Math 251 diary, fall 2008
In reverse order: the most recent material is first.

This is the seond part of the diary.
A link to later material
A link to earlier material

Wednesday, November 5

A textbook problem
Before class began I wrote a solution to problem 59 or 62 or ... in section 15.4. This was a problem which I had solved using spherical coordinates on Tuesday night, and my "solution" was ugly and inefficient. I realized later that evening that the problem could be solved using cylindrical coordinates much more efficiently. I wrote that solution on the board and apologized to the 13 students who had watched me sink into the quicksand of spherical coordinates the evening before.

Changing coordinates
Sometimes the integrand and the region of integration in a double or triple or whatever integral may show some unexpected geometric or algebraic relationships. We have been using the geometry of polar, cylindrical, and spherical coordinates for the last two lectures to compute integrals. More generally, these relationships may be exploited by a technique called changing coordinates. So let me begin with some examples which I hope will be easy.

A first example
Suppose I have two copies of the plane, R2. The left-hand copy in my pictures will have coordinates labeled with u and v and will be called the uv plane, and the right-hand copy will have coordinates labeled with x and y and will be called the xy plane. In this example, x and y will be related to u and v using the equations
Therefore the point corresponding to (0,0) in the uv plane will be mapped to (0,0) in the xy plane. Similarly, (0,1) in the uv plane will be mapped to (0,1) in the xy plane, but (1,0) in the uv plane will be mapped to (2,0) in the xy plane, because x coordinates are doubled.

If we take a blob of area in the uv plane, which I think of as dAuv, then the mapping stretches objects by doubling them in the horizontal direction. The vertical lengths stay the same. Therefore the dAxy which corresponds to the dAuv has area which is actually twice the area of dAuv:
There is an area multiplication factor of 2.

Now consider a more intricate shape in the uv plane, the unit circle centered at the origin: u2+v2=1. What shape corresponds to this circle in the xy plane? Well, since 2u=x, we know u=x/2, and v=y, so that u2+v2=1 corresponds to (x/2)2+y2=1.

We could think that the region inside the uv circle is broken up into many small pieces of area dAuv and then these are magically (?) transported to the xy plane, and they form the interior of an ellipse with horizontal semimajor axis of length 2 and vertical semiminor axis of length 1. What is the area inside the ellipse? Here is one way to compute that area:
Area of xy ellipse=∫∫ellipse in xy1 dAxy=∫∫circle in uv1 (2dAuv)=2∫∫circle in uv1 dAuv=2(π12).

Reason for =
The area of a region is just gotten by adding up the "pieces of area", the dA's, in the region. This is the double integral of 1 over the region. Here we are adding up the areas inside the ellipse in the xy plane.

Reason for =
We're changing from an integral over an xy region with dAxy to the corresponding area in uv. The corresponding region is the circle of radius 1 in uv. The integrand is very simple, just 1, so there is no need to change it. The dA's channge, however, by the previously stated area multiplication factor of 2.

Reason for =
Well, we can pull out the multiplier 2 from the integral -- it is just a constant.

Reason for =
I evaluate the area inside a unit circle by remembering that it is π(radius)2, and the radius is of course 1 here.

A second example
Here is a different relationship between two copies of the plane. In this example, x and y will be related to u and v using the equations
Here I again looked at how various points were mapping, and played with chunks of area. In this case, the geometry is related by a stretching in the vertical direction. The vertical lengths multiply by 3 going from uv to xy. The horizontal direction just stays the same.

The area of the related chunks dAuv and dAxy is still, I hope, relatively easy. Since the regions are stretched by a factor of 3, we see that

Now again I'd like to transport the uv unit circle to this xy plane. So u2+v2=1 becomes x2+(y/3)2=1 because 3v=y implies v=y/3. Now we could compute the area in the xy plane of this ellipse. It is a sequence of similar equalities:
Area of xy ellipse=∫∫ellipse in xy1 dAxy=∫∫circle in uv1 (3dAuv)=3∫∫circle in uv1 dAuv=3(π12).

The justifications for each of these equalities is much the same as what was written above. Basically, small chunks of area get stretched by 3 and the result gets stretched by 3.

Please notice that these are relatively simple stretchings and area multiplications. In more complicated situations, the area stretching will change from at different points. (Actually, exactly that happens with, say, polar coordinates. There is non-uniform stretching, the multiplication by r, which occurs.)

A third example
This is still relatively "easy" but the final result which I'll show you seems quite surprising to me. So the transformation is
(7 was used in class and is changed to 5 here, so the pictures would be a bit more manageable, maybe.)
This is an example of a shear. The shear is sort of like taking a wire framework (maybe a screen door?) and, if you could imagine all of the places where the vertical and horizontal threads cross being flexible joints, then pulling the horizontal sideways while maintaining the vertical framing. Things which are helpful to understanding a shear include experience with materials (!) and maybe a linear algebra course. Look at the geometry, which has some seemingly contradictory features.

Now what happens to u2+v2=1? Well, v gets traded for y, but since x=u+5v, we see x=u+5y and x-5y=u. The equation u2+v2=1 becomes (x-5y)2+y2=1. If you want to make things as irritating as possible, we know that (x-5y)2=x2-10xy+25y2 so that the equation is actually x2-10xy+26y2=1 (with an extra y2 coming from v2). You might not believe me if I suggested what this looks like, but if I wrote the following instruction to a silicon friend, maybe you would appreciate the picture:
> with(plots):
> implicitplot(x^2-10*x*y+26*y^2=1,x=-5..5,y=-2..2,scaling=constrained,color=black,thickness=2,grid=[50,50]);

Some discussion of the Maple command:

  • I used grid=[50,50] because as the help page states, By default a 26 by 26 grid is used, and when I just tried the bare command, this is such a strange curve that the default sampling gave a rather rough-looking object (try it and see).
  • I used scaling=constrained because if that's removed, some of the effect of the shear seems to be undone. Try it, and see what you get (a tilted ellipse with y=x as one axis, actually, and then you should explain this!).

    Since 1dAuv=dAxy, the total area is not changed at all. Therefore the area inside x2-10xy+26y2=1 is exactly equal to π12. I think if I worked diligently with dxdy integrals and used trig substitutions as they are taught in 152 I might, after a while, be able to get this result. But a whole heck of a lot of work would need to be done.

    So what's going on ...
    This result is not used as in the past examples. That is, people don't decide, "Hey, let's look at u and v and see ..." Rather, what happens is that sometimes folks realize they need to evaluate some (horrible) double/triple/whatever integral. They look at it, and see, somehow, some sort of links between the integrand and the region. They see, somehow, that everything could be described in terms of other variables. Then they reach in and use the result that follows. Note that no one I know uses this result "casually" -- they use it only if they really need it.

    The theorem
    Suppose x and y are written as functions of u and v. Then JAC, the area distortion factor, is the absolute value of a certain determinant:

        | ∂x/∂u ∂x/∂v | 
    det |             | 
        | ∂y/∂u ∂y/∂v |
    If Ruv is a region in the uv plane and Rxy is the corresponding region in the xy plane, if FUNCxy is a function written in terms of x and y, and if FUNCuv is the function rewritten in terms of u and v, then

    ∫∫RuvFUNCuv (JAC) dAuv=∫∫RxyFUNCxydAxy

    JAC is called the Jacobian. The result above, discussed in section 15.4, and particularly stated on pages 928 and 929 of the text, is called the Change of Variables Formula.

    So ...
    This theorem is difficult to work with but wonderful when you can use it. Here are two computations I showed in class.

    This example is artificial but useful as a start
    Compute ∫∫R(x-y)40(x+y)50dA where R is the rectangular region with corners (1,-1), (2,0), (0,2), and (-1,1). This is an irritating integral. But there is some not well concealed symmetry. The boundaries of rectangle can be written as x+y=2, x+y=0, x-y=-2, and x-y=2.
    It almost seems as if the integrand and the region are begging us to rewrite everything in terms of u and v where u=x-y and v=x+y. Then the region of integration can be described -2<=u<=2 and 0<=v<=2. The integrand becomes u40v50. Notice that if we add the equations u=x-y and v=x+y and divide by 2 we get x=(1/2)(u+v). If we subtract the first equation from the second and divide by 2 we get y=(1/2)(v-u).

    What's JAC?
    Since x=(1/2)(u+v) and y=(1/2)(v-u) we compute

        | ∂x/∂u ∂x/∂v |       | 1/2  1/2 | 
    det |             | = det |          | = -1/4 -1/4 = -1/2 
        | ∂y/∂u ∂y/∂v |       | 1/2 -1/2 |
    JAC is the absolute value, 1/2.

    We have in effect parameterized the xy plane with (1/2)(u+v)=x and (1/2)(v-u)=y. So everything in x and y could be written in terms of u and v. The "General Change of Variables" result becomes what follows in this case:

    ∫∫Rxy(x-y)40(x+y)50dAxy=∫v=0v=1u=-2u=2u40v50(1/2)du dv. This can be evaluated exactly easily because it is just a mess of powers of u and v. The answer is: (1/2)2·(241/41)(251/51).

    Crazy people all over ...
    Or you could just try it in Maple as it is. But we will need to break up the integral into three pieces (in either dxdy or dydx). Also, I want to learn how much time and space the computation takes, so I will use showtime.
    The instruction showtime(true); has this effect (from the Help page):
    Any Maple statement entered is evaluated normally, its result returned followed by a line numbered O1, O2, .. with the time taken and the amount of memory used being displayed.
    Here we go.
    > showtime(true);
    O1 := func:=(x-y)^40*(x+y)^50;
                                         40        50
                                  (x - y)   (x + y)
    time = 0.00, bytes = 7382
    O2 := A:=int(int(func,x=-y..2+y),y=-1..0);
    time = 0.08, bytes = 1394659
    O3 := B:=int(int(func,x=-y..2-y),y=0..1);
    time = 0.08, bytes = 1275540
    O4 := C:=int(int(func,x=-2+y..2-y),y=1..2);
    time = 0.13, bytes = 1880463
    O5 := A+B+C;
    time = 0.00, bytes = 3963
    O6 := %-(2^(41)*2^(51))/(41*51);
    time = 0.00, bytes = 3988
    So our result is the same as the rather painful direct computation, which took a total of .29 seconds. That is not terrible, but that computation gave no insight, which is bad.

    What's going on?
    If there is something common among the algebraic and geometric specifications of a double (or a triple!) integral, then we can sometimes take advantage. That's what's going on.

    Another example, but this one more realistic
    The following example could arise in thermodynamics or physical chemistry. Suppose R is the region in the first quadrant bounded by y=2x, y=4x, y=1/x, and y=3/x. Let's compute ∫∫Rx4y dA.

    Here a neat "change of variables" is a bit hidden, but maybe you can see that the boundary curves of the region are y/x=2 and y/x=4 and xy=1 and xy=3. Then you might (!) think to define u=y/x and v=xy. If you do, then uv=(y/x)(xy)=y2 so that y=u1/2v1/2. Then v=xy becomes v=x(u1/2v1/2) so that x=u-1/2v1/2. With the equations x=u-1/2v1/2 and y=u1/2v1/2 the original integrand x4y becomes u-3/2v5/2. The Jacobian computation is:

          | ∂x/∂u ∂y/∂u |       | -(1/2)u-3/2v1/2 (1/2)u-1/2v1/2 |
    det = |             | = det |                              |
          | ∂x/∂v ∂y/∂v |       | (1/2)u-1/2v-1/2 (1/2)u1/2v-1/2 |
    and this is -(1/4)u-1-(1/4)u-1. We want the absolute value so we have (1/2)(1/u). In this case, which is considerably more complicated than the others above, the amount of stretching sepends on the value of u. In the other cases we looked at previously, the stretching was the same at all points. In the real world, non-uniform stretching is more likely. (Take either a piece of taffy or a steel bar and pull at the ends. I bet that the parts near the ends stretches more than the center.) The double integral which results is ∫1324u-3/2v5/2(1/2)(1/u)du dv. The region of integration has become a rectangle, the integrand is not horrible, and the Jacobian factor is also not too bad. I won't compute this, but I hope that you see it is easy enough.

    Polar and spherical ...
    The factors involved in integration in polar/cylindrical (r drd&theta) and spherical (ρ2sin(φ) dρdφdθ) all come from Jacobian calculations. The spherical factor is the absolute value of a determinant of a 3-by-3 matrix. The computation is not fun. I've done it several times and have managed to make mistakes each time.

    Proof? Who needs a proof?
    Verification of the change of variables formula for double and on up integrals is very difficult. What's needed is knowledge of linear algebra and lots of comfort with limit processes, since the two sides of the formula are quite complicated Riemann sums. Working through the proof in detail takes about two weeks in an upper-level math course. Please believe this result, and try a few examples on your own.

    Ms. Jain's request
    The planes x=0 and y=0 and z=0 divide R3 into eight chunks. Differently put, if you remove these planes from space, you'll have eight pieces left. Each of the eight pieces is characterized by requiring that the variables x, y, and z have specific (non-zero!) signs. Ms. Jain asked how the spherical coordinates φ and θ relate to these sign restrictions.

    Look to the right. There is a very bare diagram with the spherical angles φ and θ sketched. Of course, φ is the angle between the radius vector and the positive z-axis. People usually request that φ be in the interval [0,π]. If this angle is acute, so 0<φ<π/2, then the radius vector will be above the xy-plane, no matter what the value of θ. This means z>0 exactly coincides with 0<φ<π/2. If we push the radius vector below the xy plane, then z<0 and φ will be larger than π/2. So z<0 is the same as π/2<φ<π.

    θ does not affect the sign of z at all. It interacts with the signs of x and y. So we can just look "downwards" in R3 from high up on the positive z axis. Then we might understand what we're seeing as something like usual polar coordinates (remember, the z information is carried by φ so we don't need that here). Certainly we can just read off the sign combinations of x and y by the usual quadrant information of θ.

    Sign of x Sign of y Sign of z Interval of θInterval of φ

    I hope this is helpful to other people who are trying to understand spherical coordinates.

    Monday, November 3

    Cylindrical coordinates
    This is a coordinate system that augments the r and θ of polar coordinates with z. Any problem with an axis of symmetry may be easier to understand in cylindrical coordinates. In words, the position of a point in the cylindrical coordinate system is described by its height, z, from the base coordinate plane. The foot of a perpendicular from the point to the plane then has a description in terms of an angle, θ, from an initial ray (usually the positive x-axis) and a distance, r, from the origin.
    Here are some basic axially symmetric figures:
    Cylinder Example: r=5.
    Cone Example: z=7r.
    Elliptical paraboloid Example: z=3r2.

    Some basic axially symmetric surfaces

    r=5 is the collection of points in R3 whose distance to the "axis" is 5. The axis is the z-axis, so this will be a right circular cylinder of radius 5 having the z-axis as axis of symmetry. z=7r gives a right circular cone whose axis of symmetry is the z-axis. How can you "see" this? Well, if we restrict ourselves to the slice of this surface through the xz-plane (with y=0) we get a picture sort of like what is shown. Why? Because if y=0, r=sqrt(x2+y2)=x (at least for x>0), so the result is the line shown.

    In general, since θ is not restricted, we get all the points shown as we revolve the "profile" curve around the z-axis. And this is a cone with vertex at the origin.

    z=3r2 is a paraboloid, because r2=x2+y2 and you should see, I hope, that the result is what happens when the profile curve, a parabola through the origin, is revolved around the z-axis.

    The location of Frelinghuysen Hall
    I enlightened students with these facts:
        FH's latitude is 40.50411oN and FH's longitude is 74.44836oW.
        Or, in more antique fashion (degrees/minutes/seconds), the latitude is 40o39´55.636´´ and the longitude is 74o27´5.452´´
    Do not be as confused as I am. This is not about stalactites (the down-dropping things) and stalagmites (the up-growing things).
    We discussed what latitude and longitude are. The prime meridian is a great circle (a circle whose center is the center of the earth) and it goes through Greenwich, England and the north/south poles. The longitude is the angle between that great circle and the great circle connecting FH and the north/south poles. The angle has vertex at the center of the earth. W=west in the latitude, and it means the the angle opens to the west of the prime meridian. Latitude is the angle from the intersection of the great circle describing FH's longitude with the plane of the equator, again with the vertex at the center of the earth. N=north means that we look in the northern hemisphere. Constant latitude means a "small" circle. Constant longitude means a great circle (actually semicircle). FH is located at the unique intersection on the surface of the earth of these two curves.
    I presume you know that the "23 and a half" degree tilt of the axes (north/south pole line) from the ecliptic (the plane of the earth's orbit about the sun) is responsible for seasonal variation. Nature is terrific!

