Ders çok değişkenli fonksiyonlardaki iki derslik dizinin ikincisidir. Birinci ders türev ve entegral kavramlarını geliştirmekte ve bu konulardaki problemleri temel çözme yöntemlerini sunmaktadır. Bu ders, birinci derste geliştirilen temeller üzerine daha ileri konuları işlemekte ve daha kapsamlı uygulamalar ve çözümlü örnekler sunmaktadır. Ders gerçek yaşamdan gelen uygulamaları da tanıtmaya önem veren “içerikli yaklaşımla” tasarlanmıştır. Bölümler Bölüm 1: Multivar 1'in Özeti, Dairesel Koordinatlarda Entegraller Bölüm 2: Türev Uygulamalarından Seçme Konular Bölüm 3: Çok Değişkenle Zincirleme Türev ve Jakobiyan Bölüm 4: Uzayda Yüzey ve Hacım Entegralleri Bölüm 5: Düzlemde Akı Entegralleri Bölüm 6: Düzlemde Green, Uzayda Stokes ve Green-Gauss Teoremleri Bölüm 7: Stokes ve Green-Gauss Teoremleri ve Doğanın Korunum Yasaları ----------- The course is the second of the two course sequence of calculus of multivariable functions. The first course develops the concepts of derivatives and integrals of functions of several variables, and the basic tools for doing the relevant calculations. This course builds on the foundations of the first course and introduces more advanced topics along with more advanced applications and solved problems. The course is designed with a “content-based” approach, i. e. by solving examples, as many as possible from real life situations. Bölümler Bölüm 1: Summary of Multivar I, Integral in Circular Coordinates Bölüm 2: Topics of Derivative Applications Bölüm 3: Chain Derivatives with Multi Variables and Jacobian Bölüm 4: Surface and Volume Integrals in Space Bölüm 5: Flux Integrals in the Plane Bölüm 6: Green in Plane, Stokes in Space and Green-Gauss Theorems Bölüm 7: Stokes and Green-Gauss Theorem and Nature Conservation Laws ----------- 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
Sonsuz Küçük Yüzeylerin Gösterimleri
Loading...
From the course by Koç University
Çok değişkenli Fonksiyon II: Uygulamalar / Multivariable Calculus II: Applications
9 ratings
Explore Related Videos
At Coursera, you will find the best lectures in the world. Here are some of our personalized recommendations for you
Koç University
9 ratings
From the lesson
Çok Değişkenle Zincirleme Türev ve Jakobiyan
En büyük ve en küçük değerler: yerel, mutlak ve kısıtlama altında. Kısıtlama altında en iyiyi arama (optimizasyon) ve Lagrange çarpanı yöntemi. Kısmi türevlerin uygulaması ile değişimler hesabına giriş.
Meet the Instructors
Attila AşkarProf. Dr.
Matematik Bölümü (Department of Mathematics)
Hi there.
Today in multi-storey integral 're starting to important applications.
We saw earlier two-storey integral.
The two-storey integral but in the plane We saw so on.
Here on the surface in space We'll find a two-storey integral.
What are these things?
For example, the surface of a sphere with a four-pin We know that the square since time immemorial.
How were calculated?
But even though it's only sphere in our remembrance We do not keep an even deal
With these thoughts in many ways, but We need to learn the ways of calculation.
After seeing the integral two-storey three integration of applications in floors will do.
The two-storey integral Cartesian and
circular coordinates We saw integrals.
Here we will see in space.
Again in triple integrals in Cartesian coordinates of the circular coordinates
generalization, the generalization to three dimensions cylindrical and spherical coordinates
coordinates the three-storey integration will learn.
Now let's start from the two-storey integral.
The calculation of the surface integral in space We want to find ways.
Of course, in this plane, two-storey slightly different from the integral.
We will see now.
In the space of a surface is equal to f x y z
open to function or closed with the function or
with a vector function of two variables, We call this parametric representation.
We define these three ways.
I also have a special type of surface.
We saw it in te first part.
Take a curve around an axis by switching, we obtain a special surface.
Although because these surface surfaces If two parameters are expressed,
invisible, so every second parameter point due to the same theta, theta's
same function for different values We're talking about the surface shown.
Now let's start our calculations.
As you get a surface.
On this surface a delta large and infinitely small surface
Let x and y plane and its Let's say a projection of the delta.
Of course, this is a curved surface consists of rectangular.
x and y plane of the curved surfaces
We get straight to the edges of this time is reflected as a rectangle.
This is the unit of surface Let the vector perpendicular to the surface.
Z-axis in the x and y and z coordinates of a Let the unit vector in the direction.
We call this k.
On this surface we go, In order to understand better shape.
As a simplified situation As we encounter.
Delta water surface, as here, the curvilinear tangent to the surface and the surface we see it.
These two infinitesimal terms are the same size.
And z is the unit vector perpendicular Gene k vector which is parallel to the axis,
piece of it, and this delta water surface or that the tangential surface,
When we receive the projection of the plane of x and y
in the plane of such a rectangular 're getting field.
Here is a simple geometry we make the following observations:k and n
theta is the angle between the vectors are.
We know that other vector perpendicular to k in the direction of the vector is horizontal.
Consider a vector perpendicular to n.
The angle between these two theta.
Where the theta.
Wherein the angle theta.
Now that the horizontal plane of the delta of s we calculate the projection on
When we see a result like this:Delta s times the cosine of the angle theta here.
With this account of the way to k has the inner product.
Because k with n internal multiplying a length n.
k is a length.
Times the cosine of the angle theta, delta p.
That this simple geometrical a delta equal to mind.
Therefore, the delta delta h ain, n and k is divided by the inner product.
I also take the absolute value because the internal
If the product of wide angle there could be less valuable.
However, the definition of the delta h as plus valuable space measurement.
Plus valuable space in the delta measurement.
Therefore we put this absolute value.
Here, we pass to the infinitely small icon, d s is d
k is divided by the points of inner product and the water surface
this is the integral of s will.
Now we've got four impressions.
There are three impressions, but even more precisely a fourth surface, a special type.
Impressions indication, was implicit and parametric.
In special cases, surfaces of revolution.
Now all these impressions calculation of the basic
This account comes from the difference of n.
So n open this surface equation is open surface,
Given the clear function or When given as an implicit function or
parametric when administered How come we will calculate.
A little more special for rotational surface but still there is also a case
it is important to find n.
Now here again, respectively I repeat this pattern:
As clear function, respectively When shown the surface f x y z equals.
We know that the perpendicular vector.
I have seen many times.
Less d f d x,
ie the gradient and the minus sign There are also a z-direction in three dimensions.
This i j k expressed in terms of If we're finding in this way.
This f x, d f d x short software,
y is similar as for d f d y short software.
But we need the unit vector perpendicular.
To find a unit vector perpendicular to The process, of course, we always do.
This is a non-vector around its longitudinal length If we divide the unit vector, we obtain.
The length of this vector, that's f x squared,
f y squared plus one There are a total of.
Here, too, because it is a vector, The components of these vectors.
We k internal If we take the product of k i
Due to the steep and jia there is no contribution of the first two terms.
k'yl k multiplying the length of Because of a data.
As you can see in the denominator a change is made.
Is slightly different.
Spelling seems more simple.
F x f y take forward one square frame taking the last're getting.
If we consider this information, d s is d divides k n point we knew that.
K is the point where the positive seems to be valuable.
Top of surplus value.
We are also positive from taking the square root.
So the point in the denominator where k is a plus share
f x squared plus y squared square root of f and d a'yl brings to the product.
As a simple logical question: d h min is large, d m is large?
Each time we know something The projection is less than its own length.
If the maximum horizontal This would be equal to the surface.
Here we see under the square root consists of a large number of.
d d is larger than that of s provided here, we are confirmed.
S itself is not in some problems, n times d s is also required.
See, the d s of times If we have n.
d s got here.
Both of these vectors remains the denominator gets hit.
In the denominator, see here The term in the square root there.
Here are the terms in the square root.
One denominator denominator is one simplifies to each other.
And as you can see d s n times other vectors with this vector,
but here not long units, Once d be.
