Professor Raffi Hovasapian helps students develop their Multivariable Calculus intuition with in-depth explanations of concepts before reinforcing an understanding of the material through varied examples. This course is appropriate for those who have completed single-variable calculus. Topics covered include everything from Vectors to Partial Derivatives, Lagrange Multipliers, Line Integrals, Triple Integrals, and Stokes' Theorem. Professor Hovasapian has degrees in Mathematics, Chemistry, and Classics and over 10 years of teaching experience.

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Welcome to educator.com.0000

Welcome to the first lesson of multivariable calculus at educator.com.0006

Multi variable calculus is an extraordinary branch of mathematics.0010

Those of you who are coming to this course, you have already come from a course in regular calculus.0013

What we are going to do is, we are going to take the power of calculus, and we are going to move from one dimension. 0016

We are just going to move up, to two dimensions, to three dimensions, and as it turns out, any number of dimensions.0023

That is what makes all the things that we are going to learn really, really exciting.0028

We are not limited to two and three dimensions, in general.0032

We are going to be doing most of our work in two and three dimensions, because we want to visualize things.0036

We are used to playing in space and things like that, but the results are valid for any number of dimensions.0040

So, let us jump right in and see what we can do. Welcome again.0047

We are going to start off with just some normal basics.0051

We are going to talk about points and vectors just to get ourselves going. 0053

A lot of the stuff that you may have seen before, if you have not seen it before, it is reasonably straightforward.0056

Let us start off by just defining a point in space.0063

A point in space, can be identified... Actually, I am sorry, let us start with a point on a plane, not in space. 0070

We will work ourselves into 3 space in just a minute.0082

A point on a plane can be identified with two numbers.0086

One number for the x-coordinate, and one number for the y-coordinate.0106

This is nothing that you have not been doing for years and years and years.0120

Just the normal Cartesian coordinate plane.0124

You have an X axis, and a Y axis, and when you choose a point in that plane you need 2 numbers to define where it is in that plane.0128

Let me just go ahead and finish writing off the y-coordinate. 0134

So, a quick picture, you are going to have something like this.0137

This is the Cartesian coordinate plane and let us take a number like that.0140

So, this point might be (6,2).0146

In other words you are 6 in the x direction, this is the X axis only, the X axis is horizontal, Y axis is vertical, and 2 units in the Y direction. 0147

That gives us a point in space.0157

Now, because we need two numbers to identify that point, we call that a two-dimensional space.0165

So, whenever you hear the word dimension, do not freak out, it is just a fancy word for number.0170

When we talk about a 17 dimensional space, it just means that in order to identify a space, or a point in that space, we need 17 numbers.0174

That is all it is, dimensions, numbers, they are synonymous.0181

Okay, so now let us actually talk about a point in space, the space that we think of as the space that we live in, the space that we know, 3 space.0188

So a point in 3 space, actually let us not identify it as 3 space yet, let us go nice and slow.0198

A point in space can, can be identified with 3 numbers.0210

Again, one for X, one for Y, and one for z, we generally do x-coordinate, y-coordinate, z-coordinate.0225

Let us go ahead and draw what this looks like.0239

This is going to be a standard coordinate system called the right-hand coordinate system for visualizing points in three-dimensional space.0243

What we have is something like this, slightly different.0249

This right here, this is actually the z axis, and this one that is slanted, this is actually going back and coming forward.0256

So, this z-axis and this Y-axis are actually the plane of the paper.0268

This right here is actually the X axis so, what we have done is we have taken the XY, we have flipped it down, and we have added a z-axis to it.0276

If you have some point like, I do not know, let us put a point over here, and let us call this point (3,3,2).0286

We have three numbers identifying that point.0296

That means that we have moved three units in the X direction, we have moved three units along the Y direction, and we have moved two units along the Z axis.0298

That is it, three-dimensional space, because we need three numbers to represent that point.0338

Now, analogously we can talk about any number of dimensions.0346

We can talk about five space, six space, seven space.0348

It just means we need 5, 6, 7, you know, numbers to identify a point in that space.0353

The behavior is exactly the same, even if we cannot visualize it.0360

In other words we are kind of limited to two dimensions and three dimensional representations theoretically, 0365

But algebraically, we can just do, for example, if we have a space which is (4,7,6,5,9), this is a five dimensional space, this is a point in a five dimensional space.0370

We cannot visualize it, and we cannot draw it, but we can work with it algebraically.0384

Let us go ahead and talk about a notation briefly, a notation for a space of a given dimension.0392

This is a notation that we will use reasonably regularly. 0408

We will start to use it more often when we talk about functions from say a two-dimensional space, to a one-dimensional space.0411

Taking a number that has, you know, two coordinates, and mapping it and doing something to those numbers, and coming up with an answer that ends up in a different space.0420

This notation that I am going to introduce, you will see it more later on, but I do want to introduce it right now.0432

So, R is the symbol for the real number system.0436

You will have R represents 1 space, and so when we say the space R, we are talking about one dimensional space. 0450

In other words, a line.0459

Basically, you need just one number.0460

The reason we use R is because, you notice 2 space is 2 numbers, 3 spaces is 3 numbers, 5 spaces has 5 numbers,0463

Well you are taking these numbers, the 6, the 2, the 3, the 3,0470

You are taking them from the set of real numbers so R2, when we put a little two on top, that is two space.0474

It is saying that we are taking, well, this two tells us that we need two numbers, one from the real number system, another one from the real number system to identify that.0484

So R2 is 2 space, it is the Cartesian coordinate plane.0496

Oops, we are starting to get these crazy lines again.0499

Let us see if we can write a little bit slower to avoid some of that.0502

The real number system, actually, let me go ahead and move to the next page here.0506

So, R is 1 space, R2 is 2 space, and R3 is 3 space.0511

When we talk about the space that we live in, we are actually talking about R3.0532

In other words, three numbers from the real number system to identify a point in space,0536

We will be using that more when we talk about functions again.0540

OK, these points in space are actually individual objects.0544

You can think of them just like numbers. 0550

For example, if I said I have the number 14 and the number 17,0552

These are individual mathematical objects, these points in space are also just single individual mathematical objects.0555

The difference is we need more than one number to actually describe them.0563

These numbers are the coordinates.0567

As you will see in a minute, these points in space which we are actually going to start to call vectors in a minute, they are individual objects, and we treat them like individual objects that you can treat just like numbers.0572

It is just that they happen to be made up of more than one thing.0582

It will make more sense as we start to do the problems.0587

We often call points in space, vectors, and in fact, that is probably going to be the primary term that I use.0601

It is just so ubiquitous, so common in mathematics to speak about a point in space as a vector.0613

We will go ahead and get into the geometry a little bit, but just know that it is just another term for a point in space, that is all it is.0621

So (1,-7) a point in two space is called a 2 vector, and exactly what you might think,0629

The point (3,-4,6) is a 3 vector, that is it.0644

The reason we call it a vector is, again, some of you people might have seen vectors in physics or another math course, perhaps some of your calculus courses actually talked about vectors, I do not know. 0653

Occasionally they do.0667

We call them vectors because you can actually think of a point in two ways. 0668

You can think that it is just a point in space, or you can think of it as an arrow from the origin to that point in space.0672

Let us go ahead and draw it out.0681

The geometry would be something like this.0683

So, if I had some random point, so this is the x-y plane, and if I had the point (6,2), that is a point. 0686

You know 6 along the X axis, 2 along the y-axis,0690

But it also represents an arrow, an actual direction from a point of origin in this case, the origin, in that direction.0700

This is called a vector and it is a directed line segment.0708

It is not just a point in space, it actually gives us a direction and that is why we call it a vector.0712

So there are two ways of looking at it. 0719

In general, we will be using these arrows in 2 and 3 dimensions, mostly we will be working in two-dimensions when we do the examples, 0725

Simply because we want to be able to geometrically help us see what is going on algebraically. 0729

But geometry is not mathematics, algebra is mathematics. 0731

We will be working with coordinates, and we will be using these arrows to help us sort of see what is going on.0736

Even then, eventually we will have to put the arrows aside, we want to work with just the algebra.0746

Geometry helps, it will help guide us.0750

It will give us a little of a physical intuition for what is happening.0758

The algebra is the mathematics,0761

Those are the skills that we want to develop and that we want to nurture, algebraic manipulation, not just geometric intuition.0763

Let us see, the three vector, so algebraically, we had the point (6,2), geometrically, we have this arrow.0768

Let us see, how are we going to symbolize our vectors?0796

We are actually going to have 2 or 3 different ways of symbolizing them.0800

I will try to be as consistent as possible, but I am going to at least introduce three different symbols that I am going to use for vectors. 0804

The reason that we do that is, as mathematics becomes more complex, becomes more involved, the symbolism needs to give more information. 0815

Often times, you might need more than one symbol to talk about a specific concept, because sometimes it is easier to think about it this way, sometimes it is easier to think about it this way.0820

But again, we will talk about them, we will not just drop the symbols, we will talk about what we mean so it should be reasonably self consistent.0828

So, vectors will be symbolized as follows.0840

We will often write a vector as a capital letter a. 0853

For example, (3,1,2), sometimes we will write a capital letter with an arrow over it, so (3,1,2) is the same thing.0858

Sometimes we will use a small letter with an arrow over it.0870

In this lesson ,and probably the next couple of lessons, I will probably using this notation more than any other.0874

So, again (3,1,2) they are all the same thing, this is just a vector and actually a point in space.0880

Which point in space? The point (3,1,2) three along X, one along y, two along the z.0885

We will add vectors the same way that we add numbers. 0897

Or I should say vectors, like I said are individual objects, and you can actually add them like you do numbers, there is an addition and it is actually perfectly analogous.0900

Let us write out the definition.0912

So adding vectors, let a = vector (a1,a2,a3), so these are just different components, I am using variables for them, and we will let the vector B be (b1,b2,b3).0914

Well, the vector A + B is equal to, exactly what you might think, a1 + b1, I just add the corresponding components, and I get a point, in this case, three space,0928

So, a1 + b1, a2 + b2, a3 + b3, so that is it, I take a point in space, I take another point in space, I can actually add those points in space by adding their individual coordinates, and I end up with another point in space.0956

That is what is important I start with two objects, a point in three space,0972

I do something to them and I end up with an object in three space.0976

That is actually very, very important.0981

We stayed in the same set, which is not always the case.0983

Later we might be jumping from 3 space to 5 space, from 1 space to 2 space.0985

We can really do that, and that is what is really beautiful about multivariable calculus.0990

We are no longer constrained just to work a function from x to y.0994

You take a number, a function like x squared, you put some number in it, you get another number out, number to number.0997

Now we are working from spaces in one dimension, to spaces in another dimension. Very, very exciting.1006

Let us do an example, OK, example 1.1015

We will let the vector a be (4,-7,0). 1025

Again, most of our examples are going to be in two and three dimensions to make the mathematics approachable, but it is true in any number of dimensions.1031

In fact, one of the culminations of this multivariable calculus course is when we prove, later on. 1040

We are going to be discussing Green's Theorem and Stoke's theorem.1048

Green's Theorem and Stoke's Theorem are just generalizations of the general theorem of calculus to 2 dimensions and 3 dimensions, respectively.1051

As it turns out, the fundamental theorem of calculus is true in any number of dimensions.1058

One of the most beautiful theorems in mathematics is something called the generalized Stoke's theorem.1064

That is exactly it, as it turns out, everything that you learned is not only true of the space that you know, one dimension, but is true in any number of dimensions -- that is extraordinary.1068

So (8,3,-6) is our other point, so let us go ahead and add.1078

The vector C, which is the sum of the vectors a + b is equal to, well, 4 + 8, we get 12, -7 + 3, we get -4, and 0 + -6, which is -6.1090

That is it, that is our answer.1106

OK, so now let us list some properties of vector addition. 1110

Again these properties you know of but we just want to be formal about it.1114

So vector a plus vector b, if I do that addition, and then I add another to them, it turns out that it is associative.1120

I can add them in any order, it is vector a plus the quantity, vector B + vector C. 1129

This is just like normal numbers, we are just working component wise. 1135

We are talking about an object, a vector, an actual mathematical object.1140

Okay, the vector sum A + B is equal to the vector B + A,1144

It does not matter what order I add them, the vector sum is commutative.1150

We are going to identify something called the zero vector.1155

The zero vector is the vector that has 0 as all of its components.1161

For example, the zero vector at three space would be (0,0,0).1166

We are definitely going to distinguish between the zero vector and all of these individual zeros.1176

The zero vector, I will draw a line over it, they are actually two different objects.1185

One is a point in space, that point in space has coordinates all 0.1190

That is the zero vector, remember a vector is an actual object.1195

Now, if a is equal to, let us say (a1, a2, a3), we will just work in three space,1198

Then -a = (-a1, -a2, -a3).1210

When we negate a vector, all we do is we just negate each component of that vector.1218

Last of all, if a = (a1, a2, a3), then some constant C × the vector a = well, c × a1, c × a2, c × a3. 1228

All we have done is take some constant like seven times a given vector.1251

That means we multiply each component by 7, that is all we are doing, essentially we are just distributing.1255

There is nothing strange going on here. 1260

Let us do another example very quickly. Example number two,1265

We will let a equal (7,radical(6),2) and you see, real numbers they do not have to be integers they can be any number at all.1268

They can be pi, they can be e, it could be anything.1282

I will let the vector b be equal to (2,2,0).1285

Let us do a - b the vector a - the vector b.1288

That is equal to the vector a, plus the vector -B, because we know that there is no such thing as subtraction in mathematics.1295

There is actually only one operation in mathematics that is addition.1302

Subtraction is addition of an inverse number, multiplication is actually just multiple additions.1306

Division is just the inverse of multiplication which is ultimately based on addition.1314

So there is really only one arithmetic operation or mathematic operation in mathematics it is addition.1329

Everything else is derived from it, so when we talk about one number - another number, we are talking about something plus the inverse.1332

In this case, the vector + the negative of the other vector.1337

A + -B that is equal to 7 + -2 which is 7-2, that is equal to 5, radical(6) + -2, we just write sqrt(6)-2,1339

That is a perfectly valid number, we will leave it like that, and 2-0 = 2,1351

Well that is it, a minus b is that.1358

How about if we do 7 × a, like we said, well, 7 × a, a is (7,sqrt(6),2).1362

That just means that we take 7 and multiply each of those components by 7, so we get 49, 7 radical(6), and we get 14, that is it.1367

Note, sometimes, notice we have been writing our vectors as points in space horizontally.1389

Listing the numbers horizontally component wise.1396

Sometimes we are actually going to list them vertically, it is just a notational device.1399

Later on we will begin to work with matrices. 1405

So, in multivariable calculus, you will see why sometimes we write them vertically. 1408

Just know it is the same thing when we write them this way or this way.1413

So, sometimes we will write the vector a vertically.1415

If I had the vector a, which is, let us say, a four vector (1,7,4,2), a point in four space that is going to be equivalent to (1,7,4,2).1425

The only thing you have to watch out for is you want to retain the order -- (1,7,4,2) is a specific order.1439

You cannot just write (1,2,7,4), it is not just a set of numbers. 1445

It is a list of numbers in a specific order, so vertical, horizontal, it is just a question of what is convenient for us at the time.1449

Okay, so 2 more properties.1456

Two more properties concerning vector addition. 1460

The first one says that if I take a constant, if I take two vectors a + b, and if I add them first, and I multiply them by a constant, their sum is the same as, if I can distribute this constant over each and then add them, a constant times the sum of two vectors is equal to the sum of a constant times each individual vector.1472

I can just distribute the constant. 1498

I can also do it if I have two constants instead.1502

Multiplied by a given vector I can distribute the vector over the constants. 1505

So it would become C1 × a plus c2 × a.1510

That is it, you already know all of these common properties of the real numbers, all of the behaviors are exactly the same.1517

Let us see what we have, we have done the properties.1528

I want to talk real quickly what we mean geometrically when we multiply a vector by a constant. 1534

Geometrically, multiplication of a vector by a constant, is just changing the length of a vector, and/or the direction of the vector.1542

Let us take a standard vector, say we have this vector right here, a.1579

Let us say a is the vector, the point in space, (3,2).1590

Again, we are going to be working in 2 space where we can actually visualize these things easily.1594

Well if I said 3a, you know what the algebraic value of 3a is.1600

It is just 3 × 3 which is 9, and 3 × 2 which is 6, which is (9,6).1602

So the new point is (9,6).1610

All you have done to the vector is multiplied its length by three.1614

That is 2 that is 3, so this is the point (9,6).1617

Let us see if we did -2a, now we have not only increased its length by 2, but we have actually changed its direction. 1623

We have reversed it, so in this case - 2a algebraically, well -2 ×3 is -6, -2×2 is -4. 1634

Now, we have this is -1a, and this is -2a, so now our point (-6,-4) is right there.1641

That is all you are doing.1653

Given a vector, negating that vector means reversing the direction of that vector.1660

So, if it points this way, it is going to point in the opposite way 180 degrees.1663

If you multiply a vector by a constant, if that constant is bigger than one, you are lengthening the vector, but you are keeping the direction.1668

Or, it could be smaller than one and you are taking the vector and you are shortening it.1673

Say multiply this by 1/2, now the vector becomes (3/2,1).1676

So again, you are just changing the length, that is all you are doing.1680

Okay, so this was just a basic introduction to points in space and vectors, and how we are more often than not going to just be speaking of points in space as vectors, and with a little bit of instruction in notation.1690

So thank you for joining us here with our first lesson from multivariable calculus,1699

We will see you next time, bye-bye.1702

Hello and welcome back to educator.com and welcome back to multivariable calculus. 0000

In the last lesson we introduced the notion of a function of several variables.0005

In today's lesson what we are going to do is we are going to actually talk about doing Calculus on those functions, the idea of a derivative, so let us just jump right on in.0009

So, when differentiating a function of several variables, really it is the exact same thing as a function of single variable, except what you do is you hold every other variable, you just treat it like a constant.0018

You do exactly what you have always done before. Nothing is different. 0033

So if you have x2 + y2, let us pretend that we are just taking the derivative with respect to x, and we will get to the notation in just a minute.0037

You just hold y, you just treat it like a constant, and you just differentiate with respect to x.0045

Then if you want to take the derivative of y, now this time you hold the x constant and you differentiate with respect to y.0050

Let us just do some examples, and I think it is the best way to make sense of it.0058

We will let f(x,y) equal x3, y2, so nice basic function of 2 variables, x3, y2.0071

Before I go ahead and actually do the derivatives, I just want to introduce the notation that you are going to see. 0085

You are used to seeing... you know... dx, or you are used to seeing f'(x)... those are the standard notations for single variable calculus. 0091

In multi-variable calculus you are going to see lots of different notations. There are some that are standard.0100

I am going to introduce several of them, and as it turns out, we are actually going to be using different variations of it.0106

We want to make sure that you actually recognize the different notations that are used, because again we are dealing with more variables now, so the notation becomes a little bit more cumbersome, a little bit more complicated.0116

But we try to keep it as close to this as possible.0125

If we want to differentiate with respect to x, so we might see something like this, we will see df/dx, something that looks kind of like a d, but is a little more curvy.0130

We also use, we tend to use capital letters for the derivative of functions of several variables. 0145

So dx means you are taking the derivative of the function with respect to x.0152

You will also see it as d1. Let us say that you have a function of 19 variables, well you cannot.. x,y,z,... you cannot just come up with a bunch of different, so we just say the first variable, the second variable, the third variable, the fourth variable...0157

Things like that, so da means I am differentiating the function with respect to the first variable. Everything else constant.0169

These are the shorter notations that we use. If you want to be a little bit more explicit, you can do something like d1f, or like f(x,y) if you specifically want to mention what the variables are.0178

Again, you can do something like dxf(x,y). You will also see something like this: fx or f1, something like that.0196

These are all different notations that are used to notate the partial derivative of a function of several variables with respect to whichever variable you are talking about.0209

Now, quick thing, remember when we did something like dy/dx in single variable calculus? We can actually treat this like a fraction.0220

In other words, we can move things around, move the denominator around. This notation which is analogous to this notation, this is not a fraction.0228

Symbolically we can sometimes think about it, especially later when we talk about the chain rule, but it is very, very important to remember this is not the same as that.0240

This is just a notation talking about the derivative of the function with respect to the variable x, but you cannot do the same thing with it that you do with this.0249

It is not going to be a problem though. Let us go ahead and jump right on in and see what we can do.0259

I am going to go ahead and say, I am going to go ahead and differentiate this with respect to x, so I will do dx, how is that?0265

So the derivative with respect to x, that means I hold this constant so it just shows up as y2 and I just differentiate this.0274

It becomes 3x2, and since this is a constant this y2 stays. So 3x2y2, that is the derivative, partial derivative with respect to x.0280

Again, we say partial because we are holding everything else constant.0292

Well, dx, now we are going to do the partial derivative with respect to y. Now we hold x constant, so this x3 term stays exactly as it is. 0297

We differentiate with respect to y, so this is 2y, so it becomes 2yx3. If you want you can write 2x3y, x32y, it does not matter, multiplication is commutative.0307

There are certain standards -- they generally tend to put the x before the y, but we are not concerned with cosmetics, we are concerned with the mathematics.0324

Let us do another example, there is nothing here that you have not done before. Just treat the other variable as a constant, that is it. 0332

You just have to be sort of careful when you do this, because again the functions are getting a little bit more complicated, so you just have to be careful with what is what.0339

It is perfectly common to make errors, it is not a big deal, you just check it carefully.0347

So example 2, let us go ahead and let f(x,y) equal to... this time let us do a trigonometric function, cos(x2y).0352

Well, let us see what we can do. This time we will do, I will write it as d1, so d1 means with respect to the first variable, which happens to be x. That is going to equal.. so the derivative of cos(x2y) is - sin(x2y), and of course this is a chain rule because we have a function of x,y, in that thing.0363

So now we take the derivative of what is inside with respect to x, so that is going to be × 2xy.0387

So the derivative of this, we hold the y constant, we differentiate, we get 2xy.0394

We are going to write that as -2xy × sin(x2y).0400

Now we will do d2, which is the same as the derivative with respect to y.0410

Now, again we are going to have -sin(x2y) and this time we are going to differentiate with respect to y which is x2 -- differentiating with respect to that, we are going to hold this as constant, so some constant × y, the derivative is just the constant.0415

Therefore, you end up with -x2sin(x2y). That is it. Nothing going on, nothing strange.0437

Now let us do one more example where we actually will evaluate a derivative at a certain point. 0449

You remember the derivative gives you a function. Well, that function, if you actually have a specific point, a specific point (x,y) that you put in there, it gives you a number. Same as single-variable calculus.0455

So, let us see, we have got... okay... example 3.0470

This time, we want to evaluate d1 and d2 from the previous example, so we want to evaluate this at the point (x,y) = pi,1. 0478

So, we want to know what the derivative is at the point (pi,1).0504

Well, d1 of (pi,1) is equal to -2 × pi × 1 × sin(pi2-1).0510

When you just evaluate it, put all the numbers in, the number you get is 2.7037, so you see that it is exactly like it is in single-variable calculus. 0529

d2 of (pi,1) is going to equal - pi2 × sin(pi2), and you end up with 4.2469.0540

That is it. Let us go ahead and do a function of 3 variables.0555

We have example 4, now we will let f(x) = x3 × sin(yz2).0565

Now, notice the notation that I used, slightly different notation. Up before in the previous examples, I specifically wrote out f(x,y).0583

In this case, in the argument for the function, I put a vector, a vector x.0592

Again, a vector is just a point, it is a shorthand notation for the point, in this case, (x,y,z).0598

I could have written f(x,y,z), but I want to be using notation that you will be seeing. Various different types of notation. We need to recognize them.0604

Single-variable calculus is reasonably uniform in its notation. Different books, different fields, physics, engineering, mathematics, chemistry, they tend to use a whole bunch of different notations and it is good to be familiar with them. 0612

We do not want to just stick with one notation, we want to be able to recognize others.0626

Now, let us go ahead and do, let us just do like d3, let us not do all of the d's, let us just do d3.0632

So, let us go, which is going to end up being d(z), so d3 which is equal to d(z), which is equal to df/dz, all of those notations.0643

That is going to equal, well, let us see what we have got. With respect to z, so we are differentiating with respect to z which means that we are holding everything else constant.0657

So this is going to be x3, and since z is in here it is going to be the cos(y × z2) × derivative with respect to z, which is 2yz.0668

There you go. That final answer is going to be 2x3yz × cos(yz2).0689

That is it. You just have to be really, really careful when you are doing these partial derivatives to make sure that you are holding the proper thing constant, make sure that you are taking care of the chain rule... that is it, that is all that is going on here.0700

So, now we have come to a point, I am going to introduce a definition, something called a gradient of a function of several variables, the gradient vector. 0715

This is a profoundly important definition. if you do not take anything else away from this course, by all means you want to be able to understand both algebraically, and later geometrically, what the gradient means. 0723

It comes up all the time and it is a profound importance.0735

So, definition... in fact, let us go ahead and change the color of our ink here. I am going to go ahead and do this one in blue.0740

So, definition, let f be a function from RN to R, so a function of several variables.0751

Then, the gradient of f, notated grad(f) -- we will give another notation of it, one that is a little more common, later on in this lesson -- notated grad(f) is the vector, very important, the gradient is a vector, not a number... oops, getting all kinds of crazy lines here.0764

The vector... grad(f) is equal to df/dx1, df/2, so on... df/dxn.0805

Let me tell you what this means. If I have a function of let us say 5 variables, the gradient is the vector where the components of that vectors, the first component is the derivative of f, the partial derivative of f with respect to the first variable.0828

The second component is the partial derivative with respect to the second variable, and so on.0845

The third, the 4th, the 5th component is the partial derivative with respect to the 5th variable, very important.0851

The derivative is not a number, it is a vector. It is called the gradient vector. Let us go ahead and just do an example. I think that is the best way to make sense of this.0859

I am going to go back to black here. Let us go, example 5.0870

Let f(x) = 3x2y... no, wait, what happened here, function of several variables, nope, what a minute, no let us go ahead and do it like this.0878

Yeah, that is fine, let us go ahead and do 7xy3, in fact, you know what, I tend to prefer the other notation personally.0905

I am going to go ahead and write f(x,y) explicitly = 7xy3.0917

Now, the gradient of f is equal to, well, we take the partial derivative with respect to x, so when we hold everything else constant and take the derivative we get 7y3, that is the first component.0925

The second component is the partial derivative with respect to y.0943

So, we go ahead and differentiate with respect to y holding the x constant. That is going to be 3 × 7 is 21, so it is going to be 21xy2. That is the gradient vector. 0947

What you are taking is the function of several variables. You are differentiating with respect to each variable, and the derivatives that you get respectively end up being the components of the gradient vector.0961

Let us go ahead and do another one, so example 6, let us do 3 variables now, f(x,y,z) = x × y × sin(yz).0975

Be very, very careful here, again look at the function here before you just dive right on in.0995

So, let us do df/dx, so df/dx, the derivative with respect to x is going to be... well, notice we said you hold everything else constant, so you hold the y and z constant.1000

This sine function, sin(yz), x is not in this argument. Because x is not in this argument, this whole sin(yz) is a constant and y is a constant.1013

So, the derivative of this is just plain old y × sin(yz).1025

Make sure you do not just automatically start taking the derivative and say x × cos(yz) because there is no x in the argument for the sine function. It is just a constant. 1032

You treat it like a constant. Let me make my z a little bit clearer here.1045

Now, let us do df/dy. Okay, df/dy is going to be... okay, so now you notice we have a y and we have a sin(yz) so this is going to be a product function.1050

We are going to have to differentiate the way we differentiate a product, 1 × the derivative of the other + the derivative of that one × the other. That is what it looks like.1067

So we are going to take xy, the derivative of this which is cos(yz) × derivative of what is inside there because this is a chain rule, × z.1078

Again, you just have to be very, very careful and it is a very common error, it is not a problem, that happens, that is life.1094

Plus, now let us differentiate with respect to... it's the other derivative, so it is this × the derivative of that + that × the derivative of this.1101

The other things that are constant here, remember this is our function, this is the RY, so it is these 2, so it is going to be + x sin(yz) × the derivative of y which is 1.1110

That ends up being xy, oops, let us do xyz, so that is xyz × cos(yz) + x × sin(yz).1129

I certainly hope that you are going to double check this for me. 1145

Now let us do the third partial derivative df/dz, so now we hold the x and y constant, so we end up getting... let us see, what happens... so if we hold the x and y constant, that means these are constant, so we end up with xy cos(yz) × derivative of what is inside with respect to z which is y.1150

That will equal xy2, if I am not mistaken, cos(yz), there you go.1182

This is df/dx, this is df/dy, this is df/dz. The gradient vector, the gradient of f, is equal to the first component is df/dx, the second component is df/dy, and the third component is df/dz.1190

This, that is the second component, and that is the third component. It is very, very important, the gradient is a vector.1214

Now, let us go ahead and finish off by discussing the alternate notation, the more common notation that you are going to see in especially in physics and engineering.1225

We will finish off with a couple of properties for the gradient, really simple properties, exactly as they are for first variable calculus.1239

So, let us say there is another notation for gradient, which is very common.1247

The symbol is this upside down triangle, an upside down delta, f(x,y) or just this, and it is called ∇f.1276

It is called the del operator, basically when it says you do del to f, it means you take the gradient of f, that is it.1292

It is just an alternate notation instead of writing G R A D I E N T, it is just a... again, it is just a symbol.1300

So, let us finish off some properties and they are exactly what you expect.1308

The gradient, if I have 2 functions f and g, functions of several variables, if I add the functions then take the gradient, it is the same as taking the gradient and then adding those.1314

So, gradient of f + g = gradient of f + gradient of g. That is it.1326

The gradient of a constant times a function = constant × the gradient of the function... ∇f, so if you like the del notation, I personally do not like the del notation, but that is just me, so it is ∇f + ∇g = c × ∇f.1337

This right here, when you add 2 things, it is the same as taking... so this is again, we are operating, we are operating on this function.1366

Operating on the sum of a function is the sum of the operation on that function.1377

In this case taking the gradient of a constant × a function, is the same as a constant... this is defined, is actually the definition of linearity.1382

For those of you who have done linear algebra, you would recognize this as the algebraic definition of linearity.1391

It turns out the gradient operator is a linear operator.1395

Okay, so we have introduced the gradient, again a profoundly, profoundly important notion in multi-variable calculus.1399

We will go ahead and stop the lesson here, thank you for joining us at educator.com.1406

We will see you next time, bye-bye.1409

Hello and welcome back to educator.com and welcome back to Multivariable Calculus.0000

Today, we are going to be talking about higher derivatives and mixed partial derivatives.0005

So, we introduced the notion of the partial derivative of a function of several variables.0010

You basically take the single variable derivative by holding the other variables constant, and you work through a series of partial derivatives, first derivatives. 0015

Just like for single variable, we can go ahead and take higher derivatives, second, third, fourth, and fifth, but now we can mix and match.0024

For example, we can take the first partial derivative and then we can take the derivative with respect to the other variable of that derivative, and back and forth.0030

Let us sort of just jump in with an example and see what we can do with this. 0040

Before I actually discuss that, however, I want to introduce a notion called an open set.0046

I am not going to spend a lot of time on it, I just want you to know that this idea of an open set is going to be the domain over which we are going to be defining our functions of several variables.0052

I want you to be aware of it, it is going to come up over, and over, and over again in the theorems, and the definitions -- very, very simple -- it is exactly the same as it is for single-variable calculus.0063

It is basically just a region, all the points in that region, but not including the endpoint, not including the boundaries.0072

Let us go ahead and just write the definition down and then we will move on to mixed partial derivatives.0081

So, definition. An open set in n-space -- excuse me -- is the analog, this is not really a formal definition, it is an informal definition. 0086

If you want the formal definition, you know, you can go into your books... is the analog to an open set in 1-space, which is the real number line.0108

Which is what you have been dealing with all of these years. You know that the real number line... something like this... let us say this is 0, let us say we have -1, let us say we have +1.0124

Well the open set, you remember, is the notation that has the parentheses like this instead of the closed brackets, which means that we include all the numbers between -1 and 1, but we do not include the endpoints, that is it.0134

So, basically, an open set is every point within a specified boundary, but not including the boundary points.0150

That is really all we are talking about. If you have something like... let us do a 2-space example... so this is the x,y plane, and you have some region like this.0191

Actually, you know what, let us go ahead and make it so it looks... so let us say we have some region like this, I will put dotted, something like that.0203

So this region in here, that is an open set. It does not include the boundary. That is all we are talking about.0215

Just a quick version in R3, so this is the x, the y, and the z. 0221

So we have some... let us just do a sphere. We call it a ball. Here, if you have a circle in the plane, we call it an open disc.0230

Again, the specific things that we call them are not important. The idea is the notion of an open set, or an open region.0240

Because you are not always going to get a perfect domain, you know a nice symmetrical domain like a ball or a disc.0247

In this case, let us just do that, and then maybe like that, so think of a sphere in n-space anywhere -- centered at the origin or centered at some place else -- but, not including the actual boundaries themselves, everything inside.0254

So again, these open sets... let me write this down.0268

So, these open sets form the domains over which we define our functions of several variables.0273

So, if we had some function f that is a mapping from R2 to R, let us go ahead and make sure this is a little bit more clear. 0304

So some function f, and we are taking points in R2, the vector in 2-space, a point in 2-space, we are doing something to that vector, we are spitting out a number. Let us say the domain is this.0314

Basically, all of the points in here, those are the arguments that go into the actual function. Those are the points that we operate on, if you will. That is all.0324

Just a notion that you should be aware of. It is going to come up a lot, again. All of the definitions and the theorems.0334

Okay, now let us go ahead an move on to our partial derivatives.0340

Again, we can take a second and third, fourth, fifth derivatives of functions of several variables, but now of course we are dealing with partial derivatives in the sense that we are holding one variable constant and we are differentiating with respect to the others.0345

So, let us talk about notation first. This is very, very important.0360

As you noticed from the last lesson, the notation for partial differentiation tends to become a little bit more involved because again you have more variables involved, that is what is going.0365

It is very important to be accustomed to the notation and the scientific literature and the different books that you are going to read and the different teachers that you are going to have, they tend to use notations form a lot of different sources.0373

We want to make sure that if you have not seen a particular notation, at least you can recognize was is being said.0387

We said we have... let me just write here, notation... so, we will let f be a function from R2 to R, so in this case we are going to start a function of 2 variables. 0393

So, an f(x,y) if we happen to call the variables x and y, but we do not have to it could be x1, x1, the variables themselves do not actually matter.0413

It is the notion underneath. The partial derivative of f, with respect to the first variable x is notated like that, that little modified d, df/dx.0420

We also do d1, so d1 means take the partial derivative with respect to the first variable, the first variable being x. 0433

They are written in order in the argument, that is it.0440

Again, you also have seen it as dx. Well, df/dy the partial derivative with respect to y, that is one notation, and we often write it as d2, so we are taking the partial derivative with respect to the second variable.0444

This capital D notation, that is actually very, very common in multi-variable calculus. You will this a lot also, but often times just to sort of keep the mathematics on the paper reasonably sane.0461

When you have all these df/dx's df/dy's floating around, it gets really crazy -- so, this d notation, this capital D notation - very, very, convenient.0472

Again, you will also see it as dy, specifically.0483

Now, if we want to take the derivative with respect to x of the first partial, in other words, d/dx of df/dx, it is like this.0488

It is going to be d/dx of df/dx, so the partial derivative with respect to x of the partial derivative with respect to x.0501

It is notated as d2f/dx2 completely analogous to single variable calculus.0516

The capital notation is something like this, just D1/D1f, in other words you have taken the first partial derivative, now take the partial derivative with respect to the first variable of what you just did.0523

This is also... you see it this way, d2f. This means take the partial derivative with respect to x, then take the partial derivative with respect to x again, or the first variable. 0539

Now, let us do d/dy of df/dy, that is going to be the same as d2f, dy2.0553

Again, you have got d2/d2f and this is often written -- sometimes -- squared f.0565

I personally do not like this notation myself, this d12f, d22f.0575

I like to see everything that I am doing, so this tells me that, it takes the first, I take the derivative with respect to the second variable, and then of that thing, I take the derivative with respect to the second variable. 0578

These are just differential operators. They tell me what to differentiate. That is all.0590

Now, let us go ahead and do, d/dx of df/dy.0596

Now, I have taken the partial derivative with respect to y, the second variable, now I am going to go ahead and take the partial derivative of that with respect to the first variable.0610

That becomes... actually, let me just do this one in reverse. This one I am going to write the capital D notation first, because I think it is a little bit better, and then I will write this modified D notation here.0623

So, d1d1, that means I have taken the derivative with respect to the first variable x, and now I differentiate that with respect to the second variable.0637

That is going to be the same as d/dy of df/dx, and that is d2f dy dx.0649

Clearly you can see that this notation is going to start to get really, really cumbersome very, very quickly.0664

I mean it is very beautiful aesthetically, and it is very nice to look at, and it is really important that when you do your math you at least like what you see, but it can be a little daunting.0669

Let us go ahead and specifically say, this is what we are doing first. We are moving from right to left.0682

This means you have a function f, do D1 to it, then do D2 on it. That is first, and this is second, order is very important here.0688

Now, of course we have D1, D2 of f, which means we have done D2 first, then done D1, and this is going to be d/dx of df/dy = d2f dx dy.0698

Wow, that is a lot.0717

Alright, so let us just do an example. That is the best way to make sense of this.0722

Again, you are reasonably comfortable with differentiation of basic functions. With multi-variable calculus just be a little bit more careful. 0727

You have to remember which variable you are differentiating with respect to so just go a little bit slower. That is all you have to do with Math in order to be correct.0736

Just go slow and be careful. So, example 1.0742

We will let f(x,y) = x3y2, so d1 the derivative with respect to x is equal 3x2y2, in other words I am treating y as a constant and just differentiating with respect to x.0751

D2 = x3y, this time I ended up holding x constant and differentiating with respect to the second variable, y.0772

Now, I am going to do d2 of d1, in other words I have done d1, now I am going to differentiate this thing with respect to y, the second variable.0791

This is going to give me, 2 × 3 is going to give me 6x2y.0801

Now, I am going to do D1 of D2, which I just did. I just did D2 that is x3y, then do D1 of that, which means I am going to differentiate this with respect to x.0809

I get 6x2y. We will stop and take a look at this.0823

D2D1, I did D1 first then I did D2, I got 6x2y.0828

here I did D2 first, then I got D1, then I did 6x2y and they ended up being the same. This is not a coincidence.0832

So, we will put MV, which means... take note of this... these ended up being equal.0843

Not a coincidence. Now, let us go ahead and write down the theorem that allows us to always that this will be the case.0863

Theorem. Let f be a function from R2 to R, defined on an open set. 0874

Now, we are going to be using... I just introduced the term open set, f, R2, this is often how you are going to see theorems in mathematics.0887

This is very precisely stated theorem, but it is nothing that we do not understand and know. It is just we want something that is formal and that is precise.0896

Defined on an open set, and assume D1 D2, D2 D1, exists. In other words assume that the partial derivatives actually exist, and are continuous.0906

These clearly exist, and these are clearly continuous functions. There is no problem here. Then, D1 D2 of f = D2 D1 of f.0935

Notice up here just real quickly that I just wrote, D1 D2, D2 D1, but I did not put the f. 0952

Again this is sort of a personal thing that I do. I tend to sort of minimize my notation simply because I do not like a lot of things floating around on a piece of paper.0960

The idea is that you know what you are talking about. You know that you are dealing with f, you know that D1 is the first partial derivative, you know that D2 is the second partial derivative.0969

You do not have to be that explicit. You can modify your notation depending on how much you know, how much you are comfortable with.0978

You have that freedom, do not feel that you are constrained to always write this, this, this, this. Unless you have the kind of teacher that actually demands that you write all the things out, all the x's, all the parentheses, all the y's be there.0984

Please, feel free to take some liberties with this. You are the one doing the math, you are the one that needs to be comfortable with this.0997

For me, I tend not to write the f. I know that this is not confusing at all.1004

Now, let us consider a function from R3 to R. Now let us move from 2-space to 3-space, a function of 2 variables, to a function of 3 variables.1010

Consider the function from R3 to R, or f(x,y,z).1021

That is it. Now, if we take f(y,z), so now we have D1, we will have D2, and we will have D3, partial with respect to x, partial with respect to y, partial with respect to z. 1035

Now we can sort of mix and match. Notice I can have... I can do D1, D2, D3, I can do D3, D2, D1... D2, D1, D3... D2, D3, D1... so all kinds of mixed partials are possible now.1050

Repeated applications of the theorem, which we will do an example of in just a minute, demonstrate that mixed partials are equal, mixed partial as of course 1,2,3... so no worries there... are equal regardless of order, as long as the variables with respect to which we differentiate are the same.1065

Let me just show you what that means. In other words, if I take... so D1, D2, D3 of f.1144

I have taken D3 first, then I have done D2 to that, then I have taken D1 of that.1153

Well, if I do it in different order, do... let us say, D3, D2, D1 of f... if I do D2 first, then D2, then D3... or if I do D2, D3, D1... again we are working from right to left, in other words I do D1 first, then 3 then 2. 1159

As it turns out, all of these mixed partials are the same. It does not matter which order you actually differentiate in, as long as the 1,2,3... 1,2,3... 1,2,3... as long as that is the same.1178

You can actually do it in any order. This is really, really extraordinary. I mean there is no reason to think that if you take the partial derivative of a function of several variables in... you do the mixed partial derivatives, there is no reason to believe that they should be equal and yet there it is. Really, really fantastic.1190

Okay, let us just do an example, and so it is nice to see these things sort of fall out.1209

We will let f(x,y,z) = x2y2z3.1219

Now, we are just going to do a bunch of partials. Let us do, let us start with D3, in other words z.1230

So, D3 is going to equal 3x2y2z2.1240

Now we will do D2 of D3, in other words we are going to differentiate with respect to y, this thing up here. D2 of D3.1247

That is going to equal 6x2yz2, and now we are going to do D1 of D2,D3. In other words we are going to differentiate with respect to x, this thing right here.1257

We end up getting 12xyz2.1272

So, we will do that, now let us go ahead and od it in a different order.1279

This time let us take D1, well D1 which is just Dx is just going to be 2xy2z3.1285

Now let us do D2 of D1, which is... I am going to take the derivative with respect to y of this one, of D1.1296

That is going to equal 4xyz3.1303

Now, I am going to take the derivative with respect to z of the D2 D1 that I just got.1310

I end up with 12xyz2.1317

Wow, what do you know. They are the same. Alright, now let us try another order.1322

Now let us do D1 first, we did D1 already, that is 2xy2z3.1329

Now I am going to do, instead of D2, I will do D3 next. So I will do D3 D1.1338

I am going to take the... derivative with respect to z of D1, so that is going to be 6xy2z2, I think.1344

Now, I will do D2 of the D3 D1, and I get... 12xyz2.1357

Is it the same? Yes. 12xyz2, 2xyz2, 2xyz2. I put the partial derivatives in different orders and yet it ended up being exactly the same.1368

This is very, very, very deep and extraordinary. I am just going to write that. I am just going to write "pretty amazing".1380

Now, let us see what else we can do here.1392

Regarding notation, there is just one caution that I am going to throw out there.1400

So, let us write caution: do not confuse the following.1405

Much of the problem with higher mathematics is notation. Notation, it is just one of those things. We have to have some way of representing what it is that we are actually doing.1418

As things become more complicated... well, things get more complicated, so we just want to make sure we know how we are operating.1427

What kind of math we are actually doing. Do not confuse the following.1436

If you see d/dx2 of f, that is the same as d2fdx2.1441

This is D1(D1) of f. In other words take the derivative with respect to the first variable, then take the derivative again with respect to the same variable.1455

It is not the same as this one. df/dx2, which is D1 of f squared.1470

This one says take the derivative, take the first partial derivative, or the derivative with respect to the first variable and square it as a number, or a variable, or a function.1486

This one notice, it is the d/dx, the differential operator that is squared. 1499

Here, f is actually inside, here f is outside. This one right here, this right here, not the same. This one is also written this way.1504

Remember we said often times we will write the square, D12f, so D12f is not the same as D1f2.1514

This one says differentiate with respect to f, then differentiate again with respect to the same variable.1524

This one says differentiate with respect to f, then square that number or function.1530

Two very, very different things. Another reason why I actually prefer that notation.1533

This one is reasonably clear, this one can be a bit confusing.1538

Okay, let us just do another example here. So, example 3.1543

We will let f(x,y) = sin(x2+y), and we will do... well, let us just run through them.1551

Let us run through all of them. We have D1.1569

The derivative with respect to x. We have cos(x2+y) × 2x, which we will write again... I mean you leave it like this it is not a problem but it is traditional to sort of bring all the functions forward and leave the trigonometric function to the end.1575

So, 2x × cos(x2+y).1599

D2 = cos(x2+y) × 1, which equals... because this is the derivative of the inside, right? chain rule... cos(x2+y).1607

Now, let us do this for the heck of it, let us do D12 which is the same as D1, D1.1625

So, we have D1 already, now we are going to differentiate this with respect to x. So this is going to be... I tend to pull my constants out, so I am going to pull this out, and it is going to be a power rule. x × cos(x2+y).1636

So it is going to be this × the derivative of that, and end up being -x × sin(x2+y) × 2x + that × the derivative of that, which is this × the derivative of this which is just 1. 1652

Again, I have pulled the 2 out. So, it is cos(x2+y), so again, the idea is just be really, really careful and go really slowly.1674

There is no hurry, the idea is to be right, not to finish quickly.1681

When we put this together, we end up with, so the 2, and the 2... this is a × not a minus sign.1687

You end up with -4x2 × sin(x2+y) + cos(x2+y). That is the derivative of the derivative.1696

Derivative with respect to x of the derivative with respect to x. D12, D1 D1. I really love this capital notation, and I love not writing the f's.1710

Now, let us go ahead and do D22.1722

So, D22, which is D2 of D2, when we do that, we end up with -sin(x2+y) × 1.1726

So, equals -sin(x2+y). I hope you guys are checking this.1742

Now let us do D1 D2. When we do D1 D2, we end up with, in other words, we have taken D2, now we are going to take D1 of D2.1750

-sin(x2+y) × 2x is going to be -2xsin(x2+y).1761

Now, we will do D2 of D1, in other words we have done D1 first, now we are going to do D2 of that.1783

We get -2x × sin(x2+y) × 1, which equals -2xsin(x2+y).1789

Well, what do you know, this ends up being the same as that. This mixed partial is the same as that mixed partial.1803

The order is irrelevant, but it is 1,2... 2,1.1810

That is it. So mixed partial derivatives of higher order are the same for functions of several variables provided... so the order does not matter... provided that all of the particular variables with respect to which you are differentiating are there.1815

So 2,3,1... 3,2,1... 1,2,3... 1,3,2... all of those will actually end up being equal, and I personally think that is absolutely extraordinary.1833

Thank you for joining us here at educator.com, we will see you next time for some more Multivariable Calculus. Take care, bye-bye.1842

Hello and welcome back to educator.com and multi-variable calculus.0000

Today we are going to talk about the chain rule, and you remember from single-variable calculus the chain rule just allowed you to differentiate functions that were composite functions.0005

Composite functions where something like sin(x3), something like that, you took the derivative of the sine, then you took the derivative of what was inside the argument, the x3, that became 3x2.0013

Now that we are dealing with these vector value functions, these functions of several variables and we have introduced the gradient, we can actually bring those tools to bare on differentiating a function, a composite function that involves functions of several variables.0027

So let us just jump right on in, and let me give a quick description of what it is that is going on. 0045

Again, what we want to do is not just jump right into the mathematics, we do not just want to write symbols on a page, we want to be able to understand what is happening. 0052

When there is understanding, when you can see what is going on, you can use the intuition that you have already developed to decide what goes next and where to go. That is the whole idea.0060

We want you to understand what is happening mathematically before you actually do the mathematics.0071

That is the easy part, the mathematics will come, that is just symbolic manipulation but we want to see what is going on.0076

Let us start in R2, let us start in the plane, again we are using our geometric intuition to help us guide our mathematics.0082

So, let us say we have something like this. Now let us say we have some region in... you know, here... so now, let us suppose the following things.0090

So suppose f is a function from R2 to R, so a function of 2 variables, defined on the open set u. This is u by the way. Open set u.0104

Now, also suppose that C, which is a function from R to R2 -- we are dealing with R2, okay -- R to R2, is a curve in 2-space, that passes through u.0125

Let us say this is just something like that. So passes through u. Now, here is what is great.0163

The points along the curve, the function itself is defined on this open set.0172

In other words the points in this set can be used for this function f.0177

Well, the curve happens to pass through u, so the point on the curve can be used in f.0182

That is what we are doing, so what we can do is we can actually form the composite function, what we have is some curve that has nothing to do with u and yet it happens to pass through u.0191

We also have a function that is defined on u because there is this overlap we can use the... in the function f, we can use the points along the curve.0200

That is what is really, really great here.0211

Let us go ahead and write this down, and then the points along c(t) can be used by the function f.0215

In other words, we can form the composite function of c, which is f(c).0240

Let me write it with the actual t... oh this is a capital F by the way, sorry about that... small f's big F's, hm.0270

F(c(t)), that is it, it is a composite function. Except now, instead of a composite function of single variables, it is a composite function of multi-variables.0278

In this case, we are dealing with 2. Let us just do a quick example, and then we will discuss it just a little bit more. Just to make sure we understand this concept.0289

This is a profoundly important concept, so like the gradient we definitely want to have a good sense of what is going on here.0298

We will take the time to make sure that that is the case.0307

Let us just do a quick example. Just so that we can see... so example 1.0311

Now, let c(t), let the curve equal the following: let go t3 then t2, so t3 and t2.0319

Now we will let our f, capital F of (x,y), of two variables, let us call it ln(y) and let us do cos(xy). How is that?0332

Just a... you know... nice random function. Now, f(c(t)). Notice. F is a function of two variables x and y, so its argument contains 2 things. 0345

c(t) has two things, for x we put in the t3, for y we put in the t2, into here and we form the function F as a function of t.0366

Watch what happens. F(c(t)) = f(t3,t2), right?0380

f(x,y) that just means these 2 things, so wherever I see a y I put in whatever is in here, wherever I see an x I put in what is in here.0392

Well F(c(t)) is this, it actually spits out 2 values, those 2 values, this is x, this is y.0400

We end up getting the logarithm of t2 × cos... natural logarithm × the cosine(t3t2, which equals ln(t2) × cos(t5). 0409

That is it, that is all I have done here. I have formed this composite function now, with 2 different functions that mapped to different spaces, but the space where one maps to is exactly the space that the next function needs as its domain in order to take the next step.0432

Let me actually write down this whole idea of the functions again, so C(t) is a map from R to R2.0450

In other words, it takes numbers, real numbers, and it spits out 2 vectors, a point in 2 space.0466

F(x,y) it takes a vector, a point in 2-space, and it spits out a number.0474

So what I have really done here, the composite function actually ends up being a map from R to R. That is what is happening here.0487

That is what is important to see. You can jump around from space to space, that is what makes multi-variable calculus so unbelievably powerful. 0826 That you can actually jump around from space to space like this, with well defined functions.0499

This is going to be the x,y plane, this is R2. I will write you another copy of the real number line.0513

So, this is the real number line, this is R2, 2-space, and this is the real number line.0520

C(t) maps from here to here.0527

A point to a vector... a vector in 2-space.0535

F goes from here, takes a point in 2-space, and maps to here, so this is c(t) and this is f.0540

When we find the composite function, f(c(t)), what I have now is a map from R to R. That is what this example is.0549

I have this that goes to a point in 2-space. F takes a point in 2-space and spits out some number.0566

Noticed I ended up with some function of t, ln(t2)t5 a specific value of t. This is just a single number.0572

This is what is going on here. You are just forming a composite function with curves and functions of several variables.0583

Hopefully this is reasonably clear. This is what we want to understand. This is what is happening mathematically.0593

You are mapping from one space to another, and then you are moving from that space to another space. In this case the space you end up with happens to be the space you started off with, which is the real number line.0596

So now the chain rule allows us to differentiate something like this. So now, let us go ahead and explicitly write down what the chain rule is.0607

I want you to see this simply because I want you also to start becoming accustomed to the expression of theorems, formal things, but again, it has to be based on understanding. 0618

It is a little long, but there is nothing here that is strange, so let f be a function defined and differentiable on an open set u.0630

Let c be a differentiable curve, all that means is a nice smooth curve with no whacky bumps or corners... be a differentiable curve such that the values of c(t) lie in the open set u.0666

What we were doing in the beginning of the lesson, we have an open set, we have a curve that happens to pass through that open set, therefore the points along the curve can be used for our function.0713

Then, the composite function, f(c(t)) is differentiable.0725

It is differentiable itself... as a function of t and the derivative of f(c(t)) with respect to t is equal to the gradient of f evaluated at c(t). 1250 The dot product of that vector with the vector c'(t), now you remember the gradient is a vector.0738

If I have some function, like f(x,y), the gradient is df/dx, and the second component is df/dy, I just differentiate as many variables, and that is my gradient vector.0777

So let us stop and think about what this says. If I have some function that is defined and differentiable on some open set, and c happens to be a differentiable curve that passes through that open set, in other words take some values in that open set, then the composite function f(c(t)) is also differentiable. 0794

It is differentiable as a function of t, as a single variable t and the derivative of that composite function is equal to the gradient of f at c(t) · c'(t). 0812

This is very, very, very important.0825

Now, for computations, when we actually do specific problems, we of course are going to be working with components, which is always the case.0831

With vectors we can go ahead and write out the definitions and the theorems using a shorter, more elegant notation, but when we actually do the computations with vectors we have to work with components.0841

x,y,z, whatever it is that we happen to be working with. So, let us go ahead and just sort of write out the component form of this so you see what is happening.0852

Again, the dot product is the same dot product that you know. There is nothing new here, trust what you know.0860

This is a vector, this is a vector, when you take the dot product of 2 vectors you get a number. That is what this says. You are getting a derivative. 0868

A function of t, if you evaluated a specific point of t, it is actually just a number. You are still just doing a derivative. The same thing you have been doing for years.0877

Let us just see here. So, if c(t) equals, now c1, that is the... so c(t) is a curve... c2(t), its component functions are component functions of t, just like the first example, and we have f(x1,x2).0886

This time I did not write it as x and y, I wrote it as x1 and x2, these are variables.0920

The first variable, the second variable. Then the derivative with respect to t of the f(c(t)) = well, we said it equals the gradient of f evaluated at c(t) · c'(t).0924

Okay, the gradient -- I should probably write this out -- the gradient of this function is going to be... tell you what, let me go ahead and before I write that, let me write out the gradient because I know it has been a couple of lessons since we did that. 0949

So, let me write the gradf = df/dx1, df/dx2.0968

This is a vector, the first component of which is the derivative with respect to the first variable.0982

The second component is the derivative of the function with respect to the second variable.0988

Now, c'(t) is also a vector. It is the derivative of this, c1'(t) and it is the derivative of this, c2'(t).0991

That is it, these are just functions, so now what we have is the derivative with respect to t of f(c(t)), in other words this thing right here. 1006

We said it is the gradient of f · c', this is the gradient of f, this is c'.1018

So let us see what this looks like in component form. It is... oh, you know what, I have a capital F, don't I?1025

I keep forgetting that, that small f is just so ubiquitous in most scientific literature.1032

So, we have df/dx1 × dc1/dt, that is all this is, c' is just dc1/dt.1039

It is just notation. dc2/dt. The dot product is this × that + this × that + df/dx2 × dc2/dt.1062

So that is it, that is all we are doing here. We are just doing it in component form.1090

Now personally, I think that what I have just written here is actually a little bit more confusing than just the statement of the theorem.1096

If you look at it as just the statement of the theorem, the gradient of f dotted with c', and if you know what the gradient is, you know what c' is, you know how to take derivatives.1105

You just do the dot product. This is sort of the component representation of it.1114

I personally do not like seeing all of these things because again it is notationally intensive.1121

The idea is to understand what this is, and then you can do the rest.1125

So, personally, my favorite, I still think it is great to learn it this way. Gradient · c'. Gradient of f · c'. Just keep telling yourself that about 5 or 6 times, and you will know what to do.1130

So, let us go ahead and just do an example, that is the best way to make sense of this.1145

So, let me go back to my black ink here. Actually you know what, let me go ahead and go to blue.1151

So example 2. Now, we will let our curve t be t, e(t) and t2, and we will let our function x, y, z, so we are definitely talking about a curve in 3-space, and a function of 3 variables, equals xy2z.1157

First of all, let us talk about what is going on here. We are going to form the composite function. We are going to be forming f(c(t)).1193

That is what we are going to be doing. Well, f(c(t)), x,y,z, in this case x is this thing, t, y is this thing e(t), and z is this t2.1201

We want to write everything out. Now the gradient of f, that is it, we are just going to build this step by step by step, that is all we are doing here.1220

The gradient of f is equal to, well it is the first component is the first partial, the second component is the second partial, the third component is the third partial.1231

If you like the other notation it is going to be df/dx, df/dy, and it is going to be df/dz.1244

Now, let us go ahead and actually compute that. The first partial, the derivative of this function with respect to x is y2z.1254

The derivative with respect to y is 2xyz.1264

And, the derivative with respect to z is going to be xy2, so this is my gradient of f.1271

Now, my gradient of f, evaluated at c(t), so now we will take the next step, now we will do the grad of f evaluated at c(t) which is the actual expression that is in the definition for the chain rule.1284

All that says is that take my gradient f, this thing, and I just put in the values c(t) in here.1300

Well, x is t, y is e(t), and z is t2.1308

So when I put these things into here, here is what I get.1312

y2 is just, so it is going to end up being t2e(2t), right?1323

y2 is just e(2t), z is t2, so that is t2e(2t).1329

2 × x, which is t, y which is e(t), z which is t2, I end up with 2t3e(t).1336

xy2 is t × e(t) × t2, I get t × e of... wait, e(2t), yes, there we go...1346

Okay, so that takes care of this one. That is the grad of f evaluated at c(t), now I just have to find c'.1362

That is really, really simple. C'(t). Well here is my c right here, I will just take the derivative of each one, that is it.1369

The derivative of t is 1, the derivative of e(t) is et, and the derivative of t2 is 2t.1378

Well, now I just form my dot product, so the gradient of f evaluated at c(t) dotted with c'(t), it equals this vector dotted with this vector.1390

Well the dot product is this × that, so it is t2, the dot product is not a vector, this × that so it is t2e(2t) + this × that, 2t3e(2t) + this × that + 2t2e(2t), and that is it. 1411

Let us see if there is anything that I can combine here, t2e(2t), 2t2e(2t), yes, there is.1446

So it is going to be 3t2e(2t) + 2 × t3e(2t), and that is my final answer. That is it. Let me go back.1453

I was given some, you know, a curve, and I was given a function, and I just hammered it out.1469

I took the gradient as a function, as a vector in x, y, z, I evaluated it at c(t), in other words I put these values in for x, y, z, and I got this.1480

Now it is the gradient vector expressed in t. I took the derivative of c which is c', that is easiest enough to do.1490

Then I just took the dot product of those vectors. That is it, that is all that is going on here.1498

So this happens to be the derivative of f(c(t)). That is what this is equal to. This is equal to f(c(t))... no, we want to definitely, f'(c(t)), that is it. That is all that is going on here.1501

It is just a way of differentiating a composite function. 1523

Now, you are probably asking yourself, Okay, well if I start with a function of t, t goes from R to R3, and then I take the function R3 to R so that it is a function R to R, what I actually have is a function of t, right?1526

Yes, it is just a function of t that you are differentiating. You are saying, well wait a minute, if I found f(t) up here, couldn't I just do this directly? Do I have to use the chain rule?1542

The answer is no, you do not have to use the chain rule, you can do it directly. Which one is better?1552

Well, actually it depends on the situation. It depends on the function, it depends on what you are doing, that is all it is. 1556

So, let us go ahead and actually do it directly just to confirm that you can do it directly.1562

I think it will shed a little bit more light on this relationship between the curve and the function.1566

Let us do this in blue again... so we said that f(x,y,z) is equal to xy2z.1574

We said that c(t), let us just rewrite them over again.1589

Let us see, what did we say c(t) was... t, e(t), and t2, so now let us just form f(c(t)).1595

So f(c(t) = well, xy2z, x is t, so that is t, y2 is e(2t), and z is going to be t2, I am just plugging those in, and I end up with t3e(2t).1606

Well, not if I just take... this is just a function of t, so if I just take df with respect to t, I end up getting, so it is going to be this × the derivative of that, which is going to be 2t3e(2t) + that × the derivative of that, 3t2e(2t).1633

What do you know, you end up with the same exact answer.1656

Which is better? Again it just depends on your particular situation.1660

Sometimes you want to do it directly if it makes more sense, sometimes you want to use the chain rule if it makes more sense.1667

The problem at hand we will actually decide which one is better. 1672

Ok, so that is the chain rule, thank you very much for joining us here at educator.com. We will see you next time. Take care, bye-bye.1676

Hello and welcome back to educator.com and multi-variable calculus. 0000

Today we are going to talk about a very, very important topic called the tangent plane and it is exactly what you think it is.0004

If you have a curve in space and you have some line, like in calculus when you did the derivative, you had the tangent to that curve, where this is the analog, the next level up.0010

If you have some surface in space, there is going to be some plane that is actually going to be tangent to it.0021

It is going to touch the surface at some point. We are going to devise a way using the gradient to come up with an equation for that tangent plane.0026

That tangent plane becomes very, very important throughout your studies. The whole idea of calculus is linearization.0035

When we linearize a curve, we take the tangent line. When we linearize a surface, we take the tangent plane, so it is very, very important.0042

Okay, let us just jump right on in. We are actually going to start off with an example in 2-dimensional space just to motivate it to get a clear picture of what is happening.0050

Then we will move on to three-dimensional space.0057

So, let us start with f(x,y), so a function of two variables, = x2 + y2.0064

This is just a parabaloid in 3-space.0074

Now, we will let our curve, we will let c(t) = cos(t) and sin(t).0079

As you remember, this is the parameterization for the unit circle in the plane, circle of radius 1.0090

Now, if we actually form f(c(t)), which we can because this is a mapping form R1 to R2, and this is a mapping from R2 to R1, we can go ahead and form the composite function f(c).0100

So, f(c) = f(c)... do it that way, because this is our functional notation and that is what we are used to seeing. 0113

You know what, I am actually going to start avoid writing the t's, simply to make the notation a little more transparent and clear, not quite so busy.0121

But, of course we know... we aware of the functions so we are aware of the variables we are talking about, t and x and y. 0131

But I hope it is okay if you just allow me to write f(c). Now if I put f(c), so wherever I have an x I put cos(t) so this is going to be cos2(t) + and wherever I have a y I put sin(t), so sin2(t).0140

As you know from your trig identities, cos2(t) + sin2(t), that is equal to 1.0155

So, now let us go ahead and examine the level curve. So, let us examine... and as you remember a level curve is the set of points for that function that when you put it into that function it equals a constant. 0161

It is the set of points where they equal the same number over and over and over again.0177

So, let us examine the level curve f(x,y) = 1.0184

In other words, what we are saying is, so x2 + y2 = 1.0198

Now, this... so in 3-space this thing is the parabaloid that goes up, centered at the z axis.0205

When we actually set f(x,y) equal to some constant, what it is, it is actually some curve on there.0216

Then when you look at this from above, you are actually looking down on the x, y plane.0225

When you look down the z-axis, you are looking down on the x,y plane, and what you get is this curve in 2-space, the x,y plane.0231

Well this curve, for f(x,y) = 1, happens to be the unit circle. So, the level curve happens to be the same as the parameterization, in other words the curve itself, cos(t) sin(t) happens to lie right on the level surface for f(x,y) = 1. That is what is going on here.0242

So, recall... I am actually going to write everything out here... recall a level curve for f is the set of points x such that -- x is a vector here -- set of points x such that f(x) = k, a constant. That is it.0262

Just thought I would remind us of that. In this case k = 1. That is all that is happening here. k = 1.0293

So, again, you have the level curve and you have the parameterization, the curve itself, the curve and the function are actually 2 separate things, but as it turns out, this particular curve happens to correspond, happens to lie right on the level curve for the function f(x,y) = 1.0308

Alright. Now we can go ahead and start to do some Calculus.0329

So, the gradient of f -- well, it equals df/dx and df/dy, the gradient is a vector. 0335

df/dx = 2x and df/dy = 2y.0351

The gradient of f evaluated at c, well 2x, x = cos(t), 2y, y = sin(t) because we are forming f(c).0360

That equals 2 × cos(t) and 2sin(t), so this is the gradient evaluated at c(t).0379

Now, let us do c'(t), well c'(t), that is just equal to -sin(t) and the derivative of cos is -- sorry the derivative of sin is cos(t), right?0391

So, this is the derivative of that, and this is the gradient, which is the derivative of the function and we ended up putting in the values of c(t) in there.0405

Now we have this thing, and now we have this thing. Well? We know that the d(f(c)), d(t), in other words the derivative of the composite function is equal to the gradient of f evaluated at c, dotted with c'(t).0418

Now, let me actually do this in a slightly different color. Now I am going to take this thing and this thing and I am going to form the dot product. 0437

I am going to do this × this + this × this.0443

So, we end up getting this × this is -2sin(t)cos(t).0447

Then this is + this × this, 2sin(t)cos(t) -- -2sin(t)cos(t) + 2sin(t)cos(t), that equals 0.0458

So, what I have got here is the gradient of f at c dotted wtih c'(t) = 0.0471

Anytime you have 2 vectors, when you take the dot product and it equals 0, that means they are orthogonal.0480

That is the definition of orthogonality, or if you do not like the word orthogonal, perpendicular, it is just fine.0485

As it turns out, what you have here is that any time you have some function of 2 variables, again we are working with 2 variables here for this first example. 0491

If you are going to form a composite function, in other words if the argument you put in to x and y, the values of x and y, happen to come from a parameterized curve, and if the curve happens to lie on the level curve, so if you have f(x,y) = some constant, f(c) = some constant, 0502

As it turns out, the velocity vector of the curve and the gradient vector of the function are always going to be perpendicular to each other. 0524

Pictorially, this looks like this. So this is the x,y plane, this is the unit circle, right?0532

This happens to be the level curve for x2 + y2 = 1.0543

So, x2 + y2 = 1, that is this. It also happens to be the curve that is parameterized by cos(t)sin(t).0549

The curve itself lies on this, the points that form the level curve of the function.0558

Now, the velocity vector, is this. c'(t), that is the velocity vector, so this is c', this vector right here.0565

The gradient vector is this, this is the gradient of f, they are perpendicular to each other. This will always be the case, this is not an accident.0574

Now, let us go ahead and generalize and move on to 3-space, and then we will go ahead and do a little discussion of 3-space, draw some pictures, try to make it as clear as possible, and then we will finish off with some examples.0585

So, again... well, let us just go ahead and move along here.0600

Now, let us let... should I go back to blue, it is ok, I will just leave it as blue -- you know what I think I will do this in black, excuse me.0603

Let us go back to black here. Alright, now we will let c be a mapping from R to R3, in other words a curve in 3-space.0616

A curve in 3-space... whoops, one of these days I will learn to write... we will let f be a mapping from R3 to R.0634

Okay, a function of 3 variables. Nice and simple, okay, so we can form f(c).0652

C maps from R to to R3, f maps from R3 to R, so we can take values from R3, put them into here so we can form f(c).0663

So, we can form, f(c), which is a mapping from R to R.0673

It is this R to this R. That is the composite function -- sorry to keep repeating myself, I just think it is important to keep doing that until it is completely natural.0685

We can differentiate this composite function, we can differentiate f(c).0700

So, we do d, f(c)dt, well we know what that is, that is equal to the gradient of f evaluated at c, dotted with c'(t).0712

Again, I am leaving off the t and things like that just to make the notation a little bit clearer.0730

Now let us consider the level surfaces of f. Now, consider the level surfaces of f.0735

In other words, the points in R3 where f(x,y,z) is equal to k, some constant -- let me make my k's a little better, I think these look like h's -- which is a constant.0750

This is also an important feature, and where the gradient of f(x,y,z), in other words at that point, well, the gradient everywhere does not equal 0.0780

So, when we say this... so, consider the level surfaces of f, when we had a function of 2 variables, that was a surface in 3-space, the level when that function = a constant, what we get is a level curve in 2-space.0794

Now what we are doing is we are moving onto a function of 3 variables. There we got level curves. Here, for a function of 3 variables, when f(x,y,z) is equal to some constant k, 14, what you get are a series of level surface.0812

You get surface in 3-space. So let us say x2 + y2 + z2 = 5, or x2 + y2 + z2 = 25.0828

These are spherical shells. So they are just places where the value of the function is constant and they are actually surfaces in 3-space, so we call them level surfaces.0839

In 2-space, we call them level curves. The general term is level surface.0848

This is actually really important. This gradient, not equal to 0, this basically says that the surface is smooth.0853

This is analogous to the derivative not being equal to 0, the derivative being defined.0862

Anywhere that the gradient is actually equal to 0, we can show that there are some problems there.0868

Here, where we said the gradient is not equal to 0, we just mean that it is a smooth curve, that is has no... it is not like it goes like this and then boom, there is some corner or something like that.0876

That is all this means when we say that the gradient is not equal to 0. It is just an extra feature that we take on here just to make sure that we are dealing with a surface that is well behaved.0884

Now, we say the curve c lies on the level surface if for all t, f(c(t)) is equal to k.0897

In other words, the points on the curve satisfy the equation for the level surface.0935

So, we had the function of 3 variables. We want to consider a particular level surface on f(x,y,z) = 15.0942

Well, because we form the composite function, if there was some curve that passes through R3, and if that curve happens to lie on the value, on the surface itself, in other words, the x, y, z values of the curve, such that when you put them into f(x,y,z) to form this composite function, f(c).0948

When those points satisfy the equation for the level surface. That is what we want to consider. We say that the curve lies on the level surface.0973

I will draw this out in a minute so you will actually see it.0985

In other words, the points on the curve satisfy the equation for the level surface.0988

What you are looking at is just this. Let me go ahead and draw a surface first, I will have something like this, and maybe like this, and maybe like that, and maybe like that.1018

So, here we have let us say the z-axis and I will go ahead and draw, this will be behind, of course I have that axis, and this axis.1033

This is the z, so this is the x, this is the z, and this is the y. That is basically just some surface in the x,y,z, plane... I am sorry in the x,y,z, space.1049

Now, here, if you have some curve itself, that happens to lie on the surface, well, since this is a level surface, the points (x,y,z) happen to satisfy f(x,y,z) equal a constant.1066

If the curve itself happens to lie on the surface, the points on the curve satisfy f(x,y,z) = that.1084

So f(c(t)) = the constant, that is all we are saying. That is all that is going on here.1092

Okay, now when we differentiate this, so when we differentiate f(c) = K, we get, well, on the left hand side, we get... remember? differentiating f(c), gradf(c) × c', gradf(c) · c' = 0.1099

Because, when we take the derivative of a constant, it is equal to 0.1138

So, c'(t) and gradient of f at c are perpendicular, just like we saw for... this is always going to be the case.1143

In other words, if I took that surface and I looked at it in such a way where I was just looking at the curve, I might have some curve like this... well c'(t) is going to be that vector.1164

As it turns out, so this is going to be c'(t), and this vector right here, this is going to be gradient of f at c.1176

This point right here, that is c(t), and this actually, this curve is actually on a surface.1187

All I have done is I have changed, taken the surface, so now I am looking at it like a cross-sectional view of the surface. Now I am just looking at the curve, that is all that is going on here.1194

Now, let me draw this in 3-dimensional perspective. So I have some surface, let us say something that looks like this, and I have this axis.1204

So it is very, very important. Again, we want to use geometry to help our intuition, but again this is really more about algebra.1217

Because again, when you are dealing with functions of more than 3 variables, 4, 5, 6, you can not rely on this.1227

We are just using this to help us understand what is happening physically, so that we can sort of put it together in our minds. That is all geometry is used for.1233

Essentially, this is some surface, and again we said that there is some curve that happens to lie on this level surface.1243

Well, if there is, that is going to be c'(t), and this is going to be the gradient of f at c.1250

Of course, this point right here, that is c(t) and this is perpendicular. That is all that is going on.1264

Now, I am going to draw one more picture and it is going to be the next one here, and then I will go ahead and actually write out some specific formal definitions.1277

A lot of this was just motivation. Now let us do one more picture, so, let me draw one more surface here. Something like that.1288

We have the z-axis, this comes down here, this is going to be the y-axis, and this is going to be the x-axis.1297

You know what, I will go ahead and label them. I mean at this point we should know which x's we are talking about.1308

So we have some point, this is going to be some point, now as it turns out, when you have a surface, there is more than 1 line that actually crosses that point.1313

You can have any kind of curve you want, that is the whole idea, you are not sort of stuck on one curve.1323

When you have this level surface in 3-space, there are a whole bunch of curves that pass over it, there is the curve this way, the curve this way, the curve this way.1330

If those curves that are parameterized at c(t) happen to lie on that surface, then the gradient vector and the velocity vector are always going to be perpendicular, so let me draw several curves that pass through this point.1338

So, you will have one curve that way, and another curve that way, and another curve that way. That is what is going on, as it turns out, the gradient vector is going to be perpendicular to every one of those curves.1356

This is actually very, very convenient. Now, let me call this point p, okay? Now let us write something down here. We will let p be a point on the level surface of the function f.1370

Now, let c be a curve passing through p, so some curve that happens to pass through p.1398

Well, then there is some value, let me write c(t) here, this is one of the... so c(t).1417

So let c(t) be a curve that happens to pass through t, so it is any one of these curves, an infinite number of them.1432

There is some value t0, such that the gradient of f evaluated at p, dotted with c'(t0) = 0.1439

So, we know that if a point on the curve happens to correspond, happens to actually lie on the level surface of the function, we know that the gradient vector × the velocity vector of the curve, c'(t), the dot product of those is equal to 0. We know it is perpendicular.1460

Okay, well, there are many such curves, as you can see. There are a whole bunch of curves that pass through that point.1481

Each one has its own velocity vector. So let us do another curve like that, there is another velocity vector, so I have 1, 2, 3, 4, and there is another velocity vector right there.1487

So, each one of those curves passes through that point, so there is an infinite number of them.1507

Each one has a velocity vector. Each one of those velocity vectors is perpendicular to the gradient, so what you have is this.1512

If you have some gradient vector like this, and if you have a curve passing this way, there is going to be this vector that is perpendicular. 1520

Let us say you have a curve passing this way, this vector is perpendicular.1526

As it turns out, if I take all of these vectors all the way around, what I end up with is a plane, because all of these vectors are perpendicular, they all lie in the same plane.1531

Now we can go ahead and write out our definition.1544

I hope that made sense. So here we have our gradient vector, and then here we have all of these curves with all of their velocity vectors.1553

Well, since each one of those velocity vectors is perpendicular to the gradient vector, this is perpendicular, that is perpendicular, that is perpendicular.1561

