## Diary for Math 151:04-06, fall 2003

The final exam for this course is on Monday, December 15, from 4 to 7 PM.

DateWhat happened (outline)
Monday,
October 6
I went over the fourth Question of the Day from the last lecture. More students than usual had gotten it wrong or left it incomplete. I tried to emphasize the structure of the formula, and how this led to choosing algorithms or methods. Finding such derivatives by hand takes practice and patience.

My final specific example of the quotient rule was to compute the derivative of f(x)=sin(x)/cos(x). Here top=sin(x) and bottom=cos(x), and the derivative will be (top'·bottom-bottom'·top)/(bottom)2. Notice that because of the minus sign, there is an asymmetry in the result, and this can lead to errors. f'(x)=[cos(x)·cos(x)-(-sin(x))·sin(x)]/[cos(x)]2. Here is an answer which demands simplification. The top is [cos(x)]2+[sin(x)]2 which is 1. The whole result is therefore 1/[cos(x)]2 or [sec(x)]2. Of course, this f(x) is tan(x), so we now know that the derivative of tan(x) is [sec(x)]2. Please note the parentheses. I almost always use lots of parentheses, because their use is almost required to make understandable the results of derivative algorithms.

I then put a big copy of the chart to the right on display and we discussed it for a while. Here's an html copy, which might be more readable.

```
CHIPCO
INVESTMENT DOLLARS & PRODUCTION

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

```

I discovered to my unpleasant surprise that not as many people as I had hoped were familiar with the few economic terms used above. In particular, the word "marginal" is used in a fairly technical sense. In the first table above, it refers to the approximate amount would increase per each million dollars of increase in capital investment. Therefore, for example, if \$302 million were invested, then (according to this model) chip production would be 3,040,000 (chip production at the \$300 million level) plus .28(2)(1,000) chips. The 2 comes from the additional millions of dollars of capital. The 1,000 comes from the units I use for chip production. The .28 is this "marginal" quantity. In the first table, the marginal quantity is therefore the approximate amount DELTA P/DELTA C, relating the change in chip production to the change in capital investment. It is sort of a slope, or, more likely, sort of a derivative: indeed, the use of "marginal" in economics usually means a derivative. In this model, if \$297 million were invested, the approximate expected chip production would be 3,040,000 (again, chip production at the \$300 million level) plus .28(-3)(1,000). The novelty here is the use of the minus sign, since here we are decreasing the capital investment rather than increasing it.

The second table describes a similar phenomenon, here connecting the chip amount, C, with the profit derived from these marketing and sale of these chips. For example, the profit derived from the sale of 3,000,000 chips (the first line of the second table) is \$1.2 million. If we now look at the third column, the model predicts a marginal profit of .03 (in the given units). Using this, if 3,010,000 chips are marketed (that's 10 more 1,000 units of chips) the additional profit would be .03(10) million dollars, or \$300,000. And if only 2,970,000 chips were marketed, then the profit would be 1.2 million+(.03)(-30)(1,000)million. (I think I got all the units correct.) The third column gives DELTA P/DELTA C for various amounts of chip marketing: the change in profits compared to the change in chips marketed. Of course the validity of such models can certainly be criticized, but I really wanted to show this to explain what's in the next paragraph.

The two tables linked together describe a complicated phenomenon. First we "input" capital, M (M is for money), which produces C, a certain number of chips. Then the chips are marketed (and sold, hopefully!) to obtain a certain amount of profit, P. Here we have a composition of functions. For example, suppose we were asked how much profit there is if we put in M=\$500 million. From the first table we read off C=3,100,000 chips, and from the second table we can then see that P will be 3.6 million dollars.

I hoped that this was all fairly clear. Now I asked what I thought was a difficult question. Suppose we increase M from 500 million dollars to, say, 503 million dollars. What will the model predict the profit will be? We can trace this if we are sufficiently alert. The first marginal quantity we need to consider is DELTA C/DELTA M. For M=\$500 million, this is .78. So the new chip production is old chip production + increase in chip production, and this will be 3,100,000+(.78)(3)(1,000). Now let us consider the chip/profit table. With C=3,100,000, we see that profit is supposed to be 3.6 million. But we are changing C by adding on the (relatively small) amount of .78(3)(1,000). The relevant marginal quantity here is DELTA P/DELTA C, on the row where C is 3,100(,000). The marginal amount here is .05, so that the new profit will be the old profit (3.6 million) plus (.05)(.78)(3)(1,000) million dollars. The 3 comes from perturbing the capital investment. The 1,000 comes from my weird units. The really interesting stuff is (.05)(.78): indeed, this represents the marginal profit as capital invested changes, when the capital investment is 500 million dollars. Symbolically, it might make sense written this way: DELTA P/DELTA M=DELTA P/DELTA C,·DELTA C/DELTA M. So the DELTA C's just seem to cancel out. Of course, this is more complicated than just multiplying fractions, since the fractions (the marginal stuff, the derivatives) need to be "evaluated" on the appropriate rows of the tables.

What I've done here is tried to present heuristic evidence that would allow us to believe the chain rule.

```
1. allowing or assisting to discover.
2. [Computing] proceeding to a solution by trial and error.
```
The Chain Rule Suppose that f and g are differentiable functions. The F(x)=fog(x)=f(g(x)) is differentiable, and F'(x)=f'(g(x))·g'(x).
Here o is supposed to be a little circle, and the little circle indicates composition. The tables above sort of indicated that chip production was a function of capital investment, and then that profits were a function of the chips marketed, so that profit as a function of capital investment was a composition of the two functions.

The balance of the lecture was devoted to exploiting the chain rule. There is a correct proof of the chain rule in the book. My first example was something like this (about as simple as I could imagine):
If F(x)=(x2+7)300, what will F'(x) be? Success here probably will result from recognizing that the chain rule applies.
If F(x)=f(g(x)), then g(x) is x2+7 so g'(x)=2x, and f(x) is x300 so f'(x)=300x299. Thus F'(x)=f'(g(x))g'(x)=f'(x2+7)(2x)=300(x2+7)299(2x). Whew!

But now comes the realistic comment. Hardly ever does anyone bother writing down all of these intermediate steps. That is, in practice very few f's and g's are actually identified. What happens is that people see and differentiate the outside most function (f above), put in the inner function (g) in that derivative, and then multiply by g'. For example, consider sin(ex+x2). What is its derivative? The outside function is sine, whose derivative is cosine. So I begin by writing cos(what's insidethe derivative of what's inside. The result is cos(ex+x2)·(ex+2x). This expression is a formula for the derivative of sin(ex+x2). Again, I urge you to consider the significance and necessity (!) of appropriate parentheses in these expressions. The "argument" of cosine is ex+x2 and the cosine expression is then multiplied by the expression (ex+2x).

 Yet Another Advertisement The person you are helping with the properly paired parentheses is probably yourself. Computing derivatives is only an intermediate stage in most of what we will do. If you look back over a few lines of computation and can't tell which part is a composition and which part is a multiplication, you will have cheated yourself. So please try to write carefully.

Of course the chain rule itself can be repeated. So here, for example, we can try to differentiate cos(e3x2). Its derivative is -sin(e3x2)·(e3x2)·(6x). I hope that you can pick apart the layers of the functions and their compositions. One poor metaphor for using the chain rule is that it is like peeling an onion very very carefully, layer by layer, and taking care always of the outside most layer first. Confusion is certainly possible, and that's an understatement.

Here is an interesting application of the chain rule. Suppose we want to differentiate y=sqrt(x). Well (one student in the audience immediately and correctly said, (1/2)x-1/2) here is a way to do it. Square the equation to get y2=x and then differentiate the resulting equation. I will switch to what is called Leibniz notation now. Although this notation is not my favorite, somehow it fits with this sort of computation. In Leibniz notation, the derivative of y with respect to x is dy/dx, and what I want to do is d/dx (differentiate!) the equation y2=x. The right-hand side is easy: its derivative is 1. The left-hand side needs the chain rule: it has an "outside function", squaring, and an inside function, the "unknown function" y (unknown at least as far as its derivative is concerned). The result of the chain rule is 2y(dy/dx). Since this should be equal to 1, the derivative of the right-hand side, we can solve for dy/dx, and we get dy/dx=1/(2y). But y=sqrt(x), so we may recognize that the derivative of sqrt(x) is indeed (1/2)x-1/2.

This line of approach may be extended. For example, if y=x15/4 we then know that y4=x15. Then d/dx the result to get 4y3(dy/dx)=15x14 (do not forget the dy/dx which the chain rule forces upon you!). Then solve for dy/dx. It is (15x14)/(4y3). But since y=x15/4 we know that (15x14)/(4y3)=(15x14)/(4(x15/4)3)=(15x14)/(4x(15/4)·3)=(15/4)x14-(45/4)=(15/4)x11/4. Wow! A discovery: the power rule apparently holds for rational exponents. So the derivative of sin(x4/7) must be cos(x4/7)·(4/7)x-3/7.

The trick of having an equation involving x and y, then d/dx'ing the equation and solving for dy/dx is sometimes very useful. It is called Implicit Differentiation. There are times when it is difficult or impossible to get an explicit representation for y as a function of x. Then having an implicit expression is enough, if what is wanted is some information about dy/dx. I chose a rather strange example. Suppose we look at the equation x3+y2-x=0. The picture shown here is a result of the Maple command implicitplot which allows one to plot implicitly defined functions. Of course I was waiting for students to ask why the heck anyone would ever want to plot such a curve, and I picked the curve with some intent: this is an example of an elliptic curve, and cryptographic protocols related to such curves are extremely important today (Google has about 135,000 references to elliptic curves!). This strange-looking curve is actually recognized everywhere that "communication security" is important. If x=-2 in the equation x3+y2-x=0 we see that -8+2+y2=0, so that y must be +/-sqrt(6). What is the slope of the line tangent to this curve at the point (-2,sqrt(6))? We take the equation x3+y2-x=0 and d/dx both sides. The right-hand side gives me 0 (I always prefer to differentiate constants!) and the left-hand side is 3x2+2y(dy/dx)-1 which must therefore equal 0 (the chain rule forces the appearance of dy/dx). Inserting -2 for x and sqrt(6) for y, we get 12+2sqrt(6)(dy/dx)-1=0 or dy/dx=-11/(2sqrt(6)). Indeed, if you consider what the line tangent to the elliptic curve shown must look like, it does indeed slope "down".

The Question of the day was to compute the derivative of [sin(ex)]/[(x2+3)5]
The answer uses the quotient rule and the chain rule on both the top and the bottom. It is [(cos(ex)·ex·(x2+3)5)-(5(x2+3)4·2x·sin(ex))]/[(x2+3)10].
Please notice all the darn parentheses and the correct (I hope!) minus sign on top, and the squaring of the bottom. This is involved, but it can be done and even, usually, done correctly!

These problems are due this Thursday in recitation: 3.5 #56 and 3.6 #10.

We also briefly discussed the exam, which will be given in one week.

 Our first exam will occur one week from today, on Monday, October 13, during the standard class period and in the standard class room. This is one meeting later than is on the syllabus. The workshop period on Thursday, October 9, will be devoted to review for the exam.

Students indicated that they would like the opportunity of a review session and I will schedule one for late Sunday afternoon. The exam will cover up to and including the material of this lecture (sections 3.5 and 3.6 of the text). I will try to post on the web the review problems which will be discussed in the workshop period Thursday. Also please remember to do the assigned textbook homework problems.

I will probably not allow any calculators or notes for the exam. The only "results" needed are mostly the definitions of continuity, differentiability, and the differentiation algorithms. Calculators cannot be used on the final exam, so it is probably better that "we" get used to that now.

Thursday,
October 2
Please hand in a writeup of the fourth problem next Thursday, unless you have exceptional energy and want to do the first problem.
Wednesday,
October 1
These problems are due this Thursday in recitation: 3.2 #32 and 3.4 #24 (This is correct and if what I wrote in class was different, I was wrong then. Sigh. Sorry.)

The major project for today and the next class was expanding our list of differentiation algorithms.

Qotd #1
```
/algorithm/ n.
1. [Math.] a process or set of rules used for calculation or
problem-solving, esp. with a computer.

/alligator/ n
1. a large reptile of the crocodile family native to S. America and China,
with upper teeth that lie outside the lower teeth and a head broader and
shorter than that of the crocodile.

/allegory/ n.
1. a story, play, poem, picture, etc., in which the meaning or message is
represented symbolically.
```
I noted that the reciprocal rule allows the power rule to be extended to negative integer powers, so that the derivative of 1/x33=x-33 is (-33)x-34 or, equivalently, -33/x34.

Also we deduced the quotient rule: if F(x)=f(x)/g(x) where f and g are differentiable functions, then we can write F(x)=f(x)·(1/g(x)) so that (using the product rule and the reciprocal rule), F'(x)=f'(x)·1/(g(x)+f(x)·g'(x)/[g(x)]2. This is the quotient rule. The result, F'(x), is usually written as [f'(x)·g(x)-g'(x)·f(x)]/[g(x)]2. I did a simple example and then asked

Qotd #2

Moving right along (!) I discussed the trig functions again.

Here are the standard definitions of trig functions as related to quotients of side lengths of right triangles:
sin(theta)=OPP/HYP and cos(theta)=ADJ/HYP and tan(theta)=OPP/ADJ. There are, of course, three other pairs of quotients, but need for them will be very rare in this course. Also included are two special triangles which give exact values of the trig functions at certain numbers (what is cos(35Pi/4)? what is tan(-103Pi/3)?)

Even more, we will need a kinetic view of the trig functions.

If a point moves counterclockwise around the unit circle at unit speed, the second coordinate of the point is the sine of the time that the point has been traveling. The angle is measured by the length of the intercepted arc. In this scheme (radian measurement) the full circle of 360o is 2Pi radians. This is more natural if we want to consider periodic phenomena, like motion around a circle. I want to differentiate sine, so I will analyze [f(x+h)-f(x)]/h for f(x)=sin(x). Then f(x+h)=sin(x+h)=sin(x)cos(h)+cos(x)sin(h) using the addition formula for sine. And [f(x+h)-f(x)]/h is, after some algebra, sin(x)[(cos(h)-1)/h]+cos(x)[sin(h)/h]. Consideration of the graph of sine shows that the slopes of the lines tangent to sine at various points is periodic: the slopes repeat every 2Pi. Also, these slopes are 0 at odd multiples of Pi/2 and alternate being positive and negative between the zeros. In fact, the derivative of sine looks like it should be cosine. And that's the case. The next 20 minutes of the lecture were devoted to showing that limh-->0[sin(h)/h]=1. I left the companion limit, limh-->0[(cos(h)-1)/h]=0, to students' reading the book.

Verification that limh-->0[sin(h)/h]=1
This is sometimes a handy thing to know. Also it is a limit which is often "requested" on calc 1 exams. Let me give you what I think is a fairly convincing discussion. I will look at the accompanying picture to the right. This picture shows a very small angle h, inside the unit circle (all of the radii are equal to 1).

• What is the area of triangle ABC? This is (1/2)base·height. The base is AB, which is a radius of the circle, so the length of AB is 1. The height is CD, the opposite side from the angle with measurement h and hypotenuse AC (another unit length, since another radius of the circle). Therefore sin(h) is the quotient of the length of CD divided by 1. So CD has length sin(h), and the area of triangle ABC is (1/2)sin(h)·1.
• What is the area of triangle ABE? This is (1/2)base·height. The base is AB, which is a radius of the circle, so the length of AB is 1.The height is EB, but EB/AB is tan(h), and the length of AB is 1, so EB has length tan(h). The area of triangle ABE is (1/2)tan(h)·1.
• What is the area of sector ABC? Here we need to think a bit. I fouled this up in class, accepting a rather ... incorrect suggestion from a student. I gave several examples:
The area of a whole circle of radius 7 is 72Pi.
The area of a half circle of radius 5 is 52Pi/2.
The area of a quarter circle of radius 3 is 32Pi/4.
The point of this was to convince the students that the area of a sector is directly proportional to the product of the square of the radius and the angle measurement (in radians). The constant of proportionality was (1/2): that is, the area of a circular sector of radius R and central angle Theta is (1/2)R2Theta.
Therefore the area of circular sector ABC is (1/2)12h=(1/2)h. (There's no Pi in this answer, because the Pi is already incorporated into the radian measure of the central angle!)
Now we see which of the areas is largest and which is smallest and which is middlest. (The sequence of letters "middlest" does not seem to be an English word. I am sorry.)
Smallest areaMiddlest areaLargest area
Triangle ABCSector ABDTriangle ABE
(1/2)sin(h)(1/2)h(1/2)tan(h)

Now the first two entries in the last row give us [sin(h)/h]<=1. The last two entries in the last row give us (after remembering that tan=sin/cos!) cos(h)<=[sin(h)/h]. Put them together:   cos(h)<=[sin(h)/h]<=1. We are interested in what happens as h-->0. Well, here is a valid use of version 1 of the squeeze theorem, since both cos(h) and 1 approach 1 as h-->0. So we can finally conclude that limh-->0[sin(h)/h]=1.

The function sin(h)/h occurs quite a bit when folks study vibrations of various sorts (vibrations in a beam or vibrations in an electric circuit or ...). Also the limit is sometimes really useful to know For example, on my "calculator", I just asked for sin(.0123) and got 0.0002146755. WHAT!!!??? Isn't this wrong? Isn't this way off? Well, no. I actually asked the calculator the wrong question. The calculator was set for degrees, not for radians. If you insist that your trig functions be all functions of degrees, then the derivatives will be all fouled up. In fact, the true value of sine of .0123 is actually 0.01229969, which is pretty darn close. So sine of h radians is darn close to h when h is small. And we can use this to compute other strange limits, if we have to.

Since limh-->0[sin(h)/h]=1 I know that limh-->0[sin(h)/(5h)]=5. And I know that limh-->0[sin(3h)/h]=limh-->0[3sin(3h)/(3h)]=3 since limh-->0[sin(3h)/3h]=1 (3h gets small along with h after all!)

I found an equation of the line tangent to y=sin(x) at x=Pi/3 (and, yes, I want the exact value(s) of everything). To do this, I need some information with y=f(x) and f being the sine function:
A point on the curve: (x,f(x))=(Pi/3,sin(Pi/3))=(Pi/3,sqrt(3)/2).
A slope of the tangent line: f'(x)=f'(Pi/3)=cos(Pi/3)=1/2.
An equation of the line is (y-(1/2))=(sqrt(3)/2)(x-Pi/3). I would not simplify this answer unless I had to.

I remarked that (looking again at the shape of the graphs) we can also see that the derivative of cos(x) is -sin(x) (there is a shift of Pi/2 in both graphs). Thus we get two more lines in the table of derivatives.

Qotd #3
My final gasp was to draw a quick picture, the way we all do, of sine and cosine. I distorted the bumps as almost everyone does. The bumps are actually rather flat. But "clearly" the curves intersect, and, look, look!, it seems that they intersect almost perpendicularly.
Can we check this? Well, sin(x)=cos(x) for x between 0 and Pi/2 when x=Pi/4 (that's the isosceles right triangle all the way up). Two lines will be perpendicular when the product of their slopes is -1 (or when "their slopes are negative reciprocals"). The slope of the line tangent to sine when x=Pi/4 is cos(Pi/4)=1/sqrt(2). The slope of the line tangent to cosine when x=Pi/4 is -sin(Pi/4)=-1/sqrt(2). The product of these two slopes is -1/2, not -1, so the curves do not intersect perpendicularly.

Qotd #4
The Question of the day was in several parts.
1. What is the derivative of 33x72-7ex+sqrt(117)?
The "trick" here was to force attention to the constant, sqrt(117). But its derivative is 0, even though it looks complicated.
2. What is the derivative of (100x50+ex)/(50ex-x100)?
Here the trick, if any, is to remember to copy accurately. I usually try to write the bottom (squared) first, then work on the top.
3. What is the derivative of 5x·sin(x)+17cos(x)?
4. What is the derivative of
ex(3sin(x)+5x2)

2cos(x)+[(x+1)/(x-1)]
The horror here is that uses of the product and quotient rules are buried inside a "big" use of the quotient rule. One needs to concentrate and not get lost.
The bottom of the quotient is (2cos(x)+[(x+1)/(x-1)])2. This is the square of the bottom of the original expression.
The top is:(ex(3sin(x)+5x2)+ex(3cos(x)+10x))(2cos(x)+[(x+1)/(x-1)])-(ex(3sin(x)+5x2))(-2sin(x)+[1(x-1)-1(x+1)]/(x+1)2). Structurally, this is the derivative of the top of the original expression (and that top is a product, so the product must be used) multiplied by the bottom of the original expression, and then a subtraction of the derivative of the bottom of the original expression (on which the quotient rule must be used itself, since part of that bottom is itself a quotient!) multiplied by the top of the original expression. And now we are done.

