Rectangular Green's Theorem
During the last lecture (which may feel as if it were given a century ago, due to the intervention of an exam), I verified a result called Green's Theorem for a rectangle with corners at (0,0), (a,0), (a,b), and (0,b).The verification amounted to some juggling with the Fundamental Theorem of Calculus. The result was the following:
_{Rectangular bdry}P(x,y)dx+Q(x,y)dy=_{Inside of rect}Q(x,y)/x-P(x,y)/y dA.

Silly example
I first tried a rather silly example which was something like the following. I think I took a=3 and b=2, and specified that P(x,y)=5y+Junk₁(x) and Q(x,y)=x²+Junk₁(y). Here the Junk functions were some absurd formulas: arctan(y³) or e^cos(x). I wanted to compute the line integral of P(x,y)dx+Q(x,y)dy around the boundary of the rectangle. Of course I used the version of Green's Theorem stated above, and the two Junk terms disappeared after the partial differentiations, and the double integral had integrand 2x-5, which is rather simple to compute over a rectangle.

Computation on a semicircle
This is another example created for the classroom, but it is more complex, and more resembles some real uses of Green's Theorem. Here I considered the line integral _Sy dx+x²dy where S represents the upper half of the circle of radius 3 centered at the origin. I think this integral certainly can be computed straightforwardly by a parameterization but I'd like to show another technique. First, what is the value of _Ly dx+x²dy where L is the line segment along the x-axis stretching from (-3,0) to (3,0)? Since y doesn't change on L, dy=0. Also notice that y=0 on all of L. Therefore y dx+x²dy is 0 on all of L.

So _Sy dx+x²dy=_Sy dx+x²dy+_Ly dx+x²dy=_S+Ly dx+x²dy, and the curve S+L, which is S followed by L, encloses the upper half of the area inside a circle of radius 3 centered at the origin: a semidisc. I'll call it Q.

Then a version of Green's Theorem allows us to replace _S+Ly dx+x²dy by _Q(x²)/x-y/y dA. The integrand is 2x-1, and we need to integrate it over Q.
The integral _Q2x dA is actually 0, because the region Q is symmetric with respect to the y-axis, and 2x takes equally positive and negative values on this region. So the values cancel each other. And the integral _Q-1 dA is minus one-half the area of a circle of radius 3. Therefore the original line integral, _Sy dx+x²dy, has the value -9Pi/2.

I would agree that this is an awkward, even ludicrous way of evaluating the line integral. I went through it because in a week we'll be trying computations like this in three dimensions and maybe things there won't be as clear.

The text's version of Green's Theorem
Suppose C is a positively oriented piecewise smooth simple closed curve, and D is the region bounded by C. Suppose that P(x,y) and Q(x,y) are functions with continuous first partial derivatives in D. Then:
_CP(x,y)dx+Q(x,y)dy=_DQ(x,y)/x-P(x,y)/y dA.

Definitions of terms
The statement above is complicated, and I think many of the words need some explanation.

• Closed curve
  We have encountered this phrase before. A closed curve is a parameterized curve which has an initial point (START) equal to its terminal point (END).
• Simple closed curve
  A simple closed curve is a curve with no self-intersections (crossings) except for the START=END.
  One additional remark, please: simple closed curves don't have to be so "simple" in the non-technical meaning of the word. I sketched something like what is shown to the right. It may not even be clear whether the point indicated in magenta (purple?) is inside or outside the curve. So things can really get complicated if you want to do curves more complicated than circles or rectangles.
• Positively oriented
  This is more subtle. The boundary curve C is parameterized. You can think of the parameter as time: it describes walking along the curve. And the phrase positively oriented here means that as the parameter describing C increases, the neighboring region D should be to the left. Finally, a victory for left-handed folks!
  Below are two "local" pictures of how D should look in relation to increasing values of the parameter on C. The arrows are pointing in the direction of increasing values of the parameter. Look back at the rectangle picture. You should see, I hope, that the boundary is a simple closed curve, which is positively oriented with respect to the interior. And look again at the semidisc (can't be a word!). The inside of the semidisc is to the left of the boundary as the parameter increases.
• Piecewise smooth
  This means that the functions parameterizing the boundary should be differentiable, or that they should be a finite "sum" of differentiable functions. For example, a rectangle is a "sum" of four smooth boundary parameterizations. Here the sum is understood to be in the sense of curves, where one curve followed by another is sometimes thought of as adding the curves.

Information transported from the boundary to the inside
Suppose P(x,y)=-(1/2)y and Q(x,y)=(1/2)x. Then Q(x,y)/x-P(x,y)/y just "happens" to be 1. And the double integral of 1 over the region is equal to its area. So apparently _C-(1/2)y dx+(1/2)x dy is the area of D. Let me try to make this even sharper. The curve C should be parameterized. So x=x(t) and y=y(t) for t between a and b. The line integral gets translated into the following Riemann integral:
_t=a^t=b-(1/2)y(t)x´(t)+(1/2)x(t)y´(t) dt.
So suppose you are in a race car, driving around a "closed course". Your position could be measured with a GPS device, so you would know <x(t),y(t)> (the position vector). You could also imagine that you have some idea of your velocity: <x´(t),y´(t)>. So from this "boundary data" you could compute (at least approximately) the Riemann integral _t=a^t=b-(1/2)y(t)x´(t)+(1/2)x(t)y´(t) dt and this must be the area of enclosed by the race track. This is astonishing to me.

People designed interesting mechanical linkages to compute areas based on such results. These instruments are called planimeters. People can be very clever.
Analogous results in higher dimensions would allow some knowledge of the heart based upon electrical readings on the skin. Weird, wonderful, terrific!

Another example, somewhat paradoxical
Here I will tell you what P(x,y) and Q(x,y) should be. These are certainly not randomly chosen but are specific functions which are used to model important physical phenomena.
P(x,y)=-y/(x²+y²) and Q(x,y)=x/(x²+y²)
It just happens (!!) that:
P/y=[-(x²+y²)+y(2y)]/(x²+y²)²=[y²-x²]/(x²+y²)² and Q/x=[(x²+y²)-2x(x)]/(x²+y²)²=[y²-x²]/(x²+y²)².
Therefore in this case, P_y=Q_x so that Q_x-P_y is 0.