If I actually move around and connect all of the vectors, since they are all perpendicular, they all lie in the same plane, all of the velocity vectors at that point p, they all lie in the same plane.1571

So what I have is the definition of a tangent plane to the surface at that point.1582

Now, let me go ahead and write down the definition.1590

You know what, let me write a couple of more things here just to make sure that it is specifically written down.1594

So, there are many curves on the level surface passing through p, and each has its own c'(t), 0.1601

All of these, c', all of these c'(0) form a plane.1638

So, we now make the following definition.1654

Okay... love definitions... here we go, now we finally get to the heart of the matter.1666

The tangent plane to a surface f(x) = k, notice I used the vector here instead of writing x,y,z. 1675

Vector is just the short form instead of writing x, y, z, that is the component form.1693

So the tangent plane to a surface, f(x) = k at the point p is the plane passing through p -- I am going to write and -- and perpendicular to the gradient vector of f at p.1701

Now we remember... well, if we do not remember we will recall it here... the definition how we find the equation of a plane is we basically find... well, here is the definition of a plane.1744

It is x · n = p · n.1760

Well, if you have a plane, n is the normal, it is the vector that is perpendicular to that plane. Normal, remember normal and perpendicular?1767

So if we take x, which is just the variables x,y,z,q, whatever... and dot it with that, that is going to equal the point p dotted with that.1779

If you do not recall, go back to one of the previous lessons to where we motivated this particular way of finding the equation of a plane.1789

Now we have the tangent plane to a surface, to a level surface of f, or f(x) = k at the point p, it is the plane that passes through p and is perpendicular to the gradient vector.1798

Now n, our normal vector is the actual gradient vector, and it is based on what it is that we just discussed. 1809

The fact that the gradient vector is going to be perpendicular to every single curve that happens to pass through that level surface.1816

All of the velocity vectors, they form a plane. That is what we call the tangent plane.1822

This is the general definition, here is what we are looking for.1828

x · the gradient of f at p = p · gradient of f at p.1833

This is the equation for a plane, for a tangent plane to the surface passing through the point p.1847

When we have our gradient vector, f at p, and when we have our point p, we go ahead and form this equation.1855

We multiply everything out, we take the dot product, we combine things, cancel things, whatever it is we need to do, and we are left with our equation for a plane.1862

Not let us go ahead and jump into some examples and hopefully it will make a lot more sense.1870

Let us see, alright... I will keep it in blue. So, this is example 1.1876

Okay, find the equation of the plane tangent to the surface x2 + y2 + z2 = 5.1884

Okay, at the point p, which is (1,2,1). Nice.1917

So we want to find the equation of the plane that is tangent to that surface at the point (1,2,1).1925

Now, we have our f, our function is x2 + y2 + z2.1934

Hopefully you will recognize this, this happens to be the equation of a sphere of radius sqrt(5) in 3-space. It is okay if you do not. 1945

Remember, this is x2 + y2 + z2. That is a sphere. It is moving 1 dimension up from x2 + y2 = r2, where you have a circle in the x,y plane. 1955

Let us go ahead and form the gradient, nice and easy. The gradient of f = well, the derivative with respect to x is 2x. The derivative with respect to y is 2y, the derivative with respect to z is 2z.1966

Nice and straight forward. Now when you evaluate that at p, in other words when you put (1,2,1) in for (x,y,z), we get the following.1980

This is going to equal, 2 × 1 is 2, and 2 × 2 is 4, and 2 × 1 is 2, so we get the vector (2,4,2).1988

Now we just form our equation. This thing right here. We take x... yeah, that is fine, so I will write it as x · the gradient of f at p = p · gradient of f at p.1998

This is just this. I am about to write the equation over and over again, it sort of gets me in the habit of thinking about it, remembering it, things like that.2019

Now that we actually have our vectors, now we do the component form. Whenever we do computations, we are working with components. Whenever we write out the theorems, the definitions, we keep it in form like this.2028

This is how you want to remember it, but you want to remember that vectors come as components.2038

Now, we will let x, let the variable x equal... let us just call it (x,y,z), that is not a problem... so I am going to need to go to the next page here.2045

We have what we just did, x, so which is (x,y,z) dotted with the gradient and the vector that we got was (2,4,2), that is equal to p, which was (1,2,1) dotted with (2,4,2).2062

Now we just do the dot product. This × this is 2x + 4y + 2z, make this a little more clear, = 1 × 2 is 2, 2 × 4 is 8, 1 × 2 is 2, that is equal to 12.2082

We get 2x + 4y + 2z = 12. There you go. This is the equation of the plane that passes through p and is tangent to the level surface of that function, at that point p.2101

Here is what it looks like. I am going to draw out a... yeah, let me go ahead and... so we said that x2 + y2 + z2 is equal to 5 is the sphere centered at the origin of radius sqrt(5).2118

So, it looks like this. I am going to draw it only in the first quadrant, and now I am going to go ahead and actually erase... yeah, let me make these a little bit clearer here so that you know that we are looking at a sphere.2138

So, this point... hopefully that is clear, so you have it in the first quadrant, and now the point (1,2,1) well this is the x-axis, this is the y-axis, and this is the z-axis, so 1 along the x-axis, 2 along the y and 1 along the z.2159

That is going to be some point right about there on the surface. What you want end up having is, of course, this plane that is touching that point and is tangent to it.2181

What you have is this... if I turn this around, what you would see is this curve, that part of the sphere, and you would have a tangent plane like that. That is what we found. 2196

We found the equation for that plane, hitting, that touches the surface at that point, that is it. Nice and basic.2206

Let us go ahead and do another example here. We will finish off with this one. This is example 2.2215

This time we want to find the tangent line to the curve xy2 + x3 = 10 at the point (1,2). 2227

So, notice. We said the tangent line at the curve. This whole, let me go back to black here for a second, this whole x · n = p · n, this is true in any number of dimensions. This is the general form of the equation.2264

Whether we are working in 3-space, 2-space, 15-space, that is the equation that we work.2284

Like we said before, when you have a curve, you have a tangent line. When you have a surface, you have a tangent plane.2290

Now let us move on to another dimension. When you have a surface in 4-space, well, you are going to have something that we call the hyper-plane, but we still use the same language for 3-dimensional space.2296

You are going to have a plane, you cannot draw it out, but that is what you have. This equation still works, even though we are talking about a tangent line and a curve, as opposed to a tangent plane and a surface.2309

It is the same equation. In other words, it is still going to be x · the gradient of f at p is equal to p · gradient of f at p.2320

That is what is wonderful about this. This equation is universal, and that is the whole idea. We want to generate it abstract so that it works in any number of dimensions, not just 1 or 2.2331

Okay, so let me go back to blue here. So, what we want is again, let me write it one more time. I know that it is a little redundant, but it is always good to write.2343

Gradient of f at p = p · gradient of f at p. Of course, these are that way.2354

We are going to let x = of course, (x,y), that is the component form. Well our f(x,y), or let us just say our f, let us leave off as many variables as we can, is xy2 + x3.2365

Let us go ahead and form the gradient. The gradient of f is equal to, well, it is equal to the first derivative of f, the second derivative of f, and that is going to equal y2 + 3x2.2383

That is the partial derivative of f with respect to x. We are holding y constant, so we just treat this like a constant. That is why it is y2 + 3x2.2400

Then, the second derivative, which is the derivative with respect to y, in this case, this one does not matter, we are holding x constant so that goes to 0.2409

Here it becomes 2xy. That is it.2417

Now, nice and systematic, the gradient of f evaluated at p, we go ahead and put the values of p, (1,2) into this. 2422

So, wherever we see an x we put a 1, wherever we see a y we put a 2. That equals... 22 is 4 + 3, and then 2 × 1 × 2 is 4, so we should get (7,4).2432

I hope that you are checking my arithmetic. So, when we put it back into that, now that we actually have some vectors to work with, x is (x,y).2448

The gradient of f at p is (7,4). p itself is (1,2), and the gradient of f at p is (7,4).2463

There we go. Now we do the dot product, this is 7x + 4y, 1 × 7 is 7, 2 × 4 is 8, and if I am not mistaken, we get 7x + 4y = 15.2476

There you go. Finding, in this case, this is the equation of the line that is tangent to this curve at that point. That is it. That is all that is going on.2497

You are just using the idea that the gradient vector, the velocity vector are perpendicular. That is always going to be the case.2511

Thank you for joining us here at educator.com, in the next lesson we are actually going to do more examples using the concepts that we have been studying recently.2522

Partial derivatives, tangent planes, gradients, just to make sure that we have a good technical understanding and we develop some comfort before we actually move on to the next topic which is going to be directional derivatives. 2531

So next time, further examples, take care, talk to you soon.2541

Hello and welcome back to educator.com and multi-variable calculus.0000

The last lesson we introduced the notion of the tangent plane. I thought it would be a good idea to go ahead to go ahead and do some further examples with the concepts we have been studying recently. 0004

Just more practice with gradients, and tangents, and partial derivatives and things like that just to get a better sense of what is going on.0012

Just to get a sense of the different types of problems you are actually going to run across, and to use what you know about a particular situation to reason out a problem. 0022

That is it, just for practice. Before we actually move on to the next topic, which is going to be directional derivatives.0032

Let us just go ahead and jump right on with the first example. Example 10038

Now, our task is to find a parametric equation for the line to the curve of intersection of the 2 surfaces, x2 + y2 + z2 = 49.0054

x2 + y2 = 13 at the point (3,2,-6).0112

Let us see if that was the right point here, yes, the point (3,2,-6).0134

So, this is more than a little more complicated than what we have done before.0150

As far as the expression of the question is concerned, and that is going to happen sometimes.0157

Do not let the expression of the question lead you to believe that it is any more complicated than anything that you have done before.0161

That is the real issue. You just stop and think about it and understand what is going on.0167

As it is we only have a handful of terms at our disposal already, we have partial derivatives, which we know how to do, hopefully at this point.0173

We have the idea of the gradient vector and we have this idea of a tangent plane or a tangent line.0180

Now, let us just sort of pull back and see what it is they are actually asking us to do. 0185

Find a parametric region for the line that is tangent to the curve of intersection of the 2 surfaces.0190

Here is what is happening. I will go ahead and describe it and I will write it out so that you have it for reference.0199

You know that any time 2 planes, a plane is just a surface in 3 space, it just happens to be a very simple type of surface, it is flat.0205

You know that when 1 plane meets another plane, where they intersect, it actually forms a line, right?0213

Now if I take, instead of 2 planes that are intersecting, if I actually take 2 surfaces, that are intersecting, there point of intersection actually ends up being a curve.0223

Of course we will draw all of this out in just a minute.0234

They are asking, they are saying, I have a surface, x2 + y2 = 49, and I have another x2 + y2 = 13.0239

They are going to meet in some sort of curve in 3-space. Well, at that curve, they want you to find the equation of the tangent line that hits the point of intersection and is actually tangent to the curve itself.0245

That is what they are asking. So, let us do a little bit of discussion and then we will draw some pictures and then we will actually start the problem because this is really, really important.0263

So, I will write out everything here. When 2 surfaces in R3 intersect, their points of intersection form a curve in R3.0273

Again, we will be drawing it out in just a minute so that you can actually see it.0307

Now, here is what is important. When we form the tangent plane to surface 1, and the tangent plane to which we can do, which we already know how to do.0313

The tangent plane to surface 2 at a point where they intersect, the line of intersection of these 2 planes, of these two tangent planes is, it is the line we are looking for.0342

It is the line tangent to the curve of intersection.0394

So, before I draw it out, let me just tell you what is going on once again.0408

I have a curve, I have another curve. Where they intersect is going to be some sort of another curve.0412

Well, any point along that curve where they intersect, I can actually form a plane that is tangent to one of these surfaces, and I can form the plane that is tangent to the other surface.0417

Now I have a plane, and another plane. Those 2 planes, when they intersect, they intersect in a line. Well, because they have that point p in common, that one point. 0427

The line where the 2 tangent planes intersect, that is the line that happens to be tangent to the curve of intersection. That is what we are doing.0442

We are going from the surface to the tangent plane, the intersection of the 2, that is going to be our line. This is what it looks like.0448

So, I am going to do that on the next page here.0456

Okay, so, drawings, here we go. Let me see if I can get this right.0461

So I have one surface going this way, let us just go like this, like that, like that. How is that?0467

Now I have got another surface here, so I am going to go... let us go down, let us go up like this. Let us go up here, and we will just go like that.0472

So, let me draw, let me make a little bit of an erasure here so that I can see.0489

Okay. That is what we mean. If we have one surface, and another surface, where they intersect, where they actually intersect is a curve because surfaces are curved.0502

They actually intersect at a curve. That is what is happening here.0513

Now, what I can do is I can actually form the plane that is perpendicular... the plane that is tangent... so, let us say that we have some point on there.0521

So, what we want is we want the line that is going to be tangent to that curve. That is what we are looking for.0532

Well, if I form the plane that is... so we will call this surface 2, and we will call this surface 1.0538

If I form the tangent plane at that point, just surface 2, and if I form the tangent plane to that point at surface 1, I am going to end up with 2 planes.0546

Now, let me go ahead and draw those. So, I am not going to draw on top of it, I am going to draw it over here.0553

So, we have 1 plane, we will call this the tangent through 1, and now I will go ahead and draw this one.0562

Let me make this again... little dots here, so that we know we are actually talking -- uh, it is not very good, we want to be clear. Something like that.0581

This is t2, this is the plane that is tangent to the surface along this to this surface at that point that is that.0597

This t1 right here, that is the -- let me go ahead and draw this one in 2 -- this plane is the one that is tangent to this surface at that point. It is on top of that one.0607

They intersect in a line -- let me go back to this.0621

There is a gradient vector that is going to be perpendicular to surface 2, that is going to be pointing that way.0629

There is a gradient vector that is going to be pointing this way that is perpendicular to surface 1, right? 0635

The is a velocity. This is a curve. The velocity vector is this way. The gradient vectors are perpendicular to the surface, that is the whole idea.0643

Well, same idea here, we have used the gradient vector to find these planes, so there is some vector that is going that way that is perpendicular to this plane. 0652

There is some vector that is going out that way, that is perpendicular to this plane.0661

Now, let me go ahead and take this line, and this curve together. I am going to put these 2 together. 0668

What I have is the curve of intersection, I have the line of intersection, this is the point where they meet, so this line is this line, this curve is this curve.0678

There is that vector, and there is that vector. This line is perpendicular to both vectors. It is perpendicular to that vector, and it is perpendicular to that vector.0691

Because it is perpendicular, well, we know perpendicularity means that any vector along here dotted with this is equal to 0, orthogonality.0703

This vector dotted with this is equal to 0, orthogonality. 0714

I can form 2 equations, and 2 unknowns, and I can find this vector. That is what I am doing here. Hopefully that made sense.0720

Now let us go ahead and actually do it. That was our motivation. 0726

So, I guess I can start at the bottom down here.0731

So, let us take f1, very carefully f1 is x2 + y2 + z2.0735

The gradient of f1 = well, derivative with respect to x is 2x, derivative with respect to y is 2y, derivative with respect to z is 2z.0744

Now the gradient of f1, evaluated at p is equal to (2x,2y,2z), evaluated at the point (3,2,-6).0757

This is a common notation for evaluation of a certain something at a given point. You just draw a little line and at the bottom you put whatever it is.0775

If it is a vector you just put a vector, if it is a point you just put 5.0783

That is going to equal, when I put 3 in for x I get 6, when I put 2 in there I get 4, and when I put -6 in here -- whoops, let me make this a little clearer, this is not zz, this is 2z.0787

Okay. I get -12, if I am not mistaken.0801

So, the vector (6,4,-12) is perpendicular to surface 1. 0809

That is what we found when we take the gradient and we evaluate it, the gradient vector is going to be perpendicular to the surface. Here is the surface.0821

The gradient vector is going to be perpendicular to it. That is what we found.0827

Now, let us move on to function 2.0833

f(2) = x2 + y2... okay, so now I am going to stop and just remind you of something if you do not know... well, if you know it is a reminder, if you do not, just something to be vigilant about. I am going to write this out.0837

Just because z is not in this equation does mean it is not a surface.0857

You can treat any function... so this can be f(x,y,z) = x2 + y2, z does not have to show up.0864

If you are treating this like a surface in 3-space, all it means is that z can take any value that you want. It is an infinite number of values.0875

What this actually is, is a cylinder in 3-space. The x2 + y2, so in the x,y plane you have a circle, but since z can be anything it is just a cylinder going up and down along the z-axis.0884

Because, z can be anything, we are not specifying a constraint on z.0898

So, know that you can do that, you can have a function of 3 variables, but all of the variables do not have to be there, as long as you interpret it properly.0903

Z can be anything. What you do is you get a surface, it just depends on what the problem is asking and what is going on.0913

So you have to be very, very vigilant about what is happening here. Let me actually write that down. 0920

Note that just because z is not in the equation, this can still be a function of 3 variables f(x,y,z).0929

As a surface, absence of z means z takes on whole values.0968

This is a cylinder of radius sqrt(13), centered along the z-axis, I will just write that.0998

Okay, so now let us go ahead and continue on with the mathematics. Again, it is the algebra that matters.1026

So, f(2) = x2 + y2. The gradient of f(2) = (2x,2y,0), there is no z.1033

The gradient of f(2) evaluated at the point p, is equal to the (2x,2y,0), evaluated at (3,2,-6).1046

What we get is 2 × 3 is 6, 2 × 2 is 4, we get (6,4,0).1060

So, the vector (6,4,0) is perpendicular to surface 2.1069

There we go. We found 2 vectors, one perpendicular to surface 1, one perpendicular to surface 2. They both meet at that point. Now we can do it.1075

Here we go. A vector in the direction of the line of intersection is perpendicular to both gradient vectors, right?1086

Go back to that picture... to both gradient vectors.1113

Now, we will let v = this vector. We will let v be this vector. In other words, where we had that curve, this line, we had 1 vector going this way, 1 vector going this way, this line is perpendicular to both vectors.1123

So, if we... what we want is we want to find a vector in this direction or in this direction so that we can write a parametric equation because they want the parametric equation.1153

So, we will let v be this vector, and we will call v... we will give it components of (v1,v2,v3).1162

Now, well, the vector v is perpendicular to the first gradient, so the grad of f1 evaluated at p · v = 0, and the gradient of f2 evaluated at p · v = 0.1173

So, we have (6,4,-12) · (v1,v2,v3) = 0, and (6,4,0) · (v1,v2,v3) = 0.1199

When we do this we get 6v1 + 4v2 - 12v3 = 0, and 6v1 + 4v2 + 0v3 = 0.1220

We have 2 equations, well, we have 2 equations and 3 unknowns but fortunately one of them is... so 6v1 + 4v2 = 0, because that is 0, right?1247

Essentially we have, let us go ahead and we will let v3 = 0, because again it is... this one does not have a v3, this one has a v3, so it does not matter what it is.1263

We can go ahead and set equal to anything that we want. In this case, I am going to go ahead and set it to 0, that is the easiest number.1277

So, I have got v3 = 0. Then I have 6v1 + v2 = 0, because this part and this part are the same.1285

So, I am going to get 6v1 = -4v2, that means v1 is -2/3v2.1301

Now, I can pick v2 as anything that I want, and then v1 will be contingent upon v2, and v3 is always going ot be 0.1314

Let us go ahead and take v2, so let us come up here, let v2 = 3. I think it is going to be the easiest if I let v2 = 3, the v's cancel.1321

Therefore, then v1 = -2.1340

So our vector v = v1, which is -2, v2 which is 3, v3 which is 0. Now we can write out parametric equation.1348

The line is equal to... you remember, it is the point p + some vector v × t.1362

A line passing through p in the direction of the vector p. Well, we know what p is. P is equal to (3,2,-6).1370

You notice I have decided to write it as a column vector instead of a row vector + t × v which is (-2,3,0). There you go.1379

We have 2 surfaces. Where they met was a curve. I wanted to find the equation, the parametric equation of the line tangent to that curve.1395

I found both gradients. Both gradients define tangent planes. Well, it is the intersection of those tangent planes, because they both pass through p, is going to be the line that actually passes through p and is going to be tangent to the point at p, which happens to be tangent to the surfaces.1406

That is all we did here, very, very nice. Let us go ahead and draw that this looks like, so that you can actually see it.1429

What you have is the following. Let me see if I can do this right. So, I have got... let me go ahead and let me just draw a couple of... let me draw that one, let me draw that one, and I will go ahead and... so, so this is a sphere.1437

Now, I am going to go ahead and draw the cylinder.1466

Now, let me go ahead and make this... yeah.1479

Something like that. So this was our f1, this was x2 + y2 + z2, and it was equal to 49.1498

So, this was a circle centered at the origin. I have not yet put the axes in but this is where the origin is, right there.1511

It is centered at the origin and it has a radius of 7.1517

This is f2 right here, f2, which is equal to x2 + y2 = 13. It is a cylinder of radius sqrt(13) that passes, that goes along the axis.1523

The points of intersection are these points right here. Let us just say it is like that. That is one particular point.1538

In this particular case it was what, 3,6 something like that, -12, what was our point? Oh, (3,2,-6).1546

(3,2,-6) wherever it happens to be... well, okay, you know what, let us go ahead and... if you go along the x axis 3, and you go along the y axis 2, -6, so the point is going to be probably somewhere around there.1558

That is what we have done. We have basically found the line. We found the equation of that line. That is what we have done. 1573

Where they meet is a curve. If we happen to find tangent planes to this, we can get a tangent plane that is perpendicular to the surface here, we can get a tangent that is perpendicular to the surface there. 1583

Where those -- or here -- where those tangent planes actually meet, they form a line. That is this line.1597

now, let us do another example. So, this example will go back to our idea of gradient and composite function forming f(c(t)).1603

So, let f(x,y) = e9x + 2y.1620

g(x,y) = sin(4x+y).1632

Do not let the statement of the problem intimidate you. Often times the problem is a lot easier than you think it is, but kids, they look at these equations and all of a sudden they freak out.1642

That sort of just stops them in their tracks. Read the question, it is probably a lot easier than you think. It is just equations.1652

You can do derivatives of basic questions, it is not a problem. Again, we only have a handful of things at our disposal at this point. It is not like we have a toolbox full of 150 techniques. 1658

We only have the gradient, the partial derivative and equations for lines and dot products, and the equations for line and planes. 1666

Not too much to work with, so we should be able to hammer this out. 1677

sin(4x+y). Okay, let c be a curve, such that c(0) = (0,0). Oh this is kind of nice.1680

C(0,0), okay, we are dealing with something reasonably straightforward here.1697

Given, df/dt is equal to, given df/dt at t=0, is equal to 2 and dg/dt... no, let me actually, this is not quite correct, let me give you a little bit more... d(f(c))/dt at t=0 equals 2.1704

d(g(c)), because we are forming the composite function g(c) and d(g)/dt evaluated at t=0 equals 1, our task is to find c'(t).1743

Okay, let us do it. So, we have a function. We have another function. We have a curve, and we know that the value of the curve at t = 0 is (0,0).1760

We know that the derivative, of the composite function f(c(0)) = 2, we know that the derivative of g, the composite function g(c) at t, that is equal to 1. How can we find c'(t).1772

Okay, let us just sort of see what is going on. C gradf, okay, so let us go ahead and sort of... again, you just sort of start with what you know.1787

They tell me that the d(f(c(t)))dt, so let us expand this derivative, we know that it is the gradient of f evaluated at that point, dotted with c'(t). Let us write that out.1802

So, the d(f(c(0)))dt, that is equal to the gradient of f at c(0) · c'(t).1814

We know that that equals 2, because they gave us that. Okay, well that is nice.1838

Now, let me write this a little... I want to do these in parallel so I am going to make a little more room here. I am actually going to write this down below. I am going to go like this...1843

That is equal to the gradient of f evaluated at c(0) · c'(t), and we know that that equals 2, because I want to go in parallel.1857

I know that the d(g) evaluated at c(0) dt, I know that that is equal to the gradient of g at c(0) · c'(t).1874

They tell me that that is equal to 1. Oh, okay, well this is really, really great. We have the gradient of something dotted with c'(t) = 2, and we have the gradient of something else dotted with c' = 1.1893

This suggests two equations and two unknowns, so let us just follow this through and see where this is actually going to take us.1905

So, this suggests 2 equations and 2 unknowns. The unknowns are c(t), but that is a vector and so a vector has components, x and y, that is why we have two unknowns.1911

Again, do not expect that you are just supposed to look at a problem and know where you are supposed to go. A lot of times you just have to start, and see where you go.1928

If you hit a wall, it is not a big deal, people hit walls all the time. Professional mathematicians do, scientists do all the time. You just try another path, that is the learning process. 1937

Do not think that this is supposed to be like 6th, 7th grade math where you are supposed to look at a problem and know exactly what you are supposed to do and it is going to fall out.1948

You just have to start going somewhere, based on what you know, and hopefully something will make itself clear. That is the way to do it, no worries.1956

Now, let us go ahead and actually find that and that, because we have the functions.1965

Well, we know that df/dx... let me see, what do we get, 9, 0, 2, so df/dx of the gradient of f is going to be -- I do not know if I should write it... that will be fine... so let us go ahead and do df/dx.1971

So df/dx is equal to, well when we differentiate with respect to x, the e9x+2y, we get 9 ×, let me rewrite the functions so that we have them on this page. I think it is a good idea. We want to be as clear as possible.2001

So f = e9x+2y, and let me go ahead and put g over here, so g = sin(4x+y). There we go. Now that we have them as reference.2021

So df/dx is going to equal 9 × e9x+2y, right? The derivative is this × the derivative of this, and the derivative of this with respect to x is just 9.2038

Then when we evaluate it at t = 0, we end up with 0,0 = 9.2057

So we have that number, now df/dy that is going to equal 2 × e9x+2y, right? because the derivative of this is e9x+2y × the derivative of that.2066

The derivative of this with respect to y, this is a constant, is 2.2081

So when we evaluate that at t = 0, we get 2, so now we have those numbers, and we do the same thing for g.2084

dg/dx, that is equal to... the cos = 4 × cos(4x+y), right?2094

When we evaluate that at t=0, in other words when t=0, they are saying that c(0) is (0,0). That is why I put 0 in for y and 0 in for x. Does that make sense?2121

Let me make sure that we know what is going on. Remember that we had specified that one of the things they gave us was c(0) = (0,0).2135

So over here, when I say evaluated at t=0, I am actually doing (0,0). X is 0, y is 0, because of this thing that they gave us. That is how I am evaluating it at t=0.2144

In fact, let me make it a little more clear then. Instead of doing t=0, let us just say at (0,0), because we are forming f(c(t)), right?2160

Because of the gradient right here, this gradient of f(c) at 0. c(0) is (0,0), so it is the gradient of f at (0,0).2173

Hopefully that will make it maybe a little bit more clear, sorry about that.2181

So when we do that, we end up with 4, right? When we put (0,0) in for here. cos(0) is 1. 1 × 4, okay?2185

Then, let us go back to black here, so dg/dy, that is going to equal the cos(4x+y), because the derivative of y is 1, so when we evaluate that at (0,0), that is equal to 1.2195

Now, let us let c(t), c'(t) = uv, that is our components, now we can form the... we have the gradient of f evaluated at c(0) · c'(t).2218

The gradient of f was (9,2) · uv and we said that that was equal to 2, and this was equal to 1.2256

This one, we have (4,1) · uv, and this is equal to 1, 2270

So, we get 9u + 2v = 2, and we get 4u + v = 1.2280

When we solve this, I will not go through the solution... solving, we get u = 0, v = 1, so, c'(t), actually it is c'(0) which is what we are doing.2292

We are finding c'(0). c'(0) is equal to 0 and 1.2317

There we go. Standard usage, we just followed what it is that they gave us again. The only real issue with these problems is keeping track of everything on the page.2324

All it means is you have to do a little looking, a little going back and forth, just trying to keep it all straight.2335

It is expected, I mean this is multi variable calculus, the problems are going to get a little bit more complicated. 2340

There is going to be a lot more going on, it is not just 1 derivative, you have partial derivatives, you just need to keep it all straight. So, the answer is just go slowly, that is it, just go slowly until you are comfortable with what is what and where is where. 2346

Let us go ahead and do one more example. This time you may be using the, well, let us see what the example says. Example number 3.2361

This says find the cosine of the angle between the surfaces x2 + y2 + z2 = 38, and x2 - y2 = -38, at the point (-1... let us make sure we write everything clearly... (-1,1,-6).2372

We are starting to get these stray lines again so I am going to move on to the next page after writing this.2435

So, find the cosine of the angle between the surfaces x2 + y2 + z2 = 38, and x - z2 - y2 = -38 at the point (-1,1,-6).2441

At that point, the two curves hit each other. We want to find the angle that the two curves make with each other, the two surfaces make with each other.2454

So, this is a good time to give a quick definition of what we mean by the angle between 2 surfaces.2467

The angle between 2 surfaces... nope, the angle between 2 surfaces, is defined as the angle between the normal vectors, the normal vectors to the surfaces at that point.2475

We know how to find the vector that is normal to the surface at that point. It is the gradient vector, evaluated at that point, that is the whole idea. You have a surface, the gradient vector is going to be perpendicular to that surface at that point. Very, very, nice. 2523

So, the angle between 2 surfaces, where the 2 surfaces is defined as the angle between the normal vectors to the surfaces. You have a surface, you have a normal vector; you have another surface, you have another normal vector. 2541

You have 2 normal vectors, that is the angle between the surfaces, nice and easy.2552

So, f(1) = x2 + y2 + z2.2561

Now, the gradient of f(1) = (2x,2y,2z).2570

When we evaluate at the particular point, we end up with (-2,2,-12). That is one of them.2583

So, this vector is normal to the surface at the particular point we are discussing. So the gradient of f(2) is going to be 1 - 2y - 2z, right? df/dx, df/dy, df/dz, it is a vector.2598

When we evaluate it at the particular point in question, we end up with (1,-2,12), and again I sure hope that you checked my arithmetic here and arithmetic is... well, arithmetic is arithmetic.2619

It is important for the sake of computation, but it is the map that is important. We want to understand what is going on underneath.2634

Now, we know that the cos(θ) happens to be, remember? vector a, the cos between 2 vectors a and b is the dot product of those vectors divided by the norms of the vectors, the product of the norms of the vectors. 2642

So, that is our standard. So what we have is, the gradient of f(1) dotted with the gradient of f(2)/norm(gradf(f1)) × norm(gradf(f2)). Wow, look at all of that crazy notation.2660

So, the gradient of f1 dotted with gradient of f2 is just this dotted with that, so we get -2, 2 × -2 is -4, 12 × -12 is -144, okay?2683

Then, the norm of this thing, if you do it, it is going to equal 4 + 4 + 144, so it is going to be sqrt(152), and here we are going to have 1 + 4 + 144, this is sqrt(149). 2700

So, we end up with cos(θ) = -150 over... and when I actually do that, it is going to be -150.5 + 0.9967, which is what we wanted.2720

If you want θ itself, it is going to be 4.67 degrees. There you go. 2740

Again, hopefully I did my arithmetic correct, but ultimately, arithmetic is not the real issue, we want to understand the mathematics. 2746

In this particular case we have two surfaces that are intersecting. We want to find the angle of that intersection where they happen to intersect at a given point.2753

Well, we find the vector that is normal to one surface, which is the gradient vector at that point. We find the normal to the other surface, which is the gradient vector to that surface at that point, and then we just go ahead and treat those vectors. 2761

We do our normal a · b/norm(a)norm(b), that gives us the cosine of the angle between them. 2776

Hopefully these examples have helped to again, just develop a little more of the sense of the kind of problems that you are looking at.2783

There is only a handful of techniques at your disposal at this point, but we want to get good and comfortable with this notion of gradients, and tangent planes, tangent lines, surfaces, things like that, because these are our basic tools.2793

The fundamental tools are the actual tools that keep showing up over and over and over again.2807

You have done enough mathematics at this point to realize that some of the higher end techniques do not always show up.2813

It is the things that you learn at the beginning that keep showing up over and over and over again.2821

Thank you for joining us here at educator.com, and we will see you next time here for a discussion of directional derivatives. Take care, bye-bye.2825

Hello and welcome back to educator.com and multivariable calculus.0000

So, today we are going to be talking about the directional derivative. Let us just jump right on in and start talking about it.0004

Let us go back to first... single variable calculus and let us talk about just a basic function.0012

Let f = x2, well we know that df/dx = 2x, so the derivative is a rate of change.0019

It is telling you that if I change a little bit in the x direction, that the function is going to change a certain amount in the y-direction. That is what this means, that is all it means, at a given point.0031

In this particular case, let us say the point we pick is x = 3.0045

So, at x = 3, df/dx = 2 × 3, it equals 6. That means if I am at x=3, and if I actually move to the right a little bit, or to the left a little bit, the extent to which the function is actually going to change is measured by this thing right here, the derivative.0050

That is what the derivative does, it is a rate of change.0063

Now, notice... well, let me write all of this out here so that we have it available.0077

This is a rate of change. It means the rate at which f changes with each small change in x.0084

Sometimes we get so wrapped up in actually just doing the mathematics and getting through the symbolism, that we forget what is really happening. I mean that is what it is.0117

It is a measure of how the function changes when you change the other variable at a given point.0125

Of course since functions, curves, are kind of whacky, we are not talking about a straight line for the function itself, it is going to have different rates of change at different points along that curve. That is the whole idea.0130

But the change in f has a direction, which is implicit for functions of one variable.0146

That direction is the y direction, because that is it. I mean you have yourself a graph, okay? 0178

In this particular case y = x2 looks something like this. This is x and this is y.0183

Well, at a particular point if I move a little bit in this direction, my graph is going to change, you know, by a certain amount.0191

There are only two directions, the x direction - this way, that way - and the y direction. There is really no place else to go.0200

Our domain is the x-axis. This is a function from R to R. From the real number line to the real number line.0206

I only have one direction to move in, and the direction in which the function moves, just this way... it is implicit in the fact that this is a function of a single variable.0216

Now, let us go this way. So... let me actually just write this down. So as x changes in this direction, y changes... or not y, let us call it f, I do not necessarily... this is an actual function, the value of the function, we say y but it is actually the function.0228

So f changes in this direction that is it, there are only 2 directions available, and those directions are implicit on the fact that we are dealing with a function of a single variable.0256

Again, we do not have a lot of choices. This way, this way, or this way, this way.0268

Now, let f(x,y) = x2 + y2. Okay? So, now we have a function of 2 variables. Now our domain is the x,y plane.0272

We have a whole bunch of choices for x and y. The functional value, do not think of it in terms of the z value. Think of it just as the function.0289

In other words if I have 2 here and 3 here, my function is going to be a certain value.0302

If I change x a little bit and if I change y a little bit, my function is going to change accordingly.0308

Well, we have a way of finding the derivative of the function of these 2 variables. Let me just write the mapping representation.0313

This is a function from R2 to R. From 2-space to 1-space.0320

Well, the derivative which I am going to notate as d(f), capital D, the derivative of this function, well we know what that is - that is just the gradient.0326

That is what the gradient is, the derivative of a function of several variables... equals the gradient of f, okay? 0335

That is equal to, well, we know that is df/dx, and that is df/dy. This is a vector, okay? 0343

df/dy, which in this case is going to be 2x and 2y.0354

That means, if I pick a point in 2-space from my domain, okay? Let us just actually say that this is our domain.0362

If I pick this particular point and then I take the derivative of the function, the gradient, I am going to end up with a vector at that point if I actually evaluate this.0372

In other words if I put the x in there, and the y in there, I am going to have some vector pointing in some direction.0384

So, that is what is happening here. We have a change in f as 2 directions change.0391

What this gradient vector does is it measures, as I change in the x-direction, the function is going to change a little bit, and as I change in the y direction, it is going to change a little bit.0412

Well, this gradient vector is a measure of the combined change, if you will, of the x and the y. That means if I am at this point and I move in the direction of the gradient vector, how is the function going to change in that direction.0424

Now we have a little bit of a complication... well, not necessarily a complication, well, I mean, it is a complication but it is a really great complication.0442

Now our domain is 2-dimensional. We do not have to stick with just this direction. The only reason we have the directional derivative is because the x and y axes are our reference axes. 0451

When we are dealing with coordinates we have to have a point of reference. Our coordinates x axis, y axis, y axis, whatever, those are our reference points. Our reference lines, which is why we take derivatives with respect to them.0462

But the fact of the matter is, if I am at this point, and I want to know how the function changes if I move in any direction, what if I want to move from this point, what if I want to move in that direction, or what if I want to move in this direction or that direction.0474

How is the function going to change, how is the value of the function going to change if I move in any one of these other directions instead of just the gradient.0488

The gradient is our basis if you will. It is a standard that is based on the fact that we have coordinate axes x, and y. It just gives us a nice little... but now that we have a domain, we can move it any direction that we want.0496

We are not confined to just move in one single direction, the x, like in single variable. That is what is happening here.0513

So, let me go ahead and finish off, but, so now we have a change in f as 2 directions change, x and y.0518

But, these are not the only 2 directions available for changes in the independent variables.0530

Ooh, wow, look at that crazy line. Let us make sure that we do not have those... in the independent variables x and y.0555

Again, they are independent, they do not rely on each other. It is not that x and y have to change uniformly, as one changes... one can move this way, one can move that way. They can move independently, they can change independently.0570

Now, that is why we take partial derivatives. The partial derivative as x changes, f changes, as y changes, f changes... again, the gradient is just a sort of a standard. 0583

It is the gradient, it is the vector of the partial of f with respect to x, the partial of f with respect to y, it gives us 1 direction.0597

But, maybe I do not want to know, maybe I do not want to move in that direction. I want to know how the function changes if I move from that point in another direction. That is what we are doing.0605

Let us go ahead and move to another... so, if we want to know how a function changes as we move from a given point in any direction in the x,y plane, which is R2.0618

We use the gradient vector as a base -- I will put that in quotes -- as a base, and we form something called the directional derivative -- very, very important concept, the directional derivative.0674

So, again, let me just explain what is going on, just to make sure we have an idea of what is happening here.0712

When you take the gradient of a function of several variables, at a given point you are getting a vector that emanates from that point.0718

So, again, this is 2-space, let us say we have a function of 2 variables, let us say I have a certain point... if I take the gradient, which is df/dx, df/dy, and at that point when I put the x,y values in, I am going to get some vector.0725

This vector, it is giving me the rate of change of the function as I move from this point in that direction.0740

If I move in that direction, how is the function going to change? Well, that is just a standard. The fact of the matter is now that our domain is in 2-space, we can move in any direction that we want from this point. We are not limited.0748

This is what we want. We are going to use the gradient vector to find out what the derivative is in this direction, in this direction, in this direction, in this direction, in any number of directions in the x,y plane, or in x,y,z, space, or 4 space, or 5 space.0763

As it turns out, what we are going to define is valid in any number of dimensions. That is what we are doing.0778

What if I want the derivative that way, that way, this way, it is our basis. It is of 1 direction, and it is 1 direction based on the x and y axes, because they are our reference axes. We have to have a point of reference.0784

That is all that is going on here. So let us go ahead and write down what we have got.0797

So, let a be a unit vector, and I am going to write this in capitals, be a unit vector, very, very important.0803

The direction in which we choose is going to have to be... is going to be represented by a vector. It needs to be a unit vector.0813

Now, let p be a point in R2, for right now we will go ahead and define it for R2, but this is valid for any number of dimensions. This is our p.0822

Okay, now let f, the function f, be a mapping from R2 to R, so this is a function of 2 variables and differentiable.0840

Again, the idea of differentiable meaning that you can take the partial derivatives and the partial derivatives are continuous. 0852

That is just going to be sort of implicit in the work that we do. We are not going to be really working with any functions that are not well behaved.0857

The directional derivative of f at p in the direction of a is the following. The notation is... the directional derivative in the direction of a of f at p = the gradient f at p dotted with a.0865

So, any time you dot something with the unit vector, what you are doing is you are actually forming the projection of the gradient vector onto the vector in the direction that you are interested in.0917

Now, let us go back to this thing right here. We said that this was our gradient vector. So, this was our ∇f.0929

Now, let us say I wanted to know what the directional derivative in that direction is. This is our vector a, okay? In that direction.0941

In other words, I want to know how the function changes not when I move along the gradient direction but when I move in this direction. My choice, any direction that I want.0952

What I do is I actually form the projection of the gradient vector onto a, which is... let us call this a unit vector.0961

We take a to be a unit vector, we form the dot product of the gradient of f at p · a, that is all that is. What that does, is that will give me the rate of change of the function as I move in the direction of a.0973

This is what I mean by we are using the gradient as the basis. We are using it as a base to find the derivative in any one of a million other directions, an infinite number of directions. That is what is going on here.0990

This is really, really fantastic because now we are not just limited to a derivative in 1 direction, we can choose a derivative in any direction that we want. This is profoundly, profoundly important.1003

Hopefully that makes sense. Now, okay, so this is true in... let me write this... so this is true for all n, okay? What I mean by that is RN, R, any function of any number of variables, 5, 10, 15, 30. This will always be the case. 1016

Let us see what we are going to do here... basic reference we use the gradient as our base... here is what we are doing. Let us say we have... once again, I am going to draw this. Forgive me if I am redundant, but these are very, very important and constant repetition is good.1046

At a certain point, let us say we have a gradient vector that way. Let us say this is our a. If we want to know the rate of change of the function in the direction of a, this is ∇f, remember that upside down triangle? that is the symbol for the gradient, called del.1064

We go ahead and we actually end up doing the dot product, we end up essentially forming the projection. It is the rate of change in this direction.1082

That gives the direct... the vector gives the direction. This value right here gives the rate at which it is changing.1092

Okay, let me see here, I think we are actually ready for some examples. I hope that made sense, so let us go ahead and do an example here. Example 1.1104

Example 1. So, we will let f = x - ezsin(y). This is a function of 3 variables, x, y, and z.1120

Now, f is a mapping from R3 to R.1135

Now we want, our task is to find the directional derivative, the direction of a at p, at the point p, which is -2pi and ln(3), in the direction of the vector (1,2,3).1143

This is a function from R3 to R, so it is a function of 3 variables. 1172

Basically what you have is you are going to have... let me just draw this out.1177

This is where again you want to sort of get away from the geometric, you want to use the geometric -- geometry -- to help our intuition but we really want to concentrate on the algebra.1182

There is some point, (-2pi,ln(3)), wherever the heck that is, let us just say right there, in 3-space.1193

When I take the gradient of this function, and I put the point (-2pi,ln(3)) in for the values for that gradient, I am going to get some gradient vector.1201

Now it is telling me I want the directional derivative in the direction of (1,2,3).1213

So, (1,2,3) is a vector, let us say, going in that direction. That is what I am doing. I am using my gradient as a basis to find the derivative in another direction in space, that is it.1218

Now let us go ahead and just do it. We have the gradient of f is equal to... actually, you know what, let me go ahead and write my equation because I like doing that..1234

So, the directional derivative of a, and the directional derivative of f in the direction of a at p is equal to the gradient of f at p dotted with a.1248

Again, a has to be a unit vector, here a is not a unit vector. We have to convert it to a unit vector, and it is very important that we make sure that we are working with unit vectors.1261

So, the gradient of f is equal to... well, df/dx is going to be just 1, and if I take the df/dy, that is going to be -ezcos(y), right?1270

Everything else is held constant so -ez is just a constant.1293

Then the last one, derivative with respect to z, it is going to be -ezsin(y), it looks like.1298

So, this is our gradient function, df/dx, df/dy, df/dz, okay?1313

When we put those values in, now the gradient of f evaluated at p which is (1,2,3), in other words when I put in (1,2,3,) for the (x,y,z) here, I end up with (1,3,0). I hope that you will confirm my arithmetic here because again I make a lot of arithmetic mistakes. 1322

So that is it. We have this vector here (1,3,0). That is our gradient vector. Again I am not being precise, I am just specifying a direction, that is our gradient vector at p.1342

Let us go ahead and turn a into a unit vector. a unit, well we know how to do a unit vector, we take a and we divide by its norm, in other words we just shorten it and turn it into a unit vector.1352

That equals a/1+9+9, right?1368

a is... no, 1,3, oh wait, I am sorry, this was (1,3,3), not (1,2,3)... I was going to say it was going to be 1 + 4 + 9, but it was (1,3,3) that was the point. 1377

So the norm of the vector is just this squared + this squared + this squared under the radical. 9 + 9 + 1, 9 and 9 is 18, so it equals a/sqrt(19) which equals 1/sqrt(19), 3/sqrt(19), and 3/sqrt(19).1392

Okay. Here we go. We have got it. Now we have our unit vector in the direction of a, we have our gradient, right?1419

Now, what we do is we just plug these into here, and then we actually take the dot product of this × that.1431

Let us go ahead and do that, let us do that in red. Let me just go ahead and write, so we will take (1,3,0) dotted with 1/sqrt(19), 3/sqrt(19), 3/sqrt(19), and what we end up getting is... so this times that is 1/sqrt(19) + 3 × 3 + 9/sqrt(19) + 0.1441

The answer is 10/sqrt(19). Now, let us talk about what this means. This was the direction of a.1475

That means that if I am at this point, if I start at that point and I move in the direction of a, my function is going to change by that much. That is what I am measuring. That is it. That is all that is happening.1487

This is the directional derivative. It is a measure of the rate of change of a function of several variables as you move in a direction other than the gradient.1502

if you happen to move in the direction of the gradient, perfect! You just take the norm of the gradient and that is the rate of change in the direction of the gradient.1513

In this case, this is how you do it, by using the gradient as the base. That is all that is happening here.1522

I hope that made sense. Now, let us go ahead and examine this idea of the gradient of f at p dotted with 'a' a little more closely.1529

Let me go ahead and go back to black here. So, let us examine the gradient of f at p · a, when a is a unit vector.1542

So, I am just going to subject what it is we have done to some analysis.1568

Well, we know from several lessons back, that the dot product of 2 vectors is equal to the norm, the product of the norm of those vectors × the cosine of the angle between those vectors, right?1573

If I have got some vector this way, there is some angle between them, θ.1593

Well, the norm of this vector × the norm of this vector × the cosine of the angle between them happens to equal the dot product of those vectors. It is a number.1599

The gradient of f at p dotted with a using this formula here, equals the norm of the gradient of f at p × the norm of the vector a × the cosine of the angle between them, cos(θ). 1612

Again, if this is some gradient vector, ∇f, like we did and this was a, there is going to be some θ, you know, some angle that is in between them.1635

But, a is a unit vector, so its norm is 1. a is a unit vector, so the norm of a, this number, is equal to 1.1647

What we have, since this is 1... what we have is this thing which is the directional derivative, so we have... let me write it out, yeah, it is fine.1665

So the gradient of f at p dotted with a is just equal to the norm of the gradient of f at p × cosine of the angle between them.1681

This is of course, the directional derivative in the direction of a. That is it, that is all this is.1692

However, so now let us take a look at what we have got. We are saying that the directional derivative of the function at p is equal to this right here.1703

The norm of the gradient × the cosine of the angle theta between the gradient and the particular vector a that... in whose direction we are interested in.1714

So, we did this because a is a unit vector, and we use this formula for the dot product.1724

Well, now let us see what we have got. Well, cos(θ) takes on values from -1 to 1, in other words the maximum value of cos(θ) is 1, the smallest value is -1.1732

So what does that mean? Well, that means, so if cos(θ) takes on values from 1 to -1, the maximum value of the directional derivative, in other words the max that the directional derivative can actually take, is when... 1748

Oh, let us actually write out a few so the cos(0) = 1, right? and the cos(pi/2) = 0, and the cos(pi), which is 180 degrees, equals -1. Let us just throw those out there for reference so that we remember in case we have forgotten our trigonometry.1765

So, this thing right here is going to take on its maximum value when the angle between them is going to be 0, when the cos(θ) = 1.1788

It is going to take on its minimum value when the angle between them is actually -1. What does that mean?1800

Well, that means, so if that is the gradient vector and that is a, if they are in the same direction, the directional derivative of f at p in the direction of a is at a maximum when you are actually moving in the direction of the gradient vector. 1808

That is what this is saying. It is at a minimum when you are moving opposite to that. This is the gradient vector, that is what this says.1827

Again, we came up with this particular value, the directional derivative of f in the direction of a is equal to the norm of the gradient × the cosine of the angle between the 2 directions.1837

Well, as it turns out this value takes on its maximum value when the direction happens to be the direction of the gradient. It takes on its minimum value when it is the opposite direction of the gradient.1850

Okay. Let me write all of this out, so red for this one, so the directional derivative of f at p in the direction of a is a maximum when cos(θ) = 1.1863

That is, when θ = 0, which means when the angle between the gradient of f at p and a is 0.1889

This is the most important part right here. So, the gradient of f at p is the direction of maximal increase of the function, maximal increase of f.1915

Stop and think about this again. That means that if I am at a point p, if I move in the direction of the gradient, my function is going to be changing, it is going to be growing the fastest, and the most.1939

If I move in any other direction, it is not going to be changing as much as fast. That is what this says.1957

What this confirms is that the direction of the gradient, so now we have a geometric interpretation of what the gradient vector tells us.1964

It tells us the direction in which the function is changing the fastest. If I end up going in the direction opposite the gradient, that is the direction at which the function actually decreases the fastest. 1973

Now the norm of the gradient, the actual value, the numerical value that gives us the rate of change.1989

So, the gradient is a vector. It gives us the direction of maximal increase of the function. The norm of the gradient gives us the rate at which it changes.1995

Let us write that down, the second part. The norm of the gradient of f at p is the rate of this increase.2004

Say it one more time. The gradient vector is the direction in which the function changes the greatest. If you are moving in the direction of the gradient vector, your function has the greatest change. 2030

Your norm of that gradient is the rate of that change. It is the numerical value.2044

Now, let us look at this globally. A few variables... this vector... so, let us see... yeah, let us just take a quick little picture here, so in other words..2050

For a function of 2 variables, I have a bunch of... let us say I have that direction... well, I can also move it so let us call it ∇f.2064

I can move in that direction, that direction, that direction. These are all a bunch of different a's. a's, that is it.2073

The function is going to be changing, if I start at this point and move in a given direction. If I want the greatest change of that function, I am going to move in the direction of that gradient.2079

Now, let us go ahead and do an example. Example 2.2094

A temperature distribution, so now we can actually use a nice physical situation, a temperature distribution in space -- space is given by the formula... is given by, let us not use the word formula... is given by the function temperature x,y = 12 + 4cos(x)cos(y) + 9cos(2x).2107

Let me actually change this to... no, that is fine. So a temperature distribution in space is given by a function, this thing right here.2156

All that means is that wherever I am in terms of x and y, notice that this is a temperature distribution in space. The z can be any value, the z does not change, so what we are looking at is sort of a cylinder.2167

Z is just infinite values. We are still talking about 3-space. Notice we have used 2 variables, so it is x and y that is changing here.2181

Now, so at a given point x,y, the temperature at that point is going to be t(x,y). That is what this means, a temperature distribution, at different points in space you are going to have different temperatures.2190

Now, at the point... here is our problem... at the point (pi/3,pi/3) find the direction of the greatest increase in the temperature.2204

In other words, once I get to the point (pi/3,pi/3), and if I want to move in the direction that gives me the highest and fastest increase of the temperature, what direction do I move in?2237

Well, I already know the answer to that. The direction of maximal increase is the direction of the gradient vector. That is what is nice.2249

So all I have to do is I have to find the gradient vector, nice and easy.2254

So now, let us go ahead and find the gradient of t. That is going to equal, and I hope that you will confirm this for me... It is going to be -4sin(x)cos(y) - 18sin(2x), and the y coordinate is going to be -4cos(x)sin(y).2260

Okay. It looks complicated... it is not. It is just functions, that is it. We are going to be sticking numbers in here, pi/3, it is a 60 degree angle, we already know what all of these numbers are. 2289

So now let us go ahead and evaluate this at the point (pi/3,pi/3). So, the gradient of t at (pi/3,pi/3) is going to equal... 2299

I am going to write all of these out actually... -4 × sqrt(3)/2, and again I am hoping that you will confirm my arithmetic here, -18 × sqrt(3)/2, and then this one is going to be -4 × 1/2 × sqrt(3)/2.2312

Again, arithmetic is just arithmetic, it is not as important as the mathematics. That is what is important. There will always be someone there to check your arithmetic, there will not always be someone there to check your mathematics, the concepts.2333

Okay, that is going to equal, -sqrt(3), -9sqrt(3), and then -Sqrt(3). I am actually going to go onto the next page here because i am getting some stray lines here.2345

So our final answer is 10sqrt(3) - sqrt(3). That is it, this vector. If I move in the direction of the vector, -10sqrt(3) - sqrt(3), from (pi/3,pi/3) at that point, that is the direction that is going to give me the maximal increase in the temperature.2366

Let us go ahead and write. At (pi/3,pi/3), the gradient... the greatest increase in t happens when we move in this direction. 2394

Now, the rate of this change. The rate of this increase is just the norm of that vector, of this increase is just the gradient of t at (pi/3,pi/3), and that is equal to (sqrt(3),0,3). That is it.2435

So, as I move in the direction of the gradient, temperature is going to be changing by that much for every unit change in that direction.2464

So, that is it. That is the directional derivative, and that is the geometric interpretation of the gradient.2472

Thank you so much for joining us here at educator.com, we will see you next time. Take care, bye-bye.2478

Hello and welcome back to educator.com and multi variable calculus.0000

Today's lesson is going to be a little different than the others. 0004

It is not going to be something that is necessary for any of the future work that we do.0009

This is a little bit of a rest bit, and I am going to discuss sort of a unified global view of derivatives for mappings.0014

Sort of a general approach that works for all mappings from any dimensional space to any other dimensional space, and how to deal with the derivative of those things.0020

This is just an opportunity for us to pull back, take a quick view. Again, if this is something that you are not necessarily interested in, feel free to skip this particular lesson.0032

You are not going to be missing anything, this is just for your own personal edification, for those of you who are a little more interested in sort of the global theory.0044

Now, we are not going to be doing anything theoretical as far as proofs, or getting into the nitty gritty. It is not that, it is just a way of pulling back and seeing the forest from the trees.0052

That is what we want to do. We want to make sure that we are able to what is going on and it will sort of help to make sense of all of the things that we have done so far in specific cases.0063

Again, you are absolutely welcome to skip this lesson if you want, you will not miss a thing, this is not something we are going to be using for the rest of the course.0071

It will be something that is important for those of you who go on to study, in particular, partial differential equations.0080

Again, let us just jump in and see what we can do. Okay, so let us just start with a definition.0086

So, given a differentiable mapping, f from RN to RM, so any n, any m, the derivative of this mapping is the M by N matrix, where the ith row of the matrix is the gradient of the ith coordinate function. 0093

Which will all make sense in a minute when we start doing the examples.0175

Okay, so just real briefly, let me talk about what this says. So given a differentiable mapping of f from RN to RM, let us say from R3 to R4. 0182

So, you are mapping and taking a point in 3 space and you are mapping to a point in 4-space.0193

The derivative of this mapping is the M by N matrix, where the ith row of the matrix, in other words where the third row of the matrix is the gradient of the, let us say, third coordinate function for that mapping.0198

Now, up to now, we have been doing mappings that are... we did curves, which are mapping from R to RN, right?0210

R2 to R2, R3 to R3, and we also did functions of several variables. In other words where we start with a vector, a point in let us say R3 and we are doing something to it and we spit out a number, something to R.0218

So we have gone R to RN, R3 to R, so 1 to say let us say 3, and 3 to 1.0233

We have not dealt yet with something like, for example, R2 to R3, or R3 to R2, R4 to R6. Well this is just a general approach to this.0237

We are just going to do some examples and hopefully it will give you a nice global view of what is going on and how to handle something like this.0248

Now notice, in single variable calculus, the derivative was a number. Now that we are dealing with functions of several variables, well you have already noticed that the derivative of a, of a curve happens to be a vector.0255

The derivative of a function of several variables is also a vector. The gradient vector.0269

Well now, if you have a mapping from a 1-space to another, where the dimensions of the spaces, where both of them are greater than 1, now what you get is a matrix.0275

As it turns out, a matrix, you remember from some of the work you did in high school, a matrix behaves just like a number does.0285

I mean, in a lot of ways, not always, it is not commutative, like multiplication is not commutative, but for all practical purposes, you can add numbers, you can add matrices, you can subtract numbers, you can subtract matrices. 0291

You can multiply numbers, you can multiply matrices, commutativity does not apply always, and you can divide numbers, and you can divide matrices.0303

Division of matrices is basically just multiplying by the inverse, which is really what division is of numbers.0311

In other words if I take the number 5 ×... or 5/3, I am not really dividing by 3. What I am doing is I am multiplying by the inverse of 3, which is 1/3. That is what I am really doing. 0317

It is multiplication. Division is a derived operation for multiplication.0329

So, in some sense, matrices are just like numbers, they are just bigger objects, so that is it. That is all that is happening here.0334

Let us do some examples, keeping this definition in mind, and we will see what we can do. So example number 1.0341

Example number 1. Let f be a mapping from R to R3, so a curve in 3-space, defined by f(t) = sin(t)cos(t), and let us say tan(t).0352

This is the first coordinate function, this is the second coordinate function, and this is the third coordinate function. That is it, that is all we mean when we say coordinate function.0373

So, this word right here, this coordinate function, all it means is when you are mapping to a different dimension like 3-dimensional space, you need 3 numbers.0382

Well, sin(t) is 1 number, cos(t) is another number, tan(t) is another number, for a given value of t. That gives you a point in 3-space, that is all it means. The coordinate function does this.0392

Well, based on this definition, the derivative of this f(t) is a 3 by 1 matrix. A 3 by 1 matrix. 3 rows, 1 column, well, here is what it is.0404

The derivative of f = ... we said it is the gradient of the ith coordinate function, well the first coordinate, so the first row is the gradient of the first coordinate function.0424

Well, since this function is sin(t), it only has 1 variable, it is just the derivative. It is cos(t).0438

The second row is the second... the gradient of the second coordinate function. The second coordinate function is the cos(t), it only has 1 derivative because it is a single variable t. So, this is -sin(t).0445

The third coordinate function becomes the third row, the derivative of the tangent is sec2(t).0459

There you go, that is it. I have this 1 by 3 matrix, and you already know this, we have been doing this all along, except we have been writing it this way, cos(t)-sin(t)sec2(t).0466

We have been writing is as a row and we have been separating is at columns. Now we are going to treat it more generally.0485

Any time you have a function of RN to RM, the derivative matrix is the M by N matrix, so 3 by 1, we just write it as a matrix, no longer as this list, or as this... you know... row vector, if you will.0490

Okay, let us do another example. This is actually a very, very, unified vision, a way of thinking about things, because again, the derivative of a function of a single variable is a number, is 1 function.0506

The derivative of a mapping, say from 2-space to 2-space, now you have a 2 by 2 matrix. Where that 2 by 2 matrix has 4 entries, that whole thing is a derivative, it is really kind of exciting.0524

It is just bigger objects. You are treating them exactly the same. It is a single object, but it is a matrix, it has multiple parts.0539

Okay, so let f be a mapping this time from R3 to R, so a function of 3 variables. Let me make my 3 a little more clear here. R3 to R.0546

This is going to be defined by, so f(x,y,z), we are taking a point in 3-space, and we are spitting out a number. 0563

It is x2y3z. That is our function. 0572

Well, what is the derivative? The derivative of this function is... let me see, this is 1 by 3, so it is a 1 by 3. It is a 1 by 3 matrix.0582

So, df, the derivative is the 1 by 3 matrix.0600

It is 1 row, 3 columns. Well, it is the gradient, the first row is the gradient of the first coordinate function. There is only 1 coordinate function here, because it is a mapping to R1.0606

We just take the gradient of this thing. It becomes 2xy3z.0621

Then we take the derivative with respect to y. We get 2x2y2z.0629

We take the derivative with respect to z, which is x2y2, there you go... I am sorry, that is x2y3.0637

I guess I have a love affair with the number 2 here, x2y3, there we go, that is it.0647

Notice, this is just the gradient vector that we have been doing all along, except now I do not need to put these commas here because I am treating them like a matrix.0653

This is the first entry of the matrix, this is the second entry of the matrix, and this is the third entry of the matrix.0663

Again, now we are thinking of it in terms of matrices. Yes, it happens to be a 1 by 3 matrix, and anything that has either 1 row or 1 column... you know, that we generally call them a vector... but again, this is just nomenclature. 0670

This is just names that are being thrown around. Seeing it like this is really, really nice. Let me just write: notice, this is just the gradient of f expressed as a matrix.0685

So, let us do example 3. Example 3.0728

Again, let me reiterate, there is nothing in this particular lesson that we are going to be using for the rest of the course.0735

So, by all means, if this is something that you are not interested in, it if it something that the flavor of it, you know, you just do not really like it or care for it, you are not going to be missing anything. So by all means, this is just for your edification, if you should feel so inclined.0740

Okay, example 3. Now, let f be a mapping from R2 to R2 be defined by f(x,y) = x2 + y2, and x2y2.0755

This is the first coordinate function, and this is the second coordinate function. Okay, well, we said that derivative is going to a 2 by 2 matrix.0779

Again, it is very, very important. The arrival space, the dimension of the arrival space, that is going to be the number of rows.0793

The dimension of the departure space, that is going to be the number of columns. So 2 by 2, this is actually 2 by 2, that way.0803

We are mapping this way, but the matrix is 2 by 2. Okay. 2 by 2 matrix, and the derivative looks like this. 0811

Well, we said that the first row is the gradient of the first coordinate function.0820

The first coordinate function is x2 + y2, therefore the gradient of this is, well the derivative with respect to x is 2x, the derivative with respect to y is 2y.0828

Now we go to the second coordinate function, that one right there. The derivative with respect to x is 2xy2, and the derivative with respect to y is 2x2y.0839

We have our derivative of this mapping. How incredible is that. This is really, really kind of extraordinary. It is a single object. It behaves as a single object mathematically.0850

We can actually do things with this the same way we do with any other derivative, except now we are working in 2-space instead of just 1-space. This is very, very powerful stuff.0865

So, let us go to example 4. Now, let us let f be a mapping from 2-space to 4-space, how is that? R2 to R4.0877

We will define it by the following. f(x,y) = x2 + y2 is the first coordinate function, x2 - y2 is the second coordinate function, x2y2 is the third coordinate function, and xsin(y) is the fourth coordinate function.0895

Now there is no way to make sense of this geometrically. Again, but we can make sense of it algebraically, and we can work with it algebraically. This is what makes this extraordinary.0921

Well, we know that the derivative, there is a derivative for this mapping from 1-space to another, and it is a 4 by 2 matrix. So df is a 4 by 2 matrix.0931

So, let us go ahead and write that out. I will go ahead and put the matrix bars around it in just a bit.0947

So, it is a 4 by 2 matrix. Each row of the matrix is the gradient of that coordinate function. So the first row is going to be the gradient of the first coordinate function.0955

Well, this is 2x and 2y. The second coordinate function, x2 - y2, so this is 2x and this is -2y.0968

The third row is going to be the gradient of the third coordinate function, which is 2xy2, I think.0981

2xy2, and the derivative with respect to y is 2x2y.0990

The fourth coordinate function is xsin(y), so the derivative of that with respect to x is sin(y), and the derivative with respect to y is xcos(y), I think.0997

Again, I hope that you are confirming these for me. So, this is my derivative.1009

This whole thing is my derivative, for that function. The same way that f(x) = x3, f'(x) = 3x2. This is a single object.1016

It takes on different values for different values of x. Well, this is a slightly more complicated function, this is now not a mapping from R1 to R1, this is a mapping from R2 to R4. This is the derivative. 1033

It is a single object, for different values of x and y in the given space, this thing takes on different values. It is just bigger, that is all it is. It is just bigger because our spaces are bigger. That is how you have to think about it.1043

So, let us see what we have here. Now, we will move on to the next level.1061

Well, you remember what we did -- what we have been doing actually -- for the last several lessons. We have a curve, and so we have a curve c(t), right?1070

Then we have a function... let me do this in terms of pictures... so we have a R1, 2, so this is R and we have a curve. It is a mapping from let us say R to R2.1081

Then we have a function, let us say, that is a function of two variables that maps from R2 to R.1096

Well, we can form, so this is let us say c(t), and let us say this is f(x,y). Well, we can actually form the composite function, f(c(t)), right? That is what we have been doing. 1102

Then we have been taking the derivative of that, and we have a formula for the derivative. We were doing lots of example problems for handling this derivative.1115

Now, we can go ahead and we can give a general definition for how to actually find the derivative of the composite function for a mapping from a space of any dimension to any dimension. Let us go ahead and do that.1125

So, given g, which is a mapping from let us say RN to RS, and f, which is a mapping from RS to RM, all of these different numbers, or not. They can be equal that can be different.1140

We can form the composite function... Excuse me... We can form the composite function f(g), which is a mapping from RN to RM, right?1163

RN to RS, then f takes the values of RS, maps them to RN, so what we have is a function that goes from RM to RN, right?1186

Now, the derivative of the composite function, which we will notate the same way we notate any other, df(g) -- write it like that -- is the M by N matrix.1197

Again, you are going from RN to RS, RS to RM. Well, you are actually going... the composite function goes from RN to RM, so the derivative is an M by N matrix, gotten by multiplying exactly what you think.1229

Exactly what you think, the matrix for the derivative of f, and the matrix for the derivative of g, in a given order.1253

I write in a given order because matrix multiplication is not commutative, it depends on the order in which you do it. 1271

Now we will actually write it in mathematical form. That is the derivative of f(g) = the derivative of f evaluated at g × the derivative of g.1278

So, if it is f(g), the f goes first, the g goes next. That is it, you just are actually multiplying matrices. How incredible is that?1302

You have a function which has a derivative matrix, you have another function which has a derivative matrix. If you can form the composite of those 2 functions, f(g) or g(f), you can actually find the derivative simply be multiplying the matrices of the respective derivatives. That is amazing.1311

Let us go ahead and actually do an example, and I think it will make sense. So, example 5.1329

We will let g be a mapping from R to R2, so a curve in 2-space defined by g(t) = cos(t)sin(t), so this is just a curve in 2-space.1344

We will let f be a mapping from R2 to R, be a function of several variables -- two variables in this case.1366

Defined by f(x,y) = x2 + y3.1377

So we do g first, we get this and this. These values, they are the x and the y. This is a composite function, because we are doing g first, and then the values that we get for x and y, for the first coordinate function and the second coordinate function, they go into here.1388

So basically, f(g) is actually just a function of t. So, f(g) is a mapping from R to R, R to R2, R2 to R. We are going from R to R.1405

The derivative of f(g) is a 1 by 1 matrix, which means just a function, just a number. A 1 by 1 matrix is a number, is a function, just one thing. 1425

It is equal to, so the derivative of f(g) is equal to the derivative of f evaluated at g. That matrix × the derivative of g. That is it. Now, let us do it.1439

Well, what is d(f)? Let us find d(f) first. Well, the derivative of x(f) is equal to the gradient, right? That is what the derivative is. It is a... so f is going to be a 1 by 2 matrix.1460

It is going to be... the row is going to be the gradient of this coordinate function, there is only one coordinate function, so we get 2x... let me express it as a matrix, so 2x and 3y2, how is that? That is that one.1477

Now, we evaluate at g, which means we put... this is x and this is y, right? cos(t) is x, and sin(t) is y because this is a composite function.1495

When we evaluate at g, what we are doing is we are putting in the actual coordinate functions for g in for x and y.1519

So 2 × that, you get 2cos(t), and 3sin2(t).1526

Now, let us go ahead and do dg. Well the derivative of g is a mapping from R1 to R2, so the first row is the gradient of the first coordinate function, which is cos(t), so we get -sin(t).1539

The second is the derivative of that, cos(t). So, that is that. Now all we have to do is actually multiply this and this, and we will end up with the derivative of our composite function.1558

Let us do our final... so d(f(g)) is equal to this thing. 2cos(t), 3sin2(t) × -sin(t), and this is cos(t).1573

When I multiply this matrix it is this × that + this × that, remember? You move across this way, you go down a column that way.1597

This is going to equal... let me put it a little bit lower here, so let me go ahead and put it over here... it equals -2sin(t)cos(t) + 3sin2(t)cos(t). That is my answer.1605

That is it, that is my derivative. Expressed as a function of t. That is all you are doing. 1629

Notice... I will actually do it on the next page... notice, this is just the gradient of f evaluated at g · g'.1639

It is the same as our formula that we used for finding the composite function of a function of several variables and a curve. That is it. It is the same thing.1658

So, our so-called formula, so our formula is just a special case.1672

So, the formula we have been using is just a special case of a more general method. That is it, that is all we are doing here. That is all this is about.1694

Showing you that there is something more general go on that applies to a broader range of mappings.1717

Let us see. Now, let us do one more example here. So, example 6. Example 6, alright.1726

We will let, let us see... yeah, that is fine... so we will let g be a mapping from R2 to R2. So this time R2 to R2.1740

We will take points in 2-space and we will map them to points in 2-space, defined by g(rθ) = rcos(θ)rsinθ.1754

It is a function of 2 variables and you are spitting out 2 things. So this is our x, and this is our y, if you will. First coordinate function, second coordinate function.1771

Let f be a mapping from R2 to R1 defined by f(x,y) = x2 + 4xy.1785

Let us see what we have got here. Let us draw a little picture. That is R2, that is R2, and this is R1, so g maps from R2 to points in R2. That is g.1805

f takes these points in R2 and it actually maps them to R1. Therefore we can form the composite, which is f(g), which takes points in R2 and maps them to R1. That is all that is happening here. 1820

In other words, f(g) is actually going to take Rθ and it is going to map them to some number, some function. So f(g) is a function of Rθ. 1838

G is a function of Rθ, f is a function of xy. x and y are just variables.1854

The composite function, which is just a mapping from R2 to R1, is a function of the 2 variables Rθ, that is it. That is all that is happening here.1861

Let us just run through the mathematics here. So, basically what we are going to have at the end, notice R2 to R2, R2 to R1, so f(g) is a mapping from R2 to R1.1869

Let us write that down. f(g) is a mapping from R2 to R1, so what we are going to end up having is a 1 by 2 matrix.1883

So, we are going to have df(g)/dr, it is going to be one partial derivative that we have, and the other is going to be df(g)/dθ. That is what we are going to end up with at the end, so let us just do it according to our method.1895

When we take the derivative of f it is going to equal, well f was a function, and so it is going to be a 1 by 2 matrix, and that row is going to be the gradient of the coordinate function, so we get 2x + 4y...1913

You know what? Let me go ahead and write the functions again, I want you to see them on the same page as I actually take the derivatives of.1936

So, f was f(x,y) = x2 + 4xy, and g(rθ) = rcos(θ)rsin(θ). We are forming f(g), that is what we are doing.1943

Let us go ahead and take the derivative of f. The derivative of f is going to equal the 1 by 2 matrix, and it is going to be the gradient of this coordinate function.1967

What we are going to get is 2x + 4y is our first... I will leave it as a matrix so I will not have columns -- I mean I will not have commas.1978

Then the derivative with respect to y is going to be 4x. That is our derivative of f.1989

Now we are going to take our derivative of f, and we are going to evaluate it at g. g is this.1995

That means take this and take this as our x and y and put them into here, because we are forming the composite function.2003

We end up with the following. We end up with 2rcos(θ) + 4rsin(θ), and we end up with 4x, which is 4rcos(θ). So, this is our first.2014

Now, we will do d(g). d(g) is a mapping from R2 to R2, so it is going to be a 2 by 2 matrix, so the first row is going to be the gradient of that function, the second row is going to be the gradient of that function.2036

So, it is going to be... the derivative with respect to r is going to be cos(θ), and the derivative with respect to θ is going to be -rsin(θ).2050

Now, this function, the derivative with respect to r is going to be sin(θ), and the derivative with respect to θ is going to be rcos(θ). There you go.2064

Now we will go ahead and actually do the multiplication. So, the derivative of f(g), the composite function is just equal to this thing × this thing, in that order. This one goes first.2077

We will write 2rcos(θ) + 4rsin(θ), and then we will have 4rcos(θ), so this is our matrix there.2099

Then we are going to have cos(θ) - rsin(θ), then we have sin(θ), then we have rcos(θ)... oops, there we go.2117

This is a 1 by 2 matrix. This is a 2 by 2 matrix. The middle things go away, our answer is going to be a 1 by 2 matrix, so we are going to multiply this by that, that is going to be our first entry, and then we are going to multiply this by that, that is going to be our second entry.2131

Again, I presume here that you have seen matrix multiplication at the very least. I know that you have in high school, and I know that you probably did a little bit of it in calculus. 2152

At least here or there, so I know that it is something that you are familiar with, and again, if you are not, no worries. This section is not going to show up anywhere else.2159

You are more than welcome to ignore it if you want to. Okay, so now let us do this multiplication. When we actually multiply it out, here is what we get.2166

We get the derivative of f(g) = 2rcos2(θ)... I am going to write it vertically... + 4rsin(θ)cos(θ) + 4rsin(θ)cos(θ).2178

Now I can put some things together, that is not a problem. 4rsin(θ)cos(θ), 4rsin(θ) is 8rsin(θ)cos(θ), but I will go ahead and just leave it like this.2213

Now, our second entry is going to be -2r2sin(θ)cos(θ) - 4r2sin2(θ) + 4r2cos2(θ).2222

This is what we have ended up with. Our composite function, f(g), the derivative of that is this right here, this whole thing.2247

Now, f(g) was a mapping. We said that f(g) was a mapping from R2 to R. We know that its derivative is going to be a 1 by 2 matrix. This is what we have got, a 1 by 2 matrix -- 1 row, 2 columns.2257

This right here, f(g) was a mapping, was a function of Rθ, right? That was the original departure space that left, so it is a function of Rθ.2281

Therefore, what we have when we take this derivative is d(f(g))/dR, d(f(g))/dθ. That is that one, that is that one. That is all that we have done here.2299

I hope this helped a little bit. If not, do not worry about it. It is just something that I wanted you to see. 2324

A little bit more of a global idea, something that generalizes 2 mappings of... mappings from a space of any number of dimensions to any numbers of dimensions, to demonstrate that we can still do calculus.2330

This is actually kind of extraordinary that we are not limited to a single space. We can work from two entirely, completely, different spaces, mapping from one space to the other, and we can still do calculus on this.2343

The ideas are the same. We are just taking derivatives. Now instead of a derivative being a single function, now a derivative is a matrix, and the matrix can be of any dimension.2355

In those matrix, the entries of the matrix are the actual functions that we deal with, the partial derivatives if you will.2366

Thank you for joining us here at educator.com. Next time we will go ahead and continue on with our discussion of maximum, minimum, bye-bye.2373

Hello and welcome back to educator.com and multi-variable calculus.0000

Today we are going to start out discussion of maxima and minima. Finding points in the domain where the function obtains a maximum and where it obtains a minimum, just like in single variable calculus.0004

So, a lot of the things you learned in single variable calculus are going to apply here.0016

