### Tuesday, February 8: this is lecture 7

Who's up front?
I made some nearly incoherent comments urging students to realize that during most of their college courses, they are being instructed by fanatics. Perhaps it might have been more diplomatic to say enthusiasts. Yes, your college instructors are generally enthusiasts who are experts in narrow fields of human activities, and who want to tell you all about the thrills of their subjects. Today I will be able to tell you about one of the major thrills of mathematics. If I do not succeed in telling you something interesting, please attribute this to the weakness of my efforts and not to the defects of the material. I am excited, really excited, to be given the chance to describe something so wonderful to you!

What will we do?
I mentioned in class today that almost everyone who sees the material we will begin to discuss today thinks that it is one of the high points of human intellectual achievement. We surely have prepared for this in earlier meetings of the course.

• I discussed the growth rate of trees. This is a fiction in some profound sense, because trees and other things really grow in rather irregular fashion. The idea of a growth rate is sort of an average, and it is only an idea, or perhaps an ideal. It helps us understand tree growth.
• In economics, the marginal {cost|profit|revenue} of various items is used frequently. It is hard to believe that these things mean literally and precisely what they declare. For example, if the marginal production of cars given additional capital invested is, say, 340 cars per additional \$100 million invested, does this mean that the corporation really will produce ... uhhhh ... about a third of a car more if it invests an additional hundred thousand dollars? This is sort of silly (do they stop at the back of the front tires?), but it does involve the ideas of marginal "stuff". And people make important decisions based on these ideas.
• We studied cosine and sine. And we will use radians in this course, and so do many other people. The reason we will use radians is so that the growth rates of sine and cosine will be very very simple, and not involve constants like .01745 (yuck!)
• In lots of applications, growth and decay are considered. Applications in finance (interest rates), medicine (radioactive decay of tracers, drug levels in the blood stream), and other areas are quite important. Everyone in these fields uses the exponential function, ex, and the natural log, ln(x) (log to the base e). These functions are used because their rates of change are very very nice.

A useful quote?
The text in section 3.1 introduces the rate of change idea by analyzing tangent lines to curves. "Students" are somehow intuitively supposed to know what tangent lines look like. The quote I used in class to emphasize this comes from U.S. Supreme Court Justice Potter Stewart, who supposedly remarked about pornography, "I cannot define it, but I know it when I see it." Analyzing tangent lines is sort of like that. An exact definition is difficult, and you are supposed to "know it" when you "see it."
My only quarrel is that I don't believe most students in the class have majors where they will be interested in computing tangent lines, so "tangent lines" themselves don't seem a rich source of motivation.

The first example
What is the equation of the line tangent to y=x2 when x=1?

The equation of a line (let's suppose it is not vertical!) is y=mx+b. There are two constants, m and b. Geometric data which will yield m and b are generally in one of the following forms:

• Two points on the line. Then it is "easy" to get m, the slope, as a difference of the second coordinates (the y's) divided by the difference of the first coordinates (the x's). And we get b by substituting in one of the coordinate pairs corresponding to one of the points, and "solving" for b.
• A point and a slope. Then we know m immediately, and get b by plugging in the coordinates of the given point and again "solving" for b.
In the case of "the line tangent to y=x2 when x=1", I know (because I know it when I see it!) that the line goes through the point (1,12). I write the second coordinate as 12 and not just 1 because I want to emphasize that we got it as a point on the graph of y=x2. We need to find a slope, m. In fact, because it is the slope of the tangent line, I will call it mtan.

Appoximating mtan by msec
The idea of this motivating example is containing in the picture to the right. But please note that this picture has lots of information implicit in it. The point I call P is supposed to be (1,12). O.k.: it is (1,1). The point Q is close to P: it has first coordinate 1+h. Therefore since Q is on the graph of y=x2, the second coordinate of Q must be (1+h)2. The line connecting P and Q is frequently called a secant line (it is the dashed line in the picture) and you are supposed to believe that this secant line has slope close to the slope of the tangent line to y=x2 at (1,1,) when Q is close to P. That's the same as asking that h be very small. I'll call the slope of this secant line msec. What is msec algebraically? I can get this from the points Q=(1+h,(1+h)2) and P=(1,1). So msec=[(1+h)2-1]/[(1+h)-1]. It would be nice to forget that right here in class the instructor made an embarrassing algebraic error.
I am interested in what happens as h-->0. That is, what can we say about limh-->0mh? Well, look at [(1+h)2-1]/[(1+h)-1]. The bottom +1 and -1 cancel. So the result is [(1+h)2-1]/h. If I try the "plug in" method to see what happens, I get 0/0, which is bad. This is the model situation for limits. In this case and in the others which I'll consider today algebraic transformations (which I referred to in class as algebraic massaging) will make the limit behavior clear.

So [(1+h)2-1]/h=[1+2h+h2-1]/h=[2h+h2]/h=2+h. Therefore limh-->0msec=limh-->02+h=2 and my "guess" for mtan is 2.

I now believe that the tangent line to y=x2 when x=1 goes through the point (1,1) and has slope equal to 2. So the line is (y-1)=2(x-1). Unless these is some reason or reward, I would probably leave the answer this way (yes, I agree that I could write it as y=2x-1 but why risk touching it since I might "break it": that is, make some silly arithmetic error).

In this lecture and in the following lecture, I am interested in the ideas. It turns out that for functions defined by simple formulas, computing slopes of tangent lines is a nearly mechanical process. Most of the examples of functions you will meet in the real world will reveal themselves through graphical information or numerical data. I doubt very much, for example, that a veterinarian will learn that the usefulness of a canine tranquilizer is 4x2-9x+4. Much more likely ways of conveying what's know would be a table of data or a graph. So please learn the ideas right now.

The second example
What is the equation of the line tangent to y=x2 when x=-3?

When x=-3, y=(-3)2=9. So I will put P at (-3,9). What about Q? Q will be located where x=-3+h, for some very small number h. The coordinates of Q, another point on the graph of y=x2, will be (-3+h,(-3+h)2). What information do I know about the line tangent to y=2 when x=3? I know the line passes through (-3,9), and that its slope, mtan, will be limh-->0msec where msec is the slope of the line through P and Q. I asked in class at this point in the exposition whether mtan could be 2, and I was told "Certainly not" because the sloe of the tangent line was "clearly" tilted down, so mtan should be negative.

Investigating msec
The coordinates of P and Q allow me to write a nice formula for msec: [(-3+h)2-9]/[(-3+h)-(-3)]. First on the bottom the -3's cancel. It is important to keep track of minus signs. In my computations, when I'm working by myself, I sometimes tend to lose track of them. But that's a terrible idea here, since various cancellations are important. So msec=[(-3+h)2-9]/h. As h-->0, if I try to "plug in" I get the unacceptable 0/0 again. Some algebra:

```(-3+h)2-9   (-3)2-6h+h2-9   9-6h+h2-9   -6h+h2
--------- = ------------ = --------- = ------ = -6+h
h             h            h         h```
These algebraic transformations are all valid when h is not 0. Therefore (as we say in math courses), limh-->0msec=limh-->0-6+h=-6 and this must be mtan.

I now believe that the tangent line to y=x2 when x=-3 goes through the point (-3,9) and has slope equal to -6. The slope is negative, even as predicted earlier from the picture. So the line is (y-9)=-6(x-(-3)).

The third example
What is the slope of the line tangent to y=x2 when x=x0?

I concentrate my attention on the slope because that's really the interesting information.
Now the algebra will be more intense and I will also make the situation more "interesting" by skipping a few steps. The point P has first corrdinate x0 and is on y=x2 so it is (x0,(x0)2). The point Q has first coordinate x0+h where h is some small (but non-zero!) number. Since Q is also on the curve y=x2, its coordinates are (x0+h,(x0+h)2). Now msec is [(x0+h)2-(x0)2]/h. Here I've already cancelled +x0 and -x0 from the bottom of the fraction. You must practice this, please! Here we go:

```(x0+h)2-(x0)2   (x0)2+2x0h+h2-(x0)2    2x0h+h2
------------ = ------------------- = ------- = 2x0+h
h                  h              h ```
which is valid when h is not 0. Therefore
limh-->0msec=limh-->02x0+h=2x0=m2.

The slope of the line tangent to y=x2 when x=x0 is 2x0.

I urged students to realize that when x0<0, the slope of the tangent line was negative. Hey: the tangent lines and the curve are both decreasing as we walk from left to right. Of course, when x0>0, the reverse occurs. The tangent lines and the curve both go up (increase) as we walk from left to right.
Of course, this discussion is about a simple function and simple graph. But the ideas will work in situations which are much more complicated.

A hyperbola example
y=1/x is a hyperbola. What is the slope of the line tangent to y=1/x when x=x0? We looked at the graph of y=1/x and immediately (well, fairly soon) recognized that the slope of the tangent line would always be negative. This is sort of weird but true, and will allow us to check the result we get, at least roughly. O.k, now to work. Here is what I will mumble to myself:
P ... Q ... x0 ... 1/x0 ... x0+h ... 1/[x0+h] ... msec ...
The picture may not be that helpful. I am sorry. You are not allowed to make algebraic errors in what follows. So:

```   1      1        x0-[x0+h]         -h
----  - ---     ------------      --------
x0+h     x0        [x0+h]x0        [x0+h]x0     -h
------------- = -------------- = ---------- = ---------
h                h                h      h[x0+h]x0
```
And now, since today only I am trying to be careful, I note that since h is not 0, this is the same as -1/[x0+h]x0. If h-->0, then this-->-1/[x0]2. This is the slope of the tangent line. I'll note again that your algebra needs to be really clean, neat, and correct, otherwise you will have little chance of getting the right answer.

