**Diary entries for classes from Wednesday,
September 4 to Monday, October 6**

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Wednesday, November 12 |
Please hand in solutions to problems 4.7 #26 and
4.10 #36. If you need hints, let me know.
We had a contest between two teams of students. Each team did an optimization (finding a max or finding a min) problem. Real big prizes were given out. ## The (wonderful) Optimization ContestProblemWhat is the rectangle of largest area which can be drawn inside the triangle with vertices (0,0), (3,0), and (0,2)? Why do you know it is largest?
The team which competed consisted of
They sketched the picture shown. The rectangle has opposite corners
(x,y) and (0,0), with side lengths x and y. Its area is therefore
xy. What can we say about the relationship between x and y? The point
(x,y) is on the line connecting (0,2) and (3,0), and therefore if we
find an equation for this line, x and y must satisfy this equation,
and this will be a relationship between x and y. One equation
satisfied by both (0,2) and (3,0) is x/3+y/2=1. Therefore
y=2(1-x/3). The area becomes
A(x)=x·2(1-x/3)=2x-(2/3)x - Look at A(x) and the domain, [0,3]. A(0)=0 and A(3)=0 by
direct computation, and A(x) must attain its max at either an endpoint
or a critical number. Since A(3/2) is positive, A(3/2)
*must*be the maximum value. - Look at A'(x) to the left and to the right of x=3/2. Notice that 3/2 is the only critical number of A. To the left of 3/2, A'(x)=2-(4/3)x is positive (A'(1)=2/3 for example) and to the right of 3/2, A'(x) is negative (A'(2)=-2/3). So A increases up to x=3/2 and then decreases. Therefore A(x) must have a maximum at x=3/2.
- Let's compute: A''(x)=-4/3 so A(x) is concave down, and we have a local max.
ProblemWhat is the triangle of smallest area which can be drawn with vertices at (0,0) and on the positive x- and y- axes so that (3,2) touches the edge of the triangle? Why do you know it is smallest?
The team which competed consisted of We begin with the picture shown. The triangle has height y and base length x. Its area is therefore xy/2. Is there some relationship between x and y? Look at the picture and see that (3,2) is on the same line as (x,0) and (0,y). Well, the slope of the line between (3,2) and (x,0) is -2/(x-3) and the slope of the line between (0,y) and (3,2) is (y-2)/(-3). So these slopes must be equal, and we have (-2)/(x-3)=(y-2)/(-3). Therefore y=[6/(x-3)]+2 and now A(x)=(1/2)x·([6/(x-3)]+2). What is the domain of this function? Look at the picture: x must be greater than 3. Let's find A'(x). Since
A(x)=(3x)/(x-3)+x=(3x+x How do we know we have found a smallest triangle? Here we can again find several reasons, each of which is enough to justify the conclusion. - Consider A(x) as x-->infinity. Since
A(x)=x
^{2}/(x-3), A(x) gets large (A(x)-->infinity). Also, as x-->3^{+}, A(x)-->infinity. Therefore somewhere in between there is a smallest value of A(x), and the only critical point is at x=6, so the critical point must be there. - Now a reason related to A'(x). We know
A'(x)=[x·(x-6)]/[(x-3)
^{2}]. In the interval from 3 to infinity, A'(x) has a negative sign for 3<x<6 and it has a positive sign for 6<x. Therefore A(x) is decreasing to the left of x=6 and is increasing afterwards. x=6 must represent a local (and absolute!) minimum for A(x). - Now consider A''(x). I probably wouldn't, but you can compute it
and then simplify the result: you will get
A''(x)=18/(x-3)
^{3}. Therefore if x>3, A''(x) is positive, so A(x) is concave up in our domain, and its value at x=6 must represent a minimum.
One of the teams got the first prize, and the other got the second prize. The prizes were professionally wrapped, and were incredibly precious: historically, most Kings of England would never have been able to get (or taste) anything like these prizes!
I then discussed some common difficulties which I saw in yesterday's QotD. Please look at the comments there.
And on to
Therefore all antiderivatives of, say, -7x
A silly 6 foot tall calculus instructor throws a ball straight up into
the air with initial velocity of 15 ft/sec. When does the ball hit the
ground? (We assume the calc instructor has moved away from the ball's
path, I guess).
