|Date||What happened (outline)|
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
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.
/heuristic/ adj. 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
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 inside)·the
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
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
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.
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.
|Please hand in a writeup of the fourth problem next Thursday, unless you have exceptional energy and want to do the first problem.|
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.
/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
Moving right along (!) I discussed the trig functions again.
Even more, we will need a kinetic view of the trig functions.
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.
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:
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.
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.
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
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.
Answer The answer is a huge quotient.
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.
These problems are due this Thursday in
recitation: 3.2 #32 and 3.4 #24
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:
We can do this more generally: (x+h)n=xn+nhxn-1+ h2JUNK
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.
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?
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:
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:
The Question of
the day was this:
These problems are due this Thursday in
recitation: 2.6 #44 and 2.7 #8.
I began with a
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
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.
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.
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.
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
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 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.
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
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.
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
The story of Fred and Jane (!), bugsI 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.
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.
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).
|Please hand in problem #4 of this week's
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.
do the indicated limits exist? If they do, explain why and evaluated them as well as possible. If they do not, explain why.
Part (a) asks about limx-->2[f(x)+g(x)].
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.
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.
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.
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
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.
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
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.
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"
/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.
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.
/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.
Please think about this until you understand it.
Even worse, much, much worse
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:
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:
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:
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 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:
If you read the book you will see that the graphical situation shown
here near x=-2 is described notationally by these two limit
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.
|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,
|I discussed grading for the course.
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.
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.
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.
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.
|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.
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:
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.
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 Question of the day was: suppose I know that logax=6.4 Then
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.
|Please hand in a writeup of the third workshop problem next Thursday. Please write in complete English sentences, hand in neat work, etc.|
|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:
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