Cheap check
Yes, the slope of the tangent line is -1/[x0]2 and this is always negative for all x0 in the domain of f(x)=1/x (that is, when x0 is not 0).

A hyperbola subexample
What is the equation of a line tangent to y=1/x when x=7?

Well, when x=7, f(7)=1/7. The tangent line passes through (7,1/7). The slope of the tangent line is gotten by substituting x0=7 in the formula -1/[x0]2 so the slope is -1/(49). Here is the answer as I would write it:
(y-(1/7))=[-1/(49)](x-7).

Definition of derivative
f(x) is said to be differentiable at x0 if limh-->0[f(x0+h)-f(x0]/h exists. If this limit exists, then the value of the limit is called the derivative of f(x) at x0. The usual notation for this limit is f´(x0).

The future and the far future ...

• Maybe an exam is creeping up ... I am trying to follow the syllabus closely. If you look at it, you can see an exam will jump upon us all fairly soon. You may not be able to prepare for the exam in the few hours before it growls at you, and, although I think attending class is important, class attendance alone is not sufficient. You need to actively acquire the knowledge and skills that will be tested. Please do homework. Please read the text. Try to meet weekly with other students in the course and discuss homework.
• The LRC The Rutgers Learning Resource Centers offer free tutoring on every campus. Math 135 has many students and therefore the LRC is accustomed to the questions of Math 135 students. I know a number of the LRC math tutors. They are intelligent and experienced. Please use the LRC if you need help.
• The QotD I offer the QotD as a way for me to judge "how am I doing" and as something similar for you. The questions I offer are representative of questions I might give on an exam. Therefore you might take this opportunity, free of penalty, to work on them. You can even prepare ahead of time. You can work by yourself, or you can consult with neighbors. But please try.
• Preparing for lectures Since this is an introductory course, I am trying to stick quite close to the syllabus and generally I am trying to follow the notation etc. of the text (except when I think I have a much better way!). So you can guess with little chance of error what will be discussed in the next lecture. You can spend a few minutes glancing over that section of the text. You will get more out of the lecture, and probably have the great pleasure of correcting me occasionally. Please try this.

The QotD last time
Is there a number x so that 2x+3x=4x? Give evidence supporting your answer.

• Formal and theoretical Yes, there is an x. Look at f(x)=2x+3x-4x. I compute that f(0)=1+1-1=1, and f(2)=22+32=42=4+9-16=-3. Since f(0) is positive and f(2) is negative, the Intermediate Value Theorem guarantees that this continuous function has a root inside the interval [0,2].
• Computational Yes, there is an x. Its value is (approximately) 1.50712 and I know this because I pushed the buttons on my calculator and got this as an answer.
• Graphical Yes, there is an x. I know this because  I roughly sketched y=2x+3x and y=4x and here's the sketch and you can see that the graphs intersect or cross or whatever just as I claimed.
What is not allowed and cannot be given credit is doubtful or wrong algebra. There is no simple way to write 2x+3x (exponentials just don't work that way!) so there is no way to simplify the equation. You can't do it. Such work is wrong. In particular, I do not know any way to write the solution to the equation in terms of "known" functions or constants: it is not going to be sqrt(Pi), for example. So all such work is unacceptable!

Today's QotD
What is the limit as h-->0 of [sqrt(9+h)-sqrt(9)]/h? I gave the expected value of the answer, which I think is 1/6, and emphasized that I was interested in the method, not just the answer.

HOMEWORK
Please begin to read chapter 3. I can't cover all you need to know from the sections assigned -- there just is not enough time in the lectures.
Please hand in these problems at Thursday's recitation:
2.4: 46, 47, 59
3.1: 14, 26, 34, 42, 47

Animal of the day
The hyrax. A picture of one kind of hyrax is shown to the right. This picture of a "Cape, or large-toothed rock hyrax, PROC AVIA CAPENSIS" is from http://www.americazoo.com/ goto/index/mammals/338.htm where you can learn much more about hyraxes. Aren't Cook students supposed to know about plants and animals?

Vocabulary of the day
The Oxford English Dictionary (which you can look at for free from a Rutgers terminal) states that
osculate means "To kiss (a person or thing), to salute with contact of the lips" and was first used in written English in 1656. In 1728, the word was also used to mean curves being tangent.

I also used the word obfuscate, which occurs in my teaching unfortunately too often. This word means "To cast into darkness or shadow; to cloud, obscure" especially "In figurative contexts" and its first recorded use in written English was in 1536.

### Friday, February 4: this is lecture 6

Another exciting Math 135 class, throwing off enough heat to melt the slush covering every surface outside! Today I'll discuss the exponential and log functions. Much of this material is nominally precalculus, but it is substantial and very useful.

Exponential explorations

1. Which is larger, 10100 or 10010?
The first number, 10100, is larger. In fact, 10100 is the number obtained by multiplying 10 by itself:
```      100 times!
________/\_________
/                   \
10·10·10·...·10·10·10```
and this is 1 followed by one hundred zeros.
10010 is the number obtained by multiplying 100 by itself:
```         10 times!
___________/\____________
/                         \
100·100·100·...·100·100·100```
Now if you count correctly, each multiplication by 100 gives us two more zeros, and the result is 1 followed by twenty zeros.
So the first number is larger.
2. Which is larger, 2(34) or (23)4?
Let's see, how many powers of 2 does the first number contain? It seems to have 34=3·3·3·3=9·9=81 powers of 2. It is 281. The second number is 23·23·23·23 and if we count the number of 2's here we will get 23·4=212.
I know that 281 is much larger than 212 and therefore the first number is larger.
3. Suppose that 5x=3. What is 25x?
I know that 25 is the same as 52. So I think to myself, what is the effect of squaring the equation 5x=3? Well, let's square 5x: (5x)2=52x=(52)x=(25)x. And the square of the right-hand size of the equation is 32=9. So the answer is 9.
4. Suppose that 5x=3. Is x<1/2?
If x<1/2 then 5x<51/2. Because of our assumption, this then becomes 3<51/2. If we square this inequality, we get 9<5. Hey!!! The resulting inequality is wrong. Since this string of inequalities can be reversed, the initial inequality "x<1/2" is also wrong. Therefore x>1/2.
5. Here's a followup question I didn't do this in class:
Suppose that 5x=3. Is x<3/4?
Well, if x<3/4 then 5x<53/4. Again, because of our assumption, this would mean 3<53/4. Now take the fourth power of this inequality, and get 34<53. This I can try to verify with direct numerical computation. 34 is 81 and 53 is 125. So these inequalities are all correct, and the initial assumption, that x<3/4, is correct.
Wow! Of course, the last two questions concern the equation 5x=3. The x which satisfies the equation is called the log of 3 to the base 5. x is the number of powers of 5 which bundled together make 3. The last two questions show that x is between 1/2 and 3/4. Two centuries ago this is more or less how people might have begun computing logs. Now we can just push a botton, and indeed my calculator reports that log53 is approximately .682606 (and I hope that you can use your calculator to get a similar answer).

There are exponential and logarithmic rules which you should already know. Although many of them are "wired into" calculators, the need for people to have some feeling about exponentials remains very important in many applications (compound interest is basically exponentiation, and of course pH calculations are logarithms).

 I had a silicon friend draw the graphs of =2x+3x and y=4x and here they are, and you can see that the graphs intersect or cross or whatever just as I claimed.
Exponential rulesLog rules
AB+C=AB·AC
(AB)C=ABC
A0=1
A-B=1/AB
logAB=C means B=AC
logA(B·C)=logAB+logAC
C·logAB=logABC
logA1/B=-logAB
logA1=0
I hope that you are familiar with these rules. But I need to discuss more specifically topics of interest in calculus.

Here I began by asking people to use their graphing calculators to draw y=3x, and then start zooming in. Three views are shown below:
y=3x for x in [-2,2] y=3x for x in [-.5,.5] y=3x for x in [-.02,.02]
Big things to notice
Please look now at the first picture. It has properties which are typical for exponential growth.

• Only the top half of the plane is used for the graph. 3x is always positive and is never 0.
• The point (0,1) is on the graph. (Yes, so are (1,3) and (-2,1/9), but I want to concentrate on things which are true for every graph involving exponential growth.)
• As we walk from left to right on the x-axis, the corresponding points on the graph get larger: the graph is strictly increasing.
• As we walk far to the left, the graph gets closer and closer to the x-axis. The x-axis is a horizontal asymptote on the left for y=3x.
Now I asked people to zoom in. The curvy exponential graph sort of flattens out. A further zoom yields almost a straight line, when we look at the graph on the interval [-.02,.02]. In fact, some computation shows that the slope of the (almost) line is about 1.01 or so. What the heck is this slope? We're looking at a curve which goes through (0,1) and (x,3x) when x is very small. The slope is [3x-1]/[x-0]. This is a sort of local growth rate of the curve y=3x near (0,1). In fact, I would like to consider limx-->0[3x-1]/x. We can apply various techniques. We could try to "plug in" x=0, but this yields the equivocal form 0/0. We could try experimenting, putting in small numbers for x into the formula [3x-1]/x. Or we could try to plot y=[3x-1]/x. Both of the last two suggest that the limit exists and is about 1.0896.