Now let me compute the line integral _CP(x,y)dx+Q(x,y)dy, where C is, say, the circle of radius 3 centered at (0,0). The first parameterization you should consider is this: x=3cos(t) and y=3sin(t) so that dx=-3sin(t) dt and dy=3cos(t) dt. Then x²+y²=3² because sin²+cos²=1. Thus
_C-y/(x²+y²)dx+x/(x²+y²)dy becomes _t=0^t=2Pi([-3sin(t)/3²](-3sin(t))+[3cos(t)/3²]3cos(t))dt. If the arithmetic is done correctly we need to integrate 1 from 0 to 2Pi (almost everything cancels!) and the result is 2Pi.

But but but ... the integrand on the "other side" of Green's Theorem is Q_x-P_y and we computed that to be 0. A circle certainly is a simple closed curve, positively oriented, etc. We haven't made any errors. Maybe it is true that 2Pi=0? No! That is not true.

Technicalities matter.
I've ignored the requirements on P(x,y) and Q(x,y). Here the requirements definitely are relevant. Both P(x,y) and Q(x,y), and their first partial derivatives, are quotients of polynomials, and have x²+y² in the denominator (the bottom of the fraction). That expression is 0 at the origin, (0,0). In fact, P(x,y) and Q(x,y) and their first partial derivatives are not continuous at (0,0), which I will call the "bad point". This bad point is inside the circle. The example shows that even if the hypothesis "P(x,y) and Q(x,y) are functions with continuous first partial derivatives" fials at just one point of D, the equality predicted by Green's Theorem can fail very very emphatically (I think "zero equals something non-zero" is a fairly emphatic failure).

Information transported from one curve to another, using the inside
The example considered is, however, very important in practice. Let me show why. Let's consider another path, W, which is a quadrilateral. W is a simple closed curve made of four straight line segments. It starts at (8,0), goes to (2,6), then to (7,-1), then to (0,-5), and finally back to (8,0). I would like to compute _W-y/(x²+y²)dx+x/(x²+y²)dy. Here I am not so sure I could "easily" compute the integrals which would come out of parameterizing the four line segments, and changing the x,y language to t language. At least the task would be lengthy. So let me show you another way. Again, I can't just apply Green's Theorem to W and the region inside W, since W encloses the same bad point. But I can do something very clever using Green's Theorem.

Crosscut
Let me select a point A on W and a point B on C and join them by a line segment. Now I want to describe a closed curve in the plane, and this is very very clever. I'll go first from (8,0) on W to A. Then I will go from A to B on the crosscut. Then around C in reverse (note that the picture now has the arrowhead on C reversed and the label states "-C"). Then I go from B to A, and finally, I follow the remainder of W around to (8,0). What about the integral of P(x,y)dx+Q(x,y)dy along this path? The integrals on the crosscut line segments cancel, and the integral on -C is minus the integral along C, and the remainder is the integral on W. So the result is the integral along W minus the integral along C.

Now let me split up the crosscut into two distinct line segments, one directed from W to C and one directed backwards. The points A and B become doubled, and each pair is slightly separated. Now the closed curve is a simple closed curve. The region this simple closed curve bounds does not include the origin, the bad point. Inside the region, P_y=Q_x and therefore the double integral side of the Green's Theorem equation is 0. The line integral side is also 0. This will work for any separation, no matter how small. Now the line integral will vary quite continuously as the separation --> 0. And therefore, when the separation=0, the line integral will still be 0. And so, clearly the integral along W minus the integral along C is 0 (because the crosscuts cancel!) so the integral along W is the same as the integral along C: it must be 2Pi!
That clearly may be the worst clearly in the course. I don't think an accusation of excessive rigor (too much proving: "the quality of being logically valid") would be correct in Math 251, and the statement I am asserting as clear needs proof. What I am asserting is that if one path is close to another, and if P(x,y) and Q(x,y) are, say, at least continuous, then the work along the two paths will be close. To me this is physically reasonable. I hope it is to you.

This crosscutting technique is so familiar to people who use it that it is rarely discussed in detail. In fact, any curve which goes around the origin exactly once will have the line integral of our P(x,y)dx+Q(x,y)dy equal to 2Pi for similar reasons.

A physical model?
I think the computations I've just done are difficult to understand. I mentioned the following physical model to some students after class. Line integrals can also be thought of as measuring flux, the net flow through the curve. We will investigate this situation in R³ soon. But consider the following physical setup: two parallel planes of plastic, and a hose inserted through one of the planes, pumping water in at a steady rate (the blue things in the picture are not minnows, they are water drops!). If we surround this source with some sort of circular detection "fence" then we can look at how the water goes through the fence, and from that deduce the rate at which water is being pumped in. But we could also install another detection fence. If we do the same computations for that fence, the source rate should be the same, because the hose is pumping water in steadily: its rate is not varying. The mathematical computation with the crosscuts above is a version of this physical situation. When P_y-Q_x=0, away from the hose (at the origin) the source rate is 0. Any comprehensive measurement of flow surrounding the origin will get the same answer.
Another physical setup which this P and Q model is an isolated point charge. There the work done in moving around the charge will always be the same. We will encounter this in three dimensions soon, and maybe there it will be easier to understand.

Another application (important!) of Green's Theorem
I did not get a chance to discuss this in class. When is a vector field P(x,y)i+Q(x,y)j conservative? Certainly when it is a gradient vector field: so there is a potential function, f(x,y), with f/x=P(x,y) and f/y=Q(x,y). We can try to find f(x,y) by integrating and matching the descriptions. But this can be difficult and lengthy. Differentiation is easier, and certainly if there is a potential, f(x,y), then P/y=Q/x by Clairaut's Theorem. But the example above (the isolated electric charge or the fluid flow point source) shows that we can have P/y=Q/x without having a potential. Why? If we had a potential then the integral around closed curves would be 0, but we saw that such an integral for out example may not be 0. If we had no bad points, then the integral around closed curves would be 0 using Green's Theorem. So any region where there are no bad points has the converse true:
If P/y=Q/x and there are no "bad points" for P or Q, then there is a potential.
This can sometimes be very useful because differentiation is ... easier! So people have given a name to regions which have no holes. So: a region is simply connected if it has no holes. In such a region, if the condition P/y=Q/x is correct at all points, then there is a potential, so the vector field is conservative. A disc (the inside of a circle) is simply connected. But if you remove a point, the result is not simply connected.

Why?
Take two points, A and B, in the plane and suppose that C₁ and C₂ are two paths going from A to B. If the line integrals along C₁ and C₂ are equal, then if you first integrate along C₁ and then backwards along C₂ (this is usually called, not surprisingly, -C₂) the net result will be 0. So the integral along any closed curve will be 0. On the other hand, if you know the integral around all closed curves is always 0, then you can compare integrals around two paths, say C₁ and C₂, from A to B by just looking at the work done around C₁-C₂. Since, by assumption, the integrals around closed curves are 0, the integral along C₁ must exactly be canceled by the integral along -C₂. Therefore the integral along C₁ is the same as the integral along C₂: this is path independence.