Now, instead of one variable, x, you are going to have 2, 3, sometimes more.0020

So, the problems tend to become a little bit longer, a little bit more involved, but again it is still just a max, min problem based on the derivative. Let us just jump right on in.0026

Okay. So, let me see, let me go ahead and just write a... well, let us go ahead and start with a definition. How is that?0039

I think... we do not really need a preamble here, so let f be a differentiable function -- and again by differentiable we just mean well-behaved, it has a derivative, no problem, no worries, no concerns -- on an open set.0047

Again, you recall what an open set means. It is just a set that does not have boundary points. An open set u.0075

Let p be a point in u. p is called a critical point or a critical value, we will use both terms, it is called a critical point if each partial derivative evaluated at p is equal to 0.0085

So you see this is the exact definition as for single calculus. The derivative is equal to 0 at a critical point.0130

So, let us go ahead and write what this is symbolically. We will say that is di(f) = 0.0136

So, the first derivative, second derivative, third partial, fourth partial, whatever it is.0147

Or, df/dxi evaluated at p is equal to 0. Let me actually add one more thing here. It is very... evaluating it at p, so we will actually put the p in there.0153

So, that is it. Let us just do an example. Example 1. Find the critical values of f(x,y) = x + ysin(x).0169

Let us go ahead and find the derivatives. The partial with respect to x, df/dx = 1 + ycos(x), right? We are holding y as a constant.0196

The partial derivative with respect to y is going to equal just sin(x). This one is going to be held as a constant, the x goes to 0. 0212

Let us see what we have got here. So, here, we want each of these, of course, equal to 0.0223

So 1 + ycos(x) = 0, and sin(x) = 0, so let us go ahead and do this one first.0231

sin(x) = 0, so our possible values of x are going to be 0, and π, and 2π, and 3π, and so on.0243

Now, let us see... should we do... yea, okay. Let us go ahead and take a particular value. Let us just take... so the 0, the 2π, the 4π, when we take the cosine of that, so now we are going to use these to actually solve for the y variable.0255

We have 0, 2π, 3π... I am sorry, 0, 2π, 4π, 6π, then we have the odds. π, 3π, 5π, so we have to check both possibilities because it turns out y is going to equal 2 different numbers.0281

So, let us go ahead and start with, let us say 2nπ, so when I take the evens, the 0, the 2, the 4, it is going to be... all of those can actually be written as x = 2nπ, where n is just some integer.0293

This little, funny-looking z here, that is a symbol for the integers. You know, integers, negative, 0, 1, 2, 3, 4, 5, that is it. Natural numbers including the negative versions of those negative numbers and 0.0313

This is a shorthand notation for writing the even number ones.0328

So, when we do that, we get 1 + y × cos(2nπ) = 0.0333

Well, cosine of any 2nπ, so 0, 2π, 4π, is the cosine of... it is going to be 1.0342

So you get 1 + y = 0, you are going to get y = -1.0351

Now we will go ahead and do the version for the odds. Well, the odds, the π, the 3π, the 5π and so on, that is going to be (2n+1)π.0357

That is the shorthand notation for all of the odds. Basically you just run n. So when n = 0, you have got π, when n = 1, you have got 3 π, when n = 2, you have got 5π. That is what this means.0370

So, when we do this one, we get... let me write it down here, that is okay... so we get 1 + y × cos(2n+1)π, and the cos((2n+1)π) for all n is going to be -1.0384

So what you end up with is 1 - y = 0, so y = 1.0406

So our final solution is going to look like this. Let us go ahead and write... yeah, we will write it on the next page. 0415

So, the critical values are going to be all of the points 2nπ - 1, and 2n + 1π, 1, in other words the x value is going to be 2n + 1π, and the y value is going to be 1.0423

This is for n in z, where n is any integer at all, so we have 2 different possibilities.0448

That is it. It is that simple, you just take the partial derivatives, you set them equal to 0, and now instead of just having the one equation that you solve, you are going to have 2, 3, 4, however many equations and you just have to find all of the different values.0456

You are just going to have a lot of values, so you might have one critical value.0469

It just depends. You have to use all of the resources at your disposal, all of the things you know about a function, whether it is its graph, whether it is its particular behavior, whether it is a certain limit approaching something to come up with what these critical values are.0475

So, let us go ahead and speak about the critical values a little bit. The critical values for a function need not be local max's or min's.0492

You remember from first variable calculus, just because you find a critical value, some place where the derivative equals 0, that does not necessarily mean that it is a local max or a local min.0522

There are situations where it is neither, where it is an inflection point. So, the same thing can happen here, you know in several variables. It does not have to be a local max or min, it just means that it might be. 0532

We have to subject the function to further analysis, to see if it is a local max or min, and later on if we are dealing with a closed domain if it is an absolute max or min. That is what is happening.0542

So, recall, we can have f'(x) = 0 and have a point of inflection, which is neither max nor min.0556

It looks something like this. So the graph comes like this, but you get a derivative of 0, so this point right here, I mean yeah, its derivative is 0, it is a horizontal slope, but the horizontal slope... but it is not a max or min, it is a point of inflection. That is a possibility too.0589

So, let us start with another definition here. Let f be a function on an open set u, a point p... and again these definitions, you know what they are... but again, we want to be reasonably formal in mathematics.0605

A point p is a local max for the function, if... oops, let us erase that... if there is an open ball -- and I will draw a picture in a minute -- centered at p such that f(x) < or = to f(p) for all x in that ball.0638

Let us go ahead and think about what this means. So in this particular... this is a general definition, it works in any number of dimensions.0696

So, the example I am going to use to demonstrate it is going to be in R2, in the plane, because that is the one that we are the most familiar with.0704

So, in R2, it looks like this. It looks like... so we have R2, and we have our point p... now an open ball around point p, centered at p, in this case is just an open disc.0710

If you are dealing in 3 dimensions, then you are talking about a point and there is an open ball. An open ball means, again, the boundary points are not included.0730

It is just some region around it, that you are not specifying a boundary on it. 0737

Basically it says for any x-value, in here, for any x-value that I pick around that point, if a function, if I evaluate the function at any of the x's, at any point that I pick in that open ball, if all of them are less than or equal to the value of the function at p, then that is a local max.0742

So, the graph of this, of course, if you... so if your domain is 2 dimensions, so here we are talking about a function from R2 to R.0766

So, this is our domain, we are picking points from here, we are sticking it in a function, we are doing something to it, we are spitting out a number. That number is used as the z coordinate.0776

So, a function from R2 to R, actually gives you a surface in 3-space. So if this is the domain, there is some surface on top. It is just saying that what you have is something like that.0786

That here is the point, and then every point around p forms some sort of a, like a, cone or something. It is a local maximum, not an absolute maximum because we are talking about an open domain.0795

In a minute we will actually define what we mean by absolute max and closed domains.0808

So, that is it. Again, exactly the same as first variable calculus, or single variable calculus.0812

So, let us just go ahead and formalize one last thing. So, a local min is defined the same way.0820

It is defined the same way except of course all of the functional values are > or = f(p).0830

So, you have something that looks like this where p is down here, and all of the values around it are bigger than that.0843

Let us go ahead and write down a theorem. Now, let f be a function defined, and differentiable on an open set u, and let p be a local extremum.0851

An extremum is just a single word that takes care of both max and min at the same time. So extremum, plural being extrema. So it is max or min.0894

Let p be a local extremum for f on u. Then p is a critical point.0911

This is really, really important here. This is saying that if p happens to be a local max or min, that means that it is a critical point.0924

It does not mean... this theorem does not say that if p is a critical point, then it is a local max or min, because we know that that is not true. 0934

The implication only works in one direction, not the other. Let me write this out.0944

So, note, a local extremum implies a critical point -- this double arrow in mathematics means implies -- not, definitely not. It does not mean it goes the other way. It does not mean if it is a critical point, it is a local extremum.0949

This is a very important point of logic in mathematics. Most implications -- not most, many implications only work in one direction, and we have to be very, very clear that we are not just automatically assuming that it works in the other direction.0973

Critical point does not imply a local extremum. This is not true.0990

However, what is true is something called a contrapositive of this. A local extremum implies that it is a critical point.1001

What is also true is that if it is not a critical point, then it is not a local extremum.1011

So, a implies b is the same as not b implying not a. 1018

The way theorems are actually written in mathematics, it is often the case that we do not use them as written. We often use them in another form.1029

This is one of the ones where we often use in the other direction, because when we are given a function, we actually do not know whether it is a local max or min, to decide whether it is a critical value. We are working in the other direction. 1038

We are actually looking at the function and finding the derivative and the partial derivatives, and setting them equal to 0. We are finding points that are critical values and then we are checking to see if they are local max's or min's.1050

What we can say, what this theorem does allow us to say, is that if I find, if I take the derivatives of functions, set them equal to 0 and realize there is no value in the domain, x or y, there is no point x or y where the derivative = 0, then there is no local max or min.1061

That is what is going on here. In this case, the theorem is written as if -- you know -- p is a local max or min, then it is a critical point.1080

Often, how we will be using it is if it is not a critical point, then it is not a local max or min. We will be using the contra-positive of that. 1088

So, if a implies b, that is the same as not b implying not a. That is very, very important.1097

Let us go ahead and do an example here. Example 2.1105

Let us check to see if the critical values of Example 1 are local extremum, or local max's or min's.1118

So, again, just because they are critical values, that is not a guarantee, we have to see if they are local max's or min's.1145

So we had the fact that f(x,y) = x + y × sin(x).1151

We had critical values -- what is with my h's here? -- We had critical values of (2nπ,-1) and (2n + 1π,1).1169

Defined for all n in z integers, so here we have a bunch of critical values. We have (0,-1), we have (2π,-1), (4π,-1), and then we have (π,1), (3π,1), so we have any infinite number of critical values.1192

We need to see if any of those are local max's or min's. Well, pick a value, just sort of pick on at random and work with that. Subject that one to analysis, and so that is what we are going to do.1218

So, just pick a critical point and check the value of the function -- that is what you are doing, that is how you check max's and min's -- Check the value of the function in a small region, in a small ball, a small ball or disc, around that point.1229

That is how we are going to check, right? The definition of a local max is if I have a point p and there is a certain value for the function f(p), there. 1267

If I sort of move to the left or the right, this way or that way, then the values that I get should either be all less than f(p) or all greater than f(p).1275

That will tell me if it is a local max or min, and that is what I am going to do. So, now, in subsequent lessons, in the next couple of lessons we are actually going to devise ways of handling this local max/min business more systematically.1286

But, oftentimes, in this case it just means we want to get used to analyzing the function, just sort of taking a look at it, get familiar with it, before we develop the systematic tools for deciding whether something is a local max or a local min.1300

You remember from single variable calculus, we took the second derivative to find out if it is concave down or concave up. There are analogous things like that, and we will develop them.1313

But right now, we just want to do it by observation, which is really, really important to do.1322

You know when everything is said and done, it is about how familiar you are with the function and being able to analyze it using the other tools at your disposal, all of the other things that you know.1329

So, now let us just choose 2, π, 1 with n = 0. So I am just going to pick the easiest point, so (0,-1), that is going to be our critical point p.1338

Well, let us go ahead and find what the value of the function is at (0,-1).1356

The function is x + y sin(x), so it is 0 + -1 × sin(0), which is 0.1361

So the value of the function there is 0.1371

The first thing I am going to do, since we are dealing with a 2 dimensional domain, you know we are dealing with R2... I am over here, so what I am going to do is I am actually going to... well we have x and y... well I need to check both.1375

So, what I am going to do is I am actually going to hold one of the variables fixed, and then I am just going to move the other one.1396

In other words, I am going to move a little bit in this direction, and I am going to move a little bit in that direction to see what the function does.1401

If I get my answer, great! I can stop there. If not, I am going to hold the other variable, the x fixed, and I am going to move a little bit in the y direction that way to check to see what happens to the function.1406

Hopefully, those movements will tell me what is going on. So, that is what I am going to do.1420

I am going to decide to hold... so let us hold y fixed first, and we will vary x.1425

So what I am going to do is I am going to hold y fixed, and I am just going to move a little bit this way, and a little bit that way around x.1441

Let us vary x about 0, because 0 was the point where we are, so I am going to move a little bit to the right of 0, a little bit to the left of 0, to see how the function behaves.1449

Well, let us just take... let us take x = 0.1, so let me move 0.1 in that direction. Then, f(0.1) - 1, because we are holding y fixed, it is going to end up giving me some number which is positive, 1.7 × 10-4.1461

Okay, so it is a positive number. Well, now I am going to move in the negative direction. Let us take x = -0.1.1485

Then, f(-0.1,-1), when I do that one, evaluate that, I am going to get -1.7 × 10-4.1493

Basically what has happened is this. This is the value of my function at (0,-1), okay? Looking at it from just one... I am holding y fixed, so now what I am looking at is just the x axis, this way.1506

If I move in the positive direction from 0, my function actually went up. So, the value of my function is 0, at (0,-1), so the value of my function actually became positive.1522

When I moved to the left, the value of my function became negative. So, in this particular case, this tells me everything that I need to know.1536

I do not actually need to do the next check. I do not need to hold x fixed and work on y. This right here tells me what I have is some kind of a point of inflection.1546

In order for it to be a local max or min, it would need to be like this for a local min, or like that for a local max.1556

That is not what is happening. So as I move around the point, (0,-1), in the x direction, holding -1 fixed, I end up with a point of inflection.1565

Essentially, I have turned it into a single variable problem, because I have the case where it is neither a max nor a min, and that is what this tells me. So this is neither max... I should not say neither max nor min. There is no local max or min.1576

In this particular case, there is no local max or min at these critical values. The derivative might equal 0 there, but it does not actually achieve a maximum or minimum there, because with this simple analysis, we have discovered a point of inflection, some sort of point of inflection.1599

So, that is all that is happening here. Okay. So now let us go ahead and move onto closed domains.1615

We have been talking about open sets, and now we are going to talk about closed sets, and absolute max's and min's. So, definition.1624

So we are going to throw out a couple of definitions beforehand. So a boundary point, a boundary point, b, of a set, u, is a point such that an open ball around b contains a point in u and a point not in u.1634

I know this may actually sound confusing, but mathematical definitions have to be very, very, precise and formal. Now let us draw you a picture and tell you what it actually means. It is actually very simple.1686

If I have some open region, like this, some set, and if there is some... so a boundary point is a set u... the point, such that an open ball around b -- this is b here -- well I take some open disc around b, right?1695

It contains a point in u, there is my point in u, and of course I am not in u. There is my point not in u. That is all that it means. It means that it is the point on the boundary, that is it.1719

You already know what this means intuitively, but this is a formal definition of what a boundary point is.1731

If I take any little region around it, I am going to find some points that are in u, some points that are not in u. That is the definition of a boundary.1736

Okay, very, very important to note. A boundary point does not have to belong to the set. In fact, that is the very definition of an open set.1745

An open set is a set that does not contain its boundary points. In other words, there are boundary points there, but they are not part of the set.1755

So, a boundary point does not have to belong to the set. So, a boundary point of a set does not have to belong to the set. Very, very important to know that.1765

In fact, that leads us to the next definition.1792

A set is called closed -- again, you already know all of this, a closed interval, an open interval, now instead of intervals we are talking about regions, or regions in space, or regions in 5-space, that is it.1799

So, a set is called closed if it contains all its boundary points.1814

So, let us go ahead and draw some... let me make this a little bit better so let me see... we will draw an open set, and then we will draw the closed version.1833

So, that is some open set. That is the closed version. The open set does not contain its boundary points, the closed set is everything inside including its boundary points, and of course there is... I can also think of one other set from this. 1845

I can take all of the points from on the boundary, but not on the inside or the outside.1864

So if I just talk about the points on the boundary that does not include this or this, that is another thing. We actually have 3 sets of points here.1869

We have the open set of a region, we have the closed set on that region, and then we have the boundary of that region.1876

If the set actually contains its boundary points, it is called closed. In other words, the boundary points belong to the set, that is all that is happening here.1882

Now, we are ready for one more definition, and then we can finish off with the final theorem.1895

Okay, definition. A set is called bounded, in other words the set itself is called bounded if there exists some number n, and it could be any number, it does not have to be an integer.1906

Such that, -- but we can always choose an integer -- such that for every point, x in the set, the norm of x is < or = to n.1931

So, if I have a particular set, in other words, let us say I have some set like that.1957

It consists of all of the points inside here. In this particular case I just happened to pick a closed set. So every single, these are vectors of course right? in 2-space and of course there are 2-vectors.1964

So, we can find the norm of these vectors, the norm of these points, in other words the distance they are from the origin.1979

If the norm of these, of all the points in that set, happened to all be < or = to some number, that means the set itself is bounded. That is all that it means.1985

This is a very, very important definition because often times we will be talking about sets that are not bounded. This is a very important definition. It just means that the norm of all of the points in that set are all < or = to some number. That is it.1995

There is an upper limit on the value of the points themselves.2010

Okay, now we can state our theorem.2015

Let s be a closed and bounded set, very important. Those are the hypotheses of the theorem, let s be closed and bounded.2023

Let f be a continuous function defined on s. Then, f has both a max and a min on s.2042

Okay. So let us talk about what this theorem says. So let s be a closed and bounded set, let f be a continuous function defined on f, then s has both a maximum and a minimum on s. 2081

So if we happen to be dealing with a set that is both closed and bounded, then we know that somewhere on that set, either in the interior of the set or on the boundaries of the set... that that function, there is some point that the function obtains a maximum value.2093

A highest value and a lowest value. Now this theorem does not tell me where it happens. That comes from further analysis on the particular function that we happen to be dealing with in the problem.2109

But it does tell me that an absolute max and an absolute min exist. Notice, I did not use the words absolute maximum and absolute minimum.2120

I try to keep vocabulary to a minimum. I would much rather have you understand the ideas, what is going on, but here we are talking about something that is an absolute max and an absolute min. 2129

It can happen on the interior, or it can happen on the boundary. Now, if you end up with a function where there are no critical values, let us say you have a closed and bounded set -- we are going to do some examples in a minute -- and the critical values, there is no x or y where the derivative = 0.2142

Well we know automatically, but if it is closed and bounded, the theorem has to achieve a maximum and min somewhere.2160

If there is no critical value, then there is certainly no local max or min, that automatically tells me that the maximum and minimum value of the function happens on the boundary.2166

Now, I no longer have to concern myself with inside, I just have to concern myself with what is on the boundary. That is the power of these things.2176

Okay, we are going to close off this lesson with a statement of this theorem. The next lesson what we are going to do is just a series of problems developing this theorem, working on this theorem, practicing the theorem, finding local max's and min's, finding absolute max's and min's.2184

So thank you for joining us here at educator.com, we will see you next time. Bye-bye.2196

Hello and welcome back to educator.com and to multi-variable calculus.0000

Today we are going to continue our discussion of maximum, minimum.0004

We are actually just going to do example problems. The last lesson, we finished off with the main theorem of closed and bounded sets and the existence of absolute max's and absolute min's on that particular closed and bounded set.0009

Now, we are going to just going to do some examples to get an idea of the different things that can come up when we are dealing with a particular function.0022

Really, this lesson is just about getting comfortable with all of the different tools that we have to use - all that... everything that we know, whatever it is that we can bring to solve a particular problem regarding a maximum and minimum.0030

Remembering all along that really what we are trying to find is those points where the function obtains a maximum, and obtains a minimum.0045

Ultimately this is really just a straight mathematics problem. Nothing more than that. We definitely do not want to lose the forest from the trees.0053

Okay. Let us get started. Our first example is going to be, so we will say let f(x,y) = x + y. Nice, simple function of 2 variables.0062

So, we want to find the max and min values of f on the square with coordinates + or -1, + or -1.0080

Okay, this is what this means. So, we have a square at... sorry... (1,1), (1,-1), (-1,1), (-1,-1).0118

So we have this square, and this is closed. This is definitely a closed and bounded set, so in the entire x, y plane, our domain is just what is inside here and what is on the boundary.0135

The theorem says that the function on a closed and bounded set contains a maximum and a minimum. Our job is to find that maximum and minimum, the points on there and the actual value of the function.0151

So, we have an interior and we have the boundary, so let us just go ahead and do what we do.0165

The first thing we do is, of course, take the partial derivatives, set them equal to 0. If they are equal to 0, then start with the local max's. In other words, the interior points.0170

So when I do df/dx, I get 1, and if I do df/dy, I also get 1, so as it turns out, this particular function on the entire x, y plane, there is nowhere... there is no critical point, because there is nowhere that the derivative is equal to 0.0179

What this tells me is that if there is a maximum and a minimum, which there is, it is guaranteed by the last theorem, by the last lecture, that maximum and min occur somewhere on the boundary.0203

It does not occur anywhere inside. It is going to happen somewhere on here.0214

Now, I am going to end up having to evaluate the function on the boundary. That is what is going on here.0221

Well, let us just start. So what I am going to do is I am going to call this line segment on the left, I am going to call it C1, and I am going to call this line segment here C2, and I am going to call this one C3, and I am going to call this one C4.0229

I am going to evaluate the values of the function all along all of these boundaries and see which number gives me the biggest value. That is how I do it. 0245

So, let us go ahead and let us see. Let us check the boundary points.0257

Well, let us do C1 first. So C1, every single point on here has the coordinate (-1,y), okay?0271

So, f(-1,y) has a max at (-1,1), and the value of f(-1,1) is equal to 0.0285

Well, it has a min at (-1,-1) and the value at (-1,-1) equals -2. That is what I am doing.0308

I am just evaluating, now, everything on the boundary. Once I collect all of my values of all the different values of the function on the boundaries, the max's and the min's, I am just going to pick the biggest, and I am going to pick the smallest.0325

Now, let us try C2. So on here, the max was up here, and the min was here. Everything else was just in between. Again, we do not care about what was in between, we just about the max and the min, somewhere on there.0340

So, here, all of the points along this particular line have the y coordinate equal to -1, so f(x,-1). Okay.0355

So, f of all these points that have the same y coordinate, it is the x value that varies.0372

So, f(x,-1), it has a max at 1... no, wait... 1, 1... wait a minute... where I am I? let us see... 2... 1, 1, -1, oh, you know what? I actually... oops I think I made a little mistake here. Let me go ahead and reparameterize.0377

Let me go ahead and put C2 here, C3 here, and C4 here. So C2, I am actually going to be doing this line.0411

So, same thing. Instead of x,-1, it is actually going to be (x,1). So y value is 1 and x varies as we move from left to right.0421

So the maximum is going to be at (1,1), in other words, that point, and the value at (1,1) is going to be 2 and it is going to take its minimum value on this line over there. 0433

So, it has a min at (-1,1), and f(-1,1), oh well when I put it in f, I get 0.0450

Let me go ahead and move on to the next page, and let me rewrite f again, so that I have it. It is going to be x + y, and I am also going to redraw my square real quickly, just because I would like to know what it is that I am looking at.0468

So I think we had C1, C2, C3, and C4. Okay. So now let us do C3. Now we are going to check the values along C3.0489

The coordinate of C3, the x value is going to be fixed at 1, and it is the y value that is going to vary.0500

So here, it is going to be (1,y), something like that. So f(1,y), well it has a max at (1,1) and f(1,1) is 2.0509

It has a min at (1,-1), which is this point and the value is 0. I am just doing this systematically. Just go through every single... cover the entire boundary, just make sure that all of the points are covered. 0530

Now we will do C4, which is write down there, and the coordinates for C4, well the y value is fixed at -1, and it looks like the x value that is actually going to vary. 0553

f(x,-1), that has a max at... it looks like that is the point at (1,-1), and the value of (1,-1) is equal to 0.0565

Then, it has a min at (-1,-1), which is that point. The value of (-1,-1) is equal to 2, or -2... sorry about that.0593

So, on the closed and bounded region, let us just call this s on that square, so on the closed and bounded region s, the maximum of f if I look through all of these values, the maximum is going to be 2, and the min is going to be -2. That is it.0614

Okay. Let us do another example here. Let us actually go do a different page. Example 2.0644

This time we want to find the extrema for f(x,y) = x2 + 2y2 - x.0656

Let us see what is going to happen here. Well, let us go ahead and start by doing what we always do. Now notice that in this particular case they just said find the extrema of the function f(x,y) = x2 + 2y2 - x.0680

They did not specify a domain, they did not actually specify some region. They did not say whether it was open or closed. When they do not do that, they are just saying over the entire x, y plane. 0694

That is it. When they do not specify a region, your assumption is just that it is over the entire plane. 0705

So, in this particular, we are going to be looking for absolute max, but it is actually going to end up being a local max, or an absolute min, or a local min, because we do not have a boundary that is keeping us contained. 0710

Any max or min that takes place is going to end up being some critical point. That is what is going on here.0726

So, let us do what we always do. Let us go ahead and find the... so we will do df/dx, so the partial derivative with respect to x, it looks like it is going to be 2x-1.0732

The partial derivative with respect to y... it looks like it is going to be 4y.0744

So, we set each of those equal to 0, so we end up with y = 0 as the y point, and 2x-1, x is going to equal 1/2. There you go.0753

Our critical value... so let us write, our critical value equals the point (1/2,0). That (1/2,0), the derivative = 0. It is a critical point. 0765

Now we just need to find out if it is a max, or if it is a min. So, here is where you are have to sort of stop and make some decisions about how you are going to proceed. 0779

Are you going to do the same thing that you did before? Where you take the point (1/2,0), also it is going to be some like that and we can do what we did in one of the examples from the previous lesson.0792

We can hold 1 variable fixed and move a little bit to the left, move a little bit to the right and hold y fixed, or we can hold x fixed and we will move a little bit up and a little bit down, and we will check the value of the function around that critical point to see how the function behaves.0802

To see if all of the values around it are bigger than what the value of the function is there, or less than what the value of the function is there. To see if it is actually a maximum or minimum.0821

First of all, let us go ahead and find out what the value of the function is at (1/2,0).0833

So f(1/2,0) is going to equal... 1/22 is 1/4, let us see... 0, this is 0... -1/2 equals -1/4, so the value of the function at the critical point is -1/4.0839

Here is where you stop and think about what you are going to do. Are you going to go through this process? Which you can, it is not a problem, you will get the answer. It is perfectly acceptable to do that.0859

But in this case, let us look at the function. We know the value is -1/4.0870

Well, take a look at the first two terms of this function. You have got x2 and you have got y2.0876

So, in this particular case, both of these terms, no matter what value of x you put in, no matter what value of y you put in, you are always going to end up with positive terms here.0884

Because they are positive terms, you are just subtracting this final x, you are always going to end up getting some number. In other words f(x) is always going to be bigger than f(1/2,0).0898

So these are positive. That is what is going on here. Just sort of by looking at this function and stopping and thinking about it, here is what you can do.0911

So, always positive, so... because this and this are squared terms... let me write it again actually.0921

So, f(x,y) = x2 + 2y2 - x. This is always going to be a positive term. This is always positive. Therefore, f(x,y) is always going to be > or = to 1/4, which is the value of the function there. 0931

So no matter if I move this way, this way, this way, this way, no matter which way I move, because these are both squared terms and they are both positive and it is always going to be a value that is bigger than -1/4.0963

Therefore, this is actually a minimum. In other words, all of the values around the function are going to end up being bigger... so it is going to be something like this. It is going to go that way.0975

If I hold the other variable fixed, it is going to go that way. So this is a minimum. 0988

So, min at (1/2,0) = -1/4, so this is where the minimum takes place. The value of the minimum is -1/4, so that is it.0995

Again, you are welcome to go ahead and go through the process that we did of holding a variable fixed, moving a little this way, moving a little this way, you are going to get the same answer.1013

In this case it was nice where we had a particular function, look at it, stop and think for a second, and say, Okay, well both of those terms are positive therefore at that value it is always going to be -- the value of the function is always going to be -- bigger than -1/4.1021

That is what is happening. Let us do another example. Let us see, example 3.1038

Find the max and min of f(x,y) = (x2 + y2)-1.1055

Okay, so slightly more complicated function here. In the region -- we are going to add one more bit of complication here -- in the region x-22 + y2 is less than or equal to 1.1075

Now I have not only specified a function, I have specified a region except now the region is given as an equation, so this is an equation of a circle that is centered at (2,0).1096

Let us go ahead and draw this out because we can. Any time you can actually draw a picture is is a good idea. So this is 1, this is 2, so this is the center of the circle. There is that, there is that, and there is 3.1107

So, we have a circle and it is less than or equal to 1 of radius 1, okay? It is closed, this is a closed and bounded region.1122

What we are saying here is on this region, all the interior points including the boundary points, what is the maximum value that this function obtains, and what is the minimum value that this function obtains. That is all that we are doing here.1133

Well, let us just start and see where we go. So, let us just take the df/dx.1148

When we take the first partial derivative, this is going to equal -x2 + y2 × the derivative of what is inside, which is 2x.1154

df/dy = -x2 + y2 × 2y. Again, we are just taking the derivative of this function, this is a composite function, so we take -1 × this... -2 power, sorry about that, and then of course we take the derivative of what is inside.1171

Since we are differentiating with respect to y, it is just 2y here. The first one we differentiated with respect to x.1194

Not notice, take a look at this. Well, for one thing, the function is not even defined at (0,0). You cannot even have that because this is... okay, let me rewrite this, so this is going to be (-2x/x2 + y2)2,1202

This is going to be -2y/(x2 + y2)2.1220

First of all, this function is not even defined at (0,0), however for every other value of x and y, the derivative never equals 0.1229

So, df/dx does not equal 0 anywhere on the domain, anywhere in the region.1240

Not that we were concerned with (0,0) because (0,0) is not even part of this. But anywhere on this region, the derivative does not equal 0, and neither does df/dy.1250

df/dy does not equal 0 anywhere on the region in question, because the derivative does not equal 0 anywhere on that region, we know that it is not going to take place anywhere on the interior.1260

So, the max and min of this function is going to take place on the boundary. It is the boundary that we need to concern ourselves with.1276

Here is where it starts to get kind of interesting. So, let us go through this very, very carefully.1282

Now, let me go ahead and move on to another page. I am actually going to redraw it so that I have it available for me.1290

I have this 1, 2, 3, just do a quick drawing... so, I have that. The unit circle centered at 2.1298

So, the boundary is x - 22 + y2 = 1. That is the boundary. The region was < or = to 1, but the boundary is at 1.1308

How can we find the values of the function -- let me rewrite the function -- so, find values of f(x,y), which is equal to x2 + y2... let me write it actually as a fraction so that we have positive exponents.1328

1/(x2 + y2)1. So that is actually what we are looking... how can we find the values of this function on this circle?1356

The value of this function on this circle. This is a composite function. I need to be able to take this, I need to be able to form f, so this is some... I need to be able to form f(g). This is g, this is f. I need to form the composite function.1370

Here is what I am going to do. I am going to go ahead and parameterize this circle, and express it as a function of one variable, t.1390

Let us go ahead and parameterize this circle. Well, I know what the parameterization of a circle is in general.1397

So, a general circle, the parameterization is just c(t) = cos(t)sin(t), right? That is the parameterization of a unit circle centered at the origin. But this one is not centered at the origin, it is centered at 2.1403

My parameterization for this function here, x - 22 + y2 = 1, looks like this. Equals 2 + cos(t)sin(t).1420

If you were to put 0, pi/2, 3pi/2, and 2pi into t, you will get this point, this point, this point, and this point. That is your parameterization moving in that direction.1434

This is a parameterization of this function. All I have done is taken an implicit function, and I have expressed it parametrically.1448

Now I can actually form f(c(t)). That is what is actually going on here. So, c(t) is this, this is f, I am going to form f(c(t)).1458

f(c(t)) is equal to, well, it is going to be 1/x2, well x is this. That is x, that is y, so it is going to be 2 + cos(t)2 + sin(t)2.1470

Now we are going to go ahead and actually multiply everything out. It is going to equal 1 over, well, 2 × 2 is 4, 2cos(t)2cos(t) is + 4cos(t), + cos2(t) + sin2(t). That is the denominator.1500

Well, cos2(t) + sin2(t) is the pythagorean identity. That is equal to 1, so 4 + 1.1523

So my f(c(t)) = 1/5 + 4cos(t).1530

There we go. I found a way of expressing my function as a single variable, a function of t, this is my function. This is f(x,y).1546

Now, all I need to do is find the values of t that make this a maximum, and find the values of t that make this a minimum.1557

Well, this is actually really, really simple here. Notice this cos(t).1565

Let us just... well, let us just stop here. Let us just say f is a max, the function chooses its maximum value when the denominator is the smallest and f achieves this number here.1572

Achieves a minimum value, a smallest value, when the denominator is the biggest, right?1600

This is the function, big denominator, small value. Small denominator, big value. That is what is going on here.1615

Now let us go ahead and find what this is. Well, what do we know about cos(t)?1625

Well, cos(t) has its biggest value of 1, and its smallest value at -1. It ranges between -1 and 1.1629

Therefore, it is going to achieve... so cos(t) = 1, when t=0... so we want this value here. We want to achieve the biggest and smallest value. Well, we want to achieve 1, and we want to achieve -1.1640

Okay. It is equal to 1 when we have cos(t) = 0, therefore when t = 0, that means the cos(0) is 1. That means 1 + 5 × 4 × 1. It is going to be 1/9.1666

Therefore at t=0, that is when it is going to achieve its minimum value.1693

Now the minimum value of cos(t)... cos(t) achieves -1 when it is equal to pi. So when t = pi, the cos is equal to -1.1699

Well -1 × 4 is -4, 5 -4 is 1, so now it achieves its maximum value there. So let us actually just write this out.1711

When t = 0, c(t) = 2 + cos(0)sin(0) = (3,0). That is the point at which it is going to achieve one of its extrema value.1728

Well, f(c(t)) is going to equal 1/5 + 4, which equals 1/9. That is the minimum.1752

Now, when t = pi, cos(t) = -1. c(t) = 2 + cos(pi) × sin(pi), which is equal to (1,0).1768

When I put (1,0) into that function, f(c(t)) = 1/1, because remember, we had 1/5 + 4 × cos(t). 5 + 4 × cos(t).1795

When t = 0, it is 1/9, so it is 1/9. The value of the function at the point (3,0) is equal to 1/9.1821

When t = pi, that is when it is at its smallest. The smallest value of cos is when t = pi.1833

When t = pi, well cos(pi) is -1, 4 × -1 is -4, 5 - 4 is 1. So the value of the function is 1.1840

At the point (1,0), the value of the function is 1. That is the maximum.1851

At the point (3,0), the value of the function is 1/9. That is the minimum.1859

Here is what we have done. Here is our circle. 1, 2, 3, we have a circle here. That looks like an ellipse, so let us make it look a little bit like a circle. 1863

We have this circle, so at the point (1,0), here is where it achieves its maximum value of the function. At the point (3,0), here is where the function actually achieves its minimum value. 1873

We did this by just taking our function, parameterizing it, putting it into our region, parameterizing the curve, putting that parameterization, forming (f(c(t)) like we have been doing for a while, and then we just found the values of t that make the function a maximum or a minimum.1886

The only real problem here is keeping track of all of the mathematics. There is a lot on the page.1909

There are a lot of numbers floating around, there is a lot of points floating around. The only thing that you have to do is make sure that you differentiate between the point at which the maximum or minimum is achieved.1914

In other words, these points. Those are the points at which they are achieved, and the actual value of the function at those points, that is here, and that is here.1925

Again, we want to rely on the techniques that we have at our disposal, but really what we want to rely on is our intuition and what we know about functions and how they behave and how numbers work.1935

When you get to this point, you can reason out the rest. When this is the biggest, the number is going to be the smallest. When this is the smallest, this number is going ot be the biggest.1947

You just need to go step by step and make sure that everything is there. Okay. I hope that this has worked out and everything is well.1958

Thank you for joining us here at educator.com and multivariable calculus. We will see you next time.1963

Hello and welcome back to educator.com and Multivariable Calculus.0000

Today's topic is going to be Lagrange multipliers. This is a very, very important topic.0004

Because it is very important, we are actually going to be spending several lessons on it. In this particular lesson, what I am going to do is write out the theorem and the method.0009

Then we will start with a couple of basic examples, just to get the underlying technique under our belts. Then the following lesson, I am actually going to be spending some time on just more examples, slightly more complicated, a little bit more involved. 0019

Again, this a very powerful technique for finding maxima, minima, on a restrained... the maximum and minimum of a function constrained in some sort of way.0033

So, In and of itself, it is not that difficult because, really all you are doing is you are solving a bunch of simultaneous equations. Whether it be 2 equations, 3, 4, or 5 depending on how many variables you are working with.0044

The problem with this is there just tends to be a lot of things on the page, so there is a lot to keep track of, so again, just go slowly, make sure everything is written out, do not take any short cuts, aside from that let us just jump right on in and get a sense of what is going on.0055

I am just going to start out by writing the theorem and then from the theorem I am going to use that just to write out the quick method, then we will go ahead and start the examples.0074

The theorem is going to be a bit on the long side, but it is important, so let g be a continuously differentiable function -- all that means is when you differentiate it, you get a continuous function -- continuously differentiable function, on an open set u.0085

Now, let a be the set of points in u. Let me actually give the points a name, the set if points x in u, so we are talking about a vector, a point in n space, such that g of x equals 0.0131

This is a very, very important hypothesis, but the gradient of g at x does not equal 0. So it is just really, really important that on this particular set of points that we are dealing with that this gradient of this function g not be 0. 0163

Essentially we are just saying that it is smooth. That is all. Now, let f be a continuously differentiable function on u, and let p be a point in a, such that p is an extremum, in other words it is a maximum or minimum, an extremum for f.0185

That is, let p... let us write it this way... p is an extremum for f subject to -- let us not use the word subject, let us use the word constrained, because that is exactly what we are doing for f -- constrained by g.0245

Then... and again do not worry about this, this is just a formality to write it, it is just something that we might occasionally refer to, it is the method that is important, but I want you to see the formality.0287

Then, there exists a number -- traditionally we use the Greek letter lambda for that -- such that... here is the important part... the gradient of f evaluated at p = λ × the gradient of g evaluated at p.0297

So, as far as what this means geometrically, I will be talking about that a little bit later in subsequent lessons. 0326

Probably not the next lesson, but the one after when we do... after we have done all of our examples, we are actually going to go back and think about them geometrically and I am going to talk about what this really means.0332

But again, right now I just want to get the technique under our belts. Just getting used to how we use Lagrange multipliers to find the maxima and minima on functions constrained by a certain other function, g.0343

Now, since this is the formal theorem, so these are the important parts right here, the gradient -- and again we are only concerning ourselves with situations where the gradient g at a given x is not equal to 0 -- again, smooth g.0358

Later on, if you go on into higher mathematics, you will talk about cases where this is not the case, but for right now we want to keep it nice and straight forward.0373

So this is really the important thing right here. Gradient of f evaluated at p = λ × gradient of g evaluated at p where p happens to be the max or min for the function f.0381

Now, let us describe the actual method and let us do some examples.0395

Let, and again, we are mostly going to be working in 2 and 3 variables just to keep things normal for us, so, let f(x,y,z) and g(x,y,z) be defined and continuously differentiable.0403

To find the max and min -- or min -- values of f subject to the constraint g, find the values of x, y, z, and λ that simultaneously satisfy the equation, this thing right here.0438

Gradient of f = λ × the gradient of g, and g(x,y,z). 0497

So in other words, what we are going to do is we are going to actually take... we are going to form this equation and we are going to get a series of equations from there and then we are also going to deal with the equation g(x,y,z) or g = 0.0515

When we solve those simultaneous equations, we are going to get values for x, y, z, and lambda, where the x, y, z, those are the choices that allow for the x, y, z.0525

Those are the choices that allow for max and mins. We evaluate that function at the x, y, z, to see which number is highest, which number is lowest. That is all that is going on here.0532

So let us just go ahead and jump into some examples and I think it will start to make sense, as well, when we look at examples. 0540

Again, we are going to do a large number of these examples. This is very, very important. Example 1.0549

There tends to be a lot going on with Lagrange multipliers. Lots of symbols on the page and sometimes it is not very easy to keep track of, nothing difficult, just a lot to keep track of.0560

Okay. Find the largest and smallest values that f achieves subject to g. 0572

Well, f(x,y) = the function xy, and g(x,y) = x2/8 + y2/2 = 1.0600

Let us just stop and think about what this means when we say find the largest and smallest values that f achieves subject to g. 0618

In the previous lessons when we talked about max, min. We either talked about a closed domain, or we talked about an open set. Maybe all of -- you know -- our particular space that we were dealing with. 0626

Here what we are doing is we are saying find the maximum and minimum value of the function, but we are going to put a constraint on that. We want the values of x and y to satisfy this equation also.0639

That is what we mean by a function f constrained by g.0649

In other words, this function, you know the function x, y is perfectly defined on here except... well actually no, this is defined everywhere... but here what they are saying is this is an equation of an ellipse.0653

They want us to find the values of x and y. So x and y have to be on this ellipse and then of all the points x and y on this ellipse, which ones maximize the function, which ones minimize the function. That is what we are talking about.0671

We are constraining the x and y values. We are saying that is not just any x or any y. I am going to put a constraint on the x and y. They also have to satisfy this equation, that is all that is going on.0688

In this case it is just one constraint, you can have two, three, four constraints, as many as you need depending on the problem.0699

Okay. So find the largest and smallest values of f(g) subject to g. That is our f, that is our g, well, let us just go ahead and do it.0705

What we are going to do... we are going to find the gradient of f, and then we are going to set it equal to λ × the gradient of g, and then we are going to solve a series of equations along with this equation equal to 0. 0713

Alright. So let us do the gradient of f. Let us just do this systematically. Let us see, so df/dx is equal to y, and df/dy is equal to x.0732

I am going to write this as a column vector. Again, the gradient is a vector, so instead of writing it horizontally, I am going to write it vertically. You will see why in a minute... y, x.0750

I mean it is a personal choice for me simply because I like the way it looks on a page. You can arrange it in whatever way looks good to you and what is comfortable.0760

So, y, x. So now let us do dg/dx, so let us find the gradient, alright? The partial derivative with respect to x is going to be 2x/8, which is x/4.0769

dg/dy is going to equal to 2y/2, that is equal to y.0791

This one is x/4 and y. I will write that as a column vector, so now we are going to form the gradient of f equals λ × gradient of g. So we write y, x, equals λ × well x/4y, which equals... we multiply the λ across both.0794

So we get λ × x over 4, and we get λ y. Here we go. Now we have our 2 equations.0828

Our 2 equations are... let me do this in red... this thing equal to that thing, so we have y is equal to λ × x over 4, and we have this thing equal to this thing.0838

That is why I arranged it in a column, it is just a little bit easier for me to the correspondence. The first entry - first entry, second entry - second entry.0855

I also have x = λ × y, and of course my third equation which is... remember we said the third equation in this series of equations is this, this, the gradient of f = λ × gradient of g, and of course we have the have x2/8 + y2/2 - 1 = 0.0864

We want g to equal 0, so that is why I brought the 1 over onto this side. That is all I am doing.0893

Okay. So now, let us go ahead and well, let us see what we can do. Let us solve. This is the system that we have to solve, so we need to find λ and we need to find x and we need to find y.0899

So, let us see what we have got. Let me rewrite these... y = λx/4, and I have got x = λy and I have x2/8 + y2/2 equals -1 = 0.0916

So, the x, the y and the λ they have to satisfy all three of these equations.0948

Let me go ahead and take care of this one, so let me see. 1 - λx... y, I am going to move this over... so, y - λx/4 = 0.0952

Well here, x = λ y, so I am going to go ahead and put this x into here. That gives me y - λ × λy/4 = 0.0965

I get y - λ2y/4 = 0. Let me come over here so that I can use more of the page. Let me go ahead and factor out a y.0984

I get y × 1 - λ2/4 = 0, so I have two solutions for y.0994

I have y = 0 -- excuse me... let me see here -- and 1 - λ2/4 = 0, so I get 1 = λ2/4 = 0. That means that λ + or - 2.1004

Now. I have two possibilities. I have y = 0, λ = + or - 2. Let us just deal with one case at a time. That is what you are going to be doing with these Lagrange multiplier problems. 1029

You are just going to be dealing with one case at a time. This is where it starts to get a little interesting. You have to be very careful to choose your cases properly.1041

So, case 1... we have y = 0. Well, if y = 0 this implies that x = λ × y, x = λ × 0. That means that x = 0.1047

Okay. That is one possibility. Now we have the point (0,0), but now the problem is we also have to satisfy this third equation.1072

If I put this equation x, 0, y, 0 into this third equation, x2/8 + y2/2, 0 and 0. What I end up with is... well, the thing is this point (0,0), is not on this ellipse. That is a problem. 1079

Even though we ended up with this possibility, because it is actually not on this ellipse, it is not part of the... we cannot use this point, we cannot test this point, so this one is out. We do not have to deal with that. 1097

Now we will deal with the other case. The other case where λ = + or - 2. 1109

So, case 2. λ = + or - 2. Now, let us see... from the equation... from this equation right here, x = λ × y.1116

We get x = + or - 2 × y. Okay, so now, I am going to go ahead and take this and the value of y and I am going to use this equation.1140

I am going to form, so it is going to be + or - 2y, I am basically going to put this value of x into x2 into here... squared over 8 _ y2/2 - 1 = 0.1160

Here I am going to get 4y2/8 + y2/2 - 1 = 0, and I am going to get... I am going to just multiply everything by 8 here, so I am going to end up with 4... let me do it over here... going to end up with 4y2 + 4y2 - 8 = 0.1179

So, I am going to get 8Y2 - 8 = 0, I am going to get y2 = 1, which implies that y = + or - 1.1217

Now, if y = + or - 1, and I go back to one of my original equations, that implies that x is equal to + or - 2. There we go. I have found x's and y's that actually satisfy this third equation, so these values work.1233

Now I have 4 possibilities. My first point is (+2,+1). My second point is (-2,+1). My third possibility is (+2,-1), and my fourth point is (-2,-1).1254

Now what I am going to do is I am going to evaluate the function at those points. So that is what I am doing.1277

So, f(p1) is going to equal, 2 × 1 is 2, f(p2) is going to equal -2, f(p3) is going to be -2, and f(p4) is going to equal 2.1283

That is it. Our max's are achieve there. Our mins are achieved here. A maximum at (2,1) and (-2,-1), and the minimum of that and that. That is the method of Lagrange multipliers.1303

Let us do another example. Okay. So, let us see here. Example 2... oops, let me go back to black ink, actually... well, actually, you know what? Let us do this one in blue. How is that? Okay. Example 2.1319

Find the max and min values of f, let me do it over here, f(x,y) = 3x + 4y on the circle x2 + y2 = 1.1343

okay, so one thing you should know about these max, min problems... and we will get more of a taste for this when we do some more examples in the next lesson... the problems are not written out the same way.1383

Oftentimes you are given information that looks like a Lagrange multiplier problem, but you have to extract information. In other words you have to extract what f is, extract what g is. It is not always going to say find the maximum and minimum values of f subject to g.1391

It is not going to be as explicit like this, but again we will see more examples like that.1410

okay, so find the max and min values of this function on the circle x2 + y2. So all this means is that this function, 3x + 4y is defined for the entire plane, but we want to constrain it.1415

We want to find the max and min values on the circle. So, somewhere, some point on this circle is going to maximize this function and some point on this circle is going to minimize this function.1430

So f subject to the constraint g. Okay, let us see here. What shall we do first? Well, okay. We are going to do what we always do.1441

We are going to take the gradient of f, and we are going to set it equal to λ × the gradient of g, okay?1460

We are also going to have this set of equations, and we are also going ot have g, in this case x, y = 0. That is our last equation. We have to satisfy these two equations... or this set of equations.1467

So, the gradient of f, this one is going to be... so the partial derivative with respect to x is going to be 3, the partial derivative with respect to y is 4, so we have 3, 4 = λ × gradient of g. 1483

That is going to be 2x -- let me make this a little bit better -- Is going to be 2x... and this is going to be 2y, so we have λ... or I will write 2λx and 2λy. So there you go.1500

This corresponds to that, this corresponds to that, so we have the equation 3 = 2 × λx, and we have 4 = 2 × λy.1519

Yes, that is exactly right... and of course we have the g(x,y) = 0, so we are going to get x2 + y2 - 1 = 0.1536

This is our set of three equations and three unknowns, x, y, and λ.1548

We have three equations and three unknowns, theoretically this is solvable. Now we just have to find x, y, and λ... ultimately x and y, but we have to find λ along the way. 1553

So let us see what we have got here. I think the best way to approach this is since we have this equation, and we have this, I am just going to go ahead and solve each of these equations, 1 for x and 1 for y.1563

So in this particular case, x is going to equal... well I am going to divide by 2λ, so it is going to be 3/2λ, and y, when I divide by 2λ, it is going to be 4/2λ, which is equal to 2/λ.1581

So now I have x... oops, these crazy lines showing up all over again, alright... λ, this is 4... so I have x and I have y, and now I am going to take these values of x and y, and I am actually going to put them into this equation to see what I get for λ.1600

So let us go ahead and do that. Let us do that on the next page. So, I have the function x2 + y2 + 1 = 0, and we said that x = 3/2λ.1619

This is going to be 3/2λ2 - 1 = 0. So let us go ahead and work all of this out. 9/4λ2 + 4/λ2 -1 = 0. We are not going to leave anything out here.1634

I am going to go ahead and multiply through by 4λ2, I think... no, just 4... yea, 4λ2.1658

So, when I multiply by 4λ2, I am going to get 9 over here, I am going to get 16 over here, -4λ2 = 0, just to get rid of the denominator, that is all I did. 1672

9 + 16 is 25... that equals 4λ2, so λ2 = 25/4, therefore λ = + or - 5/2. Okay, so we found λ, now it should be not a problem.1687

Now that we have found λ, well, x = 3/2λ, so when I put that in there, that is going to end up equaling... well, let us do it all, let us not miss anything here... 3/2 × + or - 5/2. It is going to end up equaling + or - 3/5.1707

Now, y = 2/λ, which is equal to 2/ + or - 5/2, which equals -4/5, and I hope I have done my arithmetic correctly.1733

Now, let us stop and think about this. We have + and - 3/5 for x, and we have + or - 4/5 for y.1750

You are probably thinking to yourself, just like the previous problem, that we have 4 points: (3/5,3/5), (3/5,-4/5), (-3/5,4/5), (-3/5,4/5). 1756

That is actually not the case. This is where you have to sort of look at other things. There is other analyses going on here.1770

x = 3/2λ, y = 2/λ. x and y have the same sign, so they are either both positive, or both negative. So, we do not have 4 points to pick, we only have 2 points to pick, 1 in the first quadrant, 1 in the third quadrant.1779

That is what is going on here. You can go ahead and use the others, it is not a problem, you will go ahead and get the answer... but you know, just something to be aware of.1793

So, that is it. Basically our points that we are going to pick are... let us see... (3/5,4/5), that is one possibility... and (-3/5,-4/5), that is the other possibility.1802

Okay. When I take... let us just call this P1, and let us call this P2... so when I take f(p1), in other words I put it back into the function 3 × 3x + 4y, 3 × 3/5... well you know what, let me work it all out. It is probably a good idea if I work it all out.1831

I will do it on the next page, so we will do f(3/5,4/5), that is going to equal 3 × 3/5 + 4 × 4/5 = 25/5... that is going to equal 5... and f(-3/5,-4/5) = 3 × -3/5 + 4 × -4/5 = -25/5 = -5.1858

So, here, at the point (3/5,4/5), it achieves a maximum -- the maximum value is 5 -- and (-3/5,-4/5), the function f achieves a minimum of -5.1899

We have found the maximum and minimum values of 3x + 4y, subject to the constraint that x and y lie on the unit circle. x2 + y2 = 1, that is what is going on.1912

So, again, in the next lesson we are going to continue on with more examples of Lagrange multipliers because we want to be very, very, very familiar with this.1926

Then, after that, we will pull back a little bit and take a look at exactly what is going on.1934

We want to make sure that you actually understand why this is the case, and why this works.1938

Again, nothing theoretical, we just want to make it plausible for you, that this is not some technique that just drops out of the sky.1943

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.1949

Hello, and welcome back to educator.com, and welcome back to multivariable calculus.0000

Last lesson we introduced vectors, points, and space.0005

Now, we are going to start operating with vectors, we are going to start doing things with vectors.0007

We are going to start doing things to them, multiplying them and seeing how long they are, so that we can really start getting into more of the deeper mathematics and the basic properties that we are going to be using over and over again throughout this course.0011

Today, we are going to talk about something called the scalar product, also known as the dot product, and the norm.0025

The norm is just a fancy word for length, the distance.0030

So, when we are talking about the norm of the vector, we are just saying "how long is that vector?"0034

We are accustomed to dealing with 2 and 3 space, but remember this is true in any dimension.0037

This is why when we say scalar product norm, we are talking about a vector in any number of dimensions, it turns out.0043

A vector in 15 space does have a length.0051

We cannot visualize that length, or a drawing of any kind, but it does exist.0054

It is a real thing.0060

Having said that, let us just jump right on in.0062

Again, so that we can continue to use our intuition, which we want to be able to use geometrically to help our algebra, we are mostly going to be working in 2 and 3 space in our examples.0067

Let us start off by defining a couple of vectors here.0080

Let a = (a1, a2, a3) and the vector b = (b1, b2, b3).0086

Okay, so define their scalar product or dot product, a · b, this is the symbol for the dot product or scalar product, = a1 × b1 + a2 × b2 + a3 × b3.0106

Now, let us just talk a little before we do the example. 0138

First of all, a product, you know that if I take two numbers, like 14 and 2, the product is 28.0142

As it turns out, I am with vectors, so now we have turned on to bigger more complicated objects in mathematics.0150

Vectors have their components as numbers, but they are treated as individual objects.0158

I can take a vector × a vector.0162

As it turns out, there is more than 1 way to multiply 2 vectors.0165

In fact there are any number of ways to multiply two vectors if you like.0169

One of the ways is the scalar product, the dot product.0173

Later on, we will learn something called the vector product, or the cross product.0176

So, now the operations start to become a little bit more complex because the objects themselves start to become a little bit more complex.0181

But, you can see, because vectors are represented by components, the definition of the products is slightly more complicated, but it is still just a multiplication of numbers added together. 0191

So, let us just do an example.0202

Again, the scalar product, given two vectors, (a1, a2, a3) and (b1, b2, b3), you are just going to take the product of the first components + the product of the second components + the product of the third components.0207

By product, we just mean numbers multiplied by numbers.0218

So, example 1.0223

We will let a = (-3,4,2), and we will let b = (4,2,1).0227

Well, the dot product, a · b is equal to -- I will write this out explicitly -- -3 × 4 + 4 × 2 + 2 × 1.0239

Then, I go ahead and equal them over here, equals -12 + 8 + 2, that is going to equal -2.0258

Okay, so we have two vectors which are points in space. 0268

We multiply those vectors by some multiplication that we call the scalar product or dot product.0272

We end up with a number.0279

Notice what we did, this is very important.0281

We took two vectors, which are not numbers, and we ended up with a number.0285

That is why in fact we call it the scalar product.0291

A scalar is just a fancy word for a number.0295

It is something that just has a magnitude, in other words, a value. 0299

A vector in of itself has a magnitude, a length, and it has a direction, but this is a scalar.0303

So, we took two vectors, we multiplied them and we ended up with a scalar.0312

That is really extraordinary.0315

Let me write that out formally.0319

Notice that this is a number, not another vector.0321

When we added 2 vectors together, we took a vector, and we took another vector, we added them together, and we ended up with a vector.0336

In other words we stayed in the space of vectors. 0341

Here we took a vector, we multiplied it by a vector, this procedure, and we ended up with a number. 0345

We did not end up with another vector, we jumped to another space, the space of real numbers.0350

I would say that that is really extraordinary. 0356

I have these in mind, I mean it is not ultimately important when we do the specific problems, but it is really sort of extraordinary to think about things globally -- that we are jumping around from space to space sometimes. 0360

Okay, so the scalar product also has some properties, let us go ahead and list those.0374

These I will number.0383

1) a·b = b·a, so if I want to take the scalar product, it is commutative, I can actually do the product in any order.0384

As it turns out later, it is not true for other vector products.0394

2) a·(b + c) = a·b + a·c, so as it turns out, the dot product is actually distributive over addition.0402

3) If I take some constant c and multiply it by vector a, and of course get another vector and I dot that with another vector, I can actually pull the constant out, dot the vectors first, then multiply it by a constant, it = c × a · b.0420

That is all that means, it means I can switch the order of constants and vector multiplication.0440

4) Okay, the 0 vector dotted by the 0 vector is equal to 0, notice the notation, 0 vector, 0 vector, 0 number, because the dot product is equal to another number, not another vector.0446

Last but not least, this is actually related to number 4, if a does not equal the 0 vector, then a dotted with itself is greater than 0 -- this is a very important thing.0460

A vector dotted with itself is a positive number. 0477

Now, there is a, well, you know what, I do not need to write this down, I can go ahead and tell you.0485

I do not want to Waste space and waste time here.0497

There is a geometric interpretation for what the dot product is.0499

In other words we can draw something out in the plane and tell you what the dot product means visually.0503

We will get to that, actually in not too far from now.0510

However, again what we want to do is we want to concentrate and we want to develop our algebraic abilities.0513

When we give a definition, which we did for scalar product, the definition was very clear. 0518

It tells you to follow this procedure to get this thing, this scalar product.0522

We want to develop algebraic capacities -- that is where the real mathematics takes place.0529

We use geometry to guide some of our intuition, obviously we do not want to throw out our intuition, we want to use it.0534

But again, for those of you who have taken a look at the linear algebra, intuition does not always guide you to mathematical truth.0542

You have to trust the mathematics.0548

At this level, that is often the case.0551

So we will talk about the geometric meaning of dot product but we will deal with it algebraically.0553

That is very important.0560

Now, let us talk about an interesting little property here.0564

You know that if I have x × y = 0, in other words two numbers equal to 0,0566

Then you know that either x or y = 0.0576

In other words you cannot have two numbers that multiplied together equal 0, one of them has to equal 0.0580

That is the whole foundation for factoring.0585

That is the whole foundation for finding the roots of polynomial equations.0589

It is the idea that you can set each factor equal to 0. 0591

You know that either x or y is 0.0595

This is not true for vectors, as it turns out.0610

In other words, this is not true for vectors.0616

In other words, there can be the vector a, which is not equal to 0, and a vector b, which is also not equal to 0,0627

In other words, a non-zero vector,0641

But, when I take a · b, I actually get 0.0646

As it turns out, there are certain properties.0652

As you can see, vectors share many of the properties of real numbers but they do not share this property. 0654

I can actually take the dot product of two vectors and get 0, without either a or b being equal to 0. 0659

Let us just do a quick demonstration of that.0667

If the vector a = (6,-2,4), and b = (2,2,-2),0670

If we do a · b, that is equal to 6 × 2 is 12, -2 × 2 is -4, and 4 × -2 is -8,0683

Look at that, we end up with 0 and neither one of these is the 0 vector.0697

Now, we are going to define and give a very important definition.0703

It will be a very simple thing to do, a very simple definition.0708

Do not let the simplicity of the definition make you think that it is not relevant.0713

It is actually profoundly relevant.0717

Define 2 vectors, a and b are said to be orthogonal, or perpendicular.0722

I am going to put perpendicular in parentheses. 0746

We definitely want to start to use the word orthogonal, because it is an algebraic definition and not a geometric definition. 0748

I will explain what that means in a minute.0754

a and b are said to be orthogonal, if a · b = 0.0757

Okay, so this is an algebraic definition.0767

It says that I have two vectors and if I take their scalar product, those two vectors are said to be orthogonal.0769

Orthogonal in 2 space and 3 space means that the vectors are perpendicular.0777

Perpendicular is a geometric notion.0782

Again from geometry, from trigonometry, when we think of things that are perpendicular, it just means that they make a right angle with each other.0785

Well, what if I am in 17 space.0793

A vector in 17 space, I have no way of picturing that.0795

If I have another vector in 17 space, if I take the dot product of those, the scalar product, and they it ends up being 0, how can I think about perpendicularity in that space? 0798

They do not actually make a 90 degree angle. 0807

A 90 degree angle is a geometric notion that only happens in the plane in 2 space.0810

As it turns out, orthogonal is the general term for perpendicularity.0815

It is true, you can go ahead and say perpendicular and orthogonal interchangeably, it is not a problem,0819

As long as you understand that it is the algebra that defines this.0825

Now we are moving away from just geometric notions of perpendicularity, of length.0828

We will still use some of these terms, but now we are generalizing those terms.0835

Algebraically if I take the dot product, if I multiply corresponding components of vectors and I end up with 0, those vectors are orthogonal.0840

Now, in 3 space, that just means they are perpendicular, but we will not be limiting ourselves to 3 space.0849

Let us do a quick example.0858

Example 2, we will let vector a = (1,1,-1) and vector b = (2,1,3).0860

The question is, are a and b orthogonal?0880

Well, let us take their dot product. a · b = 1 × 2 which is 2, 1 × 1 which is 1, -1 × 3 is -3. 0890

2 + 1 is 3, 3 - 3 is 0. 0904

The answer is yes, these vectors are orthogonal, they are perpendicular.0909

Okay, that is it, nice and easy.0911

The simplicity belies its importance. 0916

This notion of orthogonally is going to be very useful.0918

It is going to show up everywhere throughout the rest of the course0925

Now, let us talk about something called the norm.0930

We said at the beginning of the lesson that the norm is just a fancy word for the length.0938

Again, we think of the length as a physical thing. 0946

The length of the pen that I am holding, the length of a vector in 2 space, the length of a vector in 3 space, these things we can generalize. 0949

We can visualize them, we can draw them.0957

But what if we had a vector in 10 space?0962

How do you talk about the length of a vector in 10 space? 0963

Well this definition that I am about to give you, which is algebraic, is that.0968

The norm of a vector symbolized by a vertical line, a double vertical line, a vector surrounded by double vertical lines,0975

That is just a symbol, okay?0995

That is defined as the vector a dotted with itself, and then you take the square root of that.1001

That is it, this is a very important definition, it is the norm of a vector, the length of the vector, the magnitude of the vector.1020

All of those terms are going to be used.1029

You take a vector and you dot it with itself, so a · a, then you get a number, then you take the square root of that number. 1031

Remember, we said a · a, as long as a · a is positive, so you can take the square root of that positive number.1040

This is actually nothing more than the pythagorean theorem in any number of dimensions. 1048

That is all this is.1052

You know that if I have some triangle sort of facing the vector like this, let us just draw this out.1054

You know that if this is the x-coordinate and this is the y-coordinate, and this is (x,y), a vector in 2 space.1061

Well you know that this length here is nothing more than x2+y2 under a radical.1068

This is the exact same thing, except we generalized it to any number of dimensions, it is this, I promise.1075

Okay, let us see, I am going to actually elaborate just a little bit more on this.1086

What I did regarding the pythagorean theorem. 1098

Let a = (a1,a2).1101

Let a = the vector (a1,a2), then the norm of a = sqrt(a · a).1106

Let us see, let us draw this out, if that is our vector a.1120

This is a1, this is a2, right? This is a1, this is a2, the first coordinate, the second coordinate, along the x, along the y.1128

What is a · a, well, a · a = a12 + a22 and square root of that.1138

This squared and that squared under a square root.1157

So you can see it is a vector way of describing the Pythagorean Theorem. 1160

Now that we give a vector definition of it, it generalizes into any number of dimensions.1166

We are no longer locked into just x2 + y2 = z2, no more.1170

Let us do an example.1175

I think this is example number 3.1186

We will let a = (4,2,6), and the norm of a, I know that the notation can be a bit tedious, but definitely take the time to write it out.1190

Whatever you do in mathematics, do not do it in your head, write it out.1205

Being able to do something in your head is not a measure of anything, it is just a measure of being able to do something in your head.1210

This stuff tends to be complex, you want to be able to see what you are doing, write it out.1216

It takes a little extra time, but I promise it will pay huge dividends, do not write things out in your head.1220

Okay, we said it is again, a · a under the radical,1226

a · a = this times itself, so it is 4 × 4 = 16 + 2 × 2 = 4, + 6 × 6 = 36, all under the radical equals sqrt(56).1234

So for this vector, (4,2,6), the length of this vector is sqrt(56).1251

In other words, the distance from the origin to this point is sqrt(56), that is all that is.1257

Now, observe the following.1264

The norm of a vector a = the norm of the vector -a.1275

Well, this makes sense just in terms of the definition, because again, we are squaring things.1283

When we square things the negative sign goes away.1288

If you want the geometric interpretation of this, it is just this.1292

If this is our vector a, we know that -1 is the vector of the same length in the other direction, so this is -a.1295

Well the length is the same, they have the same length.1304

As it turns out, this is true, so if we negate a vector, the norm is the same.1308

If you want, you can recalculate the norm, but it is just good to sort of observe that is the case.1314

Now, let me get a clean page here.1326

Talk about the last thing that we want to talk about today.1329

Let a and b be 2 vectors.1333

In other words 2 individual vectors, not 2 vectors, not a vector in 2 space.1342

Now, we know the vectors are just another way of to specify points in space.1348

The distance between these points is the vector a - b.1381

This is a vector representation of the distance between two points.1398

We speak about a vector, a mathematical object, a directed line segment, one going this way, one going this way, they are actually points in space.1401

The distance between those two points in space is the vector a - the vector b.1410

Let us show you what that looks like geometrically, this is actually important to understand geometrically.1415

Let us just take our one vector a, and put it over here.1421

That is vector a, we will do that.1427

Now let us put vector b this way.1430

That represents this point in space, and that represents this point in space -- how do I find the distance between these two points in space? 1435

Well it is true, you are going to use the distance formula in 2 and 3 space.1444

However, as it turns out, the definition that I am about to give you for the distance between 2 points in space, or 2 vectors,1446

It is again, a generalization of the distance formula for any number of dimensions.1455

Now we are working in any number of dimensions, we are no longer limited.1459

Here is how it happens, a - b, when we are adding vectors, in this case a + -b,1462

This is just equivalent to vector a + the vector -b.1470

When we add vectors geometrically, we just take the first vector,1476

Then we take the other vector, and we can put the tail of that vector on the head of the first one, and we go some place else.1480

So the vector a - b looks like this.1490

It means do a first, then take b, which is this vector here.1494

But, since it is -b, go in the other direction.1499

That is all you are doing, if we said take the vector a + b, it would be a + b.1505

If we had another vector c, let us say we had another vector c here, if we said, what is a + b + c?1515

We would go a first, then from there you would go b in the same length and the same direction, and then you would go c, which is like that.1520

Well, in the case of a + a - b, you have a + a negative b, so you are going to be moving in the opposite direction of b in the same length,1534

That is going to give you a new vector, I will do this one with a dotted line.1545

That is the vector a - b, as it turns out, look at how long this is, that turns out to be exactly as long as this.1550

What you end up with is this parallelogram geometrically, so the distance between this point and this point, is the same as the distance between this point and this point.1560

Well, this point and this point marks the vector a - b.1575

That is what is going on here geometrically.1580

Let us go ahead and define this distance.1583

Define the distance between vector a and vector b as the following: It is the norm of the vector a - b.1590

This is the symbolism, it equals exactly what a norm is, a - b is a vector.1613

We said that the norm of the vector is the vector dotted by itself, then you take the square root.1620

So we have a - b · a - b under the square root.1627

Let us just do an example, that is what is important.1637

Example 4.1645

Let a = (-1,1,6) and b = (1,2,3,).1648

Again these are 3 vectors, points in 3 space.1660

The vector a - b equals, I am going to write all of this out explicitly, (-1,1,6) - (1,2,3). 1664

Well, that equals (-1-1,1-2,6-3) = (-2,-1,3), the vector (-2,-1,3) that is the vector a - b.1674

Since that is the vector a - b based on the definition that we just wrote, let us dot this with itself.1696

a - b · a - b, dot product with itself, again we are going to write this out explicitly.1706

(-2,-1,3) · (-2,-1,3) = -2 × -2 is 4, -1 × -1 is 1, 3 × 3 is 9, square root of that 4,1,9, 14,1719

So we end up with radical 14.1740

Therefore the norm of a-b = sqrt(14).1743

In other words, the distance between the vector a and the vector b is sqrt(14).1750

The distance between the point that a represents, and the point that b represents is the distance of 14.1756

Believe it or not, this is just the formula that I gave which is a generalization of the distance formula in any number of dimensions, expressed in vectors.1767

Again, trust the definition.1778

Hopefully the geometry, or the geometric interpretation that we gave you helps a little bit.1780

It is fine to use that geometric intuition at this point, think about it, look at it, if you want break to it up into components.1786

Write everything out until you are comfortable with this notion.1795

It is just a straight application of basic algebra.1800

The norm is this, and this distance between two points is the norm.1802

It is a norm because that is what a norm is, it is a distance, you have that definition.1808

Okay, so we have introduced scalar product, we have introduced norm, 1814

We will go ahead and stop here for today, and next time we will continue on with more vectors.1817

Thank you for joining us here at educator.com, we will join you next time, bye-bye.1822

Hello and welcome back to educator.com and Multivariable Calculus.0000

So, the last lesson, we introduced the method of Lagrange multipliers, we did a couple of basic examples.0004

In this lesson, we are just going to continue doing examples to develop a sense of how to handle this method of Lagrange multipliers.0009

Again, there is nothing particularly difficult about the method, it is pretty straight-forward. The difficulty is solving the simultaneous equations.0018

Because there is a lot going on, on the page, there are different cases when this equals 0, when this does not equal 0... so there are lots of things to keep track of.0025

These examples are going to be slightly more complicated, they are going to be handled the same way, they are just going to look like there is a lot more going on.0036

In some sense that is true, in some sense that is just the nature of Lagrange multiplier problems.0045

In any case, let us just jump right on in and see what we can do.0051

Okay. So, the first example. Example 1.0057

Find the point on the surface z2 - xy = 1, closest to the origin.0065

Notice in this particular problem, they only gave us 1 function, and they just said closest to the origin.0090

This is an example of a Lagrange multiplier or a max/min problem where we have to extract information, come up with another function for ourselves whether it is f or g, depending on which one.0097

Right now, we are not quite sure which one is f and which one is g. So, find the point on the surface closest to the origin.0107

Well, distance from the origin is just the distance function. So, distance... I will write it out, that is not a problem... so distance = x2 + y2 + z2, all under the radical.0116

As it turns out, we want the point that is closest to the origin, but the point has to be on this surface.0134

As it turns out, this is f, right here, and this is g. That is the constraint. We want to minimize, in other words, this function subject to this constraint.0144

Well, fortunately, since we are dealing with the distance function, if we minimize this square root function, it is the same as... we can also minimize the square of that: x2 + y2 + z2.0156

I do not want to deal with the square root, so I will just instead of minimizing this, I am just going to minimize the function.0170

This is actually going to be my f, right here, and this is going to be my g.0176

Let us go ahead and form what it is that we actually form. We want to form the gradient of f, we want to set it equal to λ × the gradient of g, λ is our Lagrange multiplier.0184

Again, so we are going to have this set of equations and our constraint, which is going to be z2 - xy - 1 = 0.0203

It is always going to be the constraint as a function set equal to 0. That is our other equation.0213

Now, we have to satisfy all of these simultaneously. Okay, let us just go ahead and take care of this.0222

So, the gradient of f, let us see... gradient of f, that is equal to... well, gradient of f is going to be 2x, 2y, and 2z.0230

That is going to equal... actually, let us just go ahead and do the gradient separately and then we will set it equal to each other.0245

The gradient of g is going to equal -y, that is df/dx, -x, that is df/dy, and 2z, that is df/dz of g.0255

Now, we set them equal to each other with the λ, so what we have is 2x, 2y, 2z is equal to... let me do this systematically, I do not want to get too far ahead of myself here... λ × -y, -x and 2z, which is equal to -λx, -λy, and 2λz.0268

Okay. There we go. Now, let us go to blue. This and that is one equation, this and that is another equation, this and that is a third equation, and of course we have our fourth equation. Those are our four equations and four unknowns, x, y, z, and λ.0306

Again, four equations and four unknowns. It is a bit much, but hopefully we should be able to work it out.0320

Let us see what we have got. One thing that we want... let me actually write out the equations here so that we have them in front of us... so, 2x = -λx... I am sorry, oops, I got these reversed, my apologies... this is -y, so this is y and this is x. There we go, that is a little bit better.0328

So, 2x = -λy. There you go, you can see it is very, very easy to go astray in these problems if you are not keeping track of every little thing.0359

So, -y and then we have 2y = -λx, and we have 2z = λ × 2z. Let us go ahead and actually write the other equation... z2 - xy - 1 = 0.0374

Those are our four equations. Okay. So, let us just take a look at this for a second, and see how to proceed. One of the possibilities here is... well, one thing we know, we know that λ is not going to be 0. Let us exclude this possibility.0395

λ is definitely not going to be 0 because this would imply, if λ is 0, that means 2x = 0, 2y = 0, 2z = 0, because that would imply that x, y, and z equals 0, 0, 0, but 0, 0, 0, does not satisfy this equation.0413

If I put 0's in for here, what I get is -1 = 0, which is not true, so this particular point, if λ = 0, this particular point might work, but it does not satisfy the constraint. We can eliminate that as a possibility. 0439

So that is good, λ is not equal to 0. So, our first possibility is... let me see, let us go ahead and move to the next page and let me rewrite the equation so that we have it above us.0454

So, 2x = -λy, 2y = λx, 2z = 2λz, and we have z2 - xy - 1 = 0.0469

These are our four equations that we need to solve simultaneously. The one possibility here is let us take the possibility where z is not equal to 0. Okay.0490

So, when z is not equal to 0, that means we can go λ = 2z/2z, we solve this equation right over here, and we get λ = 1.0500

When λ = 1, I can put it into this equation and this equation. Then I get 2x = -y, and I get... I have 2y = -x.0519

Well, I am going to substitute one of these in here. I am going to take this, so I am going to rewrite this one as -2y = x, and I am going to substitute this x into that equation.0540

I get 2 × -2y = -y. I get -4y = y. and the only way that this is true, is that y = 0.0556

When y = 0, that means x = 0. Now, I have y = 0, and x = 0.0574

So, let us see here. That actually is a possibility, so now if y = 0 and x = 0, now let me go ahead and take these values and put them into this equation here.0590

Now I have got z2 - 0 × 0 - 1 = 0, so I get z2 = 1, I get z equals + or - 1.0605

My possibilities... so a couple of points, are (0,0,1), and (0,0,-1), right? x and y are 0, z is + or - 1.0622

Those are two possibilities. When I evaluate them at f, I end up with 1, and I end up with 1.0636

So, those are a couple of possibilities. This was the case when z was not equal to 0.0645

Now we will go ahead and do the case for z = 0, so we are just eliminating all of the possibilities here.0652

When z = 0, what we have is, if I take this equation, the constraint equation, I get 02 - xy - 1 = 0.0661

So, I get -xy = 1, or y = -1/x.0680

This is 1 relation that I have got here. Now, let me go back to my other relations involving x and y. That is 2x = -λ y, 2y = -λx.0690

