Diary for Math 151:04-06, fall 2003
No calculators will be allowed; a formula sheet will be provided.
Diary entries for classes
from Wednesday, September 3 to Monday, October 6
Diary entries for classes
from Wednesday,
October 8 to Wednesday, November 12
| Date | What happened (outline) | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
| Wednesday, December 10 |
I discussed Monday's QotD. I mentioned that seeing x dx many
people would almost immediately try the substitution
u=x2. That doesn't exactly "work" but adjusting the
substitution to u=x2+a2 along with the
realization that the derivative of a2 is 0 since a is a
constant. I remarked that I knew perhaps three or four
ways to find an antiderivative for the integrand, which students
would learn in Math 152.
Then I computed the definite integral
I repeated that the definite integral; representated alimit of Riemann sums, and that using that representation, many quantities ofinterest in applications can be recognized as definite integrals. Then the FTC asserts that definite integrals can be computed by antiderivatives, if we happen to know the relevant antiderivative. This distinction (the definition of the definite integral and the FTC as a method of computing them) is important.
I remarked that an antiderivative of xp is
(1/(p+1))xp+1+C. This formula is valid for all p except
p=-1: we don't want to divide by 0. But how should we then compute,
say, the area under the hyperbola y=1/x. I will restrict attention to
x>0 to avoid the problem at 0. For example, the area bounded by the
x-axis andx=5 and x=7 and y=1/x is represented by
Motivated by experience with FTC, we
might decide to study the function F(w)= Look at G(w)=F(3w). By the Chain Rule, G'(w)=F'(3w)·3, and this means G'(w)=[1/(3w)]·3=1/w. So F(w) and G(w) have the same derivative. Certainly if two functions have graphs which are up/down translations of each other they have the same derivative. But in fact the converse is also true, and this follows from the MVT: if F'(w)=G'(w). then G(w)=F(w)+(some constant). Plug in w=1. Then G(1) is F(3·1)=F(3), and the equation becomes F(3)=F(1)+the conswtant. Since F(1)=0, this means the constant is F(3). So what we verified is that F(3w)=F(w)+F(3). In fact, in all of the previous discussion, we could change 3 to anything: to "v". We would prove that F(vw)=F(v)+F(w). (Yeah, yeah, I know, looks like ... something that rhymes with "frog". For example, Since F(4)=F(2·2)=F(2)+F(2), and 1/2<F(2)<1, we know that (sin ce we can double the inequality!) 1<f(4)<2. Oh my: F(w) is a continuous function with a positive derivative, so it is always increasing. F(2)<1 and F(4)>1. Therefore (use the Intermediate Value Theorem) there's is exactly one number whose "F" value is 1, and that number must be between 2 and 4. In fact, it turns out the number is 2.7182818...: it is called e. What's the point of all this? First, a pedagogical point: I wanted to review some of the most important issues of the course. Second, it is really true that hardly anyone cares about logs for the purpose of computing these days (our little silicon friends do that just fine!) but the use of logs in finding antiderivatives won't go away. (In fact, many weird anduseful functions are defined by antiderivatives!)
logs won't go away The Question of the day had two parts: how used has that QotD been and how useful has the web diary of the course been? I have not looked at the answers yet, but I am very curious. I remarked that (on average!) the QotD takes me about 3 hours a week outside of class) and the diary takes me about 5 hours a week outside of class. I hope that people will send me answers to the various sets of review problems and that, if they like, they will attend the review session, office hours, and send questions via e-mail. Recognition of attendance was distributed to appropriate individuals, along with a prize containing a prize.. | ||||||||||
| Monday, December 8 |
Area between two curvesConsider the curve defined by y=x*(x+1)*(x+2)=x3-x2-2x. What is the "finite" or bounded area between this curve and the x-axis? This question is not terribly well written, but I am trying to ask what the total area is of the region caught inside the two "bumps" shown. Is this total geometric area exactly equal to the definite integral -12x3-x2-2x dx?
