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# AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

## Wednesday, January 25, 2006

### Math can be Colorful

To be Organize means to arrange in coherent form or to have a system. Using different colors were very helpful in dealing with integration by parts. To be successful in this part of the course, we need to know how to organize our signs and understand which is which by organizing what we have. Our example for today proved this theory. We found out that the reason why we have simple errors are because we forget our signs and we work very disorganized. Truly, if your work is organized like Mr. K's it is easier to see what we still need to do and what we've done wrong. Almost all of our time was used for answering this questions for this question though not very hard in nature but very complicated in terms of bookkeeping. I'm pondering to myself if it is necessary to elaborate this question for like I said its not challenging. I now decided not to explain the group question further but rather give tips on what we need to remember in answering this questions
1. We need to understand what is given and what we need to figure out.
2. Next is to know what steps we need to take in order to go to where we want to go. Choose the right method in solving. Remember not to use a shotgun to kill a mosquito.
3. Forget not the signs
4. Be cautious with regards to finishing up the question. Finding the antiderivative is less than half way the battle.
After the group work we discussed transcendental functions which I found out are functions that is not algebraic or cannot be expressed in algebraic form like exponential functions and trigonometric functions. We really didn't have time to discuss fully the topic we need to so I giving the topic to tomorrow's scribe which is...

Prince

### Better Choices

A new way of anti - differentiating was taught today. Not all products can be integrated by reversing the Chain rule. Sometimes, using Substitution will make it look even worse than what we could've imagined. So with the complexity of some equations, the method of INTEGRATION BY PARTS is introduced.

The rule of integration by parts can be written as:
$\int f(x) g'(x)\,dx = f(x) g(x) - \int g(x) f'(x)\,dx$

According to the rule, the integral of a product of two functions may be written as a difference, one term of which is a new integral. In most cases, the new integral is simpler than the original integral. Therefore, this technique requires the importance of designating the factors f(x) and g'(x) of the original integrand in such way that f(x) simplifies under differentiation while the factor g'(x) does not become more complicated under the integration.

As what I've just mentioned, assigning the factors is a crucial aspect in performing this technique. To get better chances of heading to the correct answers, L I A T E must be taken into consideration.

L I A T E as in

L - orgarithmic
I - nverse Trig
A - lgebraic
T - rigonometric
E - xponential

It is also significant to know which method applies to a particular equation. In this way, it saves up time and makes it a lot easier.

So make better choices and always remember the past lessons and techniques that were recently learned... then your off to university... I think... well not really... but it's a good start though... *smiles*

## Tuesday, January 24, 2006

### Antiderivatives...

Another scribe that I can't seem to find the time to do.

Well here it goes.

For the past weeks we have learned to differentiate many equations. Ugly or not so ugly, we can differentiate it. But now we run the rules backwards.

Who would've thought that a derivative and an integral undo what each other does? They are both very different from one another. Unlike addition and subtraction or multiplication and division, its not so obvious that they are inverses of one another. But they are, they undo what the other does.

Keep your bank of derivatives in hand, you'll need it. But this time add a couple of new rules, for antidifferentiating. Don't forget your plus C's, why? Because the derivative of any constant is 0. A big fat ZERO. The difference now when you're antidifferentiating is that you're not only getting ONE particular function but a WHOLE FAMILY of functions.

OR
2) Make a bank of anti-derivatives.
3) + C, never forget that.

___________________________________

Natural Log function and Rational function in particular 1/x

Inverses.

Input ---> Output
its inverse is...
Output ---> Input

CORRECT?

X (domain) ---> Y (range)
its inverse is...
Y(range) ----> X (domain)

CORRECT AGAIN?

So where am I trying to lead this to?

If we look at the rational function 1/x
its domain is the reals and x does not equal to 0.

its anti-derivative should have the same domain.

This stands true, antiderivative of 1/x is ln x+c for x>0.

ln(-x)
Using chain rule.
d/dx ln(-x) = (-1)1/(-x)= 1/x

So therefore, the antiderivative of 1/x is ln absolute(x) + C.