    Spherical coordinates
    Take a point in space. We describe its position with one length and two angles. The length is the distance of the point to the origin: the length of the radius vector. The first angle, φ, is the angle from the positive z-axis to the radius vector. The second angle, θ, is the angle from the positive x-axis to the projection of the radius vector on the xy-plane. Spherical coordinates are very useful in problems with central symmetry.

    Spherical coordinates
    Take a point in space. We describe its position with one length and two angles. The length is the distance of the point to the origin: the length of the radius vector. The first angle, φ, is the angle from the positive z-axis to the radius vector. The second angle, θ, is the angle from the positive x-axis to the projection of the radius vector on the xy-plane. Spherical coordinates are very useful in problems with central symmetry. I'll work a bit with them next time.

    I deduced the following formulas (with errors but they were corrected):
          x=ρ sin(φ)cos(θ)
          y=ρ sin(φ)sin(θ)
          z=ρ cos(φ)
    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 ρ represents the distance from (x,y,z) to the origin.

    Standard restrictions on spherical coordinates
    Because the angles sort of fold over when π's and 2π's are added, most people who use spherical coordinates put some restrictions on how big/small θ and φ can be. If we only allow θ to be between 0 and 2π and only allow φ to be between 0 and π, 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

    ρ=constant gives a sphere centered at the origin. So, for example, ρ=5 is a sphere centered at the origin of radius 5. φ=constant gives a right circular cone whose axis of symmetry is the z-axis. For example, φ=π/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 π/6 (yes, 30o). The bottom half of the cone is not included because that is where φ is between π/2 and π. θ=constant gives a half-plane, with the z-axis being the edge of the half-plane. For example, θ=π/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 θ is 3π/2, and so it is not included in this object.

    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 ρ. And the total mass is the sum of the dm's. This "sum" should be a triple integral:
    Total mass=∫∫∫The whole ballC ρ dV.
    In spherical coordinates, a description of a sphere of radius R centered at the origin is easy: ρ goes from 0 to R, θ goes from 0 to 2π, and φ goes from 0 to π. 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,
    I know this is true. First, there is a discussion which is supposed to be convincing in the text (on pages 915 and 916). Second, I said it in class. Third, I actually can give an understandable argument if there is enough time later in the course. This strange multiplier is an example of what is called the Jacobian, a factor used to convert volume in one coordinate system to another. I may have time to discuss the computation later. In any case, when I use spherical coordinates, I almost never bother thinking about this weird mess, but I just write it.

    The computation
    So we have
    Total mass=∫φ=0φ=πθ=0θ=2πρ=0ρ=R(C ρ)ρ2sin(φ)dρdθdφ.
    The inner integral
    ρ=0ρ=R(C ρ)ρ2sin(φ)dρ=∫ρ=0ρ=R3sin(φ)dρ=Csin(φ)ρ4/4]ρ=0ρ=R=Csin(φ)R4/4.
    The middle integral
    θ=0θ=2πCsin(φ)R4/4 dθ=(2π C)sin(φ)R4/4=[(π C)/2]sin(φ)R4. (Just multiply by 2π, since there is no θ in the integrand.)
    The outer integral
    φ=0φ=π[(π C)/2]sin(φ)R4dφ=-[(π C)/2]cos(φ)R4]φ=0φ=π=(π 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?

    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? Here are several possible pictures of the region. 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, ρ 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 θ and φ? Here students should look closely at the definitions of θ and φ. Each of them will go from 0 to π/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 π/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/ρ5]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
    The integrand is [C/ρ3]sin(φ) after cancelling some powers.
    The inner integral
    This is an improper integral, so I will be careful.
    ρ=1ρ=BIGC/ρ3sin(φ)dρ=-C/(2ρ2)sin(φ)]ρ=1ρ=BIG=-C/(2(BIG)2)sin(φ) +C/(2(1)2)sin(φ). As BIG→∞, the term -C/(2(BIG)2)sin(φ)→0 so the improper integral ∫ρ=1ρ=∞[C/ρ3]sin(φ)dρ converges and its value is (C/2)sin(φ).
    The middle integral
    θ=0θ=π/2(C/2)sin(φ)dθ=(C/2)sin(φ)(π/2)=[(C π)/4]sin(φ).
    The outer integral
    φ=0φ=π/2[(C π)/4]sin(φ)dφ=-[(C π)/4]cos(φ)]φ=0φ=π/2=[(C π)/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.

    Wednesday, October 29

    A double integral
    Today begins three lectures where I will attempt to describe other, more sneaky methods to compute multiple integrals. These methods generally are used when specific computations are given and the regions or the integrand (the function to be integrated) share some kinds of symmetries. All of today's examples will be relevant to engineering education. I'll begin with the following problem. I want to compute a double integral, something like ∫∫Rf(x,y) dA. I will describe an f and an R.
    Let's let f(x,y)=x2+y2. That's certainly a simple enough function, just a degree 2 polynomial in x and y.

    Suppose R is the region in the xy-plane defined by these restrictions: it is in the upper half plane where y>0, and it has boundary given by y=x and y=-x (two straight lines or, actually, since they are inside a half plane, just two rays) and the circles x2+y2=2 and x2+y2=4. I think, or I hope, that this region is shown in the picture to the right.

    This would be a rather unpleasant double integral to compute as an iterated dxdy or dydx integral. I hope that you can see why. The region R is not nicely convex in either the x or y direction, and we'd need to break both of the iterated integrals into several pieces.

    I hope that there are enough accidents (?) and coincidences (??) so that you are a bit suspicious. Of course, this example is totally arranged. I hope that it makes you think of polar coordinates.

    dA in polar coordinates
    Here's a mostly emotional argument for how dA should be described in polar coordinates. Later I will be able to give a more precise derivation. Or you can look in the textbook (section 15.4) for a more careful discussion.
    Suppose I want to compute the area obtained by changing r to r+dr and θ to θ+dθ. The picture displays this area, dA, magnified a lot. As mentioned, dA is an area and has dimensions length2. If dθ and dr are very small, the area dA is approximately rectangular, and maybe the area is the product of the length of its sides. Well, one side is dr but the other side is not dθ. Angles don't have dimensions (they are ratios!) and, anyway, if you move circles centered at the origin in and out, you can see that the intercepted arcs change in length. These arcs are very short close to the origin and are longer as the radius of the circle gets bigger. In fact, the length of the intercepted arc is directly proportional to r. This length is also directly proportional to dθ: if the angle at the origin is doubled, the length of the intercepted arc is also doubled. Well, "directly proportional" means that there is some constant, uhhh ..., let's call it K, so that the length is K r dθ. What is K? In the nicest world, K would be 1 because then I would not have to worry about it any more. Well, golly, that is exactly why radian measure was invented: so this darn constant would be 1 and would not need attention.

    Comment: so what is K and what about those words?
    Why is K=1 in radians? Well, the circumference of a circle of radius r is 2π r. Here the dθ is 2π. So apparently the K is indeed 1. If you insisted on using degrees in all of calculus, then the angle for a whole circle would be 360, and for Kr dθ=K(360)r to be 2π r, you would need K=2π/360, which is approximately the obnoxious number .01745. I looked on the web, and the only other candidate for angle measurement I found was the grad, introduced in France as part of the metric system (my calculator permits angle computations in grads). There are 400 grads in a circle (I didn't know that) and therefore the constant K, if we used grads in calculus, would be 2π/400 which is approximately .01571, also obnoxious. Yes, things would be better if π were equal to 3.

    Euphemism: The expression of an unpleasant or embarrassing notion by a more inoffensive substitute
    The word "golly" is a euphemism for "God" and the word "darn" is a euphemism for "damn".

    Computing the integral
    Let's return to computing ∫∫Rx2+y2dA, if R is the region shown to the right (in the upper half plane, with the curves arcs of circles centered at the origin).
    How does one recognize that the integral is "polarish"? It is a classroom example, but the integrand has central symmetry, and so does the region. You may be helped if you recall the conversion formulas

    From r, θ to x, y      From x, y to r, θ   
    -------------------    ------------------- 
       x=r cos θ              r2=x2+y2
       y=r sin θ              tan θ=y/x
    I've given the formulas the way I most often use them. In particular, the formula for getting θ from x and y needs to be "adjusted" (by adding π) if the point whose coordinates are (x,y) is in the left half of the plane.

    I recognize (primarily from the picture, but I can also use the formulas) that R is described by π/4<=θ<=3π/4 and by 2<=r<=4 (Here I made a mistake in class -- the values of r go from 2 to 4, as in the picture, not from sqrt(2) to 2! I regret the error.) We can convert the integral into polar coordinates:

    ∫∫Rx2+y2dA= ∫π/43π/424r2 r dr dθ= ∫π/43π/424r3dr dθ= ∫π/43π/4(1/4)r4]r=2r=4dθ=(63)θ]π/43π/4=(63)3π/2.

    Of course the computation is easy. It was arranged so that after conversion to polar coordinates things would work out well. The computation in rectangular coordinates, including finding the boundaries of the integrals (there would have to be two of them) and then computing the antiderivatives, would be very tedious. This is not an entirely artificial example: it is the computation of the moment of inertia about the origin of a thin homogeneous plate in the shape of the region R.

    The earth is flat
    So here I will try to convince you by combining a valuable and truthful computation with extremely dubious logic, that the earth is flat. Please be reassured: the earth is probably not flat.

    Newton's Law of Universal Gravitation
    Suppose I have two "point masses", m1 and m2, which are a distance d apart. The magnitude of the force attracting them together is directly proportional to the product of their masses and inversely proportional to the square of the distance separating them. The constant of proportionality is usually called G (alas, not to recognize the lecturer!). Therefore the magnitude of the force is G m1m2/d2.

    Here's a quote from another site:
    ... this constant [G] was worked out by Henry Cavendish in 1798. He achieved this by measuring the gravitational attraction between two 1kg lead balls at a distance of 1 metre.
    I would guess that you would assume that this experiment must have found the attraction to be very small, in fact it was so small it was a stunning achievement to measure the tiny acceleration caused and to therefore calculate the gravitational attractive force (using Newton's second law).
    Cavendish found the force to be 6.67 x 10-11 Nm2/kg2 (or approximately 0.0000000000667 Nm2/kg2!).
    Gravity is actually much weaker than, say, magnetism. There is just a great deal of mass around, and very few magnetic monopoles.

    The plate: from description to integral
    Let me assume that the "universe" consists of an infinite flat homogeneous plate, and an external small object with a mass of m whose distance to the plate is D. What is the gravitational attraction of the object to the plate? A major part of such a problem is setting it up. The correct location of the origin and the axes can make problems much easier. In this case, I believe there are two reasonable locations for the origin: the object, or the closest point on the plane to the object. I'll use that closest point to be the origin. Of course, the xy-plane will be the plate, and therefore the coordinates of the object will be (0,0,D). The plate is homogeneous and thin. To avoid having too many letters around, let me assume that the plate is 1 unit thick (otherwise I'll just have to carry around the thickness in all of the computations, and I have a hard enough time with my own thickness, both mental and physical). Since the plate is homogeneous (the same at every point), it has a density, ρ. A small chunk of the plate ("dA") located at the point (x,y) will have mass equal to ρ dA (remember the thickness is 1, and so it is already in the formula).

    Now let us convert the ideas into more rigid "mathspeak". The magnitude of the force from the external mass to the dA piece of the plate is Gmρ dA/d2. The piece is located at (x,y), and (x,y), (0,0), and the location of the external mass are at the vertices of a right triangle. The hypoteneuse of the right triangle is d, and the leg of the triangle from the external mass to (0,0) is D. The distance from (0,0) to (x,y) is sqrt(x2+y2). Therefore d2=D2+(sqrt(x2+y2)). The square root and the square cancel. The magnitude of the force is Gmρ dA/(D2+x2+y2). Several students noticed a surprising symmetry. Since we are dealing with the whole plane, R2, the chunk of dA at (x,y) has an antipodal chunk at (-x,-y), having the same mass and the same distance to the external object. Therefore the "lateral" parts of the forces (parallel to the plane) exactly cancel out. We only need to compute the vertical component of the force.

    The vertical component of the force is the magnitude of the force multiplied by the cosine of the angle, φ, between the vertical line and the line connecting the external object to dA. But cos(φ) is D/d, which is D/sqrt(D2+x2+y2). The function to be integrated is the vertical component of the gravitational attraction between the external object and dA. This is GmρD dA/(D2+x2+y2)3/2. Since the plate is infinite, we want ∫∫All of R2Gmρ dA/(D2+x2+y2)3/2.

    Computing the integral
    Many of the letters are constants: G and m and ρ and D. We can pull them out of the integral (but we will remember them for the final result). We need to compute:
    Since this comes immediately after a discussion of polar coordinates, the student alert to pedagogical plans (how folks teach) will immediately think of converting to polar coordinates. Indeed, even those who are not so ... prescient might think: the region has symmetry around (0,0) and the integrand has that same x2+y2, so let's try polar coordinates!

    Then dA=r dr dθ, and r2=x2+y2, and all we need are the limits on the integral. For the whole plane, r should go from 0 to ∞, and θ should go from 0 to 2π. The appearance of ∞ forces me to finally acknowledge that this is an improper integral.

    00r dr dθ/(D2+r2)3/2.

    The inner (improper) integral
    I will be careful, since I am supposed to be teaching a math course.
    0r dr dθ/(D2+x2+y2)3/2=limB→∞0Br dr dθ/(D2+r2)3/2
    The r accompanying the dr is exactly what's needed to do the substitution u=D2+r2 with du=2r dr. We sort out the constant by guessing (maybe).
    0Br dr dθ/(D2+r2)3/2=-1/sqrt(D2+r2)]r=0r=B= -1/sqrt(D2+B2)-{-1/sqrt(D2+02)}
    As B→∞, the term -1/sqrt(D2+B2)→0. The other term has minus signs which cancel, and (let's say D>0) square/sqrt which cancel, so the limit is 1/D.

    The outer integral is easy: ∫0(1/d)dθ=(1/D)θ]θ=0&theta=2π=2π/D.

    But we need to multiply by the factors we pulled out. The whole answer is:

    And, therefore ...
    There is no D in the answer!. The gravitational attraction of a flat earth is constant! Now the lecturer discussed the fact that he weighs the same standing on the floor and standing on a chair. Therefore ... therefore ... the earth is flat. (And even more supporting argument: wouldn't people who wanted to lose weight climb Mt. Everest, because they would lose weight when ...).

    Discussion of the claim

  • The logic is flawed. If the earth were flat, then the weight would be the same. But other reasons could cause the force of gravity to be (or seem to be!) the same. A statement and its converse need not both be true!
  • In fact, if the earth is spherical (which it is, essentially) the gravitational attraction of the earth acts as if the mass of the earth was concentrated at the center of the earth (this can be checked using methods of this course). The radius of the earth is reported to be about 4,000 miles, which is 21,120,000 feet. Let's say that a chair is 2 feet high. The lecturer would need to be sensitive to a 1.4·10-14 change in weight (I squared the ratio, because of the inverse square law). The instructor is not that sensitive!

    This is still an interesting and useful computation. An electron is very small. If we try to analyze the attraction an electron might have to a small charged plate, even, say, 1/4 inch square, then, to the electron, the plate might as well be infinite. That is, if the electron is near the center of the plate, the edge effect hardly matters at all. And the force on the electron does not depend on distance. Such considerations occur in the design of classical capacitors, used in many devices.

    (Almost a real problem!) Moment of inertia of a cone
    Suppose a right circular cone with base radius R and height H is filled with a homogeneous substance with constant density, C. What is the moment of inertia of the cone about its axis of symmetry?
    Let me be more clear about some vocabulary.

    Right circular cone
    Take a circle (to be called the base). Put a line perpendicular to the plane of the circle through the circle's center. πck a point on this line which is not on the plane of the circle. Connect that point (called the vertex) with the edge of the circle. The solid interior to the collected line segments and the circle is called a right circular cone. The "right" refers to the right angle that the axis of symmetry makes with the base.

    Moment of inertia
    Take a little piece of mass, m, external to a line, L. The moment of inertia of m about L is defined to be Q2m where Q is the distance from m to L.
    There are many discussions of the moment of inertia on the web. One link declares that it is the "inertia with respect to rotational motion" and another reads "... the rotational analog of mass for linear motion. It appears in the relationships for the dynamics of rotational motion. The moment of inertia must be specified with respect to a chosen axis of rotation."
    I think of a small merry-go-round in a playground, and trying to push the seats around (with many noisy, small children on them). The moment of inertia measures the resistance of the merry-go-round to being pushed.

