Lecture #1 (January 19) I handed out the syllabus and other information. I discussed how the course would be graded. I discussed the aims of the course. I remarked that this was not the course for students who thought math was the single most important subject in their lives, or who would strongly need calculus to pursue an intended major (e.g., engineering or a "hard" science). Most of the students in this course historically tend to be people interested in majoring in business or biology (NOT biochemistry -- those people should take 151!) or pharmacy. I described my own aims in the course as two-fold: first, of course I had to and intended to teach calculus. But second, and equally important, I would try to explain to them WHY they certain departments "forced" them to take this course -- why the methods, the material, the vocabulary, was relevant.

I told them that there would be a quiz in every lecture, and that all they had to do to earn credit for it was to take the quiz. I also said that I would try to generaly ask serious questions about the subject matter of the lecture, and that I would grade the quiz as if it were part of a real test. Since this would take a large chunk of time, I asked them why I did it. They answered "attendance", and I replied that indeed this would reward people who came to class. But I also said that it gave a way to communicate back and forth between the lecturer and the students, and that for the students it provided a way to get some feedback on how they understood the material, and on how they could actively use the material. For the lecturer, it was equally important. I remarked that it was easy for a lecturer in such a course to be self-deceptive, and not to realize that a "perfect presentation" in fact did not communicate things to students at all well. Thus these quizzes would serve as an important check for me as well.

Students should expect quizzes in every recitation meeting. The quizzes would generally resemble some easier homework problems.

I told them that I would not be able to cover every idea from the relevant sections of the text, and that therefore they should read the book. I also remarked that they needed a graphing calculator for this course. BUT the course was NOT about use of the graphing calculator. Really, I expected them to be able to do only two things with such an instrument. They should be able to use it as a "scientific calculator", to compute, for example, (ln(14)+2^7)/53. They should also be able to enter a function such as x^3-20x^2+5 and graph it. I remarked that they should bring their calculator to every meeting of the course.

The initial parts of the course would be review. Any student who had difficulty with this material should definitely come and see me. The review would be rapid, and was not really intended to teach the material, but was mostly intended to make sure that we had a common vocabulary. They needed to know anayltic geometery, functions, and algebra.

I began with analytic geometry. They would need to know about points in the plane, distance, circles, and lines. I drew the coordinate axes, and told them I would almost always call the horizontal axis x and the vertical axis y. I plunked a point down in the plane and described what the ordered pair corresponded to in terms of ordered distances. I did an example. I then discussed distance between two points, abstractly, discussing the distance between (x_1,y_1) and (x_2,y_2). I did it right (with absolute values!) and then mentioned that the absolute values disappear because we square the results. Then I mentioned that the distance formula had a square root, and that in this course the notation "square root of a number" meant the non-negative square root of a number, so that the square root of 4 was 2, even though there were two numbers whose square was 2. I did an example.

Then I discussed the algebraic requirements for a point (x,y) to be on a (non-vertical) line connecting the points (x_1,y_1) and (x_2,y_2). I used simlar triangles. I mentioned several times that they must be familiar with elementary geometry. I did an example, and I named the slope "m", and asked what the "b" identified geometrically and was told.

Then I gave them their quiz.

Quiz #1. The quiz question was "Why are you here?" I explained that an answer to this would help me teach them. It would advise me about the kind of exmples and language which would be more appropriate for them. 80 people took this quiz, which took about 40 minutes to grade. The plurality of the students taking the quiz (about a third) named "business" as their interest. The next largest segment (about a quarter) named "biology". Then various other interests were listed, including economics and various bio-type subjects. Note that the only about two-thirds of the peple on the roster appeared to answer this quiz, and about perhaps 10 people NOT in the roster answered the quiz. 1. The quiz question was "Why are you here?" I explained that an answer to this would help me teach them. It would advise me about the kind of exmples and language which would be more appropriate for them. 80 people took this quiz, which took about 40 minutes to grade. The plurality of the students taking the quiz (about a third) named "business" as their interest. The next largest segment (about a quarter) named "biology". Then various other interests were listed, including economics and various bio-type subjects. Note that the only about two-thirds of the people on the roster appeared to answer this quiz, and about perhaps 10 people NOT in the roster answered the quiz.

Lecture #2 (January 21) I brought in a viewscreen. This device, which can be borrowed in the Undergraduate Office, allows a screen of a graphing calculator to be visible to an audience using an overhead projecter. I began by asking them to calculate (ln(14)+2^7)/53, since I had TOLD them to bring a calculator to class. About half of the sutdents had. I try hard to encourage the students to do something during class, since 80 minutes is a long time for a math lecture, and they are or want to be, rather passive. I tried to duplicate the calculation on the viewscreen. Then I graphed y=x^3-20x^2+5 in the standard window (-10<=x<=10 and -10<=y<=10) and didn't get what I wanted. You should try this. In fact, darn it!, the function I wanted to graph was y=x^3-20x+5 in both the standard window and then in the window -5<=x<=5 and -50<=y<=50. This was in order to show them that the technology could be useful, but that they needed to think while using it. So I screwed this up. I am annoyed.

Then I began discussing functions. I told them that I wanted to review the use of the words function, domain, range, and graph. I defined function using a rather obtuse (to them) and elaborate English sentence ("an unambiuous way of assigning a value to each number in ...") and wrote out a function example in the following way: this function takes a number and then squares it and then adds on to that number the value of 2 raised to that number's power. I asked them to tell me the vlaue of the function at 3 and at -1 and at .7. There was confusion. I stared at them. Some of them gave me the first two values (I used that to remark that 2^{-1} was 1/(2^1) ). For the third, there was confusion while I waited for them to use a calculator. told them that I wanted them to help me, especially by using a calculator when I needed numbers. As for the value at .7, theydarn well needed a calculator to get it. I then remarked that people commonly abbreviated the function statement using algebraic notation. So here we could write

f(x)=x^2 + 2^x

I then tried to think about the graph. I said they needed to know the graph of x^2, a parabola, which was symmetric with respect to the y-axis (if this material was not familiar then they were in the wrong class!) and the graph of 2^x (this was exponential growth, used in the discussion of compound interest or one [inaccurate and idealized!] model of bacterial growth). How to get the graph of f? Either try to "add" geometrically the two graphs I had drawn or use a graphing calculator. So I drew a graph. I remarked that the graph seemed to have one "bump" at the bottom and didn't (if the calculator output is correct) seem to "wiggle". These statements were, in fact, true, but not at all obvious. In the course we would learn vocabulary for "bumps" and "wiggles". I then asked what the domain and range were, and we had to write the range approximately. i said that finding the exact range of a function, even one given by rather simple algebraic formulas, was frequently difficult.

What problems with algebraic formulas can cause difficulties with domains? We wrote g(x)=sqrt{x^2-7} and found the domain, noting that square root in this course must always have a non-negative argument, and solving (after some thought) the rather simple implied inequality: x^2-7 >= 0. Then I wrote h(x)=1/(x^2-3x+2) and asked what the domain was. I remakred that they should be able to factor the denomiantor, and then I did it for them.

But functions derived from algebraic formulas will not occur obviously in many applications. I asked about income tax. I wanted a progressive income tax, which obeyed the following rules:

i) as income increases, tax should increase
ii) the percentage of income taxed should go up
iii) (fuzzier) the tex should appear fair.

and the tax I suggested would look like:

first $10,000: no tax
next $10,000, (or 10 to 20 thousand) a tax of 2%
next $10,000, a tax of 4%
above 30 thousand, a tax of 6%.

I remarked that my "stipulations" were rather tentative and imperfect, andI would like to make them more precise. We computed the tax on some incomes. There was one comment on what the tax should be on EXACTLY $10,000. I fixed that up. Then I said, what happens to someone who makes $9,9999 and then earns an extra $2. What should their tax be? I got two answers (as I desired!). The answers wer 2 cents OR $200.02. So we had to be more precise about what the tax was (to specify it better, more unambiguously!). So we did. We then figured out what the tax was on $25,000. The tax function had to appear to be fair, so it should not penalize people with an enormous chunk (as above) for having income near the "breakpoints" of the brackets. This was hard for them.

Then I wrote out a complete algebraic description of this function, as a piecewise defined function. I mentioned that this was actually much simpler than the real U.S. tax code. It is not clear how much experience these folks have with income axes!

I want them to be comfortable with functions as defined by formulas, defined by graphs, defined by piecewise formulas, defined by goemetrical or other considerations, and defined by tables. Then I gave them their quiz.

PLEASE NOTE: I am ALREADY "behind" the syllabus -- I did not discuss trig functions. I will go like the wind next time and catch up. Quiz #2. I wrote a table of two functions:

			x   f	g
			0   2   1
			1  -1   2
			2   0   1

I said that this table indicated that f(0)=2 and g(2)=1. This was all the information I would give them. I asked them to answer the following questions as well as possible:

What are g(f(2)), g(g(1)), f(g(2)), and f(f(1)) (the last could in fact NOT be computed!). For which x's is g(x^2)=1? For which x's is (g(x))^2=1?

Most people answered the first group of questions correctly. Many people had problems with the last two, however.

96 students took this quiz, which took about 65 minutes to grade.

Lecture #3 (January 27) Officially I have "caught up" to the syllabus. By this I mean (as I told the students) that any material which I do not discuss in class of the review sections I will assume (I must assume!) that they have worked through.

So what did I do in this lecture? First, I drew the picture of 10^x. I asked what the values of this function were when x is +/- 1 and 2 and and remarked that when x is positive, this graph really increase rapidly, and when x is negative, it is very close to the horizontal (x-) axis. I said that the graph is characteristic of compound interest and an idealized model for such things as bacterial growth.

I then drew the graph of (1/2)^x, and similarly characterized the graph and computed some values of the function. I very briefly mentioned amortization and depreciation and radioactive decay.

Then I asked how many people brought in graphing calculators. Many held them up. I asked them to graph 2^x in the window -3 I then asked the students to duplicate this with 3^x and waited while they did it. I asked first how the two curves (2^x and 3^x) would compare. This caused some consternation -- people had to think about it. But later we computed the local ratio for 3^x and got about 1.1.

I remarked that we would be doing lots of computations with local ratios, and that .7 and 1.1 were complicated numbers. I would rather have a simpler number. I suggested 1, and in order to get exponential growth with a local linear factor of 1 we need to look at a number between 2 and 3. We call that number e, and it is, I told them, approximately 2.718 ...

I wrote the standard algebraic properties of exponential functions, specialized to the base e (e.g., e^{x_1+x_2} = ...).

I graphed e^x. Then I discussed what the inverse function to e^x should look like. There was considerable confusion about this. I finally mentioned that (A,B) on the e^x graph results in (B,A) on the graph of ln, the natural log of x. I evaluated ln at e^2 and e^3 and e^{-1}. Then I asked what ln(10) was. Several people reached for their calculators, which was fine with me. Others had no idea. I remarked that ln of a number asks for the power of e which was equal to that number, and apparently since e is between 2 and 3, ln(10) should be a bit bigger than 2, which it is.

I wrote down the standard algebraic properties of log functions, specialized to ln.

I asked the following question: suppose you know that ln A = 3.5 and ln B = 1.2 and ln C=1.5, what (EXACTLY!) is ln({A^3 B}/ sqrt{C})? I waited for students to do this, and then did it myself. Some people clearly found this difficult.

I then began discussing trig functions. First, I remarked that similar triangles allowed us to do indeirect measurement. Thus one could deduce the height of a tree from the known height of a person and various distances from these objects across level ground (I drew a picture!). So proportions of sides of right triangles are important. I also remarked that only a few of these ratios would be used in this course: sin, cos, tan, and (very rarely!) sec. I defined these ratios in terms of hypoteneuse and opposite and adjacent. I also remarked that there were many many trig identites. We would not need too many of them. We'd need the Pythagorean theorem (I wrote it, and wrote the sin/cos squared identity) and I would need the addition formula for sin (I wrote that). Then I ran out of time. I will continue at the next meeting with the graphs of sin and cos and tan and their domains and ranges, which will officially conclude the review portion of the course.

Quiz #3. I drew a right triangle on the board, with one of the acute angles given as 20 degrees. The adjacent side was given to be 5 units long. Another right triangle was drawn so that one of its sides was the hypoteneuse of the first triangle, and the other side was given as 2 units long. I asked students to find the length of the hypoteneuse of the second triangle.

About a third of the students had problems with this question. Note that the definition of the trig functions along with various diagrams was on the board, as well as the Pythagorean theorem.

94 people took this quiz. I spent about an hour grading it.

Lecture #4 (January 28) I carefully sketched the graphs of sin, cos, and tan. I described the domains and ranges of each, noting that since tan was the quotient of sin and co that its domain was where cos wasn't 0, which could be seen from the graph of cos. I indicated that cos was sin dragged back by Pi/2 and compared the graphs. I remarked that this could be verified using the addition formula for sin, and then verified it. I wrote the symmetry/antisymmetry properties of sin, cos, and tan, and also listed their periodicities. I remarked that I didn't want them to know many values of these functions, but that certain values were apparent, such as where sin and cos were 0 and +1 and -1 (and the vertical asymptotes of tan!). I also remarked that the symmetries and periodicities allowed them to compute exactly such things as sin(70{1/2}Pi) and tan(-4{3/4}Pi).

I remarked that in calculus what I will do is closely study the sine function, and then use the results to deduce things about cos and tan.

I noted that very few of the people in this room would need to consider many right triangles in their future endeavors, but that almost all of them will have to analyze periodic, repeated phenomena. The analysis of such things will frequently use the trig functions.

I remarked that in the problem I had given them last time that I had specified an angle of 20 degrees. In this course I would almost always measure angles using radians. The reason for that would become apparent next week (when I expect to differentiate sin, but I didn't say that). I then quickly defined radian measure (very quickly but correctly) and asked them what the radian measure of 20 degrees was. Very few could answer. Then I asked students to make sure that their calculators were set in radian mode rather than degree mode, noting that the consequence was be disaster if they tried graphing some of the functions we'd be studying otherwise.

It was time to stop "review" and start calculus. The most important nouns in the course were certainly limit, derivative, and integral. The last two were special cases of the first, so I should better study carefully the first one first (?). I remarked that I strongly recommended reading the book because what I saw as an important part of my instructional duty was describing how limits were used as a vocabulary in applications. Therefore I would begin my study of limits by describing some "scenarios" in business and biology.

First, in business I went back to my discussion of the tax brackets which I began in lecture 2. I reminded them about the general scheme. I then asked them how to graph the tax as a function of the income. Most students found this quite difficult. I began by saying in the lowest bracket the tax was 0 so that the graph would be a horizontal line segment. In the next bracket the tax was 2%. We discussed what the graph should be for a while. Finally it was decided that the graph should be a line segment of slope .02 which I tried to draw. I then said remember that the tax on an income of $9,999 was $0, while if someone earned another $100 the tax should be ... what? Is it 2% of $10,099 which would be over $200? We agreed that would not be fair (a negative incentive, I remarked). I sketched several ine segments of slope .02. I remarked that if the line segements didn't connect with the horizontal line segment for income less than $10,000 the tax policy would not be rational. The government would either be giving people money or it would be asking for more tax than the increase in income. Therefore we want the line segments to connect with no jump or break. I wrote out an English statment to this effect, which include phrases like "close to". Then I remarked that there were many ways of phrasing this, and that notation which was briefly and less ambiguous (useful abbreviations) was widely accepted. Therefore I introduced --> for "close to" and remarked that lim_{I --> 10,000} T(I) should be 0. I did the same things with the other break points of the brackets, and sketched a reasonable graph of the tax function. I think this is all fairly sophisticated for them. Graphing and relating this to quantitative information is an important skill they need to develop.

Then I segued (?) into a biology mode. I remarked that catalogs sell trees with short descriptions which frequently include such information as "The eventual height of this tree is 40 ft and its growth rate is 3 ft/yr." I wanted to know what these numbers meant. If one took the 3 ft/yr literally, then in a century the tree would e 300 ft high, and a bit later (well, quite a bit later!) the tree would knock down the moon. I asked what a graph of the tree's growth would look like (horizontal axis in years since tree was planted, vertical axis height of tree in ft). We discussed for a while what the 40 should mean, and what the growth curve should look like.

Finally I wrote a reasonable (?) growth curve. The horizontal axis was measured in years and the vertical axis was in feet. The graph I finally drew had a horizontal asymptote of 40 feet. I discussed this, and finally interpreted it by the statement lim_{t --> infty} h(t) = 40. Then I discussed what the phrase "its growth rate is 3 ft/yr" means. I took two random points on a certain region of the graph and looked at the change in tree height and the change in time elapsed. Some of the students recognized a slope, and I said that one conventional translation for the phrase was lim_{elapsed time --> 0} {change in tree height} / {elapsed time} = 0. I remarked that we would look at many quotients of this kind.