I am going to divide the top and bottom equation, in other words I am going to divide the top equation by the bottom equation.0703

I get 2x/2y = -λy/-λx, the λ's cancel, and I get y/x.0712

So 2x/2y = y/x. The 2's actually end up cancelling, so I am just going to go ahead and cross multiply. I am going to get x2 = y2.0725

Well, let us see. x2 = y2, well I found that y = -1/x, therefore I am going to put this y into here and I am going to get x2 = -1/x2, which equals 1/x2. That implies that x4 = 1.0741

Well, if x4 = 1, that implies that x = + or - 1.0768

When x = + or - 1, I go back to any one of the equations here, and I end up with y is equal to... well, let me actually move onto the next page here... so I have got x = + or - 1.0777

x = 1 implies that y = -1, and x = -1 implies that y = 1.0798

Now I have a couple of other points. z was 0, and these are the x's and y's.0812

One possibility is (1,-1,0), and the other possibility is (-1,1,0).0818

When I evaluate these at f, I get 2.0829

There we go. We have taken care of all of the possibilities. We had the four points, we had (0,0,1), (0,0,-1), and we have (1,-1,0), (-1,1,0).0841

For the other points, we found that the values were actually 1. Here the values are 2, so, the mins occur at (0,0,1) and (0,0,-1), and the value of the function at those points is 1.0855

On that given surface, on the surface z2 - xy - 1 = 0, these 2 points on this surface are going to minimize the distance to the origin. That is what we have done here.0884

Again, as you can see, there seems to be a lot going on. Lots of cases to sort of go through. I wish that there was some sort of algorithmic approach to solving these problems other than the basic equation of setting the gradient of f = λ × gradient of g + the constraint equation equal to 0.0906

We have four equations and 4 unknowns, 5 equations and 5 unknowns. Clearly this gets more difficult. You just have lots of other cases to just sort of go through.0925

Again, successful Lagrange multipliers is more about experience than anything else. You can be reasonably systematic, but clearly you saw just with these 4 equations and 4 unknowns there is a lot to keep track of.0934

That is the only difficulty. Do not let the procedure actually interfere with your view of the mathematics.0949

Okay. Let us just do another example. That is all we can do, just keep doing examples until we start to get familiar with general notions. So, example number 2. Example 2.0958

Find the max and min of f(x,y) = x2 + 2y2 - x on the closed disc of radius 1 centered at the origin.0974

Okay. So, what we want to do is... we have the unit... now, we are considering the entire region, and it is a closed region. We are considering the interior and we are considering the boundary. That is all that is happening here.1016

So, let us see... equals 0... so this is the... our function, and we want to find the maximum and minimum values on this region.1035

We have a couple of things going on here. One of the things that we are going to do is... so this is kind of a max/min problem, and a Lagrange multiplier problem. 1055

The Lagrange multiplier problem will usually apply to the boundary here, because we have an equation for that boundary which is x2 + y2 = 1, or x2 + y2 - 1 = 0. 1072

As far as the interior is concerned, well we just treat it like any other max/min problem. We see if we can find a critical point and take it from there.1085

Let us go ahead and do that first. Let us deal with the interior, so, let us find the gradient of f and we will set it equal to 0 to see where the critical points are. 1093

The gradient of f is going to be 2x - 1. That is the derivative with respect to the first variable x, and then we have 2y2, I am sorry, this is 2y2.1105

Our derivative with respect to y is going to be 4y. Okay. Now we want to check to see where these are equal to 0, so we have 2x - 1 = 0, and we have got 4y = 0.1123

This implies that y = 0 and this implies that x = 1/2, so at the point (1/2,0), the point (1/2,0) is a critical point. 1136

Let us just go ahead and find the value of f at that point. When we go ahead and evaluate at that point, f(1/2,0), I go ahead and put it into f and I end up with -1/4.1154

That is one possible value. That is a critical point, that is on the interior. Now we can go ahead and deal with the boundary.1170

Now, let us check the boundary. So, the boundary is g(x,y) = x2 + y2 - 1, and that is going to be one of the equations, we will need to set that to 0.1180

Let us go ahead and find the gradient of g. The gradient of g is equal to, well that is going to be 2x, and that is going to be 2y.1195

So, 2x and 2y, so now we are going to go ahead and set the gradient of f = λ × gradient of g, and what we get is 2x - 1, and 4y = λ × 2x and 2y, which is equal 2λx and 2λy.1211

The two equations that we get are... let me do this one in red actually, so that I separate the equations out that I am going to be working on, again. This is all about solving simultaneous equations.1243

I get 2x = 2λx. I get 4y = 2λ × y. Let me make sure that I make everything clear so that we do not make the same mistake we made last time, so, this is 2λy.1254

And, of course we have our equation x2 + y2 - 1 = 0. That is the constraint.1273

Now I have got 3 equations and 3 unknowns, x, y, and λ.1280

So, let us deal with case 1. Case 1, let us take λ = 0. 1286

So, case 1, λ = 0. Okay. When I set λ = 0, what is actually going to end up happening is I am just going to end up getting the previous answer. 1297

That is not an issue, so let us deal with case 2.1309

case 2, where λ does not equal 0. So, one possibility for here, so, when λ does not equal to 0, now there are some sub cases that I have to consider.1315

The first sub case that I am going to consider is y = 0.1327

When I set y = 0, then I am going to get x2 + 02 - 1 = 0. I am going to get x = + or - 1.1334

Therefore, I am going to have the points (1,0) and (-1,0).1349

Well, at (1,0), when I evaluate it at f, I am going to get 0, and when I evaluate it at (-1,0), when I evaluate f, I am going to get the value 2.1355

Now I have got (1,0), (-1,0), these are the values, and I also had that other point, that (1/2,0) that I found from just working on the interior of the disc and the value was I think -1/4, or something like that, or whatever it was.1370

Now, that is the case where λ does not = 0, and the case where y does equal 0. Now we want to consider the other case where y does not = 0.1385

Again, it seems like there is a lot going on but usually you can make sense of it by just spending some time with it. So, y does not equal 0. 1400

Well, when y does not equal 0, 4y equals 2λ y. That means λ = 2. 4y/2y, let me go ahead and write that out actually. λ = 4y/2y = 2.1409

When λ = 2, this implies that 2x - 1 equals... so the equation is 2λx, right?1435

So 2x - 1 = λ = 2 = 4x, so I am going to get 2x = -1, x = -1/2.1454

So, when x = -1/2, now when I put that into my x2 + y2 - 1 = 0, when I put it into here, I am going to end up with the following.1469

It is going to be -1/22 + y2 - 1 = 0.1491

I get 1/4 + y2 - 1 = 0, I get y2 = 3/4, therefore y = + or - sqrt(3)/2.1500

Okay, so, now my other points. x is -1/2, +sqrt(3)/2, and I get -1/2 - sqrt(3)/2.1516

When I evaluate these at f, I am going to end up with 9/4 and 9/4, so we have got 9/4, 9/4, -1/2, 0, I check all of those points to see which one is the maximum, which one is the minimum. 1531

So the max takes place at the points (-1/2,sqrt(3)/2), and (-1/2,-sqrt(3)/2). Value is 9/4.1555

The minimum takes place at the original point that we found, which was... not -1/2... at 1/2, 0 and that value was -- let me see, what was that value if we can recall -- -1/4. There we go. This is our solution. I will go ahead and put f here, when we evaluate f.1580

So, at this point and this point, our function achieves a maximum and at this point, the function achieves a minimum. That is it.1610

Clearly there is a lot going on. A lot of things to sort of keep track of, but we are solving several equations in several variables. This is just the nature of the problem, the nature of the best.1618

Okay. So, in the next lesson, we are going to actually continue discussing Lagrange multipliers. We will do some more examples.1633

We will pull back a little bit so we will discuss some of the geometry of the solutions and we will try to make sense of what is actually going on.1640

Again, not theoretically, we just want this to seem reasonable to you, that we did not just drop this in your lap and say use this technique to find max/min for a function subject to this constraint. We still want this to make sense.1645

Okay. Thank you for joining us here at educator.com, we will see you next time. Bye-bye.1658

Hello and welcome back to educator.com and multi-variable calculus.0000

In today's lesson we are going to continue our discussion of Lagrange multipliers.0004

I am going to start by revisiting a previous example and describing the geometry of the solution.0008

The reason that I want to do that is so that you can see why it is that this criterion of gradient f equals some λ × the gradient of g, why it actually makes sense.0013

I just wanted you to see what the geometry actually looks like, so it does not look like it just fell out of the sky.0025

I encourage you to look at the proofs, or some of the explanations in your textbooks if you want something a little bit more rigorous and a little bit more accurate, but I do want it to make sense geometrically what it is that is happening.0031

So, let us go ahead and get started. Let us see, in the first example of the last lesson, we found... so let me just write all of this out... in the first example of lesson 20, we found the max and the min... you know what, let me not call them max and min, let me call them extrema.0044

We found the extrema values of f(x,y), and if you recall that function was equal to x × y subject to the constraint that the point (x,y) must also satisfy.0080

This function g(x,y), which was x2/8 + y2/2 = 1. We were maximizing or minimizing this function subject to the constraint that the x's and y's that actually go into this that maximize it and minimize it also happen to lie on this ellipse. That is what we were doing... equals 1.0121

Okay. We found four points that satisfied... okay, we had the point (2,1), (2,-1), (-2,1), and (-2,-1).0144

The respective f values, in other words when we put these points into x, y, the respective f values, I am just going to write them underneath... we had 2, -2, -2, and 2.0170

These two were the points where the function achieved a maximum, and these two were the points where it achieved a minimum. 0185

Let us take a look at what this looks like geometrically. So, let us look at this geometrically.0193

What you have -- I am just going to draw a little x, y plane here -- and again we are looking for points (x,y) in the x, y, plane that actually maximize this function x, y but also happen to lie on the ellipse.0209

Well, x, y, if you do the level curves of x, y... in other words x, y, equals some constant, x,y = 3, x, y = 2, x, y = 5/2, whatever. You are going to get a series of hyperbolas. That is what you are going to end up getting.0222

I am going to go ahead and draw a few. First we want to make sure that x and y actually satisfy this equation. In other words, that they lie on this ellipse.0238

Let me just go ahead and draw an ellipse here... I am not very good at drawing at ellipses, but again, the drawing itself is not that important. 0249

So, this is x2 + y2 = 2. Where you have this right here is the distance from the center to this vertex is going to be sqrt(8), and this one is going to be sqrt(2).0258

The points x,y, that lie on here, which ones are the ones that actually maximize? Well, we found these points right here. 0271

(2,1), there was one, (-2,1), (-2,-1), and (2,-1), so here is actually what it looks like geometrically... oops, not very good, let us make it a little bit better than that... here is that... here is another point.0277

So, in other words, if you were to look at this at the point (2,1), (-2,1), (-2,-1), and (2,-1), these are the points that maximize the function. These are also the points that happen to lie on this ellipse, so the constraint is satisfied. Those are the points that we found.0300

Well, geometrically, what it is, take a look at this as it turns out at this point if I were to actually take the tangent line to both curves as it turns out, they correspond. 0323

In other words, if I were to take this tangent line at this point happens to be the same for this curve, g, and for this curve, f.0338

What ends up happening is the following: the gradient, at those points, looks like this. The gradient of f will be this way, the gradient of g will be this way.0351

Because the tangent lines correspond, the gradient vector is perpendicular to the tangent line, so as it turns out they're moving in the same direction.0360

Well, one is going to be either the same length or it is going to be longer than the other. That is what the Lagrange multiplier is. It is just a scalar multiple of the gradient vector.0369

When we say that the gradient of f equals some multiplier λ × the gradient g, what we are saying is they are either moving in the same direction or the opposite direction depending on whether λ is positive or negative.0380

In other words the gradient vector g or the gradient vector of f is a scalar multiple of the gradient vector g.0394

The gradient vector g right here looks something like this. Here, so this was the max and this was a max. Notice the gradient vectors are in the same direction.0402

In the case of the mins, the gradient vectors are actually pointing in opposite directions, but it is still just a scalar multiple of the other.0413

In other words, they have the same direction either this way or this way, but they are part of the same line and they are on the same line because the tangent plane, or the tangent line itself to the curves at that point that satisfies the max and min, they are the same for both curves.0421

So, the gradient are multiples of each other. The gradient vectors are multiples of each other, and that is what is happening. That is what this actually says.0443

I just wanted to demonstrate geometrically what it is that is going on.0452

Again, it is a good idea to see that to get an idea of where it comes from, but again, it is the algebra that matters, because when you move on to higher dimensions, you are not going to quite have the geometric intuition that you can get.0456

For example, this works in 2 dimensions. If we were working with a 3-dimensional example, a function of 3 variables, we would have x, y, and z, and then we would be talking about surfaces.0470

Where you have one surface and then another surface, the point p that happens to be an extremum is going to be the point where those 2 surfaces actually touching, and the tangent plane to those surfaces is going to be the same. 0480

So the gradient vector is going to be perpendicular to that tangent plane and the gradient vector, they're either going to be in the same direction, or they are going to be in opposite directions, but they are going to lie along the same line.0491

That is it, that is all that I wanted to demonstrate. We are going to do one last example for Lagrange multipliers, except this time what we are going to do is we are going to have 2 constraints instead of 1.0502

So, we are going to try to maximize or minimize a function subject not just to 1 constraint, but another. Let us go ahead and do that example. Okay. Example 1.0512

So, let us see. A Lagrange multiplier problem with 2 constraints, so I am just going to describe how it is that we actually deal with it before we start the example.0533

A Lagrange multiplier problem with 2 constraints, we introduce 2 multipliers, that is it. The number of constraints, you just introduce another multiplier and the equation actually ends up being the same, but just with another multiplier.0561

You end up having just another equation to solve. So, introduce 2 multipliers, we will call them λ1 and λ2, so, we must solve the following system.0574

Our problem comes down to solving the gradient of f at... let me actually write it all out here, a little further to the left so that I do not have to squeeze anything in.0603

The gradient of f at the extrema = λ × the gradient of g1 at p, so now we have 2 constraints, g1 and g2 + λ2 × gradient of g2 at p. That is it.0616

We basically just add one more thing. This is the normal one, gradient f = λ × gradient g, but now since we have 2 constraints, g1 and g2, we just add one more term.0635

We also solve g1 of x, y, z. I am just going to call it point g1 of x, y, z, and again I am just using a 3-dimensional example = 0, and g2 of x, y, z = 0.0647

This system right here is the system we end up having to solve. That is it, just one more equation, and this is our one more equation and this is our one extra term.0670

So, now we will go ahead and start the problem.0680

The cone z2 = x2 + y2 is cut by the plane z = 1 + x + y in a conic section.0684

Hopefully you remember from your algebra and pre-calculus days, and perhaps even in single-variable calculus, when you have a cone in 3-space and you slice through that cone, you are going to end up getting the conic section graphs.0709

You are either going to get a circle, you are going to get an ellipse, or you are going to get a hyperbola.0725

Or, in certain, special cases, depending you are going to get a straight line. Okay... conic section, alright.0732

So, we want to find the points on this conic section, and notice we have not said anything about what this conic section is. We just know that it is a conic section.0739

We do not know whether it is a conic section, ellipse, or hyperbola.0754

So, conic section that are nearest and farthest from the origin.0758

We have this cone in space, we slice it with a plane, and there are going to be points that are on this conic section that are going to satisfy both the equation for the cone and the equation for the plane.0776

A couple of those points, or one of those points, however many, some are going to be furthest from the origin, some are going to be closest to the origin, so, that is what we are doing. 0791

Now, we need to find the function that we are going to maximize in this particular case notice this actually ends up saying nothing about.. it set it in a different form.0799

It did not give us the function that we are trying to maximize or minimize explicitly, we have to be able to extract it from the problem itself.0808

Well, since we are finding the nearest and furthest point from the origin, we are going to look to the distance function, but we are not going to deal with the distance function as it is with the radical sign. 0815

We are just going to square that distance function because if you minimize the distance, you are also minimizing the square of that distance, because again, we do not want to deal with radical signs, we want to make the math as tractable as possible.0827

Okay. SO the function that we are going to be dealing with, f itself, is going to be a function of 3 variables, and it is going to be the distance function, so x2 + y2 + z2, 0839

But again, we do not want to deal with the radical, so minimizing or maximizing this is the same as minimizing or maximizing this, just the square of it. x2 + y2 + z2, so that is our f.0853

Now our g1 of x, y, z, that is going to be x2 + y2 - z2... basically what we do is we take this equation and we move everything over to one side and set it equal to 0... - z2 = 0. That is the other equation that we have to satisfy.0863

Then of course we have a second constraint, so we also have to satisfy this equation. g(x,y,z) = 1 + x + y -z = 0. That is of course, this equation, again, with everything moved over to 1 side, set equal to 0. 0884

It always a good idea to write the equation that you are actually dealing with. The f and the particular g's, in this case we have 2 g's that we also need to satisfy.0905

Now, we have to solve the gradient of f = λ1 × the gradient of g1 + λ2 × gradient of g2... and I apologize if I kind of keep writing things over and over again.0916

I think that it is important to do that because often time in mathematics there is so much going on, we try to take shortcuts by not writing everything down and that is a bad idea.0940

Take the couple of extra seconds. If you have to rewrite an equation 2 or 3 times, that is not a problem. It is a good way to solidify it in your mind, to get a good grasp on it. You need to get in the habit of writing things out. The more shortcuts you take, the more chance that you are going to make a mistake.0948

Again, as you have already noticed with the last two lessons, these Lagrange multiplier problems, they are not difficult, there is just a lot going on, on the page, so you want to keep track of everything.0965

Okay. So, we have to solve this, that is the one equation, so we have to solve g1 = 0, and g2 = 0, so let us go ahead and come up with this one here. 0975

When we take the gradient of f, we are going to get 2x, 2y, 2z, and again, just a personal taste for me, I tend to write them in terms of column vectors instead of row vectors -- your choice, whatever works best for you -- equals λ1 × the gradient of g1, which is 2x, 2y, and -2z + λ2 × 1, 1, and -1.0986

These are the 5 equations that we are going to have to solve simultaneously.1018

The first equation, let me do this in blue. We have this equation from the first row, from the second row, and from the third row... so we have... and I am actually going to number these, I am going to refer to them as numbers instead of continuously writing them down.1022

So, equation 1, we have 2x = 2λ1x + λ2.1036

Equation 2 we have 2y = 2λ1y + λ2.1046

Then we have the third equation which is 2z = -2 × λ1z - λ2.1056

Then, of course, we have the g1 and g2, so we have x2 + y2 - z2 = 0, and we have 1 + x + y -z = 0.1064

These are the... this is the system that we have to solve simultaneously, solve simultaneously 1086

So, let us just get started. Again, when you are dealing with this many simultaneous equations, there really is no algorithmic approach to this.1097

You just have to sort of use all of the resources that you have at your disposal, try a bunch of different things... we do not want to give you the impression that just by looking at this I automatically know what to do. 1106

I mean, I have already worked this out. When I look at this on a page, I just automatically know do this, do this, do this, do this... it is not like that. We do not want to give you that impression.1118

It takes time. It takes some time, it takes some experience. I still struggle with these myself, because, again, you have got 5 equations to solve. You have different cases to consider, different things to look at... so by all means do not worry about the time that it takes and do not worry that it is difficult. 1128

On an exam of course, it is not going to be something like this. You are going to have something a little bit simpler to work with. These are just bigger problems so that you have a bigger sense of what is going on and more deep practice.1144

So, let us see. In this particular case, I am going to go ahead and put equation 1 and equation 2 together, so I am going to take equation 1.1155

When I take number 1 - number 2, I get the following... I get x - y = λ1 × x - y, okay?1169

When I take equation 2 and I add it to equation number 3, I end up with y + z = λ1 × y - z.1183

In this case what I have done is I have essentially reduced this so that now all I have is λ1 and I do not have to worry about λ2.1201

Okay. So let us take a look at the first one. So, x - y = λ1 × x - y is satisfied in 2 cases. Excuse me.1209

It is satisfied in 2 cases. The 2 cases are when x does not equal y, λ1 = 1, and the other case is when x = y, and λ1 does not equal 1. Let us take these cases individually and see what we can do.1223

Okay. So, let us do case 1 first when x is not equal to y. So, case 1.1249

So, x does not equal y. When x does not equal y, then this implies that λ = 1, right? You are just taking x - y/x - y, you are getting λ1 = 1.1257

Okay. If we put this in for one... okay, now let us take the second equation. So, put this into y + z = λ1 × y - z.1274

We get y + z equals y - z, which implies that 2z = 0, which implies that z = 0. Now, since z = 0, let us go ahead and use equation number 4. 1303

That gives x2 + y2 = 0, which implies that x = 0 and that y = 0.1322

Okay. So, now we have 0, 0, 0. We found a certain point. However, the point (0,0,0) does not satisfy equation number 5. 1338

Again, all of the equations have to be satisfied, so, we have taken these three equations, we have taken 1, 2, 3, & 4, we have managed to satisfy 1, 2, 3, & 4, but it does not satisfy equation 5.1355

So, (0,0,0) is not a viable point here. So, (0,0,0) is not a viable point to consider, so we have taken care of that case. Now, let us go ahead and deal with case 2.1367

In case 2, we said that x actually equals y, and that λ1 does not equal 1. 1382

If we put equations 4 and 5 together, let me see here, so if x = y what we end up with is the following.1393

We get z2 = x2 + y2, that is the equation 4, and z is equal to 1 + 2x, because it is 1 + x + y, y = x, therefore it is 1 + 2x.1408

Now, let me go ahead and square this, so when I square this and put this into here, I end up with 1 + 2x2 = x2 + y2, but x = y, so it equals x2 + x2, so I am going to get 1 + 4x + 4x2 = 2x2.1427

Then, I am going to move everything over and basically end up with a quadratic here, so I am going to end up with 2x2 + 4x + 1 = 0.1460

Notice, in this case I have used both equation 4 and equation 5 so that is satisfied, and this case of x = y, λ1 not equal to 1, that comes from the other equations so all of the other equations are satisfied here.1475

When I solve this, I cannot factor it so I have to use the quadratic equation. I am going to end up with x = -4 + or - sqrt(8)/4, which equals -1 + or - sqrt(2)/2. That equals x.1486

Well, it also equals y, so, y = the same thing because x = y. Alright.1510

Then, when we go ahead and we put in for z, when we put this, this x value up into this equation to solve for z, we are going to end up with the following z value.1519

You are going to get z = -1 + or - sqrt(2). I will not go ahead and do that arithmetic, but that is how I go ahead and get it. 1532

So, I have x, I have y, and we have z. Now let us go ahead and finish this up, so let us see here.1543

Let us write it this way. x = -1 + sqrt(2)/2, and it also equals -1 - sqrt(2)/2.1552

y is equal to the same thing. -1 + sqrt(2)/2, and -1 - sqrt(2)/2.1569

Our z value is equal to -1 + sqrt(2) - 1 - sqrt(2), so these are our 2 points that we found, okay? These are our 2 points. Let us call this one p1, and p2, these (x,y,z), these (x,y,z), these 2 points satisfy all of the conditions.1578

Now, okay. Now let us go ahead and write out the conclusions for this. Now, we are looking at either an ellipse or a hyperbola.1602

Statistically, when you do this you are going to get an ellipse or a hyperbola. The chances of getting a circle are kind of... you know... that would just mean a plane that is straight across like that, you have a cone and it just cuts across and cuts it in a circle. 1627

That is not what is happening here so you are definitely looking at either an ellipse or a hyperbola, okay?1642

Now, if it is an ellipse, if we have an ellipse, then, p1 is the nearest point -- in other words, it is the minimum of the function f -- is the nearest point, and p2 is the farthest point.1648

The way you get that -- again, once you get your points, you put these points back into x. Well, our function x, our function I am sorry was x2 + y2 + z2. 1677

You put in this squared + this squared + this squared, and then you get a certain value for f at this point, then you put in this squared + this squared + this squared. 1685

When you do that you will find out that this number is less than this number. That is how we do it... is farthest because f(p1) is less than f(p2).1696

So, if this is an ellipse, this is the farthest point, this is the nearest point. Okay.1709

If a hyperbola -- there is a lot to consider here, it is kind of interesting, it would be nice if we could just plug our numbers in and get a number and be done, but we have to actually consider what is happening.1716

If a hyperbola, then there is no furthest point, then there is no farthest point, because a hyperbola is asymptotic.1728

In other words, it goes on into -- you know -- a particular direction infinitely asymptotic.1756

So, p1 and p2 are the two nearest points. This is why earlier on I decided to use the word extrema instead of max/min because you never know what situation you are dealing with.1767

It is not that you are maximizing or minimizing a function, yes, within a given situation you might be maximizing or minimizing, but in this case we do not know whether we are dealing with an ellipse or a hyperbola.1784

If it is an ellipse, then yes, we have a max value and a min value, in other words a farthest point and a nearest point. 1792

If it is a hyperbola, we do not have a maximum value. What we have is 2 points that satisfy all of the conditions, but they happen to be the two points that are nearest to the origin. 1800

There are physical things to consider, there are other things to consider than just max/min, it is not just max/min -- the two nearest points. Okay.1809

Now, so, that is it. This was just an example of a Lagrange multiplier example using 2 constraints, using 2 particular functions that had to be satisfied. 1820

In this case our function that had to be satisfied was our distance function, and it had to satisfy the equation of a cone, and it had to satisfy the equation of the plane. That is it.1830

Everything else was exactly the same. You are solving the series of simultaneous equations. Once you actually get to a point where you get your points, you have to stop and actually think about what it is that is actually going on.1839

This is why we did this final analysis. We have the possibility of an ellipse, we have the possibility of a hyperbola, I am actually going to leave it to you to actually think about which one it actually is.1851

To see if you can actually use the equations that are given, the g1, the g2, and the f to figure out are you getting an ellipse, or are you getting a hyperbola.1862

So, I will leave it with this: see if you can figure out which -- and I will give you a hint, try converting the functions into cylindrical coordinates, if you happen to be familiar with cylindrical coordinates.1872

If not, do not worry about it. It is not a big deal. We do not actually have to figure out, all we wanted to do was find out which was furthest, which was closest, we found our two points, the actual problem itself is satisfied.1892

Okay. Thank you very much for joining us here at educator.com. We will see you next time. Bye-bye.1903

Hello, and welcome back to educator.com and multi variable calculus.0000

Today, we are going to introduce the concept of a line integral. Profoundly important concept, absolutely shows up everywhere. It is the heart and soul of math and science.0006

So, with that, let us just jump right on in. Okay.0016

So, this idea of a line integral, it is not only called a line integral. I also want to give you a couple of other terms that you are going to hear it referred to as.0021

It is also called a path integral, or a curve integral, so it does not matter what you call it. Again, the idea is what is going on. 0030

So, path integral, curve integral, line integral... I think curve integral personally actually is the best, because what you are doing is you are actually just integrating along a curve.0045

It makes the most sense to me, but it does not matter what is you call it.0056

Before we actually introduce the definition of the integral, I am going to talk about something called a vector field and give you that definition, give you a couple of examples and then we will go ahead and give the definition of a line integral and do some examples of that.0060

So, we have had functions that go from R to let us say R2. This is just a curve in 2-space.0075

We take a value t, and then we have some function where you actually have a coordinate, a first coordinate function is f(t), well, something like this.0095

An example would be let us say f(t) = let us say cos(t) and sin(t), so what you end up... you take 1 variable and you end up creating a vector from it. That vector traces out a curve in 2-space.0110

Okay. We have also had the following, what we have been dealing with lately. We have had a function, a multivariable function, so R2 to R, or R3 to R, or R4 to R. Some function of several variables.0131

Well, you take several variables, but you spit out an actual single variable.0143

This is just a function in 2-space and an example of that would be f(x,y) = x2 + y4, something like that. You take 2 values, you evaluate the function, and you end up with one thing. Okay.0149

So, we have talked about functions where the departure space and the arrival space are not the same dimension, for example a function f from R2 to R3, or R4 to R2.0172

We have talked about functions where you have some RN going to RM, and now we are going to talk about a special case of that.0189

Now, let us talk about where the departure space and the arrival space happen to be the same dimension.0203

Let us talk about functions f mapping from R2 to R2, or R3 to R3, and again, for our purposes R2 and R3 is what is important, but this is generally functions from RN to RN.0214

So, where the dimension of the departure space is the same of the arrival space. Let us go ahead and formalize this. Let us have a definition.0232

We are going to define something called a vector field. Okay. Let u be an open set -- and I am sure you remember what open set is, it is just a set that does not include its boundary points -- let f be a mapping which to every point of u associates a vector of the same dimension. That is it.0244

Let us say if I am working from R2 to R2, so if I have some mapping... actually you know what, let us just go ahead and do the example.0307

So, in R2, we have the vector field f, which is a mapping from R2 to R2, so in every point in R2, the x, y, plane, we associate a vector emanating from that point.0315

Let us just go ahead and do an example here. Let us start the example on the next page, so example 1.0347

So, f is a mapping from R2 to R2 defined by f(x,y) is equal to 2x and 2y, so we start with a point in 2-space, and we end up with a point in 2-space. This is the first coordinate, this is the second coordinate.0357

Well, when we actually map this, what we end up with is something that looks like this. Let us say we go to the point (1,1). 0383

If I go to the point (1,1) and I put it in here, (1,1), that means f(x,y) is (2,2). That means from here I have a vector in this direction that the x coordinate is 2 and the y coordinate is 2 and there is a vector in that direction. 0390

If I go to the point (-1,-1), okay, so, now I am over here at this point. Well, if I put (-1,-1) into here, I get (-2,-2). That is a vector pointing in that direction.0408

Over here, if I go to this point, I am going to have a vector pointing in that direction, over here I am going to have a vector pointing that way, over here I am going to have a vector pointing that way, that is it.0420

Each point in a given domain, in this case x, y, I associate a vector where the tail of that vector actually starts at that point and goes in a particular direction. 0431

The direction of that vector, well, the x coordinate is the first coordinate of the function and the second coordinate is the other coordinate, that is all that is going on here.0443

So, to each point in R2, there is a 2 vector emanating from that point, and the direction is given by the function itself.0458

Let us do another example, and this is actually a great example and we have actually been dealing with vector fields, believe it or not -- the gradient operator.0487

The gradient operator in other words, when you actually end up taking the gradient of a function, what you are actually doing is creating a vector field because the gradient is a vector.0501

At any given point, all you are doing is you are saying that once you pick a point, let us say point p, (x,y) from there is some vector that is emanating from that point. 0510

Well, the x direction, the x coordinate of that vector is the derivative in the x direction.0523

The y coordinate is the derivative in the y direction. Operator is a vector field.0527

So, given some function f which is let us say a mapping from R2 to R, so some function of 2 variables, excuse me, the gradient of f is a mapping from R2 to R2, and the gradient is defined by -- well, exactly what we have been doing all along -- it is (df/dx,df/dy). 0541

That is it. We are forming a gradient vector, and it is a vector field. It is a mapping from R2 to R2.0577

I am going to keep repeating myself a couple of times here. I hope you will forgive me. 0587

At a given point p in the domain, the gradient vector starts at that point and has df/dx... let me write that a little bit better... it has df/dx evaluated at p for its x value, for its x coordinate I should say, the vector, and df/dy evaluated at p for its y coordinate. 0595

Okay. Now a little bit more, so every point has a vector associated with it under a mapping for a vector field.0663

Plus, the domain looks like it is covered in vectors. That is why they call it a field.0703

Who knows? We have no idea what direction they are going in, it is just that every point has some vector associated with it. You can imagine if you are standing at any given point, you are going to be pushed in that direction. That is what a vector field is.0728

Okay. So, let us do another example just to make sure we have got it. Example 3.0741

f(x,y) = x2 + y2, so this is a mapping from R2 to R2, so we are looking at a vector field in 2-space.0749

Well, let us go ahead and take a look at f(2,3). So, f(2,3) is equal to... when I put 2 in here, 2 × 3 is 6, 6 × 2 is 12, 22 is 4, 32 is 9, so I have my 10, 11, 12, 13, so it is (12,13).0764

At the point (2,3), I have a vector (12,13). That is it. Let us take the point f(-1,0). Well, when I do (-1,0), -1, 0, I am going to end up with 0 for the first coordinate, -12 is 1 and I get (0,1).0787

When I go to the point (-1,0), I have a vector that is (0,1). That is that one. Let us try f(-2,-2).0810

I am going to come over to this point right here. Some are like right about there.0825

-2 × -2 is 4, and 2 × 4 is 8, so the first coordinate is 8. -22, -22, oh this is going to be (8,8), so I have a vector that is that way. That is it, that is all that you are doing. Nice and straight forward. 0831

Now, let us go ahead and make our way over to line integrals. Let me go ahead and draw a curve in space, something like that, and let me go ahead and draw a vector field -- sort of some random vectors here, and I will put a couple actually on the... so this is a vector field, and this right here is a curve in space, that is c(t).0852

Now we can start. Let f(x,y) which is equal to f1(x,y) function 1(x,y), and function 2(x,y) -- those are the coordinate functions -- be a vector field on R2. So, something like that.0886

Now, let c(t) = c1(t) and c2(t), these are just functions of t, and they are the coordinate functions of the curve, be a curve in 2-space.0911

Since c(t) passes through this vector field, because it is a curve in 2-space and we are in R2, so every single point of R2 has some vector associated with it because we have f, right?0938

As it turns out, some of the point on the curve, I can actually form vectors... in other words I can actually form f(c(t)).0955

Since c(t) passes through R2 and the vector field, we can form f(c(t)), right? 0970

A curve passes through the vector field, well there are points on the curve that are in R2, so I can actually form f(c(t)), points along the curve.0995

You know they have certain vectors associated with them which whatever direction they happen to be moving in, we do not know, because f and c are not specified.1007

Well, we can also form c'(t), in other words we can differentiate the curve, that is just the tangent vector, that is c'(t).1016

The tangent vector 2c(t) at c(t), in other words, if I have some point here, I can not only form f(c(t)), but I can also form the tangent vector.1030

So this is c'(t) and this is f(c(t)), that is it. Vector this way and then the tangent vector. The curve is moving along like that.1051

Now, let us go ahead and define if I take f(c(t)) and I dot it with c'(t), this is actually a function of t.1066

Let us do an example of that. Example 4. We will let f(x,y) = ex+y and the second coordinate function will be y3.1091

We will let c(t), the curve, equal to t and let us say cos(t). Okay?1112

Let us form f(c(t)). f(c(t)), in other words I put these 2 in for x and y, and I end up with et + cos(t), and I get cos3t.1119

Well, let us form c'(t), which is also written as dc/dt, you know more the notation that you are used to from single variable calculus.1140

The derivative of this is 1, the derivative of this is -sin(t). So, now we form... that is fine, I guess I can do it down here... f(c(t)) · c'(t).1150

When I do the dot product of this thing and this thing -- let me do it in blue -- I get e(t) + cos(t) + ... or in this case minus... minus this × that.1167

Minus sin(t)cos2(t). That is it. I do not need this little open parentheses.1189

So, that is what is important. This is our definition right here. We will start with this definition.1198

When we are given a vector field and we are given a curve with some parameterization, we can actually form f(c(t)) · c'(t), and it ends up being an actual function of t.1203

Now we get to our definition of the line integral. I am going to move onto the next page for this one.1217

So, we have... definition, if c(t) is defined on the interval where t on the interval a, b, (2,6), (5,9), whatever, then the integral of f along c from a to b is -- this is the symbol, the integral of f -- along the curve c is equal to the integral from a to b of f(c(t)) · c'(t) dt.1227

Or, we can also write it as the integral from a to b of f(c(t)), let me make this c(t) a little bit clearer here... f(c(t)) · dc/dt dt.1296

What we are saying is when you are given a vector field and when you are given a parameterized curve, you want to form f(c(t)), you want to form dc/dt, you want ot take the dot product of that, and then you want to put that in your integrand.1317

This is what is important right here. This right here. Okay? That is your integrand. It is going to end up being a function of t.1332

We can integrate that function of t. This is just this and this is just the form of integration. That is what we surround our integrand by. This is the definition of the line integral.1339

Let us talk about what it is that we are actually doing here. When we were doing integration in single variable calculus, you were actually taking a line integral except it was a very, very special line that you were moving across. It was the x axis.1350

In other words, when you took some integral of some function like this, let us say from this point to this point, what you were doing is you were integrating the function along this line, along the x axis. 1361

Well, the x axis is a very special line. Now what we are saying is well, it does not matter, we do not have to stick to just that line. I can take any line in any dimension. I can take this x axis and I can curve it this way, I can curve it this way, I can twist it this way so now I might have some curve like this.1373

I can take that function and I can integrate it along this curve, along this line, along this path. 1392

I can still do it because integration, all integration is, is allowing to come up with some number which is a function of whatever variable you are dealing with and just multiplying it by some differential and then adding it all up along the length.1398

That is what you are doing. Because f(c(t)) · c'(t) is a function of t, essentially it is just a number at any given value of t, I can add up all of those little numbers. That is all I am doing.1415

I take the function of t, multiply it by dt and I integrate it. That is all an integration is. Just a really, really long addition problem.1428

So, we can do this, and the definition of a line integral is precisely this: When you are given a vector field, when you are given a curve that is parameterized that happens to pass through that vector field, you can take f(c(t)) for that composite, and then you can dot it with the tangent vector c'(t), and you are going to get yourself an integrand.1438

You can put that integrand, you can surround it with the integral formula... you can form this integral... and you can integrate it just like you do single variable calculus. That is what this is. This is just an integral of the single variable, t. 1456

This is really, really fantastic. It is actually amazing that we can do this. So, integration in single variable calculus is insane, it is a line integral but it is a very, very special line, it is just an x-axis, it is just one direction. This is actually the same thing.1471

So, let us go ahead and do an example here. Let us see how this turns out. Let us see... should we do this one? Yeah, we will do a couple of examples. Let me go back to blue. So, example 5.1487

Let f(x,y) = xy2 and x3... and, well, let me put it below... and let c(t) = t and t2.1508

That is it. Essentially what we are doing here, let me go ahead and draw this out, this parameterization (t,t2), that is the parameterization of the standard parabola that passes through the origin y = x2. This is the parameterization.1532

It says that x is t, and y is t2, so this is the parameterization for the parabola. What we are going to be doing is we are going to be integrating this vector field along this parabola.1549

Okay. We want to find the integral of f along c or t in the integral from 0 to 3, so from 0 to 3, it is going ot travel along this path and say come up to here. Something like that. 0 to 3, the square of 3 is 9, so when we integrate the function along this curve, this is what we are going to get.1562

That is what this is, so let us just go ahead and do it. Well, the first thing we are going to do is we are going to find f(c(t)), so f(c(t)), nice and simple, we just plug them in and see what happens.1593

f(c(t)), c(t) is t and t2, so, this is the x value and this is the y value, so we are going to have t × (t2)2, and this is x3, so it is going to be t3, so that is f(c(t)).1606

That equals t3, and now we are going to form dc/dt, or c'(t), however you want to think about it. That equals... the derivative of t is 1, and the derivative of t2 is 2t.1626

When we dot these two, it is this × this + this × that, so we get f(c(t)) dc/dt t5 + 2t4 + 2t4, and that is what we are going to end up integrating. 1645

So, let us go ahead and write this out. The integral of f along c from the interval 0 to 3, it equals the integral from 0 to 3 of t5 + 2t4 dt.1674

This is a simple integration problem. You have been doing this since the first couple of weeks of calculus, so let us go ahead and do that and evaluate it. 1692

We have t6/6 + 2t5/5 + 2t5. We are evaluating it from 0 to 3 and when put these numbers in we get 729/6 + 486 -- oops, it is going to be 486 -- over 5, and that is our answer.1705

You are welcome to leave it like that. I do not know what it is that your teacher actually expects. If he actually expects a full evaluation, it is a perfectly good number, I am not going to go ahead and work that out or if he expects you just to stop at that point.1726

Once you have this, you know, evaluation is the easy part. If you do not want to evaluate it manually you can just use an algebra system like mathematica, maple, something like that, and it will do it for you.1739

This is the important part, not necessarily the evaluation. That is it. Okay. One comment before we actually do our final answer. Clearly our ability to evaluate a line integral depends on our ability to parameterize the curve.1752

Sometimes the parameterization will be given to you like these examples, sometimes you will have to come up with a parameterization yourself, and do not worry, we will be doing some examples in subsequent lessons where we actually have to come up with a parameterization for a given curve.1793

For example, in this last example, they might have said along the parabola y = x2, we would have to come up with a parameterization to see if t = (t,t2), something like that.1807

To parameterize the path, so our ability to evaluate the integral depends on our ability to parameterize the path because the definition is based on the parameterization c(t).1821

Okay, so let us do an example again. Slightly more complicated. Example 6. This time we will let our function f(x,y) = xy and we will do x2 - y3, and we will let our curve, c(t) = cos(t) and the sin(t) and in case you do not recognize that, this is a parameterization for the unit circle moving in the counterclockwise direction.1834

We want to evaluate the integral in the interval 0 to pi/2 so as t moves from 0 to pi/2, well as t moves from 0 to pi/2, we are going to move along this path right here and we are going to stop right there.1871

We are going to integrate this function along this path from here to here. That is what we are doing. Well? Let us go ahead and do it.1889

f(c(t)) = well, this is x, this is y, so we put it in there. So we get cos(t), sin(t)... let me write that a little bit better... cos(t) and sin(t).1901

Then we have x2 which is cos2(t) - y3, which is -sin3(t). Okay.1921

Now let us do c'(t). I am going to write dc/dt, so when I take the derivative of cos(t) I get -sin(t). I take the derivative of sin which is cos(t) and then I go ahead and I form the dot product, so I take this dotted with this, and I end up with the following. 1932

I am going to shorten the notation. I am just going to do f · dc, I do not want write out all the c's and t's and things like that. So f · dc, it is equal to, well, this dotted with that.1950

So, we are going to have this × this, so it is going to be -sin2... hmm, let us write this slowly, because we do not want these random lines to show up... -sin2(t)cos(t) + this × that. + cos3(t) - ... okay, actually you know what, let me go ahead and erase this and let me start on the next page. I want this particular integrant to be very, very clear.1961

So, f · d(c) = -sin2(t)cos(t) + cos3(t) - cos(t)sin3(t). 2001

When you take the dot product of those two, you are going to end up with that. That is your integrand. 2022

So, The integral of f along the particular path that we chose, in the given interval, is going to be the integral from 0 to pi/2 of - sin2(t)cos(t) + cos3(t) - cos(t)sin3(t) dt.2032

This is the important part. Being able to get here. The rest of it is just basically just technique, so I am not going to be able to integrate this manually. What I did when I did this was I just put this into my particular mathematic software. 2055

I used maple myself and it gave me the answer, nice and quick... and the answer is 1/12.2072

There you go. That is a line integral. You are doing the same thing that you did with single variable calculus except now the line is no longer just the x axis. 2078

We have taken that x axis, and now we can just integrate the function along any curve we want. In this particular case we have integrated a vector field, which is a vector from R2 to R2, or R3 to R3, or R4 to R4, where the dimension of the departure space is the dimension of the arrival space. 2089

In the next lesson I am going to talk about a -- so what we have done here is we have integrated a function from R2 to R2. We have integrated a vector field around a curve.2109

In the next lesson we are going to talk about multivariable functions. Let us say an R2 to R, where the arrival space is just the real number line. Dimension 1, what we have been dealing with, functions of several variables. 2125

You can also do line integrals of functions of several variables, not just a vector field, so we will define that in a subsequent lesson.2140

For the time being, all you have to really worry about is this basic definition. Form f(c(t)), form c'(t), take the dot product, you will end up with a function of a single variable t and then you will just integrate it, just like single variable calculus.2148

Thank you very much for joining us here at educator.com. We will see you next time, bye-bye.2164

Hello, and welcome back to educator.com and multivariable calculus. 0000

Today we are going to continue our discussion of line integrals, so, let us just jump right on in.0004

We are going to talk about notation, and as it turns out, you know, the line integral notation, there are several different notations for it. 0010

So, I am going to introduce a pretty standard notation and it is going to be the one we end up using more and more especially when we end up talking about double integrals and Green's theorem later on, which is the fundamental theorem of calculus in two dimensions.0019

So, let us introduce this notation, so that we are aware of it. Actually, you know what, I think I am going to do this lesson in blue. There we go. Okay.0032

So, notation. Okay. Let f(x,y) = f and g. So, this is a function. It is a function of 2 variables, it is a vector field. This is the first coordinate function, this is the second coordinate function, but notice I have not included the f(x,y), g(x,y), but it is implicit there.0045

We are going to start to use a slightly shortened notation, and c(t), the curve over which we integrate, or the line, is going to be some x of t, so x is a function of t and y is a function of t. Something like that.0070

Therefore, then the integral of f along c is equal to... we write it as f dx + g dy. That is the notation right here. fdx + gdy. Okay.0086

This just -- the integrand -- is just fg · dx/dy, so it actually has the advantage of sort of representing some dot product already.0106

Now, it is very, very important to remember of course that when we use notations, these are short-hand, so fdx + gdy is short for the following: let me write it up here, well, no, let me write it down here. It is short for f(x(t),y(t)) × dx/dt + g(x(t),y(t)) dy/dt dt.0128

When we multiply all of this out, this fdx + gdy is actually just short for that, which is equivalent to our more theoretical definition f(c(t)) · c'(t). Again, this is just a shorthand notation.0184

This is what it is when it is fully expanded, so, once again I am going to write this... so, again, the integral of f along c which is equal to the integral of f(c(t)) · dc/dt, c', dc/dt, just different notations... dt, forgot that little dt there... is equal to fdx + gdy.0207

This is the notation that is really, really important, but again it is just a question of notation as long as you remember what this means. This believe it or not is actually the important one. It is the definition of the line integral so that is the one that we are going to keep using. 0254

We will use this one more and more as we move towards Green's Theorem. Okay.0268

Often, a curve will be given in a non-parameterized way, so, a non-parameterized way.0276

In other words, they might say integrate this function or this vector field over the parabola as the parabola moves from this point to this point.0300

Well, you are going to have to come up with a parameterization because the line integral... the definition of the line integral involves the parameterization c(t).0309

So, you are going to have to find a way to take that curve -- which is expressed usually as a function of (x,y) equals some function of x -- and find a way to express it as a single parameter t.0317

In general, for y = f(x), let x = t and y = f(t).0328

This way, you will always get the parameterization that you need. Now that does not necessarily mean that this is the only parameterization, that is the thing about parameterizations. There is usually more than the one. 0345

The nice thing about it is that the integral itself is actually independent of the parameterization. Some parameterizations will make the problem easier, some will make it harder. There is more than one parameterization, so those are not unique. 0354

Okay. Just some general examples for the circle of radius r, x = rcos(t) and y = rsin(t). That is the parameterization for a circle. For a straight line segment between p and q, between the points p and q, the parameterization is as follow: c(t) = the point p + t × q - p.0369

We will actually see an application of this a little bit later -- where t runs from 0 to 1.0430

Again, the integral... this is the most important part... the integral is independent of the parameterization.0438

You know what, I do not like this capital letters, I will just stick to writing it out -- is independent of the parameterization chosen.0455

Profoundly important. So, two different parameterizations, same curve, same direction, you are going to get the same integral.0478

Now, when you choose a parameterization, you are specifying x a direction of traversal.0488

We will say more about this in the next lesson when we actually talk about reversing the path, integrating in one direction, integrating along another direction. You are specifying a direction of traversal.0514

In other words, if you were to parameterize that curve moving in this direction from here to here would be one parameterization, but if you wanted to move in this direction, if you wanted to move like this from here to here, it actually changes the parameterization.0528

There is a formal mathematical way of doing that and again we will see that in the next lesson when we see some examples, but for the most part what is important is that the parameterization that you use does not matter, the integral will be the same, that is what is important.0545

Let us do an example. Okay, so example 1. Find the integral of the vector field f(x,y) = (xy2,x2y) on the ellipse centered at the origin with major radius 4 and minor radius 2 from θ = pi/6 to θ = 3pi/4.0559

Okay, so find the integral of the function on the ellipse, so we are integrating along the ellipse, centered at the origin, the major radius is 4 the minor radius is 2 and we want to go from the point on the ellipse represented by pi/6 to 3pi/4.0630

Let us go ahead and draw what this looks like, and then we will do the parameterization of the ellipse.0646

We have that, major radius is 4 so this distance from there to there is 4, minor radius is 2, so something like this -- sort of like that, I am not the best ellipse drawer but something like this -- and θ = pi/6, that means this point and 3pi/4 is the 45 degree angle to this point. 0653

What we are doing is we are integrating from here to here, so let me do this in red. We are integrating along this path. Here to here. This function.0678

Let us go ahead and write our, so we have f(x), let us go ahead and write our c(t). As it turns out, you parameterize an ellipse sort of the same way you parameterize a circle.0692

You have rcos(t),rsin(t) but now the two radii are different, so it is just going to be this radius cos(t), and this radius sin(t).0705

What we have is 4cos(t) and 2sin(t), now, this particular parameterization, by choosing this parameterization I am actually moving in this direction. 0715

This is what we mean when we say any time you choose a parameterization you are specifying a direction. I am specifying a counterclockwise movement along this path.0728

If I had gone, let us say this way, let us say I wanted to parameterize from this point to this point, the parameterization would actually have to change. 0736

So, when you choose a parameterization, you are actually specifying a particular direction and it is very important that you know what that direction is.0745

You have to make sure that you understand all of the particulars of the objects that you are working with. Okay, so let us go ahead and flip the page. I am going to rewrite the parameterization of the function and then I am going to start working with it and integrate it. 0753

Let me go back to blue here. So, we have f(x,y) = (xy2,x2y), and c(t) = 4cos(t)2sin(t), and t is going to run from pi/6 to 3pi/4.0765

Okay. Let us form, as always, we form f(c(t)) · dc/dt. O0801

Okay, so, f(c(t)) = let us put -- this is x, this is y -- we put those in here, we get 4cos(t),sin2(t)... uh, no, sorry, that is not right... 4cos(t) × 2sin(t)2, and we get 4cos(t)2 × 2sin(t).0820

This is going to equal 2 × 2 is 4, this is going to be 16 × cos(t) sin2(t), and this is going to be 32cos2(t)sin(t).0853

Okay. Now we have that. Now let us form dc/dt. dc/dt = the derivative of this thing equals -4 × sin(t) and we have 2 × cos(t).0874

Now, when we form the f(c(t)) · dc/dt, it is this × that + this × that. We end up with -64cos(t)sin3(t) + this × that which is 64cos3(t)sin(t).0890

So, our integral of f over c is equal to the integral from pi/6 to 3pi/4 of this expression, -64cos(t)sin3(t) + 64cos3(t)sin(t) dt.0926

When we use our mathematical software, we end up with the number 2. That is it. I am certainly hoping that you are not trying to do this manually, it is not worth it. 0951

One of the nice things about having -- in fact, the best hting about having math software is now you are no longer confined to examples that simply for logistical reasons are simple so that you can get through them. 0960

Now you can choose any function that you want, any parameterization, and boom software gives you the answer. This is what is important, being able to set this up. This a computer can do. This is what you want to understand.0974

So let us go ahead and do another example here. A very, very important example.0987

So, example 2. Let us see here. Find the integral of the vector field f(x,y) = -y/x2 + y2 x/x2 + y2, around the circle of radius 4 from (4,0) to the point (2,2sqrt(3)).0995

In this particular case, they are actually giving us two specific points, from this point to this point.1068

So, what we are going to have to do is, in order to find the parameterization, we are going to have to find the lower and upper limits of integration. 1073

We are going to have to convert this point and this point to t values based on the parameterization that we choose. Okay.1080

Let us go ahead and do a little geometry here just to see what this looks like, so that we make sure we understand what we are doing.1086

So, we have this circle of radius 4, right? This is 4, that is 4, we know how to parameterize that, that is going to be 4cos(t), 4sin(t).1092

What we are doing is we are going from this point, the point (4,0) all the way to this point right here (2,2sqrt(3)) and we are moving in this direction, that direction, that is the parameterization.1102

Our c(t) is going to be the 4cos(t), 4sin(t)... now, when we do the integration, we need values of t because we are integrating along the curve so the t values are going to be the lower and upper limits of integration.1122

So, this point is represented by t = 0, so at the point (4,0), t = 0, so that is our lower limit of integration. Well, how about here, what is t going to be when we get to here?1143

Well, at the point (2, 2sqrt(3)), well, let us go ahead and solve, 4cos(t) is the x value here, this is the y value, let us just go ahead and solve for t.1161

So, 4cos(t) = 2, cos(t) = 2/4 = 1/2, and since we are in the first quadrant, t = pi/3.1177

So, we are going to be integrating from 0 to pi/3, that is our lower limit of integration, that is our upper limit of integration. 1194

Notice that in this case they did not give us the parameterization and the t values that we are going to traverse, they gave us 2 points and they gave us just the geometric on this unit circle. 1201

We came up with a parameterization, that should not be a problem. We need to find out what the t value is so we have to evaluate one of these to find the t values.1212

t goes to 0 as t moves up to pi/3, that covers that segment of the circle.1221

Okay. So, t is going to run from 0 all the way to pi/3. Alright, let us go ahead and solve the integral. Well, f(c(t)) when I put this parameterization into there, I end up with the following.1231

I end up with -4sin(t)/16cos2(t) + 16sin2(t), and I end up with 4cos(t)/16cos2(t) + 16sin2(t).1255

Okay. This of course is equal to -4sin(t)/16, cos2(t), sin2(t) = 1, so that factors out, and I have 4cos(t)/16. Let us get rid of these random little lines that tend to show up at the bottom of the page.1284

This is going to equal sin(t) - sin(t)/4, and... I am sorry, -4... oh, this is ridiculous, let us see... -4sin(t)/4, and 4... I have already cancelled, now I am the one that is messing up arithmetic. 4 and 16, 4 and 16, so I get -sin(t)/4 and I get cos(t)/4.1312

How is that? That is good. Okay. Now, let us do dc/dt.1351

So, dc/dt, or c'(t), that is going to equal -4sin(t) and 4cos(t), so, f(c(t)) · dc/dt is going to equal -- let me rewrite the f(c(t)) again -- f(c(t)) is going to equal (-sin(t)/4,cos(t)/4), so it is the dot product of this thing and this thing.1359

When I do that, I am going to get sin2(t) + cos2(t) and that is going to equal 1, so our final integral, the integral of f over this particular curve is going to be the integral from 0 to pi/3 of just plain old dt, 1dt.1404

Of course, that equals pi/3.1427

Here we are busy trying to solve this line integral, and we end up with notice, 0 to pi/3 and we end up with an answer for this integral is pi/3. This is not a coincidence.1433

Okay. This is a very, very important result and a very important vector field, especially for those of you in physics and engineering. You are going to see this vector field over and over and over again.1443

Let us just discuss this real quickly. So, this is a very important result and a very important vector field.1465

Here is the result. Here is why it is important: if you ever integrate this vector field along any curve, along any... and this one I definitely will put in capitals... along any curve, from point p1 to p2, and op1 and op2 make angles of θ1 and θ2, respectively, with the x-axis. I will draw this out in just a minute so that you see what happens... respectively with the x-axis, then the integral of f along that particular curve is equal to θ2 - θ1, or δθ. 1481

Here is what we mean by that. So, let us say we have this coordinate plane and let us say we have this curve that is going like this, okay? 1565

If this is point p1 and this is point p2, p1 and p2, if this is θ1 and this is -- this angle is θ1 and this angle is θ2, remember? we always measure our θs from the positive x-axis moving in a counterclockwise direction.1578

Then the integral of this particular vector field, f(x,y) -y/x2 + y2, x/x2 + y2, the integral of this vector field over any curve, no matter what curve you trace, whether it is like this or this or it could be all whacky and end up here, the difference in the angle, the integral of this vector field is always going to be the difference between θ2 and θ1, no matter what it is.1604

It does not have to start at the origin by the way, notice here this is one angle, this is the other angle. In our example, we start from the origin and we integrate up to pi/3.1636

The angle difference was pi/3, the integral was pi/3. Here, same thing. If this were 3pi/4 and this were pi/4, well, we would have 3pi/4 - pi/4, 2pi/4, the integral would be pi/2.1646

This is a profoundly important vector field. You will see it over and over and over again. 1659

It will save you a lot of time if you recognize that as you integrate this vector field over absolutely any path, where this angle arrangement is involved, you can just take the different of the angles and that is the integral.1665

Thank you for joining us here at educator.com, we will see you next time for a further discussion of line integrals. Take care, bye-bye.1678

Hello and welcome back to educator.com and multivariable calculus. 0000

The last few lessons, we have been talking about line integrals and we are going to continue our discussion of line integrals and we are going to discuss closed paths today. Let us go ahead and get started.0004

Okay. So, first of all, we will go ahead and talk about a piece wise continuous path and then we will go into closed paths.0014

If a path is given in a piecewise continuous fashion -- and I will draw this out in just a minute -- in a piecewise continuous form. 0024

In other words if you had something like that, like this, and maybe this little half circle like that, and let us say, this is curve 1, this is curve 2, this little circle here is curve 3 and this one is curve 4 and the points of continuity are here, so it is continuous, it is just called piecewise continuous.0049

Here, if this were the case, then the integral of f along c... c is the entire thing, well it is just equal to the integral of f along c1 + the integral along c2 + ... the integral of f along cn... however many pieces there are in this curve, that is it.0074

Essentially what you are going to be doing is you are going to be parameterizing each segment and then just adding up the integrals. Completely intuitive. Okay.0100

Now, let us talk about closed paths. So, we say a path is closed if the end point of cn, the last curve, is the same as the beginning point of c1, the first.0109

Let us take a couple of points like this. Let us take p1 over here, and let us take p2 over here.0150

So, we might have, let me go ahead and draw some dots so that we know what we are talking about. So, we might have something that goes like this and we are parameterizing it in this direction.0160

That is one path or we might have another one, we might have one here and there, something like that.0174

So, let us say we start over here and we move in this direction, so this will be -- let us just call this p3, let us call this p4, these are also point, so you know we start over here... so this is our start, we come this way, we come over here and then we go this way, then that way, then this way.0191

Then let me make sure to put these arrows here so that we know where we are going, and then we end up here where we started, so this is a closed path.0215

In other words, p1 and p5 are the same. That is it, a closed path. That is exactly what you think it is. It is just a closed path.0222

Okay. So, let us go ahead and do an example here of a line integral on a closed path, so example 1.0231

We want to find the integral along the path c, which is equal to c1 + c2, and we will draw this out in just a minute for the following function. 0243

So, we are going to draw it out, and from the drawing of the actual path, we are going to find a parameterization for it. We are not given the curve, we are actually going to have to come up with the parameterization ourselves for the following function.0267

f(x,y) = x2,xy2.The particular path we are going to be following is this one.0282

Okay. So, this is going to be the point (1,0), and this is going to be the point (0,1), so we are going to follow... our path is going to be like that, and then it is going to be like this.0298

So, in a counterclockwise fashion we are going to follow the circle from here to here, and then we are going to follow the line from here to here.0312

We want to know what the integral of this function is along this closed path. Let us see what we can find out.0320

Let me go ahead and mark some points so this is... oops, got that reversed, we are on the x-axis here... so this is the point (1,0), and this is the point (0,1), so we will go ahead and call this c1, and this is c2. Okay.0329

Well, the integral of f along c is just going to be the integral of f along c1 + the integral of f along c2. We are going to have to solve 2 integrals here.0344

We are going to parameterize this one, we are going to parameterize this one, we are going to do the integrals separately and that is it. So, let us go ahead and parameterize these curves.0356

So, c1, let me do this in blue actually. There we go, so c1 of t, well, c1 of t is just the quarter circle from this point to this point, so the parameterization is going to be cos(t) and sin(t), and t is going to run from 0 to pi/2. That is it. 0365

That is the standard parameterization of the circle. That is it. 0398

okay... yes, and c2, well c2(t), this is a line segment, a line segment from here to here. How we parameterize a line segment is we take the beginning point + t × the ending point, I am going to call it... let us call this p1, and let us call this p2.0403

So, p2 - p1, well p1 is the point (0,1), and notice I am writing it as a column vector simply because it makes it a little easier for me to see. I think about it this way.0428

So, (0,1) + t × p2, which is the column vector (1,0) - (0,1), so let us go ahead and solve this. 0443

We get (0,1) + t ×, well 1-0 is 1 and 0-1 is -1, so now let us go ahead and put this together.0459

That is fine, I will do it over here... so this is going to be... so the top is going to be just t, and this is going to be 1 + t × -1, so it is going to be 1 - t. That is our parameterization. If you want to see it in terms of a row vector like this, it is going to be (t, 1-t), so we actually found the parameterization for this particular line segment.0473

So, we are going to do this line integral, we are going to do this line integral for the function, and then add them together. 0502

So let us go ahead and do f(c1(t)), so f(c1(t)... well we put this into these right here. This is x, this is y, so we put them in.0508

We end up with cos3(t), then cos(t)sin2(t), so this is going to be our f(c1(t)). 0525

Now let us do our c1'(t), or I will write it as dc1/dt, in keeping with standard single variable notation.0538

So, the derivative of c1(t) is -sin(t), and then we have cos(t), there we go, and then when we take the dot product of this and this, we are going to end up with -sin(t)cos3(t) + cos2(t)sin2(t).0546

This is what we are actually going to integrate, so now the integral of... should I do it here, or should I do it on the next... that is fine, I can do it here... you know what, no, I can do it on the next page. I do not want these stray lines.0576

The integral of f along c1 is going to be the integral from 0 to pi/2 of the expression that we just had, -sin(t)cos3(t) + cos2(t)sin2(t) dt.0593

Then, when you go ahead and put this into some math software, because I am hoping that you are not actually going to do this by hand, you are going to end up with -1/4 + pi/16.0616

Just a number, that is it. That is what an integral is. Just a number. Okay. So that is our first curve.0625

Now, let us go ahead and do our second curve. So, now for c2. Again, I will go ahead and write out the parameterization. It is t and 1-t, and let me go ahead and just write f again just so we have it on this page.0632

It is going to be, I think it was going to be x3, and I think it was going to be xy2, is that correct? Yes, that is correct, so we have f. 0650

Now, let us go ahead and form f(c2(t)), so let us go ahead and put these values, this in for x and this for y, and we go ahead and work it out so you are going to end up with t3 and t × 1-t2.0661

Then we solve for this one, this is going to equal t3, we expand this out and multiply it out. We are going to get t - 2t2 + t 3, right?0680

So, expand this out, binomial, multiply, so there you go, now you have f(c(t)), or c2(t), now let us go ahead and do dc2/dt, so c2'.0696

Well that is nice and easy, that is just going to equal 1 and -1. That is nice and simple.0708

So, when we go ahead and we take the dot product of this and this, which is going to be our integrand, we end up with t - t + 2t2 - t3, which is equal to 2t2 - t3, so this is our integrand.0715

This is the thing that is going to be under the integral sign, so, the integral of f along c2 is going to equal the integral from 0 to 1, remember the parameterization for a line is really, really... for a line segment is easy because you are always going from 0 to 1. That never changes. 0742

0 to 1... of our 2t2 - t3 dt, and this one of course you can do by hand, so this becomes 2t3 over 3 - t4/4 from 0 to 1, and this gives you... I hope I did my arithmetic correct... 2/3 - 1/4.0765

Our final integral of f/c is equal to the integral of f/c1 + integral of f/c2 = -1/4 + pi/16, let me put boxes around the 2 numbers... + 2/3 - 1/4. 0792

When you add all of that together, you get -1/2, you just put the 1/4 together, I did not feel like... I am going to leave the 2/3 alone, it is just a number it does not really matter.0818

pi/16, I have never been a big fan of too much simplification myself. I believe that too much simplification actually -- well, in terms of arithmetic it is fine, but in general when you are working with math, you do not want to oversimplify too much just to make it elegant, because then you start to hide things. You want to see what is happening.0828

+ 2/3, and there you go. That is your final answer. Nice and simple, how is that?0846

Now let us talk about the reverse path, so, if we are going to integrate from one point to another along a path, what happens when we actually go from this point to this point. What do you think happens?0855

Well intuition might say it is just the reverse integral. The answer is yes, you are just going to switch signs. Let us just talk about this formally really quick.0867

So, the reverse path... the reverse path, okay... now, let c be a path -- actually I think I am going to go back to black here. I like to do my discussions in black -- the reverse path, so let c be a path traversed in a given direction. That is fine.0878

In a given direction... Then, c with a little negative sign up in the top right is the notation for the path traversed in the reverse direction, in the opposite direction, or reverse direction.0933

So, nothing strange here, just a question of notation, so if I have p and if I have q, and if I have some path where we go this way from p to q, well that is c.0964

The reverse path if we go this way, that is just c-, that is it.0975

Now, let us write out a theorem. Let f be a vector field on an open set... on some open set s... and let c be a curve in s defined on the interval a, b, so the beginning point and the ending point of the particular curve, for the parameterization.0981

Then, the integral of f along the reverse path is equal to -the integral of f along the path in one direction. That is it, you are just going to switch signs.1029

Now, this is what is important. The parameterization for the reverse path is not the same, very, very important... is not the same as for the forward path.1047

Now, formally, there is actually a way to convert a... if you have a parameterization for a particular curve, to actually get a parameterization for the reverse path, and it would go like this.1082

So, formally the parameterization of c(t) = c(a+b) = t, so let us just do an example.1095

Fortunately we are not going to have to worry too much about reverse paths and actually coming up with parameterizations for reverse paths.1110

Generally we just come up with a parameterization for a particular path that works, and if we need to go in the reverse direction, we just switch the sign of the integral. That is what is nice.1117

You do not have to actually reparameterize it and do it that way, just switch the sign. That is what this lets you do. 1125

Let us just do an example for the heck of it. So, example 2.1131

Let c(t) equal our standard cos(t)sin(t), parameterize our circle t is in 0 to 2pi.1140

Find the parameterization for traversing that circle in the opposite direction, so clockwise instead of counterclockwise.1155

So by our definition, c-t = c(a+b) - t.1168

Well, that is equal to c of, this is a, this is b, 0 + 2pi - 2 = c(2pi-t) and c(2pi-t) = cos(2pi-t)sin(2pi-t). That is it Nice and simple.1178

Basically you have this circle that you have traversed in the counter clockwise direction. That is the normal cos(t)sin(t).1210

Now if you use this parameterization, you are traversing it in this direction. If you integrate along this way, it is just the reverse of integrating along this way, you just change the sign of the integral. Good.1219

When you are given a particular curve, one parameterization might just be sort of natural. If you look at it you might say "oh, let me just do this."1241

But perhaps you are traversing it in the opposite direction. Well, you do not have to change the parameterization, just take the integral with a parameterization that you already have and just switch the sign. That is it. 1249

So, the integral of f, c- = the integral from b to a of f(c(t)) · c'(t) dt.1260

That equals negative the integral of a to b. What we have been doing of f(c(t)) this is just everything written out. Let us see c't/dt, that is it. Just change the sign.1278

If you are going to integrate from a to b, but you go from b to a, just change the sign.1292

So, now let us move on to another topic. It is still line integrals but up to this point, we have defined line integrals with respect to vector fields.1298

We said that there is some vector field, some function which is a mapping from R2 to R2, R3 to R3, well, it is possible but we have also been dealing with functions of several variables where we are mapping R2 to R, R3 to R, where you have just one function instead of a vector field.1311

Is it possible to actually integrate a function over a path, a contour the same way that we did for single variable calculus. The answer is yes.1327

Let us go ahead and define what we mean by the line integral of a function of several variables and then we will do an example.1337

So, this is going to be line integrals of functions on RN, and remember, functions again, that is the word that we set aside to use when our arrival space happens to be the real numbers, not R2 or R3, when it is just R, the real numbers.1346

So, now let us define, so let us define the integral of f along c and notice I have used a small f so generally we will us a capital f for -- oops, wow, that is a crazy line, look at that -- so generally we will use a small f for functions, a capital F for vector fields, at least I will, unless there is something in the problem where we have to do something else.1377

So, the integral of a function over the curve is equal to the integral from a to b of f(c(t)), so this part is the same except for instead of a dot product because we are not dealing with a vector is we are going to take the norm of c'(t) dt.1411

With the definition of a line integral, what we need to do is f(c(t)) · c'(t), but now we are dealing with a function not a vector field, so it is f(c(t)) still, except now you are just multiplying it by the norm of the derivative of the actual curve itself.1438

Let us just go ahead and do an example. Example 3.1456

Now, let us go ahead and let f(x,y,z), we will do a function of 3 variables, x2,y,z3.1471

So you see we plug in a vector, a three vector and we get out a number, it is a mapping from R3 to R.1483

Well, let us see, let us go ahead and let c(t), the curve along which we parameterize, let it be the spiral in 3 space defined by cos(t)sin(t) and t, and let us go ahead and let t go from 0 to 3pi/4.1491

Just to let you know what this looks like, I am going to draw out my 3-dimensional... okay, and I ill go ahead and go back that way, this way, so this thing that is coming out is the x-axis. This thing is the y-axis, and this is the z.1510

The z and the y are the ones that are in the plane, and the x-axis is the one coming out at you. 1528

This spiral, cos(t),sin(t),t, if it is just cos(t) and sin(t), it is the circle, the unit circle in the x,y plane, however with this t, now we have this... so what you are going to get is it is going to be from 0 to 3pi/4, so we are going to move along 135 degrees.1534

We are actually going to end up coming up, so if I were to take the x,y plane and look at it just from the x,y plane, I would see this curve that is going up like this. Sort of like a ramp of a parking lot that works in a spiral.1558

That is exactly what it is. It is basically a spiral because now you are also, you are not just moving along x and y, you are also moving along z for every step that you take. So, you are going like this.1576

We are going to just -- let us see what we can do with this line integral here -- okay.1588

So, f(c(t)), so let us find f(c(t)), that is going to equal -- actually I do not need this because I am not talking about a vector.1595

We have cos2(t) × y which is sin(t) × z3, so × t3, so this equals, in general we put the variable first, so t3cos2(t)sin(t).1608

Now let us go ahead and find c'(t), well c'(t) is -sin(t), cos(t) and 1.1630

Now let us go ahead and find the norm of c'(t). The length, in other words, of the tangent vector at any given moment, that is going to equal this squared + this squared + this squared all under the radical. 1642

So, sin2 + cos2 + 1 under the radical. And sin2 + cos2 is 1, 1 + 1 is 2, so we have sqrt(2).1661

The integral of f along this particular contour is equal to the integral from 0 to 3pi/4 of t3cos2(t)sin(t) × sqrt(2) dt.1673

Then when we put this into mathematical software because again I am not going to do this by hand, we get the following answer: -13pi/12 + 9/28pi3 - 41/27 + 15/32pi2.1695

Again, it is just a number expressed in terms of pi and other numbers, nothing simplified, put it in a calculator and you are just going to get a number.1721

It is possible to actually have a function of several variables and integrate that function just like it is possible, what we have been doing, you know, it is what we started off with, integrating a vector-field over a certain path.1729

That is it, everything is exactly the same except the definition is slightly different. Instead of doing a dot product, you are just multiplying by the norm of the tangent vector which is just dotting it with the vector itself.1747

That is it for line integrals here. Thank you for joining us here at educator.com and we will see you next time for a further discussion of potential functions. Bye-bye.1760

Hello and welcome back to educator.com and multivariable calculus.0000

Today we are going to start our discussion of potential functions, so let us just jump right on in.0004

We are going to start with a definition of course, of what a potential function is. This is a very, very, very important discussion, especially for those of you in physics. 0009

Well, not necessarily just for physicists but in particular in physics because most of classic physics is based on the notion of a potential function, a conservative potential function.0020

In any case, let us just jump in and see what we can do mathematically. Okay.0031

So, definition. Let f be a vector field on an open set s -- okay -- if small f is a differentiable function on s subject to... the gradient of small f happens to equal capital F... the vector field, then we say f is a potential function for f. Small f is a potential function for capital F.0037

So, what we are saying is if we are given a vector... so basically what we have been doing up to this point... at least earlier before we started the line integral discussion, we have been given functions and then we have been taking the gradient of that particular function.0117

Now what we are saying is -- and the gradient is a vector field -- what we are saying is that we are not going to start with a function, we are actually going to give you some vector field.0134

We want to know if it happens to be the gradient of some function. That is all we are saying.0144

You might think to yourself, of course if you have a vector field it comes from somewhere -- as it turns out, that is not the case.0150

This is analogous to single variable calculus when you are given the particular function, can you find the function that this is the derivative of. That is what we are asking.0157

So, if we are given a vector field first, does there exist a function such that this is the gradient of... that is what a potential function is, it is that particular function.0168

So, let us just sort of write all of this out. So, up to now, we have started with f and taken grad f to get a vector field. The vector field is the gradient.0181

If given a vector field first, can we recover a function f? That is the question.0213

If we can, then f is a potential function. f is a potential function for the given vector field.0236

The idea -- well, this one I do not have to write out -- so again, the idea of finding a potential function for a vector field is analogous to finding the integral of a function over an interval.0269

Does one exist? Precisely because the idea of taking the gradient of a particular function which we have been doing is analogous to taking the derivative. That is what we are doing when we are taking the gradient. We are taking partial derivatives.0280

But essentially a gradient is, taken as a whole, is the derivative of a function of several variables.0293

Now we are just working backwards. This is analogous to trying to find, we are just trying to integrate a gradient is essentially what we are doing. Can we integrate it?0301

If we do, does a function exist? Well, in all questions like this, that is essentially the central question. Does the potential function exist, and if it does, well, is it unique, and how can we find it? 0310

Does are sort of the 3 questions that occupy pretty much the majority of science, of mathematics. Does something exist, is it unique, and can we find it? Is there a way that we can find it, or can we just say that it exists. 0325

It is nice to be able to say that it exists, or it does not exist, but from a practical standpoint, we want to be able to find it. Okay.0339

Let us see what we have, so, let us talk about uniqueness first. That is actually the easy part, so, uniqueness.0347

So, let us start out with a definition here, so we are going to define something called a connected set. Excuse me.0361

An open set s is called connected if given 2 points, p1 and p2 in s there exists a differentiable curve contained in s, oops, let me write that a little bit better here, contained in s, which joins p1 and p2.0372

So, this is just a fancy definition for something that is intuitively clear, so let us go ahead and draw a picture.0432

We might have some set s like that, this might be p1, this might be p2, so this is p1, this is p2, is there some differentiable curve that actually connects them.0438

It could be a straight line, it could be some series of straight lines, a piecewise continuous, is there some path that actually connects them, some differentiable path that connects them. So that is a connected set.0452

That is it. Essentially it is just a set that is one piece, and if I had something like this, this is p1 this is p2, well then this particular case there is no differentiable curve that actually connects them, this is still an open set and it is still a viable open set, it is just not connected.0465

There is no path that connects them that stays in the open set. I am going to cross the boundary and I am going to be out in this no man's land. So, this is not connected. 0483

So, we need this notion of connection here. This is a not connected, and this is a connected set. Now we will go ahead and state our theorem.0495