More restatement and summary
We showed that a gradient vector field is actually path independent. Even further, if we happen to know a potential, that is, a function whose gradient is that vector field, then the integral from START to END will just be the difference: the value of the potential at the END minus the value of the potential at the START.

How to answer the previous QotD
So V is (x³-Ax²y-y³)i+(Bxy²-2x³)j. We're told that there are values of the constants A and B which make V a gradient vector field. What are these constants? Here is one approach, which also "creates" the potential function. The reason to use this approach is that (looking ahead in the problem) we'll need to use the potential function to evaluate a line integral. Suppose f(x,y) is a potential function for V. Then:
f/x=x³-Ax²y-y³
f/x=Bxy²-2x³

Let's integrate the first equation with respect to x (I put that in bold type because it is easy to get confused about what is variable and what is constant -- here x is the variable, and A and y are constants):
f(x,y)=(1/4)x⁴-(A/3)x³y-xy³+C₁(y). The function C₁(y) is an antidifferentiation "constant". It could contain any function of y (including just ordinary constants).

The second equation, integrated with respect to y, gives:
f(x,y)=(B/3)xy³-2x³y+C₂(x). The function C₂(x) is similar to the previous "constant": it can contain any function of x.

Now we have two descriptions of the potential, f(x,y):
      Description #1 (1/4)x⁴-(A/3)x³y-xy³+C₁(y)
      Description #1 (B/3)xy³-2x³y+C₂(x)
We can compare and cross-check. Everything should appear exactly once and only once in f(x,y). So the (1/4)x⁴ appearing in #1 must be somewhere in #2: ahhh, it is hiding inside C₂(x). The -(A/3)x³y in #1 must be -2x³y in #2, and therefore -A/3=-2 so that A=6. The -xy³ in #1 is (B/3)xy³ in #2, and we infer that -1=B/3 so that B=-3. There's nothing corresponding to C₁(y) in #2, so I might as well (to make my life simpler!) take C₁(y) to be 0. If I were working on an exam (or any time, actually), I think I would add suspenders to my belt and check the other way: that is, look at all the terms in #2 and make sure I can find them in #1. That is indeed true here. I combine all the information about my description and tell you that a potential for V is f(x,y)=(1/4)x⁴-2x³y-xy³.

The second part of the QotD told us that C was a curve starting at (0,2) and ending at (3,1). We were requested to find the line integral _CV·T ds. We have a potential, so this is just f(END)-f(START)=f(3,1)-f(0,2)=(1/4)3⁴-2(3³)1-3(13)-0 (all the terms of f(x,y) have an x in them and x=0 at the START).
Comment If this problem occurred on an exam I gave, I would want students to leave the answer as it is. Don't touch it -- you might break it. If you break it, then you "buy" it, and I should reduce the score.

Another way to find A and B
If we just wanted to find A and B, well, differentiation is almost always easier than antidifferentiation (that doesn't occur here, with polynomials). We can use Clairaut's Theorem (equality of mixed partial derivatives) to find A and B. Since
f/x=x³-Ax²y-y³
f/x=Bxy²-2x³
we know that (/y)f/x=(/x)f/y and therefore (/y)(x³-Ax²y-y³) must equal (/x)(Bxy²-2x³), so that -Ax²-3y²=By²-6x². Again we compare these two descriptions of polynomials. If these descriptions are to be the same, then -A=-6 and -3=B. This leads to the same values of A and B as we got before. I didn't use this method, which is frequently quicker than integrating, because I believe the person who wrote the problem: that there would be values of A and B making this vector field conservative, and that I would need the potential to complete the problem. If I just done the differentiations here, then I still would have had to integrate to get a potential.

Another way to find a potential
Sometimes in physics and engineering applications there are reasons to "prefer" one point in the plane to another. This sounds awfully silly, but ... it does occur. When this happens, "construction" of a potential can be done by paying attention to this specified point. I'll call the specified point, "the ground state". Here in this example I'll use (1,-2) as the ground state. This is certainly artificial, but the choice will illustrate the method.

If someone gives me a vector field which is guaranteed path independent, then I might want to believe the potential could be revealed by making measurements relative to the ground state. I might take paths from (1,-2) to (x,y) and call the work done f(x,y). Since the vector field is supposed to be path independent, the choice of path won't matter. I don't think I would take the silly curvy path shown. That's hard to work with. Indeed, as I mentioned in class, almost all of the paths you're likely to see in "real life" will be line segments and arcs of circles. I'd probably take the path shown, a line segment from (1,-2) to (1,y) and then a line segment from (1,y) to (x,y). An alternative suggested by students is the direct trip: from (1,-2) to (x,y) and that is certainly used some of the time, but I am lazy (I don't want to compute the darn tilted line!).

I will write this as a sum of two integrals:
f(x,y)=_{Vertical segment}(x³-6x²y-y³)dx+(-3xy²-2x³)dy  + _{Horizontal segment}(x³-6x²y-y³)dx+(-3xy²-2x³)dy

What's going on?
Why shouldn't the integral over closed curves always be equal to 0?

A computation

A realization

Example

Green

z/x
Use the chain rule. The result is f´(5cos(x)+3e^y)(-5sin(x)).

z/y
Use the chain rule. The result is f´(5cos(x)+3e^y)(3e^y).

²z/x²
Use the chain rule and the product rule beginning with the very first result about z: differentiate z_x with respect to x.
So we get f´´(5cos(x)+3e^y)(-5sin(x))²+f'(5cos(x)+3e^y)(-5cos(x)).

A more difficult example might be f(7cos(x)e^2y).

A second example
I think I then did an almost useful two variable example. Suppose that z=f(x,y), where f is a differentiable function of two variables, and x=r cos() and y=r sin().

z/r
This is (f/x)·(x/r)+(f/y)·(y/r)= (f/x)cos()+(f/y)sin()

z/

A line integral
Let's look at the curve x(t)=-cos(t) y(t)=sin(t)+4cos(t) for t in the interval [-Pi/2,Pi]. This is the curve C.
Let's look at the "force field" (vector field) F=3i-xj.

To the right is a picture of the curve. It goes from the START at (0,1) (plug in t=-Pi/2 in the equations for x(t) and y(t)) to END at (-1,4). It seems to be and is three-quarters of a tilted ellipse.