The
with(plots): F:=a->implicitplot({y=x^2,x^2+(y-a)^2=a^2},x=-1.5..1.5,y=-.1..2,color=black, thickness=2,scaling=constrained,grid=[75,75]);Then the statement F(.3) causes the first picture below to
be shown (too small!), and F(.8) gives the second picture
(too large!).
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Monday, November 10 |
Our next exam will be in one week.
I urged students to look at
the draft of the formula sheet for the second exam, and give me
comments about it. I also asked that student look at the review problems for the second exam
and, if they wish, help to assemble an
answer collection for these problems.
Next I must report the presence of my evil twin,
I then remarked that there were theoretical ways to insure we have a local max (or a local miniumum). Here are some of them:
Why
We discussed for a while how to proceed. The sketch is one I drew
after the most important decision was made, which is what should the
controlling variable be? (More technically, how should the situation
be parameterized?) In this case, there are certainly several
choices. One choice could be the angle made by the line segment with
the positive x-axis. Another choice, which we adapted, is to describe
the whole situation in terms of x, the first coordinate of the point
on the x-axis and the line segment. I called the total length,
L(x). It naturally is a sum of two "subsegments". The subsegment from
(x,0) to (3,2) has length sqrt((x-3) - There will be times in your life that computational resources will not be available: for example, our final exam (!) or when you are stranded on a desert island.
- More likely, there will be times in your life when the available computational resources are not adequate or are expensive or ... many things: and this has happened to me.
What about L'(x)? We computed, and got something like this:
L'(x)=(x-3)/sqrt((x-3) This problem is similar to numerous problems arising in physics, most immediately in connection with optics ("Snell's Law"). Sometimes you can't just "solve" in any simple way for the minimum or maximum, and numerical methods must be used for an approximate solution.
- Draw a picture, as well as I can.
- Label the picture, paying special attention to label any
interesting
*variables*. - Write any equations connecting the variables.
- Try to decide what should be the "controlling" (parameterizing) variable, and writing things in terms of that variable.
- Changing the situation to a "simple" optimization problem in one variable.
- Look at what you've got: does it make sense?
- Solve the optimization problem (are you
*sure*that you have an extreme value, and the correct kind of extreme value?). - Does your "solution" make sense in terms of your model?
The
3. Some students answered the QotD with "x=20/3". As I wrote, this answer is not responsive to what is asked: it gives the "x" but not the "f(x)". Read and answer the question, please. 4. Some (at least three) students tried to answer this question by, in effect, computing f(x) for each of the integers in [0,10] and giving me the largest result. The question as asked did not restrict consideration to integers, and therefore this approach is not correct. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Friday, November 7 |
Our next exam will be given on Monday, November 17.
## Some links to help preparing for the examOn this
page are a number of links to
pages which have practice problems with answers
completely worked out. You can glance at material you understand well,
and test yourself. You can look at worked-out problems in other
areas.
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Wednesday, November 5 |
Please do and hand in problems 4.3 #42 and 4.5 #48 on Thursday.
We sketched the graphs of a few more functions. We began with problem
#24 of section 4.5 which asked for a graph of y=ln(x
If f(x)=ln(x
If f'(x)=4x
We then discussed the rates of growth of log, and found that this
function grows very slowly for large x. For example, the function we
just graphed: if x>0, the tangent line slopes
We also discussed the logical linage between the "pointwise"
information f'(x)=0 and the existence of a local max or min. I urged
students to think about certain graphs to convince themselves that
knowing some numbers are 0 is
I think we looked at (e
Then I tried to do problem #40 from section 4.5, to sketch a graph of
y=cos(x)/(2+sin(x)). I urged people to first graph the function in a
window of width 2Pi (the function has 2Pi periodicity). It looks like
what
The
Here one dimension is x and the other is y. A relationship between
the variables x and y (a We will do more problems of this type next week. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Monday, November 3 |
Today's theme is drawing pictures of functions, especially with the help of the first and second derivatives. We know from the MVT that
It is generally much harder to check directly if a function is
increasing than to check if its derivative is positive. That's because
a serious direct check of the We also will get information from the second derivative.
- For concave up graphs, the tangent lines are below the graph and the secant line segments are above the graph.
- For concave down graphs, the tangent lines are above the graph and the secant line segments are below the graph.
We first considered e
What information can we get from the second derivative? If we use the
product and the chain rule correctly, then
f''(x)=(e
The - Compute f'(x). Where is f increasing and decreasing? Does f
have any local extreme points? If it does, where are they?