Well, between -1 and 0 the curve is on top of the x-axis. In fact,
-10x3-x2-2x dx=(by
FTC)
(1/4)x4-(1/3)x3-x2|-10=0
(the value when
x=0)-[(1/4)(-1)4-(1/3)(-1)3-(-1)2](the
value when x=-1)=-[(1/4)+(1/3)-1]=5/(12), a positive number, as it
should be.
I did a bunch of textbook problems.
What about | ||||||||||
| Wednesday, December 3 |
I began by computing the definite integral of x7 from 0 to
4, and, of course, making my first error of the lecture. Oh
well. I will try a new method of writing integrals on the web. I will
use a little picture of an integral sign. Heh, heh. Here goes:
04x7dx=(1/8)x8|x=0x=4=(1/8)48.
This computation is routine. The definite integral describes a certain
area, a roughly triangular-shaped one, bounded by the x-axis, and
y=x7, and x=4. The line x=0 hardly matters, since the
x-axis and the function graph intersect there.
I then tried another "simple" problem:
I discussed problem #2 of section 5.3 in detail. This problem gives
the graph of a curve f, and defines g by g(x)=
Numbers are numbers are numbersI then discussed numbers. I asked what people could tell me about these numbers:43 43 43 43 43 After some jeering (I felt bleak) we agreed that these numbers were all the same number. jeer
n : showing your contempt by derision [syn: {jeering}, {mockery},
{scoff}, {scoffing}]
v : laugh at with contempt and derision; "The crowd jeered at
the speaker"
bleak
adj 1: offering little or no hope; "the future looked bleak";
Then I asked what people could tell me about these numbers:35[x3]/[3+cos(x2)]dx
35[u3]/[3+cos(u2)]du
35[t3]/[3+cos(t2)]dt
35[v3]/[3+cos(v2)]dv
35[s3]/[3+cos(s2)]ds
These are all definite integrals, and they are all just numbers. Of course they are all equal. Again, the letters inside the integrand are dummy variables or bound variables. They are similar to the local variables in subroutines of a computer program, and they have no meaning outside of the integral itself. I guess that the assembly of symbols 3xxdx would make sense:
antidifferentiate, get (1/2)x2, evaluate at x and subtract
off the value at 3 to get (1/2)x2-9/2. But this is
considered bad grammar: people try not to use the same variables
inside the integral as in the limits -- it is difficult to understand.
I then tried to do some simple exercises with part of the FTC. It was
something like this: if f(x)=
We can even do one further variation. Suppose F(x)=
I tried to convince people that on the horizontal axis, if B is to the
right of A (A<B) then the distance from A to B is B-A. I then went
on to assert that on the vertical axis, if B is above A (A<B) then
the distance from A to B is B-A. It doesn't matter much what signs A
and B have: if A<B, the distance from A to B is B-A.
This was a sort of introduction to the following problem: sketch
y=5-x2 and y=2x2-7. Where do these curves
intersect? What is the area between them? We then had an interlude in which we discussed the final. /interlude/ [Theatr] a. a pause between the acts of a play.Please look at the material I have prepared to help the class study for the final. 
The Question of
the day was the following: I sketched y=x2 (a parabola) and
y=2x+3 (a line). I asked where they intersected and requested that people find
the area between these curves. | ||||||||||
| Monday, December 1 |
Today we disclose the climax of the course. I will begin with a
simple example. Our goal right now is to compute the area of the
region in the plane bounded by the x-axis, x=2, x=3, and the curve
y=x5. This is a definite integral:
S23x5dx/ Oh, I give up: this
is int(x^5,x=2..3). The last collection of symbols is easier to type,
and the traditional integral symbol isn't well indicated by S,
anyway. On the right is a picture of the region whose area I will try
to compute. Please note that although the picture I have drawn is
mostly qualitatively correct, I have certainly used very different
units on the vertical and horizontal axes (25=32 and
35=243). I will arbitrarily take the year 1600
as the dividing point in the two approaches to computing definite
integrals that I'll discuss
(the 1660's is probably more accurate historically). I'll begin with
the "old" style.
Getting the answer directly from the definition (before ~1600)Here we will go through a four-step process: subdivide, approximate, sum, and limit. This process is, in fact, quite important in many applications. It is the reason that definite integrals represent many different ideas.