The work integral _CF·T ds which I immediately translate to _C3 dx-x dy. Since we are given a parameterized C, I retranslate to t-land:
     x(t)=-cos(t) leads to dx=sin(t)dt
     y(t)=sin(t)+4cos(t) leads to dy=[cos(t)-4sin(t)]dt
The dictionary gets us _{t=-Pi/2 [START]}^t=Pi [END](-3cos(t)-{-cos(t)}[cos(t)-4sin(t)])dt and we can multiply etc.:
_t=-Pi/2^t=Pi(-3cos(t)+cos(t)²-4cos(t)sin(t))dt

Another line integral
The path I wanted here was a line segment from (0,1) to (-1,4). This path is shown to the right. In vector notation, we want (1-t)<0,1>+t<-1,4>. This takes us from t=0 at (0,1) to (-1,4) at t=1.

Understanding what we did
We showed that work done by the same vector field between the same pair of points can depend on the path traveled. This is actually realistic. If you push a marble from point A to point B, then the local conditions of the path (friction, slipperiness, etc.) can matter a great deal.
Computationally what we did was really fairly straightforward. We took the work, turned it into a P dx+Q dy integral, then parameterized and turned everything into t-land. And we computed.
Conceptually I would like to try to understand why there is a difference in the work done. I would like to understand when the work is the same.

Abstract version of the computation

What if ...
the vector field involved was a gradient vector field?

Consequences, prose version
A conservative vector field is one where the work done along paths depends only on the start and end points. Such a vector field is also called path independent. This condition is the same as asking that the integrals along closed curves (where the starting point and ending points agree) must always be zero.

If a vector field is a gradient vector field so that it is the gradient of a function called a potential of that vector field, then that vector field is conservative. The work done along a curve is equal to the value of the potential at the end of the curve minus the value of the potential at the start of the curve.

Gravitation
For example, the inverse square law we introduced is conservative because it is a gradient vector field. We can verify this by exhibiting a potential, and the validity of being a potential can be checked with two derivative computations. Then all work integrals will be "easy".

Vocabulary
Conservative Closed curves Gradient Potential

The earlier example
is not a gradient vector field. Reason #1: it is not conservative Reason #2: try to create a potential

An exam question as the QotD

HOMEWORK
Prepare the Maple assignment. Read and do the syllabus problems for 16.3. Look at the review questions and select one or two at random and write out solutions. If I don't have the solutions posted already, then send them to me, please. I will check them and post them, thank you then.

Maple
If you did not get information about the third Maple assignment, please send me e-mail immediately. Messages were sent out by midday Saturday. This assignment is somewhat longer than the previous two assignments and likely will need more time.

Exam
There will be an exam on Friday, April 7. A draft formula sheet has been prepared. Please look at it. I don't want any further mistakes to occur, and you could also advise me if you think something important is not there. Some review problems have also been prepared. I hope that students will write solutions to some of these problems. These solutions can be sent to me. I will proofread them, and post them. Everyone can benefit from this. There are a number of review problems, and they probably can't all be done the night before the exam! The sheet of review problems also contains information about what will be tested by the exam.

Irrelevance awards
I asked in the last diary entry for the identity of the individual pierced by arrows and for the painter. The answers are, respectively, St. Sebastian and Andreas Mantegna (c. 1431-1506). St. Sebastian was martyred c.288 at Rome. I found the following information which I did not know before:

During the 14th century, the random nature of infection with the Black Death caused people to liken the plague to their villages being shot by an army of nature's archers. In desparation they prayed for the intercession of a saint associated with archers, and Saint Sebastian became associated with the plague.

Four students responded from the class responded with identifications, and these four students have received professionally selected and packaged awards. Their names are
Demetrio Koloseus-Ganon and Ralitza Varlakova and Christopher Martin and Chirag Walawalkar.
I thank them for their effort and interest.
Look at these names. Isn't Rutgers wonderfully diverse?

Vector fields

• Geometrically, a vector field "is" a collection of arrows in the plane.
• Algebraically, a vector field "is" a pair of functions, f(x,y)i+g(x,y)j.

The simplest ways students discover vector fields in their studies is:
Force fields A typical example is the inverse square "law" analyzed last time.
Fluid flow A vortex is one example. Explicitly, it was -yi+xj. This sort of describes a swirling flow around the origin.
Gradient vector fields We haven't discussed this so far, and these are an important type of vector fields.

Gradient vector fields
This could be in either two or three variables. Let's suppose for simplicity we have a function F(x,y) of two variables. Then the gradient of F is F=[F/x]i+[F/y]j is a vector field: it is a function multiplied by i plus a function multiplied by j.

An example
I looked at F(x,y)=x²+y². Here I recalled that the contour lines were x²+y²=Constant. These are all circles centered at the origin. If we draw a collection of contour lines for equally spaced constants (for example, for 1 and 2 and 3 and 4 and ...) then it turns out that the lines get closer and closer together as the equally spaced constants increase. This is because the function is increasing more rapidly as x and y get larger in absolute value. The associated gradient vector field is 2xi+2yj. At (x,y), these vectors are perpendicular to the circles and point away from the origin in the direction of increase of F. As (x,y) "moves" farther from (0,0), the magnitude of the vectors increases.
To the right is a picture of this gradient vector field together with some contour lines. Previously I had defined F:=x^2+y^2;. The picture displays the output of these Maple commands:
> contourplot(F,x=-2.5..2.5,y=-2.5..2.5, thickness=2, contours=[1,2,3,4,5], grid=[40,40]);
> fieldplot(2*x,2*y],x=-2.15..2.15,y=-2.15..2.15,arrows=slim,grid=[10,10], fieldstrength=maximal(.9)color=blue,thickness=2);

Another example
This example uses a function which is less symmetric and less simple. The function is x²y-2x+y²+3y. The level curves are shown for the values -12, -10, -8, -6, -4, -2, 0, 2, 4, 6, 8, and 10: the even integers from -12 to 12. Notice that the arrows seem to be perpendicular to the level curves, and that the arrows are longer when the spacing between the contour lines (corresponding to equal differences in the constants) gets closer.

Comment
I looked at this picture and wondered what the heck happened where the arrows got really small. What does happen? Well, we can look for a critical point: F_x=2xy-2=0 and F_y=x²+3y+3=0. The first equation tells me that y=1/x and then the second equation becomes x³+3x+3=0, and I can't "solve" that. So I asked fsolve and was told there was one critical point, at (approximately) x=-.596 and y=1.678: this looks correct. What kind of critical point? (Max/min/saddle) If you look at the picture and the level curves you can probably guess. I actually computed the Hessian, and it is a saddle. Ain't math fun?