    Beginning the analysis
    Take a little piece of volume, dV, inside the cone. (Note: this is a piece of volume inside the cone. We are not just considering the surface of the cone -- the cone is filled.) The mass of this volume is C dV. Suppose Q is the distance of piece from the axis of symmetry. Then the moment of inertia of this chunk of mass about the axis of symmetry is Q2C dV. To get the moment of intertia of the whole cone we need to add up the pieces of the moment of inertia. So we need ∫∫∫The whole coneQ2C dV.
    A major decision in this and many other geometric/physical problems is where/how to put a coordinate system on the objects involved. Here almost surely people would agree that the axis of symmetry should be the z-axis. Sane human beings can disagree about where the origin should be. Some would put it at the vertex of the cone, with the base "up", and some would put the origin at the center of the base of the cone, with the vertex "up". I'll do the first alternative because I think some of the algebra will be simpler. As I mentioned in class, I drew the cone in this awkward way because I wanted people to think about how they would prefer to see it.

    Now the cone is sitting correctly (?) in the picture. The chunk of volume is at (x,y,z), and the closest point on the axis of symmetry is (x,y,0). The distance between (x,y,z) and the axis must therefore be sqrt(x2+y2). We should convert the triple integral into an iterated integral. What should be the order? Actually, it is possible to do this in any order, but the simplest way has dz on the outside. Then the z limits are clear: from 0 to H, and the slices with z=CONSTANT are also simple shapes: circles. The triple integral ∫∫∫The whole coneQ2C dV becomes the triply iterated integral ∫z=0z=H(∫∫(sqrt(x2+y2))2C dAxy)dz. I wrote dAxy to remind myself that the double integral is in the xy-plane.

    The inside double integral is: ∫∫(x2+y2)C dAxy.

    Recognition: polar
    Things are in red so that a bell will ring in your head and you will think, polar!!!. Certainly x2+y2=r2 and dAxy=r dr dθ. The limits on θ for a whole circle are 0 and 2π. The limits on r are 0 (the center of the circle) out to the radius of the circle, which I will cleverly call RAD. The double integral is then ∫θ=0θ=2πr=0r=RAD(r2)C r dr dθ.

    Look at the cone sideways and see some expected right triangles, so RAD/z=R/H and RAD=(R/H)z. The double integral becomes
    θ=0θ=2πr=0r=(R/H)zC r3 dr dθ.

    r=0r=(R/H)zC r3 dr=(C/4)r4]r=0r=(R/H)z=(C/4)R4z4/H4.
    θ=0θ=2π(C/4)R4z4/H4dθ=[(π C)/2]R4z4/H4. (Easy: no θ in the integrand so just multiply by 2π.)
    z=0z=H[(π C)/2]R4z4/H4dz=[(π C)/10]R4z5/H4]z=0z=H=[(π C)/10]R4H.

    Is this correct?
    The units of moment of inertia should be mass·(length)2. Since C is a density, C's units are mass/(length)3. And R4H is length5 so the units are correct. Sigh. What about the crazy constants (π and 10)? An engineering student who took Math 251 in a previous semester sent me e-mail about this, and here is part of his message:
    Upon consultation with my statics text, I present to you: ...
    I_x = 3/10 * m * a^2 where a is the radius of the base of the cone.
    Let's see: the statics text refers to the mass, m, of the cone. That is the density, C, multiplied by the volume. The volume of this right circular cone is (π/3)R2H. The student's "a" is our R. So the formula 3/10 * m * a^2 becomes (3/10)C[(π/3)R2H]R2 which is indeed [(π C)/10]R4H. We have confirmation by high authority: "my statics text".

    Please read about triple integrals and cylindrical and spherical coordinates in 12.7, and 15.4. If you do this, you will find the next lecture much more comprehensible. And, otherwise, there will just be too many formulas! This is almost guaranteed.

    Monday, October 27

    The ocean
    People who study the ocean (such as those affiliated with the Rutgers Institute of Marine & Coastal Sciences are interested in such aspects as the temperature and salinity and pressure and flow of the water. They employ remote sensing devices to try to record data at various depths. Then analysis of the data together with theory is used to try to predict interesting things: the weather, fishing prospects, etc. I'm going to look at a very simple model and try to link it up with things we study in this course.

    The average temperature of a box of ocean
    Consider a "box of ocean", say the region between x=a and x=b, y=c and y=d, and z=e and z=f (here a<b, c<d, and e<f). We might put some sort of measuring device at a point in this box and measure the temperature of the water at that point. One or a few temperature measurements are probably not going to give good information. If the economics (!) and the equipment and time (!) are available, many measurements should be made. One representation of the measurements might be the average: so the computation would be

    SUM of all of the temperature measurements
      The number of temperature measurements
    Considerations which might influence this "experiment" include the following:
  • The temperature measurements should be well distributed: we won't get an acceptable average temperature if the places we measure are all clumped up in one chunk of the box (the statistical phrase is that the measurements should be uniformly distributed).
  • Ideally, we would like more measurements rather than fewer: with many measurements we'd have a chance of coming up with more reliable information.

    Going abstract: the "limit"
    Let me look at the average a bit more. The discussion that follows seems very clever to me.
    Suppose that I assume that the number of observations is n3 where n is a large positive integer. Then I would have something like this:

    SUM of all of the temperature measurements
    I will multiply the top and bottom of this fraction by (b-a)(d-c)(f-e), so we would have:
    SUM of all of the temperature measurements   (b-a)(d-c)(f-e)
    ------------------------------------------ · ---------------
                        n3                       (b-a)(d-c)(f-e)
    Just consider part of this, the fraction (b-a)(d-c)(f-e)/n3. This is the same as [(b-a)/n]·[(d-c)/n]·[(f-e)/n]. If n is large, this is the same as splitting up each of the edges of the box into n equal pieces, and what we have is a very small box of the ocean. Now if we also want the points we measure to be well-distributed, then we might expect that most of the boxes will contain exactly one sample point. We can think of Temperature at that sample point)·[(b-a)/n]·[(d-c)/n]·[(f-e)/n] as T(that sample point)dx dy dz or as T(that sample point)dV where dV is this very small box inside the huge box of ocean. When we take the SUM we actually have an approximating Riemann sum to ∫∫∫box of oceanT(x,y,z) dV, which is a triple integral. Whew! The limit of such approximating sums is the triple integral, but I won't go into detail because this all parallels a similar discussion for double integrals. I don't want to forget anything: there is a factor of (b-a)(d-c)(f-e) remaining on the bottom, and this is the volume of the box.
    All of this is supposed to support the following definition:
    The average value of the temperature in the box is
    ∫∫∫box of oceanT(x,y,z) dV
        Volume of the box

    A specific example
    What if our box was bounded by x=0 and x=2, y=0 and y=3, and z=0 and z=5, and the temperature at (x,y,z) was given by the formula T(x,y,z)=x2+7yz. Then if we wanted to compute the average temperature we would convert a triple integral into a (triply) iterated integral. In this case, I see no advantage in any one of the six possible orders, so:
    020503 x2+7yz dy dz dx
    Let's compute, from the inside out: ∫03 x2+7yz dy dz dx=yx2+(7/2)y2z]y=0y=3=3x2+(63/2)z.
    053x2+(63/2)z dz=3x2z+(63/4)z2]z=0z=5=15x2+(63/4)(25).
    0215x2+(63/4)(25) dx=5x3+(63/4)(25)x]x=0x=2=40+(63/2)(25).

    If this were the 21st century, instead of 1872, we could type:

    > int(int(int(x^2+7*y*z,y=0..3),z=0..5),x=0..2);
    Incidentally, I checked and 40+(63/2)(25) is the same as (1655)/2.

    This isn't the average temperature. For that we need to divide by the volume of the box which is 2·3·5=30. The result is (331/6).

    The "moral" of this: computation of triple iterated integrals
    I don't think that there are any essential new difficulties introduced when we move from evaluating double iterated integrals to evaluated triple iterated integrals. Yes, there are more opportunities for error (50% more?) but they are not new in type. So I won't devote too much time to actual evaluation.

    Describing a volume in space
    Since the difficulties involved in computation of a triple iterated integral really are just those we've seen already with double interated integrals, I want to illustrate something that definitely seems more complicated to me: going from a description of a region in space over which we want to compute a definite integral to the corresponding iterated integrals (and there are 6=3! possible ordered for the iterated integral). Let me "integrate" (convert to iterated integrals) the function SQUIRREL over the region in space (R3) defined by y=0, z=3, and z=x2+y.
    I want to begin by sketching the region. The planes y=0 (the xz-plane) and z=3 (push the xy-pane up three units) are easy enough. The surface z=x2+y cut by y=0 and z=3 is maybe not so obvious. When y=0 we get a parabolic arc cut off at z=3 in the xz-plane. As y increases, the parabolic arc is translated up, but still cut off at z=3. In the yz-plane, when x=0, the slice is a segment of the line z=y from (0,0) (with the coordinates being y and z) to (3,3). The surface cuts the plane z=3 with the parabola 3=x2+y or y=3-x2, which opens "downward" (in the standard orientation of xy-planes).

    I've attempted to sketch the surface to the right of this description. The colors are meant to show some of the curviness. There are some extreme points which turn out to be useful in setting up iterated integrals. Those are the points (0,0,0), (0,3,3), (sqrt(3),0,3), and (-sqrt(3),0,3). These points are where each of the coordinates (x and y and z) attain maximum and minimum values on the solid region whose boundary curves were given.
    Nomenclature The surface z=x2+y is called a tilted parabolic cylinder. It is a parabolic cylinder because it results from a family of parallel lines in space which all meet the parabola z=x2 in the xz-plane. It is "tilted" because these lines are not perpendicular to the xz-plane.


    Now to the right is Maple's attempt to draw the tilted parabolic cylinder in the region of interest to us. The picture to the right is the result of using the command:

    This command did not display an immediate result on my home computer. It requested that Maple to check a three-dimensional grid of 403=64,000 points, and then compute the light and the angle, etc. I rotated and chose lighting so that I got the image displayed here. That's why "supercomputers" are needed to draw the lighting effects for Pixar, etc.

    Maple can draw ...
    ... some useful pictures for us when we want to look at double and triple integrals. Last time we looked at the iterated integral
    02x=0x=1-(1/2)y3-3x-(3/2)y dx dy
    The command plot3d(3-3*x-(3/2)*y,x=0..1-(1/2)*y,y=0..2); produces (after putting in the axes and making the view constrained) the graph shown to the right. I did not know until fairly recently that Maple had the capacity to show only pictures corresponding to double integral limits. This could be very helpful.

    A region in space
    Now back to the triple integral. There are six different orders that are possible when converting a triple integral to an iterated integral. I did three of the six orders. Let's convert ∫∫∫This regionSQUIRREL dV into various iterated triple integrals.

    dx dy dz
    I'll try this order first: ∫( ∫∫ SQUIRREL dx dy)dz.
    I've mentioned that my personal inclination in finding limits of iterated integrals is working from the outside-most limit "in". There are definitely people who are successful and do the exact opposite. I would recommend that you find your own "natural" style and try to follow that path. For me, I would look at the z limits first. For this shape, I would try to find the highest and lowest z's in the spatial region. This is not a complicated region, and we've already sketched it quite well. The highest and lowest z's are, respectively, z=0 and z=3. So we've got ∫z=0z=3(∫∫SQUIRREL dx dy)dz.

    Now let's try slicing the region by z=CONSTANT, where the CONSTANT is some unknown number between 0 and 3. This horizontal slice of the original spatial region gets us something in the xy-plane. If you were in class, you may recall that there was some effort involved in sketching the slices that are shown here. But one boundary of the sliced region is y=0, along the x-axis. The other, curved boundary, is "inherited" from z=x2+y. Now z=CONSTANT so as a curve in the xy-plane, if we write it in the standard y=function of x format, we get y=z-x2. Therefore this is a parabola (the square on x!) opening down (the minus sign). The top of the parabola (the vertex) occurs when x=0, and there y=z. The intersection(s) of the parabola with the x axis occur when y=0, and there 0=z-x2, so that x=+/-sqrt(z). The inner double integral is ∫∫ SQUIRREL dx dy. What are the bounds on the dy integral? We must look at the slice, and see what the highest and lowest values of y are on the slice. The lowest value is 0 and highest value is z: but on the slice, z is a CONSTANT. The highest value depends on z. Now we know: ∫y=0y=zSQUIRREL dx dy
    Now in the region pictured, I will slice with y=CONSTANT and see how big and how small x can be. This is a slice of a slice (maybe [slice]2?). So the boundary is given by z=x2+y, and with both z and y CONSTANT, I get x2=z-y, so that x=+/-sqrt(z-y). These will be the limits on the dx integral.

    So the answer is:
    z=0z=3y=0y=zx=-sqrt(z-y)x=+sqrt(z-y) SQUIRREL dx dy dz.

    dy dz dx
    Now ∫( ∫∫ SQUIRREL dy dz)dx.
    Examine the original picture and the limits on the outermost variable, x, should be revealed. The largest and smallest x's in this region are +/-sqrt(3), and therefore we get ∫x=-sqrt(3)x=sqrt(3)( ∫∫SQUIRREL dy dz)dx. Our task is now to slice with x=CONSTANT and try to get the other integrals' bounds.

    Again, once the "picture" is presented then much of the remainder of the work is made much easier. We spent some time in class drawing this picture. When x=CONSTANT, then certainly the slice goes through the side (on the xz-axis) so that y=0 becomes the left boundary, if we have z assigned to be the vertical coordinate and take y to be the horizontal coordinate. Also the top of the region is still z=3. The other edge is "inherited" again as the effect of the equation z=x2+y. As I mentioned in class, it is this edge which irritates my highly trained mathematical psychology (is there such a thing?). Notice that x=CONSTANT, so that z=x2+y is a straight line in the yz-plane. The slope of this line is 1. And, when y=0, z must be x2.
    The limits on the outside of the double iterated integral ∫∫SQUIRREL dy dz can now be "read off" from the picture, since the smallest value of z is x2 and the largest value is 3. Therefore we have the limits on the outside of the double iterated integral: ∫z=x2z=3SQUIRREL dy dz. Finally, the bounds on the dy integral are obtained by slicing the slice. So now z=CONSTANT also, and y goes from y=0 to the right side, which is a point on the line (it still hurts to write this when there is a square in the equation!) z=x2+y, and therefore the upper bound is y=z-x2.

    So the answer is:
    x=-sqrt(3)x=sqrt(3)z=x2z=3y=0y=z-x2 SQUIRREL dy dz dx.

    dz dx dy
    My last attempt: ∫( ∫∫ SQUIRREL dz dx)dy.
    Again, the picture shows that y in the solid region varies from 0 to 3, and we've got 2 of the 6 limits (o.k., the easiest of them): ∫y=0y=3( ∫∫ SQUIRREL dz dx)dy. The y=CONSTANT slice should give the other information.

    Again, the picture gives much of the information we need. Drawing the picture was some work. Here with y=CONSTANT, the top of the slice is caused by the plane z=3. The bottom of the slice is z=x2+y. Now since x is a variable, this is indeed a parabola. The parabola opens up (positive coefficient on the square term) and has vertex (0,y): the first coordinate is x and the second coordinate in this slice is z. The parabola intersects the line z=3 when 3=x2+y. Since y=CONSTANT, this occurs when x=+/-sqrt(3-y). The outer, x-limits, on the double integral will be x=-sqrt(3-y) and x=+sqrt(3-y). Now slice the slice, for make x=CONSTANT also. z will vary. The highest value of z will be 3 on the [slice]2. The lowest value of z is given by z=x2+y.

    The final way the poor SQUIRREL is chopped up and then summed is
    y=0y=3x=-sqrt(3-y)x=+sqrt(3-y)z=x2+yz=3SQUIRREL dz dx dy.

    First, this is a classroom example. The solid region is actually not very complicated. It is a convex region with boundaries given by low-degree polynomials. The word convex means that line segments whose ends are in the region always have the whole line segment in the region. The problem would be much more complicated if the functions defining the boundary weren't so simple, or if some of the slices weren't convex (then we'd need to split up the integrals, etc.). I remarked in class and I'll repeat here that the process of finding these limits seems to be difficult, and hard to describe -- I don't know yet of a computer program which can do it reliably.
    Here are the "answers" again:
    z=0z=3y=0y=zx=-sqrt(z-y)x=+sqrt(z-y) SQUIRREL dx dy dz
    x=-sqrt(3)x=sqrt(3)z=x2z=3y=0y=z-x2 SQUIRREL dy dz dx.
    y=0y=3x=-sqrt(3-y)x=+sqrt(3-y)z=x2+yz=3SQUIRREL dz dx dy.
    I can't immediately see that the darn limits describe the same volume in R3. Maybe you can. But you should see, just looking at the patterns of the answers, what sorts of limits are "legal" and what are not. You can only have variables in the limits if they haven't been integrated yet. For example, in the last answer, the lower limit of the innermost integral is z=x2+y, and the outside two integrals are dx and dy. I could not have a limit in, say, the middle integral of the form z=x2+y because there would be only one variable left to be integrated, and there isn't any way to "kill" both x and y. So there is a rough guide to the grammar (?) of the bounds on iterated integrals.

