This drop can say spiral, a In succeeding helix D. The difference between succeeding helix D. radius was fixed.

If you look here, along with t radius

There are an increasing spiral here, drop it It's called spiral

because it does not remain constant radius, opening going, but also

the z direction is rising, it is equal to t

zero and a is 1 and b is equal to

If you find one that's close to the corresponding values

so it turns out, is going to like, vectors that turns out.

The same way as before, where

A second thing is that, again, this time a

but this is on a spiral circle spiral

See the falling squares of x and y when you get

step onto the cylinder but a means This time linear

, not as second-degree rise.

So increasingly growing step.

You have arrived at once if you return in succeeding helix D. height there were 2 p,

twice in the neck if you return the same height I was going repeated.

Up and down the stairs of a minaret Taking your promotion

always at the same height as the stairs formed,

If this step is going to rise here, is going to increase,

gene wherein the accounts of the previous Follow your Google account.

There is a duty on the last one.

We know that the position x, t is the time

The first derivative of the velocity data, the second derivative The acceleration data.

Now i j k components that account we can

As the calculations in terms n and t We can.

Where v x is the definition of the derivative with respect to t In and

the length of the vector v, the numeric as we show.

This numerical velocity vector, which is the number if you have speed

We also quickly realized the velocity, speed The magnitude of speed.

This identity to find these results I'm waiting.

the acceleration speed

derivative with respect to time, wherein a t There are so tangential direction acceleration.

And also the perpendicular direction

There are tangential acceleration in the direction perpendicular to that in this We call centrifugal acceleration,

it already in many physics courses The result is shown.

Ever with derivatives of vectors We care.

Dealing now with the integral of the vector we want.

So far relates to derivatives Examples of what you see and

To late now subject here We stop.

In the next session of the integral of a vector will deal with.

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