Then finally I said it was time for me to display some of the limits that I was supposed to do. I strongly said that they should read their expensive textbooks.

So I asked them what lim_{x --> 2} x^2 was. After some discussion we decided it was 4. Then I asked what lim_{x --> 2} {x^2 -4}/{x-2} was. Soem further discussion followed, and we factored the top and concluded that the limit was 4.

Then I gave them their quiz.

Quiz #4. What is lim_{x --> 3} {x^2-4x+3}/{x^2-x-6}? Most people could factor the top and bottom, and get the correct answer. 96 people took this quiz, and I spent about 50 minutes grading it.

Lecture #5 (February 2) I wrote lim_{x --> A} f(x) = L and said that I would spend the period explaining this statement and just doing examples.

First I wrote: "This means that as x gets close to A then f(x) gets close to L".

I remarked that this statement was difficult to understand: exactly what does "gets close to" mean? For most of them, a very precise statement of this is not important, but they should realize that people have spent a great deal of time and effort analyzing it.

As first comment I remarked that f should be defined for x's in a deleted interval around A: so f needs to be defined for x's near A, but not necessarily at A.

As a second comment, I wrote that by selecting x close enough to A (not necessarily at A) we can make |f(x)-L| as small as ANYONE wants: I tried to emphasize that different people might have different conceptions of "small" but that the useful definition said that the difference in the lengths of f(x) and L could be made smaller than ANY conception of smallness.

Example 1: what is lim_{x --> A} ((1/x) - (1/A))/(x-A)?

If we first try to "plug in" (a technique I said works sometimes and should be thought about) then then we end up dividing by 0. This is not allowed. I also remarked (emphasized!) that we would be studying many limits of this algebraic type. The key here is to transform the function algebraically into another form where its behavior as x --> A can be recognized more easily. We worked on this transformation, and it went very slowly. These people do not love algebra. Finally we saw that it was algebraically equivalent to -1/(xA) and that as x --> 0, this --> -1/A^2, although a few people had difficulty even with this last step.

Example 2: I drew the graph of a function. It consisted of a horizontal ray, half of y=1 for x< 0, a line segment from (0,1) to (1,0), and a line segment from (1,0) to (2,2), a horizontal ray, half of y=1 for x>1. There were open circles at (0,1) and (2,2). I first tried to evaluate this function at several values. Most people seemed to understand. Then I tried to compute its limits at several numbers. Away from the breaks (where the graphical piecewise definition changed) there seemed no problem. At the breaks, we debated for quite a while. Many people want to only evaluate limits by substituting in f(A) for lim_{x --> A} f(x). Another problem exposed by this example is that many students still cannot "read" a graph and get what we would consider simple information from the graph.

Example 3: I looked at (sin x)/x as x --> 0. First I remarked that I didn't see any simple algebraic way to simplify this quotient. Then I said maybe we could experiment on it by computing the value of this quotient at, say, x=.004. I stared at the students until they (or at least some of them!) got out their calculators. They calculated and got a number near 1. I calculated and didn't because I still had my calculator in degrees. I was very abashed and remarked that this limit (which would turn out to be very important) was a serious reason that we would look at angles as measured only in radians. The value "1" was a lot easier than the other value. Then I copied almost exactly the presentation in the textbook (a wild melange of diagrams and definitions of trig functions and inequalities) which would prove the desired limit. More and more I believe that this development, which is quite incomprehensible to them, is mostly a serious waste of time. If it serves to tell them that this limit is important, that we think they should know simple facts about triangles and definitions of trig functions and radian measure, well then, o.k. (maybe). But I do not believe we can "test" them effectively on the process. or even on squeeze theorems for limits. I don't really believe that this is a worthwhile use of Math 135 class time. Oh well.

I do not think that this lecture was easy or was presented well. I am unhappy.

Then I gave them their quiz.

Quiz #5. I drew a graph of a function. It was a halfline of slope 1 going up to (0,0), and then a line segment of slope -1 going from (0,1) to (1,0) and then a halfline of slope 1 from (1,0). At 0 the function's value was 2 and at 1 its value was 1. I asked two questions:

1. For which A's does lim_{x --> A} f(x) exist? (List all such A's.)

2. For those A's for which the limit exists, list all A's for which the limit is equal to f(A).

Lecture #6 (February 4) I began by writing a number line on the board with portions of it colored red and blue. I asked students to write answers to the following questions:

a) Which numbers were red?

b) Which red numbers were to the right of at least two blue points?

The purpose of this was to introduce discussion of the quiz of the previous meeting, which students had not done well on. I wanted to emphasize the logic of the questions to students. So then I did discuss the previous meeting's quiz, trying to be very careful about limits and the language I used. I also stressed that the graph of a function contains information about the function's values, which I don't think students realize too much.

I then remarked that in a college course, the efficiency or interest of the instructor is actually not of prime importance. No matter who the instructor is or how the instructor acts, the prime source of learning will be the student's own work. I urged students again to prepare for class by reading the text. I said that I would be going very fast.

Then I wrote down a piecewise definition of a function. I wrote something like:

	    x^2 if x<1
     f(x)=  2x-1 if 1<=x < 3
	    x-3 if x>= 3

There was some confusion about what this meant, and I tried to clear up the confusion by actually computing the function's values at some numbers. Then I asked students to answer the questions from the previous meeting's quiz about this function. That is,

1. For which A's does lim_{x --> A} f(x) exist? (List all such A's.)

2. For those A's for which the limit exists, list all A's for which the limit is equal to f(A).

I left folks alone to work for a while. Then I discussed how to approach this problem. Many people suggested drawing a graph of the function, which I did slowly and carefully. I took appropriate pieces of the graphs of x^2, 2x-1, and x-3 and wrote them together on the board. I tried to discuss answers to questions 1 & 2 above very carefully. It went slowly, again with many students seeming to have difficulty with the relationship between the geometry of the graph and the behavior of the function.

I remarked that there was differing behavior to the left and to the right at the points 1 and 3, and said that people used notation to distinguish between left-hand limits and right-hand limits: x-->3^- means "left-hand limit at 3" and x-->3^+ means "right-hand limit at 3". I computed these limits for the function above.

I then defined the word of the day in the course: continuous. A function is continuous at A if lim_{x--> A} f(x) exists and is equal to f(A). I analyzed the function f already studied to see where it was continuous.

Then I studied a slightly different and more difficult problem: suppose g is a function defined by the piecewise definition

	    x^2 if x<1
     g(x)=  mx+b if 1<=x < 3
	    x-3 if x>= 3

(here m and b are NOT known!). What does the graph of g look like? I write the parabolic piece, and the right-hand linear piece, and then suggested several different line segments in the middle as possibilities for the graph of g.

I asked if we could select m and b so that g was continuous everywhere.

Most people seemed to agree that the difficulties would center around the x values of 1 and 3. At other places, g would be defined by simple functions and would likely be continuous. So I carefully analyzed g at 1 and 3. I did this by computing

g(1) and lim_{x -->1^-} g(x) and lim_{x -->1^+} g(x)

g(3) and lim_{x -->3^-} g(x) and lim_{x -->3^+} g(x)

and remarking that for g to be continuous at 1, the first group of numbers would all have to be equal and for g to be continuous at 3, the second group of numbers would have to be equal. This led to conditions on m and b: two linear equations in two unknowns. I solved these, and sketched the resulting graph.

Then I asked students their quiz question.

Quiz #6. I asked people to graph a function (and I emphasized that there were many functions!) which had the following properties:

f is continuous everywhere except -2 and 3

f(-2)=-1, lim_{x --> -2^-} f(x) = 1, lim_{x --> -2^+} f(x) = 3

f(3)=0, lim_{x --> 3^-} f(x) = 0, lim_{x --> 3^+} f(x) = 1

89 students answered this quiz (breaking the 90 barrier!). Many could not do the problem. They have substantial difficulty telling the horizontal and vertical axes apart, and thinking aobut the graph of a function.

Lecture #7 (February 9) I began the lecture by discussing with students the ways they could enhance their chances of success in Math 135. Less diplomatically, I stated that I had graded 6 quiz questions for most of the students in the room, and that the performance of many of the students was very poor. A large number of students were repeating the course, and they and others would not have the kind of success they wished if their performances continued to be as bad. I said that it was hard work to really listen to lectures in a math course.

i) Students should take notes. I do come to class rather prepared. I try to outline what I consider the most significant material and to give well-chosen examples. Students who take good notes should find that they have material which is easy and efficient to use when reviewing the course.

ii) Students should do the homework problems. My best guess is that the homework problems for each class should take an average student in the course from 3 to 4 hours to do. I strongly suggested that students work together to do the homework, and not just look at the problem statement and the answer in the back of the book and try to find ANY method of connecting the two. Working with other people tends to make folks a bit more honest about what they are doing.

iii) Students should prepare for class. The material will be going by ever more quickly. They will enhance their chances for success by briefly looking over the sections to be covered before each lecture. Yes, this is more work for students, but it is very useful.

iv) Students should ask questions. They should stay alert and not let stuff float by. This is very difficult in a class this size.


Then I began the lecture itself. I again wrote the definition of continuous function: lim_{x --> A} f(x) = f(A). I said that one very important consequence of continuity was the Intermediate Value Theorem. I wrote and diagrammed a geometric version of this result: that if the graph of a continuous function connects two points, it must pass through any intervening horizontal line at least once. I remarked very briefly on why the theorem should be true: if it were not, there would have to be a break or jump in the graph of the function.

Almost all of the functions we would consider in Math 135 would be continuous. What would be some "real world" scenarios where continuity is/is not appropriate?

#1: Travel on the Graden State Parkway, north from Mile 0 (Cape May) to Mile ~180 (Montvale?). At mile 117, I was advised, sits MIddletown. I drew a (very) vague map of New Jersey to illustrate the trip. My question was: if a car leaves Cape May and travels north to Montvale (on the road!), will it pass Middletown? We considered this. In the absence of a Star Trek-like transporter device, apparently the answer is "Yes." I then tried to draw a graph of time on the trip versus position on the highway. It was a generally increasing graph (it is inadvisable to back up too much, I remarked). I then drew Middletown's position as horizontal line and said that if we assume automobile travel is appropriately modeled by a continuous function, the function's graph must cross the line.

#2. I tried to sketch a jagged graph of the price of a stock. The graph had various corners, and then (at some time) dropped with no other information. This did not seem to be a continuous function, and might represent the stock price's repsonse to disastrous news about the company. Various possible stories were suggested by students.

As a second topic in the lecture (in an effort to catch up with the syllabus) I discussed the graph of (x^3+x)/((x-1)(x+2)). First I asked the students what the graph of this function looked like. I stared at the students until (sigh!) some of them picked up their calculators. After a while, I suggested graphing it in the window -5<=x<=5 and -10<=y<=10. They did (or at least some of them did). Some of them had trouble with parenthesizing the function properly on their calculator, as I suspected they might. I drew the graph, which seemed to include two vertical lines. The vertical lines were generated by the calculator becuase it had been constructed to believe that all the functions it would be asked to grpah would be continuous in any interval it was asked to graph them, and it tried to fill in the graph with dots using that assumption. In fact, the vertical appearing lines were NOT on the grpah, because the function was NOT continuous at 2 and -1 -- these numbers were not in the domain of the function, and the function could not be continuous at points not in its domain.

We then discussed the asymptotic behavior of this function near 1. It seems that the top looks "good" -- about 2, and the bottom looks like something small multiplied by 3. This then works out to something LARGE. But if we wanted graphical information we needed to know the sign as well, because large + means graph high up, and large - means graph high down (??).

So I discussed what happens as x-->1^+ and as x-->1^-. In the first case the fuction's values were large +, so the graph went up, and x=1 was a vertical asymptote from the right and we could write lim_{x--> 1^+} f(x)=+infty. I then wrote what happened on the other side of 1. I urged students to see what happened on both sides of -2.

Then finally I went on to the "new" topic of the lecture: rates of change/slope of tangent line/derivative.

I went back to the example previously discussed of tree growth, with a certain annual rate of growth and a fcertain eventual height. I redrew one idealized graph of tree height in feet as a function of tree age in years. I asked what would be meant by "the rate of growth of the tree at age 6 years". First, what would be the units? People replied that it seems the units would be in feet per year. Then how would they compute this growth? Possibly it could be approximated by, say, the rate of growth over, maybe the 10 lears which followed year 6. This led to the quotient (if h(t)= height in feet at time t in years)

	h(16)-h(6)  [difference in tree height]
	  16-6      [elapsed time]

This might not be a good approximated to what the tree was actually doing near year 6. A better approximation might be gotten by using only datea in the first year after year 6:


or an even better approximation might be gotten after 1/2 a year. I then asked (because there was a graph of tree growth on the board) what this could mean geometrically. I tried to draw the tree heights at the different times (with one of the times year 6) and got two dots on the graph: (6,h(6)) and (6.5,h(6.5)). Several people claimed to recognize the quotient as the slope of the line through these two points.

But even better approximations "could" be possible by choosing smaller and smaller time intervals. I alluded to folks in the chemistry labs who might be doing nanosecond experiments to get rate constants (this is almost true). In the case of a tree, small time intervals might be difficult to have significant enough height differences.

I remarked that for small time intervals, the slope of the line seemed to get close to the slope of the tangent line. The limit might BE the slope of the tangent line.

For a function with data given graphically, this is about as far as we could go. For a function given by a simple algebraic formula, one can be much more precise.

One VERY non-realistic example would be given by h(t)=t^2. Then the tree growth rate at 6 would be lim_{t --> 6} (h(t)-h(6))/(t-6). The quiz problem was to evaluate this limit.

Lecture #8 (February 11) I built on what I did last time. The growth rate of a tree was a limit of a ratio and this was the same as the slope of a tangent line as a limit of slope of secant lines and that, mathematically, this was called the derivative. The average growth rate over an interval could be measured physically if the interval were long enough (a month, a week, in the case of a tree, perhaps microseconds in the case of a chemical reaction). The limit was an idealized version of this rate. I wrote the vocabulary word of the day: derivative. While I had written derivative as lim_{t --> A} (h(t)-h(A))/(t-A) = h'(A), the textbook would write it as lim_{delta x --> 0} (f(x+delta x)-f(x))/(delta x) = f'(x) and I provided a dictionary to translate from what I had done last time to the notation of the text. I said that although I was not a fan of all of the book's choices, I would go with what was there. I particular, I remarked that, almost paradoxically, "delta x" was supposed to be read as one symbol, standing for a tiny piece of x. Many students read it as delta multiplied by x. I don't think my remark was universally understood.

What the next 3 lectures would be devoted to would be various ways of computing derivatives of functions given by simple formulas. I did an example: consider y=1/x. What is an equation of the line tangent to this curve when x=3? I carefully found a point the line must go through, and then found the derivative of 1/x at x=3. I used the information to get an equation.

Then I consider the function f(x)= x if x>=0 and 0 if x<0. I graphed the function. The graph consisted of two half lines glued together at the origin. I asked what the derivative of this function was. We decided that f'(x) "clearly" was 1 if x>0 and was 0 if x<0. The value of f'(0) was problematic. So I carefully considered the derivative of f at 0, dividing the limit up into two cases, and getting two different numbers (0 and 1) depending on whether the delta x --> 0^- or 0^+. I then deduced that the (two-sided) limit itself did not exist.

I remarked that this function was continuous, and it was an example to show that a continuous function did not have to be differentiable. I remarked that, however, a differentiable function had to be continuous.

Then I sailed into how to differentiate various formulas.

I began with considering the derivative of f(x)=x^{400}. I looked at (f(x+delta x)-f(x))/(delta x). Then I wrote

(*) (x+delta x) (x+delta x) (x+delta x) .... (x+delta x) (x+delta x)

and indicated it was 400 times, and tried to imagine how to handle this product. Certainly one resutl would be x^{400}. I then tried to imagine if a choice of a single delta x were made from each factor what sort of term I would get: (delta x)x^{399}, and how many of these terms I would get: 400 of them. And what else would be left in the expanded product? Things that would have at least two delta x's. So the result of this analysis of (*) was

x^{400}+ 400 (delta x)x^{399} + (delta x)^2 Junk

where Junk was some polynomial in delta x and x.

I then put this into the difference quotient, and did cancellations and took the limit to see that the result would be 400 x^{399}.

I remarked that I was sure about two-thirds of the people in the room knew what the answer would be, but that the definition was important as was the algebraic analysis. I repeated that an intelligent 10 year-old could be taught the rules I was showing them, but the understanding how and when to use them, the context, was difficult.