Let s be a connected set -- uhh, let us say a little more than just connected -- connected open set... okay... if f and g are two differentiable functions on s and if the gradient of f happens to equal the gradient of g for every point of s, then there exists a constant -- oh by the way this reverse e is a symbol for "there exists" -- a constant c such that f of some point in s is equal to g of some point + c for all points x in s.0509

So, basically what this means is that open connected set, if you have two functions, f and g, where if you take the gradient of f and it equals the gradient of g at every point, then essentially what we are saying is that those two functions are exactly the same up to a constant.0602

Again, this constant is not much of an issue. They are the same, they only differ by a constant. 0619

This is sort of analogous to what we did in single variable calculus. On a given interval, they have 2 functions f and g. If f = g everywhere along that interval, well then the 2 functions are the same, the only different between them is this particular constant, which does not really matter because again, when you differentiate a constant it goes to 0. 0629

For all practical purposes, they are the same. So, on an open connected set, what this theorem is ultimately saying is that a potential function is unique. Okay.0650

Now, we will go ahead and deal with existence. This is going to be the important one. So, now existence of a potential function and hopefully eventually we will talk about how to actually construct this potential function. Now, existence.0662

Let f(x,y) = f(x,y), g(x,y), so these are my coordinate functions for this vector field.0684

We want to know if and when a potential function exists for f. We want conditions that will tell us when something exists, when this potential function exists, for the vector field.0700

So, let us go ahead and give this a symbol that we are going to use over and over again. Let us call this capital pf(x,y). Potential function for the vector field f, that is what this symbol means.0730

That is, we want to know if -- again, I am going to use this reverse e -- if there exists a pf of (x,y) such that dpf dx = f, and d(pf)/dy = g. That is what we are doing. 0754

We want to know if there is a function that when I take the gradient of that function, this pf, I end up with a vector field. That is the idea of a potential function.0803

If given a vector field, is this a gradient of some function.0812

The idea is actually really, really simple in and of itself. What is going to be difficult from a practical standpoint is keeping things straight.0819

Up to now we have been going from functions to gradients. Now we are just going to drop a vector field on you and we are going to be trying to go backwards, so, take a couple of extra seconds to make sure that you have your function separated, that you know whether you are dealing with a function or a vector field f, its coordinate functions, and that these might be the derivatives, this f and g of the original function we are trying to recover.0827

There are going to be lots of f's and g's and f1's and f2's floating around. Make sure you know which direction we are moving in. That is really going to be the only difficult part, the logistics of this, in and of itself, it is not... the concept is not difficult. Just be careful in the execution.0854

So, again, so we want to know if there exists some potential function such that the partial with respect to x is equal to this, and the partial with respect to y is equal to this. okay.0871

let us see if we can come up with a test here. Let us suppose there is.0885

Suppose yes. A pf exists for a given vector field. Then, df/dy if I take the partial -- let me write these out again, up here, because I want them to be on the same page -- so RF is equal to f and g.0892

Now, suppose yes, that some potential function exists. Then, if I take the derivative of f with respect to y, that is going to equal the partial with respect to y of well, f happens to be the partial derivative with respect to x of the potential function.0930

This is just d(pf)/dx, right? because we are supposing that the potential function actually exists. The potential function is this.0953

The d/dx is f, now if I take the df/dy, it is going to be this thing. Well, that equals d2 pf dy, dx, right? Okay. So we have this thing.0965

Now, let us go ahead and do the other. Let us go ahead and differentiate g with respect to x. That is going to equal d/dx, and we said that g was equal to the partial of the potential function with respect to y, so that equals d2 pf dx, dy.0987

d2 of this, dy dx, d2 of this dx, dy, what do we know about mixed partials? that it does not matter what order you take them in. They are actually equal. These 2 are equal, because, remember? Way back in previous lessons, mixed partial derivatives are equal.1017

Because these are equal, what we get is this and this are equal.1039

So, df/dy = dg/dx. By supposing that a potential function actually exists, we actually derived a relation between the partial derivatives of the vector field, component functions of the vector field. 1049

If I have a vector field, if df/dy = dg/dx, that is a product of the fact that a potential function exists. This actually gives me a nice way of testing to see whether a potential function exists given a vector field. Okay.1066

So, now we have a test, but we have to be careful in how we implement this test, and now we are going to get into a little bit of mathematical logic. If this, then this, we are going to have to see what implies what, we are going to have to be a little extra careful.1085

Now we have a test. Okay. So we are going to express this test as a theorem. Now, let f and g be differentiable with continuous partial derivatives. 1103

This just means that they are well behaved, and again, the functions that we are going to be dealing with are well behaved, so for all practical purposes whenever I write this out in a theorem, it is not necessary to write them out but I think it is sort of good to get used to see how theorems are written.1136

You are going to see all of these things and it is important that all of these hypotheses be in place formally.1150

Partial derivatives on an open set, s in 2-space, so we are going to deal with 2-space first and we will deal with 3-space in just a minute... 2-space.1155

Okay. Here is the if-then part. If df/dy does not equal dg/dx, or using capital denotation if d2(f) does not equal d1(g), remember this capital D notation? It says take the derivative with respect to the second variable of f.1173

Take the derivative with respect to the first variable of g. In this case we have 2 variables, x and y, so the second variable is y, the first variable is x.1200

This and this are just two different notations for the same thing. Then, there is no potential function for the vector field, let me write this a little bit better, for the vector field f which is comprised of the functions f and g.1209

Okay. So, if I am given a vector field f and g, all I need to do is I need to take the derivative of f with respect to y, the derivative of g with respect to x and I need to see if they are equal.1248

If they are note equal, I can automatically conclude that there is no potential function and I can stop right there. That is what this test is telling me.1260

We came up with that by supposing that a potential function exists. If a potential function exists, then df/dy has to equal dg/dx.1270

Our test is written in the contrapositive form. We work in reverse. If the potential function exists, then this has to be true. That is the same as if this is not true, then a potential function does not exist. I am going to be talking about that in just a minute.1278

But, again, from a practical standpoint, our test is this, if I am given a vector field, one coordinate function, second coordinate function, I take the derivative, I take df with respect to dy, I take df/dy, I take dg/dx, I see if they are equal.1295

If they are not equal, then I can conclude that a potential function does not exist. However, if they are equal, that does not mean that I cannot conclude that a potential function does exist. The direction of implication is very important.1311

This is profoundly important. Be very, very careful in how you actually use these theorems. Use it as written. If df/dy does not equal dg/dx, there is no potential function.1327

So, let us just do an example real quickly. Then we will continue on with a discussion of the subtleties of the mathematical logic involved here. So, example 1.1339

Let f(x,y) = sin(x,y) -- I should write this a little bit better here -- = sin(x,y) and cos(x,y), so that is our vector field.1352

This is our f and this is our g. Well, let us go ahead and find df/dy and dg/dx.1374

df/dy = x × cos(x,y) and dg/dx = -y × sin(x,y). Well, this and this are not equal. Therefore, there does not exist a potential function.1382

In other words, this vector field does not have a potential function. So, df/dy not being equal to dg/dx implies that there does not exist a potential function for this vector field f.1413

This is how we use our test. Very, very nice. Let me go ahead and give you the analog for this in 3-space.1437

The analogous test for 3-space -- now we are going to be talking about 3 variables, we just take them 2 at a time -- is as follows.1452

The same hypotheses, of course, and plus the partial derivative part of the test is -- I am going to write this in terms of capital D notation -- D1(f2) does not equal D2(f1), D1(f3) does not equal D3(f1). 1469

Again this is where you have to be careful. Just take a couple of extra seconds to do this nice and slow.1505

D2(f3) does not equal D3(f2). Okay. Then, f(x,y,z), which is the first coordinate function f1, f2, f3, so here instead of using fg whatever, I am just using first function, second function, third function... does not have a potential function. 1507

So, let us go ahead and talk about the notation a little bit more. I am going to be using this one, but I am also going to be using the other one.1546

Okay. This says -- what I have to do is I have to check the derivative of the second coordinate function with respect to the first variable, and then I have to compare that with the derivative of the first function with respect to the second variable.1552

The derivative of the third coordinate function with respect to the first variable, I have to check to see if it is equal to the derivative of the first function with respect to the third variable.1565

I have to check the derivative of the third function with respect to the second variable, and see if it is equal to the derivative of the second function with respect to the third variable. 1575

The three variables are the first, second, and third, respectively, are x, y, and z. The three functions are f1, f2, f3.1584

So, this looks like, out here I will do this in blue, let me see if I can get this right.1595

This is going to be d -- that is plenty of room I do not need to squeeze it in here -- this is d(f2)/dx does not equal d(f1)/dy.1605

This one is d(f3)/dx does not equal d(f1)/dz, third.1623

This one is d(f3)/dy does not equal -- excuse me, d(f2)/dz.1635

Take some time to look at this carefully. First, second, third variable, x, y, z. First, second, third coordinate function of the vector field.1648

I am mixing -- it is okay, so this sometimes, some people like this notation, some people like this notation, just make sure that you pair up properly. This notation is nice because you have 1, 2, 2, 1... 1, 3, 3, 1... 2, 3, 3, 2.1658

All of the indices are taken care of so you can see that. It is just a question of getting used to this, you are probably not used to this too much, depending on what notation your teacher uses.1673

So, now let us go ahead and discuss a little bit more about how this theorem is stated, and talk about some mathematical logic.1683

So, pay close attention to how the theorem is stated. How the theorem is stated.1691

It is stated in something called contra-positive form.1718

Contra-positive form is if not a, then not b. Okay? This is equivalent to its positive form.1734

That is the standard form. If a -- oops, sorry -- if b, then a. So if I am given if b, then a, that is how we actually did it. That is how we derived... how we presumed if it has a potential function, then df/dy = dg/dx.1763

But we use it in contra-positive form because we want to have a test. If not a, then not b.1800

In other words, if df does not equal dg/dx, then we know that a potential function does not exist. These 2 are equivalent. Generally, well, actually we will talk a little bit more about it.1806

So let me write that down, so if there exists a potential function, then df/dy = dg/dx.1819

The contrapositive form is the way we used it as a test. If this does not exist, then, there is no potential function. Okay.1839

Now, the converse is not true. Let me circle this. This is the positive form. If there exists a potential function, then df/dy = dg/dx.1848

Okay. The converse is not generally true. The converse would be if df/dy = dg/dx, then a potential function exists. 1860

All I do is reverse the if/then part, I leave the if/then, except I take this over here, this over there. If I just switch the places, that is the converse, that is not generally true.1882

However if I switch places and I negate both, that is the contrapositive, that is true. It is equivalent to the positive.1893

Now let us go ahead and write this out, so, positive, we have if b, then a. That is the positive, that is the standard form.1901

The contrapositive is if not a, then not b. This is equivalent.1917

This is equivalent to the positive. That we can always use.1934

The converse is, notice, if b, then a... if a, then b... in other words, can I conclude that it works in the other direction? No. This is a completely different statement in fact -- I will not write different statement, let me write not equivalent to the positive.1940

This is not equivalent to the positive, so, now in terms of our theorem.1963

So, our theorem, the positive of our theorem says if a potential function for a vector field exists, then, d2f1 = d1f2, and so on for all of the other partials depending on how many we are talking about.1978

In order to be useful in a test, we take the contra-positive form.2006

So, the contra-positive, if d2f1 does not equal d1f2, then the potential function for f... well, let me go ahead and use some symbols... then there does not exist a potential function for f.2014

The converse, which is if d2f1 = d1f2, if I just switch this and switch this, then there exists a potential function -- I will do this in red -- not true.2045

This is true. The contra-positive is true, it is how we use it as a test. The converse, not generally true. In generally you have to actually prove the converse if it were true. In this case it is not. 2074

You are going to find potential functions where the partial, where d2f1 = d1f2, but a potential function does not exist, so I cannot necessarily conclude that this is the case.2087

So, think about it this way. I can conclude that if it is raining outside, then it is cloudy. That is an automatic implication. If it is raining, then it is cloudy.2098

Well, I can also conclude the contra-positive. If it is not cloudy, then I know that it is not raining. That is the contra-positive. Now let us go back to the positive, if it is raining, then it is cloudy, but I cannot necessarily conclude that if it is cloudy, it is raining.2108

It can be cloudy, but it cannot be remaining, that is possible. The converse and the contra-positive are two entirely different things.2123

Often, in this particular case, we have the positive that if a potential function exists, then these partials are equal.2131

We use it as a test, we use the contra-positive version to exclude the possibility. If the partials are not equal, then the potential function does not exist.2137

Be very, very careful in how you use these and do not mix the converse with the contra-positive. Just because something implies something else does not mean that it works in reverse. That is not true in general.2148

So, I am going to just say one more thing about this. Just to make sure that we are solid. Compare this with the nth term test.2162

You actually have seen this type of mathematical logic before back when you were doing infinite series, so you compare the nth term test for divergence of an infinite series... for divergence of an infinite series.2174

The positive of that theorem is this: if the series, an converges, then the limit as n approaches infinity of the nth term is equal to 0. That is the positive.2196

I know that if I have a series that converges, I know that if I take the limit of the nth term, this thing, I know that it is going to equal 0. 2217

Now, for testing purposes, we use the contra-positive. We say if the limit as n approaches infinity of the nth term does not equal 0, then the infinite series does not converge.2225

Now, the converse would be the following. It would be the limit of an as it goes to infinity = 0, then the series converges.2257

All I have done is switch places with this and this. This is not true. The harmonic series, the limit goes to 0, one over... the harmonic series was the sum of 1 over n, the limit goes to 0 but the harmonic series does not converge.2277

So, positive was fine, the contra-positive is an equivalent form of that. The converse is not. So, that is it.2290

One final statement, so for our theorem let me go back to blue here. So, for our theorem, d2(f1), in other words, the quality of the partials, d1(f2) is a necessary condition -- let me actually capitalize this, you are going to see these words used a lot, they are very, very important words in mathematics, and they should be more important in science -- is a necessary condition for existence of a potential function for f.2299

But alone, it does not suffice. It is not sufficient. So, just because I have the d2(f1) = d1(f2) that is necessary for a potential function to exist, but it is not enough for the potential function to exist.2355

I need other things to be there in order for the potential function to exist. But, if it does exist, then yes, that is certainly going to hold. So it is necessary but it is not sufficient.2386

In other words clouds are necessary for a rainy day, but they are not sufficient for a rainy day they have to be a particular type of cloud, there have to be other conditions that are met. 2395

That is what we are saying. So the words necessary and sufficient are very, very important in mathematics. Just because something is necessary does not necessarily mean that it is sufficient.2403

I will go ahead and leave it at that. Thank you for joining us here at educator.com, we will see you next time, bye-bye.2413

Hello and welcome back to educator.com and multivariable calculus.0000

Today we are going to continue our discussion of potential functions.0004

In our last lesson, we gave a condition under which we can actually test to see whether a potential function exists for a given vector field, this thing about checking the quality or nonequality, as it turns out, of the partial derivatives of the 2 coordinate functions.0008

In this lesson, so, the converse turned out not to be true. Just because df/dy equals dg/dx, we cannot necessarily conclude that a potential function exists.0025

In this lesson, we are actually going to add onto, so add some hypotheses. 0040

We are going to talk about conditions under which we can conclude that a potential function exists, so let us just get started.0046

Let us start off with a particular theorem. So, theorem, now, let f and g be differentiable on an open set s -- and of course f and g are the coordinate functions of f.0055

If s is the entire plane or a rectangle, if the partials of f and g exist and are continuous and if df/dy = dg/dx, which is equivalent to -- I will just write or -- d2f1 equals d1f2, then there does exist a potential function.0102

pf for the vector field f coordinate functions f and g. So, again, in our last lesson, we said that just because df/dy = dg/dx, we cannot necessarily conclude that a potential function exists, without certain other things being in place.0177

This theorem lays out those other things that need to be there if the df/dy = dg/dx.0199

So, what has to happen is this. Not only does df/dy have to equal dg/dx, but we have to make sure that f and g are defined over the entire plane or the particular set that we are talking about is either the entire plane or a rectangle. 0207

It is very, very important that it is a rectangle. It cannot just be some random set. It has to be a rectangle. If these hypotheses are satisfied, then we can conclude that there exists a potential function. 0227

Let us just do an example. Example 1. So, let f = 4xy and 2x2, so this is f and this is g. First and second coordinate functions.0241

So, the first set of hypotheses is definitely satisfied. As far as this vector field is differentiable over the entire plane, so that is taken care of. The entire plane is taken care of.0264

Now we want to see if these are satisfied, df/dy, so, df/dy is going to equal 4x, and we are going to take dg/dx. dg/dx = 4x.0282

Those two are equal. So, all of the hypotheses of the theorem are satisfied. df/dy = dg/dx, therefore we can conclude, so there does exist a potential function for f. Now, let us see if we can find it.0300

Let us see if we can find it. This is the important part, well, existence is important, but from a practical standpoint, actually constructing the function is what we want.0319

So, here is what we do. So, let us make sure that we know what we want here.0335

We want some potential function pf of x and y, such that d(pf) dx = f -- we can be more specific that that, we know what f is -- = 4xy, and the derivative of this potential function with respect to y is equal to 2x2.0341

That is what we want. You want to find this pf. We want to find that. Okay, let me go ahead and move on over to blue and move over to the next page. 0376

The first thing that we are going to do -- and this is a general procedure that you can always run through, this is what is really nice, we can actually construct this function by integrating one variable at a time.0385

Okay. First thing that we want to is we are going to integrate f with respect to x, so we are going to differentiate the first coordinate function with respect to x, the first variable. Okay.0396

Now, the dx = 4xy.0413

I am just going to work formally here. In other words, I am going to move symbols around, and I am just going to move symbols around... you should be able to follow the mathematics. 0419

I am going to move the dx. I cannot really do this, I mean -- really -- because this partial derivative is not the same as what we were working with in single variable calculus. 0429

Remember when we had something like dy/dx, this actually represent a division, and you can sort of move things around. Remember we said dy = f'(dx), things like that. 0438

Well, I am going to do the same sort of thing here, I am just working formally. I am just manipulating symbols, but the underlying mathematics is valid.0452

This is the situation, this dpf = 4xy. Well, I am going to go ahead and move this over here, so this is going to be -- oops -- of pf = 4xy dx.0460

Now I am just going to integrate both sides. Essentially what I have done is since I am just integrating with respect to one variable, this is just your normal dx.0487

I am just working formally so that you see where this comes from. Since I know that the potential function exists, the derivative when I take the derivative with respect to x of that potential function, I get my f which is 4xy, my first coordinate function.0496

I am just going to integrate, I am just going to work backwards. I am going to integrate with respect to x and recover a function.0510

Okay. When I integrate, I get that the potential function is equal to -- so when you integrate 4xy with respect to x, you are going to get 4x2/2, right?0518

Then this becomes y + some function of y, so remember when you integrate, this 4xy is what we are integrating. But, we are integrating only with respect to the x. 0538

This becomes 4x2/2, the y stays because the y is just a constant.0560

This is going to be some function of y, because when you differentiate this function of y which is only y and differentiate with respect to x because it is only a function of y, it is going to go to 0.0565

In other words, the x derivative is 0. So, I actually have to include that. It is not just a constant that I am adding, it is an actual function of y, because when I differentiate this with respect to x it goes to 0, because it is just a function of y, partial derivatives but y is constant.0579

So I have this. Now, I am going to go ahead and take the d dy of this.0598

So, d of pf dy = ... oh wait, let me simplify this a little bit, this is going to be 2x2 + h(y)... so when I differentiate that, this is equal to 4x -- wait a second here, what have I got, I have got 2x2 + h(y), so when I take the d dy of that, I am going to end up with... not this is 2x2y.0612

Now, when I take the d dy of this thing, this is going to be 2x2 + dh dy.0653

Set this equal to g, because that is exactly what it is. When I am taking the derivative of the potential function with respect to y, that is the same as the second coordinate function g.0670

What I have done is I have recovered the potential function in this form. If I differentiate it, I get this, but this happens to be equal to g because by definition the derivative of the potential function with respect to y is g, the second coordinate function.0684

So, let us see. So, 2x2 + dh dy = 2x2 + 0. This comes from that, differentiating the potential function, this comes from, well, just the function g.0707

dh dy is equal to 0, which implies that h(y) is a constant.0739

Because it is a constant, so our potential function for f happens to be 2x2y + c, and we can always take c = to 0.0753

So, our potential function for this vector field is equal to 2x2y.0785

Then you can confirm this by taking the partial derivative with respect to x to get the f function, and you can take the partial derivative with respect to y to get the g function.0793

Let us stop and take a look at what it is that we have done. We have taken the f, the first coordinate function, we have integrated it with respect to x. We came up with this thing right here.0804

Then, what we did is we differentiated this with respect to y to get this thing, and then we set it equal to g, which it is by definition, and we just matched things up. We ended up with this = this, and now we integrated this with respect to y.0816

That is really what we are doing. We are integrating first with respect to x, finding out what dh dy is, since it is 0, well, when you integrate this, you are going to end up with a constant. We are going to keep doing that.0835

Let us do some more examples. Let me go ahead and write the theorem in 3-space, and then we will go ahead and do some 3-space examples. Okay. Theorem in 3-space.0848

Okay. Let f = f1, f2, f3 -- you know what, I need to write these a little bit clearer here -- okay, let f = f1, f2, f3 be a continuously differentiable vector field on an open set s in R3, because we are talking about 3-space, so, in R3.0872

So, again, if f -- if s -- is all of R3 or a rectangular box in R3, that is the important part, is a rectangular box and if d2f1 = d1f2, d2f3 = d3f2, d1f3 = d3f1, then f has a potential function.0930

Again, if these partials are equal and if it happens to be defined, if our domain happens to be this connected set s which happens to be a rectangle or the entire space, then we can conclude that f has a potential function.0976

Okay. Let me go ahead and write out the other version of this thing right here, so, again, this is if f is written as f(x,y,z) = f1(x,y,z), f2(x,y,z), f3(x,y,z), so f1, f2, f3, and our three variables, first, second, third, are x, y, z, then this condition right here in the other notation looks as follows: df1dy = df2dx, df3dy = df2dz, and df3dx = df1dz. This is just the equivalent of that.0998

Okay. Let us do another example here. Example 2.1073

Let f (x,y,z) = 2x, 3y, 4z. So, this is f1, this is f2, and this is f3. Find the potential function for f... let me do this in red.1085

Just to go ahead and... we need to check to see that the partials are actually equal... let us go ahead and just do that first, so, just to confirm.1112

So, df1dy = 0, df2dx = 0, df1dz = 0, df3dx = 0, so those are equal.1125

Now, df2dz, that equals 0, and df3dy, that also equals 0. So, 0, 0, 0, 0, 0, the equality is there.1151

This particular vector field is defined over the entire space and it is continuously differentiable over the entire space, so all of the hypotheses of the theorem are satisfied, so we know a potential function exists. Let us construct that potential function using our process.1165

We are going to integrate the first function with respect to the first variable. So, again, working formally, I know that the partial derivative of this potential function f with respect to x is equal to 2x.1180

Therefore, I can go ahead and say dpf = 2x dx. I can integrate both sides and I can end up with this potential function for f is equal to x2 + some function h(y and z). 1200

Now because I am working with 3 variables, x, y, and z. I differentiated -- sorry, I integrated with respect to the variable x, so the constant of integration is some function of y and z, because if I differentiate some function of y and z, it goes to 0.1224

This is not just a constant of c, it is some function of the other 2 variables.1243

So, now, let me go ahead and take the... I am going to take d dy of x2 + h, I will just go ahead and write it that way.1250

That is equal to the derivative with respect to y of x2 is 0, so what I just get is dh dy. Okay.1266

That happens to equal, the d of the potential function with respect to y happens to equal the second coordinate function, so it equals 3y.1276

Now I can go ahead and integrate this, right? dh = 3y dy, and again, this is probably not the best way of doing this, as far as working formally, but you know what is happening. This is just the normal dy. 1289

I am integrating this function with respect to dy, and when I do that, I end up with the following.1305

I end up with h = 3y2/2 + some function, now of z, right?1312

I am doing one variable at a time, I integrated with respect to x and I ended up adding this constant, which happens to be a function of y and z. 1325

I differentiated that with respect to y, I set that equal to 3y, and now I integrate it and I ended up with now this plus a constant which happens to be a function of z alone.1334

So, my potential function now is equal to x2 + this h, which happens to be 3y2 + k, which is a function of z. 1348

Now, if I go ahead and differentiate this with respect to z. I am going to set it equal to that because that is what it is by definition.1364

So d(pf) dz, the derivative of this = 0, the derivative of this with respect to z = 0, so I get d(k) dz, well this happens to equal the 4z.1373

When I integrate that, I get k is equal to 2z2 + c, now my c is just a normal constant because I am integrating just with respect to z. I have taken care of everything else.1389

So, my final potential function is... x2 + 3y2/2 + 2z2 + some constant, but of course we can take the constant equal to 0, so this is my potential function.1409

You can confirm this by taking the partial with respect to x, partial with respect to y, partial with respect to z.1436

In other words, take the gradient of this thing and you are going to recover the vector field.1441

Okay, let us do one more example. Let us do this one in black. Example 3.1449

y3z + y, 3xy2z + x + z, and xy3 + y.1465

Okay, so that is our vector field, let us recover the potential function. Let us go ahead and make sure that all of the partials are equal.1491

So, we will do df... so this is f1, this right here is f2, so df1dy is going to equal... df1dy is going to be 3y2z + 1.1497

If I take df2dx, I get 3y2z + 1, so that checks out.1517

Now I will take df1dz = y2 and df3dx, that is equal to y3, so that checks out. 1529

If I take df2 with respect to z, I end up with 3xy2 + 1, and if I take df3 with respect to y, I get the same thing, 3xy2 + 1, and I hope that you are confirming this for me, I hope that I have done it right. 1543

Okay, so the first thing that I am going to do is I am going to integrate with respect... I am to integrate this first function with respect to the first variable, x.1568

So, I am going to... this time I am going to write it straight... y3z + y, I am going to integrate with respect to x and I end up with xy3z + xy + some function h which is a function of y and z. 1576

So this is our potential function. Now, I am going to take this thing and I am going to differentiate it with respect to y and I am going to set it equal to the second coordinate function, which his exactly what it is. 1605

So if I differentiate this with respect to y, I am going to get 3xy2z + x + dh dy. That equals 3xy2z + x + z.1623

Therefore, that is equal to that, because this is this, this is this, and so therefore we have these two equal to each other. So, dh dy = z.1652

So, I have got dh dy = z, so now I am going to integrate this with respect to y, because it is y here. 1668

I am going to integrate, so h = the integral of z with respect to y, and that is going to equal z2/2, right? -- no, not z2/2 see I made the same mistake, this is where you have to be careful, I am actually integrating with respect to y, so z is a constant.1682

This is actually going to be zy, zy + some function of z, so be very careful. As you can see, I made the mistake.1709

So, now, our potential function is equal to xy3z + xy + yz + kz.1723

Now I am going to go ahead and take d dz of this and I get xy3 + y, because this is 0, so + y + dk dz, and that is going to equal our third coordinate function, which is xy3 + y + 0.1744

Therefore, dk dz = 0, which implies that k = some constant, which we can always take to be 0.1780

So, our potential function for this particular vector field is... well... it is this thing but k is equal to a constant which we can take as 0, so what we are left with is xy3z + xy + yz. There you go, that is our potential function.1793

So, once we know the certain hypotheses are satisfied, we can go ahead and just start to construct this potential function.1824

We do it by integrating one variable at a time. We take the first coordinate function integrate with respect to x. We get our potential function, we differentiate that with respect to the next variable which is y, and we say equal to the second coordinate function.1831

We match corresponding terms, and then we integrate with respect to y. Along the way, we recover an extra constant function. We differentiate that with respect to the next variable z, we set it equal to the third coordinate function.1847

We match corresponding terms, and then we integrate with respect to z. Then we put it all together in the end to come up with our potential function. 1865

That is it. It is really that simple. Really, the only thing that you have to watch out for here is pretty much the mistake that I made myself.1872

Up here, this here we are integrating z with respect to y. z is a constant, so this is just zy, or yz. That is it, you just have to make sure that you are keeping track of the variable with respect to which you are integrating and any of the other variables inside.1881

Thank you for joining us here at educator.com, we will see you next time for the conclusion and summary of potential functions. Take care, bye-bye.1899

Hello and welcome back to educator.com and multivariable calculus.0000

Today we are going to close out, or round out the story on potential functions, and summarize everything so we have a nice set of things to follow in order to decide how potential functions behave. 0004

Okay. So, let us get started. So, let us start off with a theorem, here.0017

So, we will let f be a vector field, let f be a vector field on an open set s -- excuse me -- an open set s and let pf be a potential function for f on s.0026

Okay... continue over here... let c be a path in s connecting the points p and q. 0065

Then, the integral of f/c from point p to point q is equal to the potential function evaluated at q, the end point, minus the potential function evaluated at p, the beginning point.0087

Also, let us write this here -- also, this is the really interesting part in my opinion -- the integral is independent of the path.0109

Okay. I should actually have the path connecting p and q, I mean we know that it connects p and q, but let us just write it out anyways -- connecting p and q.0131

So, basically what this theorem says is that if you have this open set s and if you know that your particular vector field has a potential function, well instead of actually evaluating the integral itself, the way we did previously, evaluating the line integral, because in some cases it might be hard, you can just take the potential function, and evaluate the potential function at q and subtract the potential function at f.0142

This is pretty much the same thing as the fundamental theorem of calculus in single variable, this is just a version of it.0169

For some function, some vector field that emits a potential function, this is how we can actually calculate the integral instead of going into the line integral process. 0175

What is also interesting is that the integral is independent of the path, so it does not matter if I get there via straight line or a curve, or if I bounce around all over the set and eventually get there.0184

It does not matter. The integral actually ends up being the same. That is extraordinary, absolutely extraordinary and a profound importance in physics.0196

Let me actually write out what this is so that you have it here.0206

In other words, the integral of a vector field which admits a potential function over a domain is equal to the difference of the potential function evaluated at the end points, and is path independent.0210

So, if this is p and this is q, if you are moving in that direction, integrating along that path, you are going to get the same value whether you go like this or whether you go like that, whether you go like that, it does not matter. The path does not matter. 0282

The integral stays the same, simply because on this open set, the vector field admits a potential function.0309

So, the existence of a potential function is a very, very special kind of property that is related to the vector field under discussion.0315

Okay. So, let us go ahead and write out a corollary to this.0323

All of the previous hypotheses, let f be a vector field on an open set s which admits a potential function... okay, I will just write if previous hypotheses, then the integral of f around every closed path, every closed curve -- I will write path instead of curve -- in s = 0. 0337

Because a closed path, you remember, the beginning point and the end point are the same. When you evaluate them of course you are going to get a - a, b - b, it is going to be 0.0381

Okay. So... and if there exists a closed path in s such that the integral of f does not equal 0, then f does not have a potential function.0393

This is the contra-positive version of the corollary. What we wrote... this part, this is the contra-positive. 0424

So, if a vector field admits a potential function on an open set s, then the integral over every closed path in s, every closed path is going to be 0.0430

If you can find a closed path in s where the integral does not equal 0, that means the vector field does not have a potential function.0440

Again, it just depends on the direction we are working in. That is all this is -- okay... does not have a potential function -- you know what, let me not actually does not have a potential function, let me just write does not have a pf.0448

Okay. Let us do an example here. So, example 1.0468

So, in a previous lesson we had a vector field f which was equal to 4xy and 2x2.0473

Okay? That admitted a potential function, which ended up being 2x2y.0493

Well, let p be the point (-4,7) and let q be the point (3,-2) and c be any path connecting them.0501

Then, the integral of f over c is equal to, well, it is equal to the potential function evaluated at... let me see, what are our 2 points... q, right? it is the ending... so we are going from p to q.0533

So it is going to be the potential function evaluated at 3 - 2, minus the potential function evaluated at (-4,7).0550

There we go. Then when we actually do this evaluation, we end up with something like again, if I did my arithmetic correct... and by all means, please check my arithmetic but arithmetic is not important, it is the mathematics that is important.0565

I think you get something like -36 - 224, and I hope my number are right... -360... so this is the final answer, this is what is important, right here... this.0579

You do not actually have to go ahead and parameterize the path and solve the integral, you do not have to take the f(c(t)) · c'(t), evaluate that integral, just take... because it admits a potential function, you can just evaluate it as a potential function at the two end points and that is going to be your integral value.0596

Okay. So, let us see what we have. Now, as it turns out -- I will write it over here -- as it turns out, in this case, the converse of this theorem is also true.0615

So, remember that we said the converse if... if we have a theorem that says if a, then b, the converse is if b, then a, you just switch the places of b and a. 0646

Different than the contra-positive, the contra-positive, if the positive is a, then b, the contra-positive is if not a, then not b.0656

Those two are equivalent, so, if I have a theorem, the contra-positive is automatically true.0663

In this particular case, the converse is also true, but it is not true just automatically. It actually has to be proven to be true.0670

Again, we are not going to through the proof, we are not interested in the details of the proof, we are interested in the results, but as long as you differentiate between the two. 0677

A positive statement and its contrapositive come together. They are both, they are equivalent.0685

The converse may or may not be true, you have to check to see if it is. In this case it is, so let us go ahead and write it out. Let us go to the next page here.0690

It is always nice when converses are true, it always makes life a lot easier. It gives you two directions to work in depending on what your particular situation is.0702

Let f be a vector field on an open connected set -- you know, I am just going to go ahead and start presuming that the hypotheses are given -- okay, if the integral of f is path independent, then f admits a potential function.0713

In other words, a potential function exists. In other words, if you happen to have a vector field, and you happen to have two paths that you can check on two points connecting... two paths connecting two different points, and let us say they are reasonably easily parameterized, if you took the integrals and the integrals ended up being the same, you could conclude that there is a potential function.0744

You do not need anything else at that point. Again, it just depends what you have at your disposal.0767

All of these theorems are there as tools in your toolbox. You might be presented with a vector field in a particular kind of open set.0773

You can draw a conclusion. You might be presented with a vector field and a path that is easily parameterizable.0781

You might be presented with an integral value -- all kinds of things you can be presented with, we just want to present theorems that allow you to test when something has a potential function, when something does not, when you have to evaluate the line integral directly, when you have to use the potential function to evaluate it at different points, that is all that is going on here. That is why we are listing these theorems.0788

Okay, so, if f is path independent, then f admits a potential function. That is the converse of the other one. Now, let us see.0812

If the particular paths in this theorem are closed, we have the following version of the theorem.0824

Again, this is just the same theorem as this except for closed paths. We have the following version.0845

If, the integral of f = 0 for every closed path, then f has a potential function.0857

Now, the only reason that we gave this particular version of the theorem is for the sake of completeness. There is no way that you are going to test every single path.0889

There is an infinite number of paths, an infinite number of closed paths in an open set.0897

So, from a practical standpoint, this theorem has absolutely no value at all. From a theoretical standpoint, it does, of course.0901

But, again, for practical purposes, you cannot really use this in any reasonable way, because you cannot test every path. 0909

Fortunately, we have other tools at our disposal to check to see whether a potential function exists.0915

Now, let us go ahead and summarize what we have got up to here. So, let me draw a line and let me go back to black.0921

So, summary on potential functions... summary on potential functions.0932

Okay. So, let us start off with let f = fg, I am only going to state it for 2-space, but of course it is valid for n-space, be a vector field on an open connected set.0946

Okay. On an open connected s -- let us actually give a name to this set.0977

Okay. Number 1. Our first test. If d2(f) = d1(g) -- I am sorry, if it does not equal... let me write this again here -- if d2(f) does not equal d1(g), then there does not exist a potential function.0984

Now, this is the same as df/dy not equaling dg/dx.1021

Capital D notation up here, regular standard notation, more common notation for partial derivatives down here.1033

So, if d2(f) does not equal d1(g), then there does not exist a potential function, we are done.1042

Number 2. Second possibility. If d2(f) does equal d1(g) and s is a rectangle or the entire plane, then there exists a potential function.1049

I will start abbreviating it that way, and we can find it and we can find the potential function by integrating one variable at a time.1086

Remember we did several examples of that by integrating one variable at a time and again, the only thing you have to do when you are actually doing that is just be careful because there are a lot of x's and y'x and z's floating around.1102

Lots of functions -- functions, vector fields -- you just have to keep track of everything. One variable at a time.1113

So, let us have... there is a third possibility... if d2(f) = d1(g) -- I always keep putting an i there, d1(g) -- but s is not a rectangle or the entire plane, then a potential function may or may not exist.1124

Okay. Let us do case 1. So, we move onto another test, another series of, you know, that is the thing. We have this toolbox, if one thing does not work, we move onto something else, and if something does not give us a conclusive answer, we move onto a finer -- you know -- a finer version. Something else that we can test. 1166

That is what we do. We just run down the list of possibilities. So, if d2(f) = d1(g) but s is not a rectangle, then a potential function may or may not exist.1191

So, the first case is if there exists some closed path, some closed curve c such that the integral of f along c does not equal 0, then there does not exist a potential function.1202

Okay. So, this particular case says I take d2(f), it happens to equal d1(g), it is not a rectangle so I cannot conclude that there is a potential function, but if I can conclude that there is some potential path, some closed path where the integral does not equal 0, then I conclude that there is not a potential function on that domain.1229

Okay. Case 2. Now, if the integral of f around every closed path in s equals 0, then there exists a potential function.1251

I will put in parentheses... (not practical)... this one, again, strictly for theoretical purposes, you are going to have to find something else to test some other way of finding out whether a potential function exists.1281

Okay. Let us do case 3 here. Case 3. Okay. If s is, and this is the interesting one, is the entire plane from which the origin has been deleted... so if s is the entire plane from which the origin has been deleted, so our particular set s we have just taken out the origin -- deleted -- and c is the unit circle, and if the integral of f/c = 0, then their exists a potential function.1297

So, in this particular case, I happen to be dealing with a vector field that is not defined at the origin, so I actually have to remove the origin from consideration.1374

Now it is just the entire plane with the origin missing, and if I go ahead and integrate that vector field around the unit circle... just a basic unit circle, if I get an integral that is actually equal to 0, then I can conclude that a potential function exists.1385

This is actually pretty extraordinary. You are going to run across all kinds of vector fields, whether it be mathematics, engineering, or particularly in physics.1402

Simply by taking the origin out, that one point out, all of a sudden everything changes. That is really truly an extraordinary thing. that the existence of the origin as a point that is viable in our particular connected open set, it really just changes the nature of the function itself. 1413

By removing the origin, everything changes. Again, absolutely fascinating, absolutely extraordinary.1434

Let us go ahead and do an example to round this out.1442

Example 2. Okay. So, we are going to let f(x,y)... we have seen this vector field before... equal -y/x2 + y2, and x/x2 + y2.1447

Now, c(t), so this is our vector field, our curve is going to be the unit circle, and sin(t)... and we are going to take t from 0 to 2pi, that is the unit circle, and s, our open set, is the entire plane because this vector is defined over the entire plane excluding the origin.1470

Because the point (0,0), this vector field is not defined, excluding the origin.1505

Well, let us go ahead and form f(c(t)), we are going to evaluate this directly. So, f(c(t)) = -sin(t)/1, right? because if I put cos(t) for x and sin(t) for y, cos2 + sin2 is 1, standard trigonometric identity, and this is going to be cos(t)/1.1513

Now c'(t) that equals -sin(t) and cos(t). Okay.1541

So, when we form f(c(t)) · c'(t), when I dot these two I end up getting sin2(t), right? -sin(t) - sin(t), sin2(t) + cos2(t) = 1.1551

Let us see here. Is equal to 1... so now we are going to... let me do this on the next page... so now the integral of f, the integral of our vector field over our unit circle is equal to the integral from 0 to 2pi of just 1 × dt, which is dt.1574

That is going to equal t from 0 to 2pi, which equals 2pi.1597

2pi does not equal 0. This implies that there does not exist a potential function.1604

Let me write "there does not exist a potential function" for this particular vector field on this particular open set, the plane with the origin deleted.1618

If it were equal to 0, then we could conclude that a potential function exists. In this particular case, potential function does not exist for this vector field.1632

Now, if our open set changes, a potential function for this vector field may actually exist. 1641

Again, do not draw any conclusions, this is one open set. This particular open set is the entire plane with the origin deleted.1650

If I took another section of the plane that has nothing to do with the origin, maybe something in the first quadrant, something in the second quadrant, third, fourth, wherever.1659

This is specific. So do not draw any other conclusions that the problem, in other words, any value that you get, any conclusion that you come to is valid only for the particular open set that you are dealing with.1669

This particular field may or may not have a potential function if the open set is actually different, and I will let you think about that or work it out in your problem sets.1683

I am sure that this is a vector field that you will see over, and over, and over again.1694

So with that, thank you for joining us here at educator.com and multi-variable calculus, we will see you next time. Bye-bye.1698

Hello and welcome back to educator.com and multivariable calculus.0000

Today we are going to start on a new topic. We are going to be talking about double integrals. 0004

There is going to be some new notation, but in face the notation itself is not even new. It is exactly the same as what you saw in single variable calculus, just picked up from another dimension -- dimension 1, dimension 2.0012

So for all practical purposes, you have done this stuff that we are going to be doing over and over and over again.0025

Rather than belabor the point with a lot of theory, I am actually just going to give a couple of quick definitions, and then we are just going to launch right into examples.0032

We just want to be able to handle these problems and not worry about what is going on behind the scenes too much.0041

You already know a lot of what is going on. Okay. Let us just jump right on in.0048

Okay. Now, in single variable calculus what we did was we defined this integral of f(x)/some interval from a to b.0054

We had something that looked like this... the integral from a to b of f(x) dx, and this was some number and we learned a bunch of techniques for evaluating this.0063

Well, let us go ahead and talk about what some of these things mean.0077

This right here is just a value of f at a given x on the particular interval, and this ab is of course the closed interval from a to b, just something like this.0080

If there is some function, this is a, this is b, this is the function, so we are just integrating along that length. That is the important part, we are just integrating along a length.0095

Now, this dx here it is just a differential length element -- a differential is just a fancy word for small -- so differential length element, and this symbol right here... that is just the symbol for an infinite sum.0105

So, what we have done is we have a particular function, we have a bunch of x's in between a and b, we evaluate them at each of those x's, and then we multiply by the actual length, some length element along x and then we take these numbers that we get and we just add them up with this fancy technique called integration.0137

Then we get some number. That is the integral, that is all we have done.0159

Well, we can do the same thing with a function of two variables, except now, a function of 2 variables is not defined just over the x axis, it is defined over x and y.0162

So, instead of a differential length element, what you are going to have is a differential area element.0173

Because again, now you are going to have an interval from a to b along x, but you are also going to have an interval c to d along y, so now we are going to basically break up the a to b the way we did before in single variable calculus, but we are also going to break up c to d and what you end up having is a bunch of little rectangles.0180

Instead of integrating over a length, we are integrating over a length and another length.0205

Well, length × length is area, so we are integrating over the entire area. Everything else is exactly the same, the notation is exactly the same.0212

Let us go ahead and write this out. So, for a function of two variables, f(x,y), we can integrate this function over an area, which is just length × length.0220

You will see in a minute that is exactly what you are going to be doing. You are just going to be doing one integral at a time, in an iterated fashion, in a row.0272

The symbol for it is exactly the same, except you are going to use two integral signs instead of one.0282

So, double integral, f(x,y) dy dx.0288

Also written as double integral of f, I will leave off the x,y, da.0299

dy × dx, well, differential can be x, differential in the y-direction, that gives me a little bit of a square, so this square is a differential area element. That is why we have da for dy/dx, that is it.0310

This is the one that we want to concentrate on, but you will see this when you see certain theorems, certain statements along as you understand what is happening. 0327

This is a symbolic definition, so let us talk about what these mean. This is the value of f at the point (x,y).0337

This is a differential area element.0349

So, instead of having a very, very, tiny length, like single variable calculus, what we have is a really, really tiny area. One of these small rectangles or squares.0356

This, of course, is the symbol for the infinite sum -- the symbol for infinite sum -- and we use two of them to differentiate from the single variable because we are talking about two variables.0367

Later, when we do functions of three variables, you are going to see triple integral, three integral signs. It actually exists that you can keep going.0382

Okay, that is it. That is literally all that is going on here.0391

Let me go ahead and draw out one more time this thing just to make sure we completely understand.0396

So, now, because our domain is now in two dimensions, we are going to split up the x length, we are going to split up the y length, and we are just going to add the value of the function over all of these little rectangles, that is what this is.0405

Now, I am going to write out how we actually evaluate this. The practical way of finding the integral, and we do it with a single integral at a time.0423

So, let us go here. The double integral of f(x,y) dy/dx, is evaluated as -- let me go ahead and write over here -- so from a to b, from c to d, so, our interval along the x-axis is a to b, our interval along the y axis is c to d.0434

It is equivalent to taking the integral from a to b, taking the integral from c to d of f(x,y) dy, and then dx.0478

What this symbol means is that we are going to be working from the inside out, just like we do in mathematics.0495

What we are going to do is we are going back to one variable first, in this particular case, dy, f(x,y) dy, and when we do this we have to remember that y is the variable we are integrating with respect to so we keep x constant.0500

When we do this, and we evaluate this integral from c to d, we are going to get a function x.0515

Now this function of x is what we integrate from a to b and we get our final answer. That is it, this is called iterated integration and you can actually do the integral in either order. 0519

You can either do f(x,y) dy dx, or you can do f(x,y) dx dy.0531

The problem itself, the particular domain you are going to be working with, will decide for you which you are going to integrate first.0536

Again, you are just integrating one variable at a time. let us just go ahead and jump into some examples, and hopefully it will all make sense.0544

Again, you have done this over and over and over again. Now instead of just stopping after one integral, you are doing another integral after that. The only thing that you have to watch out for is keeping track of the variables.0553

If you are going to be integrating with respect to y, make sure to keep x constant and carry it forward.0563

If you are integrating with respect to x, hold y constant, and carry it forward and do the evaluation. That is actually going to be the biggest stumbling block in this. It is not going to be the mathematics itself, it is going to be keeping track of what is going on.0571

So, example 1. Find the integral of xy3 over the region x goes from 1 to 2, and y goes from 4 to 6.0585

We have this little bit of a rectangle, we have 1, we have 2, let us say 4 is here. Let us say 6 is here, so we have this little rectangle that we are going to be evaluating this function, integrating this function over this domain.0622

Well, the integral, I will just call it i. In this particular case I have a choice since I am given the endpoints of the intervals explicitly, I can do dx dy, or I can do dy dx. 0642

I am going to do dy first and save dx for last. Personal choice.0657

So I am going to integrate from 1 to 2. I am going to integrate from 4 to 6 -- we are working inside out -- xy3 dy dx.0660

okay. So, I am going to do the first integral. The inside integral first. This is going to equal the integral from 1 to 2, and it is always great to write out everything, do not keep things in your head, do not write things shorthand, write everything out.0672

It is not going to hurt if you have a couple of extra lines of mathematics, at least that way if there is a problem, you can follow it -- working your way back.0686

Now, when you integrate this, this symbol stays, we are doing this integration. We are integration with respect to y, x stays constant.0696

Well, integrate y3, the integral of y3 is y4/4, so it becomes xy4/4 evaluated from 4 to 6 and then dx... so far so good.0705

That equals the integral from 1 to 2, well when I put 6 into y and evaluate this, I am going to end up with so 64/4.0721

I get the integral of 324x -... and of course I put 4 in for y, take the fourth power, divide by 4, multiply by 6, I am going to get -64x dx.0736

That is equal to the integral from 1 to 2 of 260x dx.0752

I am going to pull the 260 out because that is just my personal preference. I like to keep the constants outside and multiply them at the end.0759

260, integral of x, dx = 260 × x2/2 evaluated from 1 to 2.0770

I just integrated this now with respect to x. I got x2/2, and when I run through this evaluation putting 2 in, subtracting putting 1 in, multiplying by 260, I end up with 390.0781

That is it. Nice and simple. Okay.0796

Let us go ahead and give a physical interpretation of what this means and it might help, it might not. In single variable calculus, we had some function and we had the interval from a to b. Well, we interpret the integral physically as the area underneath the graph. 0801

Okay. We have the same thing in a function of two variables. We said in previous lessons that a function of two variables, since it is defined over a region in the x,y plane, the value of the function can actually be used as a third variable.0821

We can graph 3-space. Basically, a function of two variables is a surface in 3-space, so let us go ahead and draw a little -- so this is the x, and this is the y, and this is the z -- so there are some surface above the x,y plane.0836

Well, basically, the integral of a function of two variables over a particular region can be interpreted as the volume of everything above that region up to the surface. That is it.0861

Again, the area underneath the graph, the volume underneath the graph. We are just moving up one dimension. That is a physical interpretation that is there for you. If you want to think about it that way, that is fine. 0884

I think it helps to -- you know -- in certain circumstances, but it is important to remember that an integral is a number. It is an algebraic property, not necessarily a physical property. It can be interpreted as such, but that is not what it is.0894

Let us go ahead and do another example here. So, example 2.0908

This time, f -- let us let f equal to x2y -- and r be the region such that x runs from 2 to 3 and y is the region that is above, well between, the functions x2 and x3.0918

Let us go ahead and draw this out. Again, we do not need a precise drawing... that is the x2. that is the x3, let us say this is 2 and this is 3.0962

We have that, we have that, so the region that we are looking at is that region right there. That is the area over which we are going to be integrating this particular function.0975

Okay. So, in this particular case, we are constrained by the nature of the problem to do our... to differentiate with respect to y first and then differentiate with respect to x. It is simply easier that way.0990

So, let us go ahead and do it.1005

So, the integral is equal to the integral from 2 to 3, of the integral, so the lower function is the x2, the upper in this case is the x3.1008

So, we are going to go from x2 to x3, this is perfectly acceptable.1023

The function that we are integrating is x2y dy dx, so we just need to make sure that we keep the order correct.1027

Well, this is going to equal the integral 2 to 3. Now when we integrate x2y with respect to y, we are holding x2 constant, so it is going to be y2/2... x2 y2/2.1041

We are evaluating it from x2 to x3, and then we are going to do the dx.1055

This is going to be the integral from 2 to 3 -- oops, let us see if we can eliminate as many of these stray lines as possible -- the integral from 2 to 3, so when we put x3 in for here, we get x into y because we are evaluating with respect to y.1060

This is going to be x6 × x2, so we are going to get x8/2 - ... and when we put x2 in for here, it is going to be x4 × x2, it is going to be x6/2 dx.1079

Right? So far so good. Let us go ahead and go to the next page.1099

When we do that integration, it is going to equal -- actually, let me write it again so we have it on this page -- 2 to 3 x8/2 - x6/2 dx = x9/18 - x7/14.1103

Right? 6, 7 × 2 is 14, 9 × 2 is 18. We are going to evaluate that from 2 to 3, and then when we do this evaluation we are going to end up somewhere in the neighborhood -- ahh its okay, I am just going to do an approximate... it is going to be somewhere in the neighborhood of about 918 when you actually run that. 1130

The number itself, I mean it is important, but it is not that important. It is the process that is important. This is where we want to come to.1153

Now, let us go ahead and do a third example. Example 3.1163

We have f(x,y) is equal to... this time, 4xy is our function, over the region bounded by y = 0, y = x and y = -x + 3. 1172

Let us go ahead and draw this region out and go ahead and see what it is we are looking at.1200

So, y = 0, that is the x axis. y = x, that is this line right here, y = -x + 3, so let us go up 3, and let us go down that way in a 45 degree angle.1207

So this is going to hit at 3 and let us go ahead and put a half-way mark here. This is actually going to be 3/2 because these have the same slope.1224

In this particular case, our region is right here, this triangle. We are going to be integrating this function over this particular triangle.1235

Of course you remember from first variable calculus, sometimes regions have to be split up simply to make the integral a little bit easier to handle.1242

In this particular case, I am going to decide to save the integration with respect to x last, the outside integral.1249

Since that is the case, I am actually going to split this into two regions. This region r, I am going to split it up into R1 and R2.1257

I am going to integrate up to here, and up to there. Then I am going to add those 2 together. The reason that I do that is the upper function from here to here is different. Here it is 0 to x, here it is 0 to -x + 3, which is why I am splitting it up.1265

Let me erase this thing. So, in this particular case, our integral over R is going to equal the integral over R1 + the integral over R2 because the integrals are additive, that is exactly right.1282

Let us go ahead and deal with the integral of -- I am going to move to the next page -- so, the integral over R1 is going to equal the integral from 0 to 3/2, that is the first half and the y value, let me draw my domain again so I have it here.1300

It is going to be that, and that, so the y value is going to go from 0 to x, and my function is 4xy dy dx.1324

That is -- excuse me -- equal to the integral from 0 to 3/2 of 2xy2, I am integrating with respect to y, I am holding that constant, y2/2, the 2 and the 4 cancel leaving me 2xy2, and I am evaluating that from 0 to x dx.1339

That is going to equal, well, 2 × the integral from 0 to 3/2... when I put x in for here, it is x2 × x is x3, so it is x3 dx.1359

This is 0, so it goes away and now that is going to equal 2 × x4/4 evaluated from 0 to 3/2.1374

When I go ahead and do that, I get 81/32, so that is the integral with respect to the first region.1389

Now we will do the integral over the second region, this region right here.1399

So, the integral over region 2 is equal to -- now we are integrating from 3/2 to 3 -- so, 3/2 to 3, and the y value is going to be 0 to -x + 3.1404

Because now the upper function is -x + 3 and again our function is 4xy dy dx.1423

Well, that is going to equal 3/2 to 3, it is going to be again 2xy2, evaluated from 0 to -x + 3.1432

When I put this in here, let us see if I have enough room, no I probably do not so let me come down here -- the integral from 0... nope, doing 3/2 to 3 now -- again, keep track of our numbers.1449

From 3/2 to 3, it is going to be 2x × -x + 32, because the 0 goes away and I am just putting this into that, dx.1464

Ohh... crazy... we do not want these crazy lines all over the place.1480

Okay. 0 dx, when I multiply this out, actually I am going ot multiply everything out here so it is fine. We will do 3/2 over 3, it is going to be 2x × x2 - 6x + 9 dx = the integral from 3/2 to 3 of 2x3 - 12 x2 + 19x dx.1490

When I go ahead and evaluate that, I use mathematical software to evaluate that, I ended up with 243 over 32, therefore the integral of f over our original region R is equal to the 81/32, the integral over the first region + 243/32, the integral of the other region = 324/32.1532

That is it. Okay. So, in this particular case, I took this region and I divided it into 2 regions and I did 2 separate integrals.1567

There is a way to actually do this as one integral by just reversing the order of integration by doing dx first and then doing the dy.1575

I will go ahead and show you what that is. I will not do the integration, but I will show you how to approach it.1584

So, let me draw the domain again. We had that, and we had this, okay? So this was 3, this was 0, we had 3/2 here, this was 3, this was y = x, or if I wrote it in terms of x, x = y.1590

Okay. This graph right here, that is the y = -x + 3. If I write it in terms of x, it is x = 3 - y.1619

Now, I am thinking about it this way. I am going to do my final integration with respect to y, which means I am going to go from 0 all the way up to 3.1630

In this particular case, there is no interference. The difference -- it says if I turn the graph this way -- the difference, this length right here, okay, that I am going to be integrating this way... this length is the difference between this function and this function.1644

In this particular case, I can go ahead and take the difference between this function and this function, and then integrate with respect to y this way, so the integral looks like this.1662

The integral from 0 to -- this is not 3, this is actually 3/2 -- 0 to, this value right here is 3/2.1681

If you solve simultaneously, this is x is 3/2, this is y is 3/2, so this y value is 3/2.1692

So 0 to 3/2, the integral... well, it is going to be... you want to keep this positive, so we are taking this function, this function, this is the lower that is the upper, so let us go 3 - y to y, 4xy, this time we are going to do dx dy.1699

That is it. So, you can do this as a single integral as long as you change your perspective and if you are integrating finally along this direction, the first integration that you do, which is going to be this way, well, this difference always stays the same in the sense that it is always going to be a difference between this function and this function.1726

That is it. The outer integral covers the entire region, I did not actually have to break this up, but the way that I chose to do it, simply because I like doing x last.1752

I had to break this region up into 2, and the reason I broke it up into 2 is because the y value is actually different depending on which function I am working with past this 3/2 mark.1764

That is it. That is our introduction to double integrals. Thank you for joining us here at educator.com, we will see you next time. Bye-bye.1778

Hello and welcome back to educator.com and multivariable calculus.0000

Today we are going to talk about polar coordinates. Now, polar coordinates, most of you have studied before, it is just a different way of representing a point in the (x,y) plane using a length and an angle as opposed ot a length and a length, the x and the y.0004

So, if you studied it before, great. This will be a quick review. If not, then this will be something really, really great.0020

Polar coordinates are actually very important and although we will be discussing polar coordinates and talking about some of the conversions that we are going to be making between rectangular and polar, really what we are concerned with is doing double integrals over regions in the (x,y) plane that are a little bit more easily expressible in terms of polar coordinates.0028

So, let us just jump right on in. Okay. Let us go ahead and make a couple of copies of the coordinates... the Cartesian coordinate system.0047

We have got one like this, and let us just take a point here... this is the point (x,y), and there is another way of expressing -- so we have 2 numbers to express a point in 2-space.0057

Well, there is another way of expressing this particular point. Instead of this distance x and this distance y -- so, let us go ahead and do this here -- what we can do is we can express it as a length R and an angle θ measured from this point.0074

That is it. That is what a polar coordinate is. It is just another way of expressing where a particular point is.0092

You are just using different numbers. This is a transformation. You are literally taking this representation of a point and converting it into this representation of a point. They represent the same point, but you are actually doing a little bit of a transformation.0099

Now, given... if you are given R and θ... and you want to convert to rectangular coordinates, well, the transformation is x = Rcos(θ) and y = Rsin(θ), you know this already. 0117

Again, you are talking about this little triangle here. If this is θ, well, this is going to be x, if this is R this is Rcos(θ, this is Rsin(θ). This is how you convert back and forth.0137

Now, going the other way... if you are given the point (x,y) and you need to convert to polar coordinates, well, R = x2 + y2, also written as R2 + x2 + y2.0151

Then θ itself is equal to the inverse tangent of y/x. So, let me write this a little bit better here. Of y/x.0171

That is it. So, if you are given one coordinate system, you can move to another coordinate system and this is very, very important in mathematics. Being able to move from one coordinate system to another, for any number of reasons.0185

Mostly because certain problems are actually expressed easier, or other times it is because it is easier to sort of visualize them physically... it could be for any number of reasons.0195

Let us just do a quick example.0206

Let the point (x,y) = 6sqrt(6)... find in terms of... ah, why say find in terms of, we will just say convert to polar coordinates, that is it... so we want to converse these to polar coordinates.0214

Okay. So, we are given (x,y), this is x, this is y, so we are going to use this set of transformations here and we want to find R and we want to find θ.0248

Again, notice, it is 2 numbers, 2 numbers, you still need 2 numbers to represent the point in 2-space.0256

Well, R = x2 + y2, so we have 62 + sqrt(6)2, all under the radical, so 36 + 6... 42, so sqrt(42) is R... nice and easy.0263

θ, well we said it is the inverse tangent of 1/x so y = sqrt(6), this is 6 and when I put it into a calculator, I get 22.2 degrees or if we can avoid some of these crazy lines here... 0.39 radians, that is it, either one is fine.0282

Hopefully you are reasonably familiar... um, not familiar, certainly you are familiar with radian measure... hopefully you are comfortable with radian measure. You definitely want to start thinking less in terms of degrees and more in terms of radian measure.0314

Again, radian measure is an actual number that you can do something with. This degree business, it is a geometric idea. You cannot really do math with 22.2 degrees.0329

When you enter 45 degrees in your calculator, your calculator converts it into radian measure and then does the math with it.0339

Let us just do another example here. This time we are going to express a particular equation in terms of polar coordinates.0347

So, example... express the circle x-22 + y2 = 4 in polar form.0359

Okay. Well, that is fine. You know we have the transformation. Wherever we say x we put in Rcos(θ), wherever we see y, we put in Rsin(θ), and we just work it out.0378

As far as simplification is concerned, you know we simplify it as far as we can within reason. So, let us go ahead and do... so x is Rcos(θ) -22 + Rsin(θ)2 = 4.0391

So, we get R2, cos2(θ) - 4Rcos(θ). We get + 4 and we get + R2, sin2(θ) = 4, so the fours go away, and we are left with... we can factor out an R2 and let us go ahead and take cos2(θ) + sin2(θ) - 4Rcos(θ) = 0.0413

Well, this is just 1, so we get R2 = 4Rcos(θ) and we can go ahead and cancel out an R. We get R - 4cos(θ), there you go.0449

So, this polar representation is, well, this particular representation is the polar representation of the Cartesian equation.0475

This is the equation of a circle, it is the circle who's center is (2,0), who's radius is 2, so this is going to be one point, that is another point, that is another point, then we are going to have 2 units over that way, this is a circle centered at the point (2,0), its polar equation is this: 4cos(θ).0485

Now, let me specify what value that θ is going to take in this particular case. Here the equation of the circle was given in explicit form, where x and y show up on one side of the equation. It is not saying that y is explicitly this function of x.0511

I mean, it is, but this is given in implicit form. In this particular case, it is saying that as θ is the independent variable and R is the dependent variable, so in this particular case, θ is going to vary over certain values.0528

Well, we know that the cos(θ), well R is a distance, right? It is a distance and a distance has to be positive.0544

So, in this particular case, θ, the values of θ are greater than -pi/2 and less than or equal to pi/2, so when θ takes on the values from -90 degrees to + 90 degrees, cos(θ) is always positive, therefore R is always positive and that is what is going on.0551

So, we specify the equation and we also specify the range of the independent variable, because we are expressing it in terms of -- you know, this explicit form -- so hopefully that makes sense.0572

All of these things are absolutely important, you have to specify what θ is going to be. Let us say that you were only taking half of a circle, well then θ is not going to go from -pi/2 to pi/2.0588

It might go from -pi/2 to 0, so you have to specify what θ is going to be.0596

So, now, let us demonstrate how we graph a function that is actually given to us in polar form.0603

What is the best way to do it? Well, use your mathematical software, or use your calculator set to polar form. That is the best way to do it. The quickest way.0634

You can do it by hand just by making a table of values of R and θ and running θ through the particular range of values, calculating what R is and connecting the dots, just like you did for (x,y).0643

Okay, so let us go ahead and take R = 4cos(θ) so we already know what it looks like, so let us see what it is like when we are given the equation, and then we graph it.0658

θ and then θ is > or = to -pi/2, and < or = to pi/2. Okay, let us just make a table of values here.0674

I am going to go ahead and make the table of values over on this side, so we have θ and we have R.0686

So, θ, when θ = -pi/2, well cos(-pi/2) is 0, 4 × 0 is 0, so R is 0. Let me go ahead and draw what it is that I am doing here... so 0... a point of length 0 is -- we are always starting from the origin, that is what we are doing -- a length of 0, so that is the point.0697

Okay. So let us go to -pi/3. -pi/3, when we put -pi/3 into this, cos(-pi/3) is 1/2, so 4 × 1/2 is 2.0723

Then we do -pi/4, we do the same thing, we just run through from -pi/2, -pi/3, -pi/4, -pi/6, 0, and then pi/6, pi/4, pi/3, all the way to pi/2, and we get a bunch of values.0738

Well, here, for this one it is going to be 2.8 and for here -- let us see -- when we keep going we are going to get 0 = 4.0757

okay, so here is what is happening. At pi/2, the length is 0. At an angle of -pi/3, which is just about here, we get a distance of 2.0768

Let us just put that there. At -pi/4, we are going to get a distance of 2.8, okay? At 0, we are going to get an R of 4, so literally what you get is when you trace out all of the angles from here all the way to here, what you are doing is you are going from -pi/2 to pi/2.0783

You are going to get a bunch of points. Well, those points are going to be precisely this circle. So, what you are going to end up getting is this.0810

This is what you are measuring. As you move from -pi/2 to pi/2, you are going to get values of R, and then when you connect all of the dots, you are going to get this circle, this particular circle, which is x-22 + y2 = 4.0820

That is all you are doing. You are just making a table of values, you are going through values of θ calculating values of R, putting dots there, and connecting the dots. That is it.0841

Ok. Now. So, what we are concerned with is taking double integrals of functions over regions in 2-space, which we have done already.0853

We want to do it in polar form. We want to do a change of coordinates. Now this whole change of coordinates thing, now-a-days in the era of very, very sophisticated mathematical software, it is probably unnecessary.0864

Ultimately what you are going to do is actually write out the integral.0878

You can just stick it into your mathematical software and it will solve it for you. You do not necessarily have to solve it by changing coordinates to polar form and then running the polar... so it does not really matter anymore.0882

In the old days, converting something to polar form made handling the integral a little bit easier, but again, now it does not really matter. It does not matter how complicated an integral you have. You just stick it into your software.0896

So, what is important is being able to actually set up the integral. You can let the software solve the integral, you do not have to worry about that.0907

Again, this is historically important. The coordinate system is important, and then -- you know, chances are that the problems that you have on your quizzes and your tests are going to be such that they are not overly complicated, but you are going to have to be able to actually convert from rectangular to polar and be able to integrate it.0915

It is important from a practical standpoint, but ultimately it is not that valuable anymore. Not like it used to be, because now you can just use math software to solve any integral, no matter how complicated.0935

So, let us see. So what we are concerned with, well, I do not need to write that down. What we are concerned with is integration.0950

So, when we are converting -- actually write these words down, I tend to write a little fast, my apologies... muddle up my writing -- when we are converting a region, in the x, y plane to polar coordinates, we have to make changes to the integral and the relationship is as follows...0961

Ok, so, let -- I am going to do this one in blue, okay... there we go -- okay, so, let t(Rθ) = the actual transformation from polar coordinates to rectangular coordinates that's Rcos(θ), Rsin(θ) and I wrote it as a column vector.1022

You can also write it as Rcos(θ), Rsin(θ), this is a transformation, so this is equal to xy, that is what we said.1055

If you are given polar coordinates, if you want to convert them to rectangular coordinates, this is the transformation that you use.1066

So, let t be the transformation Rcos(θ), Rsin(θ).1073

Now, the integral over a particular region a, in the x, y plane of f(x,y). This is just the standard double integral, dy dx, when I make the conversion into polar coordinates, it is the integral over the region a expressed in polar form of f(t(Rθ).1079

In other words I form the composition f(t) and then I multiply by R dR dθ.1110

This dy dx, this area element that we know of as dy dx, differential y, differential x, dy dx is actually equal to not dR dθ, it is actually equal dR dθ × this factor, which is the radius.1116

In order for this integral to work out when I am making the conversion, from polar to rectangular, rectangular to polar, I need to do this. I need to include it.1137

So that is it. Any time you have this integral, and you make a change, what you do is you take f(t), the transformation and then you solve this integral which will often times be easier -- we hope.1146

Okay. Now we are going to discuss what all of this means, where all of this comes from, this R dR dθ this factor here.1159

We are going to discuss where this comes from a little bit later on when we talk about the change of variables theorem. When we talk about a change of coordinates, not just from rectangular to polar, but a general change of coordinates, how we actually change the integral.1167

We are going to be talking about determinates, Jacobean matrix, Jacobean determinates, things like that and it will make more sense, but for right now we just want to develop some technical facility.1181

We want to be able to be given a particular function or a region in the x, y plane, we want to be able to integrate over that region using polar coordinates.1192

We want to develop technique first, get comfortable with that, and then we will talk about what this means.1201

So, for now, we just want to integrate, so let us just do an example. Example 3.1207

Let f(x,y) = xy.1220

Find the integral of f over the region bounded by the semi-circle who's center is the origin (0,0) and has radius = 5.1233

So, basically, so x2 + y2 = 52. That is the equation.1286

We want to do -- so let us go ahead and draw this region, oops, we do not want that, let us try this again -- so, we have a semi-circle... so this is our x2 + y2 = 5, we want to integrate this function xy over that region.1291

That is it. That is all we are doing, and the radius is 5 -- oops, sorry, 52... whoo these lines are really, really causing a lot of difficulty here... let us see x2 + y2 = 52.1328

Ok. I think I just need to write a little slower. Let us go ahead and do this.1346

Well, let us go ahead and convert this particular region into polar coordinate form, so let me draw the region again, so that we know what we are looking at.1351

We have this semi-circle, and the equation is x2 + y2 = 52.1363

So, when I put my transformation in, I get R2, cos2(θ) + R2sin2(θ) = 52, = 25.1370

Well, this just becomes R2 = 25, and it becomes R = 5.1395

So, R = 5 is the polar representation of this circle x2 + y2 = 52, this is the polar representation... and θ notice θ does not show up in this particular thing.1401

θ goes from 0 to 2pi. In other words, 5 stays and now all I am doing is taking this line which is 5 and just swinging it around to θ and I am going to trace out this whole circle.1417

In this particular case, it is not 2pi, since we are talking about a semi-circle, it is just pi. That is it. So, that gives us... this right here is the polar representation of this region.1433

Now, let us go ahead and do the integral. The integral is equal to -- we said it is the integral over a expressed in polar coordinates of f(t), right? R dR dθ.1447

Well, that is going to equal, well d(θ), θ goes from 0 to pi, so that is that one... and R goes from 0 to 5, right?1467

What we are doing is we are taking -- here is -- we are taking R from 0 all the way to 5 and then we are swinging it around.1485

We are integrating this way and then we are integrating that way, the θ. Then, f(t)... well, f was equal to xy.1493

Well, f(t), t was (Rcos(θ),Rsin(θ)), so xy is Rcos(θ) × Rsin(θ), which is R2cos(θ)sin(θ), then × R dR dθ.1505

That is it, so when we actually -- let us bring this back down here -- now, that is going to equal... we can separate this out.1532

This R2 and this R becomes R3, so this integrand actually ends up becoming -- so, we will rewrite it as 0 to pi, the integral from 0 to 5, R3,cos(θ),sin(θ), dR dθ.1547

If you want, you can separate this out from 0 to pi, sin(θ) cos(θ) dθ, 0 to 5 R3 dR -- I mean, you can solve it that way. 1572

It does not matter how you actually write it out. Again, ultimately you are just going to put this into your mathematical software. When you go ahead and solve this, I am not going to be concerned with actually solving it at this point.1590

From now on I am just going to go ahead and leave it to you to take care of actually solving the integral itself. That is secondary.1604

What is important is this right here. Being able to set up the integral, making sure that these upper and lower limits of integration are correct, making sure the integrand is correct, and making sure that your area element is there. That is what is important. Being able to set this up. The rest, a computer can take care of.1610

When we do this, we end up with 0 as it turns out. That is it.1628

We integrated the function xy over the semi-circular region. We could have just left it alone and done it in terms of x and y, but again we are trying to gain some practice in polar coordinate form.1634

We found out that the polar representation in this particular region is R = 5, as θ runs from 0 to pi, so our upper and lower limit of integration is 0 to pi. It is 0 to 5 with respect to R.1648

We did f(t), so we put in the transformation x = Rcos(θ), y = Rsin(θ), into this... that gives us the integrand. We have our area element, and the rest is just math, technique, that is it.1661

So, let us go ahead and do another example. So, example 4. 1679

Let us see. We want to find the integral of... excuse me... f(x,y) is equal to xy/x2 + y2 over the region bounded by... well, we want y > than or = to x, we want x2 + y2 > or = to 1, and x2 + y2 > or = to 2.1691

These are the boundaries of our particular region. Let us go ahead and draw what this region looks like.1737

Okay. y > or = to x, so let me go ahead and draw the line y = x, that is that line right there. This is that.1749

x2 + y2 > or = to 1, so let me go ahead and draw the unit circle... that is the unit circle -- let me make this a little bit longer.1759

And, x2 + y2 > or = to 2, so -- yes, that is going to be a circle of radius 2, rad 2. This is R2, let us just say we have this circle, a circle of radius sqrt(2).1773

Sorry about that, this is a little bit odd but you see what is going on so we want the area that is... all the x's and y's such that y > or = to x,.1792

That is bigger than the unit circle, less than the circle of radius sqrt(2), so what we want is that region right there.1802

We are going to integrate this function over that region.1816

Well, let us go ahead and find what this region is expressed in terms of polar coordinates. We know that the equation for a circle in polar coordinates, a circle of radius a is R = a, that is it.1820

In this particular case, this circle... R = 1... this circle is R = sqrt(2)... R = 1, R = sqrt(2)... and this region right here, from here onward, this is just pi over -- 45 degrees all the way 225 degrees.1837

So, R goes from 1 to sqrt(2) and θ > or = to pi/4, less than or equal to 1, 2, 3, 4, 5pi/4.1862

In this particular case, the polar representation is R goes from 1 to 2 and we are going to sweep out an angle from 45 degrees all the way to 225 degrees. That is going to fill in this region. That is it.1878

This is the polar representation of this region. Now we can go ahead and do our integral.1894

Well, let us go ahead and find f(t) first, so f(t), when I put in Rcos(θ), Rsin(θ) over all of this squared... you are going to get R2 cos(θ) sin(θ), that is the numerator, and for x2 + y2 you are going to end up... that is R2.1899

The R2's cancel, so we just get cos(θ)sin(θ), that is f(t).1920

So, our integral is equal to well, θ runs from pi/4 to 5pi/4. R runs from 1 to sqrt(2). cos(θ)sin(θ) is the integrand, and R dR dθ is the element.1927

When we go ahead and put this into mathematical software, we end up with 0 again. Okay. This is a bit of a coincidence. I know. 1955

I just happened to pick this intervals and these regions that always end up 0, it is not always going to be 0 as you will see in just a minute, we are going to get something that is not 0. This is just coincident.1963

Let us do one more. This time we will go ahead and actually express the equation already in polar form.1976

So, give us a little practice in graphing also. So, example 5. Okay.1987

Find the integral of f(x,y) = x2 over the region contained by R = 1 - cos(θ).1995

So, in this particular case, they gave us the equation of the particular region in polar coordinate form already.2021

Let us go ahead and draw out what this region is.2026

When you do a table of values, or when you put it into your calculator, you are going to end up with something that looks like this.2031

That is called a cardioid. It is called a cardioid because it looks like a heart, that is it.2042

In this particular case... this is the region that we are concerned with. We are going to be integrating over this region and we are going to be integrating this function.2048

Let us go ahead and find what R is. R is going to run... it is going to be > or = 0, < or = to well, 1 - cos(θ)... so in this case, this is an actual function not specific values just like before.2057

You know, we can have a function in the inner integral. It goes from 0 to 1-cos(θ), whatever that happens to be.2075

θ is going to run from 0... we are going to start with 0 and go all the way around, sweep all the around to 2pi. That is it.2084

These are our upper and lower limits of integration. Now let us go ahead and let us do f(t).2094

Well, x2 is R2, cos2(θ), so the integral equals... well θ is going to run from 0 to 2pi, x is going to run from 0 to 1 - cos(θ).2102