Now that is the same size d k n times
s is because it is always in, what If we let all the main formulas that account.
This general formula.
The only difference is how these n will be calculated.
k is always the same.
Equation of the surface is a closed We always see if provided with the function.
Gives the gradient vector perpendicular to the surface.
So gradient account means to the first, second,
f of the third components derivatives of x to y by z.
Here is seen a short software.
Now we want to find a unit vector that
its length perpendicular vector we obtain by dividing.
As you can see here this vector We write in terms of components.
This length of the first component squared plus
The second component squared plus third components and the square is the square root.
This Pythagorean theorem generalization to three dimensions.
n by n points such that k get the f z takes only here.
Because k components is zero, is zero, the vector a.
This inner product is zero times when he calculated f x
plus y plus zero times for once As you can see f z f z is staying.
There is no change in the denominator.
Therefore d s d n, to this k is divided by the inner product.
Here comes a share in the denominator term.
Because the absolute value of z in the denominator for f minus the value of z is possible.
We will see such examples.
Therefore, we take the absolute value of d s
By definition, plus a you might have value.
Because I measure an area.
As you can see still d s, is greater than d.
Because where p, pa, z bölsek remain here for a square root.
Here the absolute values is plus A plus
where values ??for other than Do you stand with a number of large d.
You can find such a d s.
This, as we know still no is smaller than the projection of a shape.
Or the way from the projection is larger as the length.
n d s at times may be required.
such that n d s gets hit with this by See this karekökl ??there denominator.
this is in Gay karekökl ??in d s.
Therefore, they will simplify each other and denominator remains the gradient vector.
For the absolute value of z in the denominator of this remains.
If you want to find area jobs This is the integral of s two
will receive storey integral.
Because it is a two-storey integral.
As you can see here in a previous problem
As this surface on curved surfaces
the projection of the space on account We are demoting be calculated.
In the x-y plane as we know it integral two-storey rectangular or
circular coordinates the integral to the download of the reduction.
But wherein the field,
infinitesimal area greater than one this integral is calculated by multiplying it by the number.
Now let's get a third species.
Also a third type of parametric parametric representation of the surface surface.
I will explain this way immediately.
Parametric representation of a vector function.
But the two also shown variable a vector function.
Vector functions of one variable, a We are aware that curved line in space.
This is a two-parameter two If the space variable vector function
shows a surface.
If we write this more obvious first component of x, u and v depends.
The second component y, u and v depends.
The third component z, u and v depends.
We know them.
Now we come to the following way.
Let us recall the circular coordinates, what were we doing?
that is to say u r these general the scope of representation.
We were getting r is constant.
r are increasing slightly.
Second, we find a circle.
We're getting a correct theta.
We take theta plus delta theta.
The second one is correct.
This is where the intersection of the four lines
you can see the curved edge consists of a rectangular, though.
Also in general curvilinear coordinates We are continuing the same thought.
We see this already in the Jacobian.
There is a curve in space.
If you have it curve to fix the call.
Similarly, if you secure V. We call a curve.
We slightly increase them line plus Delta,
have any lines plus the delta occurs and a rectangle with each other,
No excuse steepness, they form a curvilinear quadrangle.
Here are the areas we want to find it.
Instead edges Delta and delta vectors're getting there.
Because these infinitesimal terms The length of this curve in the same order.
As these vectors were finding.
According to the derivative of the vector x all components x,
y, z by the derivative means.
This is kept constant, v is gives the vector of time.
Similarly, we hold u constant
When the location of the position vector V is derived by the vector x.
This gives small van vector.
These delta countless respectively, number and the delta
Multiplied by the number of jobs that have curved edges of the rectangle
forming lines, The truth is that you get a straight line.
Verifies infinitesimal.
These two vectors of the two vectors product will give us space.
This mut, it of course the length field but this aspect thereof n times to give
If n is the unit vector in space its absolute value will be the length.
This delta h, where the vector s because the two vector fields
Get the vector product of the vector n we get a vector in the direction.
The size thereof,
The size of this field this The length of the vector multiplication.
D to its symbolic turning deltas
As you can see we calculate by multiplying the vector v is in u'yl
d u d v numbers, symbolic variables, The infinitesimal.