I think Rule 1 was that the derivative of x^n was n x^{n-1}. I said that the proof when n was a positive integer followed from the same method that I had just used, but the formula was also valid for other n's and the verification for such n's was more subtle.

Rule 2 was the derivative of a a sum. I "proved" this by remarking that the derivative was supposed to be a rate of change. I imagined a gerbil colony in a cage, and that the function I was studying was the number of gerbils in the cage. The derivative would represent the birth rate (or rather, the combined birth/death rate) of the gerbils. I asked what would happen if there were two cages, and what would be the birth/death rate of the total population of the gerbils in the two cages compared to the separate birth/death rates. This all seemed too graphic for some students. But I did write down that the derivative of f+g was as expected. I applied this to the sum of two spcific monomials.

Rule 3 was, I think, the derivative of a constant multiplied by a function. For this I drew again upon the gerbil analogy, and asked what would happen to the total birth/death rate if there wer 42 cages. A few people did say that the birth/death rate would be multiplied by 42. An example here was a specific constant multiplying a monomial.

Finally, I asked what the derivative of a constant was. I illustrated this by asking for the rate of changed of a colony of neutered, immortal gerbils. I thereby wanted them to see there was 0 birth/dealth rate. I think my analogy, which I was stretching a lot, might have confused them. Also the weird words might not have been too helpful.

Then I wrote a specific polynomial and found its derivative. I remarked that I really would prefer that they NOT multiply of simplify unless they really had too: so the derivative of 43 x^{37}-6x^{12}+78 is (43)(37)x^{36}-6(12)x^{11}+0.

I then sketched the graph of sin x and looked at the slope of its tangent lines, and tried to sketch a graph of the derviative of sin x. I suggested that it would be really neat if this derivative were cos x, and remarked that I would verify this next time.

I gave them their quiz. I went about 5 minutes overtime, and there's little excuse for this. I should have planned better.

Quiz #8. Suppose y=x^2+3x-1. Find an equation of the line tangent to this curve when x=2. After a few minutes I wrote 2x+3 to indicate toohe derivative. UNFORTUNATELY I went overtime during the lecture, and several people left before the quiz, and there wasn't very much time to do the quiz. BUT I'd say that less than 20% of the students were able to answer this problem correctly, indicating to me that my wonderful 2 lectures weren't too well understood and that students are not looking at the book!

86 students took this quiz. I spent about an hour grading it.

Lecture #9 (February 16) After an apology (!) for running over last time, I continued the display of standard algorithms for differentiation. I remarked that almost everything I would do was written in an equivalent manner in the textbook, and urged students to look there for maybe better or more complete versions of the material. I would cover what I felt was either important or that I could give some special (?) added insight to.

I wrote the differentiation formulations we had already seen (monomials, constants, constant multiple, addition).

I remarked that "today's quiz" would have several parts and I would periodically ask them to do things. Therefore they should reserve a page for "the quiz". I gave them their first question, which was to differentiate a polynomial.

Then I expanded sin (x+ delta x) using the addition formula for sin. I rearranged the result in the time-honored manner, and USING the limit formulas we had painfully gotten last week, concluded that IF WE USED RADIAN MEASURE FOR THE ARGUMENT TO THE SIN FUNCTION the derivative of sin was cos. I recalled for them (since I had done this very hurriedly last time near the end of class) the graphical argument showing the qualitatively, at elast, the derivative of sin "lloked like" cos (at least where the derivative was positive, negative, and zero). In fact, I concluded that the derivatve was equal to cos. I then looked at the graph of cos and tried to see what its dervative was. Interestingly it seemed to have OPPOSITE sign to sin's but, with that difference understood, the behavior was similar. I then wrote that the derivative of sin was cos and the derivative of cos was -sin.

I gave students another function to differentiate for their quiz.

Then I looked at the graph of e^x. It seemed that the derivative, again identified as the slope of the tangent line, was never 0, and in, fact, the derivative was always positive. We also noticed that as x--> -infty, the slope decreased -- still positive, but very very small. And, as x--> + infty, the slope seemed to constantly increase. I reminded studednts that in the first lecture (or was it the second?) I remarked that we would choose the number e so that the slope of the tangent line at (0,1) was exactly 1. Then I declared that e^x was the most important function and the simplest: its derivative was itself. I then used the exponential property of e^x to reduce (e^{x+delta x} - e^x)/(delta x) to e^x ((e^{0 +delta x} - e^0)/(delta x)) and that the last quantity as delta x --> 0 had limit 1 because it was the same as the slope of the secant line and we chose e so that the the limiting slope of thge secant line near 0 would be 1. In both sin and exp I emphasized that choices were made historically so that computations would be easy, not to irritate students. I remarked that, for example, the choice of definition of PH in chemistry was not as carefully made -- it had a base 10 exponential in t, and that meant that many formulas using PH had some fairly ugly numbers in them.

I gave students another function to differentiate for their quiz.

Whew. Then I said that the next "Fact" or "Rule" was, at first, a bit unintuitive. For example, look at x^2. Its derivative was 2x. Yet x^2 could also be described as x multiplied by x, and each part of that product decomposition had derivative 1. It wasn't so clear how the derivative of a product of two functions could be described in terms of the pieces of the product. In this case, I wanted to give some background to help understanding of the product rule. Or I hoped to! Anyway, what I did was try to conceived of two lengths growing. At time x and x+delta x, I wrote two lengths for each of two functions on the board: f(x) and f(x+delta x) and g(x) and g(x+delta x). Then I assembled a rectangle with one side f(x+delta x) and the other side g(x+delta x) on the board. I wrote in the rectangle another one whose sides were f(x) by g(x), and asked how to describe the difference. The difference was an L-shaped region, and I dissected this region into three rectangles, (f(x+delta x) -f(x)) by g(x) was one, and (f(x+delta x) -f(x)) by (g(x+delta x)-g(x)) was the second, and f(x) by (g(x+delta x) -g(x)) was the third. I then divided each of these by delta x and tried to show how the pieces of the first and third rectangle got the PRODUCT RULE and the second piece --> 0. I then worte the product rule and looked back at x^2 to check that it would work and tried a more complicated example. I emphasized that I did not think the product rule was intuitive in any sense.

I gave students another function to differentiate for their quiz.

Then I asked what the derivative of something like W(x)=1/(7x^{11}-3x^3+8) was. I analyzed this by looking at the function V(x)=7x^{11}-3x^3+8 and noting that V(x)W(x) = 1. I used the product rule, and concluded that W'(x)= -(77x^{10}-9x^2)/(7x^{11}-3x^3+8)^2 . It is amusing (??) to note that I made, I think, at least two computational errors in this. Sigh: I was getting tired. Then I wrote the general rule for differentiating a reciprocal of a known function.

Finally, I ended the presentation of course material by writing the QUOTIENT RULE. I remarked again that I did NOT think this was at all intuitive, and that students would need to practice it a number of times to internalize it. I did an example, and then

I gave students another function to differentiate for their quiz.


I discussed arrangements for the exam. I remarked that they needed to inform me as soon as possible and have a serious reason to miss the exam. I also said, truthfully, that no exam I would give would be calculator intensive, and that although they should definitely have a claculator with them, intense use of a calculator during an exam was not necessary. Students could also bring formula sheets to the exam, but if they found that they needed to consult the sheets frequently while preparing for the exam or while taking the exam, their preparation was not good.

I would NOT have office hours on Wednesday, but would have them on Sunday from 4 to 6. I would give out a review sheet which would be discussed (along with homework problems) in recitation on Monday. I would have a review session on Monday evening from 7 to 9 in PH 111 (that's Pharmacy 111). Answers to the review sheet questions and old exams with answers were all available on the web, and also available in print at the MSLC.

That's the story.

Quiz #9. I had people write derivatives of 5 functions given by formulas at intervals during the class. I asked them to have their answers checked by other students. I learned that students don't proofread very well! The functions were:

   5x^7-9x^4+sqrt{339} (put there to attract the error that sqrt{339} needed some special treatment!)

332 sin x + 5 cos x + (1/{10}) x^{2/3}

-9e^x+5sqrt{x}+ {Pi/6} (the sqrt here used to make people convert it to a power, and the Pi/6 again to make the point about constants!

(-3e^x+9cos x) (5x^2 -2x +.01) (again, the .01 for the same reason!)


This was VERY laborious to grade with many comments necessary. 88 students took this quiz, and I spent about 95 minutes grading it.

Lecture #10 (February 18) I began by doing some differentiation exercises: a few product rules, a few quotient rules. The last quotient rule I did was (sin x)/(cos x). After I did it, i recognized that the original function was tan x and that the derivative, after fiddling a bit with the top, was -1/((cos x)^2) or (sec x)^2. This became rule 9a: the derivative of tan x is 1/(cos x)^2 or (sec x)^2.

Then I distributed pieces of paper containing complicated numerical information, and displayed a copy of it on an overhead projector. This was an approach to the chain rule, motivated by the fact that 80% of the students had written in their initial lecture quiz that they were interested either in majoring in some aspect of business or economics. Therefore they should have had Eco 102 or be interested in it (but not that many students seemed to be familiar with the names of the instructors of the course).

I gave people the following information, which I remarked was already quite complicated although it was certainly simpler than real-world data:


Capital Invested    Chips produced      Marginal chips produced
$ in millions       1,000's of units    1,000's of units per 
                                        millions of $'s

      200               3,000                    .23  
      300               3,040                    .28
      400               3,070                    .42
      500               3,100                    .78 
      600               3,190                    .31 

CHIPCO SALES & PROFITS Chips marketed Profit gained Marginal profit 1,000's of units $'s in millions Millions of $'s per 1,000's of units 3,000 1.2 .03 3,050 2.8 .02 3,100 3.6 .05 3,150 4.9 -.01 3,200 5.1 .02

We spent some time exploring what the numbers meant in both charts. For example, how many chips would be produced (approximately, according to the chart) if the company invested 402 million doillars. Aafter a while, some folks answered 3070+(.42)*2. What if it invested 399 million? Etc. Not as much famiarization with "marginal" used as a technical adjective seemed to be present, certainly not as much as I had anticipated or would have wished.

I asked a quite complicated question. What return would an investment of 500 million dollars get the company? This meant that the students had to link the two "spreadsheets" together, and follow information from one line to the other. Then I asked what the marginal profit would be for at a capital investment of 500 million. This was VERY complicated.

First I needed to be more specific about "marginal". I defined C to be the capital investment, and P to be the number of chips produced (all units were as indicated). Then, for example, P(500) was 3,100. What about the marginal chip production? That was the "response" of the chip production to a change in capital investment. So it was (delta P)/(delta C), where the delta's meant small in the units each was measured. I then wrote pRofit, (R) as a function of Production of chips, so that R(3,100) was 3.6. And the marginal profit was (delta R)/(delta P). I noted that these marginals were the analog of derivatives, and that, indeed, each table was a list of "independent" variables, in the first column, and "dependent" variables in the second column, and the third column was some approximation of the derivative of the second column. What would happen to the profit if we changed investment from 500 to 501 million? Then chips produced change by .78, and the profit change will be (.05)*(.78). That is

	delta R        delta R   delta P
        -------    =   ------- * -------
        delta C        delta P   delta C

where all the quantities are in appropriate units and rather small with respect to those units, and where all the ratios are measured AT THE APPROPRIATE lines on their respective tables. I think I did not stress all those details enough. In the case of this "economic" data, it is clear that the fractions multiply and cancel appropriately. In the math analogue, which is with derivatives, we're looking at a much more abstract setting, with limits, etc., and we would need to be careful not to divide by 0, etc. I said that result was called the chain rule.

We then did some more experimentation with these tables and went on from there to a formal statement of the chain rule:

Fact 10: If F(x)=f(g(x)) (a composition) then F'(x)=f'(g(x))*g'(x) (a multiplication of values of the derivative function at different values of the arguments).

I then did some "simple" computational examples of this, such as:

What is the derivative of F(x)=(5x^5-9)^{300}? I remarked that I could see two strategies to compute this derivative. I could either "expand" the implied polyniomial multiplication (getting a polynomial of degree 1500, I guess, with lots of terms) or I could try the chain rule. I tried the latter: here f(x)=x^300 and g(x)=5x^5-9. I got the derivatives and reassembled the material into the correct F'(x)= 300 (5x^5 -9)^{299}*(25x^4).

Another example was something like sin(3x^9-2x^5+11). Another example was something like e^{3x^2-3}. I remarked that the last one was not too easy sometimes to recognize, because the outside-most function was "e-to-the" whose derivative was also "e-to-the".

Then I gave some data points for a function, f. They were something like

		x    f(x)  f'(x)

		0     2      3
		1    -1     -7
                2     5      4
                3     1      4

I asked what the derivative of F(x)=f(x^2) was when x=1. I wrote that F'(x)=f'(x^2)*2x, and evaluated the result when x=1 using the table. People certainly found that hard. Then I gave my quiz of the day.

Quiz #10. I asked people to find the derivative of F(x)=f(x^3)+(f(x))^4 when x=1 and f is given by data in the table. 84 people answered this quiz (several had to leave early, I think). It took me about 75 minutes to grade, and almost NO student was able to get the right answer. The quiz question was too hard for folks who had just met the chain rule. The problem was that the two pieces of the function each needed separate applications of the chain rule, and the part played by "f" in each application was different.

I wanted F'(x)=f'(x^3)*3x^2 + 4((f(x))^3 f'(x), which, when x=1 (using the values of f(1) and f'(1) in the table) yields, I think, 7. I didn't often get this answer, or, indeed, any at all!

Lecture #11 (February 23) The first in-class exam. 101 people took the exam.

Lecture #12 (February 25) This lecture would discuss some examples of implicit differentiation, an extremely useful (and used!) method for finding

 /  slopes of tangent lines
<   derivatives             .
 \  rates of change

Consider an a first example the unit circle, x^2+y^2=1. I drew the graph. Consider a point on this circle, say with x= .65. What's y? Then, realizing that the question was ambiguous, I restated it, by pointing to a point with x=.65 on the lower semicircle, and asked what y was. I stared at the class until at least a few students unlimbered their calculators, and we saw that y would be about -.76.

Then I asked what the slope of the tangent line to the circle at that point would be. I drew the tangent line. I remarked that I would show two methods to do this problem.

Method I: solve the equation for y in terms of x. This isn't too hard -- it is just a little bit of algebra. We get y=[?]sqrt{1-x^2}. The reason for the [?] is that there are two possible answers for y. We must remember that sqrt{a number} ALWAYS means the non-negative square root here and, really, everywhere. So sqrt{4} is always 2, even though -2 squared is also 4. Therefore to indicate the two possibilities, we must write y= +/- sqrt{1-x^2}. In fact, if we look at the graph of the unit circle, we can see that, inded, it is NOT the graph of a function, and that there seem to be two graphs of functions, "nice" continuous functions, concealed in it, the Top and the Bottom. So:


and we must analyze the "Bot" function to get the slope of the tangent line, by finding Bot'. SO I did that, using the chain rule and observing that Bot(x)=(1-x^2)^{1/2} so that Bot'(x)={1/2}(1-x^2)^{-1/2} (-2x). Then I plugged in x=.65 to get a value for Bot'(.65), and asserted that this was the slope of the tangent line.

Method II: This is much sneakier. I remarked that x^2+y^2=1 is an equation which is always true on the unit circle and that we could just try differentiating this equation. I did this, very slowly. I differentiated with respect to x, and the right-hand side became 0 and the left-hand side first was decomposed as a sum of two terms. The derivative of x^2 was 2x, and the derivative of y^2 was ... well, the chain rule says that this has derivative 2y multiplied by the derivative of what is "inside", which is dy/dx. So I wrote that. And I got

2x+2y (dy/dx)=0

I solved this equation for dy/dx to get -x/y. Then I said, hold it!, clearly, geometrically, the line I drew had positive slope, but the formula that I wrote had a minus sign in it! After a short discussion, we realized that the correct values of x and y for that point would indeed yield a positive slope for the formula, because x would be positive and y would be negative.

I remarked that solving for y as a function of x as we did in the first example was sometimes, in realistic situations, rather difficult. Even though we used the chain rule in both cases, the use in the second method seems simpler. In actual applications, of which there are many, it is frequently not convenient to do this solving, but people want rates of change anyway. Most people when confronted with this problem would use the second method.