(There are n terms in this sum.)
By the way, the process we are discussing here would be familiar to appropriate representatives of various classical civilizations, including Greece, China, India, and north Africa. The are wonderful medieval Chinese manuscripts which have diagrams that look exactly like our definitions of the definite integral, with much commentary written in Chinese surrounding the diagrams. One
way to analyze the elaborate sum above begins by changing it
algebraically. For example, the first term,
(2+1/n)5(1/n), is actually
([2n+1]/n)5(1/n)=(2n+1)5/n6. The
second term, (2+(2/n))5(1/n), is
([2n+2]/n)5(1/n)=(2n+2)5/n6. In each
case we can rewrite the terms so that there is a factor of
1/n6, and we can factor out (undistribute?) and the big sum
becomes:
I assure you that this result is not immediately obvious to me! I think if you gave me a while (a long while!), I could discover and verify it by hand. I don't think I would want to, though.
But now what's the next step? Divide by n6 and find the
limit as n-->infinity. This will be tedious until we recognize a
pattern. For example, let's look at the first term in that enormous
sum: 32n5(n+1). If we divide by n6 and simplify
we get 32[(n+1)/n] and this, as n-->infinity, becomes 32. Now
look please at the last term divided by n6:
-32n5/n6 which is -32/n and this -->0 as
n-->infinity. In fact, all of the terms in the enormous expression
above are polynomials, and the degree of each term is at most
6. (That's good, because otherwise the total area of the rectangular
approximation would be -->infinity as n increases.) Each term which
has degree less than 6 must -->0 as n-->infinity (hey, you could
use l'Hopital's rule or think about it!). The terms of degree
exactly equal to 6 each contribute a constant to the limit. Therefore
we can go through the whole list above and get the limiting value we
need. It will be: I don't think this specific computation has much importance, but I went through it to show how pre-1600 technology would work. I don't think they had Maple so their task would have been even more difficult: getting an appropriate summation formula would indeed have been troublesome. The new stuff! (after ~1600)
Let's summarize what we have learned about A(R):
Is this a journey worth the trip?
I then did a collection of problems from the text. They were:
| ||||||||||
| Monday, November 24 |
What will be done now is the definition of the definite integral. This
is a complicated object, and is the mathematical object which
corresponds to quantities which have properties similar to those we
discussed last time. I will concentrate on area in what follows
(because then I can draw simple pictures), but everything I'm doing is
applicable to many many many other things as well.
So again: the problem of area. In what follows I'll be looking at a region bounded by y=f(x), x=a, x=b, and the x-axis. How can we efficiently and accurately approximate the area of this region? We will use rectangles as we did last time, only now I'll give you the official names of everything. Here we go:
So for the non-continuous function f(x) we just analyzed, we can find a partition all of whose Riemann sums are very close to 9. Let us try something harder.
Definition Suppose f is a function defined on the interval
[a,b], and the following occurs: as the partition gets "finer" (so the
length of all of the subintervals --> 0) and for any choice of
sample points, the Riemann sums --> one number, then that number is
called the definite integral of f on [a,b].
Of course, there is notation: an elongated S from the sum (just
replacing the capital sigma) and an a and a b and a "dx":
Sab f(x) dx. I think of this as
a sum of very, very thin rectangles of height f(x) and "width" dx,
added up from x=a to x=b. It is a true fact (much better than a false fact!), that every continuous function has a definite integral. But there are functions which are not continuous, such as the function of example 1 above, which also have definite integrals.
I did two very examples that are "simple" if you recognize the
geometry supporting them:
The area shown is either a trapezoid or a triangle and a rectangle. In
either case, the area, the definite integral from 0 to 3, is 2.5. Half
of that area is 1.25, and the area of the triangle is 1/2, so that we
need .75 from the area of the rectangle. Since the rectangle is 1 unit
high, we easily deduce that A must be .75 more than 1, so A=1.75.
| ||||||||||
| Wednesday, November 19 |
I remarked that we would be doing something significantly surprising
intellectually in the balance of the course. So far we have studied
various nice problems involving differentiation. Things like graphing
and optimization (min/max) and slopes of tangent lines and
... stuff. It turns out that the kind of things we've been studying
have great application to what looks like a collection of completely
different problems. It is and it will be amazing!