Our integrand is R2, cos2(θ), and our area element is R dR dθ.2123

It is my recommendation to always put the R dR dθ there. Then if you want to you can go ahead and rewrite this as R2, cos3(θ).2133

It is just good practice because a lot of times you will just sort of forget the R. Make sure that it is there. 2140

Let me go ahead and put those little parentheses around it so that you know they are actually separate things. This is the integrand, this is the conversion factor, if you will.2148

Then when you put this into mathematical software, you end up with the following number, 49pi/32, so you see, it was not 0. 2155

That was just a coincidence that we kept getting integrals that were 0. 2166

Okay. That was polar coordinates and that was integration in polar coordinates.2170

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2174

Hello, and welcome back to educator.com.0000

Welcome back to multivariable Calculus.0003

Last lesson we introduced the scalar product, or dot product.0005

We also talked a little bit about norms.0008

Today, I am going to talk more about norms and vectors.0010

A lot of today, at least the first part of the lesson, is going to be basic properties.0015

Things that are pretty intuitively clear, but we want to formalize them.0020

Things that we are going to be talking about at any given moment.0024

We are going to be talking about open discs, open balls, and we want to make sure that we understand what these are, that we have been exposed to them so that we can proceed with our discussion of multivariable Calculus.0028

Okay, let us get started.0037

So let us let a be a point in the plane, remember a point is just a vector from the origin to that point. 0040

A couple ways of thinking about it, let it be a point in the plane, let us see, then the set of points such that the norm, I should say the set of points x.0052

Let me do that.0084

It is the set of points x, such that the norm, in other words the distance between x and a, if it is less than some number, it is called the open disc.0085

Let me draw a picture of what this means.0105

Now, you know this notation here, so a norm is just a distance.0111

In the previous lesson, the norm between two points is the distance between those 2 points, that is all this is.0116

So this is saying that if there is a particular point in the plane, a, and if I want to take all of the points that are less than b units away from it -- let us say less than 2 units away from it -- I basically have all of the points in a circle, everything inside of that, here is what it looks like.0125

So here is our point a, let us go ahead and label that.0145

Then let us say b is some distance, b.0151

So all of the points in this disc, all of the points in here, all of the vectors, the points x such that the distance from x to , x to a, x to a, is less than sum b.0157

Basically it gives me a circular region in the plane that I am talking about.0173

Less than b, without this less than or equal to, we call it the open disc, so it is all of the points inside that circle of radius b.0181

It actually does not include all the points on the circle.0190

Now, if we say, the set of all points x such that the norm, in other words the distance from x to a,0193

Now if we say it is less than or equal to b, it is called the closed disc.0217

The closed disc actually includes, let me draw it separately up here, we have our a, then we have our b, it includes all points in here that are less than this length b.0227

But, it also includes all points on the boundary.0246

So closed, when we talk about an open set, we mean everything up to but not that boundary.0250

Closed is everything up to and including that boundary.0256

Now, let us see, this disc and a final one.0264

The set of all, actually I am not going to write it out since I have already written it out twice.0270

Let me just go ahead and give you the notation for it.0277

If we have the distance from the points x to a, which is the norm of x - a, if we say it equals b, well, this is the circle centered at a with radius b.0278

Now we are just talking about the boundary points, nothing outside, nothing inside.0300

So we have the open disc, we have the closed disc, and we have the circle.0305

We will often be talking about one of those or all of those.0310

It is very important that we specify which region in the plane we are talking about. 0314

Now we are just going to generalize this, this is in 2 space.0318

Now we are going to generalize it to 3 space.0324

This is called the open disc, closed disc, the circle, we call it the... so for n=3, well let us say, for R3, let us be consistent, we have the closed ball and the sphere.0326

It is the same notation, the set of all x, either less than b, or less than or equal to b, or equal to b.0358

In 3 dimensions, we basically have that the picture we have is this.0373

Just do a quick picture of it, some point, and then instead of a circle, what you have is a ball.0377

That is it, just a ball, you have some center, you know this center right here is the point a, the vector a.0388

Then the radius of this ball is b.0396

So that is all it is.0401

This nomenclature of ball, closed ball, that actually applies to any number of dimensions.0404

So if I want to talk about the open ball in the 5th dimension, that is what it is.0412

It tells me there is some vector a in the center, and then all the points that are less than some specified value away from that point, we call it the open ball, the closed ball, the open sphere, depending on which set of points we are talking about.0415

Alright, let us move on.0435

We are just talking about things that we are going to be using over and over again.0436

So this is going to be a property, again this particular property is intuitive, but we might as well formalize it.0440

Let x be a number, then the norm of x × a, a vector, = x × norm of a.0450

What this says is that if I have a vector a and I multiply it by some number, let us say 15, then I take the norm of what I get. 0470

It is the same as taking the norm of the vector first, and then multiplying by the number 15 afterward.0480

This is intuitively clear, but then I will formalize it.0485

Okay, now, let us give a definition.0489

Now I am going to define something called a unit vector.0496

It is exactly what you think it is, a unit vector is a vector whose norm is one.0498

In other words, whose length is 1.0507

Formally, that is, if the norm is 1, it is a unit vector.0513

That is it, it is just a unit vector whose length is 1.0520

1 is nice because it is one of those numbers that you can use as a standard and you can multiply it by any number you want to make it as big as you want, or to make it as small as you want.0523

In mathematics it is called the identity, in mathematics it is a very important number.0535

Okay, given any vector, b, just a random vector -- let me do this notation a little differently.0537

1/norm(b), and the norm is a number, × the vector b itself = a unit vector in the direction of b.0555

This is very important, so basically if I am given some vector b of any specified length, I can actually take that vector. 0576

And, if I divide that vector by its norm, or more specifically, if I multiply it by the reciprocal of its norm as a constant in the front, remember we can do that, we can multiply vectors by constants, lengthen them, shorten them -- I can actually create a unit vector, specifically a vector of length 1 in the direction b.0586

That is what makes this very nice.0609

You have any vector, you just go ahead and you essentially divide it by its norm, and you have created a unit vector.0611

That is standard procedure for creating a unit vector in any given direction.0616

It is going to be very important when we talk about the direction of derivatives later on.0621

Now a little bit more, we say 2 vectors have the same direction.0630

Let me see, let me specify what they are, actually.0645

We say that 2 vectors a and b, have the same direction if there is some number C, such that C × a = b, or a = c ×b,0653

Where you put the c does not matter.0685

All that says if I have 2 vectors, let us say that, and that, I can find some number since they are in the same direction.0687

Basically we know that when you multiply a vector by a scalar, by some number, all you do is you lengthen it or shorten it, you scale it out.0695

You are scaling it, that is where the word scalar comes from.0702

If this is a, and this is b, well I know that if I start with a and it looks like here I maybe have to multiply it by 3 to get b.0706

If there is such a number that I can multiply by a to get b, that automatically tells me that a and b are in the same direction.0717

I do not necessarily know that, but this property allows me to say the two vectors happen to be in the same direction.0724

If I need to check they are in the same direction, I need to find some number c that satisfies this property.0732

Let us say I start with some vector b and I need to see if a happens to be in the same direction,0737

If I can take b and multiply it by 1/3, I can get a, sure enough there is some number b that if I multiply by 1/3, I end up getting the vector a.0745

That means they are in the same direction. 0755

So this is just a formal definition of when some things are in the same direction.0757

Again, what mathematics does, it takes things that are generally intuitive and it actually formalizes them.0763

It gives them specific algebraic definitions so that we can actually do something with them.0770

Geometry helps, we want to be able to draw a picture and talk about two vectors in the same direction.0775

But geometry is not math, geometry is just some sort of pictorial representation that helps you visualize something.0780

It is algebra that is math, that is where the power of mathematics comes in.0787

Now, let us move on with some more unit vectors.0795

Unit vectors are fantastic because we can basically take any vector and we can convert it to a unit vector.0799

Once we convert it to a unit vector, then we can multiply that unit vector by any number c to make it any length that we want.0807

That is the power of unit vectors, it gives us a standard, an identity vector, if you will,0814

To multiply by, that we can always use over and over again.0822

I can get any length I want by just choosing the proper c.0825

If I have a unit vector and I want a length of 15, I can just multiply it by 15.0829

If I want it to be a vector in the opposite direction to have a length of 446, I take -446 × my unit vector.0833

Let us do an example.0845

Okay, so example number 1.0850

Let the vector a = (3,2,-1), so we are working in R3, 3 dimensional space.0855

Now the norm, let us go ahead and calculate the norm, we are going to be finding the unit vector in this direction.0863

Again, we need to find the norm so that we can actually multiply by the reciprocal of the norm or divide by the norm to create the unit vector.0870

So, the norm of a, you remember from Calculus that things started to get a little notationally intensive.0878

In multivariable calculus, things are going to get even more notationally intensive.0888

Do not let the tedium of the notation throw you off, it is just notation.0890

You want to be very, very clear about what you are doing.0894

Whatever you do, do not do things in your head, write out everything.0897

This is very important.0900

When you are doing things in your head, I think it is really nice to be able to do that.0903

But the idea in mathematics is to be correct, to know what is going on, not to show somebody that you can do something in your head.0908

If you write things out, especially because this is not arithmetic anymore, this is serious upper level math, this is real mathematics.0913

You want to be able to notate everything properly.0924

I know it is tedious, but write out the notation, it equals sqrt(9) + 4 + 1, and we end up with sqrt(14).0927

So, the unit vector in the direction of a which I will symbolize like that.0945

We are often going to be throwing symbols in here that are a little less than standard, that you may not see in your book.0952

I will certainly try to be as standard as possible, I will try to use notation that you will see all the time.0960

But again, a lot of times when we are talking about a concept, we need to be able to use notations that make sense at the moment.0965

You are free to do that yourself. 0972

What is important is not the symbolism, but the underlying mathematics.0975

This is just a symbol for a unit vector in the direction of a, you can symbolize it any way you want.0977

= 1/sqrt(14) × the vector (3,2,-1).0984

Right? Because we said a unit vector is just 1/reciprocal of the norm × the vector itself.0993

Well the vector is this.1001

When we multiply a constant by a vector, we multiply everything by that constant,1005

So we get 3/sqrt(14), 2/sqrt(14) and -1/(sqrt14), this is a perfectly good vector, you can leave it like that.1009

You do not need to rationalize the denominator like they made you do in high school.1020

It is a perfectly valid number, it is a unit vector in the direction of a.1025

If you were to take the norm of this number, you would find that its norm is 1, you have created a unit vector.1033

You might want to do that yourself just to convince yourself that this is true.1039

Okay, now we are going to talk about 3 very special unit vectors.1044

So, 3 special unit vectors.1050

And we are going to be talking about our 3 dimensional space, so we have the z axis, the x-axis, and the y-axis.1060

Let me go ahead and label those.1069

Again, this is the standard right handed representation of three dimensional space, x, y, z.1074

Well, a vector in the direction of z, a vector in the direction of x, and a vector in the direction of y -- each having a length of 1 -- each of those have special names.1080

We call them, we can go ahead and say e1, or we can go ahead and say, you know what, I am going to use ex.1096

So this vector right here, the point, remember, it is a point that a unit vector represents, is (1,0,0).1106

My notation is not doing too well today, my apologies, (1,0,0), that is it.1125

Now the unit vector in the direction of y is notated as ey, sometimes notated as e2.1130

It equals (0,1,0), there is nothing in the x direction, nothing in the z direction, but a length of 1 in the y direction.1139

Of course ez is equal to (0,0,1).1150

These are mutually perpendicular, they are orthogonal.1155

In other words if you took the dot product of this, it would be 0, this and this, 0, this and this, 0.1157

That is the whole idea.1163

This is called an ortho-normal-basis, you do not have to know that.1164

Just know that they are perpendicular. 1168

Because we often talk about 3 space, we will often be using 3 space and 2 space. 1171

We just wanted to introduce this special collection of unit vectors in the direction of the coordinate axes.1176

Okay, now let us go ahead and do our general Pythagorean theorem.1184

You know our Pythagorean theorem, it is of course if you have a right triangle in the plane, a2 + b2 = c2.1188

The sum of the squares of the two shorter sides is equal to the square of the hypotenuse. 1197

As it turns out this is generally true in all dimensions.1200

I am not going to go ahead and do a proof of the theorem.1205

Again, in these set of courses, we want to be able to list the theorems, and we want to be able to use them, but we are not going to go through the proofs.1207

We want to do it in an intuitive sense, we want to get a feeling of it by doing problems and by discussing it, not necessarily by doing proofs.1218

If you want the proofs, by all means they are in your book.1225

I would certainly suggest looking at them at least occasionally.1228

At least to see what is going on if something seems like it fell out of the sky.1230

Most of these things are reasonably intuitively clear. 1232

The general Pythagorean Theorem, we are going to be using norm notation, so again a norm is a length -- if a and b are orthogonal, in other words if their dot product is equal to 0, if they are perpendicular, then the norm of a + b 2 = norm(a2) + norm(b2).1237

Again, a norm is a length.1282

It tells me that if I have a and b that are orthogonal to each other, a and b orthogonal to each other, the length of a2 + the length of b2 is equal to the length of a+b2.1284

That is that length, that is all that is going on, this is just the standard Pythagorean theorem expressed for any number of dimensions.1300

You can have R3, R5, R15, it does not matter.1308

It holds in all dimensions, that is actually pretty extraordinary that it does so.1312

Let me go ahead and draw this out a little bit more formally here.1319

If this is a and this is b, well, so this is vector a and this is the vector b, the vector a+b is this vector.1325

Remember when we said if we add vectors, we go from tail to head of one, then we take b, that direction, and we just go this way.1335

That is it, this is a + b, you start here and end up here, then you start here and you end up here.1347

So this vector right here actually is a+b.1351

Of course this is perpendicular therefore this is perpendicular, so this vector here is our a+b. 1360

Sure enough, this squared + this squared = that squared, that is all this is saying.1368

The norm of this + the norm of that = the norm of this.1372

Now we are going to introduce a very very important notion, something called a projection and we will do an example of that.1380

A projection, let us go ahead and define what we mean by that.1390

First of all I am going to draw a picture.1392

Let us say I have a vector a, and another vector b, so this is a and this is b, the order is very important here.1400

If I imagine shining a light straight down this way, this a is actually going to cast a shadow on this vector b.1410

That shadow is of course, if you shine the light straight down, it is going to be perpendicular to that. 1421

This thing right here, this vector, it is a vector along b, but its length is the actual length of the shadow that is cast by the vector a.1427

Obviously if this angle were bigger, then you would have a shorter shadow.1442

This thing right here, that is the projection.1447

It is the projection of a onto b.1450

I am going to notate it this way.1455

The projection is an actual vector.1457

The projection of a onto b.1460

This is going to be the notation that I use for my projection, it is a slightly different than you are going to use in your book but I think it makes everything clear.1464

You are saying that you are projecting the vector a onto the vector b, not the vector b onto the vector a.1472

That actually exists, you can project any vector onto another vector, but it is an entirely different vector.1480

The order matters.1484

The algebraic definition of the projection, this is the geometric, we just wanted to see what we mean by that in 2 space and 3 space, so algebraically, the project of a onto b = a · b/b · b ×, so this is a number, × the vector b itself.1487

a · b/b · b is just some number, so basically what we are doing is, again, we want to find what this vector is.1518

It is in the direction of b, so we know that this vector is some number × b, right?1528

It is basically just you are scaling out b, you either make it bigger or smaller, in this case it ends up being smaller.1535

That is all you are doing, is we want to know what vector = the projection of this vector on top of this vector.1545

The projection geometrically is what shadow does this cast on that.1554

Again, this is just some number, dot products are just numbers.1560

A number × a number, we are scaling b to find out what this is.1564

Now, let us move on to the next set.1570

The component of a along b, let me actually, so c = a · b/ b · b, all this is is just the component of the vector a along the vector b.1576

In other words, if I take a look at this vector a, it has this component this way, and another component this way.1605

It is the component of a in the direction of b, that is all it is.1613

The best way to think about it is this whole idea of shining a light and casting a shadow, that is what is going on.1618

If b happens to be a unit vector, then since the norm of b = 1, b · b, that denominator = 1.1628

Remember the definition of the norm is the square root of the dot product.1657

So if the norm is 1, and I take the square of that which is also 1, it means that that product is 1.1661

The projection of a onto b = just a · b × vector b.1672

In other words the denominator disappears, because b · b = 1.1683

So if you happen to be projecting something onto a unit vector, which looks like this, let us say that is a vector of unit 1 and this is b.1685

It is not a unit vector of 1, it is a unit vector, so it has a length of 1, my apologies.1697

If that is a, well again, if I shine a light, drop a perpendicular, that is all I get.1701

I drop a perpendicular from the end of a down onto the vector b, well, the vector b extends this way, right?1708

But this is a unit vector, so what I am looking for is that vector.1717

It has a length and it has a direction.1720

That length and that direction is given by this expression right here, if b happens to be a unit vector.1723

If not, then you just use a general vector addition.1730

So let us do an example.1735

Example number 2.1740

We will let a = (2,4,2) and b = (-3,1,3).1743

Okay, now we want to find the projection of a onto b.1755

Well we use our formula, so the projection of a onto b = a · b/ b · b × vector b itself.1763

Okay, well, a · b = 2 × -3 is -6, 4 × 1 is 4, 2 ×3 is 6, -6 + 6 is 0, so that is 4.1783

And b · b = -3 × 3 is 9, and 1 × 1 is 1, and 3 × 3 is 9, so we get 19, if I am not mistaken.1799

Therefore the projection equals 4/19 × vector b itself.1814

Well the vector b is, I am going to write it vertically, (-3,1,3).1823

Of course we have a constant times a vector, so we multiply everything right through.1834

So that is (-12/19, 4/19, 12/19), this vector (-12/19, 4/19,12/19) it is the projection of the vector (2,4,2) on (-3,1,3) in that direction.1838

That is what is going on here.1863

That is it, this is algebraic.1865

Very, very important.1868

Let us go ahead and finish it off with a nice further geometric interpretation, so let us go ahead and take our standard drawing of, this is vector a and this is vector b, and we know that this is the projection right perpendicular.1873

So this right here happens to be, this length is the norm(a), that is what the length is, what the norm is, it is a length.1890

This length right here, well it happens to be, remember c × the norm(b).1900

Basically I have taken the length of b and I have multiplied by some number c, and that c = that a · b/b · b.1908

This would be our definition.1920

So if I have the norm of b, and I multiply by this a · b / b · b, from the projection, that is the length of this thing.1923

If I, now look, I have some angle θ in between those 2 vectors, in between this vector and this vector or this vector and this vector, it actually does not really matter.1934

I can actually write cos(θ) = the adjacent/the hypotenuse of this right triangle.1945

So I end up with c × norm(b)/the norm(a).1950

C happens to be a · b/b · b, and I am multiplying that by the norm(b).1960

Then I divide all of that by the norm(a).1978

Remember the definition of the norm, remember the norm(b), or any vector, is the sqrt(b · b).1983

So b · b, if I square both sides, I get the norm squared.1997

So I am going to take b · b, and I am going to replace this with that.2004

So that equals a · b/the norm of b2 × the norm divided by the norm of a.2010

Okay, now this is squared, this is on top, this cancels this, and of course when I rearrange this because this is actually equal to a · b/norm of b × norm of a.2027

Look at that.2045

I have the cosine of the angle θ between those two vectors = the dot product of those vectors/the product of the norms of those vectors.2048

Or, another way of writing it is a · b = norm(a) × norm(b) × cos(θ).2063

In some books, possibly in your own, this is actually taken to be the definition of the dot product.2079

That is fine, it is not really a problem in and of itself.2085

However, notice we have introduced some angle θ.2093

Well the angle θ makes sense in 2 and 3 dimensions. 2096

The algebraic definition of dot product which we gave is the product of the, you know, corresponding entries added together.2100

That strictly involves a and b, it does not involve something else, and it does not involve something else called the norm.2111

I think this is fine, I think it is better to use the algebraic definition and derive this as an identity as opposed to taking this as a definition.2120

The nice thing about it is all of the other properties of the scalar product fall out easily when you use the algebraic definition as opposed to this definition.2129

Starting out with this definition, it is okay, but I think it is just better mathematically to work the other way around.2138

Notice we had just a nice straight basic definition of the dot product, and then we ended up with this identity.2149

Let us go ahead and do our final example for this lesson.2155

Example 3.2162

We will let a = (2,-2,6) and we will let b = (3,0,1).2165

We want to find θ, the angle between those vectors.2175

Well, we know that cos(θ) = a · b/norm(a) × norm(b).2190

a · b = 6 + 0 + 6, 2205

The norm of a, if I do that out is going to be sqrt(44).2211

The norm(b) is going to be sqrt(10).2215

I end up with 12/sqrt(440), which is equal to 0.5721.2218

When I take the inverse cos, or arccos, I get θ = 55.1 degrees.2228

That is it, so you have 2 vectors in 3 space, the angle between them is 55.1 degrees.2240

We used the norm, and we used the dot product.2245

We do not really talk about angles in R4 or R5, they do not really make sense.2257

We can still certainly talk about them, this value still exists, this identity is still valid, but again, we would like to make sure that we work algebraically, not just geometrically.2260

Do not try to figure intuition and make it fit the mathematics, do not try to make the mathematics fit your intuition.2275

Let the mathematics guide you.2283

At the end you have to defer to the mathematics, not your intuition.2288

Now that we are getting into higher mathematics your intuition will not always lead you in the right direction.2291

Thank you for joining us here at educator.com, we will see you next time.2295

Hello, and welcome back to educator.com and multivariable calculus.0000

So, today's lesson we are going to be talking about Green's Theorem. Absolutely fantastic theorem. Very, very important.0004

Green's theorem is the fundamental theorem of calculus in 2 dimensions. 0011

Single variable calculus, you did the fundamental theorem of calculus. In 2 dimensions it is called Green's theorem. In 3 dimensions it is called Stoke's theorem.0017

As it turns out, the fundamental theorem of calculus it actually true in any number of dimensions, and the name of that is called the generalized Stoke's theorem.0025

So, let us just jump in, and work some examples. Okay. So, let us start off with by defining what it is that we are going to be doing here.0035

So, let f(x,y) = f1(x,y) and f2(x,y), so this is just the standard vector field that we have been dealing with all along.0045

This is the first coordinate function, and this is the second coordinate function of your basic vector field.0063

Let it be a vector field -- okay, oops, go ahead and put this here -- and let c(t) be a curve.0069

Again, nothing new, we have been dealing with this all along. 0090

Now, we said that the line integral, so the integral of f along c, well that is equal to the integral from a to b, a to b is the interval over which the curve is defined, of f(c(t)) · c'(t) dt.0095

Your standard line integral that we have been doing all along. Well, we also introduced another notation for this, so I am going to go ahead and use this particular notation and the reason I am going to use it... I am sort of going to be going back and forth when I discuss line integrals with respect to Green's theorem.0119

The reason I am going to be doing that is most books, they use the notation that I am about to show you right now. We have seen it before, but we have not really used it all that much, but just to remind you, so we also introduced the notation instead of this, the integral of c, the integral of f/c... we also used the notation f dx + g dy.0136

In this particular case, since we are using f1 and f2, it is going to be f1 dx + f2 dy. That is it. It is just another notation that is used for representing the line integral.0175

The actual evaluation of the integral is this one. This is the most important part.0192

Okay. So, now I will go ahead and I will state Green's theorem.0198

Let f1 and f2 be continuous -- hmm, let me write continuously a little bit better than this -- be continuously differentiable functions over a region R and let c be the boundary of that region.0205

In other words, c is a closed curve. So, in other words, c is a closed path, a closed curve -- I will write closed path here... like that.0260

If we traverse the boundary, if we traverse c in the counterclockwise direction... this is very, very important... in the counterclockwise direction so that R, the region, is always to the left of the curve... is always left of c. 0276

Then, and here is the big one -- move to the next page to actually write this one out -- the integral f1 dx + f2 dy = the double integral over the actual region of... I am going to write this a little bit better... df2 dx - df1 dy dy dx.0316

So, let me go ahead and draw the region and I am just going to talk about it a little bit here.0362

Let us just make this region like that. This is our region R, and this is our curve c, which happens to be the boundary of that region.0367

We are traversing it in this direction. We are moving this way, and as we move this way, the region is always going to be to the left of the curve so there is orientation. It does matter actually, how this is done.0376

When this happens. When you have a given region in 2-space, and when you have these functions f1 and f2 defined over this region, and if you actually have a closed path that contains this particular region, this Green's theorem, here is what it is saying.0389

It is saying that if you take the integral, the line integral of this function along this path which we have done already, that is actually equal to the double integral of this thing that if you take the derivative of... if you take the partial derivative of the second function with respect to x, subtract the partial derivative with respect to y of the first function and if you integrate this thing that you get, okay? over the area that is actually enclosed by this curve, those 2 numbers are the same.0406

This is profoundly important. Actually it is kind of extraordinary. It is expressing a relationship between the boundary of a region and the area that that boundary contains.0439

From a purely theoretical, global standpoint, that is what this is saying.0450

It says if I -- let us say I have a line integral, but for some odd reason this line integral is really, really difficult to evaluate, but I need that line integral.0455

Well, it says it might be easier if I just actually do a double integral over the area contained by that path. That is what Green's theorem says.0465

Again, the integral of f1 dx + f2 dy, which is the integral of the particular function, a vector field, the line integral is equal to the double integral of this integrand, df2 dx - df1 dy, dy dx.0475

Instead of doing a line integral, I can do a double integral... and vice-versa. Let us say I need a double integral, if over a particular region but for some odd reason I... it is difficult or I do not want to do it... if I can some how find a boundary then I can go ahead and do the line integral instead. Green's theorem. Profoundly important.0492

Okay. Let us just go ahead and do an example here. I think that is the best way to approach this.0512

Now, we are going to have a lot to say about Green's theorem and the 3-dimensional version which is called Stoke's theorem. 0518

We are actually going to be talking about it a fair amount because this is profoundly important in engineering, in physics, in mathematics... I mean this is the fundamental theorem of calculus.0526

I mean, it is huge... but we want to develop some facility with it, and again, by gaining some facility with it we will get more comfortable with it and then once we are reasonably comfortable dealing with it technically, we can start to think about it more theoretically. That is what we want.0535

Okay. So, Example 1. Find the integral of the vector field -- I am just going to write "vf" for vector field -- of the vector field f which is equal to x2y, xy3 around the ellipse x2 + 4y2 = 16.0552

So, we have this ellipse -- it is going to look something like this -- or actually, no, I am sorry, let us go ahead and put this into standard form just so you are accustomed to seeing it.0593

This is going to be x2/42 + y2/22 = 1, that is the standard form, and yes, I apologize. 0611

It looks like that, it is going to be an ellipse that is centered at the origin. Its major focal radius is going to be 4 this way, minor focal radius is going to be 2 that way.0622

We want to find the integral of this vector field over around the ellipse, so this is a line integral we are asking for, not the region. We are going to do the region.0634

We are going to use Green's theorem, and integrate over the region that is contained by this ellipse.0644

What we want is the line integral of this vector field around the ellipse. Okay. Let us go ahead and do this.0648

Let us see. So, let us go ahead and write Green's theorem again, because it is always a great idea to write down the actual theorem itself over and over again, until you become accustomed to it. 0657

So I am going to use, the integral of f/c is going to be the double integral over a -- I will call this region that the ellipse contains a -- of d1 so this is f1, this is f2, d1(f2) - d2(f1). That is another way of writing this.0670

df2 dx - df1 dy in capital D notation, this is d1f2 - d2f1, right? Take the derivative with respect to the first variable of the second function, the derivative with respect to the second variable of the first function.0705

Okay. dy, dx, this is Green's theorem. We are not going to evaluate the line integral, we are going to go ahead and do it as an area integral.0727

Let us see what is going on here. Well, let us go ahead and first of all, see how we are going to integrate. We are going to integrate over this region.0738

Well, we need to know upper and lower limits here. Well, we know that this is going to be -4, and we know that this is 4. I am going to integrate in the x direction from -4 to 4. That is going to be my dx, the upper lower limits of integration with respect to x. 0746

As far as y is concerned, I need to find the difference from here. So, this is a function and this upper part is a function. 0763

I am going to take the... this is going to be my lower limit, that is going to be my upper limit, so I need to express... turn this equation for the ellipse into an explicit relation between y and x. 0776

Let me go ahead and do that first. So, I have got x2 + 4y2 = 16.0790

I have got y2 = 16 - x2, all divided by 4, and then when I take the square root of that, I have got + or - the square root of 16 - x2/2.0800

My upper integration is going to be sqrt(16) - x2/2, I am sorry, my lower is going to be the negative of that.0823

Let me just write what this integral actually looks like.0835

Okay. Before we do that we need to find out of course what d1(f2) - d2(f1) is.0840

Let us remind ourselves, we have x... what was our... I think we had x2y, and we had xy3, was that our...? yes, that is correct.0850

So, the derivative of f2 with respect to x is going to be y3 - the derivative with respect to y of f1, which is x2.0865

The integral of f/c is equal to -4 to 4 -sqrt(16) - x2/2 to + sqrt(16) - x2/2.0879

d1f2 is d2f1, we get y3 - x2 dy dx.0906

So, in this particular case we are using Green's theorem. They wanted the integral of our vector field around the ellipse. Well, what we decided to do is to use Green's theorem and calculate it as a double integral.0914

Well, when we go ahead and actually put this integral into our mathematical software, we get - 32pi. That is it.0927

Now what I am going to do for this particular problem, I am actually going to go back and evaluate this line integral directly, just to verify that they are actually the same. That we are going to end up with a -32pi. Let us go ahead and do that.0937

So, let us verify Green's theorem in this case by calculating the line integral directly.0949

I tell you, I love math software precisely for this reason. It really lets you concentrate just on the math, and takes care of all of the technical details for you.0974

I mean, you remember when you were in calculus... so much of the time that was spent in calculus was just algebra, and converting, converting, converting... I mean you could have 1, 2, 3, pages of just... you know... calculations and conversions.0984

It was not really calculus itself, I mean calculus was the integral. It is this that is important, the rest, we should not have to spend all of our time. Unfortunately now we have software, we do not have to spend all of our time doing it.0998

So, let us remind ourselves what it is that we have. We have our vector field, which is x2y and xy3, this is cubed, and we have our c(t), so in this particular case we need the parameterization of the ellipse.1011

Well, you remember the parameterization of the ellipse... 4cos(t), 2sin(t), or acos(t), bsin(t) where a and b are the focal radii, the major and minor focal radii of the ellipse.1029

We have to have a parameterization because that is how we solve a line integral, and of course t runs from 0 to 2pi, or another notation for that is t is from the integral from 0 to 2pi.1045

Now, let us go ahead and calculate f(c(t)). I am going to put these values into here and I am going to get 32cos2(t), sin(t), and I am going to get 32cos(t), sin3(t).1064

Again, this is f(c(t)), this is f, this is c(t), we have done this 1000 times before.1086

Now, let us do c'(t). So c'(t), which I prefer to write as dc dt, just a personal choice, we have -4sin(t), you are just taking the derivative of the curve and 2cos(t). 1092

Then of course when I take the dot product of this, and this, so... when we have f(c) dotted with c', that is going to equal -128cos2(t), sin2(t) + 64cos2(t), sinc3(t).1108

This is actually going to be our integrand, so let us go ahead and calculate this line integral. I will do it on this page here.1140

So, the integral of f around c is equal to the integral from 0 to 2pi of this thing that we just had... -128cos2(t), sin2(t) + 64cos2(t),sin3(t), very, very complicated but again, we just put it into our math software and we get -32pi.1150

This confirms it. The line integral around a closed path was equal to the double integral of this thing df2 dx - df1 dy, we will talk more about what this thing is in the next lesson, over the area enclosed by that ellipse.1181

Fantastic. Okay. So, now a more general version of Green's theorem.1202

A more general version of Green's theorem allows the boundary of a given region "a" to be composed of a finite number of curves.1223

It does not have to be a single curve, like a circle, or an ellipse, or some continuous shape.1249

It could be composed of a series of finite number of curves, and we will do a couple of examples in just a minute... composed of a finite number of curves.1261

c1, c2, all the way to cn, not just a single differentiable curve.1273

So, I will do like a quick, so we are going to have something that looks like this. That is 1, that is 2, so that, that is another... that is another... so we might have this is c1, this is c2, this is c3, c4, c5.1297

This is a perfectly good region and the boundary can be parameterized by any number of... you know... any finite number of curves.1315

Here is the best part. These curves do not actually have to be connected end to end. That is the best part. So, that is very, very important to note and I will put it down here.1322

Note these curves do not necessarily need to be connected end to end, like this particular region.1333

This is this region in a, right there. Do not need to be connected end to end, and end to end means piecewise continuous.1353

So, these are all continuous, they are not just continuous, they are piecewise continuous. It does not have to be.1368

Our next example is going to be an example of that, so let us go ahead and do example 2 here. Example 2.1376

Let us go ahead and draw out this particular region then. Something like this... if I have that circle, let us say I have a circle that is a little bit bigger.1391

Well, if the region that I am interested in, region a, happens to be the area that is actually in between those circles, well, this is c1 and this over here is going to be c2.1409

The boundary of this region is 2 curves. In this case I have 2 curves, but the curves are not connected end ot end.1424

This is perfectly valid. This is a perfectly good region, and Green's theorem actually applies, but remember what we said about Green's theorem.1432

It says that as you traverse the boundary, the region that you are interested in has to be to your left.1440

In this particular case, it says that if you traverse this particular curve, this in this direction, well yes this region is definitely to our left.1445

Here if we go in the counter clockwise direction on this region... I am sorry, along this curve... now the region is to our right.1453

All we have to do is traverse it in the reverse direction. We need to go clockwise for the inner, and when we go clockwise, when we go in this direction, now the region is actually going to be on our left and Green's theorem applies.1462

That is it. That is all that is happening here. So, let us see. In this particular case, c, the boundary is equal to c1 + c2, except c2 traversed in the opposite direction.1477

That is what this little negative sign means. Remember? We talked about that a little bit. The only difference is if you traverse something in the opposite direction, all you are doing is you are changing the sign of the integral. 1491

So it does not really matter how you traverse it, all you have to do is remember the direction in which we are traversing it, and then change the sign of the integral.1502

In this case, we have the following. We have the integral of f along c is equal to the integral of f/c1 + the integral of f/c2 -, which is equal to the integral of f along c1, because that is counterclockwise, and we just change the sign... - the integral of f along c2.1510

That is equal to well, the double integral over a of d1f2 - d2f1 dy dx. Green's theorem applies, it still applies.1543

Take the particular vector field, I form this little thing, d1f2 - d2f1, and I integrate over that particular region, in this particular case we are talking about two circles, it is probably easiest to do it in our polar form. 1563

But, again, we have math software, you can do it in any form you like. The important point is that the boundary curve of the particular region, they do not have to be connected. That is what is important.1578

Alright, so let us do an example. Example 3. We will let f equal to, again we will use our same vector field, we will use x2y and xy3, and a be the region of points xy such that x2 + y2 is greater than or equal to 1, and less than or equal to 9.1591

So, what this is saying is that we are going to integrate this vector field over the region in between the circles, the unit circle of radius 1 and the circle of radius 3. That is it.1640

Let us go ahead and draw our region. So, this right here, this stipulation... unit circle, radius 1, 9 is going to be a circle of radius 3, so here, radius = 3, here radius = 1.1652

We want to integrate the region in between there. That is it. Okay. So, let us go ahead and do this, and we said probably this was going to be easiest in polar coordinate form.1673

Simply because the equation for circles is really, really easy in polar coordinates. In this particular case, R = 1 and R = 3, that is it, and θ is 0 to 2pi, right?1684

We are sweeping out, so R goes from 1 to 3, so we are sweeping out this line out this way, all the way around, 0 to 2pi, θ is in 0 to 2pi, that is it.1701

Okay. Let us go ahead and work this problem out. The first thing we want to do is we want to find of course the d1f2 - d2f1, which is the same as df2 dx - df1 dy, so just different notations for the same thing.1715

This was the previous vector field, so we got y3 - x2.1737

Well, we want to do this in polar form so we are going to have to actually convert y3 - x2 because we are taking f(t), so let us write everything out here.1745

So, the integral of f/c is going to equal the double integral over a of y3 - x2 dy dx. That is Green's theorem.1759

Well, so this is f this is that, this is the region, so we are going to express this in polar form. We need to express this in polar form.1776

In other words we need to find this f(t), where t is x = Rcos(θ), y = Rsin(θ).1785

Therefore our f(t), this thing, when we convert it to polar form -- oops, f of capital T -- T is going to equal, well, y3 is R3sin3(θ) - x2, which is R2cos2(θ).1800

So, we have... the integral... well, we are going -- now, this is going to be the double integral over a expressed in polar form of f(t) R dR dθ. 1826

So θ is going to run from 0 to 2pi, right? we are running around, and our R goes from 1 to 3.1844

Our f(t) is R3sin3(θ) - R2cos2(θ), and then we have our area element... R dR dθ. That is our integral.1856

When we do this, we end up with -20pi.1874

That is it. We had a region, consisted of the boundary of that region, consisted of two curves, this way, and these 2 curves right here. They are not connected... Green's theorem is still valid, we can still go ahead and do the double integral... very, very nice.1882

I mean, yeah, it is fine, these are circles, so we certainly could have done the line integral ourselves. We have the line integral over this one, and then we take the line integral here, we go in the reverse direction, we change the sign of it and then we can add it and get the same number.1903

I would actually recommend that you do that to confirm that you still end up with the number -20pi, but that is the beauty of Green's theorem, I do not need to do the line integral, I can solve it as a double integral.1914

Okay. let us close off with one more example just to solidify everything in our minds.1926

So, we have example 4. Now, this time we will let f equal to 2xy and cos(xy).1934

Well, our task is to find the integral of f/c where c = the 2 curves c1 and c2, where c1 is equal to sin(x), and c2 is equal to sin(2x).1950

We would like x to range from 0 to pi, so let us go ahead and draw out what this region is.1977

So, we have the sine function, y = sin(x), this is 1 and -1, right? I am going to draw it a little bigger here, so... 2 is here, let us see... 2sin(x), actually I think I have 2x, 2sin(x)... umm, yes, I am sorry, this is not sin(2x), this is 2sin(x). There we go.1990

So, sin(2x) would actually end up giving me a different period. We are going to have the same period, so basically, I have sin(2x), and I do this one and it is something like this.2030

We want to go from 0 to pi, so from 0 to pi, so this is the region right here. The region -- so, this is one curve, this is the other curve, so basically what they are asking me to do is we want to find the line integral of this vector field around this path.2042

I have not drawn it very well but that is it. So, instead of doing the line integral, we are going to use Green's theorem and we are going to integrate this vector field over the area enclosed by that path. That is what we are doing.2075

Okay, so let us go ahead and just work it out. Well, so this is f1, and this is f2, we want to find -- let me see, let us go ahead and do df2 dx, that is going to be... so if I take this, the derivative of this with respect to x I am going to get -y × sin(xy).2090

If I take df1 dy, I am going to end up with 2x, therefore my integrand... so this minus that... so my df2 dx - df1 dy, which is my integrand under my double integral, is going to equal minus y × the sin(xy) - 2x.2121

Now we are integrating x, so our line integral of f/c is equal to... well we are going from 0 to pi, so as far as x is concerned, it is going to be 0 to pi, and as far as y is concerned, the region is between the 2 graphs, sin(x) and 2sin(x).2156

We have -y × the sin(xy) - 2x, that is our integrand and then we have dy dx, and I am actually going to let you work this out. Hopefully you have some mathematical software and see what that particular number is.2193

Again, doing it is not as important as actually being able to take a particular problem, find a region, find the boundary of that region, decide what the upper and lower limits of integration are, if you are in fact going to use the Green's theorem to solve a double integral instead of a line integral.2214

We can certainly use a line integral here, it is not a problem. We can parameterize both of these curves and do each one separately and add them, but we might as well avail ourselves of Green's theorem since it is there for us.2233

We take df2 dx - df1 dy, that is our integrand, we have our limits of integration 0 to pi, sin(x) to 2sin(x)... and that is about it. The rest is just a question of plugging it into your software and getting some number. That is it.2243

That is Green's theorem. Now we definitely are going to have more to say about Green's theorem, no doubt about it.2261

In fact, next lesson, we are going to go a little bit deeper and talk about 2 different version of Green's theorem -- something called the circulation curl version, and something called the flux divergence version.2266

Until then, thank you for joining us here at educator.com, we will see you next time. Bye-bye.2278

Hello and welcome back to educator.com and multi variable calculus. 0000

Today's topic is going to be the divergence and curl of a vector field.0004

Basically, what that means, the divergence and curl are types of derivatives for vector fields.0008

Let us just jump into some definitions and see what we can do.0014

So, we will let capital F(x,y) be a vector field, I am actually going to write everything out explicitly, so it is going to be F1(x,y), which is going to be the first coordinate function, and F2(x,y), which is the second coordinate function. 0021

I probably should have done this a little bit sooner... we had mentioned this i, j, k, notation which tends to be very popular in physics and engineering, so all this means, this is just another way of writing F1i + F2j, that is it.0045

It is just a -- you know -- unit vector notation vs. standard vector notation, that is all that is going on here.0063

Okay. So let us go ahead and write down some definitions. Definitions. 0071

The divergence of f, so the divergence of f which is also written as div(F), sometimes they put parentheses, sometimes they do not. It equals the following, I will do capital D notation and also use this standard partial derivative notation.0078

So, d1(f1) + d2(f2), which is the same as the partial derivative of the first one with respect to the first variable which usually is x, plus the partial derivative of the second coordinate function with respect to y. That is it.0099

For right now, before we actually talk about what this means, it is just that. Just symbolically.0116

If you are given a vector field, 2 coordinate functions. If you take the derivative with respect to x of the first coordinate function and add the derivative with respect to y of the second coordinate function, it gives you some number called the divergence when you evaluate it at different points.0121

That is it. Okay. We are also going to define the curl.0135

Curl of f, that is equal to d1(f2) - d2(f1), or df2 dx - df1 dy.0140

Now, you should know that curl is also called the rotation of f, so curl is also called... let me put it over here... is also called the rotation, so you will often see -- well not often, but sometimes in certain books, you will see rot(f). 0162

Now, note that the curl, this df2 dx - df2 dy, it is exactly what you see under the double integral in Green's theorem, that we learned previously.0190

Note the curl f is exactly what is under the double integral of Green's theorem.0200

Okay. So, let us just do an example of finding the divergence and curl and actually evaluating it.0224

Example 1. We will let f(x,y) = the cos(x,y), that is our first coordinate function, and our second coordinate function is going to be the sin(x2y).0231

So, this is F1, and this is F2. Okay. So, let us go ahead and move to the next page here.0254

Let me go ahead and rite the vector field one more time, so, F = cos(xy), and the sin(x2y). 0264

Now, let us go ahead and calculate the divergence. The divergence of F, we said it equals dF1 dx + dF2 dy.0278

When I take the partial derivative with respect to x of this, I end up with -y × the sin(xy), and I hope you confirm this with me... partial derivatives, you just have to go nice and slow... it is very, very easy to make a mistake because you have a lot of x's and y's floating around.0291

Then... + dF2 dy, so that is going to be + x2 × cos(x2y). That is the divergence. That is the function. It is a type of derivative is what it is... for a vector field.0311

Now, let us go ahead and calculate the curl. The curl of F is equal to, we said it was dF2 dx - dF1 dy. Let me make my 2 a little bit clearer here.0327

There we go. That is going to equal 2xycos(x2y) - (-x × sin(xy)). This becomes a plus, so I will just write this as 2xy × cos(x2y) + x × sin(xy), that is it.0346

You have your divergence here, and you have your curl here. They are not the same. Clearly they are not the same.0381

Now let us go ahead and evaluate these at a specific point, so, evaluate the divergence and curl, evaluate dif(F) and curl(F)... let us do pi/2 and 1.0390

So, when x = (pi/2,1), this is defined everywhere... this particular vector field. We want to know what the divergence and curl are at a given point.0410

Let us go ahead and do the divergence first. So, the divergence of F, evaluated at pi/2 and 1, is equal to... well, it is going to equal... I just put it into this expression for the divergence, so it is going to be -1 × sin(pi/2) + pi/22 × cos(pi/2)/4.0419

When you go ahead and simplify this out, you end up with a number -- which is all it is -- ... -2.927, so the divergence evaluated at that point is -2.927.0453

We will talk about what this number means, what the negative sign means, in just a little bit. now let us go ahead and evaluate the curl.0466

So, the curl of F at this point (pi/2, 1), again I put this into this expression here, the curl.0475

It is going to be 2 × pi/2 × 1 × cos(pi2/4), right? yes. plus pi/2 × sin(xy), which is pi/2 × 1... and this is × 1 too.0486

When I evaluate this, I go ahead and get the number -0.879. That is it. Divergence, curl, you go ahead and take care of it symbolically, you evaluate it at a certain point, that is what you are doing, you are getting a number.0519

Now, what do these numbers mean, and what is divergence and what is curl, you know, why are we using these terms divergence and curl?0535

What do these mean? What do these numbers mean, and what is div() and curl().0546

Let us go ahead and move to another page here. So, divergence, I will go ahead and give you the definition here, well not the definition, the idea... what this really means.0566

The divergence, it is a measure of the extent to which the vector field is moving away from that point.0575

When you look at the x, y, plane, a vector field is just pick any point at random, there is just going to be some arrow that is going to be emanating from that point. That is what a vector field is... you know this direction, this direction, this direction, all across the plane.0613

Well, the divergence when I calculate it, it is a measure of the extent to which the vector field is actually moving away from that point.0629

That is a positive divergence. If I have a negative divergence, it is the extent to which the vector field is actually converging on that point. Collapsing on that point. Moving into that point.0637

For example, if the vector field happens to represent the speed of some random fluid moving in a plane, it is the extent to which the fluid is actually moving away from that point, or the extent to which the fluid is moving into that point. That is what is happening.0648

So, a positive divergence implies that it is the vector field moving away, expansion. The vector field is going like that.0665

Negative divergence, that implies moving toward the point, and again, we are evaluating these divergences at specific points, like anything else, that is what it is.0686

We find the expression for the divergence, but we put the particular point in and it is telling us what is happening at that point at that instant... moving towards, which is contraction.0698

When you speak about specific vector fields in your respective engineering and physics classes, you will get a better idea of the behavior, what is happening -- electric field, heat field, fluid field -- like that.0710

So, a positive divergence, is going to be something like this. The vector field is moving away. Negative divergence, it is moving towards, at that point, at that moment.0724

Okay. There is flow away from the point, there is flow towards the point.0740

Now, the curl, this is a measure of the extent to which the vector field actually rotates around that point. The extent to which it curls around that point.0745

It is a measure of the extent to which the vector field, I will just say vf, rotates around that point. That is why you have the alternative rotation for curl.0760

So, if you are given a certain point, and what the curl measures when you take it and you evaluate it, it is a measure of the extent to which the vector field itself at that point or near that point is rotating in this direction, or rotating in that direction. 0785

How it is circulating, how it is rotating around that point.0803

A positive curl, that means rotating counterclockwise, and this is based on a convention called the right hand rule, which you probably learned in your physics classes, and a negative curl, and I will describe it in just a second... this is rotation clockwise.0808

okay. So, the right hand rule... looking down at a page, the curl is actually a vector, it represent a direction.0839

We have not represented it as a vector, and we will when we talk about Stoke's theorem in 3-dimensions, but for right now it is best to think about it in this way. 0855

So, rotating counterclockwise, based on the right hand rule, If I am looking at a vector field and a vector field happens to be rotating this way, counterclockwise... counterclockwise by thumb is pointing up, so the actual vector itself, this curl vector, is pointing out of the page, and my fingers are moving in the direction of the rotation.0865

If it is rotating clockwise, now my thumb is pointing down, so the curl vector is actually pointing down into the page, perpendicular to the page.0887

I personally do not think of curl as a vector. I mean, it is mathematically, but again it is best to sort of treat it the way we treat it formally as the definition, df2 dx - df1 dy, and just sort of remember that a positive curl is rotating counterclockwise, negative curl is rotating clockwise.0898

That is it. Okay. So, let us see what we have got here. For our example, we had a divergence of F, which was equal to -2.927, and we had a curl of F, which was equal to -0.879.0919

So, as far as the divergence being -2.927, that means the vector field at that point is actually flowing towards the point. There is contraction.0945

The curl of F was -0.879, negative curl that means at that point the vector field is actually rotating around that point clockwise.0957

Okay. So now that we have talked about divergence and curl, let us go ahead and talk about Green's theorem and the relationship between divergence and curl, the line integrals and Green's theorem itself.0969

Now we are going to state the two versions of Green's theorem. One is called the circulation curl form, one is called the flux divergence form.0980

I am going to state them. I am going to talk about the curl first, and then I am going to talk about the divergence later.0986

The two forms of curl are equivalent in the sense if you prove one, then you have proven the other. They are not the same thing in that they do not measure the same thing.0994

Clearly, you are going to see that the integrals are different. So, they are equivalent, but they are not the same thing. It is important to distinguish between the two.1000

Okay. So, it is important to start this on a new page. Well, that is okay, I can go ahead and do at least the first version here.1010

The two versions of Green's theorem. Okay. The first form is called circulation curl.1020

So, this is called the circulation curl form, and it looks like this... da -- that is ok, I do not need to... I can use c here... actually, you know what I think I am going to start on a new page here, simply because I want to avoid some of these lines that show up.1037

Let me go over here. So, the two forms of Green's theorem. One is circulation curl.1067

It says c · c' dt = curl(F) dy dx.1081

Okay. So, the circulation curl form says the following. It says that the line integral of a given vector field around a closed curve, this thing, which we already know, is equal to the double integral of the curl of the vector field over the area enclosed by that closed curve.1103

That is what this says. If I want to, I can either evaluate the line integral, or what I can do is I can integrate the curl of that vector field over the area enclosed by that path.1123

So, if I have some closed path, I can either take the line integral or I can integrate the curl of that vector field over the area enclosed by that curl. This is the circulation curl form of Green's theorem.1136

Now let us go ahead and talk about the flux divergence form. Again, what is important at this stage is... yes, we would like to have a good sense and understanding of what is going on, but we would like you to just be able to work formally.1150

We would like you to be able to construct the integrals, solve the integrals, and just work symbolically. If you do not have a complete grasp of what is going on, you will develop that as you do more problems.1164

But, definitely understand the symbolism here. That is what is important. Given F, given c, given c'... can you construct this integral, and then solve that integral?1176

The flux divergence form says the following. It says that the integral around a closed path c of F(c) this time · something called n dt, which we will talk about a little bit later... is equal to the divergence of F dy dx.1184

So, this one I am going to talk about a little bit later. This is the one that I want to concentrate on first. I want to get a sense of what the integral says, we want to get a sense of what this integral says, and then we will talk about the divergence.1210

So, notice that the circulation curl form is the one that you have actually already learned.1222

This thing right here, curl(F), well curl(F) is dF2 dx - dF1 dy, that is just the Green's theorem that you learned in a previous lesson, so this is just another way of actually writing it.1227

This is Green's theorem, but now we give it a specific name. It is called the circulation curl form. This is the circulation integral, this is the curl integral, and we will describe what these mean.1240

Now we will talk about what these mean... so given... let us say I have some closed path here and I am traversing this path in the counterclockwise direction... this is in the x, y plane, and of course I have some vector field that is defined in the x, y plane.1253

So, a vector field is just a bunch of vectors -- you know -- we do not know which way they point, some of them point this way... well because this path actually passes through the vector field, the points on this path are actually defined.1267

The vector field is defined on those points. Let us pick a specific point, and let us say... boom... we can form F(c), so here, this point is c(t), this vector right here is F(c(t))... I will just write it as F(c).1285

The vector field is defined on this x, y plane, there is a curve on the x, y plane so the point on that curve can be used in the vector field.1307

Now, we can form the following. I will write it up here. We can form c'(t), the tangent vector, we have already done it. Wherever we are given a c(t), we can form c'(t), the derivative. That just happens to be this, the tangent vector. That is c'(t).1315

We can form c'(t), which is the tangent vector to c(t) in the direction of traversal.1341

We are going in a counterclockwise direction, so the tangent vector is this way. That is it. It is just the derivative, we know this already... derivative.1367

Now, F(c) · c'(t) is the component, and I will put component in quotes, is the component of f in the direction of c' -- oops, see, this is what we did not want to happen here, with these crazy lines going here -- okay, in the direction c'(t).1376

In some sense, when we take the dot product of two vectors, you know that if one of the vectors that you are taking the dot product of is a unit vector, well that gives you the component of the big vector on top of the unit vector.1414

Well, the reason I put a component in parentheses, in quotes here, is that in some sense, this f(c) · c'(t), it is the component of this along this.1427

You will want to think of it as sort of a projection. It is not really a projection because this c'(t) is not a unit vector. So, in some sense, you want to think of it as the component of that vector in the direction of c'(t). 1444

That is just the best way of thinking about it. It is really just a number, that is all it is. 1457

Now, let us move on here. So, component, let me write component is in parentheses because in general c' is not a unit vector.1466

I will say recall, when we have some F in some u, okay... well F · u is the projection of F onto u, so it gives you the component of F in the direction of the unit vector.1496

Well, when we take F(c) · c', in sense we are taking the component of it. c' is not a unit vector, which is why I say component, but it is good to think about it that way.1514

So, like you to think of this F of c · c', which is the integrand and the line integral as the extent to which the vector field at that point is moving in the direction of c'(t).1528

This f(c) · c', it is the extent to which the vector field is actually moving in the direction of c'.1568

Now, when we take the integral of that, of all of these f(c) · c's, when we take the integral of f(c) · c' dt, we are adding all of those numbers up. 1578

That is what an integral is. You are just adding up everything around the path, so we are getting the net extent to which F is circulating.1597

Circulating along the path, and this is what is important. We had this path, well, you know we had this vector field this way, we had the tangent vector, well we can evaluate this F(c) · c' at that point, and it is the extent to which it is moving in the direction of c'.1626

Well, we can do that for every single point moving along this path. When we add all of those up, which is what we are doing when we take the integral, we can actually measure the extent to which this vector field is actually circulating around this path in a given direction.1648

That is what is happening. It is called the circulation.1662

Now, what about the double integral? What about the double integral of a of the curl of F?1668

Well, here, now we are doing a double integral, so instead of evaluating the line integral along the path, what we are doing is we are actually taking the curl of all of the points inside the region contained by that closed curve.1681

We are calculating the curl of all of these individual points, and then we are adding up all of the curls. Green's theorem says that these two things are equal.1701

We said that the curl of F is a measure of the extent to which the vector field is curling, or rotating around a point.1711

Okay. When I integrate, when I do the double integral, when I integrate all of the curls, for all of the points that are contained in that region... when I integrate all of the curls for all of the points in the region bounded by the curve, I get the net extent to which the vector field is rotating.1750

That is it. When I take the line integral, I am getting the extent to which the vector field is actually circulating around the path.1810

When I take all of the curls of all of the individual points and I add them up, some are rotating this way, some are curling this way, when I integrate them, the double integral, When I add them all up, I am going to get the next rotation of the vector field over that region.1822

The extent to which the vector field is going this way, or it is going this way. That is what I am doing. Green's theorem is telling me that those 2 numbers are actually the same.1835

That is what is amazing. Green's theorem tells me these two things are equal. Hence the name circulation curl.1847

The line integral is the circulation integral, the double integral is the curl integral... circulation, curl. In and of themselves, they are not the same thing, but they end up being equal.1870

So, we have the integral over the boundary of a region of a vector field... c' dt over a.1883

The curl of F, dy dx. The integral of a vector field around the closed curve is equal to the integral of the curl of that vector field over the area enclosed by that curve. It establishes a relationship between the boundary of a region and the region that that boundary contains.1901

This is profound, this is the fundamental theory of calculus. Okay, let us go ahead and do an example to finish it off that way.1923

So, example 2, we use the same vector field as before, so we have F is equal to... we said it was cos(xy), and sin(x2y), okay.1932

Now, let us go ahead and do our c(t), so our curve c(t) is going to be 4cos(t), 2sin(t), which is the ellipse that has a major focal radius of 4 and a minor focal radius of 2 centered at the origin.1950

Or, if I write it in Cartesian coordinates, it will be x2 + 4y2 = 16, so this is just 2 different ways of representing the same thing... this is the parameterized version, this is the Cartesian version.1973

Well we found the curl already, the curl of F we found it already, that is equal to 2xy cos(x2y) + x × sin(xy).1990

Now, let us go ahead and solve this. Green's theorem says that the integral over a closed curve of this vector field dotted with c', in other words the circulation is equal to the double integral over that of the curl of F, it is often a good idea ot repeat the formula, write it over... dy dx.2014

Well, this is going to equal, let me go ahead and draw this region here. This is an ellipse like that, and this is minus 4, this is 4, this is 2, this is -2, so x, we are going to go from -4 to 4. 2039

Again, we are solving a double integral, we need to find the region over which the double integral is.2063

x is going from -4 to 4, right? Now we are going to have to take little strips of y because we are going to go from y.2071

Now y, in this particular case, I am going to leave it in terms of x2, y2, because I am not doing the parameterization.2081

I am not doing the line integral, I am actually doing the area integral. So, x2 + 4y2 = 16. I get 4y2 = 16t - x2.2089

y2 = 16 - x2/4, so y = + or - 16 - x2/2, so y is going from -sqrt(16 - x2)/2 to + sqrt(16 -x2)/2.2103

And, of course I have my function here, which is the curl... 2xy × cos(x2y) + x × sin(xy) × dy dx.2126

Then when I end up putting this into my mathematical software, I get 0.2147

So, this is what is interesting. Remember we found the curl at a certain point. The curl was negative, so despite the fact that we had the curl of f evaluated at (pi/2,1) was equal to -0.879, the net curl equals 0.2155

In other words, this vector field over this region is not rotating this way or this way. When the curl = 0, we call the vector field irrotational.2200

This vector field is irrotational. That is it. Straight mathematics.2216

Okay. So, when we meet again next time, we will talk about the flux divergence version of Green's theorem. 2226

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2233

Hello and welcome back to educator.com and multivariable calculus.0000

In the previous lesson, we introduced these type of derivatives of a vector field called the divergence and the curl.0004

We concentrated mostly on the curl, the circulation curl form of Green's theorem.0011

Today we are going to talk about the flux divergence, and we are going to see that it is completely analogous, it just measures something else, so let us jump on in and get started.0015

Again, given our vector field... so given f = (f1,f2), I am going to leave off the x's and y's, so again we are talking about the plane so it is x and y.0025

We said that the divergence of f is equal to df1 dx + df2 dy.0043

Now, recall that we said the divergence of f at a point is a measure of the extent to which the vector field at that point is moving away from or towards the point.0058

So, the divergence at a specific point, it measures the extent to which the vector field is going away or it is coming in... diverging, contracting... or we could say diverging, antidiverging.0109

Now, we introduced the flux divergence form of Green's theorem.0127

It looked like this... of f of c · n dt is equal to the double integral over a of the divergence of f × dy dx.0157

So, let me go ahead and... you are going to see the closed curve, however I am going to start using this notation, this Da, looks like a derivative with a capital letter -- I used it in the last lesson.0176

If this is a region a, this derivative a is actually a symbol, it is just a symbol for the boundary of a.0197

The reason we use that symbol is to establish a connection between these two... I mean we know there is a connection, but we actually want to show it symbolically. 0205

So, the integral of a given vector field over the boundary of the region, is equal to the integral of the divergence of that vector field over the area enclosed by that region. That is it.0216

This is just the closed curve, the symbol for the closed curve, it is not an actual derivative. Here is where the derivative takes place.0225

Okay. Now let us go ahead and see what these integrals mean, and what these integrands mean. So, let me write that down.0234

Now, let us see -- you know what, I do not have to write that down.0241

So, now we want to talk about what the integrands mean, and what the actual integrals mean. Okay.0250

So let us go ahead and draw our curve again... like that. 0256

Okay. Let me go ahead and do it over here in fact. Let me do it over here, to the left. So these is just some closed curve, this is the region a, this is the curve c, we are traversing it in the counter clockwise direction, and we said that we had a point, there is of course this vector field... this curve is passing through this vector field, we do not know which way it is.0267

Some of the points on that curve, the vector field can actually be evaluated per point on that curve, so let us say this is f, we ended up actually forming... so this is c(t), this is f(c) and the previously lesson we ended up forming c', the tangent vector.0304

Now, we found c'. Now we are going to define... we can also form 2 vectors orthogonal to c'. There are 2 vectors orthogonal to c'.0327

One of them is this direction, it points away from the region, the other one points into the region.0355

As you are traversing the curve, the tangent vector is going to be in the direction of traversal.0364

If you go to the right of it, we call that the right normal vector. This one. Okay? The right normal vector nr. 0372

If you go to the left of the, as you are moving, if you go to the right so right normal vector, if you go to the left you are looking in the direction of the tangent, it is called the left normal vector.0381

So, orthogonal to c... so, let us say given c(t) = c1(t), c2(t) -- these are just the two coordinate functions of the curve in 2-space -- well, c'(t) = c1',c2'. You just take the derivative of each component function.0395

Now let us go ahead and define our left and right normal vectors.0428

Define nr(t), it is equal to c2' - c1'. This is the right normal vector.0433

I am given a particular curve, when I take the derivative of it, that is my tangent vector, that is my c'.0455

Now, I have c1, c2, if I switch the order of c1 and c2, make c2', c1, and make c1', -- let me write this down... so c' = c1' c2'.0460

If I switch the places of c1 and c2, make c2 x and c1 y, and if I negate the y, I get the right normal vector. That is how I do it. That is how I construct the vector.0482

The norm of this vector is the same as the norm of the tangent vector, because all I have done is switch places. I have literally created a vector that is orthogonal to the tangent vector.0493

Now, the left normal of t, well, again we switch c1 and c2, but this time we negate the first one, the first coordinate.0506

This is the left normal vector. Okay. We are only going to be concerned with the right normal vector, so we do not even have to worry about this, but I wanted you to see that you can define both.0521

You are talking about something orthogonal to the right, something orthogonal to the left. We are going to be concerned with what is orthogonal to the right, so it is the right normal vector that we are going to concern ourselves with.0536

Therefore, we are no longer going to be using this nr, we are just going to call it n.0544

So, we are only concerned with the right normal vector, so we will just call it n... n(t).0551

Okay. It is the vector to the right of c'(t) and perpendicular to it as you traverse c in the counterclockwise direction, keeping the region contained, keeping the region to your left.0583

Remember that was one of the things about Green's theorem. When we traverse a curve, the region that is contained by the curve has to be to the left. We are moving in the counter clockwise direction.0641

It always points away from the region... from the region enclosed.0655

So, given a particular closed curve, this is f, this is the tangent vector, that is the normal vector, this is c', this is f(c), it always points away.0670

We are traversing in this direction. That is it. Okay. So, actually let me draw it a little bit better here, because I am going to use this picture somewhat anyway.0683

So, I have got... that is f... that is c'... that is n, so this is c', this is n, this is f(c)... and this point of c(t), and we are traversing in this direction and this is the region a to our left.0700

f(c), the integrand of the line integral f(c) · n is the, again, component of f along n and again we said that f(c) · c', when we were talking about curl in some sense is the component of this vector along this.0728

You can think of it as the projection, and we say component because this c' is not a unit vector. Well, it is the same thing. I can project this vector onto this vector, the normal vector, and this f(c) · n is the component of f along the normal vector.0761

That is what it is. That is how you want to think about it. Okay. So, think of f(c) · n as the extent to which f is moving away from the point, but I am going to say moving across -- oops, we definitely do not want to do that... you know what, I am going to start on another page here, because I do not want these crazy random lines all over the place.0780

So, let us go ahead and draw our picture again, very, very important to have our pictures.0833

SO this is our f, that is our c', that is our n, this is our f(c)... c'.0839

This is c(t), we are travelling this way, so think of f(c) · n as a measure of the extent to which f is moving across the curve and away from the point and away from the point.0850

So, again, f is this f at this point, but it is also moving in that direction so it is a measure of the extent to which f is actually flowing, passing, moving away from that point.0900

I think this word "across" will make a little bit more sense when we actually integrate everything. 0917

But for right now, think of it as just the extent to which it is actually moving in the direction of the normal vector.0922

That is the best way of thinking about it. Now, when we form, when we take the integral... when we add up the f(c) · n for every single point around the curve, so when we take the integral of f(c) · n dt, we get the net flux or flow.0929

That is what flux means -- it just means flow in Latin -- of f across the entire curve.0967

That is what we are doing, we are measuring this... at a certain point, the vector field is moving that way, moving that way, moving this way, maybe here it is moving that way, maybe here it is moving out, here it is moving in, here it is moving out, here it is moving in, here it is moving in, in, in, out, out.0980

When I add up all of these things, the ins the outs, I get the net flow, the net out or in across that curve. That is what this measure is, that is what this integral f(c) · n measures.1002

That is why we call it the flux integral, flux means flow, it is a measure of the net flow of the vector field going out of the region or coming into the region, or in terms of the curve passing the curve. That is it. 1018

If I have some water flowing in, let us say at a bunch of points and water flowing out at a bunch of points, if I add all of those, there is going to be some measure.1032

It is going to be either positive or negative, there is going to be a net flow of water into the region, a net flow of water out of the region, or there is going to be a net flow across the boundary. There is going to be a net flow across the boundary out, that is all this is measuring.1041

Now, what about the double integral of a of the divergence of f. Okay.1059

Well, we said that the divergence of f at a point is a measure of the extent to which the vector field is expanding or contracting at that point.1073

Okay. Now, if we integrate, if we actually take... so now we are forming the divergence, we are actually forming, we are taking the divergence of all the points this time inside the closed curve, in the region, not along the curve itself. 1117

Well, when we add up all of the extents to which -- you know -- the points, the vector field is either contracting or expanding, we get the net contraction or expanding into the region or away from the region.1132

That is what is going on. Let me write that down. Let me see... if we integrate, you know this one I am actually going to move to another page here, because I know that the crazy lines are going to start again.1148

If we integrate, that is... add up -- that is what integration is, it is always good to remember what you are doing when you are integrating, you are adding things up, you are making an infinite addition -- add up all the divergences for every point in the region enclosed by the curve, we get the net expansion or contraction of the vector field over the region.1162

Green's theorem says that these two things are equivalent. Okay. So if I take the line integral, if I do the integral of f(c) · n, I am measuring... let me go ahead and draw my quick region here, remember?1239

If I have a bunch of outs and ins, I am measuring the extent... the net flow of the vector field across the boundary of that region.1258

Well, instead of doing the line integral, instead of measuring that along the line integral... I can go ahead and calculate the divergence of all of the points inside the region that is enclosed by that curve.1267

I can add all of those up and because the divergence is a measure of the extent to which its vector field is moving away or coming into a point, when I add all of those up, it is going to give me the net measure of the extent to which the vector field is moving away from the region or moving into the region.1280

Green's theorem tells me that those 2 numbers are equal.1300

So, Green's theorem, the flux divergence form, Green's theorem tells me these two values are equal. Hence, flux divergence.1306

The flux integral is a line integral, the divergence integral is an area integral. Green's theorem tells me that these things are the same.1337

So, the integral of a vector field, the integral of -- well, let me write it out first and then I will say it out loud -- so, the divergence of f dy dx, the flux of a vector field across a closed curve is equal to the integral of the divergence of that vector field over the area enclosed by that curve.1345

Let me say that again. The flux of a vector field across a closed curve is equal to the integral of the divergence of that vector field over the area enclosed by that curve. This is called the divergence theorem.1376

You will see it referred to as that. It is just the flux divergence of Green's theorem. Okay. It is not a different theorem, it is the same thing. It is equivalent, but it just measures something else.1393

You will see this referred to as the divergence theorem, but it is still just Green's theorem.1405

So, you will have a circulation curl, we have the flux divergence, they measure different things.1416

Okay. Let us go ahead and do an example. We will let f equal to... we will use the same vector field as before... so we have cos(xy), and we have sin(x2y), actually kind of a complicated vector field. 1421

We have c(t) which is as before, we will right the parameterized version which is 4cos(t), and 2sin(t).1450

Or we will write it in Cartesian, x2 + 4y2 = 16 is the ellipse centered at the origin, major focal radius 4, minor focal radius 2.1458

So, we are going to go ahead and calculate both integrals. We are going to calculate the line integral, we are going to calculate the double integral.1473

Let me go ahead and start this over here. Let me just draw out our particular region, so we remember what we are talking about here.1482

So, we have this ellipse, this is -4, this is 4, this is 2, this is -2, so we are going to go ahead and calculate the line integral and we are going to integrate in the counterclockwise direction, keeping the region to our left, and then we will go ahead and calculate the double integral.1490

Okay. So, we want to do our first one. We want to calculate the integral over the boundary of f(c) · n dt.1507

Well, f(c), that is going to equal -- let me do this a little bit below so that we have as much room as possible -- so, let us see... I have got the f(c) part, I am just going to put c into f.1520

It is going to equal cos(8) cos(t) sin(t), and I am going to get sin(32), cos2(t), sin(t).1540

Again, I hope that you will confirm this for me, and if I make a mistake it should not be a problem. Just make sure -- you know, go with what you think it is if in fact I have made a mistake, and just put it into your math software and you will get the particular answer. 1557

Okay, well we need c'(t), right? f(c), we need c'(t), so that we can form n, and then we can take the dot product.1572

So, c'(t), that is equal to -4sin(t), and 2cos(t), right? So now we form n.1582

Well, n(t), we switch these two and we negate the second. So, this equals 2cos(t) and this becomes -4sin(t), - (-4sin(t)), so it becomes 4sin(t).1592

Now let us go to blue, now I have my f(c), that is this one. I have my n(t), that is this one. Now I take the dot product of those 2.1610

So, f(c), that is it, that is all I am doing. Just plugging things in. f(c) · n = ... it is a little long... 2cos(t)cos(t)cos(t)sin(t) + 4sin(t) × sin(32)cos2(t)sin(t).1621

It is a little long. i do not want to keep writing this. I am just going to keep referring to this as z.1657

Therefore, the integral around the closed curve of f(c) · n dt, in other words, the flux of this vector field across this closed curve is going to equal the integral from 0 to 2pi of this z dt.1662

When I put it into my math software, I get the number 0.5466. There we go, it is just a number.1683

It is a negative number. Negative divergence... negative flux means that the vector field is actually not moving away from, it is not moving across the curve outward, it is moving inward.1693

In other words, it is this way. The vector field is actually contracting, fluid is flowing into the region. Fluid is not flowing out of the region. That is what this is telling me, for this particular vector field. 1708

If this particular vector field happened to represent fluid flow across this ellipse.1725

So, now let us go ahead and do the double integral just to make sure that you do actually end up getting the same number and that Green's theorem is correct... it is, but you know it is good practice because we need to actually solve the double integral.1731

So, now the double integral -- I really like this blue ink, it is nice -- double integral, okay.1743

So, we want to solve a divergence of f dy dx. 1757

Well, we calculated the divergence of f, so let us just recall what that was and if we do not recall, let us just go ahead and do it again. We have our vector field.1767

It is -y × sin(xy) + x2 × cos(x2y). Okay.1777

Now, let us go ahead and put it in. So, the double integral over a of the divergence of f dy dx.1789

Our x is going from -4 to +4.1808

Our y is moving from -16 - x2 under the radical, divided by 2 all the way up to sqrt(16) - x2/2, the positive. The lower... that is the upper. That is what y is doing.1816

Our divergence is -y × sin(xy) + x2 × cos(x2y) dy dx.1835

When I go ahead and put this into my math software, guess what number I get? 0.5466, negative 0.5466.1856

This confirms that the flux integral is equal to the divergence integral. It confirms the fact that this particular vector field is actually... it has a negative divergence. It is contracting. Fluid is flowing, there is a net flow. This is a net, remember? We are integrating.1866

At a different point, at different points inside there, you might have a positive divergence but when you add up everything, the net flow is into the region.1885

Okay. Let us go ahead and write this out. Since -0.5466 is negative, this means that there is a net flow of the vector field into the region.1896

In other words, if we have our... the vector field as a whole is contracting. Okay. Now, in the previous lesson, we found that the curl of this particular vector field, the curl of f was equal to 0.1934

So, this particular vector field is contracting into the region but it is not rotating left or right.1964

This is really, really powerful, that we can make this kind of statement about a vector field given a particular closed region. Really, really extraordinary.1972

Thank you for joining us here at educator.com for divergence and curl. We will see you next time. Bye-bye.1982

Hello and welcome back to educator.com and multi variable calculus.0000

Today I am just going to have some final comments on divergence and curl.0004

Nothing new to introduce, just some variation of notation that you will probably run across in your studies.0010

In your books, maybe other books, or perhaps some different notation that your teacher is using.0016

I want you to be aware of them and not get confused and think that they are actually other theorems. They are not.0022

It is the same thing, it is going to be just Green's theorem, the two versions of it, but you are going to see them in lots of different types of notations.0028

I am going to introduce two of them for you, the two that you are going to see most often. Okay.0034

Let us go ahead and write that down. So, there are several symbolic representations for Green's theorem.0040

Okay. I am going to present you with two of them, so I will present two of them.0065

Later when we discuss, when we move on to 3 dimensions and we talk about the divergence theorem and Stoke's theorem in 3-dimensions, I will go ahead and introduce yet another symbolic version, but I do not want to lay it on too thick all at once.0082

Okay. Now, it is really, really important to remember that they are not different theorems. They are not different theorems.0098

Just different ways of writing the same theorem.0115

Okay, now, we write the following. We write the integral c(f) -- and I am going to do the curl first, the circulation curl -- f(c) · c' dt.0133

That is what we write, that is what we have been writing. It is the definition of a line integral is what it is.0153

I like it because when you are given a particular parameterization for a curve, this parameter, we choose the letter t. The parameter actually shows up in the integral.0158

It says take the vector, form the composite function, dot it with the tangent vector, that is your integrand, and integrate it from beginning to end of the parameter. That is it.0171

It is very, very basic, it is very straight-forward. There is nothing here that is hidden.0182

What you will see is this. One of the things that you will see. You will see, this particular version: the integral of f(c) -- let me actually, should I... that is fine, I will leave it as it was... f(c) · c' dt -- what you will see is f(c) · u ds. 0186

Now, this is the same thing. Let me show you how we get this from this. Okay.0223

I am going to draw a portion of a closed curve, so it would be like that.0229

So, this is f, this is our c, and we also have our normal vector, so this is n, this is c', this is f(c), just like before, this is c(t), we are traversing it in this direction.0239

Now, here is the thing. Remember when we said that the tangent vector and the normal vector are not unit vectors? Remember I had that word component in quotes, because we are not dealing with unit vectors?0260

Well as it turns out, we can form the unit vector in the direction of c', and that is exactly what u is.0274

So, let us go ahead and work this out. u is equal to the tangent vector c', divided by its norm.0282

When we take a vector and divide it by its norm, we are creating a unit vector in that direction.0296

So, u is actually the unit vector for c', that is all it is.0302

A unit vector in the direction of c'. Again, that is all it is. So u equals this.0309