Now I asked people to do the same for y=2x. Here are some results.
y=2x for x in [-2,2] y=2x for x in [-.5,.5] y=2x for x in [-.02,.02]
The large-scale qualitative aspects of y=2x are very similar to those of y=3x. Now we zoom in, and the curve seems to flatten out. The slope of the resulting (almost) line is about .693. Indeed, we could try to graph y=2x (the green curve) and y=3x (the red curve) together, as shown to the right. You can see that both curves go through (0,1), and that the local growth rate (o.k., the "slope") of the red curve is greater than that of the green curve. We will be doing a great deal of computation of growth rates. If we had to constantly work with .693 and 1.089 these computations would be more tedious.

So what people use who do lots of computations with rates is to find a positive number A so that the "local rate" of y=Ax near (0,1) will be 1. So we will look for an A so that limx-->0[Ax-1]/x will be 1. This number will be something between 2 and 3, because 3x tilts too much (1.089) and 2x tilts too little (.693)

```"This porridge is too hot!" she exclaimed.
So, she tasted the porridge from the second bowl.
"This porridge is too cold," she said.
So, she tasted the last bowl of porridge.
"Ahhh, this porridge is just right," she said happily and she ate it all up.```
The textbook suggests taking small x's (that would appear in the limx-->0 idea) as 1/n, where n is a large positive integer. Then [A1/n-1]/(1/n) will be close to 1 when n is large. We can unroll this and get [A1/n-1] will be close to 1/n when n is large, and further unroll so that A1/n will be close to 1+(1/n). Hey, now take nth powers (all of this is not obvious but has lots of cleverness built in). Therefore the candidate for A will be approximately (1+(1/n))n.

Here is a table somewhat more complete than what I wrote in class.

 n (1+(1/n))n 10 20 30 40 50 100 250 500 1,000 2.59374 2.65329 2.67431 2.68506 2.69158 2.70481 2.71268 2.71556 2.71692

It turns out that there is exactly one number, usually called e, so that y=ex has local growth rate equal to 1 at (0,1). This number is the one which we (and virtually everyone else in the world who analyzes growth and its decreasing counterpart, decay) will use. To the right is displayed y=ex, "sandwiched" by smaller pieces of the curves y=2x and y=3x. The number e is approximately 2.71828 (I am willing to bet that you will never need to know a more accurate approximation, but if you do, here is one: 2.71828182845904523536028747135.)

Financial stuff
The textbook has information about compounding interest and what's called continuous compounding. The relevant formulas are on the top of page 89.

My sloppy lab notebook
I could imagine that I am observing a colony of bugs, and I will assume that the number of bugs in the colony increases exponentially. Such assumptions are sometimes correct for short periods of time, and reflect the biological idea that the rate of increase of the population is directly proportional to the population (if more bugs, then more bugs are born). So here is a section of my lab notebook.

9 AM23
10 AMCocoa  stain!!!!
11 AM121
Lunch  mess!! 188

Can we reconstruct the data that's been lost? Under the initial assumption, the answer is yes. If we assume that B(t), the number of bugs at time t, is Cer t, for C and r "unknown" constants, then what do we know? First, I will clarify things a bit, and measure t in hours starting from 9 AM. Thus B(0)=23. But since B(t)=Cer t, I know that B(0)=C (since e0=1). Therefore B(t)=23er t. But we also know that at 11 AM there are supposed to be 121 bugs. So B(2)=121, and in terms of the assumed formula, 23er 2=121. Now "solve" for r:
er 2=121/23
Take "logs" (actually, in terms of the way your book is written, we should take "lns", natural logs, which are logs using the base e): r 2=ln(121/23).
So r=(1/2)ln(121/23) which my calculator tells me is approximately .830148.
Now I have a "mathematical model" for my bug population:
B(t)=23e.830148 t
Under the cocoa stain is B(1)=23e.830148 1, which is about 52.75.
What time was lunch? We need to solve 188=23e.830148 t, and this gives 188/23=e.830148 t and then ln(188/23)=.830148 t so that t=[1/(.830148)]ln(188/23) which is approximately 2.5308, so lunch was slightly after 11:30.

QotD
Is there a number x so that 2x+3x=4x? Give evidence supporting your answer. This evidence could be numeric, or graphical, or even quoting some theoretical result of the course, but I'd like some evidence.

Decay
I did not have enough time in class to discuss typical decay curves. These are y=Ax where 0<A<1. They are related geometrically to growth by reflection across the vertical axis. For example, y=(1/2)x has the point (5,1/32) on it, and on y=2x the corresponding point is (-5,1/32) (I'm using A-B=1/AB.) Notice that (1/2)x=2-x. So flip all the growth curves to get decay curves. Decay curves also use only the top half of the plane, and they decrease to 0 as x-->infinity.

HOMEWORK
Finish reading the parts of chapter 2 we need, and please do some homework problems.

### Tuesday, February 1: this is lecture 5

I tried to go over some textbook problems to illustrate the result limx-->0[sin(x)]/x=1. I explained that I thought of this as saying, more or less, that sin(something small) divided by the same something small is fairly near 1, and that the algebraic manipulations which followed were all movitated by this. I also remarked that although we will use and need (tested on exams!) this result in Math 135, the utility of this result in students' future endeavors (I'm trying to be as pompous and incomprehensible as I can) is probably not large. Sigh.

Section 2.2, #23

```
sin(2x)      sin(2x)      sin(2x)
------- = ------------ = --------· 2
x         2x (1/2)        2x```
Therefore limx-->0[sin(2x)]/x=limx-->0{[sin(2x)]/(2x)}·2=1·2=2. My efforts were entirely trying to get 2x "underneath" sin(2x) as x-->0, because then I could apply the logic of sin(something small) divided by the same something small is fairly near 1." The result gives me a 2 upstairs, and this turns out to be the result of the limit computation.

Section 2.2, #25
What is limt-->0[tan(5t)]/[tan(2t)]? Again, let me show you how I tried to think through this problem.

``` [tan(5t)]     [sin(5t)/cos(5t)]      sin(5t)·cos(2t)
----------- = -------------------- = -----------------
[tan(2t)]     [sin(2t)/cos(2t)]      sin(2t)·cos(5t)```
The first algebraic "thing" I did was motivated by my intense personal dislike of compound fractions. I find them complicated and sometimes difficult to understand. Of course, it is possible to make errors when changing the compound fractions into simple fractions, so some alertness is needed. Now here is what I did with the simple fraction:
``` sin(5t)·cos(2t)     sin(5t)     cos(2t)       1
----------------- = --------- · --------- · --------
sin(2t)·cos(5t)        1        cos(5t)     sin(2t)```
Now what? Well, I have a picture of cosine installed in my brain, and therefore I know that cos(some small number) is close to 1. So I don't need to think too much about cos(2t) and cos(5t) as t-->0. But I want 5t underneath sin(5t) and 2t over sin(2t). When I put them in, I need to compensate:
``` sin(5t)     cos(2t)       1         5  sin(5t)  cos(2t)      2t
--------- · --------- · --------- = --- -------- --------- ---------
1        cos(5t)     sin(2t)     2    5t      cos(5t)   sin(2t)```
Now I can analyze the limit. The first chunk (o.k., more properly, the first quotient) gives me 5/2. The next is a sine of a small number divided by the same small number, which gives me 1. And the cosines just give me 1. And the last quotient also gives me 1. So I bet that limt-->0[tan(5t)]/[tan(2t)] will be 5/2.
Note In this course we'll usually ask for some justification answers to exam questions. What I tried to show above are, I hope, examples of adequate justifications of the answers.

A function is continuous if its graph has no jumps or breaks. For example, the crazy numbers in the U.S. tax table (Schedule X) occurred because as a consequence of public policy, the tax function should be continuous, and the graph should not have any breaks or jumps.

Suppose f(x) is defined piecewise by

```     { 2x-1 for x<=3
f(x)={
{ x2+A for x>3```
Can we find a value of A for which f(x) is continuous at x=3?

First I wanted to graph this function, since I, at least, think better when I have a picture to reason with. When x<3, f(x)=2x-1 and this is part of a straight line whose slope is 2 and whose y-intercept is -1. I just need the half-line which is to the left of x=3.

To the right of x=3, we need to think about f(x)=x2+A. Of course, y=x2+A is a parabola whose axis of symmetry is the y-axis. Its "vertex" (lowest point, the parabola opens up) is at (0,A). The graph of y=f(x) for this f(x) takes only the part of this parabola which is to the right of x=3.

The left-hand limit of f(x) at x=3 uses 2x-1, and the value of this limit is 2·3-1=5. This is also equal to the value of f(3). The right-hand limit uses x2+A, and as x-->3-, x2+A-->9+A. To make all three of these coincide (5 and 9+A and the "other" 5) we need just to choose A=-4. So this function will be continuous at x=3 exactly when A is chosen to be -4.

The last QotD
I asked for the graph of a function with the properties that

• The domain of f(x) is all real numbers, f(2)=5,
• limx-->2+f(x)=4, and
• limx-->2-f(x)=-1.
One example of such a graph is shown to the right. Note that all of the three pieces are separated. I could not make this function continuous by moving only one of the three pieces.