Please start reading chapter 5 of the text. I will cover today and Monday the material in 5.1 and 5.2, in somewhat different ways than the text does. I may even go a bit further. I began with The problem of areaI drew a blob on the board, and asked how much area is inside the blob. I told people that they could use any method they wanted to answer the question. Some students whose minds were already preconditioned by calculus gave rather strange answers, but no one gave me a direct method until Mr.Colquit remarked essentially that we could cut up the board and then weigh the blob and compare its weight to the weight of a board of known area, such as a rectangular board. So here is a direct method: cut it out!.We will try to be a bit more indirect, more sneaky, and learn properties of area which will let us approximate what we want. What properties does "area" have?
With RULES A THROUGH E we can begin an organized campaign to compute or approximate areas. For example, here is how we could compute the area of a typical blob. I will use the RULES in the analysis that follows. I would first slice the blob by line segments to get something which had at most one curvy side. I will be done if I describe to you how to compute the area of each of the pieces. I will specialize to one of the pieces, and assert that similar things can be done to all of the other pieces. So:
Now I can approximate inside this piece of the blob and outside the blob by rectangles, and since I know the area of a rectangle, I can get some estimate of the area of the piece of the blob.
Of course this estimate is not so good, but we could improve it with some sort of scheme like this:
So we can get a collection of approximations which get closer and closer and closer to the true value of the area, and maybe that's good enough. (It is, in many cases.) As I remarked, another instructor told me of a wonderful way of bringing home the idea of area: look at your hand and ask, what is its area? Giving a "good" answer to this may be difficult and involve quite a lot of approximation. Now let's look briefly at The problem of blood flowI looked at blood flowing through the heart, and asked how much blood would flow through the heart in a given amount of time, say a minute. Knowledge of this could be rather important in study of certain conditions or diseases or whatever. One rather destructive method would be to cut it up: that is, just slice into the heart and let the blood drip to fill a bucket for a minute. Well, this method might not be too satisfactory for several reasons which you can you think about yourself. But how do people answer such questions? You insert a catheter into the appropriate blood vessel (Google lists about 135,000 links in response to the search words "blood flow catheter" so there are ample references). At the end of the catheter there's a mechanism, which you can think of as a sort of paddlewheel, and the blood flow sets it spinning, and the spinning makes an electric current which is passed up the catheter and measured. How is the actual blood flow computed? It turns out that the blood flow obeys rules which are totally analogous to the area rules. RULES A THROUGH E now state, sort of:
With these observations, it is possible to make good approximations to
the blood flow using a discrete collection of flow rate measurements,
and this is what's done (without filling a bucket!).
By the way, I found a number of authentic pictures of the heart on the web and decided I didn't need to illustrate this discussion with what could be a grim and upsetting picture. And now Aliens land in Piscataway and leave a barThis is somewhat speculative. The type of bar I am referring is not one selling alcoholic beverages (!) but a long and very very heavy bar, a bar which is 100 meters long (metric aliens!) and made of a number of distinct substances. Additional conditions are that the bar is too heavy to move, that we can scrape and sample it, and that we would like to determine its total weight, or, rather, its total mass. Here maybe is a picture, with the different shapes/colors representing substances of perhaps vastly differing densities.