I continued with another example from the book. (I was rather tired, after doing the rather large chunk of grading in the last two days!) I did problem 5 on p.163, which asked students to find dy/dx if x^3-xy+y^2=4. I remarked that in this case, we could solve fo y as a function of x (the equation is a quadratic in y and we could use the quadratic formula), but we probably wouldn't want to. So I applied d/dx to the equation. The right-hand side, a constant, yielded 0, and the left-hand side was a sum of 3 terms. The derivative of x^3 is 3x^2. What about -xy? The minus sign is just a multiplicative constant, so that's not a problem. Then xy needs to be analyzed as a product, and we get -( 1y+x(dy/dx) ) as its derivative. The derivative of y^2 again calls for the chain rule, and its result is 2y(dy/dx). Thus the differentiated equation is


I then solved for dy/dx by pushing everything not involving dy/dx to the "other" side and by factoring out the coefficient of dy/dx:


I emphasized that there were lots of opportunities to make mistakes in each algebraic step. There were questions!

I did problem 15 on the same page. Here the equation is y=sin(xy). I chose this because the chain rule use was NOT going to be a simple power. I d/dx'd the equation to get:

dy/dx= cos(xy) ( 1y + x (dy/dx))

There were numerous questions. There was trouble telling what was multiplication and what was function evaluation. Then I tried to multiply out and solve for dy/dx.

dy/dx= cos(xy) ( 1y + x (dy/dx))
dy/dx= cos(xy)y + cos(xy)x (dy/dx)
(dy/dx) -cos(xy)x(dy/dx)=cos(xy)y
(1-cos(xy)x) (dy/dx)=cos(xy)y

This wasn't followed too well.

I did a last example. If y=ln(x), then e^y=x. Then d/dx this, and get e^y (dy/dx)=1. By now there was some unrest, since some of the students who were following were asking themselves, "But this is the derivative of ln so why is he ..." I persisted, and got (dy/dx)=1/(e^y) which, using the first equation, I identified as 1/x. I remarked that we had indeed verified a formula for derivative but that the same computational approach could also be used for other functions, such as the inverse trig functions (arcsin, arctan) and that's why I showed this to them.

Then I said we would move on to another much more difficult type of problem, poroblems with NARRATIVES. I wrote the following:

A bug is climbing up a vertical post standing on level ground. The but is climbing at 2 feet per minute. Another bug is standing still 3 feet away from the base of the post.

Question: when the first bug is 10 feet up the post, what is the angle that the second bug sees between the ground and the first bug? How fast is this angle changing?

I remarked that reading such narratives and trying to extract information from them was sometimes quite difficult, and could call for repeated rereading. In this case, the situation is not too complex. What should I do? I drew a picture of the post, the ground, and the bugs. I labelled the distance between the bug on the ground and the base of the post with 3 feet. I then cautioned them never to label a variable quantity with a number (people wanted to label the first bug's height 10, but I said that was not always true). I labeled the first bug's height h and the angle seen by the second bug between the ground and the first bug by Theta (also the bugs were labeled, but not named!).

We wrote an equation relating all the quantities: tan Theta = H/3. Then we found Theta when H was 10 (that is, students did - I've forgotten what the number they got was -- maybe 1.28 RADIANS). We differentiated the equation to get

 (sec(Theta))^2 dTheta/dt= 1/3(dh/dt)

I inserted 10 for dh/dt and 1.28 for Theta, and said people could solve the equation for dTheta/dt.

It was time for a quiz.

Quiz #12: students working in small groups do better in calculus courses, and this course was going to get harder. I offered to try to form small groups for students who wanted to work in them. So I asked students who wanted to work in such groups to write their two evenings that they would be free and the campuses on which they could work together, and include their e-mail addresses. I also said that if they did not want to do this, they could so indicate. Later I learned there were 76 students present and all but 8 indicated some level of interest in trying this out. I spent about 4 hours matching people and notifying them the next day.

I returned the first exam with results as reported.

Lecture #13 (March 2) I devoted the lecture to doing only a few examples of related rate problems. Below are brief descriptions of these problems. The numbers may not be exactly what I used in class but the problems are about the same as what I did in class. I went very slowly and tried to emphasize aspects of the process.

Suppose that the side length of a square is increasing at the rate of 2 inches per second. At what rate does the area of the square change when the area is 100 square inches? I drew a picture. I labeled the picture, again emphasizing as I did last time that we should use a letter for anything that is changing. I wrote an equation connecting the area of the square together with the side of the square. I differentiated the equations, and then inserted the known information. The differentiated equation needed the value of the side length when the area was 100, so I deduced that. And then I found the rate of change of the area.

I mentioned to students that when you pull a piece of taffy, the candy will lengthen, but it will also narrow at the same time. I said I wanted to propose a problem which would be a mathematical model of this. So I said that I had a piece of taffy in the shape of a circular cylinder. The volume of the cylinder will always be the same since the amount of taffy won't change. If I pull the taffy so that the length of the piece is increasing at the rate of .2 inches per second, how fast is the radius changing when the length is 10 inches and the radius of the base is 2 inches? I drew several pictures of pieces of taffy, and we discussed the necessity of having a formula, which I wrote, for the volume, V, as a function of length, L, of the taffy and the radius, R, of the taffy: V=Pi R^2 L. I asked if R would increase or decrease and what the rate would be. We discussed the situation and decided that R would decrease because V would stay constant. I differentiated the equation and got the answer.

Then I said that two people are driving away from town. Roger is driving north at 40 miles per hour, and Jane is driving west at 50 miles per hour. At what rate is the distance between them changing when Roger is 20 miles north of town and Jane is 15 miles west of town? We wrote the Pythagorean theorem to get the length of the hypoteneuse and then differentiated. We needed to get the length of the hypoteneuse at the appropriate time and went back to the original equation for it.

I emphasized in all this (mainly by giving Roger's and Jane's weight in the previous problem) that "real world" problems would have a large amount of data, and a good part of the problem would be to tell what was relevant to it.

Quiz #13: the final problem I mentioned was given as a quiz. I emphasized that students HAD to work together in this quiz, and that a working group should have from 2 to 4 students in it and at least one student should have a calculator. So: suppose the pressure P (measured in atmospheres), volume V (measured in liters) and temperature T (degrees celcius) of a certain gas obeys the following equation:

PV1.6 = k T
where k is a constant. Suppose additionally that you know:

When P=2 and V=10, then T=200. What is k? k is approximately .398.

P then increases at .2 atm/min and V decreases at .3 atm/min. Does T increase or decrease, and by how much? It increases, and the rate is about 10.4.

76 people answered this question. About half of the groups got it correct, and, in fact, the work of a half dozen of the groups was very well written. Some of the groups made horrid mistakes with the chain and product rules in differentiating the given equation. Oh well.

Lecture #14 (March 4) I first wrote the solution to the previous quiz, copying exactly the work done by a group of three students whose names I wrote. I remarked that this work was very good.

Today's lecture would mainly be a vocabulary lesson. I defined:

f has a maximum at x over its domain if f(x)>=f(at any other point in the domain).

f has a minimum at x over its domain if f(x)<=f(at any other point in the domain).

I emphasized that considerations of domain would be extremely important in applications. I then wrote the functions x2, sin x, x2 and asked what the maxima and minima were for these functions when the domain was all the real numbers. We found them. In each case I sketched a graph of the function, and I tried to support the answer with a graphical analysis, using the words "highest" and "lowest". I pointed out, very carefully, that the answer to the questions was "None" in several cases (for example, at what number does x2 have a maximum over all the real numbers?), and that there could be an infinite number of correct answers (sin x on the domain all real numbers).

Then I used the same functions but changed the domain to -Pi/3 =< x =< Pi/4. I answered the questions and we discussed the results. Some students found this difficult.

I declared that most of the next three weeks of the course would be devoted to finding out about max and min, and that a careful analysis of graphs and functions would be necessary, and that, generally, we would want to get the language to analyze (I drew a complicated graph) and to discuss its bumps (I circled them) and its wiggles. I said that today I would give the technical language for "bumps" but that wiggles would have to wait until next time.

I defined relative maxima and minima very carefully. In this and in the other cases of definitions, I tried to use the least amount of mathematical notation and to write grammatically correct and clear (!) sentences. For example:

f has a relative maximum at x if there is an interval centered at x inside the domain of f and if, in that interval, f's value at x is >= to its values at any other point in the interval.

f has a relative minimum at x if there is an interval centered at x inside the domain of f and if, in that interval, f's value at x is <= to its values at any other point in the interval.

I then asked said that f(x)=x would be the function of interest, and the domain would be the interval -13 <= x <= 45. I asked:

I remarked that until we understood a string of examples, "learning" the definitions has probably not occurred. We answered the questions above, but not without some struggle.

I then changed the question by changing the function to f(x)=x2. We answered that. I then changed the question by changing the question to f(x)=x3 -8x. People seemed startled. Of course I merely looked at them until at least a few of them unlimbered (?) their calculators. They answered the question after some time, and got the relative extrema at about +/- 1.7 . It was difficult, because one needs to focus the windows fairly well to "see" these bumps.

I asked why x=1 was not relative extremum. I was told it wasn't. I asked why. I was told it wasn't and that one could localize the window of the calculator near (1,f(1)). I said, well, can you persuade me that there will be no tiny bump of the function at x=1? There was some disquiet and no immediate suggestion. I further explained my question by saying why couldn't I claim that there was some magnification of the function's graph which would show a bump. Not possible, some said, and they declared that the graph "had" to look the way it did. I said we do have the technical equipment to make a definitive answer, and that used the derivative. We could easily compute f'(x0=3x2-8, and then conclude that the slope of the tangent line to y=f(x) when x=1 is -5. It was negative, so for numbers very near x=1 on the left, f(x) > f(1) and for numbers very near x=1 on the right5, f(x) < f(1). So there would be NO INTERVAL around x=1 in which f would have either a relative max or a relative min at x=1. By the way, I also mentioned that many texts use "local" instead of "relative" and I was more used to that also, so I apologized in advance. Therefore x=1 could not be a local extremum for f.

Where were the local extrema of f, by the way? I found that we would need f'(x)=0, and got x=-sqrt{8/3}=(approx.)1.6 so the approximatation above gotten by graphing was fairly good.

I remarked that we would need to look at lots of functions and should really try to find algebraic methods of answering the questions above, since there are limits to graphing technology.

Then I switched the question(s) above to f(x)=x3. There were some problems with this -- people insisted that 0 (where f' was 0) was a relative extremum. But I successfully (I hope) argued that 0 was not a relative extremum. I then looked at f(x)=|x|. Many people had forgotten what that looked like, so I had to draw it carefully. I did, and we answered the questions.

I then said that f has a critical number at x if either f'(x)=0 or f' did not exist at x.

I remarked that relative extrema must occur at critical numbers. I looked at the additional example of f(x)=x1/3 and we figured out where the critical numbers of f were (0) -- which took some effort.

I am now about a third of a lecture behind the syllabus, since I did not discuss Rolle's Theorem or the Mean Value Theorem. I hope to catch up next time.

Quiz #14: I asked people to sketch a graph of a continuous function with the following properties: the domain of f was [0,5], f was differentiable except at 1 f has a rel max at 1, rel min at 2, and rel max at 3. 73 people answered this question, and most got it correct. I spent less than a half-hour grading (pictures are easy!). It would have been a more effective question if I had made further stipulations, such as asking that f have its maximum in the interval at 2 and its minimum in the interval at 5. I could have found out more about what students knew.

Lecture #15 (March 9) I began with a small amount of review. I wrote again the definition of critical number: where the derivative doesn't exist (illustrated with two local pictures, a corner and a curve with a vertical tangent) or where the derivative is 0 (illustrated with horizontal tangent lines at a relative max, a relative min and a "neither" [locally like x^3]).

I wrote again the important proposition that a function has a relative max or relative minimum only at a critical numbers. Therefore I stated that looking for critical numbers is the first step in locating a relative extremum.

I mentioned that I had fall behind the syllabus and apologized. I said that I wanted to state one of the most important results in calculus, the Mean Value Theorem (MVT). I stated it as in the text: f differentiable in an interval [a,b], then there is at least one point "c" inside the interval where f'(c)=(f(b)-f(a))/(b-a).

I tried to interpret this dense algebraic/logical statement with a picture. I drew a graph of f on [a,b]. I labeled the endpoints and asked what the numbers in the equation could refer to. Several students suggested that the right-hand side could be the slope of the line through the endpoints of the graph. I suggested that f'(c) refers to the slope of the tangent line at c, and I illustrated this by drawing a tangent line at an appropriate c, and mentioning that since the slopes were equal, the two lines were parallel. I mentioned that I found the picture much more appealing than the algebraic formulation, and said that the picture "shows" why the statement is true -- just begin moving the line connecting the endpoints "up" until it last touches the graph, and there will be the point whose x-coordinate is c.

The MVT is very important in calculus, and I wanted to illustrate it with some examples, beginning with something from the textbook and moving to a more typical use in "real life". My first example was something like: suppose f(x)=x^3-2x+1 and we consider the interval [0,2]. Find a c satisfying the MVT. So I dutifully found the c by computing the slope of the endpoint line and then taking the derivative, plugging in c, and setting the result equal to the slope of the endpoint line. I remarked that it is almost surely true that NO ONE in the room (with the exception of the instructor!) would ever actually find a c in reality. I got some questions about this. I said my next example would be more realistic. I did not (unfortunately!) reinforce this example with a picture or have them get one on their calculators.

Suppose we drive on the New Jersey Turnpike (the lecture on the Intermediate Value Theorem used the Garden State, and wasn't too successful!). You begin at mile 0, which I was told was at the Delaware Memorial Bridge, and go on to the end, which I was told was at the George Washington Bridge. There was some debate about how long the Turnpike was. We "decided" that maybe it was 120 miles long for the purposes of this discussion. I proposed a revenue enhancement scenario. That is, suppose I started on the Turnpike at 9 AM at the Delaware Mem Br end and got off at 10:45 at the G W Bridge end. What could happen? I claimed that as I presented my "ticket" to the toll taker, I should get a speeding fine. My mathematical model was the following: suppose that D(t)= distance traveled in miles from the beginning of of the road where t is measured in hours since 9 AM. Then we know that D(0)=0 and D(1.75)=120. If wee use the MVT with a=0 and b=1.75, there the Theorem says that there's at least one time c where D'(c)=(120-0)/(1.75-0) = (approx.) 68.57. The units for the latter are miles per hour. I asked what the speed limit was on the Turnpike and was told that is was either 55 or 65 mph. So the speed (or more precisely in physics language, velocity) would have to be greater than 65 somewhere. In the I remarked that we can['t find c -- all that the MVT says is that there's a c somewhere. One student observed that the right-hand side of the MVT was the average velocity. I agreed, and said that's why the theorem has "Mean" in its name. It is another name for "average".

My third example involved rats. I said that I would be giving greatly simplified data about a rat colony. Imagine that you start with a rat population of 100. You also know that the birth rate of the rats in the colony varies between 7 and 12 rats per month. How many rats should one expect at the end of 10 months? Most people agreed that there would be between 170 and 220 rats. I urged a mathematical model: suppose R(t) is the number of rats in the colony at time t, where t is measured in months from the start. Then R(0)=100. What do the numbers 7 and 12 refer to? We debated this, and I remarked that we need to use the vocabulary of the course, and that the numbers referred to a rate, rats per month. Finally I declared that it meant 7 <= R'(t) <= 12. I don't think people were entirely convinced. Then I wrote (R(10)-R(0))/(10-0) and got R'(c) for some unknown c. I unrolled the equation a bit, and got an overestimate for R(10) and an underestimate for R(10) as predicted by the students. There were questions about this process, and I replied with some truth that similar estimates were commonly made to analyze the spread of diseases, etc., but that the calculations were much more intricate.

Then I passed to a "random" function, f(x)=37x^{103}+5x^{7}+3e^{5x}+7. I asked what the derivative of this function was. We found it, not without some difficulty. One irritation which was more than usually apparent during this class was that about half the students do not do anything, even when asked to compute a derivative or sketch a function or ... some way to get people more involved is necessary! Then I asked where the derivative is positive. We discussed this for a while, and realized that the function wasn't too random, because the oddness of the exponents of the polynomial meant that the derivative had even exponents, and that therefore the pieces were nonnegative, and the exponential meant that the function was always positive. People found this difficult to follow. Then I asked, which is larger, f(200) or f(100)? I remarked that the problem wasn't random, but that one much like it had been asked on the final exam in Math 135 quite recently. I said that 100 and 200 and the function itself were chosen so that it would be difficult to compute f(200) and f(100) directly with a calculator -- we'd have to do this by "thought". I compared f(200) and f(100): I wrote f(200)-f(100). I said this made me think of MVT and that this difference divided by 200-100 was f'(somewhere). Since f'(somewhere) had to be positive, and 200-100 was positive, the top of the MVT fraction had to be positive also. Therefore f(200) was greater than f(100). I remarked (truthful) that more people had gotten this question correct on the final than I had anticipated.

