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 0 answersPost by Professor Hovasapian on August 3, 2013Hi Michael,It's just because I like to avoid any excessive notation, if I can. If it is understood that a given vector is Unit, then keeping one less marking off the paper seems worth it to me. Nothing more.Hope all is well. Best wishes.Raffi 1 answerLast reply by: Professor HovasapianSat Aug 3, 2013 2:55 PMPost by michael Boocher on August 3, 2013Raffi,  You seem to be avoiding the use of the hat to indicate unit vectors? Is this for clarity issues or there something else to it?Michael.

### Final Comments on Divergence & Curl

Show that if (a,b) is a vector, then both ( − b,a) and (b, − a) are orthogonal it.
• Recall that a vector a is orthogonal to a vector b if a ×b = 0. Note that is a scalar product.
• Then (a,b) ×( − b,a) = ( − ab) + ba = 0. Also (a,b) ×(b, − a) = ab + ( − ba) = 0.
Since (a,b) ×( − b,a) = 0 and (a,b) ×( − b,a) = 0 then (a,b) is orthogonal to both ( − b,a) and (b, − a).
Show that if ( − b,a) and (b, − a) are vectors each orthogonal to the vector (a,b) then the angle between ( − b,a) and (b, − a) is 180o.
• Recall that the angle between two vectors a and b is computed by cosq = [(a ×b)/(|| b |||| a ||)].
• For a = ( − b,a) and b = (b, − a) cosq = [(a ×b)/(|| b |||| a ||)] = [(( − b,a) ×(b, − a))/(|| ( − b,a) |||| (b, − a) ||)] = [( − b2 − a2)/(√{b2 + a2} √{b2 + a2} )] = [( − (a2 + b2))/(a2 + b2)] = − 1.
Hence cosq = − 1 yields q = 180o.
Let C(t) = (t2 − 2t + 1,cos(t) − t)i) Find C′(t)
• Recall that C′(t) = ( [d/dt]x(t),[d/dt]y(t) ) for C(t) = ( x(t),y(t) ).
Hence C′(t) = (2t − 2, − sin(t) − 1).
Let C(t) = (t2 − 2t + 1,cos(t) − t)
ii) Find the right and left normal vectors to C′(t).
• We define the right normal vector to C′(t) = (x′(t),y′(t)) as C′(t)R = (y′(t), − x′(t)) and the left normal vector as C′(t)L = ( − y′(t),x′(t)).
Since x′(t) = 2t − 2 and y′(t) = − sin(t) − 1 we substitute to obtain C′(t)R = ( − sin(t) − 1,2 − 2t) and C′(t)L = (sin(t) + 1,2t − 2).
Let C(t) = (t2 − 2t + 1,cos(t) − t)
iii) Verify that the right and left normal vectors to C′(t) are orthogonal to C′(t).
• We check to see if C′(t) ×C′(t)R = 0 and C′(t) ×C′(t)L = 0.
• So C′(t) ×C′(t)R = (2t − 2, − sin(t) − 1) ×( − sin(t) − 1,2 − 2t) = (2 − 2t)(sin(t) + 1) − (2 − 2t)(sin(t) + 1) = 0.
• Similarly C′(t) ×C′(t)L = (2t − 2, − sin(t) − 1) ×(sin(t) + 1,2t − 2) = (2t − 2)(sin(t) + 1) − (2t − 2)(sin(t) + 1) = 0.
Since C′(t) ×C′(t)R = 0 and C′(t) ×C′(t)L = 0 the right and left vectors to C′(t) are orthogonal to C′(t).
Find the curl and divergence of the vector field F(x,y) = ( x + y,[1/x] + [1/y] ).
• To find the curl of F(f1(x,y),f2(x,y)) we compute curl(F) = [(df2)/dx] − [(df1)/dy], to find the divergence we compute div(F) = [(df1)/dx] + [(df2)/dy].
• Note that f1(x,y) = x + y and f2(x,y) = [1/x] + [1/y].
So curl(F) = [(df2)/dx] − [(df1)/dy] = − [1/(x2)] − 1 and div(F) = [(df1)/dx] + [(df2)/dy] = 1 − [1/(y2)]
Find the curl and divergence of the vector field F(x,y) = ( x3y2 − xy,2x2y + 2xy2 ).
• To find the curl of F(f1(x,y),f2(x,y)) we compute curl(F) = [(df2)/dx] − [(df1)/dy], to find the divergence we compute div(F) = [(df1)/dx] + [(df2)/dy].