 Here is the Maple instruction and response for question #4, with processing time about one-one hundredth of a second (actually, that may not be accurate -- it was probably less than .01 seconds, but in this case Maple likely rounded up.) The response is several fractions: apparently Maple wants to differentiate (f(x))·(1/g(x)), but it is easy to convince Maple to combine things into one fraction, if desired. This is the instruction: This is the response. You really need to look carefully at it, but Maple also has a graphic interface whose responses are presented in a more standard style. ```diff((exp(1)^x*(3*sin(x)+5*x^2))/(2*cos(x)+((x+1)/(x-1))),x); x 2 x exp(1) (3 sin(x) + 5 x ) exp(1) (3 cos(x) + 10 x) ------------------------- + ------------------------- x + 1 x + 1 2 cos(x) + ----- 2 cos(x) + ----- x - 1 x - 1 x 2 / 1 x + 1 \ exp(1) (3 sin(x) + 5 x ) |-2 sin(x) + ----- - --------| | x - 1 2| \ (x - 1) / - -------------------------------------------------------- / x + 1\2 |2 cos(x) + -----| \ x - 1/ time = 0.01```

## more Differentiation algorithms

FunctionDerivative
f(x)/g(x)[f'(x)·g(x)-g'(x)·f(x)]/[g(x)]2
sin(x)cos(x)
cos(x)-sin(x)
MORE TO COME!!!
Monday,
September 29
These problems are due this Thursday in recitation: 3.2 #32 and 3.4 #24

 Our first exam will occur two weeks from today, on Monday, October 13, during the standard class period and in the standard class room. This is one meeting later than is on the syllabus. The workshop period a week from Thursday (on October 9) will be devoted to review for the exam.

I started discussion of the differentiation algorithms. It turns out that for functions defined by generally simple formulas, there are a series of "rules" or algorithms which allow formulas for the derivatives to be written fairly easily. We will always start with the formal definition although it is comforting to recall such intuition as "f'(x0) is the slope of the line tangent to the graph of y=f(x) at x=x0" and that f' is also instantaneous velocity, etc.

We began with the very simplest sorts of functions. If f(x)=15 for all x, then surely f(x+h)=15 also, so that f(x+h)-f(x)=0 and dividing this by h also results in 0, so the derivative is always 0. This result is true for any constant function. (Also the graph is a horizontal line, and is its "own" tangent line, with slope=0.) So we are done.

Now consider f(x)=xn. We need f(x+h) using the formal definition. f(x+h)=(x+h)n and we could use the Binomial Theorem to see exactly what the "expanded" version of f(x+h) looks like (see also information about Pascal's Triangle. But we don't need very precise information. For example, let's look at n=4. Here (x+h)4 is (x+h)·(x+h)·(x+h)·(x+h). We could multiply and expand everything or we could look at the structure of things a bit. There is exactly one product which is all x's, and that product has degree 4: x4. If we knock out exactly one x and take an h instead, we would have hx3. How many such terms can we get? Well there are 4 possible h's to choose, and we only want one of them, so there is exactly 4hx3 in the expansion. Every other term has at least two h's in it. We could "collect" all those and label them h2JUNK (junk because we won't have any need here of its precise nature). So:
(x+h)4=x4   +   4hx3     +     h2JUNK
All x's  Terms with 1 h    all other terms

We can do this more generally: (x+h)n=xn+nhxn-1h2JUNK
This is exactly a restatement, by the way, of the [REMARKABLE] equation I mentioned last time: f(x+h)=f(x)+f'(x)h+Err·h where you can "see" the higher-order error terms.

But now what? We consider limh-->0(f(x+h)-f(x))/h= limh-->0[(x+h)n-xn]/h= limh-->0[xn+nhxn-1+h2JUNK-xn]h= limh-->0h[nxn-1+hJUNK]/h= limh-->0nxn-1+hJUNK=nxn-1 which is in the table.

Your textbook next studies the exponential functions ax. If a>1, this represents exponential growth. Consider the graph of an "average" exponential growth function and the slope of the tangent lines to this curve. As the point of tangency travels from left to right, the slope, which is always positive, just increases. If we could image a graph of the slope function (which is just a graph of the function y=f'(x)) it might look a great deal like ax itself. And that is the truth. A proof of this takes some effort, and could be given now, but we are supposed to run as fast as we can. So I will just write out some suggestive reasoning, following more or less what is in your text.

If f(x)=ax, then f(x+h)=ax+h=axah. The difference quotient (f(x+h)-f(x))/h becomes (ah-1)/h multiplying ax. What is the number (ah-1)/h as h-->0? I strongly recommend that you graph (2h-1)/h and (3h-1)/h and (2.71828h-1)/h for h in the interval [0,1]. The "2-graph" seems to have limit about .7 at 0, the 3-graph, about 1.16 at 0, and (WOW!) the last has limit 1 at 0.

e is defined to be the real number so that limh-->0 (eh-1)/h exists and is 1.
The immediate consequence is that
ex is its own derivative.
Of course this is a very nice, very convenient property.

Then I worked on building new functions. If F(x)=f(x)+g(x), and the derivatives of f and g exist, what can one predict about the existence and value of the derivative of F?
Since F(x)=f(x)+g(x) we know that F(x+h)=f(x+h)+g(x+h), and the difference quotient for F can be written this way:
(F(x+h)-F(x))/h=[(f(x+h)+g(x+h))-(f(x)+g(x))]/h=[f(x+h)-f(x)+g(x+h)-g(x)]/h=[f(x+h)-f(x)]/h+[g(x+h)-g(x)]/h.
And now let h-->0, and we see that the derivative of the sum is the sum of the derivatives.

Now I did a hard problem from the textbook (I think #50 of section 3.1), to show that we have gotten already to some level of achievement. The problem asks us to find the equations of the lines tangent to the parabola y=x2+x which also go through the point (2,-3). Note that although the problem statement does not request it, I would almost always begin the solution by making a sketch.

If P=(x,y) is the point of tangency on the parabola, we can solve the problem by realizing that the slope of the tangent line at P, mTAN, can be written in two different ways. First, since the tangent line goes through P and (2,-3), its slope is (y-(-3))/(x-2), which is (x2+x+3)/(x-2). But mTAN is also f'(x) if f(x)=x2+x. So mTAN=2x+1. Therefore 2x+1=(x2+x+3)/(x-2), and (2x+1)(x-2)=x2+x+3. Then 2x2-3x-2=x2+x+3 so that moving everything to one side, we get x2-4x-5=0. Since this is a problem in a textbook, the quadratic factors into (x-5)(x+1)=0. I found one of the lines, due to a brave Rugby player (Mr. Fredo). If x=5, then y=52+5=30 and the derivative is 2(5)+1=11. So the tangent line is y-30=11(x-5). We can make a cheap check: does this line go through (2,-3)? Well, -3-30=-33 and 11(2-5)=-33, so the answer is "Yes." The point of tangency is (5,30) which explains why we can't see it in the picture. You can find the equation of the other line yourself.

Now we began to discuss what is called the product rule. The statement of the product rule begins "The derivative of the product is ..." There is an expectation of simplicity and symmetry here, which should be eliminated as soon as possible. Consider x2 which is also, of course, x·x. The derivative of x is 1, and 1·1=1, but the derivative of x2 is 2x, so the product of the derivatives is not the formula we want.

If F(x)=f(x)·g(x), then F(x+h)=f(x+h)·g(x+h), so that (F(x+h)-F(x))/h=(f(x+h)·g(x+h)-f(x)·g(x))/h. Now the game is to somehow write this fraction in terms of the difference quotient of f and the difference quotient of g. Here the picture may help. It tries to show a sort of decomposition of f(x+h)·g(x+h)-f(x)·g(x). The suggestion is that f(x+h)·g(x+h)-f(x)·g(x)=(f(x+h)-f(x))·g(x)+f(x)·(g(x+h)-g(x))+(f(x+h)-f(x))·(g(x+h)-g(x)). If we now divide by h and let h-->0, then:
[(f(x+h)-f(x))·g(x)]/h-->f'(x)·g(x) and [f(x)·(g(x+h)-g(x))]/h-->f(x)·g'(x).
The blue rectangle in the corner is a curiosity. It is algebraically (f(x+h)-f(x))·(g(x+h)-g(x)) (then divided by h) In the [REMARKABLE] equation I mentioned last time: F(x+h)=f(x)+F'(x)h+Err·h the blue rectangle belongs to the Error term. So what happens is this:
[(f(x+h)-f(x))·(g(x+h)-g(x))]/h=[(f(x+h)-f(x))/h]·(g(x+h)-g(x)). The first term -->f'(x) but the second term: g(x+h)-->g(x as h-->0 since g is continuous (because differentiable functions are continuous). Therefore the blue rectangle divided by h -->f'(x)·0 which is 0: it contributes nothing to the limit. This is a bit elaborate, but I'd like to be honest when I can be (!?). So now we know that the derivative of f(x)·g(x) is f'(x)·g(x)+f(x)·g(x). This is called the product rule or, sometimes, the Leibniz rule, memorializing one of the inventors of calculus.

Examples: If f(x)=x and g(x)=x, the product rule gives us 1·x+x·1-2x, the correct answer. I also differentiated something like ex·x178. And then I differentiated 37·x23 with f(x)=37 (a constant function) and g(x)=x23. The product rule predicts that the derivative is 0·x37+37·23x22. Frequently people use this special case of the product rule without thinking about it: the derivative of a constant times a function is a constant times the derivative of the function.

Finally, gasping for energy, an old racehorse does one more differentiation rule: if F(x)=1/f(x), and f(x) is differentiable, then predict F'(x). What follows resembles algebraically very much an example we did last time, with 1/x2. So here goes:
If F(x)=1/g(x), then F(x+h)=1/g(x+h), and (F(x+h)-F(x))/h=[1/g(x+h)-1/g(x)]/h (this is a compound fraction, and I want to write it as a simple fraction, a quotient of two expressions) =[g(x)-g(x+h)]/[g(x)·g(x+h)·h]. Wow. With effort we can recognize the pieces. First, [g(x)-g(x+h)]/h-->-g'(x) as h-->0. And the "other stuff"--> 1/g(x)2 because g(x), since it is differentiable, must be continuous. This is a sort of reciprocal rule. Lots of work. I applied this to finding the derivative of 1/x207. The result is -207x206/(x207)2. In fact some algebra can be productively performed on this, and we get -207x206-2(207)=-207x-208. Since the original function is 1/x207=x-207, we can see that the power rule also holds for negative integers. Recognizing patterns is an important part of mathematics, and a very important part of being usefully lazy.

The graph of y=x3-3x is shown. The points A and B are indicated, locally a "top" and "bottom" of the curve. What are the coordinates of A and B?
A and B are points where the derivative is 0, since the tangent line is horizontal and the slope of a horizontal line is 0. If f(x)=x3-3x, then f'(x)=3x2-3, and this can be interpreted as the slope of the tangent line. f'(x)=0 when 3x2-3. Of course, this is an arranged problem, and occurs when x=+/-1. But when x=-1, f(-1)=(-1)3-3(-1)=2, so that the point A is (-1,2). And, similarly, the point B is (1,-2).

 Another Advertisement As several people noted, it is AMAZING that with just a little bit of "technology", a tiny amount of calculus, that we can exactly find the positions of A and B. I think I could do this problem without calculus, but I believe solving it would be rather difficult. So we have already achieved something.