**Answer** f'(x)=f(x)=e^{x}+f(x)=xe^{x}=(x+1)e^{x}. Since the exponential function is never 0, this is only 0 when x=-1, the only critical number of f. Testing the sign of f' is easy, since the exponential factor is always positive. To the left of -1, the function is decreasing and to the right, the function is increasing. Since f(-1)=-1/e, I know that (-1,-1/e) is a local minimum, and, indeed, in this case an absolute minimum. - Compute f''(x). Where is f concave up and concave down? Does f
have any inflection points? If it does, where are they?
**Answer** Since f'(x)=(x+1)e^{x}, we see that f''(x)=e^{x}+(x+1)e^{x}=(x+2)e^{x}. The only time this is 0 is when x=-2. To the left of -2 (for x<-2) f'' is negative (check the sign), so f is concave down. To the right of -2 (for x>-2) f'' is positive (check the sign), so f is concave up. When x=-2, concavity changes. Since f(-2)=-2/e^{2}, I know that (-2,-2/e^{2}) is the only inflection point of f. - Draw a graph of f. (I suggested using the graphing calculator, and
then confirming the picture with what you deduced with f' and f'').
**Answer** The only part of the picture I haven't completely verified is the claim that the x-axis is a left asymptote for the function. But lim_{x-->-infinity}xe^{x}=lim_{x-->-infinity}x/e^{-x}, and this second form is infinity/infinity (look: x-->**-**infinity, ao the exponential is growing also). But use L'H on this, to get lim_{x-->-infinity}1/-e^{-x}=0, verifying the asymptotic claim.**Please do and hand in problems 4.3 #40 and 4.5 #48 on Thursday.**
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Thursday, October 30 |
(Written on the thirty-first!)
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Wednesday, October 29 |
Please hand in problems 4.1: #8 and 4.2 #18 on
Thursday.
P(x) is the derivative of
Q(x)=x-3x - Q could possibly (I highly doubt it!) be 0 at
*every*point inside the interval [0,1]. In that case there are lots of points where Q'(x)=0 inside [0,1]. (This case is included to make sure that the logical analysis is complete.) - Q could possibly have some positive values inside [0,1]. In that case,
Q will have a positive maximum in [0,1]
*and*the maximum will be an "interior" maximum, that is, somewhere in between 0 and 1. We discussed this during the last lecture. At such a point, since Q is differentiable, Q'(x) will be 0. - Q could possibly have some negative values inside [0,1]. In that case,
Q will have a negative minimum in [0,1]
*and*the miniimum will be an "interior" minimum, that is, somewhere in between 0 and 1. We discussed this during the last lecture. At such a point, since Q is differentiable, Q'(x) will be 0.
In all three cases, there is at least one x between 0 and 1 which has Q'(x)=0. Since Q'(x) is the P(x) we started with, that means P(x) must have at least one root inside [0,1], as I had claimed from the beginning. You could graph P with some mysterious device, but now I am sure using "pure thought" that P has such a root (actually it
has two roots inside that interval).
It turns out that this statement is true (a detailed discussion of how to "rotate" to get a verification from Rolle's Theorem is in the text). This statement is a version of the Mean Value Theorem which is one
of the two major results of calculus, and the MVT (as I will call it)
will contribute to everything we cover in the course from now on. For
example, it turns out the MVT can be used to estimate errors in such
processes as Newton's method, and, generally, estimation of errors in
most of the calculus algorithms I know, when they are really used with
numbers begins with the MVT.
This is true because if x
If derivatives are positive in an interval, we also can learn
something. Suppose f'(x) is positive in an interval, and
x
There is a "dual" statement. If the derivative is negative, then the function values go down as we travel from left to right.
These two statements are useful because determining the signs of
functions can be substantially easier than telling that a function is
increasing. For example, if
f(x)=x
Here is all of the information combined. Things are rather complicated. There are three graphs involved: a function and its first and second derivatives. I have tried coloring parts of the background, and describing features.
On the graph of the function, the letter What is likely to be confusing is that in this exercise we began with a graph of the first derivative, then sketched the second derivative, and then sketched a candidate for the original function. I will go into further details on this next time. Please read the text!
The
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Monday, October 27 |
The math computer systems were not working for most of the afternoon,
so I am late getting this diary entry written. I thank those students,
some looking nearly drowned, who came to class today. Neither the
weather nor an impending physics test stopped them. I also wish
everyone taking the test tonight good luck.
Mathematicians have made a living for almost three centuries using what we're going to discuss!
I began by going over - What is the
lim
_{x-->infinity}[ln(x)]/[x^{1/300}]?
**Analysis and answer** Before using L'H we*must*check that some appropriate set of hypotheses are satisfied. Here certainly ln(x) gets larger when x gets large "enough" and so does x^{1/300}. That is, the limit of the top as x-->infinity is infinity, and the bottom has the same limiting behavior. It isn't hard, by the way, to see that both the top and the bottom get large. Try a big number (?), like 10^{100}. Then ln(10^{100})=100ln(10) is approximately 230, and (10^{100})^{1/300}=10^{1/3}is about 2.15. Well they do get big, but they get big "slowly" and this limit is essentially asking which of the two, the top or the bottom, gets bigger slower. In any case, this is an indeteriminate form of the type, "infinity/infinity" and we can try to use L'H. lim_{x-->infinity}[ln(x)]/[x^{1/300}]=(using L'H)=lim_{x-->infinity}[1/x]**/**[(1/300)x^{(1/300)-1}]. Let's get rid of the compound fraction in the last expression. We then need to evaluate lim_{x->}300x^{-1+1-1/300}. I think I did the exponents correctly, and the result is the limit as x-->infinity of a constant (300) multiplying a negative (-1/300) power of x. The result is certainly 0. Therefore we conclude that the limit is 0 and x^{1/300}gets bigger faster than ln(x)! But what about the numbers? We just didn't take numbers that are*big enough*to see what's happening. Try (10^{100})^{100}. A little bit of juggling with exponents will show that ln((10^{100})^{100})=10,000ln(10) which is about 23,000. And ((10^{100})^{100})^{1/300}is (10^{100})^{1/3}which is about 10^{33}. And it is true that 10^{33}is much much bigger that 23,000.*Logs grow more slowly than any positive power of x.*(See problem #67 in section 4.4.) - I looked again at
lim
_{x-->infinity}[x^{300}]/[e^{x}].
**Analysis and answer** We had done this last time, but I wanted to repeat it, to emphasize the point. You can check that this is again eligible for L'H (again, it is infinity/infinity) and then 300 uses of L'H get us the lim_{x-->infinity}[300!]/{e^{x}. The top is a HUGE constant, but it is a constant, and the bottom gets ever larger. The result is 0.*Exponential growth is faster than any polynomial growth.*(See problem #66 in section 4.4.) The results comparing log and polynomial and exponential growth are*very*important in many applications, and should essentially be in your head when you think about growth and decay. - I think this is problem #9 of section 4.4:
lim
_{t-->0}(5^{t}-3^{t})/t.
**Analysis and answer** If I "plug in" t=0, the indeterminate form 0/0 appears, and this makes the problem eligible for L'H. I need to know the derivative of a^{t}. I can memorize it (d/dt(a^{t})=a^{t}ln(a) or if clever, I can deduce it (look: a^{t}=(e^{ln(a)})^{t}since exponential and logarithm are inverse functions, and then (e^{ln(a)})^{t}=e^{[ln(a)]t}, and realize that ln(a) is just a constant, so that d/dt{e^{[ln(a)]t}}=e^{[ln(a)]t}·ln(a)=a^{t}·ln(a) by the chain rule).) So we compute: lim_{t-->0}(5^{t}-3^{t})/t=(using L'H)=lim_{t-->0}(5^{t}ln(5)-3^{t}ln(3))/1, and we can evaluate this limit by just plugging in: the answer is ln(5)-ln(3). - While the other three examples we fairly realisitic, here is a
problem I would only expect to see in a calculus class or a text or an
*exam*: what is lim_{x-->0}x^{sin x}?