Precise definition of continuous for Math 135
A function f(x) is continuous at x=a if limx-->af(x) exists and equals f(a).
This means that limits for continuous functions can be evaluated just by "plugging in", so these functions are nice and tame. There is one really important theoretical consequence of continuity which almost everyone uses unconsciously. Let me lead up to this result (the Intermediate Value Theorem) by describing driving on one of New Jersey's jewels.

The Garden State Parkway, from Cape May to Montvale and my friend Francine ...
I then discussed the Garden State Parkway. We spent quite a lot of time on the question of the length of the parkway. Mile 0 is at Cape May, while the other end, mile 172, seems to be close to Montvale. Suppose that my friend Francine leaves Cape May at 7 AM one morning, and drives north on the Garden State Parkway. Further, suppose she arrives at mile 172, the northern end, at, say, 10 AM. Must Francine at some time be at mile 135 (fairly near Busch campus)? The parkway seal below was "borrowed" from a State of New Jersey webpage.

I discussed various curves which could represent the position of Francine on the parkway in terms of miles from the start of the parkway at time t, in terms of hours elapsed from 7 AM. I tried to show that our everyday intuition lead to the graph being increasing (as you travel from left to right, the points on the graph go up). The graph can have level spots, where Francine pulls over for a rest stop. Legally Francine isn't supposed to drive backwards, though.

If we believe that motion is continuous (so Francine does not have a Star Trek transporter or other device) then the graph of Francine's position goes from (7 AM, 0 miles) to (10 AM, 172 miles) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, rests on some complicated assumptions, some of them philosophical (why should motion be continuous?). Today, though, I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. By the way, I will retain this information for later, when we analyze the rate of change of position (velocity) so that we can see whether Francine deserves a speeding ticket.

The Intermediate Value Theorem
Suppose that the function f is defined and continuous on the interval [a,b]. Then the equation f(x)=y has at least one solution for every y which is between f(a) and f(b).

Math 135 example for the Intermediate Value Theorem
Look at cos(x) and x+x3 on [0,Pi/2]. I claim that there's a number x in the interval from 0 to Pi/2 where cos(x)=x3.

I tried to analyze this first by showing you what I had in mind when I thought initially about this problem. Well, Pi/2 is approximately 1.57, and (1.57)3 is approximately 3.86 or so (yes, I used an electronic friend for this). So if you look at the picture, you should see that the curves cross. Hey! Wow!!

Now to give the answer that Math 135 will recognize. Consider the function f(x)=cos(x)-x3. Then f(0)=cos(0)-03=1 which is a positive number, and f(Pi/2)=cos(Pi/2)-[Pi/2]3=0-[Pi/2]3, and this is a negative number. The functions involved (x3 and cos(x)) are both continuous. Then the Intermediate Value Theorem applies to this function f(x) and the interval [0,Pi/2], and shows that there exists at least one number x with f(x)=0. But f(x)=0 means cos(x)-x3=0 which means in turn that cos(x)=x3.

Comment This is what is known as an "existence result". I don't think anyone would feel that the root has really been discovered or approximated in any meaningful way by what we've done here. In fact, though, exactly such ideas are used by your hand calculator to find approximate solutions to equations. The calculator first "looks" for where the graphs cross each other, and then the calculator tries to sort of "localize", and get better and better approximations to the roots. The sort of approximation works well in a surprising variety of examples. The number x=.8654740331 comes out as an approximate root of cos(x)=x3 very rapidly from a machine.

QotD
Try to explain why the equation tan(x)=x must have positive solutions.

HOMEWORK
2.2: 14, 18, 24, 43
2.3: 12, 19, 28, 43

### Friday, January 28: this is lecture 4

 lim  f(x) = L x-->a

We looked at a complicated, ferocious, horrible limit, and a nice baby lamb limit.

One example
What is the limit as x-->0 of sqrt{45x34+ [1/sqrt{20x14+16}]+57x88 }?
This might look like a ferocious limit to people. I would try not to get excited or anxious, and instead "parse" this expression systematically. When x-->0 (which I read internally as "x gets close to 0"), I look at, say, 45x34. In spite of the decorations (?) of 45 and 34, this is really a simple and reasonable function. As x gets close to 0, I think that the 34th power of x gets close to 0. And multiplying this by 45 doesn't change that situation much. So this piece of the "ferocious function" just -->0. The same is true about 57x88. But what about 1/sqrt{20x14+16}? Well, one piece of this is 20x14 which also -->0. In fact, this bewildering (?) limit is mostly just decoration: all the pieces -->0 except for the 16 which is internal to two square roots and one reciprocal. Hey: I bet that the limit exists, and that its value is sqrt{1/[sqrt(16)]}. By the way, I would be happy with this sort of answer on an exam. If you must "simplify", I guess another valid form of the answer is 1/2.

Another example
What's the limit as x-->0 of x/|x|? This is supposed to be the "baby lamb" limit, I guess, because it such minimal typography. In fact, to me this limit is much more complicated than the previous one. The absolute value signs indicate that, to be really sure of the result, I'd better analyze this limit in a piecewise fashion. So if x>0, I know that |x|=x, and then x/|x| is 1. If x<0, then |x|=-x, and x/|x|=x/{-x}=-1. So I hope you have a picture in your head of the graph of x/|x|. It is -1 to the left of 0 and +1 to the right of 0. The notation "x-->0" is supposed to indicate no bias between left and right: both sides should go to the same limit. Here that is certainly not the case. The left-hand limit is -1 and the right-hand limit is +1. Therefore the standard (two-sided) limit does not exist. This "baby lamb" limit is actually much more badly behaved than the first example, in some really profound sense.

The notation "limx-->a" denotes a two-sided limit. If we wanted a left-handed limit, what's usually written is limx-->a- and if we wanted a right-handed limit, the notation limx-->a+ is used.
So limx-->0-x/|x|=-1 and So limx-->0+x/|x|=1.
The two-sided limit exists when both one-sided limits exist and when they agree. There are more examples in your textbook.

I think I tried one more example, something like, "What is the limit of x+{|x-2|/(2-x)} as x-->-2-?" First, the notation is somewhat confusing if you are new to this game, since "-2-" has two minus signs. The minus sign in front of 2 indicates a negative number, two units to the left of 0. The minus sign up in the corner shows that I want to approach -2 from the left. So what can I do? If x-->-2-, then x<2, and what can we say about x-2 itself? Since x<2, x-2<0, so x-2 is negative and |x-2| is -(x-2). Therefore x+{|x-2|/(2-x)} becomes x+{-(x-2)/(2-x)}=x+(-1(-1))=x+1, and as x-->-2-, x+1-->-2+1=-1. The left-hand limit of x+{|x-2|/(2-x)} at -2 is -1.
You can check that x-->-2+ the limit of x+{|x-2|/(2-x)} is +1.
Serious note I've done lots of these computations. Even when I am alone, I tend to use the inequality approach that I showed above. I know that I can guess, and I know that my guess will frequently be correct. But I've learned that sometimes my guesses will not be correct, and this can be irritating.

The actual definition of limit is on the bottom of page 56 of your text. I won't discuss the definition here, but I should mention that the definition is quite precise, and the definition was arrived at after probably a century of controversy about "What is (or should be!) a limit?" People made many mistakes, and a precise definition was adapted so that fewer mistakes and misunderstandings would occur.

Section 2.2 of the text (page 61 and what follows) discuss a number of algebraic properties of limits (the limit of a sum, the limit of a product, etc.). You should read through these and use them. These rules are exactly what I used to "tame" and evaluate the initial ferocious limit in this lecture.

The QotD yesterday provides an example of how algebraic manipulation can be used. I asked people to consider

```1   1
- - -
x   2
-----
(x-2)```
as x-->2. If you try what I always try first ("plug in") the result is 0/0, which is at best equivocal (I don't know what the heck it means). We could also try graphing this mess. Or we could try some simple algebra:
```1   1      2-x
- - -    ------
x   2     (2x)        (2-x)       -1
----- = -------- = ---------- = ------
(x-2)    (x-2)      (2x)(x-2)    (2x)```
There are several comments to make about the stages in this algebraic process. First, I dislike "compound fractions" and I almost always try to get rid of them. That's why I "pushed together" the two pieces in the top of the original fraction. Then I needed to know how to change the compound fraction (two ------'s) into a simple fraction (one -------). I did that, carefully. The last step is one which is logically a bit perilous. If x is not equal to 2 (so x-2 is not 0) then I can divide the top and bottom of the fraction by x-2. I need to be careful, and I will get a minus sign on the top. But I remember that that phrase "x-->2" means "x is close to 2 but x is not equal to 2". So the original
```1   1
- - -
x   2
-----
(x-2)```
is actually (away from 2!) equal to
```  -1
------
(2x)```
. The second form is easy to understand as x-->2: it must get close to -1/4. Therefore the limit exists and its value is -1/4.

There are a few common algebraic tricks. If you practice with the textbook homework problems and with WeBWorK, you will be in good shape for what we need to do with this course.

A function you know all about
Suppose I call S(x) the squaring function. So S(x) is defined by S(x)=x2. I think you know this function well, and you know what its graph looks like (a parabola with "vertex" at the origin, opening up, symmetric with respect to the vertical axis). Certainly we know limits involving S(x). So I know that limx-->2S(x)=4 and limx-->5S(x)=25 and limx-->-3S(x)=9.