• Note that f1(x,y) = x3y2 − xy and f2(x,y) = 2x2y + 2xy2.
• So curl(F) = [(df2)/dx] − [(df1)/dy] = (4xy + 2y2) − (2x3y − x) and div(F) = [(df1)/dx] + [(df2)/dy] = (3x2y2 − y) + (2x2 + 4xy).
Hence curl(F) = − 2x3y + 2y2 + 4xy + x and div(F) = 3x2y2 + 2x2 + 4xy − y.
Evaluate the curl and divergence of the vector field F(x,y) = ( yln(x),y2 + 4x ) at (1,1).
• First we find the curl of F by computing curl(F) = [(df2)/dx] − [(df1)/dy] and the divergence by computing div(F) = [(df1)/dx] + [(df2)/dy].
• Note that f1(x,y) = yln(x) and f2(x,y) = y2 + 4x
• So curl(F) = [(df2)/dx] − [(df1)/dy] = 4 − ln(x) and div(F) = [(df1)/dx] + [(df2)/dy] = [y/x] + 2y.
• Now curl(F(1,1)) = 4 − ln(1) and div(F(1,1)) = [1/1] + 2(1).
Hence curl(F(1,1)) = 4 and div(F(1,1)) = 3.
Evaluate the curl and divergence of the vector field F(x,y) = ( xe2y, − 4x + y ) at (1,0).
• First we find the curl of F by computing curl(F) = [(df2)/dx] − [(df1)/dy] and the divergence by computing div(F) = [(df1)/dx] + [(df2)/dy].
• Note that f1(x,y) = xe2y and f2(x,y) = − 4x + y.
• So curl(F) = [(df2)/dx] − [(df1)/dy] = − 4 − 2xe2y and div(F) = [(df1)/dx] + [(df2)/dy] = e2y + 1.
• Now curl(F(1,0)) = − 4 − 2(1)e2(0) and div(F(1,0)) = e2(0) + 1.
Hence curl(F(1,0)) = − 6 and div(F(1,0)) = 2.
Evaluate the curl and divergence of the vector field F(x,y) = ( [cos(x)/2y],[sin(y)/2x] ) at ( − [(π)/6], − [(π)/3] ).
• First we find the curl of F by computing curl(F) = [(df2)/dx] − [(df1)/dy] and the divergence by computing div(F) = [(df1)/dx] + [(df2)/dy].
• Note that f1(x,y) = [cos(x)/2y] and f2(x,y) = [sin(y)/2x].
• So curl(F) = [(df2)/dx] − [(df1)/dy] = − [sin(y)/(2x2)] + [cos(x)/(2y2)] and div(F) = [(df1)/dx] + [(df2)/dy] = [( − sin(x))/2y] + [cos(y)/2x].
• Now curl( F( − [(π)/6], − [(π)/3] ) ) = − [(sin( − π \mathord/ phantom π3 3 ))/(2( − π \mathord/ phantom π6 6 )2)] + [(cos( − π \mathord/ phantom π6 6 ))/(2( − π \mathord/ phantom π3 3 )2)] and div( F( − [(π)/6], − [(π)/3] ) ) = [( − sin( − π \mathord/ phantom π6 6 ))/(2( − π \mathord/ phantom π3πace3 ))] + [(cos( − π \mathord/ phantom π3 3 ))/(2( − π \mathord/ phantom π6 6 ))].
Hence curl( F( − [(π)/6], − [(π)/3] ) ) = − [(27√3 )/(4π2)] and div( F( − [(π)/6], − [(π)/3] ) ) = − [9/(4π)].
i) Find the net curl of the vector field F(x,y) = (x + y,xy) over C described below by computing ∫C F(C) ×C′(t) dt:
• Note that ∫C F = ∫C1 F + ∫C1 F + ∫C1 F where C1 is the line segment from (0,0) to (2,2), C2 is the line segment from (2,2) to (2,0) and C3 is the line segment from (2,0) to (0,0).
• So we will have to compute ∫C1 F(C1) ×C1\cent(t) dt, ∫C2 F(C2) ×C2\cent(t) dt and ∫C3 F(C3) ×C3\cent(t) dt and sum their result to obtain ∫C F .
• Recall that we can form a parametric representation of a line segment using two points P and Q through C(t) = P + t(P − Q), 0 ≤ t ≤ 1. Note that our orientation must remain constant.
• Hence C1(t) = (2t,2t), C1\cent(t) = (2,2) and F(C1) = (4t,4t2) so that ∫C1 F(C1) ×C1\cent(t) dt = ∫01 (4t,4t2) ×(2,2) dt = [20/3]
• Similarly C2(t) = (2,2 − 2t), C2\cent(t) = (0, − 2) and F(C2) = (4 + 2t,4 − 4t) so that ∫C2 F(C2) ×C2\cent(t) dt = ∫01 (4 + 2t,4 − 4t) ×(0, − 2) dt = − 4