**Analysis and answer** Thisis a weird problem. Look at it: x-->0 and sin x-->0 also. It isn't clear to me what happens to the result of exponentiating one with the other. I would like to use L'H, and I need to somehow algebraically change this function into something that looks like a candidate for L'H: that is, something like TOP/BOTTOM which, as x-->0, will look like, say, 0/0 or infinity/infinity. There are really several ways to do this problem. I'll show one strategy here. If y=x^{sin x}, then ln(y)=ln(x^{sin x}), and then ln(y)=(sin x)ln(x). Now as x-->0, certainly ln(x)-->-infinity ("look" at the graph of ln(x) you should have installed in your head!) and certainly sin x-->0. Can I somehow change (sin x)ln(x) into a quotient? Well, here is a way: (sin x)ln(x)=[ln(x)]/[1/(sin x)]. Those of you who object to this should be advised that I'm doing it because*it works*which is about the best answer which can be given at this stage (I'm showing it to you because someone showed it to me and that person was shown it by ... all the way back maybe to the Marquis de L'H himself, maybe). Anyway, the fraction [ln(x)]/[1/(sin x)] as x-->0 does look like -infinity/infinity, so I will use L'H, and get (with some care, now, doing the derivative of the bottom!) lim_{x-->}[1/x]/[{-1/(sin x)^{2}}·cos x]=lim_{x-->0}[(sin x)^{2}]/[x·cos(x)]. This last fraction is again 0/0 as x-->0. Do L'H again, and get lim_{x-->0}[2(sin x)(cos x)]/[cos(x)-x·sin x]. All of these darn derivatives need care: for example, the derivative of the bottom uses the product rule. Now what happens as x-->0? The top is again 0 (it is 2·0·0) but the bottom is 1. So the result is finally, 0. But wait! Is it? Only if you are*very*alert will you notice that we have found lim_{x-->0}ln(y), and that it is ln(y) which has limit 0. We can*undo*ln with an exponential. So y itself has limit exp(0)=1. Wow.
**NOTE**I don't want to distress you but`Maple`'s response to the command`limit(x^(sin(x)),x=0);`is the answer`1`. Are human beings obsolete?
if the derivative
exists: functions can have corners and other "interesting" (?)
features. The other part of this statement of Fermat's Theorem
involves the term, "local maximum". I want to spend a little time on
that, and try to insure that we have a shared understanding of this
term.
Suppose a function f is defined on an interval I, and p is a point
inside I. Then p is called a So let's look at some examples. Please note that all of the examples were accompanied by pictures but for various reasons I can't provide pictures here at this time. - f(x)=-x
^{2}. Where does f have local maximums? (In this case and in all that follow, the assumed domain of the function will be "all numbers for which the function makes sense", so for this f the domain is "all real numbers".)**Answer and discussion** The graph is familiar, an upside down parabola with "top" at (0,0). As one student remarked, the function squares numbers and then puts on a - sign, so the largest value it can have is 0, and other values are negative. The only 0 of this f is at 0, and all other x's have negative values of f. So 0 is the only local maximum of f. - f(x)=384. Where does f have local maximums?
**Answer and discussion** This is a somewhat silly example, but again agreement on language is very important in what follows. The graph of this f is a horizontal line, and the interesting fact is that*every*number is a local maximum (and is also a local minimum -- weird but correct!). - f(x)=x. Where does f have local maximums?
**Answer and discussion** This function, on its*natural*domain (all real numbers) has no local maximums and no local minimums. Because if p were a local max, then to the right of p there is an x with f(x)=x>p=f(p). So p is not eligible. - This f(x) is defined piecewise. So f(x)=x if -1<x<1,
and f(x)=0 for all other numbers. Where does f have local maximums?
**Answer and discussion** A picture of this function's graph should be drawn. There's a tilted open interval (no endpoints included) going from (-1,-1) to (1,1). The rest of the graph are two closed (including endpoints) horizontal half lines. All the points with x<=-1 are local maxes, and so are the points with x>1 (but not 1 itself, because immediately to its left are higher points on the graph). Please notice that the tilted interval does*not*have any local maxes. A student asked after class why the number which is somehow "immediately before" 1 is not a local max, since it would be the highest point on the tilted line. Well, this is an interesting and subtle question. Suppose, say, that "Fred" is the number which is immediately before 1. Then thedistance from 1 to Fred is 1-Fred. What if we halve the distance? We would get the number 1-Fred/2. And f's value at this number is larger than Fred, so Fred can't be a local max. I think we really showed that such a Fred can't exist: there can't*be*a number immediately before another number! The real numbers themselves are rather weird. - f(x)=x
^{3}-x. Where does f have local maximums?**Answer and discussion** Again it is best to draw a graph of this f as we have several times before (!). The local extrema will occur where the tangent line is horizontal if we believe the "Fermat" result quoted above. So look where f'(x)=3x^{2}-1 is 0. This is when x=+/-1/sqrt(3). And a local max occurs at -1/sqrt(3).