A mild change
What if I changed one value of S(x)? That is, let me define the function T(x) by the following piecewise statement:

```     {x2 for x not equal to 7
T(x)={
{-4 if x=7```
So now T(2)=4 certainly and T(5)=25 certainly, and, with this silly definition, T(7)=-4.
What is limx-->2T(x)? Well, for x close to 2, T(x) is the same as S(x)=x2 and therefore limx-->2T(x) is 4. What is limx-->7T(x)? Well, don't jump at conclusions, please. If x is close to 7 but not equal to 7, then T(x) is the same as S(x)=x2, and therefore limx-->7T(x) exists and its value is 49. In fact, changing one value of the original function S(x) does not change any of the limits of S(x)!
The next thing I did was change two values of S(x). That didn't seem to bother folks very much. So I jumped to

Aliens invade earth and change the squaring function!
Suppose that aliens invaded the earth, and took the function S(x)=x2 and changed ten billion values of S(x). That is, they created a new function, let me call it V(x), so V(x) has the graph of S(x) except that ten billion of the points on the graph have been moved, some up, some down. The domain of the function V(x) is still "all real numbers" and V(x) squares all of the undisturbed points of its domain. Otherwise V(x) does something alien. What do we know about the limit of V(x) at, say, x=1? That is, what can we say about limx-->1V(x)? One thing you don't need to worry about is V(1). V's value at x=1 may have been changed by the aliens, but the existence and value of the limit itself does not depend on V(1). So don't worry about it. What happens as x-->1? Well, we might pass some values of S(x) which have been changed by the aliens. For example, between x=1.1 and x=1.01 we might pass by, say, a few million of these changed values. Then between x=1.01 and x=1.001 we might go by some more. Etc. But the aliens only changed ten billion or so values, and eventually we will "pass" them all. There is infinitely much room to the right and to the left of x=1. There are only finitely many values which were changed. Eventually we get close enough to x=1 so that there are no more changes to be observed, and V(x) is the "old" S(x)=x2 there. Hey: this means that limx-->1V(x) exists, and is the same as limx-->1S(x), and this limit is 12.
In fact, the same logic works at any point! limx-->wV(x) will always exist, and it will always be the same as limx-->wS(x)=w2. You may need to think this through quite a bit. I can change the graph of S(x)=x2 in any finite number of places, and I won't change the limits at all! In order to change limits, I need to change an infinite amount of data, which is quite different.

QotD Draw me a graph of a function with the following properties:
The domain of f(x) is all real numbers, f(2)=5, limx-->2+f(x)=4, and limx-->2-f(x)=-1.

I did some problems in the textbook. Let's see if I can remember:
#5
What is