    How can you check this kind of "computation"?
    Generally checking these things can be difficult and tedious. Luckily, we are in the 21st century and I have powerful friends. Well, I guess I can ask some electrons to run around. Look at the following:
    > W:=x^6*y^8*z^2;
                                          6  8  2
                                    W := x  y  z
    > int(int(int(W,x=-sqrt(z-y)..sqrt(z-y)),y=0..z),z=0..3);
                                  417942208512 3
    > int(int(int(W,y=0..z-x^2),z=x^2..3),x=-sqrt(3)..sqrt(3));
                                  417942208512 3
    > int(int(int(W,z=x^2+y..3),x=-sqrt(3-y)..sqrt(3-y)),y=0..3);
                                  417942208512 3
    I specified a "random" function, W, to replace SQUIRREL. I wanted the antiderivatives not to be a problem, so I just specified some powers of x and y and z. I asked Maple to compute the triple iterated integrals in all three ways we found. The answers are shown. They are such large and silly numbers, and they all agree exactly. I am fairly confident the bounds on the iterated integrals are correct.

    How clever? Not very clever ...
    I could have specified W=0, and then I bet the computation would have returned 0 for all of the setups, and this answer would not be very helpful. I could have specified W=1 and would have gotten (24/5)sqrt(3), the volume, for all of the setups. That would be fine. In fact, I confess that the first W that I tried was actually x3y7z2. Why wasn't this a very clever choice, and what was the answer I got? Hint: 3 is odd and this region in space is ...

    Something for you to do ...
    Choose one of the other three orders and write the triple iterated integral for that order. You can't integrate SQUIRREL without more specificity, so all you can do, and what I would like, is to write the precise bounds. Here are what I think are the answers:
    z=0z=3x=-sqrt(z)x=sqrt(z)y=0y=z-x2 SQUIRREL dy dx dz
    x=-sqrt(3)x=sqrt(3)y=0y=3-x2z=x2+yz=3SQUIRREL dz dy dx
    y=0y=3z=yz=3x=-sqrt(z-y)x=sqrt(z-y)SQUIRREL dx dz dy

    Wednesday, October 22

    Volume of a tetrahedron
    A tetrahedron is a flat-sided object with four corners. I would like to compute the volume of a tetrahedron whose corners are at (0,0,0), (1,0,0), (0,2,0), and (0,0,3). There are several ways to compute this volume, including some which need no "calculus", just vector manipulation (using the triple product formula, for example). To the right is an attempt at a picture. It shows the corners (the four vertices) and the faces, and I've made some attempt to show the four sides with differently decorated "stripes" on each. There are four flat sides. There's one on each coordinate plane (xy-, yz-, xz-) and there is a tilted face. The equation of the tilted face (the points (1,0,0) and (0,2,0) and (0,0,3) are on the tilted face) is x+(y/2)+(z/3)=1. And I got this equation totally by guessing.

    Here we'll use a double integral to find the volume of the tetrahedron here. I think of this solid as lying over a triangle in the xy-plane. The triangle is determined by (0,0), (1,0), and (0,2).The height of the solid over this triangle is z=3-3x-(3/2)y, which I got from the equation for the tilted face, just by solving it for z.

    As a double integral So the volume is ∫∫BaseHeight dA, and this is ∫∫The triangle3-3x-(3/2)y dA. I'll convert this to an iterated integral to compute it.

    47 second break for theory
    In one variable calculus, as I explained last time, the initial glimpse at the theory in back of the definite integral assumes that the function doesn't have any jumps. But real functions can jump! The functions which are met in mechanical engineering (just hit something!) can certainly look like what's shown to the right. And similarly, functions met in digital signal processing really can look like that also. They certainly can be integrated. The secret is that the jumps really don't matter very much. They can be put inside little boxes where the variation doesn't matter very much (the red boxes in the picture). So the sums defining the definite integral still approach a limit, the "correct" limit.

    Here I am apparently not even worrying about the domain. Well, this is what we could do if we had another 30 minutes to fritter away on details. I could define a function piecewise in this way:
    F(x,y)=3-3x-(3/2)y if (x,y) is in the triangle, and F(x,y)=0 if (x,y) is not in the triangle. Suppose R is any rectangle in the xy-plane which contains the triangle. The the volume of the tetrahedron would be ∫∫RF(x,y) dA. I hope that you will see this double integral is the same as the double integral over the triangle that I'll compute by looking at iterated integrals. The discontinuities of the piecewise-defined function turn out to give a perturbation of the Riemann sums which →0 as the size of the pieces →0.

    Converting to iterated integrals
    Let's write ∫∫The triangle3-3x-(3/2)y dA as a dx dy iterated integral. That means figuring out the bounds on the integrals.
    I will work from the outside in. So first I need to get the lowest and highest values of y in the triangular base:
    Lowest yHighest y???3-3x-(3/2)y dx dy
    There's a sketch of the base to the right, and the sketch declares that the Lowest y is 0 and the Highest y is 2. Now I imagine (and frequently draw, as shown on the sketch!) a very thin collection of dx by dy rectangles being added up in a row across the region. It is so thin that y is almost constant and the x's range from the leftmost edge to the rightmost edge. The left edge is certainly 0 always. But the right edge depends on y. When y is very near the bottom (y=0), the right edge is very near 1. When y is near the top (y=1), the right edge is near 0. What is the relationship between x and y on this edge? Of course the edge reflects the tilted face of the tetrahedron, which has the equation x+(y/2)+(z/3)=1. On the base, z=0, so the equation giving the tilted side of the triangular base must be x+(y/2)=1. Therefore x on the rightmost edge is given by x=1-(1/2)y. Here is the resulting iterated integral:
    02x=0x=1-(1/2)y3-3x-(3/2)y dx dy
    Even thought it is not logically necessary (because the dx dy notation does determine what variable is integrated first), I do tend to write "x=" on the limits of the inner integrals. This may save me from confusion and error as I compute.

    Computing the iterated integral
    I'll first compute the inner integral:
    x=0x=1-(1/2)y3-3x-(3/2)y dx= (antidifferentiate with respect to x, so y is a constant here!) 3x-(3/2)x2-(3/2)yx]x=0x=1-(1/2)y= 3{1-(1/2)y}-(3/2){1-(1/2)y}2-(3/2)y(1-(1/2)y)-0. The -0 comes from the lower limit, x=0. I tend to expand and "simplify" here. So we get:
    3-(3/2)y-(3/2){1-(1/2)y}2-(3/2)y(1-(1/2)y)=3-(3/2)y-(3/2){1-y+(1/4)y2}-(3/2)y+(3/4)y2= (3/2)-(3/2)y+(3/8)y2
    Now the outer integral:
    I remarked in class that, maybe it should be "clear" to me that the volume is 1, but it isn't.

    The other iterated integral
    Now, just to practice, we'll write ∫∫The triangle3-3x-(3/2)y dA as a dy dx iterated integral.
    Again, I will work from the outside in. So first I need to get the leftest (leftmost) and rightest (rightmost) values of x in the triangular base:
    Leftmost xRightmost x???3-3x-(3/2)y dy dx
    Now the base triangle is again shown to the left, but with the kind of "doodles" that I would make suitable to finding the limits of a dy dx iterated integral. The leftmost value of x is 0 and the rightmost value of x is 1. Now my dx by dy triangles form a vertical strip where x is just about constant. For the inner limits on y I need to know that the strip goes from the bottom, where y=0, to the top. The top will vary, depending on x. The equation of the boundary line for the top is the same: x+(y/2)=1. Now we need to know y as a function of x. So solve for y and get y=2-2x. That's the upper limit on the dx integral. Here is the resulting iterated integral:
    01y=0y=2-2x3-3x-(3/2)y dy dx.

    Computing the iterated integral
    I'll first compute the inner integral:
    y=0y=2-2x3-3x-(3/2)y dy= (antidifferentiate with respect to y, so x is a constant here!) 3y-3xy-(3/4)y2]y=0y=2-2x= 3(2-2x)-3x(2-2x)-(3/4)(2-2x)2-0. Again, the -0 comes from the lower limit, y=0. Now expand and simplify:
    The outer integral:
    Thank goodness, we got 1 again.

    Possible sources of error in these computations
    I'm looking ahead a little bit here. We will discuss triple integrals next time, and these are also usually computed by a transition to triple iterated integrals. There are six possible orders for iterated triple integrals. I make errors frequently. The prominent sources of error include: antidifferentiating with respect to the wrong variable, substituting for the wrong variable, and, well, general confusion. Please try to guard against these. You may make these errors, and just do the computation again, and try to keep your composure intact ("Keep cool, y'know!").
    By the way, although I wanted our first example to be as easy as possible, when I was typing up the diary notes above, I ... made several errors and had to go back and redo things. Oh well.

    Another one
    The base of a solid is the region in the first quadrant of the xy-plane bounded by the curve y=x2 and the line y=3x. The height over the xy-plane is given by z=x6y7. Find the volume of this solid.
    The double integral is ∫∫BaseHeight dA, and this is ∫∫The shapex6y7 dA.

    One iterated integral, with its computation
    We will convert this first to a dy dx integral. The outside limits come first. The most left x gets on the base is x=0. The most right x gets is x=3. We know this because we graphed the base, and found the intersection points of y=x2 and y=3x by solving 3x=x2, which has roots at x=0 and x=3. Therefore the iterated integral looks like:
    03???x6y7dy dx
    What about the limits on y? Here the sketch of the base, together with my doodles, may be useful. The vertical strip of boxes tells me that I should add up things from y=x2, the lower bound, to y=3x, the upper bound. Therefore this iterated integral is:
    03y=x2y=3xx6y7dy dx
    Now to compute the integral. The inner integral:
    This "simplifies" to (1/8)38x14-(1/8)x22. (I am using the powerful rules of exponential manipulation here!) And now the outer integral:
    ∫03 (1/8)38x14-(1/8)x22dx= (1/8)38(1/15)x15-(1/8)(1/23)x23]03=
    (1/8)38(1/15)315-(1/8)(1/23)323=(1/8)323([1/15]-[1/23]). Wow!

    The other iterated integral, with its computation
    Now for the dx dy integral. The highest and lowest values for y are 0 and 9. Therefore the integral must be:
    09???x6y7dx dy
    Now we need to consider a (fixed) y slice through the base. The left-hand side of that fixed y slice is determined by y=3x and the right-hand side of the slice is determined by y=x2. We need to know the limits on x in terms of y. So we need to know x=Left(y) and x=Right(y). That means "solving for x" in the boundary equations. This (here, in this classroom example!) is not too hard. y=3x becomes x=(1/3)y on the left, and y=x2 becomes x=sqrt(y) on the right. The positive square root gets used here because (picture!) we're in the first quadrant. The iterated integral is:
    09x=(1/3)yx=sqrt(y)x6y7dx dy
    The computation begins with the inner integral.
    This now "simplifies" (what a silly word!) to (1/7)y(21)/2-(1/7)(1/3)7y14. Now the outside:
    (1/7)(2/(23))9(23)/2-(1/7)(1/3)7(1/15)915= (1/7)(2/(23))323-(1/7)(1/3)7(1/15)330=
    (1/7)(2/(23))323-(1/7)(1/15)323=(1/7)323([2/(23)]-[1/15]) (1/7)(1/16)325-(1/7)(2/(23))323=


     1     / 1      1  \     1    / 1      2  \
    --- · | ---- - ---- | = --- ·| ---- - ---- |
     8     \ 15     23 /     7    \ 15     23 /
    The proof consists of observing that the dx dy and dy dx values of the double integral must be equal by the Fubini result, and then dividing both values by 323. (The student may, of course, verify this statement using the tools of third grade arithmetic, but the prestige of double integrals is ... [priceless?].)

    This "theorem" may be the silliest statement of the course.

    Pi? The Bible??
    Some students objected when I tried to write Pi instead of 3 as I was assembling information for the computation above. I cited the Bible as a reference. Here is information about the "history of Pi" including specific biblical citations. Since we are at a secular university, I give this not for its religious value, but for ... uhhhh ... historical context. Also, maybe ... uhhhh ... maybe the value of Pi has changed with time. Uhhhh ... also everyone should know a few digits of Pi. This reference discusses how the computation of the first trillion or so decimal digits was done. You can search the first four billion bits (binary digits) of Pi for patterns here.

    More on difficulties and on the psychology of the individual
    Please don't panic. If you want to compute a double integral, you don't need to do both iterated integrals -- just one of them. I chose to do both to show you how (I hope!).
    Notice that we needed to go from one description of the boundary curves: {y=3x, y=x2}, to another: {x=(1/3)y,x=sqrt(y)}, when we did dx dy after dy dx. "Solving" (finding a convenient form for inverse functions) may be difficult (or even impossible in terms of familiar functions).
    One last remark: I almost always try to find the bounds on iterated integrals going from the outside-most integral to the inside-most integral. Some people may find the transition from inside to outside more easy (this difference in approach will be more emphatic when we do triple integrals). You should try a series of examples and settle upon what you find most comfortable. And remember that you can always "change" to the other way.

    By the way ...

    > int(int(x^6*y^7,x=(1/3)*y..sqrt(y)),y=0..9);
    > int(int(x^6*y^7,y=x^2..3*x),x=0..3);
    So Maple gets the same answer both ways, also.

    Integrating Frog over a region
    If the integrand (the function to be integrated) is not too weird, then I hope you should be convinced that the actual antidifferentiations probably aren't the essential difficulty. The difficulty is more in setting up the iterated integrals: finding the bounds. Here is a more complicated example.

    The region R is bounded by y=x+2 and y=x2. What does this region look like? Well, this is a problem in a calculus course, so solving x2=x+2 shouldn't be impossible. In fact, this leads to x2-x-2=0 which is (x-2)(x+1)=0 so intersections occur when x=-1 (so y=(-1)2=1) and when x=2 (so y=22=4). I would like to integrate Frog over the region R:
    ∫∫RFrog dA. More precisely, I'd just like to set up the bounds of the iterated integrals which are equal to this double integral.

    Totally randomly, dx dy first
    This wasn't a random choice. The dx dy order introduces an additional kind of complexity. Consider these limits:
    Bottom of yTop of yx=Left(y)x=Right(y)Frog dx dy.
    The Top of y and Bottom of y are easy enough. In the region R, the smallest y value is 0 and the largest y value is 4. Now think about x=Left(y) and x=Right(y). The thick horizontal blue line in the sketch separates different formulas for the Left limit of x as a function of y. Below it, the Left limit is determined by the left-hand side of the parabola. Above it, the Left limit is determined by the straight line. Theoretically this does not cause any problems. But when you're actually trying to compute everything, what people usually do is separate the pieces:
    01x=Left(y)x=Right(y)Frog dx dy +14x=Left(y)x=Right(y)Frog dx dy.
    In the first iterated integral, as y goes from 0 to 1, the left and right boundaries are both given by formulas related to y=x2. Here x=+/-sqrt(y), so Left(y)=-sqrt(x) and Right(y)=+sqrt(x). In the second iterated integral, y goes from 1 to 4. Here also Right(y)=+sqrt(x), but Left(y) comes from y=x+2, so Left(y)=y-2. So the dx dy iterated integrals which are equal to the double integral are:
    01x=-sqrt(y)x=sqrt(y)Frog dx dy +14x=y-2x=sqrt(y)Frog dx dy.

    Now dy dx
    This one is much easier. I can read off the left and right extreme values of x, and then the y boundary values are given by the equations which already "present" the region R. I don't need to split up things. Here it is:
    -12y=x2y=x+2Frog dy dx.
    Almost surely, unless circumstances were very strange, I would set up the iterated integral this way and not in the dx dy way.

    The New Jersey Chorus Frog, Pseudacris feriarum kalmi, is an endangered species in some of its range.

    In my office ...
    A student visited me and we did some more Frog-like problems (one with Toad and one with Lizard. Let me show you the Lizard problem. The region we looked at was bounded by y=2-2x2 and y=x4-x2. A Maple graph with these two functions is shown to the right. Just as in the previous (Frog) example, writing the dy dx integral is quite direct. My student visitor and I wrote iterated integrals for the other order: dx dy. You can try this problem. First, though, think a bit and see how many iterated integrals will be necessary.