**Commercial break****Sponsored by the Newton-Leibniz Company, purveyors of fine ideas** This is really easy. If we didn't have calculus it would be awkward and difficult. Easy is better than hard. Good morning! We send you now back to your regular lecture, with all our sympathy!
Fermat's Theorem Suppose f is defined in an interval I, and p
is inside I. If f has a local extremum (a local max or local min)
at p, then either f'(p) does not exist or f'(p)=0.
I tried to show why, if f'(p) does exist, it would have to be 0. Here
I
I then asked how we could determine the maximum and minimum
I drew some pictures which showed how the max and min values could occur at any of the candidates mentioned by Ms. Farley's algorithm.
The
I had a perfectly splendid QotD involving a pasture along the Raritan
River with a bull in it, but people shouted at it.
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Wednesday, October 22 |
More happy volunteers computed some derivatives: Ms. Smallwood and Mr. Gehrmann and
Mr. Bruno, who had the assistance of
Mr. Rodriguez.
- Newton's method is a way to (try to)
*improve*a guess at a root of f(x)=0 when f is a differentiable function. - The formula for going from an old guess, G, to a new guess, N, is N=G-[f(G)/f'(G)].
- The geometry of the Newton's method iteration is that the line tangent
to y=f(x) (a line of slope f'(G)) is drawn at the point (G,f(G)), and
the next guess, N, is obtained by
*sliding down*this tangent line until it hits the x-axis. That intersection is the new guess, N.
A cautionNewton's method can give some surprising answers. It definitely should be used, since under good circumstances it converges very rapidly, but you should be warned that it can also misbehave.
Also notice that y=x/(1+x
I think I began with consideration of a limit like this one:
lim
Then lim
The preceding discussion is not a proof, but merely presents heuristic
evidence that the result below is true. ("heuristic" is defined in the
October 6 diary entry, and it is the appropriate word here.)
- L'H resembles the way we might like to differentiate quotients
in our dreams. Please try not to get confused:
- L'H is a method to evaluate limits.
- The quotient rule is a method to compute derivatives.
- You
**must**check the hypotheses of L'H. A simple limit like lim_{x-->0}[x^2-3x-4]/[x^{3}+5x-6] is*not*equal to lim_{x-->0}[2x-3]/[3x^{2}+5], the limit of the quotient of the derivatives. The first limit can be evaluated by direct substitution, due to the continuity and nonvanishing of the denominator: it is 4/6. The second limit can be evaluated similarly and is -3/5. These are*not*equal. You must check the hypotheses and in work submitted for this class you must*show*that you have checked them.
To analyze this kind of limit, we usually look at
lim
What is lim
The
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Monday, October 20 |
Three happy volunteers
(Ms. Brundage and
Mr. Guzman
and
Mr. Stivers)
differentiated three of the functions in the review problems for
chapter 3. I recommended that people who were not confident about
computing derivatives of functions defined by formulas show try a
bunch of these problems.
Now I considered again a
This equation was first stated in our class on September 24 which now (to me, at least!) seems like a l-o-n-g time ago. It resulted from unrolling the
formal definition of derivative. The most important qualitative aspect
of this equation is that Err-->0 as b-->0, so that the term
Err·b should go to 0 faster than first order.
^{2}, and since f'(x)=2x, the term f'(a)b is exactly
2ab. The "Err·b" term is b^{2}, and certainly when b is
small, this gets smaller faster than b. A numerical example is
something like (5.01)^{2} which we compute exactly as 25.1001,
and here a=5 and b=.01. This splits up as
25+.1+.0001=5^{2}+2(5)(.1)+(.01>^{2}, and the result
of omitting the last term is a very small error.
Richard Hamming, a very famous applied mathematician of the 20 ^{th}, declared that The purpose of computing is insight,
not numbers.So what can we learn from
these numbers? Actually, we can observe some characteristics which
will always be true, although giving a good argument for them
will have to wait a week or two.
- Consider the number of zeros in the first column and in the
last column. In the first column, the transition from one number to
the next is gotten by "adding" one 0. So the a+b becomes a+(b/10). The
last column shows the discrepancy, the difference, between the linear
approximation and the true value. Look at the number of zeros. In each
step (and this is true data!) the number of zeros increases by
*two*, so while b changes to b/10, the difference gets divided by 100. In other words, the term "Err·b" is actually sort of a constant multiplied by b^{2}: it is a second-order term. This is generally true. - What about the sign in the last column. It is always
*negative*. Why is that? The curve lies*under*the tangent, which means that the linear approximation will be an overestimate of the true value.
enormous: don't use the linear approximation when the "b" is
big. (In fact, if you needed to use a linear approximation for
sqrt(101), try a=10. Then this more appropriate linear approximation
gives 10.05
I was asked if I could give a quantitative estimate for the error, and again I muttered that I will need to wait a week or two but I would then be able to show one.