```      z2+z-3
lim --------
z-->1   z+1```
Here I can just "plug in". There are no divisions by 0 to worry about. The answer is -1/2.
#11
```        4-u2
lim  --------
u-->-2  2+u```
Of course here the top can be factored, and we get (2-u)(2+u) so that, away from where 2+u is 0 (so u should not be equal to -2) we can divide top and bottom by 2+u and we get 2-u as the result. Now as u-->-2, 2-u-->4. So the limit exists and its value is 4.

Limits and inequalities
This was not well done in class. I rushed too much. I regret the rush. Covering lots of stuff rapidly is much worse than doing a few things well.

 f(x)<=g(x) If I know that f(x)<=g(x) for all (relevant) values of x, and if I know that the limits exist, then limx-->af(x)<=limx-->ag(x). I think a picture supplies some substantiation, but I'll need to use this later in the course, and will return to it then.

 I also will need the Squeeze Theorem. f(x)<=g(x)af(x) exists and limx-->ah(x) exists (the limits for the top and bottom functions) These two limits are equal. then limx-->ag(x), the middle limit, exists and equals the common values of the other two limits. Here the top and bottom functions are usually going to be something nice, and the middle function will be something weird.

This early in the course the major use of the Squeeze Theorem is to analyze how sin(h)/h "looks" when h is a very small number. If you have a picture of sine then you know that sin(h) as h-->0 goes to 0. So sin(h)/h is definitely a "0/0" situation, and it is therefore one whose behavior can't be predicted, and whose limiting value, even if we assume the limit exists!, can't be predicted. Please believe me when I declare that sin(h)/h is an important 0/0 situation.

I did not use the Squeeze Theorem. For that, please see pages 67-68 of the text. I asked students whose last names began with the first half of the alphabet to use their graphing calculators to sketch sin(h)/h with radians, and I asked the students in the second half of the alphabet to sketch sin(h)/h in degree mode.

Degrees: the "top" is about .01745 Radians: the "top" is exactly at 1
This is part of the graph of sin(h)/h where I have forced the graphing program to evaluate sine in degrees. Look closely, please, at what happens as h-->0. The values of the function seem to be around .01745, and indeed, if you insist upon working with sine in degrees, the limit will be that rather unpleasant number. Here is a picture of sin(h)/h where h is measured in what turns out to be much more natural units, "radians". The word is funny, but it just means the length of a piece of circular arc whose radius is 1 and whose angle is designated. This is a sort of "native" measurement, and you should please consult the textbook for a picture explaining why sine of a very small angle in radians is quite close to that angle itself. This is very neat, and very nice to use in practice.

In this course I would rather cope with 1 than with .01745, so we will always use radian measure in calcuous applications.

### Tuesday, January 25: this is lecture 3

Brave students volunteered and did the first part of the previous QotD. I will not name them because then they may be pursued endlessly by fans. And, anyway, they didn't totally volunteer, but rather were gently persuaded.
Exercise E
Consider E(x), which is defined by E(x)=1/K(x), where K(x) was my pet function.

BEGIN: PEDAGOGICAL RANT
Most of the students in this course will sometimes need some of the ideas covered by the course. But if the stated majors or intended majors are correct, essentially all of the students in the course will need to be able to read and understand the graph of a function. This is, perhaps, a pre-calculus skill, but the kind of understanding I mean incorporates much sophistication. Being able to sketch E(x) with comprehension of what's going on is one example of what I mean by "read and understand the graph of a function." You'll need this. Please learn it.
END: PEDAGOGICAL RANT

We have some precise information about K(x) in terms of four specific points on its graph. I'd first try to exploit this precise information as well as I could. For example, we know that K(1)=2 since (1,2) is on the graph. Therefore, E(1)=1/K(1)=1/2, so that (1,1/2) is on the graph of E(x). Also K(2)=4, so E(2)=1/K(2)=1/4, and (2,1/4) is on the graph of E(x). Let me pause here, and ask if we can infer more information for x's between 1 and 2. If I understand what the graph indicates, I think that K(x) gets bigger (the "technical" word, I guess, is increases) as x moves from 1 to 2. I know, darn it, that the idea of x moving is silly (what the heck, there's the graph, what is "moving" about that picture?). But I do think that way, and I find it useful sometimes, and maybe you might also. So x goes from 1 to 2, and y, on the graph of K(x), goes from 2 to 4. What about 1/y? When y is positive, 1/y, the reciprocal of y, is also positive. As y increases, 1/y decreases. So the "chunk" of the graph of E(x) for x between 1 and 2 decreases from 1/2 to 1/4.

Now the nice precise piece of information: E(3). Well, E(3)=1/K(3)=1/0 ALARM!. We don't divide by 0 in Math 135. In fact, here is the initial segment of

 The official list of good manners in Math 135 Do not divide by 0. Do not take square roots of negative numbers.

These manners must be minded when we worry about the domains of complicated algebraic functions. That won't often be the object of the game, so don't get maniacal.

But E(3) has no value! We can, however, deduce much about the behavior of E(x) by observing the qualitative aspects of the graph of K(x). The graph goes from 4 to 0 as x goes from 2 to 3. The y-values decrease and are all positive. What about their reciprocals (1 over the y-values)? Again, they will be positive. And 1/{a smaller number} (if the numbers are positive) gives {a larger number}. So the values of E(x) increase on the interval from 2 to 3. How high do they go? Well, that's a more subtle question. Look again at the given graph of K(x). If you believe it, and "look" closely to the left of x=3, you should see the graph diving down towards 0. IF you thought, for example, that the graph of E(x) would never get bigger than, say, 100, that would be you believed 1/K(x) would be less that 100 so that K(x) itself would be greater than 1/100. (This is what I meant by sketching "with comprehension" -- this is work!) But, golly, I understand that the graph of K(x) does get closer and closer and closer to 0, and doesn't stop at 1/100. This means that E(x) should get bigger and bigger and bigger: it should have no upper bound, and, in fact, x=3 should be a vertical asymptote of the graph of y=E(x). Whew!

What happens in the last chunk, for 3 to 4? Since K(4)=-1, E(4)=1/-1=-1. Because K(x) is negative in 3<x<4, E(x) must be negative also. But you've got to again think carefully and critically. Let me make x move "left" from 4 in this case, move left from 4 to 3. K(x) will then move from -1 towards 0, always being negative. 1/K(x) will be negative, and its magnitude (size!) will be the reverse of K(x)'s: K(x) small <==> E(x) large. So as x moves from 4 to 3, K(x) moves from -1 to 0 and E(x) "moves" from -1 to ... very large negative. How large? Well, golly, as large as "you" want: E(x) will have no lower bound, and will "tend towards" what some folks call "minus infinity". Again, the graph of E(x) has the line x=3 as a vertical asymptote on the right and on the left of x=3.

Finally, we read off the domain and range of E(x). The domain was [1,3) and (3,4]. The range was [1/4,infinity) and (-infinity,-1]. Sketching E(x) was a difficult problem.

 Exercise F Consider F(x), as defined by F(x)=K(1/x). Here all of the precise information in the table of K(x) can be used to produce "precise information" (points on the graph of F(x)). The only secret is to realize that in order to get the desired inputs to the K(x) function, we need to input the reciprocal numbers to the F(x) function.``` x | F(x) ------------------------------------ 1/4 | F(1/4)=K((1/(1/4)))=K(4)=-1 1/3 | F(1/3)=K((1/(1/3)))=K(3)=0 1/2 | F(1/2)=K((1/(1/2)))=K(2)=4 1/1 | F(1/1)=K((1/(1/1)))=K(1)=2``` Yes, I know that 1/1 is just 1 but it seemed more natural to me to write it as 1/1. We can now graph these 4 dots and try to interpolate the remainder of the graph of F(x). The logic is parallel to but not identical with what we've just done. For example, as x "travels" from 1/4 to 1/3, 1/x goes from 4 to 3 (backwards?), and K(x)'s values go from -1 to 0, increasing. Each of the other intervals is similar. The domain of F(x) is [1/4,1/1]. The range is [-1,4], the same as the original range of K(x). So the "picture" is flipped, left to right, and then there is some strange proportioning going on: the stretching/compressing isn't the same in different intervals.
Complaints
You may object to the transformations I have made to K(x)'s graphical information. I will respect your objections, but to me they mostly show naivety, a lack of experience. In particular, applications I have seen in biological contexts do much more complicated "things" than what we've discussed here. But you will learn by doing more examples.

Mostly this course is about myths and, not lies, precisely, not exaggerations, but certainly idealizations of reality, which may not exist in any strict sense. I discuss this briefly before we fling ourselves into the ideal swamp of calculus.

• Tree growth rates Poor Quercus imbrecaria, whose life I began by discussing. This was a tree whose growth rate was, let us say, 1 foot per year. What the heck does this really mean? I don't anyone in class believes this means a real tree grows 1 foot per year, or that it would grow an inch each month, or maybe 1/365 foot each day (and be smart enough to know to grow more slowly in leap years!). Well, more or less, if you do lots of measurements, and if H(t) is the height in feet at time t measured in years since germination, then, errr, maybe [H(t2)-H(t1)]/[t2-t1] should be about 1 when t1 and t2 are close. More or less.
• Marginal cost Most car manufacturing companies today are huge. They make millions of cars. One concept, a sort of a rate concept, which is foundational in economics, is called marginal cost. An extremely good definition of this was offered by a student today in class and I would really like to give her name, but, darn it!, I don't know who you folks are! So let's see if I can do this correctly: suppose D is the amount in dollars a car company invests to produce P(D) cars. So, for example, D might be \$1.2 billion, and that might produce 5,000,000 cars. In a formula, with usual math notation, that's P(1,200,000,000)=5,000,000. How much would it cost to produce 1 more car? To a certain extent, that is a ridiculous question. I mean, it reminds me of, let us say, a very generous eater looking over at a thin person eating a lunch of an apple and a carrot and saying, "Hey, I drip more than that, kid!" There is no realistic way to know actually what it would cost to make 5,000,001 cars, I think. But that ficticious (?!) amount is called the marginal cost, and people work hard to understand this number, and to compute it. It turns out that marginal cost is also a sort of rate, and we'll come back to it. By the way,Google's first definition of marginal cost is:
The increase or decrease in a firm's total cost of production as a result of changing production by one unit.
I hope that helps.
• Rates and drugs If any of you have gotten any sort of prescription medication and looked closely at the informational enclosures, you will fill a bewildering array of information. A number of students in the class indicated they were interested in pre-med or pre-vet studies. Part of med/vet school is trying to learn, understand, and then be able to use even much more information than this about how drugs are absorbed in biological systems, about different rates of drug efficacy, etc. All of the "rates" though, are gotten by combining some sorts of numerical averages with theoretical analysis, and most are really more or less fictional. Wow! Why should people care?

All of my warnings are correct. More or less, all of these rates are indeed fictitious. But everyone who works in those areas uses these rates as "truth" and the uses are almost always very convenient, and very useful, indeed. So the ideas turn out to be useful fictions about rates of change. I'm going to try to follow the text now and briefly postpone a direct approach to rates by looking at something that might appear to be simpler.

Functions on the edge
What can we say about the behavior of functions near where they are not defined? This is a bit silly, as written, and I need to be much more precise. But first let me give two examples, one geometric and one algebraic, and then try for a general description.

A geometric example
To the right is the graph of the function Z(x). I choose, by the way, to specify only one point exactly on this coordinate plane, and that point, (2,1), is supposed to be the one point which is a hole in the graph!

So you are supposed to see, in this case, a function whose domain is all of the real numbers except for x=2. Aside from that the graph is not remarkable at all. My question for you to ponder is: what happens to the values of Z(x) as x moves towards 2? Again with the "kinetic" words: I am sorry, but that's how I think of it. O.k., rephrasing, if you think of Z(x) as a machine with an input and an output (no, not a biological model with an input and an output, too easy a joke!) then we could imagine putting in values of x close to 2 and inspecting the values of Z(x) which come out, and trying to see if there is any "simple" sense which can be made of them. Please, the picture and the question are probably (I hope!) almost too easy. I hope that you look at the picture and almost can declare that the values of Z(x) for x close to 2 (but not x=2!) are close to 1. That's the answer I want.

Important!
One thing I could use to confuse you: I could have given Z(x) a value at x=2. Below that's the violet (magenta?) dot in the graphs. Then (this is important!) any value I might choose to assign to Z(2) won't change the answer to the question I originally asked, "What happens to the values of Z(x) as x moves towards 2?" or to the thought experiment, "Put in values of x close to 2 but not equal to 2 and inspect the results." So the answer would be the same if the graph of Z(x) were any of the pictures below. I did not emphasize this enough in class.  Same answer and same answer and again, the same answer!

An algebraic example
I defined the function Q(x) by the equation Q(x)=[sqrt(x)-2]/[x-4]. Now and in the remainder of the course, when a function is defined by an algebraic formula, the domain to be used is "all the x's that allow everything to work". In this case, keeping in mind the restrictions imposed by our manners stated previously, we see that because of the square root, we can only have x>=0. But also x-4 should not be 0. So the domain of Q(x) should be [0,4) and (4,infinity). What happens to Q(x) when x gets close to 4? We could try various approaches. One might be to plug in numbers, not that there's anything wrong with that. I don't especially want to compute values of Q(x) because that seems so ... "dirty". Wait. I have a silicon friend who has volunteered. My friend gives me the following evidence:

```  x     |    Q(x)
-------------------
3     |  .26795
3.5   |  .25834
3.9   |  .25158
3.99  |  .25016
5     |  .23607
4.5   |  .24265
4.1   |  .24846
4.01  |  .24984```
The first 4 lines of the table give x's getting close to 4 from the left, and the next 4 lines five x's getting close to 4 from the right. Please look for the patterns, and see if the values of Q(x) seem to be getting close to "something".

We will use algebraic manipulations to be even more certain that the values are getting close to a unique "something", and to identify with precision what the "something" is. There are various tricks, almost all of which have been in use for hundreds of years. Here is one version that works with Q(x):

```      [sqrt(x)-2]   [sqrt(x)+2]    [{sqrt(x)}2-22]        [x-4]             1
Q(x)= ----------- · ----------- = ---------------- = ---------------- = -----------
[x-4]      [sqrt(x)+2]   [x-4][sqrt(x)+2]   [x-4][sqrt(x)+2]   [sqrt(x)+2]
```
Therefore, if x is not equal to 4, Q(x) can also be described by the formula 1/[sqrt(x)+2]. I am trying to be almost excruciatingly careful here. The reason I added the phrase "if x is not equal to 4" is that in the algebra above we canceled x-4 on the top and bottom. This step is only valid if the formula canceled is not 0.

The advantage to having an alternative description of Q(x) is that maybe this one is not so sensitive to behavior near 4. Indeed, if we look at 1/[sqrt(x)+2], I hope it is not difficult for you to see that when x creeps (?) close to 4, this formula creeps close to 1/4. So Q(x) gets close to 1/4 when x gets close to 4, and changing the formula algebraically is one way to see this.

More from my silicon friend
O.k., my pal is electronic. But let us see what else it could tell us. I asked for my friend's response to a request for Q(4.000000001) (yeah, that's 4 plus a little bit, where the little bit is 1/1,000,000,000). If the algebra is correct, we'd hope that the response is quite close to 1/4. The actual answer I got is 0. (Huh?) On the other side, when Q(3.99999999) was requested, the answer was .2, again not very close to 1/4. What is going on? The computations are being done with about 10 decimal digits of accuracy. I'm asking much to much of the program to give me accurate answers at the very edge of its powers. You may need to keep this in mind when you use electronic computational help.  The wonder of human vision There's another way to use computational data. Tabular data is nice, but the human brain, I have been told, has an enormous amount of its processing power devoted to analysis of images. We can ask that machine produce a picture of Q(x). One picture is shown to the right. The "machine" creates the picture by taking a bunch of data points and connecting the dots. The chance that one of the data points is a sample at exactly x=4 is small, so the machine isn't likely to run into much trouble. And human eyes like the picture, and the picture certainly seems to confirm that when x gets close to 4, Q(x) gets close to 1/4.

Limits!!!
Well, I wrote three "excitement marks". The phenomena we have been discussing are called limits. There are standard phrases to say, and standard ways to write what we have been doing.

In the first example, we would say that, "The limit of Z(x) as x approaches 2 is 1." The notation used is limx-->2Z(x)=1. I like the little arrow underneath. In the second example, the notation is limx-->4Q(x)=1/4.

The whole limit idea is supposed to encapsulate the behavior of a function near, but not necessarily at the value. Here's where I recommend you begin reading chapter 2 of the text book.

Taxes, again?
I tried to revisit our two-bracket example of a tax function. So:

```      { (1/10)x if x<=10,000
T(x)= {
{ (2/10)x if x>10,000```
What can you tell me about limx-->10,000T(x)? We discussed this politely, as always (why does the guy in the front of the room always talk so loud, though?). If x is near 10,000, and x is above 10,000, then T(x) will be near 2,000. If x is near 10,000, and x is below 10,000, then T(x) will be near 1,000. So there does not seem to be one unique number that describes any sort of clustering tendency for T(x) when x is near 10,000. Therefore, I will say that limx-->10,000T(x) does not exist.

Yes, there are "one-sided" limits: look at the text, and I will mention them next time. But the genuine article, the two-sided limit, the one that's supposed to somehow, all by itself, represent T(x) for x near 10,000, does not exist. I remind you that this T(x) formula was one we had to fix, because it also did not adequately reflect certain societal requirements. Tricky, tricky!

QotD
Look at the quotient

```1   1
- - -
x   2
-----
(x-2)```
and use algebra to try to find the limit as x-->2 of this formula.
Several students came up after class with very good questions inspired by this QotD. Wasn't I trying to teach them how to "compute" 0/0 and, if so, should it be 1, or shouldn't it be ... what? This is certainly something that you could consider. But I am not dividing by 0, I am just trying to understand what happens to the values of certain functions near certain "defects" in their domains. More next time!

HOMEWORK
1.3: 12, 50, 61, 64
2.1: 1, 2, 5, 6
You will need to read 2.1 to understand the notation in some statements of the problems.
Word of the day Efficacy: capacity or power to produce a desired effect.

### Friday, January 21: this is lecture 2

The QotD (Question of the Day) I asked last time was, "Why are you taking this course?" I urged people to write clear responses.

24 students indicated they were taking the course because it was "required". 32 wrote specific majors which needed calculus. Another 25 just remarked that Math 135 was needed by some unspecified major.

One brave student wrote that the course would "help me be successful", and another was taking the course "to get an A": I guess these were the optimists.

The idealists were few. A few students wrote such statements as, calculus is "kinda fun", "I enjoy math", and, better, "I love math".

So the overwhelming number of students are in Math 135 due to some outside compulsion: a lovely situation.

My job here is to teach math, and, I hope, to show you that the material is interesting and relevant. Let us see if I succeed. I certainly hope I will succeed.

I wrote Schedule X of the U.S. income tax form 1040 on the board as a function (look at the end of the last lecture). It is quite fascinating, and the details are complicated to me. By the way, I will mention Schedule X to any one who complains later in the course about the complexity of a computation! Understanding T(x), the Federal tax owed, as a function of the individual's taxable income takes some effort. I first wanted to assure myself that people understood how to compute T(x), so we discussed what the tax owed on an income of \$35,000 would be: T(35000) would be computed using the third line of the table. It would be 4000 plus 25% (a quarter) of the amount over 29050, so 4000+(1/4)(35000-29050) which a calculator tells me is 4000+1487.5, or \$5487.50. But as we consider the information defining T(x), there seem to be a bunch of almost random-looking numbers scattered. I mean, why is there 14,325.00? Why is there \$35,717.00? The formulas for T(x) are all pieces of straight lines, which is why such a function is frequently called a piecewise-linear function.

Since I thought the intricacies of the real thing were perhaps too much for us, I decided to try to analyze a toy example.

Toy taxes
We could begin with a flat tax. Here the idea is that the tax rate stays the same for everyone. For example, we could ask that people owe 10% of their income: there are no tax brackets. Thus in this scheme, T(x)=(1/10)x for all x>=0. Taxes don't just earn revenue for government, of course. They are also frequently used as a way of encouraging certain social policies. Therefore many localities don't have a flat tax. The belief is that, somehow, if x is large, then a higher proportion of x should be demanded as tax. Such a tax system is frequently called progressive.

For example, we could complicate our original flat tax by asking that the tax rate be 10% for x's less than (or equal to!) 10,000, and the rate be 20% for incomes above 10,000. This seems imply that if we want to write a piecewise specification for this T(x), it might be

```      { (1/10)x if x<=10,000
T(x)= {
{ (2/10)x if x>10,000```
Is this a satisfactory formula? Well, what would the tax be for an income of 9,999? I guess T(9999)=999.9. If income, x, increase by, say, two bucks, what would the tax be using this scheme? T(10001), would be (2/10)(10001)=2000.2. Let's see: earning two more dollars increases the tax to be paid by over a thousand dollars! That doesn't seem sane or fair (of course, tax policy need be neither sane nor fair!). But here is a social rule that people like: if the income changes a small amount, then probably the tax owed should only change a small amount. The problem might be easier to see on a graph of this T(x). I cheated a bit on this graph. The horizontal axis and vertical axis don't have equal units.

So there is a jump, and this jump is not good, because T(9999) and T(10001) should be close. How can we fix this, yet still have the tax rates the way we want? Geometrically we can drag "down" the part of the graph that is associated with x>10000. Algebraically we need to do something, and maybe the U.S. tax code specifications can help us in the algebraic formulation. We want a tax rate of 20% on income above \$10,000. At \$10,000, the tax owed would be \$1,000. So maybe the specification should be written
```      { (1/10)x if x<=10,000
T(x)= {
{ (2/10)(x-10,000)+1000 if x>10,000```
Now the graph looks nicer: there is no break, there is no jump in the graph itself.

We could now discuss the effects of changing income slightly with this T(x). In economics, most of what is described here has the adjective "marginal" attached to it, for a change on the margin, or on the edge, a tiny change. If your income were securely inside the lowest bracket, say x=4,000 or x=5,000, what additional tax would be due on an extra \$1 of income? I think ten cents (10% of that extra dollar). Or if you made a dollar less, then you would owe 10% less, ten cents less. Things are simple. If your income were \$30,000, though, the tax rate would be 20%. Small changes in your income, a dollar more or less, would "filter through" as a 20% change in the tax owed. So the tax rate is allowed to increase (progressive income tax). Is there a place where things are not so simple? If you really think about it, there is, for this T(x), one value of x for which the rate is not neatly defined, and that's \$10,000. Here if you increase your income by a little bit, you will owe 20% of that increase, but if you decrease your income by a little bit, you will get "back" 10% of that decrease. So something funny, something "kinky" (as I called it in class) happens here. We don't allow this to happen to T(x) itself, but we do allow it to happen to the rate. In mathspeak, in the language of this course, we will actually say that the function T(x) is continuous (no jumps or breaks) while the graph of T(x) has no nice tangent line when x=10,000 (this has to do with the derivative).

Before leaving the toy tax, I asked how to specify T(x) if we wanted a third rate, say 30% for incomes, x's, which are greater than 20,000. Again, some thought is required. When x is less than 20,000 and gets close to 20,000, then T(x) is given by (2/10)(x-10,000)+1000, so T(x) is getting close to (2/10)(20,000-10,000)+1,000, and this is 3,000. If we wanted to have three brackets, algebraically the formula would be

```      { (1/10)x if x<=10,000
T(x)= { (2/10)(x-10,000)+1000 if x>10,000 and x<=20,000
{ (3/10)(x-20,000)+3000 if x>20,000```

I think you should now be able to see how all the weird numbers in Schedule X come about. Of course, I am not commenting on the rates, on the number of brackets, etc. Here is one reference listing the highest and lowest U.S. marginal income tax rates and brackets over the entire history of that tax (since 1913).

Well, back to Math 135. The due date for WeBWorK assignments for these sections (21, 22, and 23) will be Tuesdays at 9 PM. I'll have an office hour after class on Tuesdays on the Douglass campus. I should spend most of the 4th period in the Math Department's "satellite office", Chem 101. That's on the first floor of the building next to the Ruth Adams building. Please note that you don't need to do all of the WeBWorK problems in one session. The program will remember your progress. Also please try to keep up with the syllabus and suggested homework problems. You really won't be able to review them all immediately before an exam!

The trig functions
Sine and cosine are sort of the same, one is just shifted over a bit compared to the other. The tangent function is the quotient of sine over cosine. Wait: that's not enough. Let's see:
Here is a graph of sine. From the graph the following can be seen:

• The domain of sine is all numbers.
• The range of sine is the closed interval (an interval including its endpoints) from -1 to 1.
• The sine function is odd: antisymmetric with respect to the vertical axis: sin(-x)=-sin(x).
• The sine function is periodic. This is really most important, since this property explains why functions whose initial definition involves triangles are used to analyze periodic changes in all sorts of phenomena: temperature, interest rates, etc.
The proportions of the graph are interesting and easy to foul up. The first positive root of sine is at Pi, which is about 3.1415. Thus the initial "bump" over the horizontal axis actually has proportion (horizontal to vertical) of about 3 to 1. Like everyone else, when I try to draw the curve rapidly, I frequently exaggerate the proportions so that the bumps seem steeper. But they really are quite flat.

I believe in radians as the true scale for inputs to the trig functions. You should know the following coordinates of points on the sine curve:
(0,0)    (Pi/6,1/2)    (Pi/4)    (Pi/3,sqrt(3)/2)    (Pi/2,1)
and from these you should be able to deduce lots of other values of sine by symmetry and periodicity, since sine is periodic with period 2Pi: sin(x+2Pi)=sin(x). What are exact values of sin((7/4)Pi) and sin(-(5/2)Pi) and sin((13/6)Pi)?
Pedagogical note After this course is over, I think forgetting these exact values and freeing the few brain cells devoted to that task should be an immediate goal. I believe that very few occasions will require you to know the sine of Pi/6. But I am not in charge of this course, and exams will demand exact answers to certain questions. Sigh.

If you can demonstrate that you have 360 fingers and toes, I will attempt to get you an exemption to these requirements. Until then, please ... Also, as we will see later, doing calculus with degrees turns out to be much, much more awkward than doing calculus with radians.

Here is a graph of cosine. Please make sure that you know the corresponding properties of cosine, such as:

• The domain of cosine is all numbers.
• The range of sine is the closed interval (an interval including its endpoints) from -1 to 1.
• The cosine function is even: symmetric with respect to the vertical axis: cos(-x)=cos(x).
• The cosine function is periodic with period 2Pi: cos(x+2Pi)=cos(x) for all real numbers, x.
Again, certain exact values of cosine should be known.

There are many formulas from trig and precalc involving the sine and cosine functions. I think we will generally reference only

• sin2+cos2=1 (this is quite useful, from the Pythagoras Theorem).
• sin(a+b)=sin(a)cos(b)+sin(b)cos(a)

I don't think you should forget every other formula but I doubt I will use others.

The tangent function is the quotient of sine divided by cosine. Since cosine equals 0 when its input is an odd multiple of Pi/2, we should expect weird behavior near odd multiples of Pi/2. The result is a collection of vertical asymptotes. Etc., etc.
You are supposed to know all about these functions already. Read the (expensive) text if you need more review,and do the homework problems!

The QotD (Question of the Day) was to sketch the graphs of 1/K(x) and K(1/x). I think this is fairly difficult, so I asked that people work in groups of 2 to 4 individuals. As soon as I read the answers I will return here and discuss them.

HOMEWORK
Please read the sections of the book related to the first two lectures, and at least look at the problems. Most of the exams will resemble textbook problems. Also, spend a little time on WeBWorK.

### Tuesday, January 18: this is lecture 1

I began the class by writing some information about Quercus imbrecaria, the Shingle Oak. To the right is a picture of one of these. This is a species of tree, I have read, which is a "medium to large oak reaching 50 to 60 feet" and it has "slow to medium growth of 12 to 18 inches per year".
First, I decided to simplify the data a bit: let's assume that the stated growth rate is 1 foot per year (also making the units a bit simpler!) and that the ultimate height is 60 feet. I asked what these numbers meant. Students replies and my calm (!) responses and comments allowed us to have a nice discussion.

Growth rate If the growth rate is 1 foot per year, that means the tree will germinate from a seed and grow a foot in the first year. (Maybe, but this a bit startling.) We can extrapolate a bit more. The distance from the Earth to the Moon is about 240,000 miles, and if the tree grows 1 foot a year, then (since there are 5,280 feet in a mile) the tree will hit the Moon in about a billion and a quarter years. That's a tragedy. I show to the right a sort of impression of what this might look like.

Well, of course just using the information "1 foot per year" as I did above is a bit simple-minded. The reality is that if H(t) is the height in feet of the tree after t years of growth, and if at germination we assert that H(0)=0, then initially growth is slow, and for large t's, the growth rate will be much less that 1 foot per year. "Growth rate" in the time interval between t1 and t2 (where t1<t2) will be the difference in the heights divided by the time span, or [H(t2)-H(t1)]/[t2-t1]. In reality the assertion that the growth rate is 1 foot per year means that for t1 and t2 somewhat away from 0 and not very large, this quotient will be approximately 1. Graphically, of course, the quotient [H(t2)-H(t1)]/[t2-t1] is the slope of the line connecting two points on the tree growth curve. It may be more than 1 (a good year, lots of moisture, sunlight, and required nourishment) or less than 1 (lousy weather, etc.).

What does a graph of the tree height look like? With the numbers 1 and 60 in mind, I suspect we may think that the collection of points whose coordinates are (t,H(t)) may look as I display it. I actually doubt very much the graph is as simple as this: "piecewise linear". I don't think that the tree grows until it gets to its "ultimate height" and then just stays there. We will need more precise language to analyze such questions.

I asked what the graph would look like. I got suggestions involving "just any graph" (so maybe the tree height could be negative, or get smaller as t gets larger?), or exponential (that would mean the tree got really big really fast without any "ultimate" height. The suggestion was made that the tree had logarithmic growth. Here we needed to think a bit. I asked if log functions were bounded. For example, consider log10. I know that log10(1,000)=3, and I can even find a number W so that log10(W)=7: take W=10,000,000 (seven zeros). I asked if we could find a number V so that log10(V)>238. We thought for a while. I finally suggested V=103,070. This will work. So log10 is not bounded and couldn't be a model of tree growth. In fact, tree growth probably looks more like the graph displayed to the right here. It is a complicated function, and even so, it is really greatly simplified from reality, because I don't think that the growth is ever as smooth as what is shown. Later in the course I will try to return to this model, equipped with some vocabulary and notation, and get a function that models the growth shown (the logistic function).

### What is this course?

The course is not a "pre-professional" course for math or CS or engineering majors. Those students should take Math 151. It is a course meant for students in certain biological sciences and intending to major in business areas. I will try to show the value of the vocabulary and techniques of calculus. This is my job, but I can succeed only with your good will and effort.

Grading I'll give two in-class exams (100 points each) and there will be a uniform (not written by me!) course final (200 points). Tentatively, I will make up a score of 150 points for the following activities: WeBWorK, textbook homework problems, quizzes in recitation, and activities in lecture.

Now, the world's quickest review of precalculus. (Look at sections 1.1 and 1.2 of the text.) This is supposed to be a review. If the material is not familiar, you must do many of the problems in the first few sections of the text.

• For which x's is |3x-5|<7? I unwrapped the absolute value to see that then -7<3x-5<7, so that -2<3x<12 and thus -2/5<x<4. Please notice that these steps are all reversible so that we have indeed "diagnosed" what the x's must be. This is the open interval (-2/5,4), which I sketched.
• What is the distance from (3,-2) to (7,5)? I tried to draw an appropriate right triangle with corner at (7,-2) to show that the distance would be sqrt[(7-3)2+(5-(-2))2].
• What restrictions on x and y would make the point (x,y) have distance 4 from (7,5)? Here we got sqrt[(7-x)2+(5-y)2]=4.
• For which x and y would (x,y) be on the straight line connecting (3,-2) and (7,5)? I mentioned that one neat form of most straight lines was y=mx+b (m is the slope, and b is the y-intercept). Only vertical lines (x=something) do not have this form. What's m in this case? It is [5-(-2)]/(7-3)=7/4. So y=(7/4)x+b. We can find b by plugging in one of the points, say (7,5): 5=(7/4)7+b, and b=-29/4. The line is y=(7/4)x-29/4. (If you worry about the answer [as I do] you can try (3,-2) in y=(7/4)x+b and get -2=(7/4)3+b which also gives b=-29/4. Now I feel slightly more confident!)
• Which of the equations, y=(7/4)x-29/4 (a straight line), and sqrt[(7-x)2+(5-y)2]=4 (a circle) defines the graph of a function? If (a,b) is on the graph of a function, "a" specifies the input (the domain variable), and "b" specifies the output (the range variable), and the output should be unique, so to be the graph of a function means that the Vertical Line Test must be fulfilled: a vertical line should intersect the graph at most once. Any circle is not the graph of a function, and this (non-vertical!) straight line is.
I wrote the beginning of the specification of a much more complicated function, part of the U.S. tax rate schedules for 2004:
```Schedule X Single

If taxable income is
over-- 	        But not over-- 	The tax is:
\$0 	        \$7,150 	        10% of the amount over \$0
\$7,150 	        \$29,050 	\$715.00 plus 15% of the amount over 7,150
\$29,050 	\$70,350 	\$4,000.00 plus 25% of the amount over 29,050
\$70,350 	\$146,750 	\$14,325.00 plus 28% of the amount over 70,350
\$146,750 	\$319,100 	\$35,717.00 plus 33% of the amount over 146,750
\$319,100 	no limit 	\$92,592.50 plus 35% of the amount over 319,100
```
I would like to try to analyze this function next time.

The Question of the Day was "Why are you taking this course?" I will report on the answers later.

HOMEWORK
Look at WeBWorK. Look at the text. Do the homework problems. On Thursday, please hand in 1.1: 30, 42 and 1.2: 12, 40.