In general ...
What was written above is generally true about a gradient vector field and the contour lines or curves (contour surfaces in R³). That is, the gradient vector field "arrows" are perpendicular to the level curves. They "point" towards increasing values of the function. Their magnitude (the length of the arrows) is a measure of the density (?) of the contour lines: if the function is rapidly changing then the length will be longer.
By the way, if F=V, then F is called a potential of the vector field, V. The word and concepts connected with it are important in physics. I'll discuss more about this during the next class.

Is the vortex flow a gradient vector field?
Look at -yi+xj. Is this vector field a gradient vector field? If it is, can we find a potential for it? If it is not, can we explain why? This turns out to be a serious and interesting question, because of things we will learn very soon (Friday). Also it has some interesting physical consequences. Well, here is one way to decide the answer.
Suppose F(x,y) is a potential for -yi+xj. That means

      /y 
Fx=-y -----> Fxy=-1
      /x 
Fy=x  -----> Fyx=1

Line integrals
We need an additional technical tool to investigate vector fields: line integrals. Line integrals will allow the computation of work for force fields, and flux for fluid flow, and something for gradient vector fields. Work and flux and other things are important physical quantities and turn out to have nice mathematical properties. So here we go.

Mass of a wire
I'll introduce line integrals using the metaphor of the mass of a wire. The integral calculus mantra is: chop up, approximate, sum, limit.
Here I'll assume that there is a wire sitting in R². Its geometry is simple, with a constant cross-sectional area (assumed to be 1 in the measurement system used). The density of the wire will vary according to the position on the wire: at some points the density will be high, and the wire heavy, and at other points, the density will be low, and the wire rather light. The task is to create some technical tool which will represent the total mass of the wire. By the way, the accompanying picture looks like a particularly ugly worm or snake. I am sorry.

Suppose I cut the wire up into lots of little chunks. How little? Well, the length of each chunk will be ds, a tiny piece of arc length (we discussed this in several lectures given late January, long, long ago). If I assume that the density D(x,y) varies only a small amount because the chunk of the wire is very small, then dm, the mass of this piece, is nearly D(x,y) ds (remember I made the cross-sectional area equal to 1). Further, I can add up these pieces of mass to get the total mass. I can take the limit as the number of pieces gets large and the length of the pieces gets small. The result is that the mass of the wire should be given by _The wireD(x,y) ds.
This integral is called a line integral, where the word line doesn't mean "straight line" but is used in the sense "A thin continuous mark, as that made by a pen, pencil, or brush applied to a surface." Now I need to show you what this means. So I will compute a very artificial example.

Silly example
My example was the following: the wire follows the part of the circle of radius 3 which is in the first quadrant. The density of the wire at the point (x,y) is 7y+5. Find the mass of the wire. Well, I will take the integral _The wireD(x,y) ds and parameterize everything in sight. To me the natural parameterization of an arc of a circle uses essentially the angle from the center. The text likes to use t here, so I will parameterize this quarter circle with x=3cos(t) and y=3cos(t). The interval of this parameterization is [0,Pi/2]. Now ds is sqrt([dx/dt]²+[dy/dt]²) dt. The square root stuff is the magnitude of the velocity vector, the speed. And ds=SPEED·dt is a translation of distance=rate·time of course. In this case, dx/dt=-3sin(t) and dy/dt=3cos(t) so that sqrt([dx/dt]²+[dy/dt]²) just happens (!!!) to simplify to 3. And ds=3 dt. What about the density? Since D(x,y)=-7y+5, we know that the density is 7(3sin(t))+5. And the integral should go from 0 to Pi/2. Therefore
The mass=_The wireD(x,y) ds=_t=0^t=Pi/2{21sin(t)+5}3 dt=-63cos(t)+15t]_t=0^t=Pi/2=63+(15/2)Pi.

This is a totally insignificant, physically unrealistic (to me) computation. The only thing I had fun with is drawing the picture, which with its varying colors is supposed to suggest the increasing density of the wire as y increases. There is one very important fact which should be mentioned now:

Independence of parameterization
In this line integral and in all line integrals, the result will be the same no matter which parameterization is used. Verification of this uses the one variable chain rule and is not too interesting right now. So let me go on.

Random example
I then featured a really random example. I asked students for some random (positive) integers which I then used to construct an example very much like the one presented in what follows. The previous example was arranged so that ds was nice. I now tried to compute something like the mass of a wire where the wire was sitting on the curve y=x⁴ from, say, (0,0) to (2, 16), and the density is D(x,y)=x⁴y⁷. I not being totally idiotic here. Any computational strategy in calculus should be able to handle low-degree polynomials fairly efficiently, even by hand. So here's what we did.
A simple parameterization of a graph of a function is just to use the independent variable as the parameter. So I used x=t and y=t⁴ and then the density x⁴y⁷ became t⁴(t⁴)⁷=t²⁸. This isn't too bad, but I am saving the worst for last, and in this computation the worst is ds. So dx/dt=1 and dy/dt=4t³, and therefore ds/dt=sqrt(1+16t⁶). The line integral for this "mass" translates in t-land as follows:
_The wireD(x,y) ds=_t=0^t=2t²⁸sqrt(1+16t⁶)dt. This integral cannot be computed exactly in terms of standard "familiar" functions. I found to my amazement that Maple does have a storehouse of weird functions which it can use to "evaluate" this integral. But then I have no feeling for the functions it used and certainly not for the values of the functions.

ds is the obstacle
What is horrible about most line integrals is ds. Almost no ds except for those included in textbook problems lead to familiar antiderivatives. At the same time, important quantities such as work and flux are defined as mathematical objects in terms of line integrals, and the general expectation is that these quantities can be computed exactly in terms of familiar functions. To me this is an excellent example of the psychological phenomenon known as cognitive dissonance:

Cognitive dissonance arises from conflicting cognitions. Cognitive dissonance is the perception of incompatibility between two cognitions, which for the purpose of cognitive dissonance theory can be defined as any element of knowledge, attitude, emotion, belief or value, as well as a goal, plan, or an interest. In brief, the theory of cognitive dissonance holds that contradicting cognitions serve as a driving force that compels the mind to acquire or invent new thoughts or beliefs, or to modify existing beliefs, so as to minimize the amount of dissonance (conflict) between cognitions.

Work
Let's consider the physical concept called work. This is "force" times "distance", loosely, but we're going to need to be a bit more specific. For example, consider a mass being pushed up a frictionless triangle. If the triangle is steeper, then more force is needed. In the picture, the force of gravity is directed down. The blue arrows on each inclined plane (triangle) show the force component needed to move the box up that triangle.
So instead of dieting, avoid steep triangles and your weight will decrease ... no no no ... this is just more bad information from the lecturer!
Anyway, what really matters is the component of the force in the direction of motion (we could take the dot product of the force and the displacement also).