• And C3(t) = (2 − 2t,0), C3\cent(t) = ( − 2,0) and F(C3) = (2 − 2t,0) so that ∫C3 F(C3) ×C3\cent(t) dt = ∫01 (2 − 2t,0) ×( − 2,0) dt = − 3
Thus ∫C F = ∫C1 F + ∫C1 F + ∫C1 F = [20/3] − 4 − 3 = − [1/3].
ii) Find the net curl of the vector field F(x,y) = (x + y,xy) over C described below by computing dA
• We can integrate our region A by letting dA = dydx. Note that this region is between the lines y = 0 and y = x.
• Our intervals of integration are y ∈ [0,x] and x ∈ [0,2] and so dA = ∫020x curl(F) dydx.
• Now, curl(F) = [(df2)/dx] − [(df1)/dy] = y − 1 so ∫020x curl(F) dydx = ∫020x (y − 1) dydx.
• Integrating yields ∫020x (y − 1) dydx = ∫02 ( [(x2)/2] − x ) dx = [1/3].
• Note that our version of Green's Theorem to find the net curl applies only when the orientation is counterclockwise, we compensate by taking the opposite of our result.
Thus dA = − dA = − [1/3]. Note that ∫C F(C) ×C′(t) dt = dA.
i) Find the net flow of the vector field F(x,y) = (x + y,xy) over C described below by computing ∫C F(C) ×N dt.
• Note that ∫C F = ∫C1 F + ∫C1 F + ∫C1 F where C1 is the line segment from (0,0) to (2,0), C2 is the line segment from (2,0) to (2,2) and C3 is the line segment from (2,2) to (0,0).
• So we will have to compute ∫C1 F(C1) ×N(t) dt, ∫C2 F(C2) ×N(t) dt and ∫C3 F(C3) ×N(t) dt and sum their result to obtain ∫C F .
• Recall that we can form a parametric representation of a line segment using two points P and Q through C(t) = P + t(P − Q), 0 ≤ t ≤ 1. Note that our orientation must remain constant.
• Also recall that our normal vector N must be to the right of a counterclockwise orientation. In our case we form N from C′(t) = (x(t),y(t)) by finding the right normal vector C′(t)R = (y′(t), − x′(t)).
• Hence C1(t) = (2t,0), C1\cent(t) = (2,0), N = C′(t)R = (0, − 2) and F(C1) = (2t,0) so that ∫C1 F(C1) ×N dt = ∫01 (2t,0) ×(0, − 2) dt = 0
• Similarly C2(t) = (2,2t), C2\cent(t) = (0,2), N = C′(t)R = (2,0) and F(C2) = (2 + 2t,4t) so that ∫C2 F(C2) ×N dt = ∫01 (2 + 2t,4t) ×(2,0) dt = 6
• And C3(t) = (2 − 2t,2 − 2t), C3\cent(t) = ( − 2, − 2), N = C′(t)R = ( − 2,2) and F(C3) = (4 − 4t,4 − 8t + 4t2) so that ∫C3 F(C3) ×N dt = ∫01 (4 − 4t,4 − 8t + 4t2) ×( − 2,2) dt = − [4/3]
Thus ∫C F = ∫C1 F + ∫C1 F + ∫C1 F = 0 + 6 − [4/3] = [14/3].
ii) Find the net flow of the vector field F(x,y) = (x + y,xy) over C described below by computing dA
• We can integrate our region A by letting dA = dydx. Note that this region is between the lines y = 0 and y = x.
• Our intervals of integration are y ∈ [0,x] and x ∈ [0,2] and so dA = ∫020x div(F) dydx.
• Now, div(F) = [(df1)/dx] + [(df2)/dy] = 1 + x so ∫020x div(F) dydx = ∫020x (1 + x) dydx.
• Integrating yields ∫020x (1 + x) dydx = ∫02 ( x + x2 ) dx = [14/3].
Thus dA = [14/3]. Note that ∫C F(C) ×C′(t) dt = dA.

*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.

### Final Comments on Divergence & Curl

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

• Intro 0:00
• Final Comments on Divergence and Curl 0:37
• Several Symbolic Representations for Green's Theorem
• Circulation-Curl
• Flux Divergence
• Closing Comments on Divergence and Curl

### Transcription: Final Comments on Divergence & Curl

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