    I think three pieces are needed. Notice that y=x4-x2 is the same as x4-x2-y=0 and this is (x2)2-x2-y=0, a quadratic in x2. So you can "solve" for x2 using the quadratic formula, and then take square roots of the resulting answers to get 4 possible values of x for each value of y. Certainly this is a mess, but this is possible to do without extravagantly advanced methods.

    A problem from a calculus textbook
    This problem shows one further "wrinkle" that can occur with double or triple or any kind of "multiple" integral. Here's the statement:
    Evaluate the double integral ∫∫Dex/ydA, where D={(x,y)|1<=y<-2, y<x<=y3}.
    As to why this problem introduces a new kind of complexity, I invite you to ask Maple to integrate ex/y both dx and dy. That is, what are the respective antiderivatives? The x antiderivative is yex/y but ... there is no y antiderivative in terms of familiar functions. So if we want to get an answer to the textbook's problem, we'd better first do dx and then leave dy until later. The dx dy double integral is easy enough to write, because the region D is described suitably:
    12x=yx=y3ex/ydx dy
    The inner antidifferentiation gives ∫x=yx=y3ex/ydx=yex/y]x=yx=y3=yey3/y-ye1=yey2-ey and then ∫12yey2-ey dy is just (1/2)ey2-(e/2)y2]12=(1/2)e4-2e.
    Comment I do know some examples, in "real" applications, not textbooks, where looking at the order of the iterated integrals changes something really nasty into a function which can be handled routinely. So, although this is a textbook/class example, it does show an idea which may be useful.

    Monday, October 20

    Quick review of the definite integral in calc 1
    Suppose we have a nice function f(x) of one variable defined on and interval, [a,b]. We might want to find the area under y=f(x) on that interval, althogh that is more of an excuse to define the definite integral than a real ambition. Here is one approach to the definition. Take a large positive integer n and divide [a,b] into n equal parts each of width x=(b-a)/n. In each subinterval choose a sample point, say qj in the jth subinterval. Compute the sum ∑j=1nf(qj)x. This is a Riemann sum approximating the definite integral. As n-->infinity, any sequence of sums is supposed to approach a unique limit, and that limit is the definite integral. But there are many choices of sample points, and maybe it isn't clear that the choices don't influence the limit. So what if we take another point pj in the jth subinterval as a sample point? Then since the width is Δx, certainly |qj-pj|≤x. The Mean Value Theorem (let me assume that f is differentiable) then says there's some constant, C, so that |f(qj)-f(pj)|≤Cx. (I'm not too interested in the details here -- we're only skimming!) But we can estimate the difference when different choices of sample points are made:
    |j=1nf(qj)x - ∑j=1nf(pj)x|≤nCxx.
    The "n" comes because there are n pieces in the sum. One of the x's occurs because that's a common factor in both sums. The other comes from the MVT estimate we just stated. But now, since x=(b-a)/n, we see that the estimate of the difference is C(b-a)2/n. Two of the n's cancel. We are left with an n on the bottom, and this means that the Riemann sums do get closer as n→∞, no matter what choice of sample point is made.

    Theorem For any choices of sample points, as n→∞, the Riemann sums→a unique limit, the definite integral of f from a to b, and written ∫abf(x) dx.

    Of course the integral sign and the dx's are notation to remind people of the approximating sums. We could approximate the definite integral with Riemann sums, and of course there are many other numerical approximation schemes. But the champion method is:

    FTC If F´=f, then ∫abf(x) dx=F(b)-F(a).

    Defining the double integral
    In fact the definition more or less parallels the single integral definition. I'll follow the text closely here. We begin with a nice function (say, continuous) defined on a rectangle R in R2 with bouindaries x=a, x=b, y=c, and y=d. Chop up the area of the rectangle into a bunch of chunks. In each chunk, choose a sample point. Compute the corresponding Riemann sum, the sum over all the chunks of f's value at the sample point multiplied by the area of the chunk. If f(x,y)>0 on R, then this Riemann sum approximates the volume under z=f(x,y) and over R. Then it's true that as the maximum size of the chunks→0, the Riemann sums→a unique limit. This limit is the double integral over R of f(x,y): ∫∫Rf(x,y)dA. This is a mathematical abstraction of the volume. The volumes computed as a result of formulas in earlier calculus (solids of revolution, solids with simple cross-sections, etc.) take advantage of symmetry. The theoretical tool defined here allows us to compute volumes without any simple kinds of symmetry. As I mentioned in class, numerical computations of double integrals (and other higher-dimensional creatures) are sometimes necessary, but the computations get much more intricate than the simple ideas shown in calc 2.
    Almost all the computations of volumes that I've made in my life have occurred as a result of teaching third semester calculus. Maybe the following is a bit more interesting.

    Mass of a plate
    Maybe this is a more realistic "scenario". Suppose you are given a thin rectangular metal plate with an unknown density distribution. Therefore this is not necessarily a homogeneous thin plate. The plate is too heavy or too unwieldy to weigh directly, and you need to estimate the total mass. Also the mass distribution -- the density -- is not necessarily given by a simple formula. What maybe could be done is tiny Samples taken at various parts of the plate, according to some method (maybe depending on accessibility or expense or ... anything). Then these samples could have their density measured, and maybe then, after dividing the plate (thoughtwise!) into pieces, the sample densities could be multiplied by the areas of the pieces. The sum of these products would then be an estimate for the mass in the plate. (It is a Riemann sum.) If a better (more accurate?) estimate was wanted, maybe then use more sample points, smaller areas, etc. The process is exactly the same mathematics as the definition of the double integral.

    Well, you might work for Schlumberger and your sample points might cost three to five million dollars each as you try to investigate the oil or gas quantities of some region. You'd then really think a bit about the whole process. And that's what they do.

    Some very simple examples of double integrals
    Example A The rectangle R is defined by x=3 and x=7 and y=5 and y=8. The function is f(x,y)=700. Then ∫∫Rf(x,y)dA is 700(7-3)(8-5). Of course, I am using the fact that the volume described by the double integral is the volume of a rectangular solid with edge dimensions 700 and 7-3 and 8-5.
    Example B Here I took f(x,y) to be 5-x2+y2, definitely a more complicated function than the previous example's. The rectangle I took was defined by x=-3, x=3, y=-3, and y=3. Let's temporarily discard the 5 and concentrate on -x2+y2. If I interchange x and y the sign of the function's value changes. But the rectangular domain is symmetric about (0,0), so the net value (+'s and -'s cancelling!) of the double integral of -x2+y2 over the rectangle is 0 (!). Now I "integrate" the 5, and the result is 5(3-(-3))(3-(-3)). So we did have a more complicated function but the choice of domain made the hard part of the function drop out.

    Basic properties
    The basic properties of the double integral are exactly like those of the 1-dimensional version.
    • Linearity If f and g are both nice functions defined on the rectangle R, then ∫∫Rf(x,y)+g(x,y)DA=∫∫Rf(x,y)dA+∫∫Rg(x,y)dA; if k is a constant, then ∫∫Rkf(x,y)dA=k∫∫Rf(x,y)dA.
    • Order If f(x,y)≤g(x,y) for all points (x,y) in the rectangle, then ∫∫Rf(x,y)dA≤∫∫Rg(x,y)dA.

    How to compute?
    Maybe all this theory is very nice, but let me show you the way most double integrals are computed. One method of chopping up a rectangle is to use vertical and horizontal lines, parallel to the sides. So we get a grid of subrectangles, each x by y, with both 's very small. In addition to choosing this special chopping strategy we could also decide to add up the contributions (the f(sample point)xy) in an orderly manner. So, for example, we could add up the contributions from the lowest "row" first. In that row, since y is small, y hardly varies at all. The sum of a row sure looks like the definite integral with respect to x only for "that" value of y. The same can be done for each row. When the row sums are done, we now have the y's to worry about. But this is a y integral. Of course, a completely symmetric procedure can be used in the other order: first dy, with x held constant, and then dx. Again what's here is not a "proof" but I hope the discussion supports the following result.

    Fubini's Theorem
    Suppose f(x,y) is a continuous function in a rectangle R defined by x=a, x=b, y=c, and y=d. Then
    ∫∫Rf(x,y)dA=∫cdabf(x,y)dx dy=∫abcdf(x,y)dy dx.
    Comments The first "creature" in the equation above is called a double integral. The two others are officially called iterated integrals: iterated means "repeated". Technically and precisely these integrals are different creatures. Also please note that the outside integration limits go with the outside d-variable -- sometimes this can be confusing. When it is, I write things like "x=a" instead of just "a" so I don't confuse myself.

    Example 1
    Let me try to compute ∫∫R x3y7dA where R is the rectangle defined by x=1, x=4, y=2 and y=5. The Fubini Theorem allows me to "trade in" the double integral for an iterated integral, either dx dy or dy dx. There are some occasions where one order or the other might be preferable (we'll see this later) but here I don't think that happens. So:
    ∫∫Rx3y7dA=∫2514x3y7dx dy.

    I'll begin by computing the inner integral:
    14x3y7dx=(1/4)x4y7]14. In this antidifferentiation, the y7 is a constant (this is, not surprizingly, exactly the inverse of partial differentiation). Then we evaluate and get (1/4)44y7-(1/4)14y7=(255/4)y7. I remarked in class that I sometimes lose my way in these computations, and need to write (1/4)x4y7]x=1x=4 to insure that I remember to substitute for the correct variable.

    Now ∫25(255/4)y7dy=(255/4)(1/8)y8]25=(255/4)(1/8)58-(255/4)(1/8)28.

    My silicon buddy ...
    A report from Maple:

    > int(int(x^3*y^7,x=1..4),y=2..5);

    Example 2
    Let's try a random function: f(x,y)=sqrt(3x+8y). Well, this isn't so random (as you'll see!). I just wish I had thought of it in class. The rectangle I have in mind has these boundary lines: x=0 and x=3 and y=0 and y=2. In the previous example we converted the double integral into a dx dy iterated integral. Let me try a dy dx order here. I think that either order, again, is about the same amount of work.

    So let's try:
    ∫∫Rsqrt(3x+8y)dA=∫0302sqrt(3x+8y) dy dx.

    The inner integral is ∫02x3sqrt(3x+8y) dy. We need a "dy" antiderivative of sqrt(3x+8y). Here we can really get confused! To me writing the function as (3x+8y)1/2 makes the problem easier. I guess that the antiderivative will be something close to (3x+8y)3/2. Well, but I need to multiply by stuff to get rid of the various constants. For example, I need to multiply by (2/3) because of the power. And I need to multiply by (1/8) because of the coefficient of y that the chain rule will push out. So the answer is (2/3)(1/8)(3x+8y)3/2, and we must substitute:
    (2/3)(1/8)(3x+8y)3/2]y=0y=2= (2/3)(1/8)(3x+16)3/2-(2/3)(1/8)(3x+0)3/2, and this is (1/12)(3x+16)3/2-(1/12)(3x)3/2.

    And now the outer integral:
    03(1/12)(3x+16)3/2-(1/12)(3x)3/2dx=(1/12)(1/3)(2/5)(3x+16)5/2-(1/12)(1/3)(2/5)(3x)5/2]03= {(1/90)(3·3+16)5/2-(1/90)(3·3)5/2} -{(1/90)(3·0+16)5/2-(1/90)(3·0)5/2}= (1/90)(255/2-95/2-165/2+0)= (1/90)(55-35-45)= (1/90)(3125-243-1024)=(1858/90)=(929/45).
    Well, y'see, everything was chosen so that the final answer would have no square roots. Isn't that wonderful!

    Or, in the 21st century ...

    > int(int(sqrt(3*x+8*y),x=0..3),y=0..2);

    Exams returned
    The exams were returned. Grading information is available, a version of the exam can be inspected, as well as some answers to tha version.

    Wednesday, October 13

    We studied the following problem from 1 variable calculus:
    Consider the ellipse x2+5y2=1. Find the rectangle of largest area inscribed in this ellipse with sides parallel to the coordinate axes. Of course this turns into: maximize 4xy (the objective function) subject to x2+5y2=1 (the constraint). Consideration of the geometry (varying rectangles) suggests that there is indeed a "biggest" rectangle, somewhere.
    How the "heck" does a calc 1 student solve this problem since the function to be maximized, 4xy, has two variables. I suggested the following methods of solution:
    1. Reduction of dimension, simple version
      Since x2+5y2=1, we know that y=sqrt((1-x2)/5). Then the area is F(x)=4x·sqrt((1-x2)/5). The domain for this function is 0=<x=<1. General theory from one variable calculus states that max/min are obtained at end points or critical points. But F(0)=F(1)=1. So the max is gotten where F'(x)=0. We computed this. Of course, in a random situation, it may be very difficult to solve (effectively!) for one of the variables in terms of the other.
    2. Reduction of dimension by parameterization
      We could make an inspired "guess": try x=cos(theta) and y=sin(theta)/sqrt(5). Then the pair (x,y) is on the ellipse, and since the max is obtained somewhere in the first quadrant, we are left with maximizing 4xy=(4/sqrt(5))cos(theta)sin(theta)=(2/sqrt(5))sin(2theta) for theta between 0 and Pi/2. This can be solved almost "by inspection": just take theta to be Pi/4. The max value is then 4/sqrt(5). Of course, in a "random" situation it may be very difficult to get nice parameterizations.
    3. A weird way to do the problem
      I had Maple sketch some level curves of 4xy, the objective function, and compare them with the constraint curve x2+5y2=1. Here is the result of these Maple commands.
      B:=implicitplot(x^2+5*y^2=1, x=-4..4,y=-4..4,color=blue,
      thickness=2, scaling=constrained, grid=[80,80]): 
      The picture is shown to the right.
      A close-up view
      Suppose you consider a level curve of the objective function that crosses the constraint curve, as shown. One math word which applies to this situation is that the two curves are transversal. So we have 4xy=C crossing x2+5y2=1. What happens if we "wiggle" C a little bit, so we consider 4xy=C+epsilon and 4xy=C-epsilon. Now it seems reasonable (4xy is certainly continuous, so its values don't hop around or break or anything) that these level curves are close to 4xy=C. These level curves must also cross the constraint curve. That means the function 4xy has values C+epsilon and C-epsilon on the constraint curve. (The level curves are exactly where that function takes on its values!) Since there are both larger and smaller values of 4xy on the constraint curve, C can't be an extreme value (either max or min) for 4xy on x2+5y2=1.

      Local picture near a level curve
      corresponding to a non-extreme value
      Another close-up view
      This seems to imply, if you examine the picture closely, that the largest (and the smallest) values of 4xy will be at points on the ellipse where the ellipse will be tangent to level curves of the constraint, x2+5y2=1. If the level curves of the objective function are not tangent, then we will be able to vary the values of the constant generating that contour and get bigger and smaller values of the objective function on the constraint curve.

      If the level curves are tangent then the normal vectors of the constraint curve (∇f at that point) and the objective function (∇g) at that point will both be perpendicular to the same line (in three dimensions it would be a tangent plane). These gradient vectors may not be exactly the same vector, but one of them must be a scalar multiple of the other.

      Local picture near a level curve
      corresponding to an extreme value
      Now the algebraic side
      If f(x,y)=4xy and g(x,y)=x2+5y2=1, then at such points (extreme values of the objective function on the constraint curve), there is some real number λ so that ∇g=λ ∇f (everyone uses this Greek letter) because the tangent lines are the same, and therefore the normal vectors must be parallel: one must be a scalar multiple of the other. This one vector equation in R2 gives two scalar equations, one for each component of the vectors:
      This, together with the constraint equation g(x,y)=x2+5y2=1 gives a system of 3 equations in 3 unknowns. We can solve this by, for example, solving for λ in each of the first two equations and setting them equal. We need to watch out for spurious solutions or evasions of solutions. These may occur when we divide by certain variables. This gave us another way to solve the maximization problem, a method which is more in the spirit of several variable calculus. It turns out that this strange idea is actually quite useful in "real world" problems. The method is called Lagrange multipliers and is discussed in section 14.8 of the text. The method is used extensively in economics and in many areas of engineering.