But it was time for more vocabulary. So I defined what it meant to say a function f was increasing on an interval and decreasing on an interval. I wrote out the definitions carefully, and remarked again that unless they knew and understood these terms, much of the rest of the course would be incomprehensible to them. I drew pictures of each word, and remarked that moving from left to right meant that an in(de)creasing function went up(down).

I said that the "random" function above was always increasing. More typical was x^2, which was a parabola. Its derivative was 2x which changed sign at 0.

Quiz #15: I then asked students to look at f(x)=x^3-3x, with derivative f'(x)=3x^2-3. Where is f'(x)=0 and where is f'(x)>0 and where is f'(x)<0? I remarked that they didn't have to "solve" inequalities -- they only needed to see where f' was 0 and "test" the sign of f' inside every subinterval (three of them!). This wasn't a successful test. I hadn't led up to it correctly and I don't think people understood what I wanted and why, even thought at the time I hopped it was clear. 74 people took this quiz, which took about 40 minutes to grade. Only about a third of the class was able to answer it correctly.

Lecture #16 (March 11) I asked what the graph of f(x)=x^3 -3x looked like. i stared at folks until at least some of them pulled out their calculators and I got the graph I wanted, say for x between -5 and 5 and y between -10 and 10.

I indicated that the picture had information in it and wanted to discover other ways of getting that information. I remarked that laast time we saw:

I then computed f'(x) and found out where it was 0: at +/- 1. I remarked that this had been what I asked them to do last time. Then I said we could "construct" a number line for f' and see where f' was positive and negative. Since this f' is continuous, we need only check the sign of f' in each subinterval at a convenient point. There were questions about this which I hope I answered adequately, using the Intermediate Value Theorem as a reference. I then "tested" f'(0). It was -3. I asked what we knew about f'(100,000) and f'(-100,000) and learned that they were positive. I tried to connect the sign behavior of f' with the {in/de}creasing behavior of f in the graph.

Then I said that I would try another "random" example. I remarked that I would be using the output of the program Maple, which was available on Eden, and had lots of nice abilities. Here was another polynomial which I wanted to graph (the information displayed below was shown on an overhead):

> F;

      16   30  17          18   4060  19           20         21
1/16 x   - -- x   + 145/6 x   - ---- x   + 5481/4 x   - 6786 x
           17                    19

       593775  22   2035800  23              24           25
     + ------ x   - ------- x   + 1950975/8 x   - 572286 x
         22           23

                  26              27               28            29
     + 2311155/2 x   - 6069700/3 x   + 12356175/4 x   - 4129650 x

                  30   155117520  31   145422675  32   39919950  33
     + 9694845/2 x   - --------- x   + --------- x   - -------- x
                          31              32              11

       86493225  34            35              36   14307150  37
     + -------- x   - 1560780 x   + 3338335/4 x   - -------- x
          34                                           37

       5852925  38          39             40   142506  41
     + ------- x   - 52200 x   + 118755/8 x   - ------ x
         38                                       41

               42   4060  43   435  44        45         46
     + 1305/2 x   - ---- x   + --- x   - 2/3 x   + 1/46 x
                     43        44

So this F(x) was a polynomial of degree 46. I did know one value of it: F(0)=0. That's not much to start with. Then I looked at the derivative, which I had Maple compute:

> DF:=diff(F,x);

       15       16        17         18          19           20
DF := x   - 30 x   + 435 x   - 4060 x   + 27405 x   - 142506 x

               21            22            23             24
     + 593775 x   - 2035800 x   + 5852925 x   - 14307150 x

                 25             26             27              28
     + 30045015 x   - 54627300 x   + 86493225 x   - 119759850 x

                  29              30              31              32
     + 145422675 x   - 155117520 x   + 145422675 x   - 119759850 x

                 33             34             35             36
     + 86493225 x   - 54627300 x   + 30045015 x   - 14307150 x

                37            38           39           40
     + 5852925 x   - 2035800 x   + 593775 x   - 142506 x

              41         42        43       44    45
     + 27405 x   - 4060 x   + 435 x   - 30 x   + x

Still not too much to get excited about. The problem of graphing F looks very difficult. I remarked that Maple can also, in fact, graph functions, but that THIS function, due to the large numbers and sign alterations, was a bit difficult to handle. Then I revealed that this wasn't exactly a "random" function, by asking Maple to factor the derivative.

> factor(DF);

                                   30  15
                            (x - 1)   x

That changed the nature of the problem. Now I quickly constructed a number line for F'(x) and saw that F'(x)=0 when x=0 or 1 only. We "tested" the signs of F' in each subinterval: F'(1/2) was positive, and F'(100,000) was positive and F'(-100,000) was negative. This wasn't completely obvious until we noticed the even & oddness of the exponents in F'(x). I was then able to draw a qualitatively correct graph of y=F(x), including information about {in/de}crease, minimum, and horizontal tangent lines. It was a bit difficult to convince people that a horizontal tangent line could cross the graph of the function.

I then stated that I wanted to do something somewhat more intricate. I displayed on an overhead a bar chart which I got from the business section of the N.Y. Times. The information in it was approximately as follows:

           |                                          |
           |            MARGIN LOANS                  |
VERTICAL   |                                   51%    |
AXIS       |                                   XX     |
LABELED    |                                   XX     |
% ANNUAL   |                                   XX     |
GROWTH     |                       38%         XX     |
RATE       |              30%      XX          XX     |
           |        27%   XX       XX    28%   XX     |
           |  25%   xx    XX       XX    XX    XX     |
           |  XX    XX    XX       XX    XX    XX     |    14.0%
           |  XX    XX    XX       XX    XX    XX     |     XX   7.9%
           |  XX    XX    XX       XX    XX    XX     |     XX    xx   4.7%
           |  XX    XX    XX       XX    XX    XX     |     XX    XX    XX
             1995  1996  1997     1995  1996  1997         1995  1996  1997   


I asked what this all meant. I vaguely indicated what margin loans were. We thought for quite a while. We realized that all of these quantities were increasing all the time but that the rates of increase were changing. Then I tried to draw graphs of the quantities (total margin loans in the first two cases, and total consumer loans in the third case). That is, I tried to crasw curves that would use the data in the bar graphs. It was slow going, because we had to realize that the heights of the bars reflected the slope of the tangent lines. I did draw some suitable graphs, but slowly. Then we saw that the first case above had a graph bending up, the last case had a graph bending down, and the middle case had a wiggly graph. In all cases, though, the graph was increasing.

I had another vocabulary lesson, and defined concave up (respectively, down) in an interval as the derivative increasing (respectively, decreasing). I remarked that this could be checked by looking at the derivatives derivative -- the second derivative. I also defined point of inflection.

I went back and considered x^3-3x again, computed f''(x)=6x and found out where the function was concave up and concave down, and displayed this on the same graph which had the {in/de}creasing behavior, and also indicated that 0 was an inflection point. I remarked that this graph showed that the "creasing" behavior didn't really have much connection with concavity. This is subtle. I urged people very strongly to do the homework problems.

Quiz #15: I displayed on the overhead the graph of the function which had appeared on the first exam, in problem 7. I asked people to answer three questions. First, where are there points of inflection? The second one was, where was the graph concave up? The third, where was it concave down? Unlike many previous quizzes, I certainly allowed people enough time and urged them to consult with their neighbors. A total of 59 students answered this quiz (golly, it is two days before "vacation", I remarked sarcastically, and New Jersey is such a big state to go home in). Only about 40% answered all the questions correctly -- but the graph displayed (which is on the web in exam #1) is complicated.

Lecture #17 (March 23)

I summarized some important ideas on the board:

If f' is positive on an interval, then f is increasing on the interval.

If f' is negative on an interval, then f is decreasing on the interval.

If f'' is positive on an interval, then f is concave up on the interval.

If f' is negative on an interval, then f is concave down on the interval.
I also wrote a list of "vocabulary terms" which people should known, including critical number, (relative) maximum and minimum, and inflection point. I remarked again that "speaking the language" was vital to understand what was going on in the course.

I then wrote the principal topic of the week: drawing (good) pictures of functions defined by formulas. I got a courageous (really!) and very good question: WHY should "we" (= the students) be able to do this? The question actually disconcerted me a bit (maybe a good thing!), and it also made me uneasy that I was not really teaching, which to me includes trying to convince people that what I was doing was useful. I thought for a while, and remarked that, first of all, I do it because it is fun and because it is pretty. I remarked that I would not expect many of them to see that point of view, at least for a while. Then I said that most of the students in the room would have to deal with situations where something intricate was modelled by a complicated function (I recalled the long polynomial which I put up on the overhead before vacation) and that most people found such formulas very hard to understand, but they found visual information much easier to assimilate. And the purpose of what I was doing here was to show them how to create good pictures and to understand the pictures. I hope that was a useful answer.

I asked students what step 0 (that's STEP ZERO, the first thing to do) should be when asked to graph a function. There were blank looks. O.k., then, I said, let's graph the function f(x)=e^x/(1+e^x). I looked at the students until at least some of them were digging for their calculators. Then I wrote that step 0 should be, whenever practical, get a rough idea on a graphing utility.

I waited while students graphed the curve. I then asked for one output, and got a wonderfully unintelligible result. Someone had graphed the function in the standard window, -10<=x<=10 and -10<=y<=10. I sketched on the board as well as I could what the calculator actually showed, which was just a few straight line segments. I remarked that the function I wanted to analyze wasn't written down at random, but rather was a function (cleaned up quite a lot!) used in portraying enzymatic reactions in biochemistry. So it really would be useful to try to understand it.

The first calculator-drawn graph was not very successful. I tried to differentiate the function. I got after some flailing around a derivative which we saw was always positive so that the function was always increasing. I asked for another try at graphing it, and got another only slightly better result, a graph from I think -5 to 5. Then I computed the second derivative, and seeing the sign of this function too still more effort, but we finally found that the second derivative was positive for x<0 and negative for x>0. It had, as a student pointed out, an inflection point at 0. I sketched a better version, and finally got a picture on one student's calculator for -1<=x<=1 and 0<=y<=2 which looked pretty good.

I remarked that this was a common misuse of lots of graphing software: they worked by digitizing an image and the image has a certain "resolution", a certain number of pixels, and unless the window was chosen nicely there might be a real problem in finding a nice picture which was revealing of interesting details. I don't think that this comment, delived too fast, was understood by many students.

I remarked that the function's graph in this case was dominated by two horizontal asymptotes, y=0 and y=1. I said that these were revealed by studying the limits of the function as x--> +infinity and x-->-infinity. I analyzed these limits. I said that e^x-->0 as x-->-infinity. The students were not happy. I then looked at what happens x-->+infinity. I deduced the approach to left and right horizontal asymptotes.

Then I tried some examples which were much simpler, d with rational functions. I discussed the limit as x-->+infinity of (x^3+9x+100)/(x^3-3x+7). I handled this by dividing the top and bottom of the fraction by x^3 (actually I multiplied the top and bottom by 1/(x^3) ) and then letting x_->+infinity. I discussed the limit as x-->-infinity of something like (x^2+2x+1)/(x^4-3x^3-1), here by dividing the top and bottom by x^4. Finally, I looked at just the limit of x^2 (or x^2/1!) as x-->+infinity. I remarked that thiswas one of the few times I would differ from the book. I said that most people woudl write that x^2 goit large and stayed large when x got large, and they would abbreviate this statement by writing limit of x^2 as x-->infinity is infinity. The text would say that the limit did not exist.

Quiz #16: Consider the function (e^(3x)+9)/(e^(5x)+2). What is the limit of this function as x-->-infinity and as x-->+infinity? 74 people answered this question, and it took about 45 minutes to grade. 15 people answered it correctly. In looking at the +infinity limit, I urged people to multiply the top and bottom by e^(-3x). About 15 people incorrectly "simplified" (e^(3x)+9)*e^(-3x) to 9e^(-3x). Most people could do the -infinity limit but found the algebra hint not sufficient.

Lecture #18 (March 25) I began by asking:

Which gets bigger faster, x^(100) or e^x?

I wanted to start some discussion about what the heck the question meant. I also wanted to be asked why anyone would be interested in such a question. We did have a little interchange about this. I remarked that in modeling various processes from biology and chemistry and finance, the "correct" or apropriate function needs to be selected. Part of the selection process might be (is!) how fast the function grows, and questions such as I asked are very relevant. I also remarked that what I was about to do was not in the text.

We evaluated both functions at x=3, and e^x's value was about 20, but x^(100)'s value was 5 times 10^(47).Of course, a few students already remaked that I was trying to mislead them, and I responded that it was rather difficult to answer my question (dominating the center board) by trying values. I said that one needed to do something rather subtle.

Consider the quotient, x^(100)/e^x. I want to study the asymptotic behavior of this as x-->+infinity. First I did some algebra on the bottom part of the function. e^x=e^((1)x)=e^((1/2)x+(1/2)x)=e^((1/2)x)*e^((1/2)x). A portion of the students seemed to follow this only with difficulty. Then I took the original quotient, x^(100)/e^x, and rewrote it as a product of x^(100)/e^((1/2)x) times 1/e^((1/2)x). I remarked to students that what I was doing was quite intricate and not immediately motivated.

Then I tried to analyze the graph of x^(100)/e^((1/2)x). We were able to say quite easily that this function was never negative, so the graph always was in the upper half plane. I wanted more information so I suggested we differentiate this function. With some effort, we got (100x^(99)*e^((1/2)x)-x^(100)*e^((1/2)x)*(1/2))/((e^((1/2)x))^2). I noted that the bottom was always positive, and what can we say about the top? We factored the top: (x^99)*e^((1/2)x)*(100-(1/2)x). We were able to "read" that this was 0 when x=0 and when x=200. I emphasized that I was interested in what happens when x is LARGE. So there is a critical number when x=200. What about the sign of the derivative? We saw that for x>200, the sign was negative. I then concluded that the function was decreasing. I emphasized this by trying to sketch x^(100)/e^((1/2)x). I put a dot when x=200 and y= 200^(100)/e^((1/2)100), and then I asked where I should put a dot when x=300. There was some confusion, but after a while I was advised to put the dot below the other but above the x-axis. I safely concluded that "decreasing" might be understood here.

Back to the main show. The function x^(100)/e^x = ( x^(100)/e^((1/2)x) ) * 1/e^(1/2)x)) =< (200)^(100)/e^((1/2)200) 1/e^((1/2)x) for x>200. The second factor --> 0 as x-->+infinity, and therefore I can finally conclude that e^x grows larger faster than x^(100). I emphansized that this was for x sufficiently large, but this was what is actually observed in "real life" -- typically, in chemical reactions, x represents time, and the scale is such that 10^4 or or 10^5 occurs very soon.

A general fact I urged upon the students was that any exponential growth is eventually faster than any polynomial growth. For example, .0000345e^(.0000123x) grows fastever (eventually!) than 5,432,100x^(3,123,789). I said that this was sometimes hard to see with calculations with specific "small" x's, but that it was indeed true!

The remainder of class was devoted to analyzing examples of graphs of functions given by fairly simple formulas, indeed, by problems in the book. The first example I analyzed was 3x^4+4x^3, which is problem #7 in section 3.6. I wanted a good picture of this. What's the first step, I asked? Finally a number of people responded by taking out their graphing calculator. I got a fairly good picture and asked people about the picture. How many local extrema does the graph (seem to) have? How many inflection points does the graph (seem to) have? I asked how we could confirm/refute these statements.

We found f' and analyzed it. We got information about {in/de}crease and critical numbers. We found f'' and analyzed it. We got information about concave {up/down} and inflection points. We confirmed the picture. I had in earlier material covered tried to tell students that both graphical and computational output of calculators could not be completely trusted, therefore I wanted to confirm calculator output for a few examples. I finally asked which was larger, f(10,000) or f(15,000)? We discussed this, and answered it, but not without some effort (and not by direct computation!)

I did problem #23 from section 3.6: graphing the function e^(3x)*(2-x). I went through the same procedure as above. This all takes quite a while! I had a calculator graph, I "guessed" at critical numbers and types, and at points of inflection. I confirmed these guesses, and intervals of increase and decrease and concave up and down. Again, manipulation of f' and f'' to get information took a while: deciding where they are 0 and positive and negative.

I quickly tried to sketch ln(1+x^2), again finding f' and f'': using the chain rule and then the quotient rule took time. I tried to verify what the calculator graph looked like.