Now let me go ahead and multiple through, I get the following.0325

I get c'(t) = u × norm(c'(t)). It is just simple algebra I have moved over, but the norm is equal to ds dt.0327

Okay. This ds dt... s is the length of the curve, ds is a differential length element for the curve... it is a very, very tiny length element of the curve.0352

This is a derivative, so the norm here is ds dt. When you take the derivative of a curve, you get the tangent vector.0366

When you take the norm of that tangent vector, you are going to get some number. What that number measures is a rate of change. It is the rate at which the length of the curve is changing as you change the parameter.0376

That is what a derivative is, it is not just a slope, it is also a rate of change. 0390

So this ds dt, which happens to equal the norm of the tangent vector at a given point, it is measuring how much the curve is actually lengthening as you do make some differential change in the parameter t.0394

This is ds dt, so when I put this into here, I get the following.0409

So, c'(t) actually equal to u... this is this, ds dt.0416

So, now when I put this c' = u ds dt into here, here is what I get: now the integral of f(c) · c' dt, I am just going to replace them with other symbols = to the integral of f(c) ·... well we said that c' is equal to u ds dt.0432

Now we will put that in and -- let us see, I will go to blue -- we put this in, it is here, we did c', that is this thing, and now we will put in our dt.0467

Well dt and dt cancel, leaving you with the integral over c of f(c) · u ds.0480

It is a way of actually expressing the circulation integral, the line integral, the definition of a line integral using unit vector notation and instead of the parameter t, which comes from the parameterization of the curve, what you are doing is you are using this differentiable length element of the curve itself.0489

You are using the length of the curve to parameterize. That is what this is. That is all this is, it is just a different way of writing this using unit vector notation and a differentiable length element as opposed to the actual parameter.0513

I personally prefer this because I like to see the parameter in my definition of a line integral. When I am doing the line integral, I do not care for this very much.0529

It is a little obscure, this, everything is exactly where it needs to be.0539

This u needs to be a unit vector. Well, in order for u to be a unit vector, I have to take my vector which is c', and I have to divide by its norm, it is just an extra step that is unnecessary.0545

Again, this actually tells me what to do. This, take this, dot these two and integrate.0556

But, you will see it like that. So, there you go. That is it. So, you will see the following.0563

You will see -- let me write this a little better here -- so the circulation curl form of Green's theorem, also looks like this.0574

This is what you will also see. The integral over the path of f(c) · n ds is equal to the integral over the region over the curl of f dy dx.0607

You will see this -- oops, not n, what am I saying... I have not gotten there yet -- dot u, there we go. Unit vector notation.0629

There we go, you will see that. Sometimes they will not even put the c. Sometimes you will see it as the integral of f · u ds, that is it.0639

Same thing, just a different way of writing it. I wanted you to be aware because you will see this in your books. I wanted you to see where it came from.0652

Okay. There we go. Now, for flux you do the same thing. For flux, we write the integral over c of f(c) · n dt.0660

Our flux integral was -- you know -- we take the c', we form the right normal derivative, we dot it with the right normal derivative, this is how we write it.0681

It includes the parameterization. There is no change that we have to make to it, we can just use this as is. 0689

Now let us go ahead and define n. There is a small n, which equals n(t)... it is just the unit vector in the direction of the right normal vector, so I take n(t), and I divide by its norm. That is it.0699

The unit vector in the direction of n, and again, let us remind ourselves. This is a curve... if this is f, n is here, c' is here, f(c) is here, this is c'(t), we are traversing this way. 0717

It is just a unit vector... that is it. Small n, that is all this is.0745

Well, so if... I am messing up my u's and n's... so if n = n(t)/norm(n(t)), okay... this implies that n(t) = n × the norm of n(t), I just multiplied through, this is just a number.0750

But, the norm of n(t) = norm(c'(t)), this n... all I have done is I have switched c1 and c2 and I have negated the second one. The norm is still the same. 0781

Well, that equals ds dt, that is the rate of change of s as you change t.0799

So, capital N(t) = n × ds dt. Now we go ahead and substitute this into our integral.0805

The integral of f(c) · n dt, we take this right here, we put it into here, we put it into the integral of f(c) · n ds dt dt, the dt's cancel, and we are left with f(c) · n ds.0819

There you go. That is it. This is just this, expressed in unit vector notation and instead of the parameter t, it is using a differential length element of the curve itself, s.0846

Okay, so you will see this. So, you will see Green's theorem as at over da of (f(c)) · n ds is equal to the double integral over the region a of the divergence of f, notice the right side is the same... divergence of f dy dx, and sometimes they leave the c off to make it even shorter... they write it as f · n ds.0860

Lowercase letters, unit vector, that is all it is. It is just the unit vector version of what we already know.0896

Now, again, I mean ultimately, it is a question of taste and again these are not the only symbolic representations of Green's theorem that you will see.0905

You are going to see others that use special vector operators, dot products, cross products, things like that. We will get to those, and we will talk about them, but again what is important is... it is important to understand the symbolism but you want to understand what is going on... which is the reason I prefer this symbolism and the analogous symbolism for the circulation integral.0915

I like it because the parameter for the curve that we are dealing with is actually part of the integrand.0945

I have not changed anything. f is there, I have c, I can get n, I can get c', I can take a dot product, and I can integrate. Everything is visible.0952

When I see it this way, n, yeah, it is a unit vector... and true it is better to think of a particular length, an actual component when you take the dot product of a vector, with a unit vector you are actually going to get the actual component along that vector.0965

It is true -- you know -- everything has its... it is a personal thing, it is a completely personal taste. What is really important is that you understand the underlying mathematics.0983

My personal taste is what it is that we have been doing all along. I do not care for unit vector notation, because I like to see my parameter.0992

So, that is it. That takes care of it for divergence and curl and we look forward to seeing you next time.1000

Thank you for joining us here at educator.com. Bye-bye.1005

Hello and welcome back to educator.com and multi variable calculus.0000

So, today's topic we are going to start talking about triple integrals and the nice thing about it is there is actually nothing new to learn.0004

You are just integrating three times. That is it.0011

Let us just jump right on in and do a quick couple of words and then we will just get right to the examples, because that is what is important.0015

So, for single integrals, we integrated over a length. Either a length on the x axis, or for a line integral, some path.0024

But we integrate along a length. For a double integral, we integrate it over an area, a 2 dimensional region. Either a square or some other region.0031

Well, for triple integrals, we are now integrating over a region in 3-space, which is a volume. That is all it is. We are still going to just integrate one variable at a time. 0041

So, let us just start with an example and it will make sense.0052

The important thing with these triple integrals and double integrals and all integrals is being able to construct the integral... taking a look at what your region is, being able to see this variable is going from this number to this number, this variable is going from this number to this number... that is what is important.0059

Now, when you take your test, of course, the integrals that you have to deal with are not going to be all together that complicated. They cannot be, because you only have a limited amount of time to solve it.0076

What your professors want to see is that you can construct the integral. That is what is important.0085

When you are doing your homework, or other projects, mathematical software can do the integral for you, but it cannot tell you what to integrate. That is your job. That is what is important.0090

That is the skill that you want to develop. The integration part, that is incidental. Anyone can do that.0100

You want to be able to develop a skill, so when you are doing these problems, do not worry about the integration so much, worry about the region. What region am I integrating over, what are my values, what is my lower limit of integration, what is my upper limit of integration. That is where you want your emphasis to be.0106

So, example 1. You know what, I think I am going to use blue ink because I like it. Alright, example 1.0124

Okay. Find the volume of the region bounded by... again with these triple integral problems, the problems themselves can be worded in all kinds of ways... sometimes they will give you the equations explicitly, sometimes they will say bounded by this, bounded by that... you have to work out the equation.0138

So, there is no one way that you are going to see these integration problems... region bounded by the xy plane, the yz plane, the xy plane -- I am sorry, the xz plane, I have xy, I have yz, I should have xz... there we go -- the xz plane, and the sphere of radius 3 centered at the origin.0168

In this particular case, they did not really give us a lot of information. There is no real equations, they are just giving me things by which it is bounded. 0222

They are expecting me to know that -- you know -- the equation for the radius of a sphere of radius 3, the xy, the yz, the xz plane.0228

So, let us talk about what this looks like. I have this... let me draw my 3-dimensional xyz coordinates.0236

This... and this... and this... and this... so, this is going to be my y axis, this is my x axis, and this is my z axis.0249

The xy plane, that is this thing right here. The yz plane, that is going to be this right here. The yz plane, and then of course the xz plane.0258

So, I am bounded from below, from this side, from this side, and I am bounded by the sphere of radius 3. 0274

Well, from the origin, I have a sphere of radius 3, so I am going to have that, and that, and that.0280

Basically, it is the portion of the sphere only in the first octant of three dimensional space. That is it, that is my particular region, that is the region that I am looking at... this sphere in the first octant.0290

My volume, well, by definition the volume of a 3... using triple integrals... the thing that we are integrating, the function that we are integrating is 1 dx dy dz over this region. 0308

That is the definition of volume. In other words, there is no function in the integrand. The function 1 will suffice... that is the function that you are integrating.0330

Now, what we need to do is we need to find the region. Our upper and lower limits of integration.0339

Okay. So, let us go ahead and draw the xy plane, let us just look at the xy plane first.0348

So, this is y and this is x. Well, we have this sphere, right? So, the z axis is coming straight out and going straight down.0355

Well this is x2 + y2 = 32, that is the equation of a circle in the xy plane.0368

So, now, what I am going to do... 0 and 3, so as far as the x variable is concerned, x is going to go from 0 to 3.0376

The y variable, well, y is changing. The height is changing, so y is going to run from 0 to this equation, which I have to express explicitly in terms of x, which becomes sqrt(9) - x2, right?0390

If I take x2 + y2 = 9, and I move the x over, I get 9-x2, I take the square root, I get + or -, - gives the part below, the + gives me that part.0417

So, x goes from 0 to 3, y goes from 0 to 9 - x2. Now z.0428

Well, z, the equation for the sphere is x2 + y2 + z2 = 9.0435

Well, I need everything about the xz axis, so z is going to be 0, it is going to be the lower limit of integration, and it is going to go all the way to this sphere.0447

z2 = 9 - x2 - y2, so z = + or - 9 - x2 - y2.0460

I take the + version, because I do not want everything below the xz plane, I want everything above it.0473

I need to go up so it is going to be sqrt(9) - x2 - y2.0482

This is what is important. I need to be able to express the region. I am integrating x from 0 to 3, I am integrating y from 0 to 9 - x2, and I am integrating z from 0 to 9 - x2 = y2 under the radical.0493

In other words, I go in the x direction, go in the y direction, then I integrate in the z direction, depending on what my boundary is.0508

So, now I have my integral, so the volume -- excuse me -- is going to equal the integral from 0 to 3, that is going to be my x, is it always good to keep track of what is going first, second, third, because we are going to work your way... working inside out.0519

So, the integral... y is going from 0 to sqrt(9) - x2, that is y.0541

z is going to go from 0 to sqrt(9) - x2 - y2, and of course our function is 1.0553

We are going to do dz, dy, dx, right? Inner to outer. xyz, zyx, we are integrating in that direction.0563

Well, when you put this into your software, you get 9/2 pi.0572

That is it. That is the volume of that region. Okay. Let us do another example. Let us see here... so, example 2.0579

Example 2. Find the integral of f of xyz = xy2z3, over the region bounded by the xy plane, the function y = x2, the function y = x, and z = x2 + y2.0595

This time they gave us some specific equations to work with. Let us go ahead and draw out this region.0645

Again, it is always best to start with things that are easiest. In this case they gave me y = x2, y = x, so let us go ahead and draw out the xy plane first.0653

So, xy plane. Okay. So, this is x, this is y, y = x2, that is going to be that thing. y = x, that is going to be that thing. So, this is 0 and 1, that is where they meet.0663

This is the particular region. So, as far as -- we have taken... xy plane, y = x2, y=x... so this xy plane means that the z value is going to be 0 and then the upper value of 0 is going to be x2 + y2, so we are going to integrate along x from 0 to 1.0683

We are going to integrate along y from x2 to x, and then we are going to integrate up z from 0 to x2 + y2, so just to draw out what this looks like... let me draw it out in 3-dimensions.0710

So, again, this is y, this is x, and this is z. Well, in the xy plane, we have the parabola, we have that, this is this region right here.0726

Now, of course, we have the paraboloid, so everything over this up to this surface. That is what we are doing. That is our region, and we are going to integrate this function over that region.0743

Let us go ahead and see what x and y and z are, and again it is always good to write them out specifically.0762

So x, it is going to go from 0 to 1, y is going to go from x2 to x, right? the x2 is the lower function, x is the upper function... and z, that one, is going to -- these are not equal signs, these are colons... colon, colon, colon -- z is going to go from 0, because it is bounded by the xy plane all the way up to x2 + y2.0770

There you go, so our integral is going to equal 0 to 1 for x, x2 to x for y, 0 to x2 + y2... we have x, we have y2, we have z3, and now this was x, this was y, this was z.0806

So, now, we are going to integrate dz dy dx.0837

Again, the actual order itself does not matter, it just depends on what the problem is at hand. Sometimes it might be easier to do the dy first, or the dz first, or the dx first, it does not really matter.0844

What you have to be able to do is look at your region and see what is going to give you the easiest integration, that is it.0856

Then when we go ahead and put this into our math software, we get this really whacky number, but it is just a number... 6,401/411,840. That is it. Perfectly valid number.0863

So, the integral of this function over this particular region of space is this number. That is it. What is important is being able to construct this integral.0877

If on a test you are pressed for time and you cannot actually solve the integral, move on, most of your points are going to come from being able to construct this integral. I promise you. There is no teacher that is going to punish you simply because you could not do the integral, but you are going to get lots of points taken off if you cannot construct the integral.0888

They understand that. Be able to this and you will get most of your points.0907

Let us go ahead and evaluate an integral by hand, just so you see what this single step manual process looks like, as opposed to putting something into mathematical software.0912

I just want you to be able to see it one time, at least. So, this one, we will do by hand. So, example 3, and of course the function cannot be that complicated, because we cannot be here for 30 minutes solving an integral.0923

Okay. So, f(x,y,z), it is going to be a rather simple function... we are just going to go with x. That is it.0937

The region is the same region as the previous example.0949

Okay. So our integral is going to look like this. Our integral is going to be from 0 to 1, and it is going to be from x2 to x, and it is going to be from 0 to x2 + y2.0967

This time it is going to be just x, and again we are going dz dy dx, because we did x, y, z, moving from inside out.0983

Okay. Let us go ahead and deal with the first integral here. This is going to be this integral, right here, and I am going to go ahead and pull the x out, because here the x is constant. I am integrating with respect to z first. I am going to do this in red.0992

I am integrating with respect to z first, so x is constant, I am just going to pull it out of the integral, so we get 0 to x2 + y2 of just plain old dz.1009

That is going to equal z going from 0 to x2 + y2, and my answer is x2 + y2.1017

Well, there is this x here, so I will put it over here, and I will go ahead and multiply it out. I could carry the x, I could keep carrying it back, it is not a problem. I do not have to multiply it out, it is just my personal taste.1026

I like to go ahead and deal with what I can deal with right away, rather than pulling it out as a constant.1041

Again, there is no one way of solving these. It is just a question of personal taste, so I will go ahead and multiply this out so I end up with x3 + y2 -- oops I will write it as xy2, how is that.1047

So, x3 + xy2. That is it.1063

Now, this thing is going to be my new integrand for my second variable. So, taking care of that one. Now I am going to go ahead and drop this in here, so I am going to get the integral from x2 to x of x3 + xy2, and this time we are integrating with respect to y, so dy.1071

Well, again, x is a constant. I am integrating one variable at a time, so this becomes x3y + x × y3/3, and this time I am going from x2 to x.1093

Now when I put these values in, I am putting these in for y, okay? x is a constant. Remember these values, I am integrating with respect to y, so this x goes into y.1112

So, when I do this, I get the following. I get x4 + x4/3 - x5 + x7/3, which ends up equaling x4 + x4/3 - x5 - x7/3. That is my new integrand for my last integral.1126

Now I am going to integrate from 0 to 1 of, well, x4 + x4/3 - x5 - x7/3 dx.1166

I end up with the following: I should get, x5/5 + x5/15 - x6/6 -x8/24, and I am integrating from 0 to 1.1185

When I put 1 and 0 into there, I get the number 7/120. That is it.1208

When you are doing this by hand, you just take it one variable at a time. Remembering to hold all of the other variables constant. That is all you are doing. Just integrate slowly and carefully.1217

Now, are you going to make mistakes? Of course. There are a bunch of symbols floating around, there are a bunch of numbers floating around. 1228

You want to do as many problems as possible, just to become accustomed to actually doing these. They do take a little bit of time, but again, you are not going to get anything complicated on a test just because you do not have the time.1233

I mean look, it took nearly 5 minutes to just do this integral. You do not have that 5 minutes. Constructing the integral, that is what is important.1247

Okay. So, let us do a final example here. So, this is going to be example 4.1255

We will let our function, f(x,y,z) = some trigonometric function this time... xz × cos(xy). That is going to be the function we integrate.1265

We want to find the integral of f over the region bounded by y = sin(x), y = 2sin(x), x2 + y2 + z2 = 4, and z greater than -- oops, sorry, not greater than or equal to -- and z < or = 0.1280

Okay. So, these are our boundaries. y = sin(x), y = 2sin(x), x2 + y2 + z2 = 4, and z > or = 0. 1330

Okay, we are going to have everything below the xy plane, because z is going to be negative, so let us go ahead and deal with the xy plane first. Let us just see what this looks like.1342

So, we have y = sin(x), so that is going to equal that graph right there... y = 2sin(x), that is going to be that graph right there -- sorry, my hand is a little shaky right there but you know what the graph looks like.1357

It looks like x is going to run from, so x is going to run from 0 to pi, from here to here, because we are looking at this region in the xy plane.1375

z is less than 0, so it is going to be everything below this plane and x2 + y2 + z2 = 4... that is the sphere of radius 2.1390

that is what you have got. Now, let me draw this in 3-dimensions as well as I can. It is kind of difficult.1403

Again, you do not necessarily have to see it. If you have all of this information, if you have the kind of mind that can sort of pick out what is going where, you do not need to draw it out.1410

Remember, this is -- oops, crazy lines again... I love those things -- so this is x, this is y, this is z, so y = x so we have that is sin(x), that is sin(2x), so this is the region... then of course we have the...everything below that. 1420

So we have the sphere -- it is a little weird, I know... you know what, I am not doing a very good job of drawing it but you know what is happening.1450

It is basically this region, everything below the xy plane touching the sphere x2 + y2 + z2 = 4.1471

Okay. Now let us go ahead and recall what is going where.1482

x, we said is moving from 0 all the way to pi. Well, y is moving from sin(x) to sin(x), so sin(x) to 2sin(x), that is lower and upper, and z... well z is going from -sqrt(4) - x2 - y2 up to 0.1488

The lower limit of integration is the actual sphere itself, and it is going up to 0, because it is below the xy plane.1527

So, this is what is important. Now our integral is going to equal the integral from 0 to pi, the integral from -- oops, not 0 -- the integral from sin(x) to sin(2x), and the integral from -sqrt(4) - x2 - y2 up to 0.1537

Of course we have our function which is xz cos(xy), and again this is x, this is y, this is z... so we are working out... so we are going to do dz dy and dx.1571

When we put this in our software we get the following number. 13.657. Again, Perfectly valid number.1590

So, when you are doing these integrals, this is what you want. We want to be able to construct the integral.1601

We want to be able to integrate along x, you are going to integrate along y, then you are going to integrate along z... x, y, z, three things.1606

You need to be able to choose these things. That is what is important. The rest is just integration.1619

Okay. So, that is our introduction to triple integrals. In the next couple of lessons, we are going to be talking about cylindrical coordinates and spherical coordinates, and triple integrals in those particular coordinate systems.1627

Thank you so much for joining us here at educator.com, we will see you next time. Bye-bye.1641

Hello and welcome back to educator.com and multivariable calculus.0000

Today's topic is going to be cylindrical and spherical coordinates. It is just two alternative ways to describe points in 3 space. That is all.0004

Because again, just like polar coordinates, there are certain regions that are just expressed a little bit easier in cylindrical or spherical coordinates. That is all it is. 0012

Again, with the introduction of mathematical software, the real power behind this -- these coordinates systems, you do not really... they do not play as important a role as they used to.0022

But, of course, it is part of the mathematical curriculum. It is part of the history, so we study it.0035

Let us go ahead and get started. Okay. So, polar coordinates, which was the R, θ in two space, and the plane, that generalizes to 3-space two ways, cylindrical and spherical coordinates.0040

We are going to do cylindrical first. So, we have cylindrical coordinates.0057

Let us see. Given R, θ, z, again we need three numbers for 3 space, the relationship between the Cartesian coordinates and the cylindrical is x = Rcos(θ) which was the same as polar.0070

y = Rsin(θ) so the x and y are the same as polar, and z is just equal to z. So, z is just equal to z.0091

If you want to think about it, the polar coordinates gives you what is happening in the xy plane, and the z is just straight up and down, which is why they call it cylindrical coordinates, because Rcos(θ), Rsin(θ) defines the unit circle, a circle of radius R.0097

Then if you just take z infinitely up and down, what you get is a cylinder, that is why it is called cylindrical coordinates. We are not going to worry too much about going the other way.0115

Given xyz, how do you gate R, θ, z, well, I mean you can get that because again, this is the same as polar, this is the same as polar, and that is the same. 0125

Really what we are concerned with is going from R, θ, z to x, y, z.0137

So, let us go ahead and draw this out so we see what it is we are looking at. Just so we know if that is a point out there, drop a perpendicular down to the xy plane, this is y, this is x, this is z.0141

We have this thing right here, well this angle is the θ, okay? So, this is the point x, y, z, and this right here, that is R.0162

Well, again, this point is the polar coordinate of the xy, and then you have the z going up. That is it. So, R, θ z. That is all there is.0176

So, let us go ahead and write the transformation as -- oh, also, these, so R is greater than or equal to 0, and θ is > or = 0, < or = to 2pi, all the way around the circle... and z is > or = negative infinity and < or = positive infinity. z can take on any value it wants.0192

z is < or = positive infinity. z is arbitrary. Okay, we can write this transformation as c(r), θ, z = Rcos(θ), Rsin(θ), z.0222

You will see the transformation that way, and that is exactly what it is, a transformation. You are converting one coordinate system to another coordinate system.0243

In this case, given Rθz, you want to express -- this is your x, this is your y, and this is your z -- okay, now when integrating a region in 3-space, or integrating over a region in 3-space, when we want to do a change of coordinate system, that is where this is important.0250

So, let us go ahead and write the integral formula here. So, when integrating over a region in 3-space, upon transformation -- or conversion if you will -- upon transformation the triple integral becomes the following.0269

So, triple integral over a region R, of the function that is expressed as a function of x, y, z, in Cartesian... dx dy dz... it is equal to the triple integral of R under the transformation c... c is for cylindrical, f(c), x, y, z, -- so wherever you see x, y, z, you put Rcos(θ), Rsin(θ) and z.0311

So, you form the composite function f(c) and the differential volume element, R, dR, dθ, dz. It just depends on the region as to which one of these is going to go first. 0343

This dx dy dz, that is the volume element in Cartesian... R dR dθ dz is the volume element, the equivalent volume element under transformation for the cylindrical coordinate system. That is it, that is all that is going on here.0360

Okay. Let us do an example. Let us see, let me go ahead and start the example on the next page so that we have everything on one page. Let me do this in blue.0375

So, example 1... excuse me... let R be the region bounded by the circle of radius 2 in the xy plane.0390

Let z be > or = 0, < or = x2 + y2, and the plane x + y = 0.0426

Okay. let us see what this region looks like. Let us go ahead and draw out the xy plane here. It is often best to work in the xy plane here, and it is often best to work in the xy plane and then consider of course z being coming out and going down. It is the best way to think about it.0443

So, this is y -- oops, sorry, we are not drawing the 3-dimensional system -- this is x, and this is y, so the circle of radius 2 in the xy plane, well, let us just go ahead and draw this circle right here, so that is a circle of radius 2, and let us leave z alone for a second.0459

The plane x + y = 0. Well, at the plane x + y = 0, notice z does not show up here, so z can take on any value, so it is going to be infinitely up and infinitely down, but it is saying that z is > or = 0, so we are concerned with everything above the xy plane.0479

So if you are looking like this, where this is the z axis, this is the y, and this is the x, it is everything above that. So here, we get y = -x, well y = -x is this line right here.0496

So, imagine instead of just the line that that is the plane. That is going to be parallel to the z axis. So, now, z, x2 + y2, well, x2 + y2... z = x2 + y2 is just the paraboloid around the z axis, that is it.0515

That is all that is happening, so it is going like that. So, the base that you have is the circle of radius 2 and let us see here... we want -- so we have the circle of radius 2... good, that is one of the bounds... it is everything in here, and it is going to be everything above the xy plane.0538

That is going to be that one. So, our region, oh actually I think I forgot to take care of 1 other thing. Looks like there is one other constraint here. We are going to have y > or = 0.0567

What we are looking at is this region right here, in the xy plane, and then of course up above the xy plane, z is going to go up and it is going to go all the way up to x2 + y2. That is it.0586

So, if I were to draw out the zy axis, for example, so let us say this is z and this is y, what you are going to have is this paraboloid and there is some region under here in the xy plane and that is going to be this region.0605

It is going to be everything up, so this is like 2, it is going to be everything above that. So, that is it. Think of this as the base of R volume, R solid, and this is the z version, so this is the base, this is going to be the height. The height is going to go up to this point.0621

So, let us go ahead and see what our R, θ, and z is going to be. R is going to go from 0 all the way to 2.0645

It is going to go from 0 to 2, θ is going to go... so now we integrate along R and then we are going to sweep this out an angle of 135 degrees, or 3pi/4, so θ goes from 0 to 3pi/4, and z, well z is going to run from 0 all the way to x2 + y2.0658

But, we have to express x2 + y2 in terms of polar coordinates, so it needs to be in terms of R and θ. So, z actually is going from 0 all the way to R2.0691

That is it. That is our region, and these are our upper and lower limits of integration, so, let us go ahead and do this. So, let us see, now, what function are we integrating?0710

We will let the function of xyz be equal to x + y - z, so let us form our f(c).0724

Well, x is Rcos(θ), y is Rsin(θ) and then -z, just put the transformation, the cylindrical transformation into the function for xyz. 0739

Now we go ahead and form our integral. So, the integral is equal to... we will go ahead and do R first, so R goes from 0 to 2, so I will put that here... R.0757

R θ is going to go from 0 to 3pi/4, right? and then z is θ, and z is going to go from 0 all the way to R2.0770

Then the function is Rcos(θ) + Rsin(θ) - z.0785

Then we have R dz dR -- nope, sorry -- R dz dθ dR, because R θ z, z θ R, we are working out.0793

If we did everything correct, and the mathematical software works properly, the number that I got was -4pi + 32/5 + 32sqrt(2)/5. There we go.0814

So, again, the integral part is... what is important is being able to construct this integral. This is what we want. Being able to take a particular problem and convert it into an integral. Being able to find out what the region is... in 3-space.0834

Then finding out what the upper and lower limits of integration with respect to each variable are. 0851

In this particular case, the radius went from 0 to 2, and then we sweep out this radius... sweep it out this way, an angle of 135 degrees, which is 3pi/4, and then of course we take the z value.0859

z goes from 0 to R2, or x2 + y2, because you are going, the z value is changing. That is it. You just put everything in, make sure you form the composite function, make sure you have the proper volume element, and the rest is just a question of integration.0876

That is all. The important thing is being able to find what that region is.0893

Now, I am going to go ahead and actually express this in terms of the Cartesian coordinates, just to show you what the Cartesian integral would look like.0902

So, let me do this in blue... Now... the Cartesian integral would look like this.0912

Let me draw my region again, and my region was... so this was the y axis, this was the x axis, we had this circle... we had this thing, so we were looking at this region right here, right?0931

As it turns out in this particular case, I am going to break this up into 2 regions. I am going to call this region a, and this region b, so I am going to do 2 integrals.0950

Let us go ahead and the actual integral itself was going to be the integral over a and the integral over b, so let us talk about a first.0960

a, x is going from, well, this point, the x value of this point where you have this line which is y = -x and where it meets the circle of radius 2.0971

So, it is going to be -2/sqrt(2) all the way to 0, so x is going to go from here to here.0985

Now, the y is going to go from, well, you are going to have this little strip so it is going to go from -x all the way to 4 - x2, under the radical, because the equation for this circle is 4 - x2 under the radical, when expressed as a function... y as a function of x.0995

Because this circle is x2 + y2 = 22. Therefore, y = 4 - x2, all under the radical, and of course z is going to go from 0 all the way to x2 + y2.1021

Those are our first set of upper and lower limits of integration for this particular region a, this little triangular sector.1039

Now for b, well b, now x is going to go from 0 to 2, so x is going to run from 0 to 2, y is still going to run this time... this is our strip, so it is going to run from 0 all the way to 4 - x2 under the radical.1048

It goes from 0 to sqrt(4) - x2, and z is of course still... goes from 0 to x2 + y2.1067

When I put this together, the integral in Cartesian coordinates equals 0 -- oops, the integral of a -- so it is going to be -2/sqrt(2) to 0, the integral from -x to 4 -x2 under the radical, 0 to x2 + y2, and of course our function was x + y - z.1078

This is going to be... this is z, this is y, this is x, so we are going to do dz, dy, dx.1109

Then plus the integral from 0 to 2, the integral from 0 to 4 minus -- oops, this is all under the radical -- 4 - x2, and the integral from 0 to x2 + y2 of x + y - z.1118

Again, dz dy dx, so, this in Cartesian coordinates, the integral would look like this. Again, software can handle it, but again, as you can see, this looks reasonably complicated. Obviously you would not want to do this by hand.1141

I mean it might be okay, I am not exactly sure but clearly the cylindrical coordinates in this particular case work out a lot easier, so but ultimately it is a question of choice. 1159

If you have this software available, it is not really much of a problem. As long as you understand, again, this is the important part. Being able to describe the region, the upper and lower limits of integration. That is what is important.1171

Now, let us go ahead and start talking about the other coordinate system. The cylindrical... I am sorry, we just did the cylindrical, this is the spherical coordinate system. 1186

Again, it is just another way of expressing a point in 3-space, so let me go back to black for this one.1193

So, we have, let me put it over here, so this is our xyz coordinate system, this is y, this is z, this is x, and we have some point and when we a perpendicular to it, to the xy plane, we have this right here.1201

Okay. So, this one is the one that is most analogous to the polar coordinate. So, we have 3 variables, we have ρ, which they use the Greek letter ρ instead of R, because they reserve R for polar coordinates.1220

This ρ, is the length from the origin to the point. Okay. Now this angle, θ, is the same... it is the same θ that it makes in the xy plane, and then there is one more angle here.1241

It is this angle. We will go ahead and erase this. This angle right here that it makes with the z axis, this is the angle φ, and of course this right here is the ρ.1256

So, what you have is the following: a transformation given ρ, φ, and θ, if you want to transform them, the x value is equal to ρ, × sin(φ) × cos(θ).1274

The y value equals ρ × sin(φ_ × sin(θ), and the z value equals ρ × cos(φ). 1301

So, our three variables are ρ, φ, and θ. Now, ρ is going to be greater than or equal to 0, it is a length. φ is going to run from... it is going to be > or = to 0, and < or = to pi, 180 degrees.1314

θ is going to run > or = to 0, and < or = 2pi. This is the spherical coordinate system. Basically what is happening is this. What you are doing is you are starting on the x... well, you know what, let me write a few things and I will describe what is actually going on here.1329

So, this point, (x,y,z)... it consists of a length along the z axis and then a certain angle downward from the z axis, and then of course you are going to rotate this an angle θ away from the x axis.1350

So, let me write out the reason. It is called spherical, and I think it will make a lot more sense where this ρ, φ, θ stuff comes from... it is called spherical coordinates.1370

I am going to draw out the zy coordinate system, so this is z, and this is y.1390

z here and y here, so the x axis is coming out and going down. Here is what we do.1400

The first thing, here is how I think about it, I think it is the best way to think about it. Go up a distance ρ along the z axis.1409

We will go up -- I am going to do this one in blue -- so we will go up a distance ρ along the z axis.1426

Now, swing down... actually you know what, let me go ahead and make this the x -- so I am going to change that, I am going to make that the x axis.1436

So now the second thing I want to do, I want to swing down... in other words I am going to take this point, and I am going to swing it down in this direction.1453

Through an angle of 180 degrees. I am going to take this thing and I am going to swing it down so now it is down here.1465

That is our φ. Through φ = pi, 180 degrees.1476

Now, I am on the x axis, so now I am going to take this semi-circle and I am going to swing it around the z axis all the way around. A full circle, that is my θ, 2 pi. 1482

When I do that, I am going to end up sweeping out a sphere, that is where the spherical coordinate comes from.1498

So, sweep this semi-circle around the z axis, θ = 2pi, and you are going to get the sphere of radius ρ. 1504

This is the best way to think about it. If you are given a 3-dimensional coordinate system. Let us say... ρ is just go up along the z axis, and then swing down along the zx plane, a certain angle φ, and then θ is, you swing around this way and that will give you the point somewhere.1527

That is the whole idea. So, here or all the way down -- from your perspective that is the z axis -- this is the zx plane, you are going to get a semi circle, and then you are going to swing this semi circle around.1549

I do not know if you have ever held a chain, if you sort of swung it around this way it sweeps out a sphere. That is why it is called a spherical coordinate system. That is it, that is all that is going on here.1565

Now, let us go ahead and describe the transformation. So, given... yeah, that is fine I can do it on this page. 1574

Given, some function of f(x,y,z) and the transformation, the spherical transformation, we will call it s, which is just a change of coordinates... change of coordinates.1588

When we say transformation, we are just talking about a change of coordinates.1608

We will do it this way... s(ρ,φ,θ) = ... and I am going to write this in column vector form... ρ, sin(φ), cos(θ), ρ, sin(φ) sin(θ), and ρ cos(φ).1615

Given some function, hence the transformation which is this thing right here... that is the transformation, that is the change of coordinates.1645

Given that, the integral is going to be as follows: the triple integral over a region R of f(x,y,z), I am going to make this a little bigger and a little clearer... so the triple integral of f/R and f is a function of x,y,z dx dy dz is equal to the transformed region R under the transformation s, f(s) -- we form the composite function -- and then of course the volume element.1655

ρ2, sin(φ), ρ2, sin(φ), dρ, dφ, dθ, in the proper order... depending on the particular region that you are given. That is it.1700

So, this is our conversion. Very, very important. This f under the transformation becomes the f(s), that is the composite function. When I put the transformation into the x,y, and z.1720

This volume element in Cartesian three space becomes this column element in R, φ, θ space. Cannot forget this ρ2sin(φ).1733

In other words, when I take this volume element and I transform it, I have to multiply it by this factor. That is what this is.1746

Okay. Now let us go ahead and do an example here. Let us do example 2.1765

Let f(x,y,z) = xyz. We will let R, the region, be the section of the unit sphere in the first octant. In the first octant.1777

In this case, the description that they gave you of the region... they did not give it to you as an equation, they did not tell you it is bounded by this, bounded by this, bounded by this... this time they just sort of told you qualitatively it is the part of the sphere in the first octant.1807

You have to figure out what the upper and lower limits of integration are.1819

So, our task is to find the integral of f over the region R. Okay. Let us just take a look at our region.1824

I am going to go ahead and draw out the full 8 octants. So, this is the first octant over here, this is z, this is y, this is x. this is z...1835

We have the unit sphere there, there, and there.1850

That is it. It is the section of the sphere that is just in the first octant, essentially 1/8 of the sphere.1861

Well, let us go ahead and find... so let us convert this... and since we are dealing with a sphere, a sphere is ideal for dealing with spherical coordinates because the equation of a sphere is just ρ... just some radius.1868

let us go ahead and find f(s) first... so, f(s) so we can put it into our integral formula here.1883

So, f(s), well, f is xyz, x is Rsin(φ)cos(θ), y is Rsin(φ)sin(θ), z is ρcos(φ), so we just put them in and multiply them out. 1891

So, xyz = ρsin(φ)cos(θ) × ρsin(φ)sin(θ) × ρcos(φ).1908

I end up with -- I want to simplify as much as possible -- so, ρ3sin2(φ), cos(θ)sin(θ)cos(φ).1924

Looks complicated but it is easily handle-able. Now we will go ahead and do the integral.1940

Well, let us see. First of all let us go ahead and talk about... so ρ, let us see, this is the unit sphere, so ρ is going to go from 0 to 1.1945

φ, well φ is in the first octant, so it is going to be from 0 to pi/2. It is going to swing down through pi/2, 90 degrees, and θ is going ot go from 0 to pi/2, so 0 to pi/2.1962

Now we are ready for our integral. Our integral = ... that is okay, I will go ahead and do it there... so 0 to 1, that is going to be our ρ, we have 0 to pi/2, let us make that our φ, so I will go ahead and write ρ and φ.1983

It is always good to write the variables underneath. Then our θ is going to be 0 again, to pi/2.2003

Then of course we have our function, so ρ3sin2(φ), just have to go slow, make sure everything is there, cos(θ)... and again, be very careful with these particularly because you are dealing with 2 angles, φ and θ, do not confuse them -- which I do all the time.2013

I can give you the advice, but I make the mistake all the time myself... cos(θ)sin(θ)cos(φ), that is the function, the f(s) part.2032

Now we have to do the transformed volume element, so ρ2sin(φ), and then we did θ first so we do dθ, dφ, and dρ. That is it.2047

When you put this into your mathematical software, you end up with 1/48.2062

That is all. Straight, simple, you just need to be able to take a look at a particular region, try to express it, that region in terms of... in this particular case... spherical coordinates, and decide this part right here.2067

What is going to be happening? ρ is going to go from what to what, φ is going to go from what to what, θ is going to go from what to what. That is all.2087

In this particular case, the spherical region, this is the best way to deal with it rather than delaying with Cartesian coordinates.2098

The integral for Cartesian coordinates is incredibly complicated. It has a lot of radicals in it, I do not even know if it is solved by elementary methods, you might just have to rely on numerical methods to solve it if you were actually going to use the Cartesian coordinate system.2105

So, that is it. That is spherical coordinates, cylindrical coordinates, hopefully we will get a chance to do more practice in the context of other concepts.2118

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2129

Hello and welcome back to educator.com and multi variable calculus.0000

Today's topic is going to be very, very important.0004

We are going to be talking about parameterizing surfaces and then we are going to begin to discuss the cross product.0008

Now we are moving away from 2-dimensions, and we are moving into 3-dimensions.0012

We are going to revisit Green's theorem and the fundamental theorem of calculus in its 3-dimensional form, Stoke's theorem, divergence theorem, things like that. 0018

So, what we are going to do today is lay out the foundations, how to describe a surface, how to think about a surface, this thing called a cross product, what it means, so, let us just go ahead and get started.0029

Okay. So, let us go quickly back to the parameterization of a line.0041

A line or a curve is a 1-dimensional object, so the x axis, or the y axis, you just need one number... 5, 7, to tell you where you are along that line.0048

Well, it is a one dimensional object, so to describe a curve parametrically, we require one parameter. That is it.0075

We have been calling that parameter t. So, one dimensional object, a line or a curve, all you require is one parameter to describe that curve parametrically.0101

An example would be... example... c(t) = let us say t, t2, and t3.0113

Okay. We have one parameter, that parameter is t. Well, we have 3-coordinate functions, 1, 2, 3.0126

This means it is a curve in 3-space. If I had 7-coordinate functions, this would describe a curve in 7-space, you could not describe it geometrically because we cannot picture a 7-dimensional space, but mathematically it is a perfectly valid object. That is it.0150

So, a curve, a 1-dimensional object, requires 1 parameter. I can define it in any dimensional space that I want to. 0172

Okay, now a surface, which is exactly what you think it is intuitively, a surface is -- actually, let me write one other thing, qualify it one other way -- a surface, whether flat or curved... so if I take a flat piece of paper and curve it, it is a surface but it is still a 2-dimensional object... flat or curved is a 2-dimensional object.0178

So, let us see, to describe it parametrically requires -- you guessed it -- two parameters.0220

I think I definitely need to write a little bit better here... two parameters. Let us call these parameters t and u. 0246

As it turns out, the parameters that you choose, it is going to depend on the problem itself, sometimes it is going to be φ θ, sometimes it is going to be xθ, sometimes it is actually going to be the xy that you happen to be dealing with in your function, but just for generic purposes, let us just call them t and u.0258

So, a good way to think about this, you know... two parameters, surface... well, you are already familiar with the idea of one parameter and a curve... well, think of a surface this way.0274

We have this, this, let us say that, and that. Well, that is 1 curve in the surface, let us say there is another curve in the surface, right?0285

If this is a surface, if you hold one point you can follow a single curve, that is one parameter and you can go in another way along that surface, that is another parameter.0301

You can think of it as once you -- you know -- sweep out one curve, and then you can go ahead and add another parameter and take that curve and sweep it this way.0311

So, you sweep something this way, and then you sweep that curve along this way to form the surface, that is what is going on. That is it.0322

Okay, so let us do an example of this one. An example might be s(t,u) = t and u and let us say t2 + u2.0330

In this particular case this happens to be the parameterization of a paraboloid that is going up along the z axis, the standard paraboloid that you are used to.0347

So, notice here we have two parameters, we have t and we have u. So, we have 2 parameters.0358

Two parameters t and u, and again we have 3 coordinate functions, this one, this one, and this one. Each one a function of t and u.0369

Now, granted the u does not show up here, the t does not show up here, the u and t show up here, that does not matter, it is still considered a function of the two variables t and u, t and u, t and u.0381

So the three coordinate functions mean the same thing that they did before. This is a surface in 3-space.0390

If I wanted to take some function of t and u, but if I had 8 coordinate functions, I would be defining a surface in 8-space.0400

Obviously we do not know what the hell that looks like, but there it is. It is possible mathematically.0409

So, 3-coordinate functions means a surface in 3-space, and of course that is what we are going to be concerning ourselves with. A 2-dimensional object in a 3-dimensional space.0416

In other words, a surface in Euclidian 3-space. Okay.0428

So, let us go ahead and just one thing I should say. Recall when we talk about the parameterization of a line, parameterizations are not unique. There is more than one parameterization for the same object.0436

So, if I had a paraboloid, I might have one parameterization, somebody else might have another parameterization, in the long run it does not matter. You are going to end up getting the same answer. 0450

The final answer does not depend on the parameterization. The parameterization is just an artifact. It is something that we use to be able to deal with the object.0460

When you integrate things, when you calculate certain mathematical quantities, those numbers are actually the same because they are deeper intrinsic properties of the objects themselves. 0469

That is what is amazing. So, a parameterization is just a representation. It is not just the object itself. You are just representing the object. It is not unique.0480

So, parameterizations are not unique.0490

Okay. So, let us just do an example and I think this is the best way to do it. You are going to see lots of parameterizations for lots of different objects. It just depends on what it is that you are dealing with.0502

We are just going to throw out some basic ones, so let us just parameterize this here of radius r.0513

So, we have a sphere of a given radius r, I am going to go ahead and use the letter p for parameterization. The same we used c for a curve, c(t), I am going to use p, p(t,u)... you can use any letter you want, it does not matter.0523

Let us use p for parameterization. If you want you can use s for surface.0543

Okay. So, the parameterization for a sphere, as it turns out, t and u is the following. It is rsin(t)cos(u), rsin(t)sin(u), and rcos(t), where t runs from 0 to pi, and u runs from 0 to 2pi.0557

Does this look familiar? It should. It is basically spherical coordinates except now we fixed r, we set a specific radius for a given sphere, and then parameters are t and u, so it is t and u that vary. That is all that is going on here.0588

Instead of using t and u here, let us just go ahead and use t and θ because that is what we are accustomed to and that is what... this spherical coordinate is. 0604

So the parameterization for a sphere, of given radius, is a function of the 2 parameters φ and θ, and it is equal to rsin(φ), cos(θ), rsin(φ)sin(θ), and rcos(φ). There you go.0617

That is your parameterization for a sphere. At least it is one of the parameterizations, it is not a unique parameterization... it is probably the one that is used most often because it is very, very easy.0640

Now we will do another example, so example 2. Now I am going to give you another parameterization for a sphere. So, another key for the sphere of radius r.0653

Okay, well we know that the Cartesian equation for a sphere of radius r is x2 + y2 + z2 = r2.0675

It is just a generalization in the 3-dimensions for the equation for a circle.0689

So, here is what I am going to do, I am going to go ahead and actually solve for z, explicitly.0694

When I move everything over, I am going to get the following: z = + or - sqrt(r2 - x2 -- y2).0701

This is actually really, really, really convenient. Here is what I am going to do.0715

I am going to say t = x, u = y, then my parameterization in terms of the generic variables, t and u, actually ends up being the following.0720

t,u, and r2 - t2 - u2, all under the radical.0734

So, this is another parameterization for the sphere, except this time when you are given the sphere in terms of Cartesian coordinates, what you can do -- if you can half solve this implicit equation explicitly for the variable z, you can set the first parameter equal to something, the second parameter equal to the other variable, and you can just put the equation that you solve explicitly for z as your third coordinate function.0744

I mean, clearly you do not need the t and the u, they are just generic variables, so why do we not just stick with the x and y?0774

So, we can rewrite our parameterization as x and y, well, that is equal to, well x is the first parameter, y is the second parameter and the function itself, the function of x and y when you solve for the third variable, that is the third coordinate function.0780

We get r2 - x2 - y2 under the radical.0799

This is our other parameterization for the sphere, that is it. 2 parameterizations, they describe the same object, it is not unique. Now, this is true in general.0806

It is really very nice. So, it is actually very, very convenient, so that any time you are given an equation in x, y, and z, you automatically have a parameterization.0818

So, let us say true in general.0829

What is true is the following. If you are given z = to some function of x and y, well, then -- oops, here we go with these lines again -- then, your parameterization of x and y is equal to... well x is your first parameter, and y is your second parameter, and the actual function itself as a function of x and y is your third parameter.0840

So, again, you are given a function, you could be given the function explicitly, you know... maybe you are given something like z = sqrt(xy), something like that, that is explicit, or you could be given the function implicitly like this.0874

x2 + y2 = z2, in which case if you can solve for z, then great. You can go ahead and use this parameterization. 0890

If you cannot solve for z, then you are going to have to search for another parameterization. Something with a t and a u, some other variables.0897

So, let us do another example here... let me go... let me go to blue, how is that?0907

So, example number 3. This time, I am going to parameterize a paraboloid.0916

Again, that is just a 3-dimensional parabola, it goes up and around the z axis. Well the first parameterization for a paraboloid happens to be the following. 0928

I am going to write p1, and it is going to be the parameters t and θ.0940

Well, the parameterization is tcos(θ), tsin(θ), and t2. This will give you, as you vary t and you vary θ... let us make sure we have the certain values... t > or = to 0, of course, and θ > or = to 0 and < or = to 2pi, so as you vary t and θ, you are going to end up with this surface.0946

Again, you are going to be given several different parameterizations.0976

You are either going to be told what this is, or you are going to be asked to describe what this is.0983

This is something that you just have to spend a little bit of time with. Pick some values of t, pick some values of θ, stop and think about where these points end up in 3-dimensional space.0990

That is really the only way to wrap your mind around this geometrically. If you are somebody that absolutely needs pictures in order to make the math come together.1000

If not, if the pictures do not matter, then do not worry about it, because ultimately we are just going to be operating on these functions the way we do -- you know -- everything else. 1011

We are going to be taking derivatives, and second derivatives, and partial derivatives, so if you just want to treat them as mathematical objects that have no picture associated with them, that is not a problem at all. Whatever works best for you.1020

Okay. So, that is one parameterization right there. Well, let us go ahead and use our thing that we just discussed a minute ago. Something being true and general.1032

The paraboloid, in terms of x and y is given explicitly by the following. x2 + y2, so if I want another parameterization, let us call this one p2, and this we will call x and y.1042

Well, it is equal to the following. x is the first parameter, y is the second parameter, very, very convenient and of course, x2 + y2.1056

Now, you notice, this is the same thing as earlier in the lesson, when we had t and u.1066

We said t, u, t2 + u2, we do not need t and u, we can just use the variables x and y that we are familiar with.1070

So, this is a second parameterization of the paraboloid, and again, when you start doing the math with this, your final answer is going to be the same.1079

These are just two representations of the same object. Okay.1086

Now, let us do a final example. Example 4, a very, very important example. Something from single variable calculus. Remember when we did surfaces of revolution? We had some graph in the xy plane and we decided to rotate that graph either along the x axis or the y axis.1092

Well, in this case, now we are going to talk about how to parameterize some surface described as a surface of revolution.1115

So, a surface of revolution and our axis of revolution is going to always be the z axis. So, a surface of revolution around the z axis.1122

Okay, now, we will let f be a function of 1 variable, let us go ahead and call that variable x for x > or = some value x1, < or = some value x2, like for example if x is between 3 and 7, something like that. Okay.1143

Then, the surface of revolution generated by rotating the function z = f(x) around the z axis, is the following.1174

Your parameterization is given by p, the two variables, the two parameters are going to be x and θ, always, and you are going to have the first coordinate function is xcos(θ), the second coordinate function is going to be xsin(θ), and the third coordinate function is just going to be your f(x).1213

So, what you are doing is you are actually taking the z axis -- oops, let me get rid of this so it does not look like a triangle -- and you are taking x here.1235

You can think of the y -- you know -- as going in and out of the page, and into the page, well if you are going to have some function, let us say, something that looks like that, function of one variable, z as a function of x, your normal single variable calculus.1248

If you take this and you rotate it around the z axis, now you are going to end up with this flat surface that is above the xy plane. So, what this parameterization does is it gives you the x coordinate... I will do this in red... this arrangement right here, it gives you the x-coordinate of a point on the surface, it gives you the y coordinate of a point on the surface, and it gives you the z coordinate of a point on the surface.1260

This is just a way of parameterizing a surface of revolution. That is it. Around the z axis.1294

Okay. So, let us do an example. Our specific example will be the following. We will let f(x), or actually let us call it... yea, that is fine, we can call it f(x), let f(x) = sqrt(x)... and let us just say sqrt(x) is > or = 1, and < or = 3.1302

Let us go ahead and draw this out. This is going to be something like that, okay? From 1 to 3. This is the x axis, this is the z axis.1331

When I take this thing and I rotate it around the z axis, I am going to end up with something, some surface that looks like that... around the z axis, the parameterization for that is the following.1347

The parameterization in terms of x and θ = xcos(θ), xsin(θ), sqrt(x). That is it.1368

Very, very, very common. Many of the problems that you have will be surfaces of revolution. This will always give you a parameterization. Nothing more, nothing less. Okay.1383

Now let us move on and develop a little bit more. I am going to introduce something called the cross product, and we are actually going to use this cross product in our next lesson when we talk about more properties of the surface.1398

In particular the tangent plane to the surface, the normal vector to the surface, things like that, but let us go ahead and define it before we do anything else.1413

So, let us see, given 2 vectors a and b, well we already define the dot product, the scalar product, right?1422

We already defined a · b, we said that if a has... is a1, a2, a3, and the vector b has coordinates b1, b2, and b3, well we define their dot product as a1b1 + a2b2 + a3b3... we have been dealing with this forever now.1439

Now, let us go ahead and define something called the cross product, or the vector product. Now, define -- I will do this in blue, I will go back to blue here -- so define the cross product, also called the vector product. 1475

The reason it is called the vector product is because you actually get a vector. Now when you multiply two vectors under the cross product operation, you are going to get a vector. Here you multiply this under the dot product operation, you got a scalar, a number. So, vector product.1498

I am going to write out the definition, but then I am going to give you a symbolic way of actually solving it because the definition is actually a big unwieldy.1516

So given these 2 vectors a and b, a cross b, again, that is why we call it cross... we use the × signal... is equal to a2b3 - a3b2, that is the first coordinate of the vector, a3b1 -a1b3, that is the second coordinate, and a1b2 - a2b1, that is the third coordinate.1526

In terms of i,j,k, notation, you get the following... equals a2b3 - a3b2i + a3b1 - a1b3j, the unit vector in the y direction, i is the unit vector in the x direction, and that plus the final one, a1b2 - a2b1k. That is it.1560

That is the definition of the cross product. Now, I am going to go ahead and give you a way of doing this symbolically given 2 vectors a and b.1600

Okay. Now, what I am about to describe is definitely just symbolic. It is a way of remembering this so that you do not have to remember this, all of these indices 2, 3, 3, 1, you know, 2, 1.1609

Okay. a cross b, so a × b is equal to the determinant -- let me go ahead and draw a matrix first before I... okay, we write i, j, and k on the top, we have a1, a2, a3, b1, b2, b3 -- we form this matrix and then we take the determinant of this matrix. 1623

Let me write it out in determinant notation, i, j, k, a1, a2, a3, b1, b2, b3. This is symbolic.1660

When you take a 3 by 3 determinant, you are going to get a number, but again this is just a symbolic way of finding the cross product.1677

Let us just do an example here. So, this is example 5. We will let a = the vector (3,1,-5), and we will let b, the vector, be (2,-2,6).1685

So, we want to find a × b, well, let us make our little determinant here, so we have i, we have j, and we have k.1709

We have (3,1,-5), (2,-2,6), and we are going to take this determinant. 1725

Again, with determinants and matrices, the biggest problem is going to be algebra, the positive and negative signs. What you are multiplying, there are a whole bunch of numbers floating around so just go slowly.1733

Are you going to make mistakes? Yes, you are going to make mistakes. I make them all the time with this stuff, because again, it is just arithmetic. Arithmetic can be tedious and annoying.1742

So, let us remember, when we do this we are actually expanding this determinant along the first row. That is why we have the i, j, k, so we do not just expand randomly, the way you would with any determinant.1754

You can choose the row or column of your choice to expand along, but here we are definitely expanding along the first row.1767

So, let us recall +, -, +, -, +, -, +, -, +, when you are expanding along the first row, your first term, you are going to have 3 terms.1775

First term has a + sign, second term has a - sign, third term has a + sign. You have to account for all of this in addition to all of the pluses and minuses of the numbers.1791

So, let us go ahead and do this one in red. So, for i, we are going to have the following. We are going to have 1 × 6, so let us actually write all of this out, so 1 × 6 - (-5) × -2, that is going to be i.1803

Now, -, I will expand along j. We are going to have 3 × 6 - (-5) × 2, and that is going to be j.1834

Then we have +, see this +, -, +, that comes from here, here, and here, and I keep everything separate. Do not do this in your head. Write it all out.1850

3 × 6 -- no, this is that way so now I have 3 × -2 -- 3 × -2 - 1 × 2... and this is k.1861

Okay. Hopefully I did all of this right. We have 6 - 10, so this is going to be -4i... 3 × 6 is 18, and this is -10, but this is - (-10), so it is + 10, so 18 + 10 is going to be 28... so we are going to have -28j.1877

Then this is going to be -6 - 2, this is going to be -8, so it is going to be - 8 + -8, is a -8k. There you go. This is our vector.1909

Given (3,1,-5), (2,-2,6), their vector product, their cross product is the vector (-4i,-28j,-8k), or if you prefer this... so the i,j,k stuff tends to be popular with the engineers and physicists in terms of coordinate functions, you get the vector (-4,-28,-8), which is more characteristic of mathematical work.1922

Okay. Now, here is the interesting part. a cross b -- let me go back to black ink here -- let us do this, now the vector a cross b, it is a vector, it is perpendicular, it is orthogonal, perpendicular to both a and b.1950

Its direction is given by the right hand rule.1981

The right hand rule is as follows: so, we have got a cross b, so the order of a vector product is actually very, very important. Now, the first part, your fingers, your 4 fingers, they point in the direction of vector a, the first term in the multiplication.1998

2. What you do is you swing your fingers, you curl your fingers, you swing them in or curl them in, swing your fingers in the direction of b. Towards b, the second.2034

3. Whichever direction your thumb is pointing, when you actually do that, the direction of your thumb is the direction of a cross b.2053

What you have is something like this. So, if you have a vector -- let us say this is that and this is that, let us call this vector a and this is vector b -- well, let us say these, both of these vectors are in the plane right here.2092

Well, if I take a, my fingers are that way and I swing them in the direction of b, my thumb is point up.2107

As it turns out, the vector a cross b is actually pointing out of the page. Well, there is another vector. If I do the other direction, if I change the order and do b cross a, swing it in the direction of a, now my thumb is pointing down. So that would be down this way.2117

So, I am going to say this is a cross b, and this is perpendicular there, and it is perpendicular there.2135

In the other direction, that would give me b -- excuse me -- cross a, so in normal multiplication, 2 × 3 is equal to 3 × 6 -- I am sorry, 2 × 3 is equal to 3 × 2, and the dot product is also commutative. Cross product is not commutative.2144

You get the vector of the same magnitude, same length, but opposite direction. That is all that is going on here. Given 2 vectors, the first one, swing the fingers towards the direction of the second one, the direction that the thumb is pointing in, that is the direction that the actual a cross b vector is pointing in.2161

It is perpendicular to both. That is what is important. Okay. Let us just do an example here.2180

Oh, one last thing I should let you know. Remember we had the formula a · b = norm(a) × norm(b) × cos(angle between them), something like that if this is the angle θ.2187

Well, there is a similar formula for the cross product, and it is good to know.2210

a × b = norm(a) × norm(b), sure enough × sin(angle between them), so sin(θ).2216

This formula is also good to know. Okay. Let us do an example here.2230

Okay. Example 6.2238

So, let us say we have a is the vector (1,2,1), and let us say b is the vector (1,4,1). Let us go ahead and draw this out just for the heck of it.2248

So we have got this vector here, let us call that one b, let us call that one a, so this is vector a, this is vector b, we know that the cross product is going to be a vector that is going to be perpendicular to both.2265

It is going to be coming out of the page if in fact -- well, in this particular case these are 2 vectors in 3-space, so we are just going to draw them out like this -- this particular vector which is a cross b, we do not know if that actually comes out of the page or not because we are dealing with 3-space, but it is perpendicular to both a and b.2279

So, let us go ahead and do our symbolic calculation. So a cross b = i,j,k, and we have (1,2,1), (1,4,1).2298

When we actually do this, expand it along the first row, we end up with following with -2i + 0j -- and I hope that you will confirm this for me because it is very, very possible that I made an arithmetic error -- and in terms of regular coordinates it is (2,0,2).2318

That is our vector. The vector itself is (2,0,2), and now let us go ahead and find its norm.2336

So, the norm of a cross b, well that is equal to 22 + 02 + 22 all under the radical.2344

That equals sqrt(8), so the length of this vector is sqrt(8), its direction is perpendicular to both vectors. It is a vector. It has a magnitude and it has a direction.2356

The dot product was a scalar product, it is only a magnitude, it is only a number.2370

Okay. So, let us go ahead and find the angle between those 2 vectors. So norm(a) = sqrt(6), because it is 12 + 22 + 12, and norm(b) = sqrt(18), right? 12 + 42 + 12.2375

So, now, we have sin(θ) = -- I am going to rearrange that formula -- = a cross b... actually, you know what I am going to go ahead and stop there. I am not going to do that.2402

So, that is what is important here... being able to find the cross product given the vectors a and b, realizing that the cross product itself is a vector and the direction of that vector is going to be given by the right hand rule, where you take the first vector a cross b, the first, swing it into the second.2421

The direction that the thumb is pointing, that is the direction. The first direction is b cross a, and you can find the norm of the vector like you would the norm of any other vector, and it also turns out if I draw a little parallelogram... complete the parallelogram for the vectors a and b.2441

This particular parallelogram that is spanned by the vectors a and b, that actually happened... this cross product, the magnitude of the cross product happens to be the area of that parallelogram.2462

Of course we will talk more about that next time when we discuss surface area and things like that. 2476

Okay, so that is it for today's lesson, parameterizing surfaces and cross products.2482

Thank you for joining us here at educator.com, and we will see you next time.2487

Hello and welcome back to educator.com and multi variable calculus.0000

In today's lesson, we are going to discuss the tangent plane and the normal vector to a surface.0004

Remember we described how to parameterize a surface with two parameters, now we are going to talk about the plane that is going to be tangent to it, the same way if you have a curve, you have a tangent line, if you have a surface you have a tangent plane.0010

We are going to define the normal vector, the vector that sticks away from the surface and we are going to set the stage for the integration of functions and vector fields over surfaces.0022

So, let us get started. The first thing I am going to do is draw a little picture here, so that we have something that we are working with.0035

I am going to draw your normal xyz coordinate plane, actually you know what, I think I am going to draw the surface first, and then I am going to put the axes on top of that.0040

So, let me go something like this, and this, and this, and that.0052

So, let me draw one line that goes like that and goes along the surface that way, and let us say that we have another line that goes along the surface that way.0062

Our axes, let us go ahead and put them here... this is going to go like that, and this is going to go over like that, so this is the y, this is the x, and this is the z, so we have this surface in 3-space.0071

okay. So, let us go ahead and start talking about what is going on. We have been using the letter p for parameterization of a surface, so let p(t,u) = well, we have f1, which is a function of (t,u), these are the three coordinate functions.0089

We have f2 which is a function of (t,u), and we have f3 which is a function of t and u.0108

So, Let this be a parameterization for a surface.0115

Okay. Excuse me... now, if we pick a specific value of t, if we hold it constant and if we hold it constant and we just vary the variable u, what you end up with is a curve along the surface.0132

If we end up holding u constant, pick a specific value of u and then just vary t, again we get another curve. Essentially you just get a function of 1 variable.0153

Really what we are doing when we parameterize this surface is we are making a curve this way, and then we are parameterizing, we are taking that curve and we are sweeping it out to create the rest of the surface. That is what is happening.0165

So, imagine this curve, and then imagine sweeping it back and forth in this direction and this direction, and that gives you the parameterized surface.0178

So, if we hold one of those constant, we actually get curves along the surface. So, let me write that down.0189

If we hold... if we pick a value for t, hold it constant and vary u, we get a curve... so let us call that one c1.0196

It is actually just a function of u, but it is actually the parameterization when we are holding t constant. That is what it is.0236

We can do the same for the other ones. So, this is c1(u), let us just call this curve over here c2(t), so, if we hold u constant and vary t we get a second curve.0247

This is actually a really, really good way of thinking about surfaces. Think about them as a curve in this direction, and a curve in this direction and the curves cross and it will give you a sense of what is happening on that curve near that point.0276

So, this point right here, that is the p(t,u) for some value of t and u.0290

Again, if you are there and if you happen to hold t constant but you are going to vary u, you are going to move in this direction and in this direction. 0296

If you are going to hold u constant, you are going to move that way and that way. If you vary both of them, you are going to vary all over the surface. That is what is happening.0304

So, you get a second curve. We will call this one c2 and it is a function of t, but it is still just the actual parameterization t and u, where one of them is held constant.0312

Okay. Now, here is what is nice, we know that we can differentiate curves. Well, differentiating these curves, taking c1' and c2' is the same as just taking the partial derivative of the parameterization with respect to the particular variable that you are varying.0324

What you end up with when you take the derivative is you get the tangent vector to the curve.0342

So, let me go ahead and write this down and then I am going to redraw the picture and I am going to redraw some tangent vectors on top of it.0348

We can differentiate these curves because we have been doing so all along.0355

Well, c1'(u) is nothing more than the partial of the parameterization with respect to u, and c2'(t) is nothing more than the partial of the parameterization with respect to t.0369

We know how to do partial derivatives, we have been doing them the entire course. So, that is it. That is all that is going on. Hold one variable constant, vary the other one, and you can take partials.0387

What you get are the following. Let me redraw this surface here, i am just going to draw the surface without the axes.0399

Let me draw it over here and give me a little bit more room to work. We have something like this, like that, like that, like that and we said we had a curve here and we have this curve here... like this.0405

So, when we take... so we pick a particular point, this point is our p(t,u), if we hold one of the variables constant and we take the derivative of it, it is like taking the derivative of the curve.0417

What you get is this vector, the tangent vector. Let us just call this one dp du.0432

If you move along this curve, where you are varying t holding u constant, you get a vector this way. You get the tangent to the curve at that point.0441

This is dp dt. I hope this makes sense. You have got these 2 curves and at a particular point, you can take the partial of the parameterization you are going to get two tangent vectors that are the tangents to the curves the way we have done before.0451

Now, these are... let me actually write all of this down. So these are just the tangent vectors to the curves.0469

Well, these two vectors, they span a plane, that is what it is. Any time you have two vectors, what you have is the plane that they cover.0495

These two vectors span a plane. This plane is tangent to the surface, right? it is tangent to the surface at pt(u). 0508

So, if I have a particular point, or a particular t and a particular u, if I take the partial with respect to u I get a tangent vector.0536

If I take the partial with respect to t at that point, I get another tangent vector. Well these two tangent vectors span a plane, and that plane is actually the plane that is tangent to the surface at that point. That is what is going on.0545

I have got a surface, I have got a vector this way, I have got a vector this way, this vector spans a plane, this plane is what is tangent to this surface at that point. Pretty fantastic.0557

Okay. Now, here is where it gets great. So, if dp du -- oops, not cross yet -- and dp dt are non 0 vectors, which for all of our problems they will be, then the cross product dp dt cross dp du, or perhaps in the other direction depending on the orientation that we want for the surface is a vector perpendicular to orthogonal to both dp du and dp dt.0572

So, given two vectors, if I take their cross product, I get another vector. This particular vector happens to be perpendicular to this one and this one. It is perpendicular to each. 0630

That is very, very convenient because now we have a plane that touches a surface, and at that point, we are going to create a normal vector like this.0639

Well, this normal vector is going to be at that point the normal vector to the surface, the same way we had a normal vector to a curve. 0650

Now, it just happens to be two curves that cross each other, those 2 curves that cross each other define a surface at that point... you have a normal that is... you have a vector that is normal to both curves. It is a vector that is normal to the surface.0658

We are constructing a tangent plane and we are constructing the normal vector at that point. That is what we are doing.0671

So, let me draw this time, let me draw it on top of this. So, we have dp du, actually you know what, let me do it one more time... so, let me go this way, make it a little bigger, so we have this is 1, this is 2, so we have... that is 1 tangent vector, that is another tangent vector.0678

Let us call this one dp du, and this is going to be dp dt. Now let me go to blue. Now when I take the cross product of dp du cross dp dt, and remember the right hand rule? I am going to get a vector that points this way.0707

It is going to be perpendicular to that, and it is going to be perpendicular to that.0726

So, this vector right here, it is dp du cross dp dt.0732

Now it is specifically dp du cross dp dt, if I went dp dt cross dp du, I would get the vector that is going in the other direction, that is actually going into the surface.0742

We want the one that is going away from the surface. That is a question of orientation.0754

We make that choice when we are doing these problems. We always want to pick the vector that is pointing away from the surface. You do so just by choosing which comes first, the dp du or the dp dt.0758

Okay. SO that is it, this is what we have created. Nice.0772

Let us go ahead and let us see, so let me just make sure we have got everything, dp du, dp dt. Let us go ahead and label this point, this is the point p of t, u, okay.0776

Now, this vector right here, we actually call it the vector n, for normal, capital N, and if you want to put a vector sign on top of it you can. I am going to do both. Sometimes I am going to have it on top, sometimes I am going to have the vector arrow now on top of a capital letter. 0790

Generally, when I do lower case letters and it is a vector, I will always put the vector mark on top, but normally for capital letters, I will not have the vector notation unless there is a point of confusion.0810

So, let us go ahead and define N. N is the normal vector, that is also a function of t and u, it is dp du cross dp dt.0824

You have a parameterization p, you take the partial with respect to t, you take the partial with respect to u, you are going to get 2 vectors, you take the cross product of that vector and what that gives you is the normal vector n. So this is N.0843

The normal vector to the surface and again let me write, making sure we choose the orientation such that N points away from the surface.0859

Now, again, for the most part this is not going to be too much of a problem.0891

When you are doing these problems, when you are dealing with a particular surface you are going to be dealing mostly with familiar surfaces and you are going to be able to sort of look at, once you come up with N, you are going to be able to see which direction the vector is pointing in.0896

If it is pointing away from the surface, or if it is pointing into the surface, and you can adjust accordingly.0910

By adjusting accordingly, all you are really doing is taking a negative sign, so dp du cross dp dt is just negative dp dt cross dp du.0917

When you are actually doing these problems, the only thing that changes is the sign. If you end up with a negative and you want the positive, you just change the sign.0926

What that means is you have taken the opposite orientation. The magnitudes and numbers do not actually change. There is nothing that you... ultimately there is nothing that you really have to do.0935

You are going to end up with the same number. It is just the sign might be different.0944

That is not really a problem for most applications. Okay. So, hopefully this is clear.0949

Now, we also define one other thing. Once we actually find N, so N is not in general a unit vector. It is not a vector of length 1, we can make it a unit vector by dividing by its norm, so we also define that thing.0958

So, let me go ahead and go to the next page and define that. So, we also define small n, that is just n divided by its norm.0972

Well, n is dp du cross dp dt, and it is divided by the norm of the vector which is dp du cross dp dt, because dp du cross dp dt is a vector.0990

This is the unit vector -- excuse me -- in the direction of n.1012

That is it. So, be very, very careful. Now that we have introduced the cross product and we have been dealing -- you know -- extensively with the dot product, you are going to get them confused. I get them confused.1030

In fact, yesterday when I was working on a problem, I ended up taking a dot product when I meant to take the cross product. In some of the things that we are going to be dealing with, like integrals of a vector field, integrals of a function over a surface, you are going to have a mix.1041

You are going to have dot products and you are going to have cross products, so just be very, very careful. There is going to be a lot of notation on a page. 1055

Just be very clear about what you are doing and -- you know -- go slowly and go carefully. When you want to take a dot product instead of a cross product, when you want to take a cross product instead of a dot product. 1063

Dot product is a number, it is a scalar. Cross product is a vector, you actually get a vector.1074

So, let us just do an example and hopefully everything will start to come together here. So, example number 1. We will take the sphere of radius r, a given radius.1080

Well we have a parameterization for a sphere of radius r, so let us go with this particular parameterization. φ and θ.1100

That is going to equal rsin(φ), cos(θ), rsin(φ) sin(θ).1108

Now remember, we have two parameters. r is not a parameter, r is a constant, we have chosen the radius that we want because we are not talking about the volume, the whole thing, we are talking just about the surface.1119

We have r -- oops, make it a little bit clearer -- this is, I know we know what it is but -- we have rcos(φ), and of course φ is going to run from 0 to pi, and θ is going to run from 0 to 2pi.1130

That is our parameterization. Let us go ahead and construct our particular tangent plane and our normal vector.1148

Let us take dp dφ, so if I take dp dφ, well, I just -- you know -- take the derivative of this with respect to φ derivative of this with respect to φ, derivative of that with respect to φ, I should end up with the following.1158

Again I hope that you are going to confirm this for me because I do tend to make a lot of little errors.1175

So, rcos(φ), cos(θ), and then we have rcos(φ)sin(θ), again we are doing it with respect to φ so the θ stays the same, and this is going to be -rsin(φ).1182

There you go, that is dp dφ, well now let us go ahead and do dp dθ, so again when we differentiate this, this is 3-coordinate functions. We have a vector.1203

Now dp dθ, this is going to be -rcos -- oops, not cos(φ), see again, there you go, this is the original right here... let me actually circle that one... this is our parameterization, so you have a lot of things floating around so let us make sure -- so rsin(φ)cos(θ).1216

This is going to be -rsin(φ)sin(θ), and if I take the derivative of this, this is going to be rsin(φ)cos(θ), and there is no θ here so the derivative with respect to that is just plain old 0.1237

Okay. So now I have the two partials of the parameterization and I am going to go ahead and take the cross product of these two partials... dp dφ, dp dθ.1263

Again, it is just a lot of notational tedium, not a problem. Of course you are dealing with something that is reasonably complex. It is a surface in 3-space, there is a lot going on.1273

So dp dφ cross dp dθ, and again we are going to go ahead and use our symbolic representation to solve this. We are going to have i here, and I will put j here, and I will put k here, and now I have got this vector... the dp dφ.1285

So this is going to be rcos(φ)cos(θ)... this is rcos(φ)sin(θ), this is -rsin(φ), and here we have -rsin(φ)sin(θ).1308

This is rsin(φ)cos(θ), and this is 0, thank god. I have one thing that is going well here.1331

So, we are going to end up taking the determinant of that. Oops, got our crazy little lines here towards the bottom of the page again... hopefully the will not interfere with what it is that we are doing.1341

That is that, and we are going to expand it of course along the first row, so when I do that I am going to... well hopefully you are ok with determinant expansions, so let me write down what it is that you actually get here, and I am going to write it on the next page.1350

So, just to show you, the first one is going to be this × that - this × that.1367

The second is going to be this × that - this × that, with a negative sign in front because this is +, -, +.1374

The third one is going to be the third component function... is going to be this × that - this × that... okay?1382

Let me write it all out so that we actually see what we are dealing with here.1390

We are going to have r2sin2φcos(θ), and this is going to be the first coordinate function i + r2sin(φ)sin(θ), when you actually work out all of the negatives and positive signs, all of the negative and positive coefficients, this is what you end up getting, + r2cos(φ)sin(φ)cos2(θ) + r2sin(φ)cos(φ)sin2(θ) × k. 1396

Well sin2(θ) and cos2(θ)... uhh... sin2 + cos2 = 1, so our final, and I am going to write this in normal notation without the i, j, k, okay? 1453

You have the following vector. You have r2sin2(φ)cos(θ), you have r2sin2(φ)sin(θ), and you have r2sin(φ)cos(φ).1468

That is our normal vector. So this is n. This equals n. This is n, the normal vector to the surface of the sphere.1497

Okay. Now let us go ahead and find the norm of n. Now we are going to end up taking the norm of this vector, now let me go to blue, so now that we have n, we are going to take the norm of this.1522

Well, the norm of the vector is this squared + this squared + this squared, the 3 components under the square root sign.1536

So, the norm of n is going to equal r4sin4cos2(θ)... actually you know what, I do not know if I want to go through this whole thing... I would much rather just, ahh it is okay, I will go ahead and go through it. It is not a problem, it is nice to sort of see everything.1543

Okay. r4sin4sin2(θ) + r4sin2(φ)cos2(φ), this is going to be sin4(φ), I have to put an angle in there... this is also going to be sin4(φ). Okay... and all of this is of course under the radical.1574

When we simplify some things we end up with r4sin4(φ) + r4sin2(φ)cos2(φ), and all of this is under the radical.1605

I can pull out an r4 and I can pull out a sin2, so this is going to be r4sin2(φ) × sin2(φ) + cos2(φ), which is equal to 1.1623

So, this is equal to r4sin2(φ), and the solution to this is r2, you can just take the square root. The absolute of sin(φ). 1645

When you take the square root of something that is squared underneath, it is the absolute... but since φ is > or = 0, and < or = pi, the sin(φ) is always going to be positive, so the absolute of sin(φ) is just plain old sin(φ)... we can get rid of the absolute value signs.1657