    Example #1
    Here the constraint is x2+xy+y2=1, and the function to be maximized, the objective function, is x2+y2. The picture corresponding to this situation is shown to the right. The bigger circles correspond to larger values of the objective function.
    Suppose that T(x,y)=x2+y2 were the temperature in a thin metal plate with shape the interior of x2+xy+y2=1, where will the plate be hottest or coldest? I remind you that in this "heat" language the level curves or contour lines are called isothermals.
    Well, local extrema only occur at critical points, and only (0,0) is a c.p. That, easily, is the coldest point in the plate. But where is the hottest point? It must be on the edge, and it will NOT be a local extremum, but only an extremum for a constrained maximization. We seek therefore the extrema on the boundary using Lagrange multipliers.
    Compute the gradients, etc. Then the multiplier equations and the constraint equation are:

    Again we can solve with (2x+y)/(2x)=(2y+x)/(2y) so x=+/-y (and possible special cases of x or y being 0). And so the temperature is going to be T(x,y)=2 or 2/3 since x2+xy+y2=1 gives x2=1 or x2=1/3. There are no solutions with x or y equal 0, because if one of them is 0 then the other is also 0 (using the two multiplier equations) and the point (0,0) does not satisfy the third equation. Here is a picture of these special isothermals T(x,y)=2 and T(x,y)=2/3, and the constraint.

    Fan mail for the Lagrange multiplier method
    I think it is wonderful that a relatively small amount of algebraic effort can produce such a lovely geometric result (the specific circles centered at (0,0) which are also tangent to the ellipse). This reassures me that things algebraic and geometric both reflect the same reality.

    Example #2
    Find the maximum and minimum values of 3x-4y+5z on the unit sphere x2+y2+z2=1. Here is perhaps a more complicated picture, with the constraint (the unit sphere) and five planes representing where f(x,y,z)=3x-4y+5z=-8 and -3 and 1 and 5 and 9. The picture is supposed to help you understand that max/min occur where the planes will be tangent to the sphere.

    The system of Lagrange multiplier equations (three of them here, since we are in R3) together with the constraint follows.

    The left-hand sides are the components of ∇(x2+y2+z2) and the right-hand sides are λ multiplying the components of ∇(3x-4y+5z). You can solve for x and y and z in terms of λ, and substitute these values in the constraint equation, getting λ=+/-(2/sqrt(50)). Then 3x-4y+5z turns out to be (for the two choices of λ, generating two candidates for where extreme values take place) sqrt(50) and -sqrt(50). Here a final picture of the constraint and the two planes given by 3x-4y+5z=+/-sqrt(50).

    Proofs, etc.: the dual (?) nature of math
    I learned and "liked" Lagrange multipliers in a several variable calculus course, just as I hope you are. The justification for the method was more or less what I have shown you. So I knew it was "true". But I never saw a "proof" of the Lagrange multiplier method until my second year of grad school. Sigh. It really isn't that difficult to prove. Maybe I didn't (even as an apprentice professional mathematician!) feel the need to prove such a lovely idea.

    More examples
    I discussed in detail some more complicated examples when I taught an honors version of this course. Here is a link to an algebraic analysis of the examples. The examples are complicated, and some complicated pictures of them can be viewed here.

    Wednesday, October 8

    Review of 1 variable
    I try not to work hard, so I thought maybe a quick review of extreme value material from 1 variable calculus would be useful. The names of ideas to recall include these:
    maximum, minimum, absolute maximum, absolute minimum, local maximum, local minimum.

    Fermat's fact
    What I called "Fermat's fact" was the following wonderful observation in one-variable calculus:
    If f is differentiable at x0 and if f´(x0) is not 0, then f does not have an extreme value at x0.

    The picture shows a "proof" (well, I hope fairly convincing to a picture person). If there is a tilt in the tangent line, then there are both higher and lower values near x0. If x0 is either kind of extreme value (max/min), then we see that f´(x0) cannot be 0.

    Critical number
    Therefore the following definition is written.
    x0 is a critical number of the function f if either f is not differentiable at x0 or f´(x0)=0.
    For simplicity in this discussion I'll assume that f is defined in some interval that has x0 inside it (in the interior).

    Identifying ("classifying") the type of critical point
    Last time we saw the 68th derivative test. The idea was to use Taylor's Theorem and then use the (crazy) hypotheses to make a big chunk of the Taylor polynomial vanish. The function then qualitatively resembled the next term of the Taylor series.

    And now let's look at more than 1 dimension: maximum, minimum, absolute maximum, absolute minimum, local maximum, local minimum. These words and phrases mean more of less then same, but the max/min situation in more than one variable is much more complicated. Some examples will be useful.

    The simple pictures with simple formulas
    Here are some pictures and some formulas.
    Discussion and formulasThe pictures
    A function defined on all of R2 with a local (and absolute) minimum is f(x,y)=x2+y2. The graph of this function is a surface called a paraboloid. It is a nice, smooth "cup" opening up. Vertical slices through (0,0) are all parabolas opening up and the contour lines are circles.
    The red dot is the critical point and the brown plane is the tangent plane at that point (the xy-plane).
    The simplest local and absolute strict maximum is, of course, just the reflection of the previous example, done with minus signs algebraically. So here f(x,y)=-x2-y2, and (0,0) provides a strict maximum. The graph is a paraboloid whose axis of symmetry is again the z-axis. This graph opens "down".
    A saddle
    The function f(x,y)=-x2+y2 gives a nice example of a saddle point. The xz-slice (where y=0) shows the curve z=-x2 and the yz-slice (where x=0) shows z=y2. Each has a (strict) extreme point at 0. One is a max and one is a min. Such behavior is called a saddle point. Perhaps the behavior most similar in one variable calculus would be that of the function x3 (an inflection point). But in 2 and more variables the local situation can be much more complicated.
    Here the surface is more complicated, and my picture is certainly not so good. But the tangent plane and critical point are the same. The tangent plane cuts through the surface (similar to the way a tangent line at an inflection point in 1 variable calculus cuts through the graph of a curve).

    Ridiculous interesting (?) fact
    The surface z=-x2+y2 is what's called a ruled surface. The word "rule" comes from an older version of English, and here means "straight line". Every point on the surface is on a straight line which sits entirely in the surface. I didn't believe this when I was told it for the first time, and maybe you don't either. For example, the point (1,2,3) is on this surface (since 3=-12+22) and the line whose parametric equation is x=1+t, y=2-t, and z=3-6t goes through that point and sits entirely on the surface. Such surfaces have applications in computer graphics.
    You can verify algebraically that the line whose parametric equations are given above is on the surface, or you can look at the picture to the right, which shows the saddle together with the line: nearly silly.

    A surface with corners?
    I wanted to give an example of a function which might make you think a bit. Consider f(x,y)=|x+3y|+|5x-y|. This is a strange function, but it is easily computed. For example, f(4,-2)=|4+3(-2)|+|5(4)-(-2)|=10+22=32. So computationally, this is an easy function. To the right is the result of this Maple command:
    plot3d(|x+3*y|+|5*x-y|,x=-3..3,y=-3..3, axes=normal,grid=[75,75])
    The grid option makes the program sample at 75·75 points in a rectangular arrangement. This is much more than the default value, and will usually make the sketch of the surface more accurate. It can take much more computing time and storage, however.

    I claim that this function has an absolute minimum at (0,0). Why? Certainly f(0,0)=0. And if the value of f equals 0, then (since absolute values are always non-negative) both x+3y and 5x-y must be 0. The only solution of the system of equations x+3y=0 and 5x-y=0 is (0,0) (this is either "high school math" or sophisticated linear algebra) so f(x,y)>0 for all (x,y) which are not (0,0).

    Using Fermat's fact here
    If a point is a local extreme point of some function f in several variables, and if that function is differentiable at that point, then all of the first partial derivatives of the function must be 0 at that point. If that's not true, just "slice" the function at that point in the direction of the derivative which is not 0. The one variable Fermat fact implies that the function does not have an extreme value (max or min) at the point in one variable, and therefore the function in several variables has both higher and lower values near the point. Therefore (whew!):
    An extreme point must be a critical point.
    Our functions will almost always be differentiable (not like the graph with corners above), so our functions will have their extreme values where ∇f=0.

    Definition of critical point
    Suppose f is a function of n variables. Then f has a critical point at p in Rn if either ∇f doesn't exist at p (so at least one of the first partial derivatives fails to exist) or ∇f(p)=0 (the zero vector, remember!).
    In this course, almost all the functions we'll consider will be differentiable. This doesn't mean that non-differentiable functions (functions with jumps or corners) are not important or interesting in mathematics and its applications (again: linear optimization, shock waves in physical phenomena). Just learning to use the tools for higher dimensional analysis of differentiable functions is a big enough task.

    Another instructor's final exam question
    Suppose f(x,y,z,w)=3x6+8y12+55z8+9w64. What are the critical points of f and what type (max, min, saddle) are they?
    As I remarked, this problem seems a bit forbidding. Bug fx3·6x5. The only way this is 0 is for x to be 0. Similar remarks for fy, fz, and fw imply that the only critical point for this function is (0,0,0,0).

    What type of critical point is (0,0,0,0)? Well, f(0,0,0,0)=0. And if any of the coordinates in (x,y,z,w) is not 0, then (since we have only even powers!) f(x,y,z,w)>0. Therefore this critical point is an absolute minimum.

    Suppose z=f(x,y), and f is differentiable. What is the geometric meaning of "(x0,y0) is a critical point of f"? Since ∇f(x0,y0)=0, both of the first partial derivatives are 0. Therefore z=f(x0,y0) (that is, z=a constant) is the tangent plane to z=f(x,y) at the point (x0,y0,f(x0,y0)). The "flat" plane through the point, parallel to the xy-coordinate plane, is tangent to the surface. This can be difficult to "see" in a graph, though.

    Monkey saddle
    The examples already shown are the standard critical points for functions of two variables. But there are many, many other kinds of critical points. The graph z=x3-3xy2 shows one of them. Again the origin, (0,0), is the only critical point. This is because zx=3x2-3y2 and zy=-6xy. For zy to be equal to 0, either x or y must be 0. But then use zx=0 to conclude that the other variable must be 0 also. For this function, the xy-plane is the tangent plane at the origin because (0,0,0) is on the graph. This critical point's local behavior is up/down repeated three times (at equally spaced 120o angular intervals) if you walk around the surface in a small circle centered at the origin. The critical point is called a monkey saddle because, presumably, a monkey could sit on it with spaces for two legs and a tail to hang down.
    Critical points of more than one variable can have many, many different local pictures, and there has been a great deal of effort expended trying to understand them.

    Taylor's Theorem in two variables
    Here is a result: f(x,y)=f(x0,y0)+
         fx(x0,y0)(x-x0)+fy(x0,y0)(y-y0) +
              [1/2!](fxx(x0,y0)(x-x0)2+fyx(x0,y0)(y-y0)(x-x0)+fxy(x0,y0)(x-x0)(y-y0)+fyy(x0,y0)(y-y0)2) +

    Here the first line is the constant term, the second line has the first-order (degree 1) terms, and the third line consists of the second-order (degree 2) terms. Of course, the third line (because mixed partials are equal) is actually only fxx(x0,y0)(x-x0)2+2fxy(x0,y0)(x-x0)(y-y0)+fyy(x0,y0)(y-y0)2). If (x0,y0) is a critical point, then the linear (first order, degree 1) terms are 0. We can expect and hope that the shape of the degree 2 terms maybe looks like f near (x0,y0). The problem is that these degree 2 terms can themselves be a bit hard to understand. Here are pictures of three polynomials of degree 2 in x and y.

    I think this one is a saddle.
    Another saddle (there are lots of saddles in this business!).
    And this seems to be a maximum.

    Now a second derivative test in two variables
    There's one second derivative test which is usually "given" to students in a third semester calculus course. It is a bit complicated. The test essentially results from computing the second directional derivative at the critical point and seeing how to ensure that this result is always positive (or always negative or ...). That together with results from one variable calculus (on concavity) will insure some kinds of local behavior near the critical point. There is a description in the book, but I want to concentrate on stating the result and then giving some examples. That is enough of a task!

    Second derivative test for two variables
    Suppose f(x,y) is a differentiable function, and (x0,y0) is a critical point. That is, fx(x0,y0)=0 and fy(x0,y0)=0. Then compute D=fxx(x0,y0)fyy(x0,y0) -(fxy(x0,y0))2. Here we go:

    If D>0 and if fxx(x0,y0)>0 then f has a local minimum at (x0,y0).
    and if fxx(x0,y0)<0 then f has a local maximum at (x0,y0).
    If D<0 then f has a saddle point at (x0,y0).
    If D=0 then this second derivative test supplies no information.

    Please note that when D>0, the signs of fxx and fyy must agree (as I said in class, this is not obvious!) so that you can check either of these numbers. The textbook calls D the discriminant. It is also called the Hessian in some places. Also I mentioned that I think of D as the determinant of this:

    ( fxx  fxy )
    ( fyx  fyy )
    Looking at D this way turns out to lead to second derivative tests in more than 2 variables.

    Problem 19 of section 14.7
    We are asked to "find the critical points of the function. Then use the Second Derivative Test ..." and the function in problem 19 is f(x,y)=x-y2-ln(x+y).

    Let's find the c.p.'s. Since fx=1-(1/[x+y]) and fy=-2y-(1/[x+y]) we knw that 1-(1/[x+y])=0 and therefore x+y-1=0. In fact, x=1-y. Use this in the second equation, -2y-(1/[x+y])=0 by substituting for x. The result is -2y-(1/[1-y+y])=0 so that -2y-1=0 and y must be -1/2. Since x=1-y, x=1-(-1/2)=3/2. The only critical point for this f is (3/2,-1/2).

    Important note Solving non-linear equations can really be quite difficult, and each collection of such equations can present different "challenges". Any method that gets a solution is a good method, and there's likely to be more than one good method!

    Now let's use the Second Derivative Test. We know fx and fy. So we can compute fxx=1/(x+y)2, fxy=1/(x+y)2, fyx=1/(x+y)2, and fyy=-2+1/(x+y)2.
    A silly note about a habit of mine Since I sometimes make mistakes computing derivatives, I usually compute both fxy and fyx and quickly see that they're equal. This provides a small check in this computation.
    At the critical point, (3/2,-1/2), D=fxxfyy-[fxy]2=1·(-1)-12=-2, so this critical point is a saddle point. Luckily this is an odd-numbered problem in the textbook, and I can quickly check (as I just did) that this answer is correct!

    Euler's example
    Leonhard Euler (1707–1783) was a great and very prolific mathematician. He published Institutiones Calculi Differentialis (In English, Methods of the Differential Calculus) in 1755. It was an influential text, and was the first source of criteria for discovering local extrema of functions of several variables. In it Euler investigated the following specific example: V=x3+y2-3xy +(3/2)x. He asserted that V has a minimum at both (1,3/2) and (1/2,3/4). Was Euler correct? (My source for this information is A History of Mathematics by Victor J. Katz, Harper Collins, 1993, p.517.)

    Let's compute. Vx=3x2-3y+{3/2} and Vy=2y-3x. Critical points occur where both Vx and Vy are 0. There 2y=3x or y= {3/2}x, so the Vx condition becomes: 3x2-{9/2}x+{3/2}=0 or 6x2-9x+3=0 which factors (even then textbook problems were predictable!) into (2x-1)(3x-3)=0. The critical points are as Euler asserted: (1,{3/2}) and ({1/2},{3/4}). Logically just checking that Euler's points are critical points is not enough -- we should check that he found all critical points, which we did.

    Now to test the type of the critical points: Vxx=6x, Vxy=-3, Vyx=-3, and Vyy=2. So the discriminant is the determinant of

    (6x   -3) 
    (-3    2)
    which is 12x-9. At (1,{3/2}) this is 12-9>0, and Vxx=6>0, so this critical point is a local minimum. At ({1/2},{3/4}), the discriminant becomes is 12·{1/2}-9=-3<0, which makes this critical point a saddle point. Euler was wrong!

    Testing the monkey saddle
    This shows another weakness of the Second Derivative Test. If z=x3-3xy2, then we saw that the only c.p. was (0,0). Here zx=3x2-3y2 and zy=-6xy, so that zxx=6x, zxy=-6y, zyx=-6y, and zyy=-6x. When x=0 and y=0, all of the second partial derivatives are 0, so that D=0. The Second Derivative Test returns no information. This test says "saddle" only when the function has a second-order saddle point -- any other type of saddle behavior forces D=0. A second-order saddle goes up/down/up/down as you walk around the critical point, and the second-order saddle has D<0. So in a number of ways, the Second Derivative Test has problems. It can also be difficult to compute by hand, for example, as some examples below indicate.

    Two more functions
    There are two amazing and disconcerting examples. At least, to me these problems are both amazing ("surprise greatly; overwhelm with wonder" -- well, at least the first) and disconcerting ("disturb the composure of; agitate; fluster" -- certainly they show me I don't understand too well what can happen in "space"). The results show some huge differences between 1 and 2 dimensions.

    One strange example
    The function f(x,y)=-(x2-1)2-(x2y-x-1)2 is given. This is not the world's most horrible function. It is "only" a polynomial of degree 6. Let me find the critical points.