Since the bar is geometrically very simple, we can just take a sample scraping somewhere on it, find the density, and approximate the total mass by multiplying the density recorded by the geometric shape's volume. But this is a very rough approximation, since there are different substances and the bar could be complicated. So what we could do is periodically sample the bar, find the densities, and multiply by chunks of the volume. It seems likely that we will get a better estimate of the total mass if we take more samples. The alert student will observe that in these imaginary measurements and computations we are being guided by metallurgical (I checked the spelling!) analogs of RULES A THROUGH E. Look at these rules, and try to formulate which they would be for density/mass computations. Finally A last look at Francine
In this case, Francine is traveling north on the Garden State
Parkway. She leaves mile 0 in Cape May at 9:00 AM. You have only the
following "equipment" (it is another "thought experiment"):You can look at the speedometer on her car and you can look at your watch. You are asked to tell where she is (what mile on the GSP) at 10:00 AM. What procedure can you follow? Here is one: look at the speedometer at, say, 9:30, record the miles per hour, and multiply by 1 (hour, that is). This gives an estimate of her position. A more "proactive" (means: "creating or controlling a situation by taking the initiative") method would be to take several measurements (say, record the speed every 5 minutes) and then multiply each speed reading by 5/60, and add these up. This would likely give a better estimate of the position she will be in. Again, if you are alert, you will see in all this we are again using the analogs of RULES A THROUGH E. Here the quantity to be computed is distance, and we are approximating distance by a sum of "rectangles" where we assume constant speed and multiply by a time interval.
Big newsAny quantity which obeys the analogs of RULES A THROUGH E can be computed (either approximated efficiently, or in a large number of cases, computed exactly) using the methods we will learn. There are many many such quantities, ranging from such physical "objects" as force and work to quantities in economics. And it is really amazing (to me, at least!) that what we've done so far can be applied to this problem. It is a bit difficult to "see" derivatives and slopes in all of these problems, without being shown where to look.So the idea will be to take a quantity, break it up into small, more easily approximated parts, and then to add these approximations up, and the resulting sum will give a good approximation (or, even, sometimes, an equality!) to what we want to compute. This is a method of disintegration and integration: taking apart and putting back together. On Monday I will go into the technical details of the method, and there will be much notation to struggle with. But if you can "internalize" the ideas of what I'm describing here, you will see that the result is worthwhile.
Exam 2 returnedI returned and discussed the results of the second exam. I apologized for the excess length, but not for the level of the problems. The difficulty of the problems was probably appropriate, but the length was not: I think I had about one problem too many -- I should have had 8 rather than 9. But I additionally remarked that the performance of students on some of the exam was clearly below the level expected. Students must learn how to study.
I forgot to ask a Question of the day! |
Here it becomes useful to observe that f is a graph of
g'. Thus, to the left of 1, g' is positive and to the right, it is
negative, so 1 is a local max. Similarly, 1 is a local min.
How about if f(x)=
We divide the interval from 2 to 3 into n different equal parts. Here
you should think of n as some very large positive integer. (The
picture shows n=6, not terribly large -- imagine n=10,000!) Then each
piece of the interval will have width (3-2)/n=1/n. We will approximate
the area over the first subdivision, from 2 to 2+1/n, by a rectangle
whose height is the value of the function at the right-hand
endpoint. Similar things happen if we choose the left-hand endpoint or
the middle or ... any choice, in fact. So the right-hand height is
(2+1/n)5, and the area of the rectangle over the first
subinterval is
(2+1/n)5(1/n)
(area is
height·width). What about the second rectangle? It has the same
width but the right-hand endpoint's height is
(2+(2/n))5. So the second rectangle's area is
(2+(2/n))5(1/n). And the third rectangle's area is
(2+(3/n))5(1/n). Etc. What does this Etc. mean? We
are supposed to find the sum of all the rectangles and there are n of
them. So we have:
Here I will use the notation in the textbook. Also I will refer to what
follows as FTC.
The Question of
the day was a bit involved. I first found the area under
y=x2 from x=0 to x=2 by quickly guessing an antiderivative
and taking its value at 2 and subtracting its value at 1: so start with
(1/3)x3 and get 8/3-0/3=8/3. Now look at the box this curve
is in. The box with corners at (0,0) and (2,4) has area 8. Therefore
the other part of the box has area 8-8/3=(16)/3. The QotD:
write that area as a definite integral with respect to y (and
rearrange the boundary equation so that it will read
"x=something"). Compute the integral, and make sure you get (16)/3.
The Question of
the day was the following: what's shown is the graph of a
function, f(x), made up of two line segments. First compute the
definite integral of f from 0 to 3. Then find A so that the definite
integral from 0 to A is half of the total area.