## Magic?I asked how a calculator computes square roots. Most of the answers I got were, "By pushing the button," but I repeated my question. Here are my suggested answers.- Since World War II smaller and smaller genetically modified people have been raised. I think in '47 the two foot high person was the standard of excellence in this direction, and then in '53, the first 1/4 inch person ... and then in the 70's really teensy people were created and put inside calculators and they just used clever guessing to get square roots. They were also put on super-amphetamines to work really fast.
- If your calculator can manipulate 10 decimal digit numbers, then
(since each number position can hold one of 10 choices of digits) the
calculator can only be asked to find the square root of
10
^{10}numbers. These square roots were all computed in 1975 in a massive research project similar to sequencing the human genome. The cost was about 47 billion dollars to compute all these square roots, but the task was finally done when Horace and Myrna Whiteapple of Frog's Throat, Georgia, announced ten digits of the square root of 37:**6.0827625302**. - The calculator uses ancient Babylonian mysteries to compute each square root very efficiently.
very
dramatically presented. Sigh. And #2 was (correctly!)
criticized because there wasn't enough storage space (and actually it
would take rather a lot of time to compute those square roots). The
correct answer is actually #3.
^{2}>A) then B>sqrt(A) and
(B/A)>(1/sqrt(A)) so that (A/B)<sqrt(A). We are averaging
numbers that are on either side of sqrt(A). So maybe we will get
closer. The amazing result is that we get very much closer, very
fast. Here is an example, again computed by Maple:
I wanted to compute the square root of 2. As is well-known (but not by
me!) this is 1.4142135623730950488 to 20-digit accuracy. My first
guess, A, was 3. Here is a table of next guesses using the Babylonian
method, and the difference between the guess and the true value of
sqrt(2):
Notice how rapidly the sequence of guesses converges to sqrt(2): here the number of 0's in the difference between the true value and the guesses seems to double at each step (this is true in
general, which is why this method is so nice to use).
I don't know how the Babylonians came up with this method, but I can
give you an explanation in the context of this course. Consider the
function f(x)=x As far as I know this is the method which is used for square roots in most computational devices. It is rapid and easy to program, and the error analysis (not presented here) is not hard.
This "scheme" of forming a new guess from the old is called Newton's Method.
The 1.537690539, 1.307076219, 1.261601801, 1.259923288, 1.259921050The sixth iteration turns out to be correct to 12 decimal places (so the error is less than 10 ^{-12}).
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Thursday, October 16 |
Another workshop.
Please hand in a writeup of the third question. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Wednesday, October 15 |
These problems are due
tomorrow in workshop: 3.8 #50 and 3.10
#24.
I reviewed the derivatives of arcsin x: this is
1/sqrt(1-x
If y=a
Finally I started on today's topic,
The graph that is displayed was generated by
## The exams come back ...I returned the exams. Here is a discussion of the results. Please realize that there will be another exam (100 points) and a final (200 points) and more points from the QotD, textbook problems, and workshop problems. I urge students to study. | ||||||||||||||||||||||||||||||||||||||||||||||||||||||||||

Sunday (!), October 12 |
About 40 to 50 students showed up. I tried to outline the
course so far:
- "Review" of precalculus: functions, graphs, analytic geometry
- Limits: the definition, graphical interpretation, "intuition", algebraic manipulation to detect limits.
- Continuity: definition, graphical interpretation, Intermediate Value Theorem, continuity of familiar functions
- The derivative: a rate of change, slope of a tangent line. Formal definition. The derivative as a function.
- Differentiation: the algorithms and the deriviatives of the common functions.
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Wednesday, October 8 |
I asked Ms. Brundage to do problem 21
in section 3.4: find an equation for a line tangent to y=tan(x) at the
point (Pi/4,1). We need to know the slope of the tangent line, and
that slope is the derivative of tan(x) at x=Pi/4. But tan's derivative
is (sec(x))^{2}, so the slope needed is
(sec(Pi/4))^{2} which is equal to 2. You can check (and you
should know!) the wonderful special
triangles. Therefore the line desired is y-1=2(x-Pi/4).