    Well, fx=-2(x2-1)2x-2(x2y-x-1)(2xy-1) and fy=-3(x2y-x-1)x2.

    Consider the equation fy=0 first. Well, maybe x=0. Then fy=0 and fx=0-2(-1)(-1) is not 0. So this doesn't get me any critical points.

    Now to get fy=0 we can ask that x2y-x-1=0. Then fx=0 becomes (x2-1)2x=0 since the other piece becomes 0. Then either x=0 or x=1 or x=-1. Whew! If x=0, x2y-x-1=0 becomes -1=0: false. If x=1, x2y-x-1=0 becomes y-2=0 so y=2. Therefore (1,2) is a critical point. If x=-1, x2y-x-1=0 becomes y=0, and (-1,0) is a critical point.

    If you don't like this logical torture, try the following:

    > f:=-(x^2-1)^2-(x^2*y-x-1)^2;
                                   2     2     2           2
                           f := -(x  - 1)  - (x  y - x - 1)
    > solve({diff(f,x),diff(f,y)});
                            {x = 1, y = 2}, {x = -1, y = 0}
    Yup, two critical points. Below are two very local pictures of the graphs near the critical points.

    The pictures certainly shouldn't be convincing evidence, but they do seem to support the assertion that the function has local minimums at both critical points! (We can verify this assertion with the second derivative test but this is too much work for me right now.) Why do I find this disconcerting? Well, imagine we walk from one peak to another (shown to the right, the blue "trail"). Shouldn't we somehow pass through a saddle? Well, in fact, no, we don't need to: maybe the lowest point on the blue trail is not a critical point -- the tangent plane to the surface at that point may be tilted. In this example, the tangent plane is always tilted at every point except the two peaks. Shouldn't we somehow pass through a saddle?

    The situation in 1 variable calculus is considerably different.
    If I have two local maxes (and, yeah, if the function is continuous, differentiable, etc.: nice) then there must be a local min between them.

    Another strange example
    Here f(x,y)=3xey-x3-e3y. Then we compute fx=3ey-3x2 and fy=3xey-3e3y. If fy=0, then ey3(x-e2y)=0 so x=e2y (the other factor, ey, is a value of the exponential function and is never 0). Then fx=3ey-3x2=0 leads to 3ey-3(e2y)2=0 or ey-e4y=0 or ey(1-e3y)=0. Since exp is never 0, we need 1=e3y and this occurs only when y=0. Therefore this function has only one critical point, at (1,0). My friend does this, by the way:
    > f:=3*x*exp(y)-x^3-exp(3*y);
                            f := 3 x exp(y) - x  - exp(3 y)
    > solve({diff(f,x),diff(f,y)});
                                   2                                 2
     {x = 1, y = 0}, {x = RootOf(_Z  + _Z + 1), y = ln(-1 - RootOf(_Z  + _Z + 1))}
    Since I know that z2+z+1 has no real roots (the discriminant is 12-4·1·1=-3<0) this function has exactly one critical point. And the formula for the function isn't really that horrible, either.

    The left graph below is a local picture of the critical point. This seems to convincingly support the assertion that (1,0) is a local strict maximum of the function. (We can verify this assertion with the second derivative test but I am too lazy.) In the graph on the right, x goes from -5 to 5 and y varies just between -.05 and .05: therefore y is just about 0, and 3xey-x3-e3y is just about 3x-x3-1. Certainly this shows that the function has no absolute max or min.

    The situation in 1 variable calculus is considerably different.
    Suppose we have a function defined on all of the real numbers (o.k., a differentiable function) which has exactly one critical point and that critical point is a local maximum. Then that local maximum is a global, absolute maximum for the function. What the heck is happening in several variables? I just don't understand, that is, really understand beyond superficially.

    Monday, October 6

    The handout
    I tried to give each student a copy of this handout. I sincerely thank the gentleman (name?) seated towards the back who gave some late entering students copies.

    Do we understand the handout?
    I remarked that I used unusual variable names and some new notation. The new (?) variable names and notation represent my effort to get you to think about the logic behind all the notation. As we will see later, sometimes the notation in this subject seems almost intentionally designed to obstruct understanding!

    The first table on the handout had the information below. f and g are declared to be differentiable functions of two variables.

      M     N  f(M,N) D1f(M,N) D2f(M,N)  g(M,N) D1g(M,N) D2g(M,N)  
     -1    -2     6      4        0         3      8       1
     -1     2     2     -2        1        -5      7       6  
      1    -1    -2     -5        4        -2      9       4 
      1     2     5     -7        6        -1     -2       7
      2     1     0     -1       -2        -3      7       4
    So M and N specify inputs to the functions f and g. Therefore f's value when, say, M=1 and N=2, is 5: f(1,2)=5. What are the D1 and D2 columns? This is another notation for partial derivatives, notation which some people prefer when there might be confusion about how the variables are named. D1 would refer to the partial derivative with respect to the first variable (frequently we have called this x) and D2 is the second variable (usually called y). Therefore in more traditional notation, ∂f/∂x(1,2)=-7 and ∂g/∂y(-1,2)=6.

    The second table on the handout referred to values to two differentiable functions of one variable.

      v  h(V)  h´(v)  k(v)  k´(v)
     -2   5     2      3     5
      0   0     2     -2     7
      1   1     3      2    -1
      2  -1     4      4    -2
    Maybe this table (in spite of the choice of variable name: v!) is a bit easier to understand. The value of the function k at input 1 is 2, and k's derivative at 1 has value -1. Also, h(0)=2 and h´(0)=3.

    Doing the problems
    A sequence of four problems were given. Let's try them.

    Club suit:
    If S(t)=h(k(t)), compute S(1) and S´(1).
    This is one variable calculus. But let me try to think about it a bit first:
    Thinking about it
    I can think about the function S as a sort of box, which takes inputs and processes them in some fashion, and produces outputs. The S box also has some internal structure. The input first is sent to box representing the k function, and then the output from that (sub?)box is sent to the h function. If we follow through (using values from the second table) we can see that 1 "changes" to 2 and then to -1.

    What about the derivative? The derivative is a multiplier of a tiny change in the input. It signals the first order change in the output compared to the input. In the case of S, if we "kick" the input by c (think of c as a small number) then 1+c is fed into the k box. The output will be approximately (neglecting H.O.T., higher order terms) 2 (the old output) +k´(1)c, which is 2+(-1)c. Now feed in 2+(-1)c. If c is small, (-1)c will be small. The output from the h box will be -1 (the old output, what h "does" to 2) plus a change. The first order part of the change will be a proportionality constant, h´(2), multiplying the kick that is passed to the h box. The kick passed to the h box is (-1)c, so the compounded effect is that h's new output (approximately, first order) is -1 (the old value of h's output) plus h´(2)(-1)c=4(-1)c.

    Now go "up" a logical level. The input, 1, to S was kicked to 1+c. The S output, to first order, is -1 (the old output) plus 4(-1)c. Therefore the derivative of the S box at 1 is 4(-1), since the derivative is the multiplier of the kick.

    The diagram below is supposed to be visual "support" of the preceding discussion.

    A formula
    If S(t)=h(k(t)), the one variable chain rule states that S´(t)=h´(k(t))k´(t), so S´(1)=h´(k(1))k´(1)=4(-1). Formulas are good!

    Diamond suit:
    If W(t)=f(h(t),k(t)), compute W(1) and W´(1).
    Thinking about it
    Now the W box has a different structure. The input is split (bifurcated -- what's the point of being in an academic environment if a silly, uncommon word isn't used in place of one that would be understood!) into two, and each is fed separately into h and k boxes. The outputs, now in order, are carefully put into f. The output from the f box is then pushed outside as the value of W.
    To compute W(1), we find h(1)=1 and k(1)=2, and then compute f(1,2)=5. Easy (?)..

    What about the derivative? Suppose we kick 1 to 1+c. The response of the one variable boxes, h and k, should not be difficult to understand: the outputs, linearized, are 2+k´(1)c=2+(-1)c and 1+h´(1)c=-1+3c, respectively. It is important to remember which output is which! Now feed this into f. The multiplier for perturbations in the first variable is D1f(2,1), so the effect of the change in the first variable adds on D1f(2,1)(-1)c to the output. The second variable contributes in proportion to its perturbation, with the constant of proportionality being D2f(1,2), so D2f(1,2)(3)c gets added on. If we look up the numbers and do arithmetic, we can see that the total (linearized) effect (neglecting higher order errors!) is 5 (the old output) plus 25c. Therefore the output of the W box seems to indicate that the derivative is 25. I don't think I can do this sort of computation correctly unless I understand the linearization idea -- that there are contributions from changes in both input variables which must be added to get the total first order change in the output variable.

    Again, here is a diagram which may help some people understand the preceding discussion.

    A formula
    O.k., if W(t)=f(h(t),k(t)), we will label the variables in f: the first variable is x and the second variable is y. Then we follow through the changes and use the chain rule:
    This is a fine result, but if we need to evaluate it, we'd better remember that
    (we need to know at which specific numbers the functions should be evaluated!) and now you should see the numbers that appeared above.

    Heart suit:
    If Q(x,y)=f(h(x),g(x,y)), compute Q(1,2) and ∂D/∂x(1,2) and ∂D/∂y(1,2).
    Thinking about it
    Certainly Q(1,2)=f(h(1),g(1,2))=f(1,-1)=-2: easy enough. Now to get the derivative with respect to "x", the first variable, let's perturb or kick 1 to 1+c. The effect filters through h as (linearized!) 1+h´(1)c which is 1+3c. If we kick 1 in g but hold the second variable constant at 2, then the output, to first order, is -1 (the old output) plus D1g(1,2)c. This is -1+(-1)c.

    Now the input to f is, in order, 1+3c and -1+(-1)c. There are changes to both variables. So we need to use a linear approximation in both variables. The output from f (which is what is reported as the output from Q) will be -2+D1f(1,-1)(3)c+Df(1,-1)(-1)c=-2+(-5)(3)c+4(-1)c=-2+(-19)c. Again I need to use linearization here. There are changes to both input variables of f. Therefore the proportionality factor is -19, and this is the requested value of ∂Q/∂x(1,2).

    A formula
    So if Q(x,y)=f(h(x),g(x,y)), then I think that ∂Q/∂x=(∂f/∂x)h´(x)+(∂f/∂y)(∂g/∂x). There's still the question about how to get values, and in fact, in more detail, this chain rule reads:
    Look at this miserable equation! The traditional notation (notation which I confess I usually use, except when I am getting confused) declares that the x derivative of Q needs values of the y derivative of f in its computation. This is miserable, ludicrous, horrible ... etc.

    If we want the y derivative, then we could compute this: ∂Q/∂y(x,y)=0h´(x)+(∂f/∂y)(h(x),g(x,y))(∂g/∂y)(x,y). The 0 is there because there is no y involvement in the first variable of Q. Now insert x=1 and y=2, and read off from the information given that the value is 4·7=28.

    And even worse ...
    This is all horrible. I will admit to you that I usually try to use formulas and only rarely (like most other human beings) try thinking. But sometimes thinking is needed. For example, those who like formulas might contemplate this task:
    What is the partial derivative with respect to x of P(x,y)=f(h(y),g(y,x))? Notice that I "swapped" the variables in g. I think the partial derivative with respect to x will be ∂P/∂x(x,y)=0+(∂f/∂y)(h(x),g(y,x))(∂g/∂y)(y,x). Some people might find this notation very objectionable (I do!): look, two derivatives with respect to y multiplied turn into the derivative with respect to x!

    Spade suit:
    If C(t)=h(f(t,2-3t)), compute C(1) and C´(1).
    Thinking about it
    Certainly C(1)=h(f(1,2-3·1))=h(f(1,-1))=h(-2)=5: no worries.

    The derivative is maybe a bit more difficult. I think that C´(1) will involve first the derivative of the outside most function, h. This is a function of only (!) one variable, so we just get h´(the inside value), that is, h´(f(1,-1))=h´(-2)=4. But we've got to multiply this by the appropriate result of considering f's linearization.

    The linearization of f tells me I should write something like D1f(1,-1)(∂t/∂t)+D2f(1,-1)(∂{2-3t}/∂t). If you wish traditional notation, this is (∂f/∂x)(1,-1)(∂t/∂t)+(∂f/∂x)(1,-1)(∂{2-3t}/∂t). We look up the numbers in the table and evaluate the t derivatives. The result is (-5)(1)+(4)(-3). Don't forget to multiply by 4 from the derivative of h! The official answer is 4((-5)(1)+(4)(-3)) which is -68. I hope this is correct -- I don't think I did it in class and I make errors more frequently when no one is watching.

    A formula?
    If C(t)=h(f(t,2-3t)), then (I hope!) C´(t)=h´(f(t,2-3t))((∂f/∂x)(t,2-3t)(1)+(∂f/∂x)(t,2-3t)(-3)). The 1 and -3 come from ∂t/∂t and ∂{2-3t}/∂t respectively.

    The wave equation
    Suppose we have a homogeneous string overlaying part of the x-axis. The function g(x,t) will represent the following: the height of that part of the string which is "over" position x at time t. We saw (in class, in your head: think guitar or violin or rubber band) that g(x,t) is a function of both x and t. It turns out that under some very weak assumptions (I am omitting units which multiply things by a constant) the partial derivatives of the function g(x,t) have the following strange property:

    2g   ∂2g
    --- = ---
    ∂x2   ∂t2
    In words, two space derivatives of g must equal two time derivatives of g.

    This is not intuitively obvious, at least to me. I bet that some engineers and physicists would assert that the equation is intuitive and clear. The equation turns out to be a very good model of physical vibrations, for at least small displacements (just like Hooke's Law does describe springs, but not if you try to stretch an ordinary rubber band 10 feet!). Here are two links to reasoning which shows how this equation follows from physical ideas:
         At the University of British Columbia This uses Newton's Law and force considerations.
         A Wikipedian entry This uses Newton's Law and Hooke's Law more directly.

    One solution
    So I'd like to get a function which is a solution of the Wave Equation. I think I wrote something like this: g(x,t)=38(x-t)53. I claimed this was a solution. How can I verify my claim? Well, I need to compute some partial derivatives. I'll use subscript notation for these partial derivatives (yeah, people have all sorts of strange notation, and they won't stop using it just because, say, I don't like it!):
         gx=38·53(x-t)52 and then gxx=38·53·52(x-t)51.
         gt=38·53(x-t)52(-1) and then gtt=38·53·52(x-t)51(-1)2.
    The -1's in the t derivatives come from using the Chain Rule. Since (-1)2=1, indeed gxx=gtt. So this g is a solution to the wave equation. I am not asserting, by the way, that this g(x,t) is a model of a physically realizable solution!

    Many solutions
    Suppose f(blah) is any differentiable function of the variable blah. Then I claim that g(x,t)=f(x-t) is always a solution to the Wave Equation. Here is the verification:
         gx=f´(x-t)(1) and then gxx=f´´(x-t)(1)2.
         gt=f´(x-t)(-1) and then gtt=f´´(x-t)(-1)2.
    I have been very careful and written some stuff you may consider not needed. The 1's and -1's come from the Chain Rule. And since 12=(-1)2 we know that gxx=gtt for any eligible function f of one variable. In fact, f(x-t) represents a wave traveling from left to right. Similarly, if h(blah) is another one-variable differentiable function, then h(x+t) is also a solution of the Wave Equation (here I need 12=12 only) and this is a wave traveling left. In fact, it turns out that all solutions of the one-dimensional Wave Equation can be written in this way: g(x,t)=f(x-t)+h(x+t) (superposition of right and left traveling waves). I did try to show in class that the wiggles move both ways.

    To the right is a picture created by Maple of the function g(x,t)={1/(1+.25(x-t)2)}+{.3/(1+.4(2+x+t)2}. The picture shows this function over the interval [-12,12] on the x-axis, and t varies from -7 to 7. The picture was created using the animate command.
    The wave traveling to the right has profile given by f(v)=1/(1+.35v2) and the wave traveling to the left has profile given by h(v)=.3/(1+.4(2+v)2. I hope you enjoy watching them.

    Crazy computations in thermodynamics and physical chemistry
    I wanted to briefly indicate some horrible computations which come up in applications. The computations are not really horrible, but the notation makes them look quite weird.

    Implicit functions, two dimensions
    I first began with a return to a 1 variable calculus situation:
    Suppose F(x,y) is a differentiable function of 2 variables, and the equation F(x,y)=0 defines y implicitly as a function of x. What is dy/dx in terms of F and "things" related to F?
    So take the equation F(x,y)=0 and d/dx this equation. The right-hand side is 0, and the left gives you:
         ∂F/∂x(dx/dx)+∂F/∂y(dy/dx) by the chain rule.
    Certainly dx/dx is 1, and dy/dx is what we want, so we can "solve" for it in the equation ∂F/∂x+∂F/∂y(dy/dx)=0. This means:

    A formula!

    dy      ∂F/∂x 
    -- = – ------- 
    dx      ∂F/∂y

    I think an example is needed here before we go on. Let's look at:
    Calc 1 problem: find dy/dx if y3-7xy2+4x5-6=0.