Quiz #17: I asked the students to sketch a graph of a function with the following properties: the domain is all numbers EXCEPT x=-3, x=5, and x=7; the function increased on (-infinity,-3) and on (5,7); the function decreased on (-3,5) and on (7,infinity); the graph has horizontal asymptotes at y=-2 and y=4 and vertical asymptotes at x=-3, x=5, and x=7. NO STUDENT gave an entirely correct sketch! Students showed great reluctance to have graphs cross horizontal asymptotes. They showed also that they were reluctant to make curves bend the way they should for the vertical asymptotes. Much remains to be "internalized" by them about graphing. 73 people took this quiz. It only took 35 minutes to grade. I think I'd better go over it at the next class!

Lecture #19 (March 30) I went over the previous day's quiz very carefully since people had not done well on it. I tried hard to explain some of the choices which vertical and horizontal asymptotes allowed, and then how the graph of the answer was restricted by the stipulation of increasing and decreasing behavior. I assured people that graphs of functions could cross horizontal asymptotes (I illustrated this by a curve going off to +infinity wiggling back and forth with diminishing amplitude -- I suggested that people go and push down the bumper of an old car and watch it wiggle). I also remarked several times to them that it was sometimes almost necessary to makes "mistakes" -- to repeatedly revise the graph until it satisfied all the criteria.

Then I began optimization problems. I remarked that math departments had made a fortune from this subject over the last three centuries. Basically we'd be given a paragraph, a "narrative", with information about a quantity, such as area/weight/profit/revenue, which we would want to maximize or minimize. There would be other conditions on the variables involved. A major difficulty would be to translate the paragraph into a calculus problem. I remarked that all I would do this week is examples of this process.

Example 1: Suppose a rectangle has one side on the x-axis, and two vertices above the axis on the curve y=16-x^4. Find the largest such rectangle.

How should we begin? People suggested that I draw a picture of the curve involved, and after realizing that it would look sort of like an upside-down rectangle, I did. Then I asked for more a No one pointed out that the original problem was not terribly well-stated (I realized that -- I should have asked for the rectangle whose area is largest: the "largest such rectangle" is ambiguous, and could refer to the perimeter or to the diagonal or ... lots of things). We found the area of a rectangle, and then tried to draw various candidate rectangles. It was not clear how to find the rectangle of largest area. If (x,y) is a corner of the rectangle, and if (x,y) is in the first quadrant, then the rectangle will have area (2x)y. x and y are related by the equation y=16-x^4, so the area will be 2x(16-x^4). I asked which x's are allowed. For example, is x=100 a good value? Then we decided that 0,=x<=20 was the domain for the function A(x)=2x(16-x^4).

How can we find the maximum of A(x)=2x(16-x^4) on the domain 0<=x<=20? Where can such maxima occur? After some discussion we realized that they can occur either at endpoints or at critical numbers. A(0)=0 and A(20)=20. We looked for critical numbers, knowing that A(x)=32x-2x^5. A'(x)=32-10x^4, so to make A'(x)=0 we must have x=+/-(32/10)^(1/4). We dropped the - sign because that x is outside of our domain of interest. A's value at this x was 2((32/10)^(1/4))(1-((32/10)^(1/4))^4. This was clearly a positive number, so I remarked that we were done -- we had found the largest rectangle, and it had dimensions 2((32/10)^(1/4)) by (16-((32/10)^(1/4))^4. There was some grumbling, but I said that we knew the max had to either at endpoints or critical numbers, and the values of A at those three points were 0 and 0 and a positive number, so clearly the last one was the maximum value.

I also said that we could add to their vocabulary. Phrases that are frequently used in economics in situations like that are "objective function" and "constraint". The objective function was a function which had to be minimized or maximized. Here that function was 2xy. The constraint was a relationship among the variables of the objective function (here y=16-x^4) or a restriction on the domain of the variables of the objective function (here 0<=x<=20).

Example 2: Two non-negative numbers have sum 20. How should we choose the numbers so that the sum of the cube of one of them plus the square of the other is a minimum? There didn't seem to be a useful picture to draw. So we decided that x+y=20 with x>=0 and y>=0. We needed to minimize x^3+y^2. This we reduced to a calculus problem: find the minimum of Frog(x)=x^3+(20-x)^2 with 0<=x<=20.

Frog(0)=400 and Frog(20)=8000. To be sure of finding the minimum, we need to look at critical numbers. Frog'(x)=3x^2+2(20-x)(-1). When is 3x^2+2x-40=0? Somewhat painfully I remarked that you can't always factor things easily, and that it might be more useful to just use the quadratic formula. So we got as roots (-2+/-sqrt{4-4(3)(-4)})/(2(3)). These are (-2+/-22)/6. The negative root again is irrelevant -- not in our domain. The positive root is 10/6. But which value is minimum? We computed Frog(10/6) and got (approximately) 315, so that's the minimum value.

Somehow things went very slowly and not very well. I did not have enough time to consider other ways of ensuring that 10/6 gave a minimum value (by looking at Frog'(0) which would be negative and Frog'(20)>0, so that 10/6 would have to be a minimum OR by considering Frog''(10/6) and seeing that it would have to be >0, also guaranteeing a minimum). I shold have discussed this.

I remarked that our second exam would be in exactly two weeks.

Quiz #18: I asked students to identify the objective function and the constraint in the second problem I did. I thought this would be straightforward, but I am glad I asked the question. Only about half the class was able to answer it correctly. Others interchanged the answers, or simply misstated them. Technical jargon ("vocabulary") never seems to be "easy". This took about 35 minutes to grade and 71 people answered the quiz.

Lecture #20 (April 1) This was a much better lecture than Tuesday's because ... because ... well, perhaps it couldn't be much worse! Sometimes things just don't work out too well. At least that is how I began the lecture, apologizing, and hoping for improvement. I said I would continue doing examples of optimization problems. I began with the following:

What is the rectangle of largest area which has two sides on the positive x- and y-axes and one corner on the line 2x+3y=6?

First we drew some rectangles, just to get a feeling for the problem. Then I asked how to make the geometry more algebraic: how can we introduce variables and try to get a function and turn this into a more routine calculus question. We decided that the corner of the rectangle on the tilted line could be called (x,y) and then the area of the retangle would be xy, so we needed to maximize A=xy. A constraint relating the two variables x and y was given by the line 2x+3y=6, so that y=2-(2/3)x, and thus A=x(2-(2/3)x). Which x's would be allowed, I asked? Could we have x=10,000 or x=-10,000? We would be guided in this by the original problem, and by the diagram which we had sketched. It showed that candidate x's could vary from 0 to 3. Now we had a "pure" calculus problem.

What is the maximum of A(x)=x(2-(2/3)x) when 0<=x<=3? I suggested that we just hop on the bus (?) and find the critical numbers of A(x). I multiplied out and got A(x)=2x-(2/3)x^2, so that A'(x)=2-(4/3)x. SInce A'(x) exists everywhere in the domain, the only critical numbers occur when A'(x)=0. The only critical number of this function was x=6/4=3/2. Then we can find y, etc. So I declared that we had found the rectangle whose area was maximum.

Why? That is, why have we found a rectangle of maximum area? I said that I'd like to go over three strategies, any one of which should convince people that a rectangle of maximum area has actually been found. The reason for doing this is that each of the strategies might, under different circumstances, be more or less easy to use in certain problems, and we might as well see what they are like in a simple "toy" problem.

We can look at the function. The maximum of A must occur at an end point or a critical number. It isn't very hard at all to evaluate A(0) and A(3). Both A(0) and A(3) are 0. What about A(3/2)? Well, whatever it is exactly (it is, exactly, (3/2)(3-(2/3)(3/2)) ). which is a POSITIVE number. Now the maximum value must be among these function values: 0 and 0 and (3/2)(3-(2/3)(3/2))>0. So the last one must represent the rectangle of largest area.

We can look at the first derivative. Since A'(x)=2-(4/3)x, we can draw a "sign line" for A'. It will look like this:

      Here A' is      0 when      Here A' is
      positive        x=6/4       negative

       Here A is                  Here A is
       increasing                 decreasing

and we see that the value of A when x=6/4 must be the largest value of A, so it must be a maximum.

We can also look at the second derivative. A''(x)=-4/3. Since this is negative always, we see that the function A(x) must be concave down always, and therefore, the critical number where the tangent line is horizontal must be a maximum -- the graph must lie below the tangent line.

I tried to emphasize that in practice any one of these three types of reasoning may or may not be easier than the others. I remarked that in many applications in economic theory, the second derivative approach is used. In our textbook, the first derivative discussion I did was called the First Derivative Test and the the next one was called the Second Derivative Test. I threfore proposed that the approach using just the function values (which is probably the one I would use, at least in this example) should be called the Zeroth Derivative Test. (!).

I moved on to another example: I think it was something like this. Suppose a line passes through the point (2,3) and cuts off a triangle in the first quadrant. Which line will cut off the triangle which has smallest area?

The setup here was much more complicated. Drawing an appropriate picture, and then convincing people that the picture and the problem were meaningful took a chunk of time. I used a student's drawing of this problem, and got a triangle. The base was called x and the height was called y so that the area was (1/2)xy. But howcould we get a constraint? We drew a vertical line segment from the point (2,3) to the x-axis, and observed that there were similar triangles all over the place. Then ratios of corresponding sides were equal, so we got something like (3/(x-2))=y/x, so that y=3x/(x-2) and the area function was (1/2)(3x/(x-2))x. This is not as nice as a quadratic polynomial. Not only that, but finding the appropriate domain of this function was subtle. In fact, the domain is 2 So I needed to find the minimum of (1/2)xy=(1/2)x((3x)/(x-2)) on the domain x>2. I found the derivative of A(x)=3x^2/(2x-4), and it was A'(x)=(3x^2-12x) /(2x-4)^2. Candidates for critical numbers seem to be 0 and 4 (where the top is 0) and 2 (where the bottom is 0 so that the derivative does not exist). In fact, we can exclude 0 and 2 because they are not in our domain. x=4 is the only useful critical number, so that must be the minimum.

I remarked (giving a truthful example from history!) that asserting any likely critical number had behavior which we wanted (was, for example, a minimum rather than a max!) was dangerous in real life, because sometimes real things don't behave the way we think our models do. So how can we confirm that x=4 is a minimum?

We can study the function. But there are no endpoints, and instead, we can look at the limit as x-->2^+ and x-->+infinity of A(x). In each case after some study the limits turn out to be +infinity. Therefore the function goes "up" towards the edges of its domain, so we can conclude that we've found a minimum.

We can study the derivative. A'(1) is easily computed, and turns out to be negative. Therefore the function A(x) is decreasing in the interval of x's from 2 to 4. A'(at, say, 10,000) turns out to be positive, so A(x) is increasing in the interval 2 We could also study the second derivative which I declined to do in class. We could compute A''(4). It probably is positive which would show that the function is concave up at 4, and therefore has a minimum there. In fact I just computed the value (which I didn't do in class!) and the value is 3/2. This is a perfectly good way to conclude that A(4) is a minimum, but I would rather use one of the other strategies in practice.

A last problem: your significant other is drowning 1/2 mile offshore. You are 2 miles down a straight beach, and would like to rescue this person as soon as possible. You can run at 6 miles per hour, and swim at 2 miles per hour. What's the quickest way you can reach the drowning person? I remarked that this was close to a real problem, and people are quite interested in finding the proper (fastest, cheapest, whatever) strategy. Not much time remained, so all I could do was draw a diagram and indicate that a number of different strategies existed, and that it was not clear how to find the least time method.

Quiz #19: I drew a curve going through (0,0). The curve was in the fourth qudrant and in the second quadrant, so it was positive for x<0 and negative for x>0. First I said that the curve was a graph of f', and asked people to sketch a possible graph of f. Then I said that the curve was a graph of g'', and asked people to sketch a graph of g. I expected in the first case a relative max at 0, and in the second case a point of inflection at 0, with concavity changing from concave up to concave down at 0. 63 people answered this quiz. It took about 35 minutes to grade. About half the class got both questions correct, and almost everyone got at least one of the graphs correct. Here "correct" means, of course, qualitatively, since we certainly can't sketch a totally accurate graph based only on the loose information given.

Lecture #21 (April 6)
I completed the reduction of the last optimization problem I stated in the last class. That is, I changed it into a problem about a function and an interval. I remarked that one could run along the beach for x miles, and then going swimming on a straight line to the person 1/2 mile off shore. There's a right triangle whose legs are 2-x and 1/2. so the hypoteneuse would by (by Pythagoras) sqrt((2-x)^2 +(1/2)^2). How much time is taken by swimming and how much time by running? Since distance = (rate)(time), for running I have x=6(time), so the running time is (1/6)x. For the swimming time I have sqrt((2-x)^2 +(1/2)^2)=2(time), so the swimming time is (1/2)sqrt((2-x)^2 +(1/2)^2). The total time needed is therefore (1/6)x+(1/2)sqrt((2-x)^2 +(1/2)^2), and the domain for this problem is 0<=x<=2. This function and this problem is more realistic than most (much more complicated minimization problems are considered in "real life"). I confessed that I would probably try to solve this problem by graphing the function on the interval 0<=x<=2, and I suggested that students try this. They should know before they even begin that the function will be minimized either at endpoints or critical numbers. In this case the graph is actually a bit subtle, and it takes some hunting to see that there is actually a minimizing critical number quite close to 2 (it is, I think, between 1.9 and 2).

Our next exam would be in a week. I will hand out review problems and there are answers to these problems on the web. Also next Monday's recitation would be devoted to these problems. And I'll have a review session Monday evening at 7:40 PM in ARC 110.

I recalled to people the chart I had used a month or so ago to discuss composition of functions and the chain rule. I wrote again on the board the first line of the first chart:


Capital Invested    Chips produced      Marginal chips produced
$ in millions       1,000's of units    1,000's of units per 
                                        millions of $'s

      200               3,000                    .23  

I tried to discuss what the ".23" represented. Some questioning elicited the admission that the word "marginal" was used in Economics 102. So for this example, I first tried to get an estimate of chip production if the capital investment were $203 million. After a while, we decided that the chip production would be 3,000.69 =3,000 + .23(3) 1,000's of units. then if $196 million were invested, we decided that chip production would be about $2,999.08=3,000+.23(-4) 1,000's of units. The .23 represents the local response of the system to small perturbations of capital investment.

More generally I remarked that this use of the derivative (because that's what the adjective "marginal" means) is a very important idea. So what IS the idea? In calculus, we learn that the limit as delta x --> 0 of (f(x+delta x)-f(x))/(delta x) is f'(x). What does this mean? Well, it should mean that when delta x is very small, the quotient (f(x+delta x)-f(x))/(delta x) is very close to f'(x). I'll use the word "approx" to mean "very close" So we know
(f(x+delta x)-f(x))/(delta x) approx f'(x).
We can multiply by delta x and get
f(x+delta x)-f(x) approx f'(x)(delta x)
We can add f(x) and get
f(x+delta x) approx f(x)+f'(x)(delta x)
and this is what is used in many applications. Indeed, in the financial application discussed above, f is a function whose domain variable represents the amount of capital invested, and whose range variable is the chip production (all the variables are expressed in the appropriate units). x is 200 and delta x is, first, 3, and then, second, -4.

So the numbers above translate to the following approximation of chip production for $196 million:

3,000      +    .23      (-4)
f(x)            f'(x)   (delta x)

I then said that maybe we should do some textbook problems and see how things work in simpler examples. So what happens if f(x)=x^5 and we know easily that 2^5=32. What does our approximation give for (2.03)^5? Well, f'(x)=5x^4, so f'(2)=80, and the formula f(x+delta x) approx f(x)+f'(x)(delta x) becomes (2.03)^5 approx f(2)+f'(2)(.03)= 32+80(.03)= 34.4 . The true value of (2.03)^5 is 34.473088, so the approximation isn't too bad. I then remarked that I already knew whether the approximation would be an over- or under- estimate of the true value, because I had a geometric idea of what the approximating formula represented.

f(x+ delta x) computed exactly was the y-value for x+delta x on the graph of f. But we can get close to that by looking at the tangent line to the curve at the point (x,f(x)). That line had slope f'(x). By moving to the right a distance of delta x, the height of the tangent line moved f'(x)(delta x). Several explanations: f'(x) is rise/run, so if run is delta x, the rise is f'(x)(delta x); or consider the slope to be the tangent of the angle the line makes with the horizontal direction and examine a small triangle whose horizontal leg is delta x).

The error between the approximation (which the text calls the differential but which I prefer to call the tangent line approximation) can be looked at as caused by the bending of the curve. In this case the curve x^5 seems to bend up. We can verify this by computing f''(x)=20x^3, and noting that f''(2) is positive (only the sign is of interest right now) so that the curve is certainly concave up near x=2.

