Math 135 diary, first part, spring 2005

Tuesday, February 8: this is lecture 7

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

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

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

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

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

In the case of "the line tangent to y=x2 when x=1", I know (because I know it when I see it!) that the line goes through the point (1,12). I write the second coordinate as 12 and not just 1 because I want to emphasize that we got it as a point on the graph of y=x2. We need to find a slope, m. In fact, because it is the slope of the tangent line, I will call it mtan.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

The future and the far future ...

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

Possible answers: