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
Vektörlerin İç Çarpımı ve Vektör Çarpımı
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Çok değişkenli Fonksiyon I: Kavramlar / Multivariable Calculus I: Concepts
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Genel Konular ve Düzlemdeki Vektörler
Fonksiyon kavramı: girdi – çıktı, bir değerin diğerine gönderimi, çizit, ve dönüşüm gösterimleri. Çok değişkenli fonksiyonların sınıflandırılması: uzayda eğriler, yüzeyler ve vektör alanları. Düzlemde karteziyen ve dairesel koordinatların, uzayda karteziyen, silindir ve küresel koordinatların tanıtılması. Fonksiyonların açık, kapalı ve parametrelerle gösterilmesi. Vektörler: düzlemde geometriden cebire. Düzlemde toplama, bir sayıyla çarpma, iç çarpım ve vektör çarpımı. Bu işlemlerin üç boyuta genellenmesi ve üçlü vektör çarpımları. Bu kavramların geometrideki anlamları ve uygulamaları. Uzayda doğrular ve düzlemler.
Conheça os instrutores
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Our previous session of vectors I have seen two basic operations.
Addition and multiplication by a number.
Geometry parallel edges thereof diagonal
and by collection of the vector in the same direction changing the length
If we stand up c changing the c'yl We saw a number in shock.
And also that the numbers
algebraically how these processes We have seen how to do.
We collect in the collection of components.
The first component of the first vector, the second vector
second components, the first component We collect.
In a number of components that the shock We stand by number.
It came up with a simple rule.
Now we will see more of two processes.
This inner product and vector product operations we say.
What you need them?
As far as needed.
Our goal is to understand our environment,
to show our observations, to tell.
For the identification of two basic shapes grip is needed.
Length of BI them.
BI is one angle.
Now let's start from the most simple.
Let us be given two points.
points a and b.
These coordinate sets of the start of Let us unite to it.
And we combine among themselves.
Thus, we have obtained a triangle.
Wherein the angle theta between two vectors Let's just say.
Our goal this means that the angle theta to calculate.
Means to calculate the cosine of the angle B
Or if you know if you know the sine is sufficient.
To know that respect.
So we want to do and b points
this angle in terms of coordinates Can we calculate?
I once we know the length of the OA.
This is the length of x.
We know the length of the OB.
He length y.
Account these coordinates in terms of able.
The EU is the vector y minus x.
Because the collection of rules learned.
x, y when I go to the EU I find.
So I remove y from x EU remains backwards.
The length of this.
Yet we know as bi.
It since high school, middle school age 've seen.
E, we know from geometry of a triangle edge
square of the length of the adjacent edges thereof lengths
square of the product of these two times two times the cosine of theta, so the formula is here.
EU frame of square of the length of OA plus the square of OB
OA OB times minus two times the Open be multiplied by the cosine.
So if we know the length of We can find the cosine of theta.
OA and OB and the EU's length we know we can write.
Now the size of the vector immediately 'll decide the terms.
I once e, just to look at this.
X, y minus the EU.
x the length.
So the EU squared minus x the length of the year Or squared x minus y squared.
Perhaps in terms of symmetry x minus y frame is a little easier to write.
OA of the length of the square frame of x, y frame length
minus two times the length of the x, y has a length times the cosine of theta.
So we have omitted an important step b.
We passed Geometry vectors.
After passing into the vectors with the components We can now proceed.
Here the size of the length of y in X,
multiplied times the cosine of theta as the inner product We define.
We will see why this one is worth bi.
Because we know them.
the length of x, y or something long.
So this size gives the cosine theta
b compact size, since it is We define.
Now that the inner product of two times two times x y
and convenience in writing, to provide short We brought it here for description.
But conceptually an important concept in will be.
As you can see the length of x.
E, x according to the length of the Pythagorean theorem The length of the first edge,
The length of the second edge, the square frames iii giving.
The length of the vector is going on here.
Similarly, the size of the components of y of squares
minus two times x minus y.
x of y components of the old differences.
This vector and the length of these components sum of squares.
Now see how we do it simplification will be.
Where x has a square.
If x is a square frame open this one next year the next frame.
x next two frames.
y next two frames.
All of them are here.
So to simplify them.
But back than two terms remains.
Because when we open this square x y merger There's one product.
x is the product of the two by two years.
Every one will be minus two times.
There are also negative at the beginning of b.
Therefore, two times x plus one year a
One of the first two years, and two from x the second comes from.
This means a very simple structure, symmetric index structure.
y in X, where two of the inner product If we simplify
Even if the product of the sum of components is equal.
The first component of the vector of the other The first component
other vector with the second component in the second We collect components.
We have come to a conclusion very symmetrical.
Here we find geometry or we have defined
algebraically equivalent results found We're going.
This supremely important.
Just seeing that we proceed step b in X, y
Once this aspect of the inner product cosine theta was multiplied.
Divide to take them painted.
Means that x y is the cosine of theta
to the length of the inner product of x and y is divided.
Up to know.
Just found here.
This component of the product.
The length of the x in the denominator.
given the length of x.
x One x squared plus two squared square root.
Pita, a simple longitudinal Pythagorean theorem.
In the first class we learned a lot issues.
Here the squares of the second component We won a very important step here b.
The geometry of a significant magnitude vector more
rather in terms of the points components have achieved.
Vek, the center coordinates of these points By combining the teams get to the start of
we have vector components or
equivalent of these points coordinates.
Now the question may come to mind.
Trigono, we know from simple trigonometry.
Such as sine cosine threshold of each other,
as complementary size We can think of.
Inner product gave us the cosine theta.
A sine theta likewise wonder describes a process does significantly?
The answer is yes.
Now we can easily see.
A second process and an important to identify.
We found the cosine theta iii.
Product of components, the sum of x and y to the length of the division.
Get rid of the denominator.
Bring it multiply.
E, on both sides.
x and y the length of the cosine of theta were found.
On the other hand we know that the sine-squared A minus cosine squared theta theta.
Because we know that the square cosine theta plus sine of theta is equal to a square.
Now take it by the square of height multiply.
We hit on the frame size.
Why are we doing?
Because here the cosine squared stone, there is.
Is wise for her to be checked.
Now that we hit the take.
We hit on the right side.
Let's open.
The first is the product of the square of length negative again
cosine squared times the square of height product.
Now we know.
Now let's write it again.
As lengths.
x the length of x squared plus x two square one.
The length of y y y a square two squares.
Longitudinal always Pythagorean theorem.
Less E, this, other than the inner product Not at all.
E, we know what the inner product.
Components of the product.
We also take this writing.
So it's an x, y, a, x two, two of the year We are writing to take, but there are square.
We take the square.
Now, here, here, here's the only cosine the cosine square.
Now when we open them a As in previous i.e. cosine
In finding that the inner product of the sine theta vBulletin Are there any simplification sinus as well.
See the next four terms here.
Multiplying the first and the second with the first the second
multiplying the second with the first and second Multiplying the first.
Here's the next three terms.
Birnc the square, a second square of the All of the product.
See something interesting occurs.
First customer gets hit with the first term x a square, a square y.
Get here by the square of a square in the x y a the same square.
This means that it will go.
Again, let the second term.
When we get two square two squares of x y See also here that the second with the second of
The second term from the square two squares y squared x two squares, this term is going well.
There were two terms here they are staying.
two squares of x and x is a square two squares y a y ie square cross-terms stays.
x an y one, two x two years.
Terms checkered fell.
We are writing you take them.
a square x, y, y is a two square two squares of x square.
There's also a term here.
This term of the multiplication of the two two times.
Have a minus sign in front of, So minus 2 times.
x A, the product of the whole thing.
x a, y two x two, a year.
The same one that none of these four terms the non-multiplied by four terms.
Just something catches our attention.
This is because he looks like a square is a square plus
a product of a squared plus a minus.
Indeed, an x, y, see here two There are cross-terms.
Here are the cross terms.
x is a two-year, one x two years.
Minus two times the product of all of the.
Indeed, we see that it is equal.
In a simple structure has been extraordinary.
Of course there are frames on the left.
There are squares on the right.
We're getting when you get the square root of the square.
x y is the length and wherein the length of the sine theta the multiplication.
Here's the right size by dividing the sine theta and we get the following are achieved.
See to it tetayl cosine Let's compare.
The above components are multiplied, Binary binary.
Here you have a total.
Here bi cons but overall there are similarities.
If denominators the same.
x times y is the length of the height, times the length of x y is the same size as the denominator completely.
So one of the sine cosine parity sense of completeness, full
wherein components complement each other terms is also evident.
Already in mathematics, in any branch If there is something symmetry
in another place in the mold to do another emerges.
But here we have two size process When plane
When trading under the rug to something We tried.
To generalize to three dimensions because it we want.
Our goal is a three-dimensional environment because it We are.
Events of this kind occurs in the region.
Obviously right now, of course, see possible
but to generalize it to three dimensions for
x and y are vector cross product already
multiplied by the ratio comes from being called a vector requires.
This vector water.
X, y forms a plane.
elimizle of x to y in the right way Let's say you have the direction vector.
Perpendicular to this vector.
X and y in the plane, perpendicular to the plane of vector k can say.
Because i and j in the plane of said unit vectors.
This means that k is perpendicular to b nice choice.
These vectors are denoted by n.
This because in three passes, we want to generalize.
In this algebraic representation in the following length were found.
x times y is the length of the length of the sine of theta We have seen that these components.
But it is also multiplied by n in the plane of the random
not quite the vector x, i and j perpendicular to unit vector.
That there is one.
From x to y in this rule so that when indicating the direction of the vector
and thumb of your right hand thumb is defined by this.
It's a choice.
But you could also define the left-hand rule I suppose most people historically
the right to use their right hand more hand rule has accepted the case.
You could also accept the left-hand rule.
Very consistent in her mathematics could make.
Now compile them a bit.
What at first as the geometrical Let's see what comes to meaning.
There was the cosine of the inner product.
Points A and B to bring Gene Let us unite to start.
After this initial terms combine theta, B, from B
If we pull a strut perpendicular to OA here We're creating a triangle.
Why are we doing this?
Because we need a cosine.
See Ob, the length of the cosine If you hit you will find tetayl OH.
This means that showing a projection.
In this numerical formulary If we write OH, of course,
The arrival of letters was interesting Turkish terms
OB length times the cosine of theta.
The projection of x on y in OH call.
Here is an international term projections I used to be.
Projections are already in Turkish is being used.
Showing that way.
The projection of x on y, that OB
y is the length times the cosine of theta We know.
We know what is the cosine of theta.
Divided by the length of the inner product, here As you can see y
a denominator seems to be in two places at in the denominator.
Simplify them, the x on y inner product divided by the length of the projection of x.
If y'all projections on what it takes is coming.
Similarly to the projection on the OB You could buy.
Following the intent of the projection.
Here, a parallel light from an infinite source
if you submit a projector that OA OB in the direction
would be in line with projected shadow, OH it
such a parallel here to the projector Looking at similar parallel here
if you look at a projector OA OB
The shadow on the JV would be.
Here's called projection for him already It's called projection in Turkish.
Therefore, OG, OE times cosine theta be here again because of the right triangle.
And here is the projection of y on x as Og'y are describing.
Again he crossed the length of x that cosine theta We want to be able to write in terms of the inner product.
As you can see here again as x times inner product be divided by x divided by y.
x and y remain simplifies this time.
As you can see a completely symmetric representation.
Denominator of X, y is the inner product.
If the projections made in the denominator to which he it seems.
Here, the projection is made with respect to x.
He is seen.
It's a bit more simple and meaningful structure can bring,
Because I looked, see where the x's own
There longitudinal section, here the y's own There longitudinal section.
So, based on its longitudinal section of x u
If we say that this is the unit vector in the x direction shows.
Shows a long vector.
Similar as its longitudinal y
The ratio of the unit vector in the y direction shows.
So the projections, projections can wrote, see
x where x is divided by the height-to- Because it is Y, Keren
When we get there from here y times x is going on.
So from here
If we look at the projection of y j, If we look at the projection of y to y
has itself, on the projection of x When we look at x's.
Which way, which is projected onto the vector
If the unit vector in the rate of work done enters.
Already simple, light and shadow, the shadow
If you look at the size of this analogy it Although so that
does not change much even if the length of the shadow, Therefore,
projection of the vector in the direction of your length is not critical.
Wherein the length direction of the projection does not appear, Projection
On your vector, where the In a similar manner, thus significantly
and elegant, simple results have We're going.
Now the geometric meaning of inner product 've seen.
I wonder if the geometric mean vector multiplication Do you have?
Let's get this sewing again here.
So let's get to strut on the OB.
See sinus future here.
We multiply the length of OA sinus sinus This is not the time where the cosine
time and across sinus future planting to give.
So it came to the opposite angle to find long
multiplying the side neck for sinus is required.
This we know from simple geometry.
If we write this, that means that AG planting height,
this side of the height of the edge length x OA of the length of the sinus
multiplying, in terms of these vectors One type of x because x is the length of the length.
The size of the vector x, times the sine of theta.
Now that the bottom edge parallel with OB height
If we multiply this parallel sided AG's We can find the area.
He's off again, we come to OACB When the parallel
edged with OBI AG is the product of the area.
We know OB, OB y height, He AG I have found here at the height.
x times sine of theta.
But now we see that x multiplied by the length of
Multiply the length of the sine of theta y, y in X, is size.
So a geometric vector multiplication gives the size of
and this is a useful size, you can say If I knew what will be the parallel edges
but more complex shapes that can be How to parallel edged work as a child
By combining small particles more it's complicated
way as it would have obtained geometry
will allow us to understand.
Here also we see that the vector multiplication have a geometric meaning.
This is the area of parallel edges.
We were with a line from the EU that the EU line
two equal parts in a parallel sided bölsek interests.
Thus, in the area of the triangle OAB we find.
One moment parallel edges Order is half the battle.
Now we define the inner product.
We define the vector product, one of them gave a number.
One of them gave a vector.
These features slightly We want to promote.
They also, it's also still a natural
because it is part of the next subject at a time will do.
Opinion until goodbye.
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