We did another example, an effort to estimate sqrt(15.7). I quickly sketched sqrt(x) and asked if the tangent line approximation would be an over- or under- estimate. I was told that it would be an overestimate. Here f(x)=sqrt(x)=x^(.5), so f(16)=4. f'(x)=.5x^(-.5), so f'(2)=1/8. The tangent line approximation was 4+(1/8)(-.3)=3.9625. The true value of sqrt(15.7) is about 3.9623226, a little bit smaller. We check the concavity by computing f''(x)=(.5) (-.5)x^(-1.5), so that f''(16) is indeed negative, so the curve is concave down, and the tangent line lies above the curve.

I then gave the quiz, but I didn't say enough about the tangent line approximation. I should have remarked on its weakness: delta x must be reasonably small in order for the approximation to be useful. I'll say something about this next time. I also didn't discuss the connection between the approximation and the Mean Value Theorem.

Quiz #20: I asked people to compute (8.05)^(1/3) using the differential or tangent line approximation scheme. I said they'd need to find an appropriate f(x) (I had in mind x^(1/3)) and x (2) and delta x (I wanted .05 for this). I also asked them to explain whether the approximation they got was an over- or under- estimate using the second derivative. 70 students took this quiz, which I spent about 55 minutes grading. The grading was a bit tedious because several answers needed to be checked. In addition, I wanted people to write a sentence explaining the relationship between the sign of the second derivative and the approximation. More than half the students seemed to understand how to do this problem.

Lecture #22 (April 7)
I went over the last quiz in detail. I commented that the ideas were very important but that I didn't want to mislead the students on the applicability of the method. Sometimes it wouldn't work too well, and the statements I had made about it so far weren't much help. For example, I asked what the cube root of 1000 was. We decided it should be 10, and then I showed what would happen if we used the tangent line approximation which came from the quiz.

Then 1000^(1/3) = (8+992)^(1/3) = 2+(1/12)(992), which seems to be about 84. Most people would not call 84 an accurate approximation of 10. I tried to indicate on the graph of y=x^(1/3) the discrepancy. Concavity certainly explained the sign of the difference between the approximation and the true value, but the size, the magnitude, needed more sophisticated ideas than what we had here. I said that the Mean Value Theorem could be used to explore the difference, but that the difference was (essentially) proportional to (delta x)^2, so that if delta x were very small, then the square would force the error to be smaller. If delta x were large, however, then the square could make the error huge. But in real life, it is also important to consider units. The difference "992" above seems large, but if we had used it in the CHIPCO application, where the units in the domain were millions of dollars, 992 would actually be a fairly small delta x. Worrying about the size of the error and whether the approximation method was applicable to a specific situation was actually quite a difficult thing.

I went on to something (apparently!) completely different. Calculus had originated in getting mathematical treatments of physics problems. One standard physical setup was travel along a straight line (I drew a horizontal axis). If t=time, the function considered in a simple physical model might be position on the line, s(t). The derivative of this, s'(t), was called speed, or more properly, velocity, v(t). The second derivative of speed was acceleration, a(t).

Motivated by these definitions, I asked: if you travel at 3 mph for 2 hours, where are you? 3 here was +3, so that travel was in the left-to-right direction. I was told that "you" were at 6. I remarked that the problem as given couldn't be answered, because I hadn't stated where we had started from. So there's a subtlety. Then I wanted to try a harder problem: suppose v(t)=3t+5. What can one say about s(t)? We need to guess an antiderivative of 3t+5. We needed some coaxing, but to get a t we clearly needed to start with some sort of t^2, but then when t^2 is differentiated, 2t results. How can we get rid of the 2 and insert a 3? Students suggested that I write (3/2)t^2 so I did. To get the 5 a similar process resulted in 5t, so one answer for s(t) was (3/2)t^2+5t. I asked whether there were other answers, and was given a few nice examples such as (3/2)t^2+5t+10 etc. I asked if (3/2)t^2+5t+cos(e^sqrt(t))-sin(5t^3) could also be an answer. After some laughter people told me that it couldn't be and I asked why. I did say that we could (using the chain rule) differentiate the ugly thing I had written and that it did seem quite unlikely there would be so much cancellation that we'd get 0. But maybe I wasn't sure.

What functions have derivative 3t+5. I imagined that I had two of them, F(t) and G(t), so that F'(t)=3t+5 and G'(t)=3t+5. How can I compare F and G? I decided to look at the difference. I called Diff(t) the difference, F(t)-G(t). What did I know about Diff(t)? If I differentiated, then Diff'(t)=F'(t)-G'(t)=(3t+5)-(3t+5)=0 always. What can we then say about Diff(t)? Some suggestions were made.

One answer Since the slope of the tangent line to y=Diff(t) was always horizontal (that's what the derivative=0 always means) clearly the graph would have to be a horizontal line.

Another answer Since Diff(t) may represent some physical motion, saying that Diff'(t)=0 always asserts that there's no movement. Therefore the position can't change, and Diff(t) would have to be constant.

And yet another I agreed with these methods of analysis, but said that many people found them inadequate. I asserted that most of the students in the room would not be using calculus to deal with physical problems, and even graphs can be deceptive. I proposed another method of dealing with Diff(t) that many people found more satisfactory. I used the Mean Value Theorem to write (Diff(t_1)-Diff(t_2))/(t_1-t_2)=D'(somewhere in between), and since the D' value was always 0, I knew that the top of the fraction on the left-hand side would be 0. So Diff(t_1)-Diff(t_2)=0, and therefore Diff(t_1)=Diff(t_2) for any t_1 and t_2, so that the function's values are always the same: it would have to be constant.

So the most general function with derivative 3t+5 was (3/2)t^2+5t+C where C was an unknown constant. I called the process I just illustrated, antidifferentiation.

I did a bunch of antiderivative examples. I think I did things like 4t^11+5t^2-9 and 5sqrt(t)+1/t (here the trick was rewriting sqrt(t) as t^(1/2) and also recognizing that 1/t could NOT be antidifferentiated as a power -- we needed to use the ln function). I also found antiderivatives of something like 5cos(t)-3sin(t) (being careful of signs!) and something like 74e^t, always with a "+C" in each.

I then declared I was going to do my one physics problem of the semester. I wrote something like the following paragraph:

Fred is standing on the top of a 100-ft tall building. Fred throws a ball straight up in the air at 20 feet per second. Gravity pulls the ball down at 16 ft/sec^2.(Yeah, I got the gravity of the earth wrong -- this should have been 35 ft/sec^2!). My questions: how high will the ball get, and when will the ball hit the ground?

As a calculus problem this is s''(t)=-16 (we had to think about the sign here, and determine that - means down and + means up), with the additional information that s'(0)=+20 and s(0)=100. I then took one antiderivative of s'' to get -16t+C, so that s(t)=-16t+C. s(0)=-16(0)+C=20, so C=20 here. Therefore we have s'(t)=-16t+20. Another antiderivative gets us s(t)=-8t^2+20t+C, and setting t=0, we get s(0)=-8(0)^2+20(0)+C=100, so that s(t) MUST be -8t^2+20t+100. Whew! This was a lot. I didn't finish the problem because I was running out of time. I would have had to find out where s'(t)=0 to get the highest the ball can be, and I would have had to solve the quadratic equation s(t)=0 to find out when the ball would hit the ground. I did say that the conditions s(0)=100 and s'(0)=20 were called initial conditions.

Quiz #21: If f'(x)=x^3 and f(1)=2, what is f(x)? 69 people answered this quiz, the majority answered it correctly, and it took about 45 minutes to grade.

Lecture #23 (April 13) The second in-class exam. 93 people took the exam.

Lecture #24 (April 15)
I began by saying that there were 5 lectures left in the semester, almost 20% of the course lecture time, and that there were important topics still be be covered, and almost a quarter (!) of the points on the final would ask about material in this section.

We've been studying differential calculus, and now I wanted to start integral calculus. At first it would seem like such a huge change in problems and ideas that we could not be doing the same subject, but a rather large surprise would occur later.

My first example was suggested by Joe Rosenstein of the Rutgers Math department as a way of stimulating discussion of ideas related to integral calculus. I asked: What is the area of my hand?

There was, of course, confusion and laughter. The first was caused by my asking such a low-level question, and the latter by its absurd nature. I responded that it was actually quite difficult to deal with this question. It had direct counterparts in medicine where people might want to know areas of bodies in order to estimate fluid balance, etc. I remarked that one should try to answer the question sim0ply and directly.

We discussed the question for a while. FIrst we enclosed a traced image of my hand on the board with a large square, thus getting an overestimate. Then we put a small square inside the central palm area of my hand's shape, getting an underestimate. Then there was little progress for a while. I wanted people to "refine" these initial estimates, because the gap between them was quite large. Finding better estimates was more difficult than I had hoped. but we did cut out some triangles and rectangles from the big square, and we did add some triangles and rectangles to the enclosed square. There were "mistakes" on the way: for example, students wanted to just throw an approximatuing rectangle on the hand, with some of it inside and some of it outside the had's area. That looked tempting but I remakred that we couldn't be sure that the amount contributed by area outside could balance what was thrown out from the inside. I also remarked that I only knew the area of rather simple shapes, such as rectangles and squares and triangles and circles, and I wanted to build up from there.

Writing a detailed protocol for measuring the area of my hand would be quite difficult. Basically, though, we wanted to create an underestimate by including rectangles which took up area previously unmeasured, and which did not overlap with other rectangles and which did not go outside the hand. For the overestimate, I wanted to throw out rectangular areas which did not include area inside the hand and which didn't overlap. All the rectangles involved should somehow be "as large as possible" subject to the constraints identified. This was all rather difficult to discuss.

I aksed students to imagine a more general problem, that of trying to find the area of a big irregular blob (I drew one on the board). I "dissected" the blod withan assortment of straight lines so that what I had left was a collection of pieces each of which could be described in the following way:

The area bounded by the x-axis, two vertical lines (x=a and x=b), and the graph of a function y=f(x) where f was positive between a and b. I asked how we could hope to find that area. If we could, then we could answer the apparantly more general problem I posed above.

I suggested a specific example which I sketched: the area enclosed by y=1/x, the x-axis, and the line x=1 and x=2. I remarked that there were many ways to try to organize the computation I would describe, but that I would restrict myself to the method cited in their text. I divided the interval [1,2] into three equal parts and erected vertical lines at x=4/3 and x=5/3. I drew rectangles at various places, and got an overestimate of the area with the area of three enclosing rectangles. As quiz, I asked people to find the corresponding underestimate.

I did not go into the error in such estimates, which would have allowed me to describe a way of getting an estimate of the true area with as good an accuracy as desired.

I wanted to allow enough time for the return of the second exam whose grades are discussed elsewhere.

Quiz #22: I asked for the underestimate described above. 64 students answered this quiz. It took about 20 minutes to grade, and almost everyone answered it correctly.

Lecture #25 (April 20)
I rewrote the problem which was discussed at the end of the previous lecture: that is, analyzing how to compute the area enclosed by y=1/x, the x-axis, the line x=1 and the line x=2. I called this area, A. I again divided the interval [1,2] into three equal parts and obtained the estimates:
(1/3) (1/(1 and 1/3) +(1/3) (1/(1 and 2/3)+ (1/3) (1/2) < A < (1/3) (1/(1) + (1/3) (1/(1 and 1/3) + (1/3) (1/(1 and 2/3)
I asked how we could improve these estimates. I was told that we could break up the interval [1,2] into more pieces. I tried 10 pieces. I carefully graphed y=1/x on this interval, divided the interval [1,2] into 10 equal pieces, erected vertical lines at each division point, and drew horizontal lines depicting the upper sums and the lower sums. I said that I wouldn't be doing the arithmetic to get the exact numbers for any of these sums, but I was interested in getting an estimate of the error, the difference between the over- and under-estimates.

I took the little rectangles between the sums and moved them all horizontally into one stack. They fitted into a rectangular stack which was 1/(10) wide and 1-(1/2) high. So the error, the difference between the over- and under-estimates, was apparently at most (1/2)(1/10)= 1/(20). I remarked that the reasoning I was using was certainly made easier by the fact that the function I chose had quite a simple graph, but that similar reasoning could be used for a function whose graph was more wiggly. I asked if anyone could tell me a way to find a sum which would approximate the true value of A with an error of less than 1/(100,000).

We discussed this. I emphasized that I didn't want the actual sum, just a "protocol", a clear description of the method, which would approximate the sum closely. After a while, we got the description. We divided the interval [1,2] into 50,000 equal pieces (the first description omitted the word "equal" and we added that in after more thought), and take over- and under-estimates corresponding to this division of the interval. The error could be trapped in 50,000 little boxes, all of which could be stacked in rectangle which would be 1/(50,000) wide by 1/2 high, and this stack would have area 1/(100,000), so the error would be less than that.

We moved on to talk about blood flow. I asked how one could estimate how much blood flows through the heart, drawing an especially hideous picture on the board. I suggested the following.

The Aztec protocol: cut open a person's chest, and sever an artery leading from the heart. Hold a bucket under the artery for 20 minutes. Measure how much blood there is in the bucket. Of course, this picture has the unpleasant consequence of needing to discard the person after it is done.

I tried to explain how one might actually measure such blood flow. One might insert carefully a sensor into the blood stream at some accessible site, and slowly and carefully push it up towards the heart. Once it is there, one could measure the blood flow. for example, if we wanted blood flow during a 20 minute period, we could do this, and get a reading (making it up!) of 37 cubic centimeters/minute. How much blood could we then estimate flowed through the heart during 20 minutes? The answer is surely (37)(20). How could one get a more accurate estimate? I was told that we could take more readings. So what if we knew the following:

Time in minutes    5     10   15    20
  since start 

Blood flow in      20    37   25    22
 cubic cm's

Apparently a corresponding estimate would be (20)(5)+(37)(5)+(25)(15)+(22)(20). Each "5" is written because it is our assumption that, for the purposes of this approximation, the blood flow is constant during the corresponding period. (A alternate explanation involving averaging the blood flow, was also suggested. I didn't pursue that, because I wanted to get to the method described in the text.) With more frequent measurements, one could suspect that the estimate of blood flow might become more accurate. What if the following measurements were made?

Time in minutes    5    7    10     12    14     15    20
  since start 

Blood flow in      20   25   37     12    -4     25    22
 cubic cm's

I was questioned about the "-4" and what it meant and how likely it represented "reality". I explained that sometimes the heart may not pump too well, and sometimes there may even be leakage backwards. If one were interested in the net amount of blood the heart pumps through the system, well, then, sometimes there would be a negative reading. If we had the measurements above, a (presumably) more accurate estimation of the net blood flow would be (20)(5)+(25)(7)+(37)(3)+(12)(2)+(-4)(2)+(25)(1)+(22)(5). I drew a picture of the data: that is, I plotted the data, and I interpolated a continuous curve of blood flow in cc's over 20 minutes using this data. I drew some corresponding rectangles, one below the horizontal axis.

Then I defined Riemann sum, partition, and sample points as in the book. I did an example of a Riemann sum for f(x)=x^2, partitioning the interval [2,9] into 2<4<5<7<9 with sample points 2.5, 4.5,6.9, and 7: (2.5)^2(2)+(4.5)^2(1)+(6.9)^2(2)+(7)^2(2). I announced that it was time for our quiz.

Quiz #23: I wrote the following table:

   x    sqrt(x)

   1     1
   2     1.41
   3     1.73
   4     2
   5     2.24
   6     2.45
   7     2.65

I said I wanted Riemann sums corresponding to the partition 1<3<'5<7. As sample points: if the last name began with the letters A to K, choose left-hand endpoints of the subintervals (value: approx. 9.94). If the last name began with J to P, choose right-hand endpoints of the subintervals (value: approx. 13.24). If the last name began with Q to Z, choose midpoints of the subintervals (value: approx. 11.72). 55 people took this quiz, and only about half of them got it right. The most common mistake was using the function x^2 instead of sqrt(x), because (I guess) a Riemann sum for x^2 was already on the board! Other people gave the wrong number of terms for the sum, etc. The results were disappointing to me.

Lecture #26 (April 22)
I discussed quite a lot today. First, although yesterday's quiz had been done fairly well, i reviewed it and drew careful pictures (which I deliberately had not done when I first gave the problem, to emphasize the sheerly numerical aspect of it). I remarked that the three answers were all estimates of the area enclosed by y=sqrt(x), x=1, x=7, and the x-axis.