Work with a varying vector field along a curve
Again, chop up, approximate, sum, limit.
I would like to describe how to compute the work done against a varying force field F as we travel from p to q (in R² but the same ideas will work in R³, also) along a curve, C. We chop up C into small pieces (the red lines are the chopping places). One of them is displayed under a "magnifying glass" (sort of) in the picture. The piece is so small that it is almost a straight line, and its length is ds. The piece is also so small that the vector field, F, is almost constant near the piece. Remember that a unit tangent vector, pointing in the direction of the piece, is called T (go back and think of January!). Then the part of the force field along the curve is F·T and the piece of the work will be approximately F·Tds. All of the work will be the integral along C of this quantity. Therefore the work is _CF·Tds.

A computation
Let us "test" this definition with a random computation: well, sort of "random". C will be the curve y=x⁴ and p will be the point (0,0) and q will be the point (2,16). I will have the force field be x²y³i+xy⁵j. Now let us compute. We need to change everything into t-land, where I will choose a most routine parameter: x=t so that y=t⁴.

ds
As before, the speed becomes
sqrt([dx/dt]²+[dy/dt]²)=sqrt([1]²+[4t³]²)=sqrt(1+16t⁶) which is horrible enough. Thus ds=sqrt(1+16t⁶)dt.

F
At the point (x,y), F is x²y³i+xy⁵j so that at (t,t⁴) on the curve, F is t²(t⁴)³i+t(t⁴)⁵j=t¹⁴i+t²¹j.

T
Now comes almost the miracle. In the movie Shakespeare in Love, one character states, "The natural condition is one of insurmountable obstacles on the road to imminent disaster." He then says almost immediately, "Strangely enough , it all turns out well." When asked "How", the character replies, "I don't know. It's a mystery." So, if not a miracle, let me show you the little mystery here.
The unit tangent vector, T, is a vector in the direction of the curve. The position vector of the curve is <t,t⁴> so that the velocity vector is <1,4t³>. But we need a unit vector to get the projection of F in the direction of the curve. Divide by the magnitude of <1,4t³>. Therefore, T=<1,4t³>/sqrt(1+16t⁶).

Assembling the work integral
_CF·Tds=_t=0^t=2(t¹⁴i+t²¹j)· (<1,4t³>/sqrt(1+16t⁶))sqrt(1+16t⁶)dt. This is _t=0^t=2t¹⁴+4t²⁴ dt. which with totally routine polynomial calculations can be evaluated: it is 2¹⁵/15+(4/25)2²⁵.

What happens?
The speed comes in to squeeze down the velocity vector to get the unit tangent vector. The speed comes in as the factor which multiplies dt to get ds. The two appearances of the speed cancel. They will always cancel!.

Using the notation to help
Here is how people use notation to guide their way through the computation. No one computes the speed (the square root stuff) when computing work because it will cancel. So:
Problem statement
Compute the work if F(x,y)=x²y³i+xy⁵j and the curve is y=x⁴ going from (0,0) to (2,16).
A solution
So I initially write _CF·Tds but then I immediately change to _Cx²y³dx+xy⁵dy. I evaluate this line integral again by changing everything to t-land. So x=t and y=t⁴ and dx=1dt and dy=4t³dt, Also, x²y³=t²(t⁴)³=t¹⁴ and xy⁵=t(t⁴)⁵=t²¹. Therefore x²y³dx+xy⁵dy=t¹⁴dt+t²¹4t³dt.
Therefore _Cx²y³dx+xy⁵dy= _t=0 [START]^t=2 [END] t¹⁴+4t²⁴dt. And the result will be the same. Almost everyone uses this notation, and never bothers with computing T and ds and then canceling, etc. Of course, the ideas are important: the physical quantity we are computing has certain properties (wait until Friday!) which make this computation extremely interesting.

QotD
What is the work done if F is the vector field x²i+yj is and the curve is x=t and y=t³-t as the parameter goes from t=0 to t=1? The answer turns out to be 1/3.

Caution!!!
I just read students' answers to the QotD. Many people did this problem completely correctly. Many others clearly had the ideas correct but made one or two or three algebra or arithmetic errors. This sort of thing can really hurt you on exams!

The word iatrogenic refers to illnesses or other problems "caused by medical examination or treatment." Please don't let your solutions make the problems worse!

HOMEWORK
Keep reading chapter 16. I will discuss conservative vector fields, etc. (16.3) on Friday. Please be there.

Maple?
I hope that either today or tomorrow information about the third Maple lab will be sent to students. This lab will be worth 20 points (!). Please watch for e-mail.

Spherical coordinates
I drew the same picture I showed last time. I deduced the following formulas (with errors but they were corrected):
      x=(rho)sin(phi)cos()
      y=(rho)sin(phi)cos()
      z=(rho)cos(phi)
I remarked that it was useful to know that such formulas existed, but that I had rarely used them. One result that I have used frequently comes from the fact that rho represents the distance from (x,y,z) to the origin.
      x²+y²+z²=rho².

Standard restrictions on spherical coordinates
Because the angles sort of fold over when Pi's and 2Pi's are added, most people who use spherical coordinates put some restrictions on how big/small and phi can be. If we only allow to be between 0 and 2Pi and only allow phi to be between 0 and Pi, then there will be unique spherical coordinates for every point in R³. So I will generally work with these restrictions.

Some shapes in spherical coordinates

rho=constant gives a sphere centered at the origin. So, for example, rho=5 is a sphere centered at the origin of radius 5.	phi=constant gives a right circular cone whose axis of symmetry is the z-axis. For example, phi=Pi/6 is a cone with vertex at the origin and whose axis of symmetry is the positive z-axis. The angle between the positive z-axis and any of the cone's "generators" (lines from the vertex on the surface of the cone) iw Pi/6 (yes, 30o). The bottom half of the cone is not included because that is where phi is between Pi/2 and Pi.	=constant gives a half-plane, with the z-axis being the edge of the half-plane. For example, =Pi/4 gives a half-plane which is perpendicular to the half-line y=x (x>0) in the xy-plane. The other half of the plane is where is 3Pi/2, and so it is not included in this obhect.

Integral #1
A spherical region of radius R is filled with material whose density is directly proportional to the distance from the origin. What is its mass?
This is not very realistic. The center is light and fluffy and the outer edge is heavy and tough (my kind of cooking?). The density is supposed to interpolate linearly between these extremes. Maybe the appropriate assignment would be to build an object of this type.