So, the norm of this vector is equal to r2sin(φ)... there we go. We have everything that we need to deal with the normal vector if we want to take this vector and divide it by this thing, we end up getting the unit vector in the direction of n, and that is it.1684

Given a parameterization, take dp dφ, take dp dθ, take the cross product, and then take the norm if you need the norm. You do not always need the norm, but we thought that we would go through the process. That is all that is going on here.1705

So let us go ahead and take a look at what this actually looks like. Let me draw out a circle.1718

Now, I am going to go ahead and give you a... I am going to pick a curve here, it is going to be one curve, and I am going to pick another one... so another curve.1732

So, where they meet, this is going to be some p(t) -- oh, p(φ,θ). This is going to be some point on the sphere... p(φ,θ).1744

Okay. Now, let me go ahead. So if I have... so this curve right here, if I pick a θ... if I hold θ fixed and I vary φ, I am going to get this curve right here. That is this one.1759

So, hold θ fixed and vary φ and of course this other curve is then the opposite.1778

If I hold φ fixed and vary θ, I get this curve. Well, if this is the one where φ is varying, I can get... and if I take that... this is dp that is dp dφ.1793

Well if this is the one where θ is varying, I get that vector. This is dp dθ.1817

When I take the cross product of those two, I end up with a vector that is perpendicular to both. It is perpendicular to this, and it is perpendicular to that.1825

This is my vector n, this is the cross product of dp dφ, dp dθ, this way and it points out of the sphere.1838

There is the center of the sphere. That is it. That is all that I have done. Very nice.1850

Okay, let us do another example. Just to sharpen up our skills here, so example 2 we will let the parameterization of t and θ be equal to tcos(θ), tsin(θ) and t2. This happens to be the parameterization for a paraboloid.1857

Of course t is going to be > or = 0, and θ is going to run greater than 0 and less than 2pi.1885

So, let us go ahead and calculate things. We do not need to -- actually, we do not even need to worry about what the surface is.1895

It is nice to know what it is, but really we are just working algebraically, so let us just go ahead and do the calculus, do the math.1902

So, if we take dp dθ), let us do θ first. It does not matter what you take first, you are going to end up with -tsin(θ), you are going to end up with tcos(θ), and you are going to end up with 0.1910

Let us go ahead and take dp dt, that is going to equal cos(θ)sin(θ) and 2t.1934

Now, when I go ahead and I take dp dθ cross dp dt, this vector cross that vector, I am going to end up with my normal vector.1948

Well, that is going to equal i,j,k, my symbolic representation - tsin(θ), this is going to be tcos(θ), this is going to be cos(θ), this is going to be sin(θ), and this is going to be 2t. I take the symbolic determinant of that expanded along the first row.1967

So, now I am going to move to the next page. When I do that I will go ahead and let you confirm what it is that I end up getting. I end up with the following.1994

I end up with 2t2cos(θ)i + 2t2sin(θ)j + t × k, and expressed in regular list notation, without the i,j,k, I end up with 2t2cos(θ), 2t2sin(θ, and t. This is my n.2004

That is my normal vector. Now, if I want to take the norm, well, not a problem. square of this, square of this, square of this, square of this... all under the square root sign.2045

You end up with 4t4cos2(θ) + 4t4sin2(θ), and we like it when we sin2 and cos2 because that is going to equal 1 + t2, all under the radical, and the cos2 and the sin2, I factor out the 4t4.2060

I end up with 4t4 + t2 all under the radical. This gives me the norm. The length of that vector at a particular value of t.2082

My normal vector is this one. Okay? It is a vector.2097

This is the norm of that vector. It is the magnitude of that vector. It is the length of that vector.2104

Now, let us see what this looks like. I am going to draw a paraboloid here, so we have something like this and of course this is a 3-dimensional paraboloid, but I will not put the axes on there.2111

Well, there is going to be some particular curve that goes this way as you hold one of them... if you fix θ and vary t, you are going to get this thing.2125

If you fix t and vary θ, you are going to get that thing.2135

Well, this point right here, this is a particular p(tθ). You are going to get a tangent vector that way, this is going to be dp dt.2140

You are going to get a tangent vector that way. That is the dp dθ, and when I take dp dθ, cross dp dt, I end up with my normal vector to the surface.2157

This is a vector as the surface changes, this normal vector is always going to be perpendicular to it... no matter which, that is the whole idea.2174

So, it is perpendicular to that one and it is perpendicular to that.2185

This happens to be our n, which again is dp dθ cross dp dt.2190

If you end up doing dp dt cross dp dθ, what you end up with is the vector going into the paraboloid.2199

It is still normal to the surface, but it is just pointing in. That is it. This way instead of this way.2206

It is just a question of orientation, positive, negative, that is all it is, the only thing that changes is the sign. 2213

Okay. So, that is tangent plane and normal vector to a surface.2219

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2223

Hello and welcome back to educator.com and multi-variable calculus.0000

So, today's lesson we are going to be talking about surface area. How to find the area of the surface given a particular parameterization.0004

From there we are going to jump off into subsequent lessons. We are going to be talking about integrals of functions and vector fields over given surfaces, but we want to start with the notion of surface areas.0013

So, let us just go ahead and jump in. Okay. So, in the last few lessons, we have introduced some concepts, so let us go ahead and recall real quickly.0026

We introduced this concept of a cross product, so given a vector a and a vector b, we can form a cross b and we used it to construct the normal vector to a surface.0035

Well, we also said that the magnitude of a cross b, well, was just the norm of a cross b.0052

I do not need to write that. The magnitude is equal to the norm of a cross b, so it's length, basically.0072

We also introduced another formula... it's the norm(a) × norm(b) × sin(θ), the sine of the angle between them.0081

I am not sure if we will have a lot of opportunity to use this particular formula. It may come up, it may not, depending on a particular example that we are doing, but mostly we are concerned with constructing the normal vector and then finding its norm.0094

That is what is important. Then, once we have the vector, finding a norm is really, really easy.0108

Okay. Now, we also said that a cross b, that this magnitude is the area... and we said it real quickly, but it will be really, really important here -- but we also said that this magnitude is the area of the parallelogram that is spanned by the vectors a and b.0112

So, if we are given a vector, given a couple of vectors like this, let us just say this is a and this is b, well, this... when we take those vectors and we close it out and form the parallelogram, well when we take a cross b, we end up with this vector which is perpendicular to both, right?0160

That is the a cross b, well this parallelogram that is actually spanned by a and b, the area of that parallelogram happens to be the norm of this vector. That is the thing.0181

So, there is an association between the vector itself and the area. So, there is some number we can attach to it and that number is the norm of the cross product.0195

Okay. So, now we can go ahead and jump into our notion of surface area.0207

So, now, given a surface in 3-space and the notions we are about to develop are complete analogs of the lower dimensional cases.0213

We work with curves, we define line integrals, we define the length of a curve, now we are dealing with a surface, we are going to define a surface area which is analogous to the length of a curve.0228

It is going to be the exact same thing. You are going to see that the integral is going to look the same, and that is what is nice.0239

So, if you ever lose your way, just sort of... if you know the lower dimensional case and feel comfortable with that, you can sort of extract the higher dimensional cases.0245

Okay. Given a surface in 3-space and a parameterization, which is the important part... and a parameterization... let us call that p, we found dp dt cross dp du.0254

So, now let me draw a surface here. This one... it is very, very important that we draw this one properly, so that is the surface, and we have a point -- I am going to draw one vector like that, and I am going to draw another vector like that.0287

Then, I am going to go that way, and I am going to -- oops, close this one out, let me just color this in here... so we have -- okay, so this point is our p(t,u) for some value of t and u.0312

This vector right here, let us call it dp du, and let us call this vector right here... that is the dp dt, okay?0349

Well dp dt and dp du, they span this particular parallelogram, okay?0365

So, dp dt and dp du span a parallelogram -- all of a sudden I do not know how to spell parallel -- parallelogram, whose area, we said, was the norm of that cross product. The area is... the norm of dp dt cross dp du, right? So this parallelogram has the area of a cross product.0373

Now, notice that this parallelogram right here, the reason I have drawn this little thing on the surface, what you have is this parallelogram which is the plane, so this parallelogram is part of the plane that is tangent to the surface at this point.0413

This colored is the shadow of this parallelogram if you happen to be shining a light from on top. So, we wanted to give the sense that there is this shadow, which is also in the shape of a rectangle on the surface.0428

Well, the idea is if you take these parallelograms small enough, what you end up actually getting is... you are adding up all of the surface area elements on the surface itself. That is essentially what we are doing.0444

I just wanted you to see this. This is just the projection, the shadow of the parallelogram on the surface.0458

Okay. So now we are ready to define our notion of surface area. 0466

So, define surface area is equal to, we are going to use this symbol dΣ, that is the symbolic representation of that, and the double integral itself is the double integral over the entire surface of dp dt cross dp du with respect to t and with respect to u. This right here.0475

So, what we are doing is when we form dp dt and dp du, when we take their cross product we get the vector. When we take the magnitude of that vector, we get the area of the parallelogram.0515

Well, what we do is we take the area of the parallelogram and we basically add up all of the areas of the parallelogram, the differential elements and we end up getting the surface area.0529

This is what is important here. I am going to talk about this a little bit more, but we just want to be able to take a look at that.0539

So, the surface area given a parameterization, it is going to be the double integral of the normal vector, or the norm of dp dt cross dp du, integrated with respect to t and with respect to u. That is it. That is the definition of surface area.0547

Okay. Now, let us discuss this a little bit.0567

So, symbolically, ds, when we use the symbol ds, that is equal to dp dt cross dp du dt du.0573

This is a differential area element. Okay. 0590

When we integrate, first with respect to t, then with respect to u, we have integrated over the entire surface, right? That is what we are doing.0608

When we do an integral, we can only go one variable at a time. We go along t first, then we go along u, and we have covered the entire surface. We have integrated over the entire surface.0635

Okay. Now, recall -- oops, let me do this one in blue -- recall the following. We said that curve length, we define curve length as the integral from a to b of the norm of the tangent vector, with respect to t.0652

We take the tangent vector and then we integrate it along the curve. Well, this is a complete analog. Except now, we are not dealing with a curve, we are dealing with a surface.0681

So, we are going to integrate along one curve, and then we are going to take that curve and we are going to integrate along the other parameter, and we are going to cover the entire surface. That is what we are doing.0694

This is the analog for a two-dimensional object, which a surface is, a two-dimensional object.0705

Okay. So, this area integral, I hope I am not belaboring too many points, I just want to make sure that there is a good intuitive sense of what is going on.0720

I mean, it is true ultimately that we are going to give you the definitions and we are going to lay these integrals out and then just sort of see if we can follow... take a look at what you need in the integrand, then fill in the blanks and integrate, but it is important to know where these integrands come from. Why it is that they take the form that they do.0732

So, we want to help develop a sense of intuition. We are still dealing with 2 and 3-space, so we still have the geometric intuition, we might as well use it.0751

Okay, so the area integral is a double integral. What we are doing, well not essentially, what we are doing explicitly, what we are doing is finding the length of p(t,u) holding one variable constant, then integrating this curve along the other variable -- I am going to draw this out in just a minute -- until we have covered the surface.0761

What we are doing, essentially, is finding the length of a curve and then we are going to integrate that curve, that whole curve. We are going to sweep it out along the other variable until we have covered the surface. What we have is something like this.0845

So if this is our surface, what we do when we take the first integral is we find essentially that curve length, and then we integrate in this direction, the other parameter.0858

What we do is we swing this, we sweep this curve out until we have covered the entire surface. So we move in that direction.0870

That is all that we are doing. It is like two curve integrals, one after the other. That is it.0878

Okay. So, let us go ahead and do an example, and again, when you are doing your examples, when you are doing your problems for your homework or whatever it happens to be, it is always best to write out what the integral is... the definition just to get use to writing it out symbolically.0885

Very, very important. So, let us do example 1.0901

Find the area of a sphere of radius r, find the surface area of a sphere of radius r. Well, you already know what that is from geometry, but the idea is where did that formula come from.0911

You know, when you learned it we just sort of dropped it on you, now we are going to show you where the formula actually comes from.0931

Okay. So, we have the parameterization, we have p(φ) and θ, so let us use the standard parameterization, and we know that this equals rsin(φ)cos(θ), rsin(φ)sin(θ), oops, that is sin(θ), and then we have rcos(φ), well we already found what the norm was.0939

We already, in the previous lesson, we did this, we already found the norm of dp dφ cross dp dθ.0969

That was just the norm of the normal vector, and that was r2 × sin(φ).0984

So, our ds element, dΣ, that is equal to this thing dφdθ = r2sin(φ)dφdθ.0994

Well, we know that φ, for a complete sphere is going to be > or = 0 and < or = pi, which is 180 degrees.1018

θ is going to run all the way from 0 to 2pi, which is 360 degrees, so, and like I just said a minute ago, let me go ahead and write the integral symbolically so we get used to writing it.1027

So, we are taking the integral over s of the differential area element. We are adding a differential area element and we are going to integrate, we are going to add all of the differential area elements over the surface.1042

Again, that is what integration is. Do not forget that. It is very important to remember that all you are doing is you are adding something. It is an infinite sum. The idea of integration is a particular technique, but it is just addition.1056

As it turns out in mathematics, all you can do to something is add things. That is it.1067

So, that is equal to the integral over s of -- alright -- dp dφ cross dp dθ, dφ dθ1073

Now we are actually going to write out the integral itself, and that is going to equal the integral from 0 to 2pi, so let us go ahead and do θ last.1090

It is going to be from 0 to pi, so this is going to be φ, and then we have this thing.1100

This thing is that, or ds is this whole thing.1107

So, we have r2, sin(φ), and we are going to be integrating with respect to φ first, then with respect to θ next, and when you go ahead and solve this integral... what you get is 4pi r2, which is exactly the formula you learned in geometry.1115

This is where it comes from. It is based on the definition of surface area given a particular parameterization which is given by this. That is the definition of surface area. That is it.1137

Okay. Let us do another example here... excuse me.1151

Alright, so, example number 2. We want to find the area of the paraboloid z = x2 + y2, so now it is given... this paraboloid is not given to us in parameterized form.1159

It is given to us in just a straight Cartesian fashion... z = some function of x and y... for z > or = 0, and < or = 5.1188

So, we are trying to find the area of... so we have this paraboloid and z is > or = 0, so obviously it is going to be above the xy plane, and up to a length... up to a height of 5.1211

So, we want the surface area of that paraboloid.1230

Real quickly before I get into this problem, I want to talk about the notion of length, area, and volume.1234

Length is -- you know what length is -- and there is a physical way of actually measuring length.1243

If I said what is the length of some curve, well what you can do is you can take a string, run it along a curve, make sure that it is there, and then pull the string taught and you can measure the length of that string.1250

There is a physical representation, a physical manifestation of length.1263

As far as volume is concerned, if you want me to measure the volume of some strange looking object, what you do is you have a cup of water, you measure the height... you measure how much... you read off the graduation and then what you do is you insert this object into the water until it is just -- you know -- under the surface, and of course the water level is going to rise.1267

Well, the difference in water level gives you the volume of the object. Unfortunately for area, for surface area, there is no physical manifestation, so in some sense, area is a very, very odd thing.1289

Length we can deal with physically, volume we can deal with physically, area we cannot really deal with physically.1302

Area is strictly a mathematical notion. I just wanted you to be aware of that. It is very, very curious that that is the case.1308

Okay, so let us go ahead and finish this problem. So we are given our equation that way, so let us go ahead and do our parameterization.1317

So, our parameterization, when we are given z = f(x,y), well that is equal to x and y and the function itself x2 + y2, so this is our parameterization.1326

Coordinate function 1, number 2, and number 3. Okay.1339

Let us find dp dx, so this is going to be a vector. This is going to be (1,0,2x), I am just differentiating this.1344

let us go ahead and find dp dy, so dp dy, that is going to equal (0,1,2y), excuse me.1355

Now we need to find the cross product of these two vectors, so dp dx cross dp dy, and we are going to go ahead and use our i,j,k, symbolism. We have (1,0,2x), and we have (0,1,2y).1366

When you expand this determinant you are going to end up with the following. You are going to end up with -2x,-2y, and 1.1387

In this particular case, I ended up choosing the opposite orientation. In this particular case, I ended up choosing such that this normal vector -- this is the normal vector -- is actually pointing into the paraboloid as opposed to out of it. 1396

It is not really going to be a problem because when we take the norm of this, the norm is the same whether it is pointing in or pointing out, but it just so happens that I noticed that I have the incorrect orientation.1412

Later, that might be a problem, but for right now, it is not... and we will deal with it when we get to it.1424

Okay. So, this is our vector. Now we need the norm of this vector. So, the norm of dp dx cross dp dy = norm of this.1430

Well, it equal 4x2 + 4y2 + 1 under the radical.1448

So, our differential area element that we are going to integrate, ds is equal to the radical 4x2 + 4y2 + 1 dy dx.1457

That is what we are going to integrate. Now, for... now for z = x2 + y2, we said that we wanted z < or = 5, so x2 + y2 < or = 5.1474

Well, if that is 5, then the disc that we are looking at, this length right here is actually going to end up being sqrt(5), and this length right here is going to be sqrt(5).1499

So, what we are going to be doing is we are going to be integrating... this is x and this is y, the z axis is going like this.1514

We are integrating over a region of the xy plane, right? dy dx, so we need to know what x goes from and we need to know what y goes from.1522

In this particular case, z = x2 + y2 < 5.1534

Well, that means this circle is in the xy plane, that is this way, and of course the paraboloid is going up this way.1543

Well, if z itself is 5, well then this radius, our circle underneath has a radical of sqrt(5), simply based on the parameterization... x2 + y2.1552

I hope that that makes sense. So, in this particular case, our x is going to run from - sqrt(5) all the way to +sqrt(5).1567

It is going to go from this point to this point... excuse me... and our y value is going to go from... our y, the height is going to change... so y is going to run from -5 - x2 all the way to sqrt(5) - x2.1583

That comes from if I move this x2 over here, and I take this square root, I get y explicitly in terms of x, because remember, again, I am integrating with respect to y and I am integrating with respect to x, so I have to know what x does and what y does.1606

So, my area integral ends up being as follows: area = the integral from -sqrt(5) to sqrt(5) -- excuse me -- the integral - 5 - x2 under the radical, to sqrt(5) - x2 under the radical.1627

Of course I have my function, the norm of the cross product, which was 4x2 + 4y2 + 1 under the radical, dy dx.1653

Because this is y and this is x, so I am integrating that first.1667

When I put this into my mathematical software I get the number 49.864. There we go. That is the surface area of the paraboloid, that has a height of 5.1672

Okay. Now we did this in Cartesian coordinates. Now let us go ahead and run this integral in polar coordinates, just to see what it looks like. Just for the sake of some practice.1688

Okay, now, in polar coordinates -- forgive me, I seem to have something in my throat here -- in polar coordinates, it looks like this.1698

Again, we have a circle and we have this paraboloid on top and of course it is on top of this circle.1724

The radius is sqrt(5), so r is going to run from 0 to sqrt(5), because now we are going to be integrating r out this way, and then we are going to sweep this line around... all the way around the circle... to pi.1741

So θ is going to run from 0 to 2pi, so these are our lower and upper limits of integration. All I am doing is I am taking this region and I am expressing it in terms of r and θ. I am converting to polar coordinates.1765

When I convert to polar coordinates, I have to make the appropriate changes to the integral. Now, this is my -- let me go to red here -- this is my integrand. 1780

I have to express this integrand in terms of r and θ.1794

Well, the conversion, the transformation is x = rcos(θ) for polar coordinates and y = rsin(θ), and when you put rcos(θ), rsin(θ) into this, you end up with the following.1798

So, 4x2 + 4y2 + 1 in polar form is.. .when you put these into here and solve, what you get is sqrt(4)r2 + 1.1813

That is our integrand. Okay. Now, dy dx... remember when we convert to polar coordinates, the differential area element is equal to r, dr, dθ, there is this extra factor of r.1836

Now let us go ahead and put everything together. Our area in terms of polar coordinates.1856

I am going to go ahead and do r last. Go from 0 to 2pi, this thing is that thing, it is 4r2 + 1, and now dy dx is equal to r dr dθ.1863

So I write r, dr, dθ. When I go ahead and put this into my mathematical software, I get the number 49.866. That is exactly write. This is .864, this is .866.1883

These had to be done numerically, there was no simply symbolic way of doing it to come up with nice closed form expression.1897

That is nice. The software will do that for us. That is it.1904

All I have done is, I was given a particular surface, I did the Cartesian xy, I found the integral this way, I wanted you to see what it looked like in polar coordinates, it is not necessarily easier, it is just a question of what is easier for you.1910

Or perhaps... and again, it is going to depend on the problem. 1928

We cannot say do it this way or do it this way, but when you do make a conversion from a particular coordinate system, and you solve the integral in another coordinate system, you have to remember that the differential area element, or the differential volume element if you are dealing with triple integrals... it also undergoes a transformation.1931

That is it. For every object that you have in one integral, you are going to have a corresponding object in the other integral. Just keep things straight.1951

Okay. Well, thank you for joining us here at educator.com to discuss surface area, we will see you next time for a discussion of surface integrals. Take care, bye-bye.1960

Hello and welcome back to educator.com and multivariable calculus. 0000

Today's topic is going to be surface integrals. Last lesson we introduced the notion of surface area, and now we are just going to add onto that.0004

However, before we discuss surface integrals, I just wanted to say one final word regarding surface area.0012

Especially with respect to the last problem that we did. Let me go ahead and talk about that for just a minute or two and then we will jump into surface integrals.0018

Okay, so, we had this... so the last example in the previous lesson... we had this function and it was given to us as z = some function of x and y. It was given explicitly as a function of two variables. This surface in 3-space.0027

Now, if you have that, if you are given z = f(x,y), of course you know that the parameterization that... for that... you can form the parameterization of x,y and that is going to be x, y, and of course the function of x, y.0045

That is the basic parameterization when you are given this thing explicitly. Well, if you were to just take this parameterization and run a surface area problem, run through it generically, without actually having a specific function in mind.0078

If you just took the partial with respect to x, the partial with respect to y. If you took the cross product, you are going to end up with this.0095

You are going to end up with an area equal to sqrt(1+df/dx2 + df/dy2) dy dx.0103

If you actually run this problem generically with this parameterization, this is the surface area integral that you get. This is the integral that you actually learn in single variable calculus, and it is also going to be one of the formulas that shows up in your book for multi variable calculus.0130

This is a very specific formula for a specific way that a function is given to you. If your function is given to you explicitly as a function of x and y, you can go ahead and use this formula directly.0149

However, I think it is better not to have to learn specific formulas for specific situations. What is the best is the learn it the way that we did learn it, the way that we defined it. Given a particular parameterization, you want to be able to form that integral.0160

Therefore, you will always end up with this if in the particular case that you are dealing with, you are given a function explicitly as opposed to given a parameterization.0173

So, it is better to have the general integral for surface area, that way you will always be right. You will always end up where you need to end up... surface area for any parameterization, and that is what is important... parameterization, because it is the parameterization that actually controls how the integral looks.0185

Oh, again, the area, as a recap, is equal to the double integral over s, and we use Σ for the area -- that is the symbol -- and the actual mathematics is the norm of dp dt cross dp du dt du. Where this ds is this thing.0223

So, this is the symbolic representation of it. This is the actual mathematics, that says take dp dt, the parameterization, take dp du, take the cross product of those things and then you are going to get a vector... take the norm of that vector which is a number and then integrate that number over the surface and that will give you the surface area. That is it. Nothing more.0257

You will always... and then for a particular situation... if you do this, you will end up with this when you are given that and when you have a parameterization like that.0277

Okay. So, let us go ahead and start our surface integrals. Let us start on the next page. So, let us see what we have.0287

We previously defined... well, I do not need to write all of this out... we previously defined two types of line integrals.0297

We did an integral -- excuse me -- of a function over a curve, so we did the integral of a function over a curve and we also did the integral of the vector field over a curve... a path, a line... something like that.0309

So, now we are going to do the same for surfaces. So, now we are going to just move one up. A line is a one dimensional object, a surface is just a two dimensional object. We are just moving up one dimension.0348

Now, we just do the same for surfaces. We are going to define a... the integral of a function over a given surface... and we are going to define the integral of a vector field over a given surface and just see that they are perfectly analogous in terms of actual notation.0362

Let s be a surface parameterized by some parameterization p, t, and u, and of course t and u are generic variables. 0392

Now, let f(x,y,z) be a function defined on s. So, real quickly, in case you need to see what a picture looks like, so you have got some surface in 3-space, and it is going to be parameterized by some parameterization, and since it is in 3-space, the surface itself, the points on that surface are in 3-space, therefore we can form if we have some function of three variables, x,y,z, it can take that function on that point, on that surface.0413

That is all that is happening. Exactly, exactly the same for line integrals. Okay. So, now let us go ahead and do our definition. Then, the integral, so this one is the integral of a function defined over a surface, f(x,y,z), it is a function of 3 variables.0458

It takes a vector, 3 variables, and it spits out a number.0480

The integral of f, small f, over s is, well, symbolically we have f dΣ, and mathematically it is equal to f(p(t,u)).0484

In other words we have formed the composite function of f of the parameterization p, just like in a line integral we formed f(c(t)).0509

× dp dt cross dp du, dt du. Notice this is exactly the same as the surface area except you know you have this extra component f. That is the symbolic -- and really what you are doing -- is you are taking f(p), f of the parameterization.0524

The composite function of the function of the parameterization and you are multiplying it by the norm of the cross product of the normal vector.0546

Then, this is a number, and of course you are integrating this number over 1 variable, over the other variable, in other words over the surface. That is all that is happening here.0555

When f = 1, when f is identically 1, well, we simply recover the integral for surface area integral. That is it0568

We defined surface area, now we are going to stick a function in there. After this, we are going to stick a vector field in there. Let us do an example.0589

So, example 1. Well, let us let f(x,y,z) = xsin(z). Okay? and s be the surface z = x3 + y, and we are going to specifically say x > or = -1, and less than or = 3, and y is > or = 0, and < or = 3.0597

We are going to specify the x and y, we are going to specify the domain. This is our function, it is given explicitly in terms of a function of x and y, so we know how to form the parameterization when we need to in a minute.0643

We have the function of 3 variables, the y does not show up here but it still is a function of 3 variables. It is a surface, y is arbitrary, so now let us go ahead and solve for the integral of this function over this surface.0657

Okay, we are going to use this thing right here.0670

So, let us go ahead and I think I am going to actually write down one more time up here in the corner... that is going to be the integral of f(p), so I am going to write it this way, f(p) × norm(n), and of course n is the cross product of the partial of p with respect to t, that big long cross product thing.0675

So dt du, just so I have it as a reference. Okay.0707

Well, so let us go ahead and list our parameterization. These are actually pretty easy in the sense that you just have to run through the process.0713

Again, running through the process is easy enough to do, it is just a question of keeping track of everything that is going on.0721

There are going to be a lot of symbols, a lot of numbers floating around, just be careful with what you do.0730

So, our parameterization, it is going to be in terms of x and y, so we have x, we have y, and then we have z, which is the function x3 + y. So that is that.0734

Now, let us go ahead and calculate dp dx, so dp dx is going to be (1,0,3x2) and dp dy is going to be (0,1,1)... there we go.0749

Now we need to form the cross product of these two, so dp dx cross dp dy, and we are going to use our symbolic determinant i,j,k... (1,0,3x2) -- excuse me -- (0,1,1). We are going to form the determinant of that. 0780

When I actually expand -- I am not going to go through the expansion, but I hope you are going to confirm for me, because I could very well have made some arithmetic errors -- 3x2,-1, and 1. This is n, right?0804

This is the normal vector the surface at a given point. That is what the cross product of the two partials is. That is n.0823

Now, what we want to do. We need the norm(n), so let us go ahead and take the norm of this, so the norm(n), well, it is just going to equal this squared + this squared + this squared all under the radical.0830

So it is going to be 9x4 + 1 + 1 under the radical, which equals 9x4 + 2 under the radical, okay?0846

Let me go ahead and put a little circle around that... we are going to need that, that is this one.0857

Okay. Now, we also need f(t), okay, well f is -- let me go back to black -- so f, that was equal to x × sin(z), well the parameterization p of x,y, we said was xy and the function x3 + y.0863

So f(p), f of the parameterization... well f is just x, z is x3 + y, so it is x × sin(x3 + y). There you go.0893

Now we have this... and now let us go ahead and put it all together. So, the integral is equal to, well, let us see, let us go ahead and put... so we are going to do dy dx, so let us go ahead and do x as the outer integral, and we said that x is going to be > -1, < or = 3, so it is going to be from -1 to 3.0909

The y value is going to go from 0 to 3, right? We said y was going to be > 0 and < 3.0943

f(p), well we have x × sin(x3 + y), that is that, and then if we multiply that by the norm, which is sqrt(9x4 + 2), and then of course we decided to do y here, then x, so it is going to be dy dx. There you go. You are just literally plugging it in.0955

Find this, you find this, you put them next to each other, you put it into your math software, or if you want to do it by hand that is fine.0983

You do not want to do this by hand. When you solve this integral, you get the number 3.8527, to five decimal places, or to 4 decimal places... 5 digits. There you go. That is it.0990

The important thing here, as always, is being able to construct the integral. Not being able to solve the integral. You need to be able to construct the integral. Run through it piece by piece.1005

You have the parameterization, find the normal vector, find the norm of the normal vector, and then form f(p), and then just put it into the integral and solve the integral, and be able to extract the upper and lower limits of integration.1014

The surface over which you are actually integrating. The domain of the parameterization. The t and the u. In this case the x and the y. Okay.1032

So, now that we have seen the definition of a function, the integral of a function over a surface, let us go ahead and do the integral of a vector field over a surface.1041

For this one, let us see, we have the integral of a vector field over a surface.1052

So, let me go ahead and draw a surface real quickly. So, we have something like this, like that, like that, like that. So this is some surface and of course there is some point p here, and there is some -- you know -- again, this is a vector field, now you have a surface in 3-space.1072

Now the vector field is a 3-vector field. In other words, for every point, this particular vector field is going to be a function from R3 to R3.1090

In other words, for every point, in 3-space, there is some vector that has 3 components emanating from that point. That is what this is. This is f right here.1103

Well, a surface is in 3-space, so all of the points on that surface are subject to the vector field. You can form the composite function f of the parameterization. That is what is happening here.1115

So, we know that we can form the normal vector, right? This is the normal vector and we said that the normal vector is equal to dp dt cross dp du, and we have a function, a vector field which has composite functions f1, f2, f3.1128

Well, again, just like the vector field over a curve, you can think of what we are going to be doing is we are going to be integrating... well, let me just go ahead and write the definition and then we will tell you what this is.1158

So, the definition of the integral of a vector field over a surface is the following.1175

We have the integral over s -- f -- · ds, this is the symbol for it, here is the mathematics.1182

The integral over s... f(p), it is the composite function dotted with n dt du.1193

This right here, that is the quote component of f along n. It is the component of f in the direction of n. "Component" in quotes, because n is not usually a unit vector.1207

I can go ahead and make it a unit vector by dividing by its norm, but that is not it. This is how you want to think of it.1233

When you are putting this integral together, what you want to do is you want to take your parameterization, you want to form the composite f(p), that is going to be a vector because this is a vector field.1241

f(p) is going to be a vector, well n is also going to be a vector. You want to take the dot product of those two vectors, and that is going to be a number.1252

After you have a number, you can integrate. You can only add numbers. You cannot -- you can add vectors, but you are not really adding vectors, what you are doing is you are adding components.1257

Here, when we integrate something, a function, a vector field, whatever it is, we are adding numbers. That is why this dot product shows up.1266

We need to add a number. That is it. This is the definition of the integral of a vector field over a surface. Let me give you the expanded version.1273

s f(p) dotted with dp dt cross dp du dt du. Notice, in this case, the norm does not show up for the integral of a vector field..1285

The reason the norm does not show up is because this is a vector, therefore I need it to dot it with the vector itself, not its norm and this dot product is what gives me the number.1304

This is the symbol f · ds. This ds, that is this part right here.1314

It is different than the Σ. The dΣ, that had the norm of this. The ds, the capital s, that has the actual cross product, the vector itself, not the norm of the vector. Okay.1323

Let us go ahead and do an example. Example 2.1337

Okay. We will let f, our vector field of x, y, z... we will let it equal to xy, xz, and yz.1346

We will let s be the surface parameterized as follows: p(t) Θ = tcos(Θ), tsin(Θ), Θ.1362

Where t is going to run from 0 to 1, and Θ is going to run from 0 to 3pi.1401

Okay. What we want you to do is to find the integral of f over this surface. Great. Should be no problem at all.1416

So, let us go ahead and -- I actually should of started this on another page, it is not a problem -- I will just rewrite everything that I need to rewrite.1431

Let me go ahead and rewrite the formula itself. Always a good idea. So, I need to go -- let me do it over here -- so the integral of f(p) dotted with n dt du. Okay.1439

Well, let us go ahead and form f(p). Actually, you know what, I am going to go ahead and write f and p over again.1462

f = xy, xz, and yz... and my parameterization of t and Θ is going to be tcos(Θ), tsin(Θ), and Θ.1470

Therefore my f(p), so I am just going to put... and this is my x, this is my y, this is my z... let me go ahead and go to blue ink here.1490

x,y,z... wherever I see x I put this, wherever I see y I put this, wherever I see z I put this.1504

So, I end up with the following. I end up with t2sin(Θ)cos(Θ), Θtcos(Θ), and Θtsin(Θ).1510

I certainly hope that you are all confirming this for me. Now, I go ahead and I take dp dt.1533

Again, just running through it. I need this, I need this, I just have to go through this step to get what I need.1540

So, I take the partial of the parameterization with respect to t, and I end up with cos(Θ), sin(Θ), and 0.1549

I am going to take the partial of the parameterization with respect to Θ, and that is going to be -tsin(Θ), it is going to be tcos(Θ), if I am not mistaken, and it should be 1.1559

Okay. Now of course I have to form the cross product of these 2, because that it going to give me n.1574

So, dp dt cross dp d(Θ), that is going to be my symbolic determinant.1581

So I have got cos(Θ), sin(Θ), 0, -tsin(Θ), tcos(Θ), and 1.1592

When I form this determinant, I end up with the following: I end up with sin(Θ), cos(Θ), and t. Good.1605

Okay. So this is going to equal n. This is my n. So now I have n and I have f(p), now I am ready to go ahead and construct my integral.1620

So, my integral is equal to, and of course they gave me the range. They said t goes from 0 to 1, and the Θ goes from 0 to 3pi, so I go ahead and I have my upper and lower limits of integration.1630

Let me go ahead and do t first, that is going to go from 0 to 1, and let me go ahead and do Θ next, that is going to go from 0 to 3pi.1646

Now, I have my f(p), which is right here, and I have my n, which is right here.1659

I need to form the dot product of f(p) · n, so I have to form... I could do this × this + this × this + this × this. That is what I am doing.1669

So, my integrand is going to look like this. t2sin2(Θ)cos(Θ) + this × this + Θtcos2(Θ) + Θt2sin(Θ).1682

I did Θ first, so this is going to be dΘ, and dt is going to be last. 1712

When I solve this I end up with 9/8 pi2 + pi. That is the integral of that particular vector field over this particular surface with that particular parameterization. That is it. It is that simple. 1719

Okay. Here is where we have to give a caveat. Now, let me go ahead and actually write this out... "Be very careful and vigilant."1739

Here is why. You have both these problems, surface integrals, particular vector fields with surface integrals, you have both dot products and cross products in the same problem.1756

You have both dot products and cross products in this integral.1774

Right? The definition of the integral... it contains a cross product, which is the definition of n, the normal vector, and the integral itself contains a dot product. You are dotting f(p) with n.1788

So, just keep track of everything very carefully. I actually made the mistake... I still make the mistake all the time. I did when I was doing this problem the first time.1800

I have to make sure that if I am taking a cross product and if I have two vectors and I need their cross product, to do their cross product, do not dot product them1810

If I need a dot product, take their dot product, do not take the cross product. The dot product gives you a scalar, a number. The cross product gives you a vector. Very important to keep track of those.1818

Yes, so keep track of everything very carefully. Keep track of things carefully, that is the take home lesson here.1831

Okay. Well, let us go ahead and do another example and we will finish off with this. So, let us see. Example 3.1842

Alright. Let f(x,y,z) = let us do something slightly more complicated here. We will do sin(x), we will do sin(y2), and we will do sin(z). Maybe it just looks complicated, we will see how it actually turns out.1855

So, that is our vector field. We want to find the integral of f over the surface z = x2 + y2, where z runs from 1 all the way to 3.1880

Let us go ahead and draw a picture of this first to see what this looks like. So, we know what this surface looks like, z = x2 + y2, that is just a paraboloid.1911

So, that is just the paraboloid that is going around the z axis, for z from 1 to 3, that means... so z1 is going to be right about here, and 3 is going to be right about here.1921

We are looking at that surface right about there. We are looking at this region right here. Basically, a piece of the paraboloid. That is it.1939

Okay. Well, let us go ahead and see what we can do. So, let us go ahead and see what this actually looks like, well, let us do the parameterization first.1949

So, we are given this function explicitly, so we know our parameterization is going to be a function of x and y, so we are going to have x, we are going to have y, we are going to have the function itself... x2 + y2. That is our parameterization. 1960

Well, let us see what this looks like in the xy plane. Well, z = 1, that is going to be a circle of radius 1, and here it equals 3, for z =3, let us see if I go ahead and project this down, and I go ahead and project this down, I am looking at this region right here, as far as the parameterization domain, that is the whole idea.1977

So, here my... this radius right here is equal to 1. This radius here is equal to sqrt(3), and the reason is because if this is sqrt(3), sqrt(3)2 is what gives me the 3.2006

They said z < or = to 3, so when you project this down onto the x-y plane, the circle that forms the boundary, the shadow here when you drop this perpendicular, it is sqrt(3), not 3. z is 3. It goes from 1 to 3.2025

So, the parameterization runs from 1 to sqrt(3). Just wanted to make sure that you guys get that.2043

We have our parameterization, so dp dx, well that is going to equal (1,0,2x).2053

Well we go ahead and form dp dy, that is going to equal (0,1,2y).2065

Well, we go ahead and form n, which is of course n which is the cross product of dp dx cross dp dy. We are doing a cross product here.2074

Now the cross product involves the symbolic determinant, I am not going to go ahead and do it here, but I hope that you will actually confirm this.2084

What you are going to end up with here is (-2x,-2y,1), good.2090

Okay. Now, let us go ahead and do... so we have n, now we are going to have to form f(p), and so f(p), the composite function.2104

Well, the composite function, the parameterization is x, y, x2 +y2, so f(p) is going to equal the sin(x), the sin(y), and the sin(x2 + y2).2119

Of course we need to form the dot product now, f(p) dotted with n, so it is going to be this thing dotted with n, and we said that n was -- let me go ahead and write it again here -- n = (-2x,-2y,1).2144

Okay. So, this dotted with that, you are going to end up with -2x × sin(x), -2y × sin(y), and 1 × sin(z), which is x2 + y2, there you go. That is my integrand.2169

Let us do this in polar coordinates. The reason you want to do this in polar coordinates is instead of in terms of x and y is because we have a nice circular region in the x, y, plane. In our domain of parameterization the x y, we have a nice circular region.2205

In fact, we have a region between two circles, which is even more important that we do it in terms of polar coordinates. Right?2218

So when we look at this in the x y plane, this is y, this is x, we have this region right here, where r is going from 1 to sqrt(3), and Θ is going from 0 to 2pi, all the way around. This actually goes all the way around.2226

I just did a... this is our domain of parameterization -- the x and y.2247

Now, let us go ahead and see what we can do here. So, in polar form, well, this is our integrand wherever we see x we need to put rcos(Θ), wherever we see y we need to put rsin(Θ), right?2257

Because, that is the whole polar coordinate transformation. So, go ahead and write that down.2276

So x = rcos(Θ), y = rsin(Θ), so whenever we convert something in x, y to polar coordinates, we are going to have to... so let us go ahead and write f(p) · n2284

In polar form it looks like this, it is going to be -2rcos(Θ), sin(Θ) -- no, sorry, let me get this right, -2rcos(Θ) × sin(rcos(Θ)) - 2rsin(Θ) × sin(rsin(Θ) + sin(x2 + y2), when I put these in I am going to get r2 sin(r2).2308

Well, r runs from 1 to sqrt(3), from 1 to sqrt(3), that is the radius and Θ) is going to run from 0 to 2pi, so r is going to run from 1 to sqrt(3), and we are going to sweep this line out all the way around 2pi.2354

Therefore, you get... so, of course we have to remember that dy dx, whenever we are doing a change of coordinates R dr dΘ, right? The differential area element. That is the conversion, so we have to put that in.2379

So, here we go. The integral is equal to... let us see if we can fit it all in here... let us go ahead and do it over here.2399

The integral from 1 to sqrt(3), just going to make this a little bit bigger, the integral from 1 to sqrt(3), the integral from 0 to 2pi, of this whole thing right here.2413

Okay, you know what, I am just going to write it all in again... -2rcos(Θ)sin(rcos(Θ)) - 2rsin(Θ)sin(rsin(Θ)) + sin(r2) × r dΘ dr, because this is Θ, this is r -- oops, I definitely do not want these random lines here.2439

Up there, dΘ dr, because we are doing the Θ first and the dr last... we are going inside to out.2484

Of course when you solve this integral, when you put it into your math software, you are going to get -... = -4.6667. There you go.2491

Now, you probably noticed -- you know -- why is this a negative number? Well, the reason this is a negative number, if you caught it earlier on, when I form the dp dt cross dp du, remember we said that a surface has 2 normals.2510

It has one normal that actually sticks away from the surface, and it has another one that goes towards the surface. It just depends on how you do your cross product. I did dp dx cross dp dy, that ended up giving me this one.2524

Well, this one, if I did dp dy cross dp dx, I would have gotten this normal.2539

I would have gotten the positive one. The only difference between the two normals is that one is positive and one is negative.2547

Ultimately, it does not really matter. I mean, yes, the integral here is... you are going to get the same number, it is just going to be a positive 4.6667 instead of a negative 4.6667.2555

If you want to keep track of that and at the end just sort of drop this negative sign, if you can sort of realize what is going on, this is... I would not say it is a minor point.2567

I mean definitely the way that this integral is defined, the way the whole idea of a surface integral is defined or surface area is defined is with that normal vector pointing away from the surface... away... in other words, the surface curves away from that normal. 2577

Because the idea is you want to be able to form a tangent plane here, right? A tangent plane this way, not that way.2594

So... but again, the difference is really just one of a negative sign, so if you can keep track of that, that is what I mean when I say it is a minor point. It is not a minor point, we still want this one instead of this one.2601

Time and experience and doing some problems getting accustomed to what the surfaces look like, to what 3-dimensional space looks like, to how certain curves and surfaces behave... that will give you a sense of when you can recognize when something is a certain way.2614

Other than that, if you want, the mathematics is still correct... I would not worry so much about the sign of the number.2629

Okay. So, as a quick recap here, so let us recap what we have done.2638

Alright. The integral of f(d(Σ)) = the integral of f(p) × norm(n) dt du, this is the integral of a function over a surface and let me go ahead and write where it... n = dp dt cross dp du.2647

Again, dp du, they just differ by... it is a negative sign.2692

Now, the integral of f · ds... recognize these, these are the symbolic representations of these integrals.2699

The integral of a function over a surface, the integral of a vector field over a surface. Okay.2709

That is equal to the integral over s of f(p), this part is the same, f(p) F(p), small f, capital F, · n.2715

Not times the norm of n, dot n, dt du.2726

This happens to be the integral of a vector field a surface. These are the things that you want to know. This is the definition of the integral. You can solve every problem by using this right here.2736

You find n, you find dp dt, you find dp du, you have to work from a parameterization, you form the cross product, that is n.2755

In the case of a function, you take the norm of n. If it is not a function, if it is a vector field, you just keep it as regular n. You form f(p), you form F(p), here you multiply the two, here oyu take the dot product of the two, and then you integrate depending on what t and y are, what particular domain of parameterization you are dealing with.2766

That is it. Integral of a function over a surface, integral of a vector field over a surface. Entirely analogous to the same definitions for line integrals.2785

The integral of a function over a curve, the integral of a vector field over a curve. You have just moved up one dimension, and as you can imagine, you can keep moving up one dimension at a time. It is true in any number of dimensions.2797

Thank you for joining us here at educator.com, we will see you next time. Bye-bye.2808

Hello and welcome back to educator.com0000

Welcome back to multivariable Calculus.0003

Last lesson, we introduced the notion of a projection.0005

We also introduced this notion of the angle between two vectors.0008

Remember a · b = the norm of a × norm of b × cos(θ).0012

We finished off an example and I would like to do one more example of that.0018

But this time with vectors that are not exactly located, that start from the origin.0022

Vectors that are just sort of random in space.0027

So, we have got... let us let a = the vector (1,2,3).0040

We will let b = (2,1,-4).0048

We will let c = (1,1,1).0055

I will go ahead and close that out.0061

We want to find the angle θ.0066

Between the vector ab, and the vector ac.0073

Again, a vector is just a point in space.0082

From this point to this point is a vector, and from this point to this point is another vector.0085

So we want to find the angle between those two vectors.0093

So, the vector ab.0097

Remember when we said we had two vectors, the vector that has one as a beginning point and the other as an ending point, that is equal to the ending point vector - the beginning point vector.0100

So it actually equals b - a.0113

If you look back a lesson ago, or two lessons, you will find it in there.0115

The vector b - a = (1,3,-7), just this - that.0121

b - a... oh, you know what... oh this is -2, that is why.0130

There you go, so (1,-2,3), (2,1,-4), and (1,1,1), there we go.0135

And -4 - 3 is -7.0143

And the vector ac = the vector c - a.0146

So that will equal (0,3,-2).0153

Let us check that real quickly.0156

1-1=0, 1--2=3 1 - 3 = -2, that is correct.0158

Now we can do our cos(θ) = ... now here our are vectors, these are the 2 vectors right here that we are finding the angle between.0166

So it is going to be the vector ab · ac/norm(ab) × norm(ac).0170

Okay, when we do ab · ac, we end up with 0+9+14.0188

When we run the norm of these vectors, we end up with sqrt(59) and sqrt(13), so it will = 23/... when we multiply this and take the radical, over 27.69480198

When I take arccosine of that, I get θ = 33.8 degrees.0217

There we go.0227

Let us see what this actually looks like in 3 space.0230

I am not going to do the actual vectors, I am just going to kind of do the random vectors.0235

This is the vector a.0239

Well, let us say this is the vector b, and let us say that this is the vector c.0242

Now, the vector ab, this is a and this b, and this is c, it is just these vectors here.0250

Those vectors are just 3 different points.0260

ab is that vector right there, and ac is that vector right there.0268

We are taking these vectors and we are creating other vectors.0277

This vector is b - a, and this vector is c - a.0282

So this is the vector ab, and this is the vector ac.0286

This θ is the angle between them, that is all we did, so we have to be very careful when we read these.0292

If they say a vector a and a vector b, and the angle between them, or if say they give us another point, we have to know which vector they are talking about.0303

Be very careful about how you read these, do not just jump into the problem.0307

Okay, let us go ahead and state a couple of theorems.0311

Very important theorems.0316

We are just going to state them again, we are not going to prove them, but we just want to be aware of them because they will come up from time to time.0318

They are very important actually.0325

They are important for theoretical reasons actually, but we do at least want to be aware of them.0328

So, let a and b be 2 vectors in RN.0334

Again, we are being as general as possible.0346

We will also use the notation instead of actually writing all of this out, we will often say a and b are members of RN.0350

We will often just do it that way.0359

Then, a · b, the absolute value of a · b, is less than or equal to norm(a) × norm(b).0366

This is called the Koshi Shwartz inequality.0381

It is a very important inequality in mathematics.0390

Not only does it hold in N space, any number of dimension of Euclidian space, it actually even holds more generally than that.0395

It actually even holds when you talk about a function space. or any other kind of space.0404

It is a very beautiful theorem.0409

Let us just talk about what it means.0412

It says if you take two vectors in RN, and you take their dot product, you know that you end up with a number.0414

It says that that number is going to be... that number has an upper limit... in other words, when you take the absolute value of that, the magnitude of that number is going to be less than the product of the norms of the two vectors involved in that scalar product.0420

Again, very important so we will just be aware of it.0436

Now, let us move on to the second theorem, which will probably make more sense geometrically.0444

Again, let a and b, ok let me draw these a little bit better here.0450

Let a and b be in RN, then the norm of a + b, in other words if I take a + b, and then I take the norm of that, it is going to be < or = the norm of a separately, and adding it to the norm of b.0461

This is called the triangle inequality.0485

We will go ahead and give you a nice geometric interpretation of this, so that you can see what it is.0490

It makes complete sense. You actually know this. It makes complete sense. 0495

I mean you actually know this from... intuitively you know what this means from the time you are 5, 6, 7 years old.0503

You just looked at some triangular shape and you knew intuitively what this meant, but again, we just want to formalize things.0504

So this says that if I have 2 vectors and I add them, and I take their norm, the length of that vector, the magnitude, the length of that has an upper limit.0510

That limit is the sum of the two norms taken individually.0520

Let us just work in 2 space here.0526

If I have some vector a, and I have another vector b, well I know a + b is going to be this vector here.0530

It just means you go to a and then do b, so this vector is a + b, and this is b, and this is a.0538

I get a triangle right here, this little triangle right here.0548

This triangle inequality says that if I add the length of one side and the length of the other side, it is going ot be bigger than the length of the third side.0553

That has to be true, we do not know this just from geometry, but we know this from looking at it.0565

If the length of the third side were somehow equal to the length of this and this, you would not have a triangle. 0568

What you would have is, one side, one side, and then on the third side, you would basically just have a triangle that collapses on itself.0575

If however for some odd reason the length of one side and then the length of the other side is somehow longer than the length of the third side, you do not even have a triangle at all.0582

You do not even have something that connects to any one side.0588

That is all this says.0594

It says that, for a triangle, we know that the sum of the two sides has to equal bigger than the third side, that is all this says.0597

This is just a generalization into any number of dimensions.0605

So again, geometry helps to see it, it is the algebra that is important.0608

This is true in any number of dimensions.0614

We are going to start talking about... we are going to give the definition of a parametric line.0620

Later on, when we start talking about functions of several variables, we are going to define a function parametrically.0627

This notion of a parametric definition is profoundly important.0633

We will talk about it a little bit now.0639

I will actually talk about it broadly and theoretically, but without proofs or anything, I will take some time to discuss what this means in the next lesson.0641

Right now, let us just get a sense of what is going on here.0650

Definition.0654

A parametric equation, also a representation, in fact I prefer the word representation so a parametric representation of a straight line.0659

We are only talking about a straight line so far.0678

The straight line passing through a point p, which is a vector in the direction of a vector "a" is x = the point p + some parameter t × the vector a, where t can be any number. Any real number.0681

Okay, what this means is this x right here, it means that if I start at p, and if I go in the direction of a any length, I am going to end up at a different point, right?0728

Right, all those different points that I end up, that is what the x is.0739

All those points in this direction and that direction, they form the line.0745

I will draw a picture in just a minute.0748

Just to let you know, the vector x that you get, or the points in space that make up this line, they are functions of t.0752

So it equals p + t × vector a.0762

Now let us draw a picture and see what it is we mean, this is very, very, important.0768

You must absolutely understand what a parametric representation of an equation, in general of a function is.0775

You are going to need as we go along later in the course, you are going to need to actually write down... we are going to give you some equation in x,y, or maybe x,y,z, and you have to give the parametric representation.0785

It is very important that you understand what we mean by this thing called a parameter.0794

Like t, why it is we write things this way.0798

Let us see if we can develop some intuition.0800

This is our point p and we know that our line is going to be passing through this.0805

Well, we want to pass, we want the line to be in the direction a, so let us just pick some random vector a, that is a.0808

So we know the line is going to pass through here, and it is going to be in this direction.0817

In other words, the line is going to be that line right there, going this way and going that way.0823

We want to be able to write an equation for that.0830

Watch what happens.0833

This is the equation that I gave, right?0834

Let us just do a couple of examples for t.0837

Well, when t = 0, if I put t = 0 in here, t = 0 and 0 × a = 0, therefore our point is just our vector p, so the point is there, no problem.0840

Now let us take t = 1, well now, if I put 1 in here, that means it is going to be p, our new point x is going to be p + 1 × vector a.0855

I start here and I do just one length of vector a in that direction.0870

Now my new point is right there.0874

So it is going to be p + the vector a.0877

If I do t = 2, that is going to be p + 2 × the vector a.0881

That means I start at p, then I go one length of vector a, and then I go another length of vector a, that puts me at that point there.0891

Well, what if I do t = -6?0901

That means my new point along that line is going to be... start at p, and since it is - 6, that means -6a.0905

So start a p, then go in the direction of a, but opposite, -6.0916

So it is going to be 1,2,3,4,5,6 that puts me at that point. 0921

Now, if I take all values of t, t = - infinity to positive infinity, .6 sqrt(13), -pi, I get all of these points.0929

I get all of these points along that line.0940

What I have done is now I have an equation for that line, with respect to one parameter.0943

It is no longer y as a function of x. 0950

What I have done is I have freed myself from thinking of things in a coordinate system, y as a function of x, two variables.0953

Now I have created an equation in one variable t, a parameter.0961

I have picked a starting point, I have picked a direction, and now as t goes from - infinity to + infinity, it represents all of these points.0967

That is the power of a parametric representation.0975

It frees me from expressing one variable in terms of another variable.0980

I can recover it if I want, but it totally frees me from that.0984

Now I can treat each variable separately as a function of t, you will see that later on.0988

We just said, the vector x(t) is some starting point + numbers × some vector a which is the direction of the line.0998

Let us do this in component form, x is just a vector.1015

Let us just take components in 3 space.1017

In 3 space, this would just be x,y,z.1020

I am going to write these vertically, is equal to some p.1026

Let us say p1,p2,p3 + t × a1,a2,a3.1030

This is the vector form, and this is the component form.1043

When we do a specific problem, we work in components.1048

When we work theoretically, algebraically, we are not doing numbers, we work in vector form... that is the best way to do it.1050

Now, let us give another definition.1058

Let a and b be vectors in RN.1062

Then, the line segment between a and b is... we will call it segment t... and again, we are still talking about vectors here.1070

it is equal to a + t × b - a.1095

This is a parametric representation for the line segment between the vector a and b, between the point a and b.1100

Let us just... this is the definition we gave, let us just try a couple values and see if this makes sense.1110

If I let t = 0, then s(0) = a + 0, t = 0, so that is equal to a.1116

What happens when t = 1?1131

Well s(1) = a + 1 × b - a, which is distribute the 1 and we get + b - a, a's cancel... and I get b.1136

Oh, I forgot to say t > or = 0 and < or = to 1.1150

The line segment between two points, two vectors, is this equation when t takes on the values 0 and 1 and everything between.1156

At 0, it is the point a, at 1, it is the point b.1167

At every point in between, 0.1,0.2,0.5,0.6,0.7,2/3, things like that, you are going to get all the points in between.1173

So here is what it looks like.1181

This is the point a, the vector a.1185

Oops, we have our stray lines again, this is the point b, the vector b.1191

So this is a, and this is b, a, b. 1196

We want to find a parametric representation for this line segment in between them, this is what we want.1202

Well, this is the equation as we set it up.1210

If I start here, that is my starting point, right... this is the same as the other parametric, the definition we gave was just some starting point + t × some direction vector. 1214

Well, the direction vector is this direction.1226

This direction vector is equal to b - a, there is nothing different here.1228

Now, basically what happened is we set t=0, we get this point. t = 1, we get this point, well all the points in between is when t = everything between 0 and 1.1235

Literally, what you are doing is you are starting here and you are moving along this direction, you are just moving in this direction taking all the points.1248

Eventually, if you take all of the values between 0 and 1, you end up with all of the points in that line segment.1256

So again, this is the definition of the line segment between 2 points, a and b.1270

Okay, let us go ahead and do some examples.1279

Let us do example 1.1283

Actually, let me go back to black here.1286

Now, we will let a = (1,3,5), and we will let b = (1,-3,-5).1290

Our problem in this case is to find the coordinates of the point, 1/5 of the way from a to b.1302

So we want to find the coordinates of the point which is 1/5 of the way from a to b.1332

We have some... it is a line segment from a to b... we want to find the coordinate, the distance, we want to find the coordinate of that point.1337

So pictorially we are looking at something like this.1347

We said if this is a, and this is, that is the line segment, 1/5, we want to find that coordinate.1350

In other words, we want to find that vector, not this length, we want to find this vector.1357

Here is how we do it.1363

We said that the line segment t = a + t × b - a, when t > or = to 0, < or = to 1.1365

In this case t = 1/5, this is the starting point.1380

1/5 of the way here, so here we are going to take t = 1/5, this is what we want.1385

Well, let us just do the general version.1393

a = (1,3,5), I am going to write these as vertical vectors, + t × vector b - a.1395

Well, b - a, 1-1 is 0, -3 -3 is 6, -5 -5 is - 10, so this is our generic equation for this line segment in coordinate form.1406

That is the vector form, this sis the coordinate form, this is the equation, this is our parametric equation, it is a function of t.1419

Now, we know t = 1/5, so it is going to equal (1,3,5) + 1/5(0,-6,-10).1430

That is going to equal... I will do it down here.1443

It equals (1,3,5) + (1/5 × 0 is 0, 1/5 × -6, is -6/5, and 1/5 × -10 is -2).1448

Now, when I add those together I end up with 1+0 is 1, 3 + -6/5... when you do the math it end sup being 9/5, and 5 - 2 is 3.1465

There we go.1482

The coordinates of this point is (1, 9/5, 3).1484

They are the coordinates of the point that is 1/5 of the distance from the point (1,3,5) to the point (1,-3,-5).1490

Notice how when we are solving a specific problem, we work in coordinates.1500

This is the vector representation, when we actually do the problem, we work in coordinates.1506

We actually put the vectors in.1512

I wrote them vertically because it makes more sense for me when I put them in, you can do horizontally, it is whatever works for you.1514

Okay, so let us do another example here.1520

Example 2.1528

Find a parametric representation for the line passing through the vector a, which is (2,4,6), and the vector b... let us get some room here... and b equals (-1,-2,6).1530

Let us make sure we understand what it is this is saying.1575

This is saying find a parametric representation for the line passing through a and b.1580

We want to find a parametric line that passes through a and b, notice it does not say passing a in the direction of b, it says passing through a and through b, these are two different points.1585

Well, the direction from a to b, or form b to a, it does not matter which way you do it, let us call that vector c.1596

So, c = the direction, so it is going to pass through a and b, I can choose either one as my starting point.1605

I am going to go ahead and choose a as my starting point, so it is going to go in the direction b, ab.1615

So C... actually let me write it at, yea it is fine, we will go and say that... this is my direction vector c, it is going to = b - a.1620

When we have two points, the direction, this direction is the vector, that vector minus this vector.1630

The arrow, the head minus the tail.1635

Therefore, we are going to have x(t) = our starting point a, so again we just want to write out the generic form so we know what it is we are working with, + t × vector c.1640

That is the direction we are going in.1657

That equals a + t × b - a.1659

As you can see, this looks like the line segment definition, except now we want the entire line, not the segment, so we are not going to restrict our values of t between 0 and 1.1667

We are going to let t be any value at all.1677

So let us go ahead and put our coordinates in.1679

a = (2,4,6) + t × b -a, now let us take b - a.1680

-2 - 1 is -3, -2 - 4 is -6 , 6 - 6 is 0, and boom, there you go, it was as easy as that.1690

We just have to make sure we understand what it is asking.1705

It is passing through these two points, which means you pick one point to start from, and the direction is from a to b, or from b to a, it does no matter what direction you take or what parameterization you use as long as you know what it is asking.1708

As long as you have a direction vector, the proper direction vector.1718

Notice that b is not the direction, that is the direction vector, not b, it is passing through b and a.1727

Sorry if you think I am belaboring the point, I understand, forgive me if I do that but it is very important to get this right.1736

Example 3.1745

Let a = (-1,2)... this time we are working in 2 space... and we will let b = (2,6).1749

Wow, now these crazy lines are coming up all over the place.1764

A line passing through a in the direction of b is, parametrically... ok now that we are getting to the bottom of the page these lines are going to show up, so I am actually going to skip and go over one.1770

So we have x(t) = a + t × b -- which is going to equal a, which is (-1,2) -- + t × (2,6).1800

Let me write this in component form.1820

The vector x is just a vector, so it is some point (x,y) = -1 + 2t, 2 + 6t.1824

There you go, so now I have the component version of this, now let me separate these.1840

x = -1 + 2t, y... let me do this, y = 2 + 6t, let me solve for t.1845

This is x +1/2 = t.1860

This is y-2/6 = t, the t's are equal to each other, it is one parameter, this is the power of it, it is one parameter.1866

So let me set these equal to each other.1875

x+1/2 = y-2/6, let me cross multiply.1878

I get 6x+6 = 2y-4.1886

Let me bring the y over. I get 6 -2y = -10, look at that.1890

I have taken my parametric equation and now I have recovered the standard form of the equation that you are used to ever since middle school and high school1895

An expression of the line in terms of the two variables x and y.1903

We can do this in 2 space because we have just the 2 coordinates, two equations we can set equal to each other.1908

The problem is, when we try a parametric representation in lets say 3 space, we cannot solve for t and set these things equal to each other and come up with some, you know, equation like this.1913

Now that we are actually moving into higher dimensions, the parametric representation of a line is the only representation of a line.1925

It is the best, this whole idea of working with x's and y's, it is good, and it is fine for what we did, but now we need to move on to a more powerful tool.1936

We need to move on to parametric representations of lines.1948

The parametric representation of a line is valid in any number of dimensions. 1954

I no longer need to work with x and y, we are moving up, we are becoming more sophisticated with our tools.1955

That is why we introduced this notion of a parametric equation.1960

That is it, so again, if we need to recover it for dimension 2, that is fine.1968

This is certainly doable, you solve for t and you do this.1973

But we are going to be working strictly with parametric equations because it is a more powerful tool and it works in any number of dimensions. 1980

In fact, parametric representation is the only representation that works in any number of dimensions greater than 2. 1985

Okay, thank you for joining us here at educator.com1992

We will continue our discussion of parametric representations and lines next time. Bye-bye.1995

Hello and welcome back to educator.com and multi variable calculus.0000

Today's lesson we are going to be talking about divergence and curl in 3-space.0004

So we have talked about divergence and curl before, we did it for 2-space when we discussed Green's Theorem.0008

Now we are going to talk about it in 3-space. We are going to introduce some new symbolism that will take care of all cases all at once.0016

Let us just go ahead and jump right on in. Okay.0024

So, we will let f(x,y,z) = f1(x,y,z), f2(x,y,z), and f3(x,y,z).0028

This is just a highly explicit way of representing a vector field. A vector field, these are the coordinate functions, the coordinate functions are functions of x,y,z themselves. I have just written everything out.0051

So, let this be a vector field on an open set s in R3, in 3-space, 3-dimensional space.0064

Okay. That is... and I know you know what this is but I am just going to repeat it for the sake of being complete.0084

For each point in s, there is a vector given by f pointing in some direction away from that point... pointing in some direction... starting at that point. That is it.0095

So, if I take some 3-space, so in this particular case let us just say that our open set s happens to be all of 3-dimensional space.0140

If I pick a point at random, well if I pick a point and I put that point in for f, I am going to get a vector going this way, maybe for this point I have a vector this way, maybe for this point a vector this way, that way, that way, that way, that way, could be any number of things. That is a vector field in 3-space. That is all it is.0150

Okay, let us define the divergence of f.0171

The divergence of f, which is symbolized as div(f) = df1 dx + df2 dy + df3 dz, or in terms of capital D notation, D1f1 + D2f2, + D3f3. Okay.0180

It is a scalar, it is a number. It is a scalar. A number at a particular point, Df dx + Df2 dy + Df3 dz at a particular point you put that particular point into this thing and it is just going to spit out a number.0218

Okay. That is the divergence in 3-space. It is just the analog of the divergence in 2-space.0246

So the curl is the... it is the analog of the curl in 2-space that we discussed previously, but it is a little bit more notationally complicated.0253

So, let us go ahead and write it out, and then we will go ahead and give you a symbolic way of representing it. 0262

So, curl of f... define the other definition, the curl of the vector field... is equal to... this one I am going to do the capital D notation first, and then I will do the regular partial derivative notation.0267

It is going to be D2f3 - D3f2, D3f1 - D1f3, D1f2 - D2f1. Notice this is not a scalar. This is a vector. The curl of a vector field gives you another vector field.0282

In terms of partial derivative notation, this looks like this. So, D2 of f3, this is Df3 dy - Df2 dz.0316

This is Df1 dz - Df3 dx.0335

This is Df2 dx - Df1 dy.0346

I hope to heaven I have got all of those correct. Okay. That is the curl. It is a vector.0356

So, when you are given a vector field f, when you take the divergence of it, you want to put a number at a given point.0367

When you take the vector field f, the curl of that vector field, what you end up with is another vector field. It is a vector at a given point.0374

Okay. Now we are going to introduce something called the Del operator.0386

The del operator is a symbolic way of simplifying the calculations. That is really what it is. And making things just look more elegant. The del operator, okay, it is usually an upside down triangle, or you can just write del.0392

So, the definition of the operator is the following. For the time being, the notion of an operator, an operator is just something that tells you to do something to something else.0413

In other words, we talk about the differential operator d dx. So, for example you knew what this is, d dx. If I apply this differential operator to some function f, d dx of f, well, that is just df dx.0423

That just means take the derivative of it. The integral operator. The integral operator says integrate on the function f, you are going to end up getting something else, so that is all an operator is.0440

It is a fancy term that says do something to this function. Well, the del operator is the same thing.0450

It is an operator that says do this to a given function, except it has multiple parts. These are individual operators. The differential, the integral operators. The del operator actually is written symbolically in the form of a vector.0456

So, but other than that you treat it exactly the same way as you do anything else. It says do something to something.0473

As it turns out, the del operator is a differential operator. It is a partial differential operator. You will see what we mean in just a minute.0480

So, let us go ahead and write out the symbol and then do some examples and of course everything will make sense.0487

Let us see, so this, or del, is equivalent to d dx, d dy, d dz, or D1, D2, D3, notice there is no function here because it is an operator. 0493

I have to choose a function and then say do this to it. That is the whole idea. We write it as a vector because it is going to operate as a vector on another vector. That is the whole idea.0518

So, now, let us go ahead and write what we mean by divergence and curl symbolically, using this del operator notation.0534

Okay. So, symbolically, and again, this is all symbolic. Symbolically, the divergence of f = the del operator dotted with the vector field, f.0544

Well, the del operator dotted with the vector field f, is equal to... well, the del operator is D1, D2, D3, that is my symbolic operator for del, dotted with well... f is f1, f2, f3.0568

Well, I know what a dot product is. It is just this × this, this × this, this × this, added together.0585

Except I am multiplying these two numbers, this is a symbolic vector operation. It says do d1 to f1, in other words, take the derivative of f1 with respect to x. Take the derivative of f2 with respect to y, take the derivative of f3 with respect to z. This is a symbolic notion.0591

We are symbolizing using the idea of a vector, that is what an operator is. So, this is equal to... so, it is d1f1 + d2f2 + d3f3, except this is not multiplication, this d1f1... this is a differential operator.0614

It says take the derivative of f1 with respect to the first variable, this says take the derivative of f2 with respect to the second variable, this says take the derivative of f3 with respect to the 3rd variable. I hope this makes sense.0637

Now the curl of f. Okay. This is going to be kind of interesting. Let me go to the next page.0651

So, the curl of f. Now you can memorize this any way you want to. If you want to just go back to what I initially wrote on the first page of this lesson, when I defined divergence and curl, you are more than welcome to remember curl in that way if you want to remember the indices -- 23 32, 13 31, 12 21.0657

That is fine, or you can remember it this way, symbolically. 0677

So, the curl of f, it is defined as del cross f, the del operator crossed with the vector field. Well we know what a cross product is. We have been dealing with it symbolically. It is the symbolic determinant i,j,k, and now del cross f... it is like this is a vector, this is a vector.0682

Well, the first vector is in the second row, so we will just write D1 D2 D3, and f is just f1 f2 f3.0705

We will symbolically take the determinant of that. When you do that, you end up with the following.0720

You end up with (D2f3 - D3f2)i - (D1f3 - D3f1)j + (D1f2 - D2f1)k. i... j... k... this is the first component function, second component function, third component function.0726

What you end up with is exactly what we had earlier. You get D2f3 - D3f2, this - turns this into a negative, turns this into a positive... what you end up with is D3f1 - D1f3 and you get D1f2 - D2f1.0784

This is the vector representation, this is the i,j,k, representation. This is the curl.0813

So, all I have done is I have taken this... I have created this thing called the del operator, given it a symbol like a vector, and I have been able to define the divergence and curl in terms of the two vector operations that I have.0819

The divergence of the vector field f is equal to del · f, and then curl of the vector field f is equal to del cross f.0835

This is the symbolic way of keeping things straight. So, let us just do some examples and I think it will make sense here. So, Example 1.0843

We will let our vector field f = the first component function is x2, the second component function is xy, and our third component function is going to be ex,y,z.0859

Okay. So, the divergence of f, divergence is always going to be easier because you are just sort of taking partial derivatives 1 by 1. So, the divergence, the partial of this with respect to x is going to be 2x, + the partial of this with respect to y, so this is going to be cos(xy) × x, so it is going to be + x × cos(xy), and the partial of this with respect to z is just going to be ecy.0875

There you go. That is the divergence. It is a scalar, not a vector. 0906

Now, let us go ahead and form the curl of f. The curl of f, well, we said that the curl of f = del cross f, so let us go ahead and form our symbolic determinant here... i, j, k.0915

We have d1, d2, d3, or if you want, let us go ahead this time for this particular example, let us use our partial differential and our d dx, d dy, it does not really matter.0934

So, let us do i, j, k, so we have d dx, d dy, an d dz, that is the del vector operator, and then we have f, which is x2, sin(xy), and exyz. We want to form this determinant.0950

Okay. So, let us see what we get. When we expand along the first row, we get the derivative of this with respect to y - the derivative of this with respect to z, so I end up getting the derivative of this with respect to y is exz, ex × z - 0 × i - ... because remember it is + - +, alternating sign.0978

The derivative of this with respect to x - the derivative of this with respect to z.1008

So the derivative of this with respect to x is exyz is e2 × yz - 0, and that is going to be j.1014

Of course our last one k is going to be the derivative of this with respect to x - the derivative of this with respect to y.1025

So, the derivative of this with respect to x is going to be y × cos(xy) - 0 × k, so is the... so I get this, this, and this... and so I end up with ex × z is my first component function of my curl, ex × yz is the second component function of my curl, and y × cos(xy) is the third component of my curl vector. There we go.1032

This is a scalar divergence... curl is a vector at a given point.1071

If I take all of the points, it gives me a vector field. That is it.1078

Okay. Let us do another example. So, example 2.1085

f(x,y,z) = x2yz, xy3z, and xyz4. Okay, well, let us see what we have got. 1095

The divergence of f, we said, is equal to del · f, well del · f is d1f1 + d2f2 + d3f3, but we are not multiplying the d and the f, this means take the derivative of f.1122

Well, the derivative of this with respect to x is going to be 2xyz + this one, the derivative with respect to y is going to be + 3xy2z, and the derivative with respect to z of this one is going to be +4xyz3. This is our divergence of the particular vector field.1141

Okay. So, now let us go ahead and do the curl of the vector field. The curl of this vector field, well it is equal to del cross f, and del cross f, well, is symbolic.1169

It is going to be i, j, k... let us do capital D notation here... D1, D2, D3, and we have x2yz, we have xy3z, and we have xyz4.1187

Okay. So, now let us go ahead and expand along the first row. It is going to be the derivative with respect to y of this - the derivative with respect to z of this. The derivative with respect to y of this is xz4.1208

The derivative with respect to z of this is xy3. This is the i component... - , now we go to the next one.1227

The derivative with respect to x of this - the derivative with respect to z of this, derivative with respect to the first variable which is x, derivative with respect to the third variable which is z. That is what is going on here.1241

This × this, this × this, except it is not times, it is symbolic. It means operate on this. 1253

Okay. So, the derivative with respect to x of this is yz4 - the derivative with respect to z of this, x2y. This is the j component.1262

Okay, we are almost done. Now, the derivative with respect to x of this - the derivative with respect to y of this. 1273

So, the derivative with respect to x of this is going to be y3z - the derivative with respect to y of this is going to be x2z.1282

Again, I hope that you are confirming this for me. There are lots of x's, y's, z's, i's, j's, k's, floating around. y3z, there you go.1294

I will go ahead and actually leave it in this form. That is the first coordinate function of the curl, that is the second coordinate function of the curl, and that is the third coordinate function of the curl.1304

Actually, you know what, let me go ahead and write it out. It is not a problem.1319

So, I will go ahead and erase these stray lines here.1320

So, we have got xz4 - xy3, that is the first component function, we have yz4 -x2y, and then we have y3z - x2z, notice the divergence is a scalar, the curl is a vector. 1329

It has three component functions... an x, a y and a z. At a given point (x,y,z), there is some vector pointing in some direction and away from that point. That is the whole idea.1359

Again, it is all based on this notion of what we call an operator. It is just a symbolic way of telling you what to do to a given function.1370

It is a unifying scheme, so you had this thing. You can take the divergence and curl of a vector field. We want to be able to express that in terms that we know.1380

Well, if we gave this... we called it a del operator, we have it a symbolic representation s D1, D2, D3, as a symbolic vector... If we do del · f, we get divergence of f, if we get del cross f, we get the curl of f. That is it. It is just a unifying scheme. 1390

Okay, thank you for joining us here at educator.com for divergence and curl. We will see for a discussion of the divergence theorem in 3-space. Take care, bye-bye.1410

Hello and welcome back to educator.com and multi variable calculus.0000

Today's topic is going to be the divergence theorem in 3-space. When we did Green's theorem, we had -- remember -- two different forms of Green's theorem, we had the circulation curl which was the original form, and then we had the flux divergence.0004

We talked about what each integral meant, what flux meant, what divergence meant, well, the divergence theorem in 3-space is just the flux divergence form of Green's theorem, which is a 2-dimensional theorem kicked up one dimension.0017

That is exactly what we are doing. We are just moving up one dimension at a time. So, that is it. Let us just jump in and see what we can do here.0030

Okay. So, let us just remind ourselves of the flux divergence version of Green's theorem was as follows: so, I will write Green's -- oops, let us make it a little bit clearer here -- Green's flux divergence theorem... I will just write Green's flux divergence.0039

It said this in terms of symbols. That over Δ a, we have f(c) · n dt = a divergence of f dy dx. This was a symbolic representation