## Differentiation algorithms

FunctionDerivative
FunctionDerivative
f(x) limh-->0(f(x+h)-f(x))/h This is the formal definition
Constant0
xn when n is a positive integernxn-1
ex (Here e is approx. 2.71828)ex
f(x)+g(x)f'(x)+g'(x)
f(x)·g(x)f'(x)·g(x)+f(x)·g'(x)
(Const)f(x)(Const)f'(x)
1/g(x)-g'(x)/[g(x)]2
MORE TO COME!!!
Wednesday,
September 24
These problems are due this Thursday in recitation: 2.6 #44 and 2.7 #8.

I began with a

 Typical Calculus Question Graph the function f(x)=1/x2. Find an equation of the line tangent to the curve shown at the point when x=2.
I first tried to get information about y=1/x2. We surely could plot the function, one point after the next, but that's tedious. Or we could use a silicon friend, but there might be circumstances where that's not possible. So, bare-handed, we looked at y=1/x2.

First we noted that the top of the fraction was 1, so that y can never be 0. Then we saw the 2 in the exponent. That meant the bottom was always positive, so y must always be positive. Therefore no part of the graph can be in the lower half of the plane. Also, since the square of -a and the square of a are the same, the graph must be symmetric with respect to the y-axis. Finally, asymptotic properties: as x-->infinity (of if x-->-infinity) then y-->0 so that the x-axis is a horizontal asymptote. Further, as x-->0, 1/x2-->+infinity, so that the y-axis is a vertical asymptote of the graph.

Let's work on the tangent line question. When x=3, y=1/32=1/9. To find an equation of the line we will need a point on the line (here that will be (3,1/9)) and the slope of the line. We will try to approximate the slope of the tangent line, mTAN, by the slope of a secant line, mSEC. We have done several problems before in this course using this method. In the accompanying pictures, the graph of f is shown in black. The tangent line is in red, and the secant line is in green. The dashed blue circle is intended to show an implied magnified view.

I will use the notation of your text. There mSEC is gotten by finding the slope of the line connecting the points where x=3 and x=3+h, where we think of h as "small". So since the secant line connects (3,1/9) and (3+h,1/(3+h)2), so mSEC=[1/(3+h)2-1/9]/[(3+h)-3]. We want to discover what happens as h-->0. If I try my secret approach which is more or less, plug in, disaster occurs: 0/0. We must transform this quotient algebraically in order to discover what happens. Here we go: [1/(3+h)2-1/9]/[(3+h)-3]=[1/(3+h)2-1/9]/h. Let us look at the top of this fraction: 1/(3+h)2-1/9=[9-(3+h)2]/[(3+h)2·9]. But (3+h)2=9+6h+h2 so the top becomes just [9-(9+6h+h2)]/[(3+h)2·9]=(-6h-h2)/[(3+h)2·9]. We should not forget that this is all divided by h, which was the bottom of the original fraction. Therefore, since (A/B)/C=A/(B·C), we know that mSEC=(-6h-h2)/[(3+h)2·9·h] and now we can cancel h's top and bottom. There are many ways to make algebraic errors in all this! So finally mSEC=(-6-h)/[(3+h)2·9] and as a function of h we can easily see what happens as h-->0. The quotient --> -6/(32·9)=-6/81.

Therefore the slope, mTAN is -6/81, and an equation for the tangent line is (y-1/9)=(-6/81)(x-3). I remarked that this was a fine answer to the question asked, and if, for example, I were to ask such a question on an exam, then this response, exactly as given, would be a valid, correct answer. One doesn't need to simplify anything.

There is a good deal of vocabulary associated with what we just did. Depending upon the application, mSEC could be the average rate of change or the average velocity. mTAN would be the instantaneous rate of change or the instantaneous velocity. And mTAN is also called the derivative of 1/x2 at x=3.

So here is the real formal
Definition A function f is differentiable at x if the limit limh-->0(f(x+h)-f(x))/h exists. If the limit exists, then the value of the limit is called the derivative of f at x, and is written f'(x).

Example: f(x)=1/x2. Then the difference quotient (f(x+h)-f(x))/h is [1/(x+h)2-1/x2]/h. Combining the fractions just as before yields [x2-(x+h)2]/[h·(x+h)2·x2]. The top becomes x2-(x+h)2=x2-(x2+2xh+h2)=-2xh+h2. The giant fraction is then [-2xh+h2]/[h·(x+h)2·x2] and, again, an h can be lost (remember, in this limit computation, h is not 0). The difference quotient is then [-2xh+h2]/[(x+h)2·x2] and we can compute the limit as h-->0 of this (so the limit exists!). The value of the limit is f'(x)=-2x/x4=-2/x3.

 Advertisement It turns out that if a function is defined by a "simple" algebraic formula, there are algorithms (rules) which make the transition from f to f' fairly easy. Please read ahead in chapter 3 about this, or else the next week in class could be very confusing.

What else? Many people in class claimed to have seen what I just went over. I want to be sure to point out some important other interpretations, one from geometry, and one from a computational point of view.

Geometry
We can look at a graph of y=f(x) and imagine a sequence of "zooms" centered at a point. The dashed blue circles represent the domain of the zoom, and the dotted blue circles are supposed to represent what happens after the zoom to the dashed circles.

The more one zooms in, the more the curve, f(x), with tangent line, start to resemble each other. So a curve representing a differentiable function is locally linear: you can magnify the graph so it must look like a non-vertical line, and the slope of the line is the derivative of the function at that point.
Unfortunate technical detail
There are some nice curves (for example, circles) where we can zoom in on any point and the curve gradually begins to look more and more linear. However, in the case of the circle, there are two points (on the horizontal diameter) where the tangent lines are vertical, and therefore have no slope. The function(s) involved will not have derivatives at those points. There are ways of avoiding this difficulty which will be explained later.

There are functions which are not locally linear, "clearly". Several examples were suggested by Mr. Morris. The one on the left is y=x2/3 which has a cusp at 0. The one on the right is y=|x| which has a corner at 0. I analyzed y=|x|, but for both I think that the zooming idea is not too hard to imagine -- magnify the graphs centered at (0,0) and you will "see" that they don't ever look close to straight lines.

If f(x)=|x|, the difference quotient (f(0+h)-f(0))/h becomes just |h|/h. We may analyze this separately for h-->0+ and for h-->0-. For the right-hand limit, |h|=h since h is positive. The difference quotient is then h/h and this is 1, so the right-hand limit is 1. For the left-hand limit, |h|=-h since h is negative. The difference quotient is then -h/h and this is -1, so the left-hand limit is -1. The "whole" limit does not exist, since the left- and right-hand limits don't agree. You should be able to "see" the different limits in the slopes of the two sides of the absolute value graph.

Computation
If f is differentiable at x, then the official definition states that limh-->0(f(x+h)-f(x))/h exists and is called f'(x). So f'(x)=limh-->0(f(x+h)-f(x))/h exists. If we remove the "phrase" limh-->0 then in general the two sides are not equal. The limit states that some approximation gets better and better. In fact, we should really write
[(f(x+h)-f(x))/h]=f'(x)+Err, where Err is some sort of Error term which will depend on the function, f, and on x and on h (probably not on the latitude and probably not on the longitude). I just know that as h-->0 then Err-->0. We can unroll this equation to get

 [REMARKABLE]   f(x+h)=f(x)+f'(x)h+Err·h

What does this equation "say"? If we think about a function as a machine, then it says that if we perturb the input to a differentiable function f by changing x to x+h, then the output has some sort of structure: the old output, f(x), and a change in the output which takes the change in the input, h, and multiplies it by f'(x) (the derivative here appears as a multiplier of an input disturbance). The last term, Err·h, is sort of a higher order term: if h is small and Err is small, then the product is likely to be smaller. The most important change in the output is because of the multiplier effect on the input. We will investigate this important equation further, and try to see the numerical consequences of it.

Continuity and differentiability
We saw that |x| is a function which is continuous at all points but is not differentiable at 0. The [REMARKABLE] equation above shows that as h-->0, then f(x+h) must -->f(x). So any function which is differentiable must be continuous. The example of |x| shows that the reverse implication is not always true.
This is an important observation which I forgot to point out in class.

The Question of the day was: use the formal definition of derivative to determine for which x's is the function f(x)=5x3 differentiable, and to find f'(x) for those x's.
I was asked for a formula for (A+B)3 and gave the answer A3+3A2B+3AB2+B3. This can then be used to analyze the difference quotient ... and, as before, the h underneath (?) drops out, and as h-->0, we get f'(x)=15x2. It is important to recognize that f(x+h) is 5(x+h)3.

Monday,
September 22

I began by confessing that I was the demon who wanted people to write workshops neatly, and staple them, etc. The standards are easily available. Please honor your own efforts by presenting it well. This is something which your workplace will want, also.

Then I discussed a problem I had not put on the last workshop. It concerned the functions F(x)=x3 and G(x)=4cos(7x+5)+8sin(x2-9)+6. I can pretend that I understand the function F fairly well, but G is defined by a confusing mess of a formula. The domain of G is all real numbers so there is no problem of domain. But the combinations (the composition!) of x2 with sine -- that is horrible. The result is no longer a periodic function. The x2 makes quicker and quicker "trips" between -1 and +1 as x gets larger. What's to the right is just a picture of sin(x2 showing this phenomenon. So the graph of G will indeed be terrible. I wanted to solve F(x)=G(x). Well, how terrible is G? A student pointed out that G's values (its output, but not its input) can be roughly described quite easily. The first term, 4cos(7x+5), will surely be between -4 and 4. The second term, the "troublesome" 8sin(x2-9), will be between -8 and 8. The last term is just 6. We can add up these three sets of restrictions, and see that for all x, -6<=G(x)<=18. This is nice. That means the graph of y=G(x) must lie only between the lines y=-6 and y=18.
The graph of F(x)=x3 is, by contrast, very big when |x| is very big. So the equation F(x)=G(x) is really an example of BIG STUFF=negligible stuff. Now I will take advantage of the fact that I understand the BIG STUFF (x3) very well, and that the BIG STUFF has very large positive and negative values. The solutions we want are the same as the solutions of F(x)-G(x)=0. If we insert x=10, the value of F(x)-G(x) is 1000-(something between -6 and 18), and this is certainly positive. When x=-10, the value of F(x)-G(x) is similarly certainly negative. (Yes, this is an invented example, but situations like this occur often enough in practice to justify displaying it in class.) Since F(x)-G(x) is a continuous function having both positive and negative values, therefore by the Intermediate Value Theorem, F(x)-G(x) must be 0 at some x. Here is a picture drawn by Maple of F(x) and G(x) (I think you can tell the curves apart!). The intersections of the curves show the roots of F(x)=G(x).