It s the delta area instead of lowercase d times
with s, defined infinite small area icon.
Now you can see if these accounts do v is a vector product as u'yl
gene vectors we know from section like i j k will write in the first row.
u were the partial derivative of x by u.
If you have the same gene of the vector x partial derivatives of the components according to v.
That's when he calculated the determinant i
Because these vectors k j A vector will be multiplied.
Ju ju to its components x, y and z j, the j z say.
Because we see in this vector multiplication As we meet here again Jacobian.
I was hit by the term wherein the dual Jacobian.
y and z components and v formed by the partial
derivatives of the components i'l We say this because xj.
J'l component will be a minus.
i.e., the first row and the j
again by the second column separates into two a determinant of duality going on.
Again, this Jacobian structure.
Because the x and z coordinates by u derivatives of gene x and z coordinates
derivatives thereof according to v j We say this year because components j'l.
As a component in the K calc It is getting in the way.
Jacobian of a circle circular Panini
Jacobian coordinates new account While this gene were welcomed by z'yl.
Because no components z'yinc For them to be zero,
j x and j y is zero was happening to her.
Vector j do it so it Let's show the value of the jar with s.
Absolute value here as well.
Of course, here plus the square of the first The second component squared plus third
components and their squared We also know that the square root.
Now that two of our accounts options are emerging.
One of them is horizontal gene ie in the plane with coordinates calculation
The calculations made using d.
d in the horizontal plane of the elements.
d S is the area of ??a curved surface in space, infinitely small space.
Please see n where n Once we know d s.
This j x j y j z consists of.
But if you want to find the length of the unit, we divide the length of this vector.
So this vector multiplication The value of the determinant
us the direction of the vector but here shows
j x j y j z component derived vectors of a size that is not guaranteed.
Therefore take this vector We need to partition the longitudinal
Let's get a unit vector.
This unit vectors we obtain here.
N and k we always need more Previous eg, as in approaches.
Get this inner product of k components zero zero one.
It gets hit with the zero times j x plus zero times, one time j y j z.
Therefore remains just above z j.
J in the denominator of the length of this j s, this value remains.
Now let's see, let's find d s.
d p, d's, where n and k in denominator.
Therefore, j s share of this income in the denominator,
To the point that we put k denominator.
j as a share of our income.
Here we need to take the absolute value.
Because the negative z j can be valuable.
But in many applications of coordinates When selecting the appropriate sequence of z j
plus the immediate future A case in many cases.
Now, as we see here is We found again in terms of s d.
There are a large number of genes than here.
Because it is supposed to be greater than d s d.
A projection shape itself to be smaller.
But here a second more options are emerging.
Because we u and v curvilinear We know the coordinates.
Therefore, d s is n times d u d v is j times
Because we know is that we have also p.
account the length of this side is let s say.
where n is the length of one means that the d s j d u d v is the length of this is the product.
We know the size of e j.
j p. Therefore, j j x squared plus y squared j z
We find that the square root of the square.
So here curvilinear with the parametric coordinates
I gave up half way through representation Let's say we could projection.
In this way we can work.
Or whereby the curvilinear coordinates We know and mostly curvilinear
she coordinates the native surface are the coordinates of the appropriate representation.
It continues with coordinates We also have the opportunity to.
When it comes to surfaces of revolution,
will begin at once with the question: Why surfaces of revolution important?
In nature, technology also A very common form of modeling.
These transactions.
get function f r z equals.
In other words, the circular coordinates We're forgetting theta for now.
This way we rotate around the z axis As you can see when you look
are faced with a structure in the following way.
In three-dimensional space we obtain a surface.
A curvilinear surface, and characterized in it is the surface of revolution.
Here it any theta if you receive this in the section vertical theta,
theta function independent of z
Because all these will be the same way.
Now it still curvilinear coordinates I can think of a special case.
Here r and r plus delta
If we see is that
and between the del, theta plus delta of each relevant R each relevant R and r
If we increase the current vertical will be two curved lines.
is that in the first run.
When we say that the above is r plus the delta.
The intersection of the four lines again.