Don't forget that absolute sign or the + C as they have great distinctions to your answer.

=D

Sorry for the late post.

## Monday, January 23, 2006

### The method of substitution

We did the integration using the chain rule yesterday. Today, we had some questions on the board. Firstly, we used chain role solved first 3 questions. But if only use chain role, we wouldn't able to solve some problems like 4 and 5, so some new stuff for today. That's the method of substitution.
Three steps: Substitution, antidifferential, resubstitute
Exp:S cos (3x) 3 dx
Let u = 3x
du = 3dx
S cos (u)du= sin u+ C
=sin 3x + C
Another one:
S X/(rt (x +7) dx )
Let u = rt (x+7)
du = 1/2 (x+7) ^-1/2 dx
du = 1/ (2 rt (x+7)) dx
2du = 1/( rt (x+7) ) dx

u= rt (x+7)
u^2 = x +7
u^2-7 = x

S(u^2-7)* 2 du = 2 S( u^2-7) du = 2 ((u^3/3 – 7u))
=(2/3 u^3 -14u+c)
=2/3 (rt (x+7))^3 -14 rt (x+9) + C

Next scribe: Ara

## Sunday, January 22, 2006

### Scribe for Friday

Friday, first we answer some questions in the Iflurtz questionaire for Valentines Day. Then for our class we start by answering this question. S (x+x^2)^2 dx.
To answer this question is that you make it to a simpler equation first before you antideffrientiate it. So it would be S (x^2+2x^2+x^4) dx. Now its easy to antideffrientiate it, so it would become x^3/3+x^4/2+x^5/5 + C.

Then Mr. K showed us this F(x)= f(g(x))
F'(x)= f'(g(x)) * g'(x)

So if were given this: S 3x^2 sin (x^3) dx we can say that it is equals to this - cos (x^3)+ C, because of the relationship above. Its like undoing the chain rule.

However, if the 3 is not there we can put the 3 there but in order to make it equals to 1 we should multiply it with 1/3. And we can use this relationship: S k f(x) dx= K S f(x) dx. So if this is given S x^2 sin (x^3) dx, we can put 3 in front of x^2 so that it would look like its derivative. But we should put 1/3 in front of S too to make it balance. 1/3 S 3x^2 sin(x^3)dx then the answer would become - 1/3 cos (x^3) +C .

And that's what we learn last friday....the next scribe is xun...

### Sunday Punting Practice

Like sokoban the target is to push objects (in this case punt-discs or 'pucks') around a maze to cover various targets. In a punt maze however the pusher slides forward tilt-style until it hits an obstacle, and a puck that gets struck will be punted forward a matching distance.'

'Aim: Use the black cross as a pusher to 'punt' the yellow pucks onto the blue targets.
Movement: Use the arrow buttons provided to move the pusher (black cross). The pusher will run in a straight line until it hits a wall or a yellow puck. If it hits a puck the puck will be punted forward a matching distance.'

(Thanks again to Think Again!)

## Thursday, January 19, 2006

### Why Should I Learn Math?

This is taken from an article (Math Will Rock Your World) from Business Week. A few snippets:

Y'wanna get a really interesting job working with people on lots of interesting things?

But just look at where the mathematicians are now. They're helping to map out advertising campaigns, they're changing the nature of research in newsrooms and in biology labs, and they're enabling marketers to forge new one-on-one relationships with customers. As this occurs, more of the economy falls into the realm of numbers. Says James R. Schatz, chief of the mathematics research group at the National Security Agency: "There has never been a better time to be a mathematician."

Learn math!

How'd ya like a six figure salary?

...new math grads land with six-figure salaries and rich stock deals. Tom Leighton, an entrepreneur and applied math professor at Massachusetts Institute of Technology, says: "All of my students have standing offers at Yahoo! (YHOO) and Google (GOOG)."