I then asked the Question of the day: Suppose that f(x) is defined piecewise in the following way: f(x)=2-x2 for x>=3; f(x)=ax+b for 1<x<3; f(x)=x2+2x for x<=1. Please select a and b so that f is always continuous.
Here one must "match" left- and right-handed limits at 1 and 3 in order to get information. The limits at x=1 give a+b=3 and the limits at x=3 give 3a+b=-7. These two linear equations can be solved to give the answers a=-5 and b=8. Most people seem to have gotten these answers.

These problems are due this Thursday in recitation: 2.6 #44 and 2.7 #8.

I then tried to see what I could learn about the asymptotic behavior of Q(x)=[x(x-1)]/[(x+3)(x-4)2] using only little or no technology. This will help us understand how such functions behave, and will serve as a useful support later when we have more elaborate methods for graphing functions. I have a Maple plot of y=Q(x) in the window where x is in [-10,10] and y is in [-15,15]. We know already that something "funny" (?) may occur near x=-3 and near x=4. I tried to analyze slowly what happened.
Near x=-3, the top of Q(x) becomes about 12, and the bottom is (x+3)(-7)2, which is 49(x+3). As x-->-3- (from the left) then x<-3 so x+3<0, and x+3 is small and negative. Therefore Q(x) is the quotient of 12/(49[SMALL/NEGATIVE]) and this is a large negative number. The textbook would like us to write this as limx-->-3-=-infinity. It takes some practice to see that this means a vertical asymptote, and that y goes down as x approaches -3 from the left. A similar analysis as x-->-3+ (from the right) establishes that limx-->-3+=+infinity. x=-3 is a vertical asymptote of the graph of y=Q(x).
Near x=4, the top is about 12, and the bottom is 7(x-4)2. Because of the square, the behavior of the quotient is the same from both sides. The bottom becomes small and positive, and the quotient is 12/(7[SMALL/POSITIVE]) which is a large positive number. Again, the text's notation for this is limx-->4=+infinity. And x=4 is a vertical asymptote of y=Q(x).

Other information we can get readily from this nicely factored form of Q(x) is that Q(x)=0 only at 0 and 1. Since Q(x) is continuous we can conclude that Q(x) has only one sign (positive or negative) inside the "unit interval", (0,1). In fact, further analysis of the sign changes shows that Q is negative inside that interval. And there is a suitable graph to show you (the horizontal spread is much narrower than the preceding graphs).
Finally I wanted to determine what happens to Q(x) when x is very large, that is, when x-->+infinity. Here I can take advantage of the precise form of Q(x) again. I divided the top and bottom of Q(x) by x2 and got this: Q(x)=[1-(1/x)]/[(x-3)(1-(4/x))2]. As x-->+infinity, the top of Q(x)-->1 and the bottom is really approximately x-3. So the quotient-->0, and limx-->+infinityQ(x)=0. A similar analysis as x-->-infinity also gives 0. Of course this is a result of the fact that the degree of the bottom of Q(x) is higher than the degree of the top. So we get information about asymptotic information when x-->+infinity and x-->-infinity which might be useful.

### The story of Fred and Jane (!), bugs

I expressed my sincere admiration for those students who were fluent in more than one language, and remarked that a goal of mine in this course is to add even more fluency to the linguistic achievements of students, by making "math", and more specifically the calculus dialect of math, a language they will be quick to use. So I went over last week's workshop problem, which investigated the motion of Fred and Jane. Fred moved uniformly up the y-axis, so his position at time t (t positive) was at (0,t). Jane moved on the curve y=sqrt(x) so that her first coordinate was at x=t. Therefore Jane's position was at (t,sqrt(t)). I was interested in the long-term behavior of the line segment from Fred to Jane. In this case, looking at specific t's (t=2 or t=4 or whatever) tells relatively little. The "long-term behavior" should be translated into math language as t-->infinity.
Length
The distance from Fred to Jane is the length of the line segment. This is sqrt((t-sqrt(t))2+t2). While this is a complicated expression, to obtain asymptotics about its behavior as t-->infinity all we need to notice is that both squared terms are non-negative, and certainly the t2 alone goes to infinity. And taking the square root of that won't change the asymptotic behavior, so that the limt-->infinityLENGTH must be infinity. So, in my "narrative" accompanying my computations I would remark that the length grows without any bound. I might remark that it gets large and stays large.
Slope
We can get the slope by taking the difference of the second coordinates and dividing by the difference of the first coordinates (of Fred's position and Jane's position). The result is (t-sqrt(t))/(-t). If we divide top and bottom by t, the result is (1-1/sqrt(t))/(-1). Now as t-->infinity, 1/sqrt(t)-->0, and the result is that the limit exists and is -1. I think this is not completely obvious! limt-->infinitySLOPE=-1. The long-term behavior is that the line segment connecting the two bugs has slope getting more and more like the "antidiagonal" (if that is a word).
Thursday,
September 11
Please hand in problem #4 of this week's workshop problems.
Wednesday,
September 17
I began the lecture by responding to the concern expressed by Mr. Soang in e-mail to me. He respectfully suggested that I go over some textbook problems. I thank him for this suggestion and tried to go over some textbook problems.

I discussed some parts of (section 2.3) problem #2. Here are the relevant graphs.

The general instructions for this multipart problem were approximately:
do the indicated limits exist? If they do, explain why and evaluated them as well as possible. If they do not, explain why.

For this we need some limit laws. (This sounds ominous.) So we need to analyze limx-->2f(x) and limx-->2g(x). The f limit is 2 (even though f(2) itself is 1 since (2,1) is on the graph of f!). We see that limit by tracing the graph when x is near 2. This is almost the best I can do to describe the process, which to me seems a combination of the visual and the kinetic: feel it, see it, etc. As for the g limit, its value seems "clearly" (well, about as clearly as the f limit) is 0. So by the limit laws, the requested answer is 2+0=2.

Part (d) asks about limx-->-1f(x)/g(x). Again we separately analyze the limits: limx-->-1f(x) is -1: that is, when x is close to -1, f(x) gets close and stays close to -1. The situation for g is more interesting relative to the limit laws, since limx-->-1g(x) is 0. Now the situation is -1/0. The limit laws don't apply since the bottom (I have forgotten the word "denominator") is 0. This means we must look at the situation more closely, and not just abandon it.
What happens on the left?
If x is close to -1 but less than -1, f(x) is close to -1, and g(x) is close to 0 and negative. I get the information about g by looking closely at the graph of g "immediately" to the left of x=-1. The quotient is -1/(small negative #) and this is a large positive number.
What happens on the right?
If x is close to -1 but greater than -1, f(x) is close to -1, and g(x) is close to 0 and positive. I get the information about g by looking closely at the graph of g "immediately" to the right of x=-1. The quotient is -1/(small positive #) and this is a large negative number.
The behavior of the quotient is rather different on both sides of -1, so no one unique limit exists: the limit limx-->-1f(x)/g(x) does not exist. The textbook would actually describe this situation in more detail as follows:
limx-->-1-f(x)/g(x)=+infinity
limx-->-1+f(x)/g(x)=-infinity
The need for similar analysis will occur often when we systematically want to graph functions.

Problem 11 of section 2.3 asks for analysis and possible evaluation of the limx-->2(x2+x-6)/(x-2). Here I confessed what I actually do in the back of my head when I am asked about a limit defined by a fairly simple formula, as in this case. I just "plug in" x=2 and get 0/0, which is a bit interesting. I need to analyze the situation algebraically. Please note that this "plugging in" strategy is something I really only apply for fairly simple formulas. It won't work for graphically given information (see the pictures above, for example). And it would be very laborious for intricate formulas, where I look for better strategies. Anyway, here (x2+x-6)/(x-2)=[(x-2)·(x+3)]/(x-2)=x+3. There is some need to be careful of logic here. The last equality holds exactly when x is not equal to 2. But that is the situation which holds here, since in dealing with "limx-->2" we do not look at what happens when x=2. Anyway, I do know that x+3-->5 as x-->2, so we are done.

I started to do problem 33 of the textbook but got distracted and "invented" another problem, perhaps more interesting. If one takes an old car, and pushes down on one corner of the body and then releases it, the car will vibrate up and down for a while, with diminishing amplitude. I hope anyway the amplitude won't increase (I think energy is conserved, and likely there is friction, so the amplitude will decrease -- this is far from an ideal pendulum!). This type of "motion" is called damped oscillation. It occurs also, for example, in simple LRC circuits. Or in a spring vibrating in jello. The result is that the equilibrium position is the limit as time-->infinity. Of course, in "real life", I think one would be able to feel the up and down vibrations of the car only a few times. That the limit is infinity is an example of the squeeze theorem, another limit law.

For a specific example of a kind of function to which we could apply the squeeze theorem, consider D(x)=e-3x cos(4x). The exponential factor, as x-->infinity, has a negative constant. This function an example of exponential decay. The function cos(4x) wiggles back and forth. Certainly cos(4x) is bounded -- there is a horizontal strip in the plane which contains the function: say the strip whose boundaries are y=10 and y=-10 (in fact, of course, |cos(4x)| is always less than or equal to 1, but I want to emphasize that any bound is o.k., and we don't need to find the best possible bound, whatever that may be!). Since limx-->infinitye-3x is 0 and the other factor is bounded, the squeeze theorem applies and the resulting function, here D(x), has limit 0 as x-->infinity. This is a horizontal asymptote on the graph of D.
The picture here was drawn by Maple. In order to show the oscillations better, I had to make the "damping" slower. So this is a graph of the function e.3xcos(4x), and of the upper envelope (e-.3x) and the lower envelope (e-.3x).

I then did problem 4 from section 2.4: the PRECISE definition of the limit. This is graphical, and I'm getting way over quota for drawing graphs in this diary entry! The trick here was finding a good delta which would work on both sides of x, and we decided that one good delta was .7, and, in fact, as I indicated, any positive number less than .7 would be a valid answer. However, if this were an exam problem, I would also request an explanation of why the answer was correct. And this would count for most of the credit in the problem.

I looked at problem 41: how close to -3 do we have to take x so that 1/[(x+3)4]>10,000? We cross-multiplied, and got 1/104>(x+3)4=|x-(-3)|4, where |x-(-3)| represents the distance from x to -3. And if we take "fourth roots" (?) we see that x must be closer to -3 than .1 (one-tenth). I remarked that problem 42 essentially asked what limit this computation applied to, and the answer was:limx-->-31/(x+3)4=+infinity. This is a vertical asymptote on the graph of y=1/(x+3)4.