As we have previously obtained a one curvilinear surface, we find the infinitesimal surface.
We need to find the area of ??this surface.
Now the equation of surfaces of revolution was it.
In fact, the two can be easily parameterized representation can go to.
vector x, x is the vector times the cosine of theta.
y components of r times the sine of theta.
z component is also given here already.
f, r.
As you can see we surface to surface of revolution
two parameters equation r and theta in terms of the parameters can write.
Now we carry out the same operations that it
In a previous parametric u r my representation,
If we choose V as well as theta, We will also be vectors d.times.d.
See here the derivative with respect to r.
Here comes derivative of f by r.
R linearly around their As appears by itself by r
We'll take derivatives.
Only cosine and sine remains.
Theta derivative of x by the second vector.
Here you can see remains constant as is.
Cosine of theta minus sine theta.
Kosin, sine cosine of theta-theta.
Since there is also theta f r its derivative with respect to theta zero.
Now these two vectors If we take the vector product,
Repeat the previous process by As you can see these vectors is generated.
I did not write here for a long time.
Because of this, in the third term
be zero, the derivative with respect to theta accounts for a little easy going.
This is f prime minus cosine theta f r sine theta and a base turns.
This size calculator
If we here that the delta h The magnitude of the vector product.
This means that the magnitude of vector product here are a three-component vector.
So it's a plus, the first vector,
first component squared plus square and the square root of the latter.
But there is a very interesting structure here, of course.
Sine squared plus cosine squared theta wherein for Base multiplier same for
home base as a plus for the r squared times are achieved.
There are delta delta theta.
One is always came out here.
This shape
R z is equal to that for the shape As we can fully rotate
A theta two theta value We can rotate up to.
Therefore, this surface From the area of ??a strip element
We found that the first to find d s d theta integral're getting on.
But here theta does not appear.
Therefore, the only theta A two minus theta is staying.
Here are the features of a surface of revolution is here.
The two-storey integral directly coming down to an integral single storey.
If you get a full acre two pins will be two theta.
Theta will also be zero.
Here comes two pi factor.
And here we see a slightly different Go is infinitely small area,
This happens because the area of ??a strip of theta
one to two theta theta When returning the values ??of r
d r r plus the
See the following, we find the area of ??the strip, horizontal strip.
Because we do it on the field theta account We are a full acres and comes with two pins.
Nobody in a partial transformation Open coming up here.
We find the area of ??this strip.
Therefore, just run backwards remains an integral over.
Now then it is parsed We'll start problems.
Single-resolved problems before Before we do this calculation,
We found a compilation of the results Let's see here as fast.
D s six different ways we have found.
Therefore surface calculations six different ways can be done.
In the former the function of the surface,
function is on that surface We chose it as a function.
In this way are then d s.
The overall structure.
thereafter
Our choice of a closed surface It was shown with the function.
It closed with the function of representation The point here is to divide the
k so this is part of the three product we obtain the following expression.
Therefore find bi s In the first case we'd like following
integration will account field.
This two-story on plane We're going to have integral transform.
Here in the plane, two-storey We're going to have integral transform.
As parametric functions when given in the plane again
here are integral relegates multiplied by this coefficient.
That's how he finds the coefficients From this inner product so that
Jacobins occurs.
u i j k x and y by derivatives based on components
we have achieved as a vector of type.
We e, where this option
in addition to curvilinear coordinates We can choose to account.
As you can see he u and v directly on the surface,
calculating a curved surface on comes equivalent geometrically.
Surface of a rotating face if you have a well that
Instead of a two-storey single integral integral to the story in this way is down.
As you can see in the first four One general representation.
This outdoor function individually,
implicit functions and parametric function to display it, but with x,
y onto the plane taking the results obtained.
Now we're standing here today.
Our next session with several examples of this
Results obtained We will use and the examples
I also interesting in its own right I think we have achieved both of these
seeing the structure of the application will solidify concepts.
Please goodbye until the next session.
Explore our Catalog
Join for free and get personalized recommendations, updates and offers.
Coursera provides universal access to the world’s best education,
partnering with top universities and organizations to offer courses online.
© 2019 Coursera Inc. All rights reserved.