    Calc 1 solution to Calc 1 problem We d/dx everything, being careful to remember that y=y(x) mysteriously. Then:
    3y2y´(x)-7y2-(7x)2yy´(x)+20x4=0, and now we solve for y´(x). We get:
    y´(x)(3y2-(7x)2)-7y2+20x4=0 so that y´(x)=-(-7y2+20x4)/(3y2-(7x)2).

    New technology (?) solution to Calc 1 problem We will use the formula above. Here F(x,y)=y3-7xy2+4x5-6 so that
    ∂F∂x=-7y2+20x4 and ∂F∂y=3y2-(7x)2y+0 and the formula gives
    dy/dx=-(∂F∂x)/(∂F∂y)=-(-7y2+20x4)/(3y2-(7x)2y+0) which is of course the same answer! And you can look at see the same pieces occurring, so the world is not so crazy.

    The darn formula, though, is a bit mysterious. If you try to understand the form (?) of the formula, the ∂x and ∂y might seem in the wrong place and there might be an extra minus sign ... and ... and ... the notation is terrible!

    P and V and T
    Do you know about gas laws? For a gas, there are the quantities P (pressure) and V (volume) and T (temperature). A gas law might be a function of three variables which relates these quantities:
    If we assume that the function is differentiable and that each one of the quantities is implicitly defined as a function of the other two by the function, something funny happens. Let me show you.

    Suppose that G(V,P,T)=0 implicitly defines V as a function of P and T. Let's compute ∂V/∂P. Here T is constant, and sometimes in thermodynamics the quantity is called (∂V/∂P)T just to remind people that T is constant. We will ∂/∂V the equation G(V,P,T)=0.
    I use the chain rule, and the result is:
    But ∂P/∂P must be 1 (the derivative of something with respect to itself) and ∂T/∂P must be 0 (because T is constant!). Therefore we can solve for ∂P/∂V just as we got dy/dx before and get:

    So far so good. But in fact we can find other partials in a similar way:
    ∂T/∂V=–(∂G/∂V)/(∂G/∂V). Now clearly (NOT AT ALL CLEARLY!):
    because when we multiply all these expressions together the fractions all cancel and we are left with -1. Why is this true physically and what does it mean? Take physical chemistry, take thermo, etc., and find out. But the notation is horrible and, for me, makes things harder to state and understand.

    The 68th derivative test
    For the final few minutes of the class, I returned to 1 variable calculus just to prepare you for Wednesday's class. I reminded people about local maxs and mins in one variable. Everyone (they were too tired to argue, I think!). People claimed to remember the first and second derivative tests. But they did not know the glorious

    The 68th Derivative Test
    Suppose f(x) is a differentiable function of 1 variable, and we know that
         f´(x0)=0, f´´(x0)=0, f(3)(x0)=0, f(4)(x0)=0, ..., f(67)(x0)=0.
  • If f(68)(x0)>0, then f has a local minimum at x0.
  • If f(68)(x0)<0, then f has a local maximum at x0.
  • If f(68)(x0)=0, the glorious 68th derivative test returns no information.
  • I don't think any sane person would care about such a thing, but I wanted to think about it just a little bit, because similar thoughts will help with max/min in several variables.

    Why would such a "test" be correct? Here is some reasoning which may help you understand. Taylor's Theorem states that f(x)=f(x0)+f´(x0)(x-x0)+[f´´(x0)/2](x-x0)2+[f(3)(x0)/3!](x-x0)3+...[f(67)(x0)/67!](x-x0)67+[f(68)(blah)/68!](x-x0)68 and blah means some number between x and x0. Think of x as close to x0, so blah is close to x0 also. Because of the green hypotheses (?) in the 68th derivative test, lots of stuff drops out. In fact, we know that f(x)=f(x0)+[f(68)(blah)/68!](x-x0)68. I bet that if blah is close enough to x0, the sign of f(68)(blah) is the same as the sign of f(68)(x0). Well, if the darn sign is positive, then the function sort of looks like
         f(x)≈f(x0)+(something positive)(x-x0)68
    68 is an even power, so qualitatively the graph is a sort of cup opening upwards. There: f has a local min at x0.

    What I used was Taylor's Theorem to get f equal to a polynomial plus an error term. The hypotheses made the polynomial just about disappear. Then I looked carefully at the error term to see what that graph looked like, and f "inherited" the local behavior of that graph. So ... we will discuss what happens in 2 dimensions next time.

    Wednesday, October 1

    The spaceship in a nebula
    My online dictionary states that a nebula is "a cloud of gas and dust, sometimes glowing and sometimes appearing as a dark silhouette against other glowing matter." So we could pilot a spaceship through a nebula. We might be concerned about the physical effects of the nebula, for example, the temperature. I'll assume that the spaceship measures temperature at the tip of its front. A point in the nebula will be located with rectangular coordinates, (x,y,z). The temperature at that point will be T(x,y,z). The rocket will fly a path so that at time t its location will be <x(t),y(t),z(t)>.
    From this we can see that the temperature measured at the rocket at time t is T(t)=T(x(t),y(t),z(t)), and this is a composition. First we find out where the spaceship is at time t, and then we compute the temperature at that point.

    Computing dT/dt
    I would like to compute and understand the rate of change of the temperature, T, with respect to time. It turns out that this is a significant computation. Now T is a number and t is a number and T is a function (complication: there are intermediate variables x, y, and z) of t. So the derivative of T will involve taking its value at t+Δt and looking for the linearization multiplier. That is what we declared earlier. So here we go. I will try to accompany the steps of this somewhat elaborate manipulation with explanations.

                            We are "kicking" the time variable a little bit, and we would like to examine the change in the T variable.
      We use the fact that each of the components of the position vector are differentiable, so each function value at t+Δt can be replaced by the original value of the function, a multiplier (the derivative) which multiplies the disturbance, and higher order terms. If I were being careful, I would use different notation for each of the H.O.T.'s, but in practice people don't do that too often. You'll see why soon.
     This is the differentiability of T. The changes in the inputs to T (there are three inputs) are passed outside. What happens? The linearization idea says that the changes are each multiplied by an appropriate partial derivative. And this isn't really exact (the numerical example showed this!) so there is also a H.O.T. from the differentiability of T. (Complicated? Sure is.)
    T(x,y,z)+(∂T/∂x)(x´(t)Δt)+(∂T/∂y)(y´(t)Δt)+(∂T/∂z)(z´(t)Δt)+ (∂T/∂x)H.O.T.+ (∂T/∂y)H.O.T.+ (∂T/∂z)H.O.T.+ H.O.T.
     Now I did some multiplication and rearrangement. I pushed everything involving any of the Higher Order Terms to the end.
      Here is the important step, and it sort of asks you to think a bit about the ideas of calculus. All of the terms with H.O.T. are actually, all added together, just another big H.O.T.
     This is the last rearrangement. Please, I hope you have the patience to see what has happened. We have the "old" unperturbed value of T(x(t),y(t),z(t)), and then we have a mess (not really, as you'll see) multiplying Δt, and then, finally, we have a whole bunch of things which logically are inside the H.O.T.

    Now what? If f(v) is a function of one variable, then f(v+Δv)=f(v)+f´(v)Δv+H.O.T. identifies the derivative of f by what happens to small perturbations. The red stuff is the derivative, because it is the multiplier of the small perturbation of the input.

    We carried out an elaborate analysis of how the temperature function changes. Now look at the result again:
    The multiplier of Δt is dT/dt, so we see that


    But this formula is not the point of the discussion. Look at the equation and think a bit. The luck and glory is recognizing that the mess on the righthand side is a dot product. In fact, look:

    dT    / ∂T   ∂T    ∂T\    / dx   dy   dz \
    -- =   --- , --- , ---  ·   -- , -- , --
    dt    \ ∂x   ∂x   ∂x /    \ dt   dt   dt /  
                    Left                              Right
    There are certainly many ways to organize this as a dot product, but this way turns out to give some insights that amaze me.

    The vector on the right-hand side is one we've looked at when discussing curves. It is the derivative of r(t), the position vector, so it is v(t), the velocity vector. This vector deals with the spaceship and its motion.

    This vector seems to be "new": it is the vector of all the first partial derivatives of T in order. This is called the gradient of T and is frequently written ∇T and is sometimes also called grad T. The upside-down triangle (or upside down Δ) is sometimes called "del" or "nabla". This vector can be computed only from the nebula information.

    Thinking about the temperature derivative this way "decouples" (yeah, a word that's used) the influence of the nebula from that of the spaceship.

    Now I will try to make a sequence of observations which might help people understand the excitement I feel thinking about gradient.

    Observation 1
    Let's imagine two spaceship trips through the nebula. Now these trips (voyages?) may be completely different except that at the time the two spaceships pass through the point (x,y,z), the spaceships have the same velocity vectors: that is, the spaceships are heading in the same direction and at the same speed. Their v(t)'s are the same. Then the rate of change of the temperature, dT/dt, that the two spaceships measure is exactly the same.

    I asked students if they could deduce this from the physical and geometric aspects of the "scenario". I don't think I can. As a math fact goes, this is nearly obvious: since the v(t)'s are the same, the right-hand side doesn't change, and the nebula's temperature function is the same, so the left-hand vector ∇T doesn't change. Therefore the dot product, which computes dT/dt, is the same. But ... but ... what the heck ... can you "see" this physically? This is not the temperature at the point, but the rate of change of the temperature: the rate of change is the same if the velocity vectors are the same.

    Observation 2
    Now r´(t)=v(t), the velocity vector. Let me call the unit tangent vector u(t) here (in the discussion of curvature it was called T(t) but that will just be too darn confusing). Then v(t) is the same as (ds/dt)u(t) where ds/dt is the speed and u(t) is the unit tangent vector. In the formula ∇T·r´(t) the ds/dt effect just "filters out" of the dot product. If you travel twice as fast on the same path, then the rate of change of the temperature with respect to time is just doubled. So this is easy to understand. But the more subtle aspect is what happens as the direction changes.

    Observation 3
    Here I will suppose that ds/dt=1 for simplicity. Again, call the unit tangent vector, u, for unit vector. Then what can we say about ∇T·r´(t)? It is (ds/dt)∇T·u, or just (since I'm assuming unit speed) ∇T·u. But, hey, the dot product is also ||∇T|| ||u||cos(θ). Since cos(θ) is between -1 and +1, I now know that dT/dt is between -||∇T|| and +||∇T||.

    How could we choose u so that dT/dt is largest? We need to make cos(θ) equal to +1. Therefore we need θ to be 0, and u should be a unit vector in the direction of ∇T. That is, choose u to be ∇T/||∇T||. To make the rate of change as much negative as possible, choose u to be -∇T/||∇T||, and then dT/dt will be -||∇T||.

    An example (?)
    Here is an example. If T(x,y,z)=x2eyz-5z3 then since ∇T=<∂T/∂x,∂T/∂y,∂T/∂z>, we compute:
    As far as I know this function and this computation has no great or special "meaning".

    A better example (!)
    Im my kitchen I have just finished backing my famous chocolate brownie pie and I left the oven door slightly open. Also I managed to forget to close the refrigerator. As a result, the contour lines of temperature could like what is shown to the right. In what direction should I go (I am the little green man in the picture!) to most rapidly increase the temperature? In the direction of the gradient, which will point towards the oven. I will most rapidly decrease the temperature by traveling in the opposite direction, towards the source of the cold.

    Observation 4
    I could imagine that spaceship travels through the nebula on an isothermal surface. An isothermal is a collection of points where the temperature is all the same. We have seen this already: T(x,y,z)=C is a level surface (dimension 3) or level curve (dimension 2) or a contour {surface|curve}. But if the spaceship travels on such a surface, then the rate of change of the temperature must be 0. But then ∇T·v=0. This means that the velocity vector is perpendicular to the gradient. But then in turn this means that the gradient vector is perpendicular to the level surface, and it is perpendicular to the tangent plane of the level surface. In the kitchen, I would walk perpendicular to the contour lines to increase or decrease temperature most rapidly. I would walk along the contour lines if I wanted no rate of change of temperature.

    Back to the example
    Let me look more closely at the example with T(x,y,z)=x2eyz-5z3 when x=3 and y=2 and z=1. Well, T(3,2,1)=9e-3. And ∇T=<2xeyz-5z3,x2eyz-5z3(z),x2eyz-5z3(y-15z2)> becomes ∇T(3,2,1)=<6e-3,9e-3(1),9e-3(-13)>=<6e-3,9e-3(z),-117e-3>

    Now forget all that, and solve the following geometric problem:
    What is the equation of a line tangent to the surface x2eyz-5z3=9e-3 at the point (3,2,1)?
    This could be, indeed, I claim, this is a hard problem. But if we now disobey my urging ("forget all that") I can tell you that ∇T(3,2,1) is perpendicular to the surface and to its tangent plane at (3,2,1). So I can write the answer, since I know a point and a normal vector to the plane requested:
    I think that solving such a problem so efficiently is really remarkable.

    Topographic maps
    A topographic map shows contour lines. Frequently while hiking people mind want to find the most direct route to the "top" (a mountain peak) or to the "bottom" (a creek?). They know by experience that the most direct route, only looking at the map, that is, only the geometry of the situation, would be to walk as nearly as possibly perpendicular to the contour lines.
    This can be adapted into computational strategies for finding maxes and mins. If you can readily compute your function's gradient, then find maximums by going in the direction of the gradient. This is hill climbing. Find minimums by going opposite the direction of the gradient. This is the method of steepest descent. Of course these computational ideas don't always work, and there are many implementation details to worry about, but the general strategy is valuable.

    Here's a neater example. Consider the ellipsoid (egg) x2+2y2+3z2=9. The point (2,1,1) is on this ellipsoid. What is the equation of a plane tangent to the ellipsoid at (2,1,1)? Well, the gradient of the function x2+2y2+3z2 is <2x,4y,6z> and at (2,1,1) this is <4,4,6>. The equation of the tangent plane is 4(x-2)+4(y-1)+6(z-1)=0.

    To the right is a Maple picture made by the commands which follow. I hope that the picture helps to convince you that the plane is the tangent plane.


    Contour map example
    This kind of came up spontaneously in class (I try to have a well-prepared, well-thought-out lecture, but sometimes things come up). So to the right is a made-up example of a contour map. The lines shown are at height multiplies of 100 feet. I hope that you can see (?) there is sort of one part of the map which deserves the name "peak" (in the upper left-hand corner). This map is a bit more complicated than the one I drew in class. I clain there is also a part of the map which could be named "the pits" (probably at least two of them!) and that's inside the two loops at the 300 foot height.
    Hill climbing -- a way to get to the maximum
    Now suppose we are at the point P on the map. How should we hike in order to get uphill as fast as possible? I claim that the directions shown will do that. Notice that I am not always walking towards "the peak", but I just want to always walk perpendicular to the contour lines -- in the direction of the gradient of the height function.
    The method of steepest descent -- how to get to the minimum
    This is a way people use to try to get to minima of functions in several variables. We start at the point Q and move in the direction opposite the gradient of the height function. In what I'm showing here, we take some sort of fixed step size (more sophisticated versions vary the step size) and after that step we keep going until we hit another contour line. Then we recompute the direction (minus the gradient) and step again. You can see, I hope, that this path is headed for one of "the pits".
    These methods don't always work and don't always work efficiently, but they are easy to understand, they can be done in many dimensions, and the programs run fast.
    The most change
    I could also ask where there is the most change in height in a fixed distance. These are contour lines corresponding to equal changes in height. The heights will change most rapidly (as a function of the position on the map) when the contour lines are most closely spaced. This also corresponds, of course, to the region in the plane where ||∇ height|| is largest: the largest magnitude of the gradient vector. I think the region indicated (in magenta) is the region in this rather simple contour chart.

    I forgot to mention this definition. Please forgive me!
    Directional derivative
    If u is a unit vector, then the directional derivative of T at (x,y,z) in the direction u is the rate of change of T at unit speed in the direction u (at the point). The textbook's notation for this is DuT(x,y,z) and the preceding discussion should convince you that the directional derivative's value is ∇T(x,y,z)·u.
    more notation, more words ... this is so terrific!!! (so academic)

    I started a discussion of the more general Chain Rule which I'll continue next time.

    Maintained by and last modified 10/8/2008.