## Limit Laws

Limits and arithmetic
Suppose limx-->af(x)=b and limx-->af(x)=c. Then
• limx-->af(x)+g(x) exists and is b+c.
• limx-->af(x)·g(x) exists and is b·c.
• limx-->af(x)-g(x) exists and is b-c.
• If c is not 0, limx-->af(x)/g(x) must exist and is b/c.
Warning This situation arises frequently in beginning calculus (the derivative is an example). If c is not 0, you cannot conclude that the limit of the quotient doesn't exist. It may or it may not: you must investigate further!
Limits and order
• If f(x) is always non-negative, and if limx-->af(x) exists then the limiting value must be non-negative.
Warning Look at x2 near 0. For non-zero x this is positive, but the limiting value at 0 is 0. So a "strict" inequality doesn't persist for limits, just the weaker ones.
• If f(x) is always non-positive, and if limx-->af(x) exists then the limiting value must be non-positive.
• The Squeeze Theorem
Version 1 If you know that f(x)<=g(x)<=h(x), and you know that both limx-->af(x) and limx-->ah(x) exist and are equal, then limx-->ag(x) (caught in the middle!) must exist, and its value must be the common value of the other two limits.
Version 2 (Most common application) Suppose you know that limx-->af(x) exists and is 0. If you also know that a function g is bounded near a (so there is some number M with |g(x)|<M) then limx-->af(x)·g(x) exists and is 0.
This result is used a number of times in applications. Limits which are 0 can swallow up bounded things. You could think of a function as bounded if its graph can be confined in a thick horizontal strip in the plane. You don't have to necessarily know the exact height of the strip -- that isn't required. You just need to know that such a strip exists. For example, I know that 5sin(2x)+11cos(1/(x2+1)) is never bigger or smaller than 5+11=16 or -16, so this function is bounded with M=16. I know this because cosine and sine are never bigger than 1 in absolute value, and the argument (the "input") doesn't matter. The worst estimate is gotten when there is no cancellation, and that worst case would just give me 16 as the biggest answer.

We went on to section 2.5, and tried problem 3. This gave another graph, and we were asked to list where the function whose graph is shown is not continuous (this is called discontinuous) and then where it was continuous from the right and continuous from the left or neither. We discussed this problem after I wrote a definition of continuity.

A function f is continuous at a if

1. a is in the domain of f and
2. limx-->af(x) exists and the value of this limit is f(a).
This is one of the important definitions of the course. Continuous functions are those whose limits can be evaluated by "plugging in". The textbook gives some variations on this definition in the following ways. Recall that limx-->af(x) exists exactly when limx-->a-f(x) and limx-->a+f(x) both exist and their values coincide. If only one of the two limits exist, and if that limit equals f(a), then (for the - case) the function is continuous from the left and (for the + case) the function is continuous from the right. Here are some pictures.

Many functions given by simple formulas (polynomials, rational functions, trig functions, exponential and logarithmic functions) are continuous in their domains.

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 Ho-Ho-Kus. 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.

 Time Location(in miles on GSP) 7 AM Some time? 10 AM 0 135 172

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 (7,0) to (10,172) and therefore the graph must have on it at least one point with coordinate description (*,135). All of this, by the way, is a complex bunch of assumptions. I guess today I believe that motion is continuous, and therefore at sometime Francine must be at Mile 135. By the way, we should 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).

The textbook problems due tomorrow are 2.3 #38 (use the Squeeze Theorem!) and 2.5 #40 in addition to the workshop problem.

The Question of the day was to draw the graph of a continuous function defined on [-1,3] so that f(-1)=-1 and f(3)=2 and f has exactly two roots. Of course, it is possible to interpret this question in several ways if one knows what the phrase "double roots" means. Sigh. I just expected something like what's shown, something simple.

Monday,
September 15
I began by finding the slope of the line tangent to the graph of y=sqrt(x) at the point where x=A. If we wanted the tangent line, then we know the line must go through the point (A,sqrt(A)) since that point is on the graph. The tangent line is in red on the graph. We need mTAN, which we will call the slope of the tangent line. The classical method is to approximate this slope by the slope of a "secant" line, which is green on the graph. This secant line passes through the two points P=(A,sqrt(A)) and Q=(A+h,sqrt(A+h)) on the graph. The strategy is to compute mSEC, the slope of the secant line, and then to examine the algebra closely as Q gets close to P try to discern what happens to the slope, since mSEC should "approach" mTAN then.
```/discern/ v.tr.
1. perceive clearly with the mind or the senses.
2. make out by thought or by gazing, listening, etc.
```
Now mSEC is the difference of the second coordinates of Q and P divided by the difference of the first coordinates. This is (sqrt(A+h)-sqrt(A))/h. Things to remark:
1. The square root of a sum has no simple relationship with the square roots of the parts of the sum. For example, as was suggested by a student, sqrt(1+1)=sqrt(2) which is approximately 1.4, while sqrt(1)+sqrt(1) is 2. Note that 2 and 1.4 are not equal.
2. If we think about this when h is small, we see that we will have a quotient of something small by something small. This situation is very unstable computationally, and even if you try your calculator on it for h very small you may not be enlightened.
``` /enlighten/ v.tr.
1. (often foll. by "on") instruct or inform (about a subject).
2. (esp. as "enlightened" adj.) free from prejudice or superstition.
3. [rhet.] or [poet.]
a. shed light on (an object).
b. give spiritual insight to (a person).
```
So we will try to transform the quotient algebraically to get a more tractable formula.
```/tractable/ adj.
1. (of a person) easily handled; manageable; docile.
2. (of material etc.) pliant, malleable.
```
So take (sqrt(A+h)-sqrt(A))/h and multiply top and bottom by sqrt(A+h)+sqrt(A). I was told that this is the "conjugate". In any case, since 1=(sqrt(A+h)+sqrt(A))/(sqrt(A+h)+sqrt(A)) it won't change the value.
(sqrt(A+h)-sqrt(A))/h=[(sqrt(A+h)-sqrt(A))/h]·(sqrt(A+h)+sqrt(A))/(sqrt(A+h)+sqrt(A))=[(sqrt(A+h))2-(sqrt(A))2]/[h·(sqrt(A+h)+sqrt(A))].
Then the top, which is (sqrt(A+h))2-(sqrt(A))2, becomes just h. And by coincidence (not really, not at all!) there is also an h in the bottom. We can cancel these. So the quotient, mSEC, becomes just 1/(sqrt(A+h)+sqrt(A)). The behavior of this is relatively easy to understand as h gets small. A+h-->A, so sqrt(A+h)-->sqrt(A), so sqrt(A+h)+sqrt(A)-->2sqrt(A), so mSEC-->1/(2sqrt(A)) and this must be mTAN.

EDITORIAL
from the management

Some students have already objected to the rather tedious analysis above. They've said, "Why don't you do this the short way?" Well, o.k., there is a sort of short way. If functions are defined by simple formulas, there is a set of algorithms to go from one formula to another. My very serious response is that's not the content of the subject. If you log onto eden and type the word maple at a "command line" and then type diff(sqrt(x),x); you'll get the answer I just wrote above. If that's what you think the subject is about, then you could buy a copy of Maple or Mathematica or Derive and stay home. It is certainly true that learning the algorithms of calculus is part of the course, but learning what's going on and why is the reason you're in Math 151. I hope to be successful in explaining what's going on and why here. You will need to work, and buy into the truth that the content of the course is more than the manipulative aspect of it (or else you could be replaced by a computer program!).
 (possibly) The Management
```/algorithm/ n.
1. [Math.] a process or set of rules used for calculation or
problem-solving, esp. with a computer.
```
I then discussed problem 3 of last week's workshop. The evil aliens changed one million points on the graph of S(x)=x2 to obtain a function V. A part of the graph of V might look like what's shown at the right. The dashed vertical line segments are drawn to help you understand how the points are moved from the graph of S to create this graph. Then we discussed what the limit of V(x) as x-->c should be. Limits as x-->c only deal with the asymptotic behavior of the function near c but not at c. Therefore, the aliens who changed only a few (hey, a million points on the real line is only a few!) values of S haven't disturbed the limiting properties. If you get close enough to an undisturbed point on V, the graph looks locally like the graph of S, so the limits of V and S will agree. If you get close enough to a disturbed point, well then, the graph of V and the graph of S away from the disturbed point will agree locally near the disturbed point, so again the limits of V and S will agree.

Even worse, much, much worse
A brave student from India, Pankaj Bajaj, spoke to me after class. After the usual complaints (addressed in the editorial above) he and I discussed a much worse menace, the Infinite Aliens. Unlike the finite aliens above, these could disturb the graph enough to change some limits. For example, the infinite aliens could change S on all of the points x=1/n where n is an integer. If they made the values of the new function 308, then the limit as x-->0 wouldn't exist at all. So if there is lots of change possible (here "lots" is more precisely infinitely much) then the limits could change. That is a complication I did not want to bring up in the workshop problem.

I then went from a graphical representation of functions to a more "mechanical" interpretation. Here the function was a box, with inputs and outputs. I tried to describe more precisely what limx-->af(x)=b might mean. The method I wanted to use involved what I called input and output tolerances.

The first example, admittedly rather simple, was the function f(x)=3x+7. Here f(1)=10. In real life, it is probably very unusual to get exactly the value or values you want. I ludicrously tried to use the example of manufacturing screws: the manufacturer may want to produced an ideal size, and reject screws that aren't that size. But really there would be some acceptable tolerance, otherwise the pile of rejections might be the total output! Now I asked is there so input tolerance so that if |x-1|<that input tolerance, then |f(x)-10|<1/5, so that 1/5 is the output tolerance. I untwisted the inequalities, and we learned that:
If |x-1|<1/15, then |f(x)-10|<1/5.
In fact, by looking at the process of going from one inequality to another, we saw that if, for example, we wanted to guarantee outputs within 1/107, a rather narrower output tolerance, then:
If |x-1|<1/(3·107), then |f(x)-10|<1/107.
But of course this function is indeed rather simple.

Suppose f(x)=1/x. This is a non-linear function, and the amount of "stretching" varies. For example, I observed that although the distance between 100 and 101 is the same as the distance between 1 and 2, the "stretch" between f(100)=1/100 and f(101)=1/101 is less than 10-4 while the difference f(1)-f(2) is 1/2. In the case of 3x+7, the stretch is always a factor of 3. I asked people to use a calculator to determine an input tolerance which would guarantee that if |x-1|<input tolerance, then |f(x)-f(1)|<1/5. This "exercise" was not too successful.

The actual formal definition of limx-->af(x)=b uses the ideas of input and output tolerances. The traditional name for "output tolerance" is epsilon and the traditional name for "input tolerance" is delta. So the definition would go something like:
Given epsilon>0, there is delta>0 so that
if 0<|x-a|<delta, then |f(x)-b|<epsilon.
The reason for < between 0 and |x-a| is that limits are officially not supposed look at the value at the number, just near the number. If |x-a| is not 0, then x can't equal a.

