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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 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.
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)
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)( 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 /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
imagine):
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(e
Of course the chain rule itself can be repeated. So here, for example, we can try to differentiate cos(e ^{3x2}). Its
derivative is
-sin(e^{3x2})·(e^{3x2})·(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
This line of approach may be extended. For example, if
y=x
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
The
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.
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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/x^{33}=x^{-33}
is (-33)x^{-34} or, equivalently, -33/x^{34}.
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)] Qotd #2 Moving right along (!) I discussed the trig functions again. 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. measured by the
length of the intercepted arc. In this scheme (radian measurement) the
full circle of 360^{o} 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
lim_{h-->0}[sin(h)/h]=1. I left the companion limit,
lim_{h-->0}[(cos(h)-1)/h]=0, to students' reading the book.
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 7^{2}Pi. The area of a half circle of radius 5 is 5^{2}Pi/2. The area of a quarter circle of radius 3 is 3^{2}Pi/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)R^{2}Theta. Therefore the area of circular sector ABC is (1/2)1^{2}h=(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 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
lim_{h-->0}[sin(h)/h]=1.
Since lim
I found an equation of the line tangent to y=sin(x) at x=Pi/3 (and,
I remarked that (looking again at the shape of the graphs) we can also
see that the derivative of cos(x) is Qotd #3 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 Question of
the day was in several parts.1. What is the derivative of 33x ^{72}-7e^{x}+sqrt(117)?The "trick" here was to force attention to the constant, sqrt(117). But its derivative is 0, even though it looks complicated. Answer
72·33x^{71}-7e^{x}+0.
2. What is the derivative of (100x ^{50}+e^{x})/(50e^{x}-x^{100})?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. Answer
[(50·100x^{49}+e^{x})(50e^{x}-x^{100})-(50e^{x}-100x^{99})(100x^{50}+e^{x})]/(50e^{x}-x^{100})^{2}
3. What is the derivative of 5x·sin(x)+17cos(x)? Answer 5sin(x)+4x·cos(x)-17sin(x).
4. What is the derivative of e ^{x}(3sin(x)+5x^{2})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.
Answer The answer is a huge quotient.The bottom of the quotient is (2cos(x)+[(x+1)/(x-1)]). This is the square of the bottom
of the original expression.^{2}The top is: (e^{x}(3sin(x)+5x^{2})+e^{x}(3cos(x)+10x))(2cos(x)+[(x+1)/(x-1)])-(e^{x}(3sin(x)+5x^{2}))(-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.
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Monday, September 29 |
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'(x
We began with the very simplest sorts of functions. If f(x)=15 for
Now consider f(x)=x
We can do this more generally: (x+h)
But now what? We consider lim
Your textbook next studies the exponential functions a
If f(x)=a _{h-->0}
(e^{h}-1)/h exists and is 1.^{x} is its own derivative.
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
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, m
Now we began to discuss what is called
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
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 e
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/x
The
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Wednesday, September 24 |
These problems are due this Thursday in
recitation: 2.6 #44 and 2.7 #8.
I began with a
^{2}. 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/x^{2}.
First we noted that the top of the fraction was 1, so that y can
Let's work on the tangent line question. When x=3,
y=1/3
I will use the notation of your text. There m
Therefore the slope, m
There is a good deal of vocabulary associated with what we just
did. Depending upon the application, m
So here is the real formal
Example: f(x)=1/x
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 detailThere 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
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
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
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.
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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
I then asked the
I then tried to see what I could learn about the asymptotic behavior
of Q(x)=[x(x-1)]/[(x+3)(x-4)
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).
## 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 intomath language as t-->infinity.
LengthThe distance from Fred to Jane is the length of the line segment. This is sqrt((t-sqrt(t)) ^{2}+t^{2}). 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 t^{2} alone goes to
infinity. And taking the square root of that won't change the
asymptotic behavior, so that the lim_{t-->infinity}LENGTH
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.
SlopeWe 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! lim_{t-->infinity}SLOPE=-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).
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Thursday, September 11 |
Please hand in problem #4 of this week's
workshop problems.
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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.
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 lim
Part (d) asks about lim
Problem 11 of section 2.3 asks for analysis and possible evaluation of
the lim
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
For a specific example of a kind of function to which we could apply the
squeeze theorem, consider D(x)=e
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
I looked at problem 41: how close to -3 do we have to take x so that
1/[(x+3) +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
A function f is - a is in the domain of f and
- lim
_{x-->a}f(x) exists and the value of this limit is f(a).
_{x-->a}f(x) exists exactly
when lim_{x-->a-}f(x) and
lim_{x-->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.
If we believe that motion is
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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
m_{TAN}, 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
m_{SEC}, 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
m_{SEC} should "approach"
m_{TAN} then.
/discern/ v.tr. 1. perceive clearly with the mind or the senses. 2. make out by thought or by gazing, listening, etc.Now m _{SEC} 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, m_{SEC}, 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
m_{SEC}-->1/(2sqrt(A)) and
this must be
m_{TAN}.
/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)=x ^{2} 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.
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
lim
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
Suppose f(x)=1/x. This is a
The actual formal definition of lim 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 lim
_{x-->3}H(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 lim
_{x-->-4}H(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 class ended with consideration of the function
R(x)=(x-2)/(x
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: 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,
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Wednesday, September 10 |
I discussed grading for the course.
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
So I started off with tree growth. I described a situation where two
"facts" about a dwarf apple tree are supplied: the tree's
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 _{t-->infinity}H(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
lim_{delta 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)=3t 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=3x
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
Problems to hand in tomorrow at recitation: 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
2
^{x} 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: a We wrote down but did not linger over a bunch of exponential formulas: - a
^{b+c}=a^{b}a^{c} - a
^{-b}=1/(a^{b}) - (a
^{b})^{c}=a^{bc} - a
^{0}=1
I then commented on
The inverses of exponential functions are called logarithmic
functions. So y=log
^{x} eventually is bigger and stays
bigger than, say, x^{10,000} even though when x=10, the
exponential function has "only" 5 digits (3^{10}=59,049) and
10^{10,000} 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
The - What is log
_{a}(x^{3})? - 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
"log_{a}x=6.4" really mean? It means there are 6.4 powers of a
"in" x. Written as an equation, it means that a^{6.4}=x. The
first question asks what are the number of powers of a "in"
x^{3}. Well, if a^{6.4}=x then we could cube (raise
to the third power) this equation. We would get
(a^{6.4})^{3}=x^{3}. But repeated exponentials
multiply, so that a^{6.4·3}=x^{3}, and since
6.4·3=19.2, we conclude that
log_{a}(x^{3})=19.2. If you are the type of person
that likes to memorize formulas (not that there's anything wrong with
that) then
log_{a}(x^{3})=3log_{a}x=3·6.4=19.2.
The other question is more ... interesting. We know a ^{6.4}=x
and we need to know something relating a^{4} and 4. Well,
"golly" (?), by using some of the exponential rules, we see that
(a^{4})^{something}=a^{6.4}. This means
4·something=6.4 so that something=1.6, and
therefore
(a^{4})^{1.6}=x. There are 1.6 powers of a^{4}
"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.
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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
The So for the
**Appendix A, #28: Solve the inequality in terms of intervals and illustrate the solution set on the real number line: x**^{2}<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. |

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Maintained by
greenfie@math.rutgers.edu and last modified 9/2/2003.
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