More generally, I drew a wiggly picture of y=f(x) from x=a to x=b, positive for only of its domain. I subdivided [a,b] with a fairly "fine" partition, I selected random sample points, I drew rectangles of heights f(each sample point) and width the length of the subinterval (remarking that there were negative heights where f was negative). Then I said that the total SIGNED area was the Riemann sum. This complicated creature depends on the partition, the sample points, and the function. I also finally wrote for the first and only time in the course a formal summation using SIGMA (capital sigma!): SIGMA_j=1^j=n f(c_j)(x_j-x_{j1}) and remarked that people might see similar things in textbooks. The SIGMA came from an abbreviation for Sum.

It is true that as the number of subdivisions in the partition increased, as the width of largest subinterval so defined decreased to 0, then the Riemann sums (with any choice of sample points, any, at all, no matter how arbitrary) would approach a unique limit. I indicated how on my drawing the boxes etc. would get to be a really fine approximation to a certain SIGNED area (I had one chunk below the x-axis and two chunks above it): the areas of the two "above" chunks minus the area of the "below" chunk.

The name of this unique limiting number was "the definite integral of f from x=a to x=b". I remarked that "integral" meant "putting together in one piece" (related to "integration"). I wrote the standard notation for this. Here this will be written int_x=a^x=b f(x) dx and again in this notation the integral sign ("int") is a lengthened "S" and is an abbreviation for sum. The "f(x) dx" was supposed to be a reminder of the slim rectangles which were added up, each of height f(x) and width "dx". The a and b were to tell people to add these areas up from x=a to x=b.

So this is formidable notation. Why should Math 135 students study such a thing (again, I reminded students that I told them on the first day of class that this question is legitimate and they should ask it any time something new or complicated comes up. I gave a few reasons.
1. Compute definite integrals in order to compute signed areas (!). This is a somewhat false reason, because I don't really think that many of the students in the class would in the coming years need to compute many areas.
2. Compute fluid flows, like blood flow as discussed in the last lecture. In this case, the graph of f I had been using was rapidly replaced by the graph of blood flow as a function of time. I used the same shape curve, and we decided that the "positive area" area represented flow in the correct direction, while the bump of "negative area" represented backflow. So that total integral represents the net fluid flow.
3. This is a harder example from finance, which I certainly didn't go into. It turns out that the definite integral can represent the future value of an income stream: I asked people to imagine that they were getting a sequence of payments, and that they could deposit or invest these payments, and that the investments would pay a varying amount of "interest". The value in the future of this sequence of payments and its reinvested proceeds is call the future value of the income stream. I remarked (truthfully) that this situation could be modeled using a definite integral. In this case, the positive "area" would represent a profit (approximately) while the negative "area" would be a loss. This is all fairly technical, and students interested in finance would definitely see more of it.

I discussed some simple properties of integrals, illustrating each one with a picture.
Fact 1: If a Fact 2: If f(x) < g(x) on the interval [a,b], then int_a^b f(x)dx < int_a^b g(x)dx. Again, "true" by picture.

How about some examples? I computed several "fake" examples. A student inquired about my use of the adjective, so I erased it, but maintained that the word would be justified by the selection of examples.
Example 1: int_712^712 x^3 cos( e^sqrt{x^4+7} ) dx. Huh. People seemed stunned (as I had hoped!) by the ugliness of the formula. But we decided anyway that the integral should be 0 because there wasn't too much area in a line segment.
Example 2: int_3^11 4 dx. There was some confusion, but here the f(x) was the simple function 4, whose graph is a horizontal line. The definite integral represented the area of a rectangular region, and its value was (11-3)4=32.
Example 3: int_0^{2 Pi} sin x dx. We decided this had to be 0 because the + and - areas canceled.

What about going back and trying to compute something not so clear, namely, int_1^7 sqrt{x} dx? We had our estimates. I said I could think about 3 strategies. The first was to do much more numerical work. This was certainly possible, and, indeed, with graphing calculators available (each having the computing power fo two tons of equipment 20 years ago!) this is certainly practical. A second approach would mean using various elaborate algebraic tricks (illustrated in the book) for computing certain special Riemann sums. This is difficult, and I also didn't want to do it. Instead, I wanted to try a third approach. The merit of the third approach is that it represented certainly one of the ten best mathematical ideas I knew, and maybe was a valid contender for one of the ten best ideas known to the human race. I urged people to pay close attention, because in the next 20 minutes I would be doing something really remarkable.

I said that the way to solve this problem efficiently was to try to solve a more difficult problem. Instead of looking at the area from 1 to 7 under sqrt{x}, consider the area from 1 to Q. Let's call this area, A(Q), and try to systematically investigate the function A(Q), even though what we really want to find right now is A(7). We considered the function for a while and learned we could exactly and easily compute one value of the function: A(1)=0 (the area from 1 to 1 -- well, there isn't much!). We also saw that as Q increased, A(Q) seemed to increase. But what else could we do with A(Q)? Since this was a calculus course, it was suggested that we try to differentiate A(Q). So I wrote (A(Q+deltaQ)-A(Q))/(deltaQ) and tried to analyze this. I drew lots of pictures, and decided that the top was the area of a "sliver", bounded by the x-axis, the lines x=Q and x+q+deltaQ, and the curve y=sqrt{x}. We could try to estimate the area of the sliver. We could take a box larger than this area, which was deltaQ wide and sqrt{Q+deltaQ} tall, and this would be an overestimate. An underestimate is supplied by a smaller box, with height sqrt{Q} and width deltaQ. SO the difference quotient obeyed the inequality:

  (area of           true value of the        (area of
 small box)         difference quotient      large box)

sqrt{Q}deltaQ        (A(Q+deltaQ)-A(Q)      sqrt{Q+deltaQ}deltaQ
-------------        -----------------      --------------------
   deltaQ                 deltaQ                    deltaQ

Divide by deltaQ, and take the limit as deltaQ-->0. Both the over- and under-estimates approach sqrt{Q}, so now we know that A'(Q)=sqrt{Q}. But our previous work on differential equations allows us to conclude then that A(Q)=(2/3)Q^{3/2}+C (I went too fast here). Since A(1)=0, if we insert Q=1 in the formula we get (2/3)1^{3/2}+C=0, so that C=-(2/3)Q^{3/2}. Therefore A(Q)=(2/3)Q^{3/2}-(2/3)1^{3/2}. Now A(7) which is the original quantity desired can be gotten by just plugging in: A(7)=(2/3)7^{3/2}-(2/3)1^{3/2}. Indeed, the value of this expression as reported by calculator falls right into the middle of the range of the three approximations reported earlier.

This was all an example of one version of the
Fundamental Theorem of Calculus If you know that F'=f, then int_a^b f(x)dx=F(b)-F(a).

In our example, f(x)=sqrt{x}. F(x)=(2/3)x^{3/2}, a=1, and b=7. There was some discussion of this amazing fact, and then I sketch the curve y=x^2 with the lines x=2 and x=3, and asked them to find int_2^3 x^2 dx.

Quiz #24: 59 people took this quiz, and about three-quarters of them got the correct answer. It took about 25 minutes to grade.

Lecture #27 (April 27)
I again wrote the time and place of the final, and the time and place for my review session for the final. I announced that we would do evaluations today, and that therefore I'd like to give the quiz in two parts during the lecture.

I wrote the first part of the quiz: I drew a graph of a function with a horizontal line segment at height -1 from x=0 to x=2, followed by a line segment from (2,-1) to (3,0), followed by an upper semicircle centered at (4,0) with radius 1. I asked people to compute the definite integral of this function from 1 to 4. After a few minutes I advised them that the answer should be negative.

I reviewed upon request the fact that the definite integral gives what is called a "signed area": areas above the x-axis are counted positively and below the x-axis are counted negatively. Actually, I merely copied the picture from the formula sheet onto the board and discussed it.

How should one compute definite integrals? Here are a few methods.
i) If the graph is made up of sufficiently simple geometric shapes, just get the areas and count them algebraically. That's what I expected people to do with the first quiz question I asked them.
ii) Try to approximate the definite integral with a bunch of very fine Riemann sums, and see what the limit seems to be. You'd better have some computational help available if this approach is chosen.
iii) Use the Fundamental Theorem of Calculus (FTC), or try to. Most people do this first. This result is: if you know (somehow) that F'=f, then the definite integral of f from x=a to x=b is F(b)-F(a).

I applied FTC to find the area enclosed by y=x^3, x=1, x=2, and the x-axis. I guessed an antiderivative and did the computation. My "guess" was (1/4)x^4+117. There was a bit of objection, but I used it anyway and got "the" answer. I remarked that other people might use a different antiderivative, like (1/4)x^4+23, but they would get the same answer. In fact, we know from earlier work in the course that any antiderivative of x^3 was (1/4)x^4+C, and the "+C" would cancel when used to evaluate a definite integral. I remarked that in evaluating a definite integral, people frequently left of the "+C" because it would just cancel and not affect the answer. I remarked that many people find it comforting to imagine that they divide the area I drew on the board into lots of thin strips, each "dx" wide and x^3 hide, whose area (almost a rectangle) was x^3 dx, and then think of adding up (Summing, with a capital S so that we'd think of the integral sign) the areas of these strips from x=1 to x=2.

Then I computed the definite integral of sin x from 0 to 2PI. I got 0, and I asked if I should then conclude that there is no area under the curve at all. We discussed this, and some students persuaded me that we had done this problem before (!) and concluded that the area would be 0 because the + part exactly balanced the - part. So I asked what was the area under one bump. We had to figure out what the limits on the definite integral were (0 and PI) but could then compute it: 2. I remarked that this answer seemed approximately correct if you looked carefully at the graph. I was also asked what the - sign was in the antiderivative of sin x, because there was some confusion about derivatives and antiderivatives.

Then I drew a graph of y=16-x^4 on the board (an upside down parabola, I was told by a student). I said that the second part of their quiz was to compute the area inside the bump of this curve. I left them to do this for a while, and wandered around the room. After I bit I went up to my graph, which was totally unlabeled, and asked if I should add anything. I was told to put in -2 and 2 for the x-intercepts of the graph, which I did. I wandered some more, and then asked what the answer was. Several people responded and at least a few of the answers were the same, so I put that on the board also.

I went on. I did a bunch of problems from the book (section 4.4):
#21: the definite integral of 1 + sin x from 0 to Pi. I simplified the answer the way the text would.
#23: the definite integral of (sec x)^2 from -Pi/6 to Pi/6. I simplified the answer the way the text would.
#27: the definite integral of 2x+ (1/x) from 1 to e. I simplified the answer the way the text would.
#29: the definite integral of e^x + 6 from 0 to 2. I simplified the answer the way the text would.
#35: a graph of y=x-x^2 was given, and the part above the x-axis was shaded. The problem was to find the area of the shaded region. I set up the definite integral and evaluated it.
#37: a graph of y=(3-x)sqrt(x) was given, and the part above the x-axis was shaded. The problem was to find the area of the shaded region. I set up the definite integral. We transformed the function's formula into 3x^.5-x^1.5 by distributing the multiplication, and computed the definite integral.

Finally I wanted to evaluated the integral of (I think!) e^(7x) from 3 to 5. I "guessed" at the answer for the antiderivative as (1/7)e^(7x). I verified the guess by differentiating the formula, although many people were unhappy about the chain rule, still. Then I computed the definite integral.

I illustrated the method of substitution for the indefinite integral of e^(7x), with u=7x, du/dx=7, du=7dx, (1/7)du=dx, so that int e^(7x) dx changed to int (1/7) e^u du which i computed as (1/7)e^u+C and then (1/7)e^(7x)+C. I remarked that for indefinite integrals it was customary to put in the "+C". I said that I would spend most of Thursday's lecture discussing the method of substitution and giving examples. I discussed quite a lot today. First, although yesterday's quiz had been done fairly well, i reviewed it and drew careful pictures (which I deliberately had not done when I first gave the problem, to emphasize the sheerly numerical aspect of it). I remarked that the three answers were all estimates of the area enclosed by y=sqrt(x), x=1, x=7, and the x-axis.

Quiz #25: 55 people took this quiz. About half of them got both questions correct. Varied mistakes were made in both parts by other people. The first question seemed to be more difficult: actually finding areas of the simple shapes involved, and then putting + and - on the areas seemed hard. The quiz took about 55 minutes to grade.

Lecture #28 (April 28)
I began the last lecture by again writing the date, time, and place of the final exam and review session on the board. I also reminded students that there would be another recitation tomorrow because Friday was following Monday's schedule.

Last week I verified the Fundamental Theorem of Calculus by writing the area function from 1 to x of sqrt(t) and differentiating this function. The wonderful result was that this was equal to sqrt(x). Another aspect of the Fundamental Theorem of Calculus records this: if f is a continuous function, and if A(x) is the integral of f from a fixed number a to x, then the derivative of A(x) is f(x). I offered the example of integral from 312 to x of (t^3)sqrt(exp(t^2)+cos(t)) dt. Then A'(x)= (x^3)sqrt(exp(x^2)+cos(x)). I remarked that the instructors of the course expected students to know this fact and would test for it, but that in context it can show important aspects of a problem.

For example, here is a much simplified example from economics. Suppose you start a business with a capital of, say, $50,000. How will the capital change? Clearly you will have Income, and this will be either a profit (+) or a loss (-). In fact, capital at time now = $50,000 + integral from starting time to now of income. The rate of change of capital is equal to income, using the different version of FTC stated above. More complex statements can assume the shape of the FTC. But I returned to the use we have been making, evaluating definite integrals by finding antiderivatives.

I asked what the integral of sqrt(7x+9) from 0 to 1 was. We substituted, using u=7x+9, du=7dx, so (1/7) du=dx, and int sqrt(7x+9) dx= int sqrt(u) (1/7) du = (1/7) int sqrt(u) du = (1/7) int u^(1/2) du = (1/7) u^(3/2)/(3/2) +C = (1/7) (7x+9)^(3/2)/(3/2) +C. I remarked that it was conventional to put the "+C" on an indefinite integral, but that most people omitted it when evaluating a definite integral, since constants would get subtracted and play no part in the final answer. Since we have an antiderivative of sqrt(7x+9), we can evaluate the definite integral. It is (1/7) (7x+9)^(3/2)/(3/2) at 1 - (1/7) (7x+9)^(3/2)/(3/2) at 0 which is (sigh!) (1/7)(16)^(3/2)/(3/2)-(1/7)9^(3/2)/(3/2) = (1/7)(2/3)(64-27)=74/21. I remarked that sometimes they would need to "simplify" answers in exactly that way, but that ordinarily a calculator approximation would be sufficient. Also, an antiderivative can always be checked by differentiating, and we did this, of course needing the chain rule at a vital part of the process.

I remarked that most of the rest of the examples I did would be antiderivatives. I'd avoid the tedious evaluation and simplification stages.

I did a sequence of problems from section 4.5. I did also again remark that watching me do these problems was a poor substitute for students' practicing by doing a number of them.
#13: int 5x (1-x^2)^(1/3) dx. Solved with u=1-x^2.
#12: int x(4x^2+3)^3 dx. Solved with u=4x^2+3.
#17: int x^2/(1+x^3)^2 dx. Solved with u=1+x^3.
#49: int x^2e^(-x^3) dx. Solved with u=-x^3.
int sin(17x) (cos(17x))^5 dx. Solved with u = sin(17x). In retrospect, I might have made the exposition clearer by doing s sequence of two substitutions, first u=17x then v=sin u. I also remarked that of course this example which I "invented" in order to answer a student's request for something involving the trig functions, was certainly special, and arranged so that substitution would work out easily. That certainly won't alway's be the case, but it works often enough so that in practice it is the first thing tried in many cases.
int (x)sqrt(x+5) dx. I substituted u=x+5, and got to integrate (u-5)sqrt(u) du which I handled by distributing the multiplication to get int u^(3/2)-5u^(1/2) du. Then I antidifferentiated, and substituted back to get a complicated answer. I remarked that there were several substitutions which might have worked here, and that the "proceedings" could get very complicated.
#19 int x^2/(1+x^3) dx. Solved with u=1+x^3 and recalling what ln was used for in calculus terms.
int (sin x)/(cos x) dx. Solve with u=cos x, so du=-sin x dx and the antiderivative is -ln(cos)+C. Of course I now stated that I had found the integral of tan x, and it was usually written ln(1/(cos x))+C=ln(sec x)+C.

I ended a bit early but gave perhaps a more challenging quiz than usual.

Quiz #26: What is the exact value of the integral from 0 to 2 of x^2*sqrt(x^3+1)? I asked people to find the antiderivative and then actually s"simplify" the answer. I remarked that such a problem could very well be on the final exam in this course. I would grade it and try to get it back to them tomorrow. After a while, I got the answer from some students and wrote it on the board: (52)/9 54 people took this quiz. More than half seemed to get the right answer, showing their work. Some still seemed to have absolutely no idea what was going on. It took about 50 minutes to grade.

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