I then returned to a picture. I drew a graph of a function H, and it looked something like what's at the right. I also made at least one silly mistake. Oh well. I asked:

• What is H(3)?
H(3)=3 since the point (3,3) is on the graph. (A point is on the graph exactly when it is (w,H(w)) for some w, so once you know a point is on the graph, that gives a value of the function.)
• Does limx-->3H(x) exist? If it does, what is its value?
The limit does exist. As x gets close to 3 (not at 3 -- please ignore the value at 3 when considering this question) then H(x), the height over the horizontal axis, seems to get close to 1. Therefore "convinced" (?) by this graphical evidence, I assert the limit exists, and that its value should be 1.
• What is H(-4)?
Again we can "read off" H(-4)=1.
• Does limx-->-4H(x) exist? If it does, what is its value?
Here the situation is more complicated. The limit from the left is ... well, the graph looks like the values of H are getting close to -2. From the right, the values look like they're getting close to 4. For the limit to exist, the outputs should be close to one specific value. That does not happen here. Therefore the limit does not exist.

I tried to illustrate the fact that limits don't exist and that strange things can happen by slowly bending a piece of chalk. The "response" by the piece of chalk, if you look at it and feel it closely, was to deform with the pressure until [poof!] the chalk breaks. So real phenomena can be quite complicated, even "everyday" things. In the graph above, there is some sort of limiting behavior, although the limiting behavior is different on the left and on the right. There is additional notation about this. We could say (for the function H whose graph is displayed):
The left-hand limit at -4 of H exists and is -2. The notation used for this is limx-->-4-H(x)=-2.
The right-hand limit at -4 of H exists and is -4. The notation used for this is limx-->-4+H(x)=4.
The superscripts - and + can sometimes be rather confusing! Please read the book and practice!.

The class ended with consideration of the function R(x)=(x-2)/(x2-4). This is a rational function (a quotient of two polynomials). It has "problems" at +2 and -2. I asked if the limits of R existed at those numbers, and this turned out to be the Question of the day! I encouraged and expected that people would try to graph this function. The graph, as drawn by the program Maple, is at the right. You can see that apparently the limit at 2 exists and the value of the limit ... well, it isn't too clear. In fact, a little bit of algebra helps:
(x-2)/(x2-4)=(x-2)/[(x-2)·(x+2)]=1/(x+2).
The last equality is true if x is not equal to 2, and since we are looking at a limit as x-->2, this requirement is fulfilled. Therefore the function away from 2 "is" 1/(x+2), and this has limit equal to 1/4 at 2. What about at -2? The algebra doesn't clarify this, and the graph makes clear that the function doesn't tend to one unique number as x-->-2. Therefore the limit as x-->-2 does not exist.

If you read the book you will see that the graphical situation shown here near x=-2 is described notationally by these two limit statements:
limx-->-2-R(x)=-infinity and limx-->-2+R(x)=+infinity.
The sort of graphical behavior shown here is typical for rational functions.
The Maple picture shown was gotten by typing the command plot(f(x),x=-4..4,y=-6..6,discont=true);
The silly colors occur because I don't know how to handle some parts of the program as it "exports" pictures.

I remarked that we were "covering" the material of sections 2.1--2.5 of the textbook. The reason for the quotes is that I believe I will only really talk about 30 to 40 per cent of this material, and you will need to work very very diligently to learn what you will need in this course. In particular, doing this will in effect force you to become proficient with the type of algebraic manipulations you need in this course.

I can't write the standard symbol for "infinity" (sort of a sideways 8) in html. So you may enjoy reading about where some of this came from.

I announced four problems and solutions to two of these will be collected Thursday. Sigh. I don't have the problem numbers with me now.

Thursday,
September 11
Please hand in problem #4 of this week's workshop problems although I really enjoy problem #3, since it "stretches" one's intuition. Also the strange constant in the preceding workshop problem is, I think,
 17sqrt(3)96
but I will check this as I read through the writeups of section 4.
Wednesday,
September 10
I discussed grading for the course.
First inclass exam (announced in advance)100 points
Second inclass exam (announced in advance)100 points
Final exam (Same for all sections)200 points
 Textbook homework (best 10 grades) Other Workshop writeups (best 10 grades) Class participation
150 points
A total of550 points

Strong suggestion I urged students to try to work in groups for two hours, say twice a week. Use the workshops to talk to other students, and use the supplied e-mail addresses to communicate. There is substantial statistical evidence to suggest that students who work together in groups do better in courses.

We now begin the central work in the course. Chapter 2 defines and discusses some of the central topics of the course: limits, continuity, and differentiability. It is my happy job to explain these ideas to you in the next week or so. I suggest again that a useful thing might be to spend 5 or 10 minutes before each lecture skimming the sections of the book I am likely to cover during that time. I will try generally to stay with the schedule (this is lecture #3).

So I started off with tree growth. I described a situation where two "facts" about a dwarf apple tree are supplied: the tree's rate of growth (which I think I gave as 4 inches per year) and the tree's ultimate height (given as 12 feet). I do think that these quantities are usually supplied when trees are described.

 What do these quantities mean? This is actually a fairly subtle question. If the rate of growth is actually always 4 inches per year, then, since there are 12 inches in a foot and 5,280 feet in a mile, and (I guess!) about 240,000 miles between the earth and the moon, doesn't the growth rate imply that the tree must hit the moon in about [how many?] millions of years? To the right is a hypothetical picture of this lovely situation. We discussed whether this really accurately represented what the given growth rate means, and decided it did not. Here is another interpretation, slightly more realistic. A student suggested that the tree grew at the given rate until it reached its ultimate height. So the graph should look like what's displayed on the right. The "rate" of growth corresponds to the slope of the central segment of the graph: 4 inches per year means that 36 years would "ideally" go by until the tree reached its "ultimate" height (12 feet is 36 times 1/3 since 1/3 is 4/12 of a foot per year). Please note that the horizontal axis units are in years, with time measured from when the tree began growing. The vertical axis is height of the tree in inches. What's shown is the graph of a function, H(t), the height of the tree at time t. The graph is what is called "piecewise linear": three line segments. The non-horizontal part of the graph has slope 1/3. Does the tree (any real tree?) grow like this? Probably not.

More subtle, more subtle: as one person suggested, all of these numbers are averages. The "12 feet" is an average for this population (exact figures are not likely in the real world!). And the tree probably never exactly stops growing. And when the tree is near its ultimate height it probably grows only slowly. And the growth is not steady (the growth rate is an average also, over all sorts of circumstances involving water and fertilizer etc.). And the growth is slow on an absolute scale initially. So all together the tree growth probably more looks like the graph below. The tree initially grows fairly slowly (region to the right of the first vertical line which isn't the vertical axis). Then it grows "steadily" for a while, at the stated growth rate. And then in the region to the right of the second vertical line, growth slows as the tree gets close to its ultimate height. The graph is a sort of logistic curve: a graph which is quite common in many chemical and biological applications.

In math language, the graph of H(t) has a horizontal asymptote at height 12. We can repeat this is more abstract math language: the limit of H(t) as t-->infinity is 12. Or limt-->infinityH(t)=12.
The growth rate in the central region compares the height of the tree over time. So tree growth over an interval of time divided by the duration of that interval is approximately 4 inches per year. Or (H(t+a little bit of time)-H(t))/(t+a little bit of time-t) is close to 4 (units ignored). The average growth rate is close to 4, and as the duration of the interval is decreased, we could hope that the growth rate gets close to 4 (in reality as the duration gets much smaller the growth rate probably hops around quite a lot, because of local chemical and climatic variations, but, sigh, let's ignore that). In traditional math language, a little bit of a variable is frequently replaced by the notation "delta (triangle) variable". So we have (H(t+delta t)-H(t))/delta t should get close to 4 when delta t-->0. This is the derivative, written H'(t): I guess we could call it the instantaneous rate of growth of H at t. Or we could write it as limdelta t-->0(H(t+delta t)-H(t))/delta t=H'(t). The H'(t) is the standard notation for the derivative of H, the local rate of growth of H.

We will spend much of the next week or two trying to understand and systematize what I just described. I then tried to discuss another situation, the coiling of biological macromolecules as a function of pH. This, it turns out, is not simple. So after discussing a few facts (don't let the pH of your body get far from 7.2, for example!) I tried a more traditional example.

We considered the motion of a particle. Suppose the position of a particle at time t on a line is given by S(t)=3t2. What can we say about the average velocity of the particle near t=1? We computed this for the interval [1,1.2], I think. It was the quotient (S(1.2)-S(1))/(1.2-1). And then I asked about the average velocity over the interval [1,1.005], etc. (This is more-or-less an example in the book.) I then asked what the average velocity was for the interval [1,1+h]. This was (S(1+h)-S(1))/(1+h-1)=(3(1+h)2-3)/h=(3+6h+3h2-3)/h and remarkably (I don't know. It's a mystery.) things cancel in amazing ways. The 3's drop out, and the h's up and down cancel, and we have 6+h as the average velocity in the interval [1,1+h]. Now as h-->0 remarkably this gets close to 6. Indeed (!) in physics 6 is called the instantaneous velocity of S(t)=3t2 at t=1. Whew!

The function S(t) is quite simple, but the general procedure turns out to be possible for a very wide variety of functions given by formulas. And interpreting the results gives lots of interesting results.

I then tried to consider the problem of find the slope of the tangent line to y=3x2 at x=1. We found the slope of the secant line, msecant, by looking at the points (1,3) and (1+h,3(1+h)2). Then as h-->0, msecant-->mtan, the slope of the tangent line. The algebraic manipulations are exactly the same as what we just did for the transition from average velocity to instantaneous velocity, so if we can do all this in one case we can surely do it in the other.

I then tried to find the slope of the line tangent to y=sqrt(x) when x=4. This means finding the slope of the secant line, the line joining (4,2) to (4+h,sqrt(4+h)). This slope is (sqrt(4+h)-2)/((4+h)-4). The 2 in the top is better written sqrt(4) (amazing!) and then we need to analyze (sqrt(4+h)-sqrt(4))/h. A suggestion was made to multiply top-and-bottom by sqrt(4+h)+sqrt(4) and see if a miracle happens. It does, and as h-->0 the result will -->1/4, the slope of the tangent line to y=sqrt(x) at x=4. And this was the Question of the day! The solution is to multiply on top: (sqrt(4+h)-2)·(sqrt(4+h)+2) gives 4+h-4 (hey: (A-B)·(A+B) is A2-B2) and the 4's cancel so the top is h. The bottom is h(sqrt(4+h)+2). The quotient is h/(h(sqrt(4+h)+2)) and then the h's top and bottom cancel and we are left with 1/(sqrt(4+h)+2). As h-->0, this -->1/(sqrt(4)+2) which is 1/4.