Finally, I remarked: - There would be a review session in our regular classroom on Sunday, October 12, at 4:30 PM.
- There is a review sheet of problems which will be gone over in workshop tomorrow.
- I would like to urge students to read the questions carefully, and write their answers with some care. I would like to read answers to the questions I actually ask, and not imaginary or invented questions. I cannot read students' minds about what they would or should have put on papers.
- I have now graded workshops for all three sections. Some students are clearly getting the idea, and can write good narratives to accompany their computations. Some do not write such statements, and therefore might be at a disadvantage when I ask for sentences describing an answer on the exam.
## Some links to help preparing for the examHere are some pages which have practice problems with answers completely worked out. You can glance at material you understand well, and test yourself. You can look at worked-out problems in other areas.- Using the definition of derivative to find the derivative of a function.
- The product rule
- The quotient rule
- Trig functions and the product rule
- Trig functions and the quotient rule
- The chain rule
- Quiz on differentiation of formulas
- Evaluating limits, mostly using algebra
- Limits using graphical information
- Piecewise defined functions and continuity
- A quiz about continuity
- Horizontal asymptotes
- Vertical asymptotes
- y=arcsin(x)
`The inverse function` - sin(y)=sin(arcsin(x))
`Using the inverseness` - sin(y)=x
`Recognition of inverseness` - cos(y)(dy/dx)=1
`Implicit differentiation` - dy/dx=1/cos(y)
`Solving for dy/dx` - Since sin(y)=x, x
^{2}+(cos(y))^{2}=1, and so cos(y)=**+/-**sqrt(1-x^{2}). Which sign to take? In the range we are considering between -Pi/2 and Pi/2, cosine is positive. Also the slope of the tangent line is positive for arcsine. Thus we will take the + sign. And cos(y)=sqrt(1-x^{2}).`Solving for dy/dx stuff in x` - Thus arcsin'(x)=1/sqrt(1-x
^{2})`Statement of formula`
^{2}). Notice that for -1<x<1,
this is positive, and that the derivative formula is not valid
at +/-1 as we predicted from the geometric evidence.
Example: The derivative of arcsin(5x
- y=arctan(x)
`The inverse function` - tan(y)=tan(arctan(x))
`Using the inverseness` - tan(y)=x
`Recognition of inverseness` - (sec(y))
^{2}(dy/dx)=1`Implicit differentiation` - dy/dx=1/(sec(y))
^{2}`Solving for dy/dx` - Since tan(y)=x, x
^{2}+1=(sec(y))^{2}(much easier than arcsine!).`Solving for dy/dx stuff in x` - Thus arctan'(x)=1/(1+x
^{2})`Statement of formula`
^{(5x4)}) is
1/(1+(e^{(5x4)})^{2}·
e^{(5x4)}·5·4x^{3}: somewhat
of a mess, along with several uses of the chain rule.
- y=ln(x)
`The inverse function` - exp(y)=exp(ln(x))
`Using the inverseness` - exp(y)=x
`Recognition of inverseness` - exp(y)(dy/dx)=1
`Implicit differentiation` - dy/dx=1/(exp(y))
^{2}`Solving for dy/dx` - Since exp(y)=x, we're done! This is even easier still
`Solving for dy/dx stuff in x` - Thus ln'(x)=1/x
`Statement of formula`
^{sin(x)} by doing this:
ln(y)=ln(x^{sin(x)})=sin(x)ln(x), so that we can d/dx the
equation and get:
(1/y)(dy/dx)=(product rule!)cos(x)ln(x)+sin(x)/x, so that
dy/dx=y[cos(x)ln(x)+sin(x)/x]=x^{sin(x)}[cos(x)ln(x)+sin(x)/x].
Next time: a
The
I asked students to find dy/dx if
Let me try to "disassemble" this function. For example, what is the
derivative of arctan(e Now what is the derivative of ln(x+3)? This is perhaps a simpler use of the chain rule, and this derivative is 1/(x+3)·1.
What is the derivative of ln(x+3)/arctan(e (arctan(e^{x})^{2}
Now continue to go backwards. We are asked for the derivative of
I hope this is comprehensible and again I apologize for throwing such
an absurd thing at people. See you Sunday, if you would like to be
there. Or else Monday |

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