The math setup
Take a small piece of volume, dV, in the sphere. The corresponding piece of mass, dm, is related to dV by dm=(density)dV. We know that the "density is directly proportional to the distance from the origin." Place the origin of the coordinate system at the center of the sphere. So there is some constant C>0 so density=C rho. And the total mass is the sum of the dm's. This "sum" should be a triple integral:
Total mass=_{The whole ball}C rho dV.
In spherical coordinates, a description of a sphere of radius R centered at the origin is easy: rho goes from 0 to R, goes from 0 to 2Pi, and phi goes from 0 to Pi. We just use the agreed upon ranges for the angles to sweep out a whole sphere. There is one sticky point, however.

dV in spherical coordinates
We need to convert dV to spherical coordinates. In fact,
      dV=(rho)²sin(phi)d(rho)dd(phi)
I know this is true. First, there is a diagram which is supposed to be convincing in the text (figure 8, page 1035). Second, I said it in class. Third, I actually can give an understandable argument if there is enough time later in the course. In any case, when I use spherical coordinates, I almost never bother thinking about this weird mess, but I just write it.
Vocabulary Several students commented that the "weird mess" is called the Jacobian, which is how area or volume is distorted with a change of coordinates. Again, if there is time I'll discuss Jacobians later in the course (see 15.9 for more information right now, if you'd like).

The computation
So we have
Total mass=_phi=0^phi=Pi₌₀^=2Pi_rho=0^rho=R(C rho)(rho)²sin(phi)d(rho)dd(phi).
The inner integral
_rho=0^rho=R(C rho)(rho)²sin(phi)d(rho)=_rho=0^rho=RC(rho)³sin(phi)d(rho)=Csin(phi)(rho)⁴/4]_rho=0^rho=R=Csin(phi)R⁴/4.
The middle integral
₌₀^=2PiCsin(phi)R⁴/4 d=(2Pi C)sin(phi)R⁴/4=[(Pi C)/2]sin(phi)R⁴. (Just multiply by 2Pi, since there is no in the integrand.)
The outer integral
_phi=0^phi=Pi[(Pi C)/2]sin(phi)R⁴d(phi)=-[(Pi C)/2]cos(phi)R⁴]_phi=0^phi=Pi=(Pi C)R⁴.
I don't know any way to check this answer. Build a model? Weigh it?

Is this silly?
Well, yes, it is silly. The problem is invented and certainly designed exactly for spherical coordinates. But I would not use spherical coordinates, which definitely have peculiarities (look at the pictures above and look at the expression for dV) unless both the region and the integrand can both be described in a nice fashion with spherical coordinates. I won't use this coordinate system otherwise. (Could you imagine using spherical coordinates to describe a cube?)

Integral #2
Consider the region in the first octant consisting of points whose distance to the origin is at least 1. Imagine that this is filled with material whose density is inversely proportional to the fifth power of the distance to the origin. What is the mass of this object?

Translating
All of R³ is divided into eight parts by the coordinate planes: x=0, y=0, and z=0. Each part is called an octant. While the corresponding regions in the plane (the quadrants) have individual designations, the only octant that is named is the first: the octant where x>0 and y>0 and z>0. In this first octant, I'm excluding points whose distance to the origin is less than 1. What does the remaining region look like? I think the best picture I drew in class was one looking at the octant from the back, so here is a version of that picture. In this picture (sort of the corner of a rectangular box), a spherical "bite" has been taken out of the corner. The bite is centered at the vertex (the origin) and has radius 1. Wow!

To the right is a more oblique view of the octant with the bite. The nice thing about this region is that it can be described very briefly in terms of spherical coordinates. Certainly, rho will go from 1 (as close to the origin as the bite will let us get) out to ... out to ... infinity (an improper integral!). What about and phi? Here students should look closely at the definitions of and phi. Each of them will go from 0 to Pi/2. This is best confirmed by taking "angles" with vertex at the origin and a side along the x- (respectively, z-axis) and then opening the second side of the angle to an aperture of Pi/2 (I think "aperture" means the angle's opening).
By the way, as I remarked in class, this is actually a more realistic example than the first integral. Things like this do occur in electricity and magnetism.

The computation
Again dm=(density)dV=[C/(rho)⁵]dV because "density is inversely proportional to the fifth power of the distance to the origin." And we know the limits from the discussion above, so the total mass is
_phi=1^phi=Pi/2₌₀^=Pi/2_rho=1^rho=infinity[C/(rho)⁵](rho)²sin(phi)d(rho)dd(phi).
The integrand is [C/(rho)³]sin(phi) after cancelling some powers.
The inner integral
This is an improper integral, so I will be careful.
_rho=1^rho=BIGC/(rho)³sin(phi)d(rho)=-C/(2(rho)²)sin(phi)]_rho=1^rho=BIG=-C/(2(BIG)²)sin(phi) +C/(2(1)²)sin(phi). As BIG-->infinity, the term -C/(2(BIG)²)sin(phi)-->0 so the improper integral _rho=1^rho=infinityC/(rho)³sin(phi)d(rho) converges and its value is (C/2)sin(phi).
The middle integral
₌₀^=Pi/2(C/2)sin(phi)d=(C/2)sin(phi)(Pi/2)=[(C Pi)/4]sin(phi).
The outer integral
_phi=0^phi=Pi/2[(C Pi)/4]sin(phi)d(phi)=-[(C Pi)/4]cos(phi)]_phi=0^phi=Pi/2=[(C Pi)/4]
Again, I will admit that I don't know any way to check this answer. When such an integral comes from a real physical problem, there is frequently some way to see if the final answer is reasonable.

Further defense of silly (the same defense)
I would only use this technique, I hope!, where both the region and the integrand are suitable. So, although the problems may have seemed silly, they are the sort of applications which might occur. We will need integration in spherical coordinates a few times later in the course.

Where are we? Where are we going?
About 75% of the course has gone by. Much of what we've done will be very useful for many of the students, but, perhaps, some of what we've done will be neeeded less often (I am trying to be "diplomatic"). The last portion of the course, about what is frequently called Vector Calculus, can help students understand many complicated "things" they will likely encounter.

For example, consider a big "chunk" of "stuff". Look at the heat in the chunk. Sometimes some of the heat might originate inside the object, and sometimes there might be cooling. And there are also boundary effects. If there's something really hot near part of the boundary, heat will flow in around that part, and, similarly, there might be cool things (?) near some other part of the boundary, and heat will flow out that part of the boundary. This is all very complicated. Somehow, there should be some way of balancing the heat sources and sinks inside the region, and the boundary flow of heat, and expressing this in some neat way. We'll try to understand this.