 The answer to the workshop problem due tomorrow turns out to be some weird number multiplying x2. I did the problem after class today with Michael Handel, a student from Warren County and section 4. But the weird number turns out to be truly weird. I will not torture you by writing it here. The true aim of this problem is for you to get an answer and explain the answer. Please do this.

Problems to hand in tomorrow at recitation: #24 on p.79 of the review problems for chapter 1; #6 of section 2.1 on this Thursday.

I will try to inform you of these problems ahead of time in the future.

Monday,
September 8
I discussed the importance of practice which in the context of this course means taking responsibility for a larger proportion of your own education, as compared, to, say, high school. You will need to work problems. I strongly suggest that you try to form workgroups with other students in the course (that's why I typed all those lists -- so you could get in touch with each other). There has been much investigation of factors leading to student success in such courses as Math 151. The major factor is students studying with other students. You tend to stay more honest then, and cover more material, and, if you have questions, you can try to explain the solutions to each other. I strongly suggest again that you form a group of 3 to 7 students and meet several times a week, and go over all of the assigned homework problems. Please think about this and do it!

I tried to use technology (in this case, a graphing calculator hooked up to a viewscreen hooked up in turn to an overhead projector -- a wonderful thing, all those wires!) in order to look at a graph of 2x and 3x. Both of these graphs go through (0,1) and we "zoomed in" on the point (0,1). I wanted you to observe what happened. It isn't always true that calculators show essentially correct pictures, but here it is. As we zoomed in the curves became more and more like straight lines, tilted straight lines, to be sure, but fairly flat. These curves had no jumps or breaks (this will be called by the technical word continuity later in the course). And also, under magnification, the curves looked more and more like straight lines. Another way of saying this is that the functions were locally linear near (0,1). Again there will be a technical word for this (differentiability) later. In this course we will be mostly interested in rates of change, and the most familiar geometric interpretation of the rate of change will be the slope of the tangent line. 2x seems to have a tangent line at (0,1) with slope about .7 (I will justify this later -- it is not supposed to be clear now) and 3x seems to have a tangent line with slope about 1.1. Here we need a leap of "faith" (it seems silly, but more or less it is faith in the subject or intuition or whatever). There must be an a between 2 and 3 so that the graph of ax at (0,1) has slope=1. This would be the easiest slope to deal with. It turns out (we must justify this later) that there is such a number. The number is called e, and this number, between 2 and 3, is approximately 2.71828.

 e (approximately 2.71828) is selected so that ex has a tangent line with slope=1 at (0,1).
We then discussed exponential growth (ax with a>1): this models cell division, bacterial growth, chain reactions, etc. As you travel from left to right on this graph, the height (the y-value) increases steadily and remarkably rapidly. You should have a mental picture of this graph, please. I remark additionally that the word "model" there is used in a very restricted sense. I don't think that if a bacteria under certain circumstances doubles in the number of individuals every 5 minutes that it will fill the known universe in ... I guess a few days. "Models" must be examined critically, and their construction and interpretation and validity can be difficult.

We discussed exponential decay: ax when 0<a<1. This sort of function models radioactive decay, cooling, discharge of certain electrical circuits, etc. It is the reflection of growth across the y-axis.

We wrote down but did not linger over a bunch of exponential formulas:

• ab+c=abac
• a-b=1/(ab)
• (ab)c=abc
• a0=1
These are valid for positive a's. In this course we will not allow the "base" of an exponential to be 0 or negative.

I then commented on Inverse Functions where input and output are switched. Such a candidate (!) for a function must satisfy the vertical line text, and this may not always occur. (Heck, even the simplest circle is not the graph of a function!) A function must give a unique output for every input, but one output may arise from several different inputs. The simplest example of this is the function F(x)=x2. Here F(2)=4=F(-2), so the candidate for the inverse function (whose graph is a sort of sideways parabola opening to the right) doesn't satisfy the vertical line text. Usually we restrict ourselves in this case to non-negative numbers. The inverse will be sqrt(x), whose domain and range are non-negative numbers. The graph of an exponential function, made up of the coordinates of points interchanged, is the result of the graph of the original function flipped over the "main diagonal", y=x.

The inverses of exponential functions are called logarithmic functions. So y=logax exactly when ay=x. So y is the "number of powers of a in x". I think I discussed and looked at the graphs of 4x and log4x.
Logarithm functions have a whole bunch of properties corresponding to the exponential properties listed above, for example: logabc=(logab)+logac). In "ancient" times these were very useful for computations, since such equations convert multiplications to additions, and additions are easier. Now we don't need computational help of that type (we have our silicon buddies) but it turns out that the modeling and graphical aspects of the exp and log functions are still very important.

 Exponential growth is very fast growth. Logarithmic growth is very slow growth.
I will need to explain those statements, but they are good to have in mind. In particular, the statements mean something like this (which is nearly unbelievable): 3x eventually is bigger and stays bigger than, say, x10,000 even though when x=10, the exponential function has "only" 5 digits (310=59,049) and 1010,000 has, of course, 10,000 digits. So much more later on this. I add, as I did not in class, that this contrast in bigness is very important when modeling chemical and biochemical reactions.

I rapidly discussed sine and cosine. The unit circle , the world's best circle is x2+y2=1. This is the collection of points (x,y) whose distance (the radius) from (0,0) (the center) is 1. If we think about a point moving at "unit speed" around the circle in the counterclockwise direction (this is called the positive direction and if we think of the radial line from (x,y) to (0,0), it has an angle which we could call theta with the positive x-axis. Then a little bit of triangle trigonometry says that the coordinate of the point (x,y) is actually (cos(theta),sin(theta)). Following a point on the circle "kinetically" (as the point moves) it is [almost] easy to see what the domain and range of sine and cosine are. So cosine and sine are more or less the "same" function (one lags the other by Pi/2). And these functions both have range in the interval [-1,1] (includes endpoints). And these functions are periodic with period 2Pi.Mostly I will try to use radian measure for angles, because that is more easily understood when analyzing uniform motion around the circle. I will "slip" and use degrees sometime. I drew some sort of picture of the graphs of sine and cosine, and talked a little bit about their domain and range.

 The functions sine and cosine both have domain all of the real numbers. Their range is [-1,1]. They are periodic (they repeat) in every interval of length 2Pi (the length of the unit circle). Pi is about 3.14159, so that the graphs are actually fairly "flat" (ratio of about 3:1).

The Question of the day was: suppose I know that logax=6.4 Then

• What is loga(x3)?
• What is log(a4)x?

This was a bit vicious. I remark that, YES, you are supposed to know and be able to use things you have known before. Please practice! So what does "logax=6.4" really mean? It means there are 6.4 powers of a "in" x. Written as an equation, it means that a6.4=x. The first question asks what are the number of powers of a "in" x3. Well, if a6.4=x then we could cube (raise to the third power) this equation. We would get (a6.4)3=x3. But repeated exponentials multiply, so that a6.4·3=x3, and since 6.4·3=19.2, we conclude that loga(x3)=19.2. If you are the type of person that likes to memorize formulas (not that there's anything wrong with that) then loga(x3)=3logax=3·6.4=19.2.
The other question is more ... interesting. We know a6.4=x and we need to know something relating a4 and 4. Well, "golly" (?), by using some of the exponential rules, we see that (a4)something=a6.4. This means 4·something=6.4 so that something=1.6, and therefore (a4)1.6=x. There are 1.6 powers of a4 "in" x. So log(a4)x must be 1.6 and we have done this problem by "pure thought".
I hope that you were able to do the first part of the QotD, which should have been almost routine. The second part needed more understanding.
Thursday,
September 4
Please hand in a writeup of the third workshop problem next Thursday. Please write in complete English sentences, hand in neat work, etc.
Wednesday,
September 3
I talked about Who are you? and Who am I? and What is this [class]?

We reviewed some of the material of precalculus: absolute value, associated manipulations, intervals, distance on a line, the plane as ordered pairs, distance between points in the plane, the condition that (x,y) be on a line determined by (a,b) and (c,d) (the equation of a line), slope, y-intercept, functions: functions determined by data, the "metaphor" of a function which turns inputs into outputs (with the collection of allowable inputs called the domain, and the collection of all outputs called the range), functions determined by graphs and ways of making new functions, and finally functions determined by formulas. It is very important that students read the textbook and try appropriate homework problems.

The Question of the Day (to be abbreviated QotD from now on) concerned the function A whose graph is shown here. We determined that the domain of A was the closed interval from -1 to 3: this is [-1,3]. The range of A was the closed interval from -2 to 2: this is [-2,2]. We can tell this is the graph of a function by applying the vertical line test: any vertical line hits the graph at most one time. This means there is a unique output for each input. The domain consists of those numbers c for which the vertical line x=c hits the graph shown. The range consists of those numbers d for which the line y=d hits the graph shown.

So for the QotD I asked people to sketch the graph of the function E determined by the equation E(x)=1/A(x), and to describe the domain and the range of E. This is a difficult question. I think one way to handle it is to look first at the domain: which x's can we "feed into" 1/A(x) so that the result makes sense. Now we should not divide by 0 (in this class or in any other class, I think!). So we should not "feed in" x=-1 or x=2 to the function E. We've got to throw those x's out. So the domain of A, which is [-1,3] gets altered to create a rather weird domain of E: the interval (-1,2) (an "open interval" not containing either end point) and the interval (2,3] (a "half-open interval" just containing one end point). This is fairly strange unless you remember things like the domain of secant (!) etc. What about the graph of E in the interval (-1,2)? Well, the height (the second coordinate) of the graph of A goes from (near) 0 on the positive side on up to 2 and then down to (near) 0 on the positive side. Since E(x) is 1 over A(x), we need to think about how "one over" behaves. "one over" takes small positive numbers to large positive numbers. And, at least for positive numbers, it reverses inequalities. For example, 2<3 and 1/2>1/3. So "pure thought" (a difficult exercise) says that if we want to graph E on the (-1,2) part of its domain, it will have vertical asymptotes (going to "+infinity" at x=-1 and x=2, and dip down to a lowest point at x=0 when E(0)=1/2. Whew! This is hard work. On the part of the domain we haven't considered yet, that is, (2,3], the values of A are negative so that "one over" will also be negative. And "one over" a small negative value is a large negative value (the sign stays the same, but the magnitude changes). Therefore E has a vertical asymptote at -2 from the right, and this asymptote goes to "-infinity". Otherwise the graph just goes on up to E(3)=1/A(3)=-1/2. The result is the graph that is shown. So the domain of E is the two intervals (-1,2) and (2,3]. The range of E is the two intervals [1/2,+infinity) and the interval (-infinity,-1/2].

Please hand in these two textbook problems:

• Appendix A, #28: Solve the inequality in terms of intervals and illustrate the solution set on the real number line: x2<2x+8.
• Section 1.1, #50: Express the surface area of the cube as a function of its volume.
in recitation on Thursday, September 4.