The Fundamental Theorem of Calculus, one of the wonderful results of calc 1, is something like this:
If F´=f, then _x=a^x=bf(x) dx=F(b)-F(a)
This is a wonderful result. Think about it. It states that some sort of stuff inside the interval [a,b] (we could think about height times width, f(x)dx, or some accumulation of "stuff": f(x) inside [a,b] and added up over the whole interval, [a,b]) is equal to an edge quantity. There isn't much "edge" to an interval. Here the edge quantity is F(b)-F(a). You can, with some effort, see things as some sort of balance between interior accumulation and stuff in and out the edges. The remainder of the course is an effort to generalize this to dimensions 2 (Green's Theorem) and 3 (Divergence Theorem) and dimension (sort of) 2.5: Stokes' Theorem and Gauss's Theorem. These results provide a language for considering edge effects compared to inside effects for things like heat flow and fluid flow and ... lots of other phenomena. So please listen and look carefully. We still need to learn a few more vocabulary notions.

Vector fields
I'm going to begin with two dimensional vector fields, since the pictures are easier to draw. Those of you who have seen Euler's method in calculus will remember bunches of arrows in the plane. So in the most elementary sense, a vector field is such a "bunch of arrows". In fact, it is an arrow, a vector, sitting at every point, (x,y).

To the right is a "picture" of part of a vector field. The picture was produced by Maple using a command I learned only recently (there are thousands of commands, and I believe I know less than a tenth of one percent of them). This command is in the plots package. The picture was produced by
fieldplot([x^2+10*y,3*x-y^2], x=1..5,y=1..5, arrows=slim, grid=[10,10], fieldstrength=maximal(.9), thickness=2);
This vector field is the vector (x²+10y)i+(3x-y²)j at the point (x,y). The Maple command sketches a few of the arrows ([10,10] gives 100 of them). Sometimes such a picture of a vector field can be very helpful.

Models?
Vector fields are mathematical models of some basic physical phenomena. Two that are immediate are force fields and fluid flows.
The force field (gravity, electromagnetism, etc.) might be the simplest. The "arrow" at every point denotes the presence of a "force" which can act on the proper kind of object (gravitation: an object with mass, magnetism, an object with magnetic "stuff", etc.). In Newton's law of gravitation, we could think of a powerful mass, M, (the sun?) at the center of the universe, and other very small masses, m's, scattered around. Ignore interactions between the m's. Each m will be attracted to the M with a force of magnitude GmM/(dist)² where "dist" is the distance from m to M. The direction of the attraction is from m to M. To get a vector field, just sketch an arrow in the direction of the force, with a length proportional to the magnitude of the force. Then to complete the creation (?) of the force field, just remove all the little m's and leave the arrows. This is quite an idea, really a big leap of imagination. I haven't seen any of these arrows, but imagining them is sometimes quite helpful. A quote from Albert Einstein describing radio may be appropriate here:

You see, wire telegraph is a kind of a very, very long cat. You pull his tail in New York and his head is meowing in Los Angeles. Do you understand this? And radio operates exactly the same way: you send signals here, they receive them there. The only difference is that there is no cat.

More detail about the inverse square law Suppose we stand at (x,y) with a mass m, and the origin has a mass, M. Then the gravitational attraction is a force of magnitude GmM/(x2+y2). What's on the bootm is the square of the distance from (x,y) to (0,0). I'll forget the m now (erasing the mass m and leaving the arrow). So we need a force of magnitude GM/(x2+y2). The direction should be from (x,y) to (0,0): the direction of -xi-yj. But that vector has length sqrt(x2+y2). So to get a vector of the correct magnitude and direction we need to divide by sqrt(x2+y2) (that creates a unit vector pointing in the correct direction) and then multiply by GM/(x2+y2). So an inverse square law pointing at the origin is -[x/(x2+y2)3/2]i-[y/(x2+y2)3/2]j. Is this complicated enough? It can be difficult to see the inverse square law under all this algebra. To the right is a picture of this inverse square vector field in the first quadrant. The magnitude of the vector field decreases rapidly away from the origin.
Fluid flow: vortex flow The velocity field of a fluid flow may be less familiar. Here we could think of water flowing in between two parallel planes, close together. At a point we could insert a drop of ink, and look a short time later. The ink might be stretched into a short segment. This would be the velocity vector of the fluid at that point. It could change from one point to another, both in length and in direction. The length of the segment will (hopefully!) depend on the speed of the fluid, and the direction of the line segment will be the direction of the fluid flow.
I gave an example, which turned out to be difficult to explain (this means "I didn't do it well"), of a vortex: a counterclockwise velocity field of a fluid spinning with uniform angular velocity. Suppose the vector field was Ai+Bj at the point (x,y). This "vortex" would need to be perpendicular to the circle centered at (0,0) going through (x,y). That means the vector field would be perpendicular to the radius vector, xi+yj. We can check perpendicularity with the dot product: Ax+By must be 0. But the magnitude of the vector, sqrt(A2+B2), also should increase directly in proportion to the distance of (x,y) to the origin, so the fluid flow will have constant angular velocity. After some computation we saw that this would have to be -Kyi+Kxj, with K positive to keep this counterclockwise.
Other flows: 3i+0j (uniform flow to the right) Please note that adjustments were made to the fieldstrength option in this fieldplot command and others to avoid having the arrows overlap each other and to display better some of the other vector fields.
3xi+0j, a flow which doesn't move on the y-axis, but flows away from the y-axis otherwise, and faster, the farther the fluid was away from the axis.
Associated to a fluid flow are streamlines, which would describe how a fluid actually flows (the velocity vector field only would give what's called the "instantaneous" flow). Such streamlines are investigated and can sometimes be described exactly by solving differential equations. So, what is a vector field? Geometrically, a vector field "is" a collection of arrows in the plane. Algebraically, a vector field "is" a pair of functions, f(x,y)i+g(x,y)j. HOMEWORK Please read about cylindrical and spherical coordinates and begin reading chapter 16, at least 16.1 and 16.2. Hand in at the next recitation: 15.8: 8, 19; 16.2: 3, 11; hand in a plot of the vector field ((10/x)+x2)i+((10/y)+y2)j with 100 vectors in the region 1/2<x<5 and 1/2<y<5. Hint Don't do this by hand!	Not this kind of arrow! Identify the historical figure and the painter and win BIG!

Maintained by greenfie@math.rutgers.edu and last modified 3/24/2006.