Learn math.

D'ya wanna to work on the biggest most cutting edge issues of our day?

This mathematical modeling of humanity promises to be one of the great undertakings of the 21st century. It will grow in scope to include much of the physical world as mathematicians get their hands on new flows of data .... "We turn the world of content into math, and we turn you into math," says Howard Kaushansky, CEO of Boulder (Colo.)-based Umbria Inc., a company that uses math to analyze marketing trends online.

Learn math.

Y'wanna make one of the most significant contributions to the betterment of humanity?

"The next Jonas Salk will be a mathematician, not a doctor."

Learn math.

What are the implications for k-12 education?

Outfitting students with the right quantitative skills is a crucial test facing school boards and education ministries worldwide. This is especially true in America. The U.S. has long leaned on foreigners to provide math talent in universities and corporate research labs. Even in the post-September 11 world, where it is harder for foreigners to get student visas, an estimated half of the 20,000 math grad students now in the U.S. are foreign-born. A similar pattern holds for many other math-based professions, from computer science to engineering.

The challenge facing the U.S. now is twofold. On one hand, the country must breed more top-notch mathematicians at home, especially as foreigners find greater opportunities abroad. This will require revamping education, engaging more girls and ethnic minorities in math, and boosting the number of students who make it through calculus, the gateway for math-based disciplines. "It's critical to the future of our technological society," says Michael Sipser, head of the mathematics department at Massachusetts Institute of Technology. At the same time, school districts must cultivate greater math savvy among the broader population to prepare it for a business world in which numbers will pop up continuously. This may well involve extending the math curriculum to include more applied subjects such as statistics.

"But I don't like math. Besides, I don't need it. I'm going into the humanities or business!"

As mathematicians expand their domain into the humanities, they're working with new data, much of it untested. "It's very possible for people to misplace faith in numbers," says Craig Silverstein, director of technology at Google. The antidote at Google and elsewhere is to put mathematicians on teams with specialists from other disciplines, including the social sciences.

Just as mathematicians need to grapple with human quirks and mysteries, managers and entrepreneurs must bone up on mathematics. Midcareer managers can delegate much of this work to their staffers. But they still must understand enough about math to question the assumptions behind the numbers. "Now it's easier for people to bamboozle someone by having analysis based on lots of data and graphs," says Paul C. Pfleiderer, a finance professor at the Stanford Graduate School of Business. "We have to train people in business to spot a bogus argument."

Yes, it's a magnificent time to know math.

'Nuff said.

## Wednesday, January 18, 2006

### Integral facts

I
nverse of a derivative

Not an antiderivative for it is conceptually different

The infinite sum of small rectangles in an interval

Early use of integration was by Archimedes by finding area of a circle

Generates practically a number not another function

Riemann sum is a method used to approximate values of integral

Accumulation of quantities such as area under the curve or volume displaced

Leibniz introduced the standard long “s” notation of integral

## Tuesday, January 17, 2006

### Learn for a week

WE finished this chapter in just like - a SNAP! -

I can't believe that I have actually ended this topic with a COMPLETE - let me emphasize more on that - ABSOLUTELY EXECUTED set of homework exercises plus an agreeable understanding of what integrals and derivatives are. For the VERY FIRST TIME...

WHOOHOOO!!!

That feels OH so good! Starting the year with a no-absent, no-late attendance... haha! I wish this would go on 'til the end of the next semester. -- i wish? :D

Quite a big information for such a small chapter. It was basically all about ACCUMULATION FUNCTIONS. These functions are called as such because they are integrals evaluated by accumulating the area under the graph of the integrand function f.

What I've gained throughout this short episode:
• review on INTEGRALS and the FUNDAMENTAL THEOREM OF CALCULUS Part I
• SECOND FUNDAMENTAL THEOREM OF CALCULUS
• ACCUMULATION FUNCTIONS

and best of all

• ways on getting the area or the integral of a function in a given interval using the calculator - that is so cool! if I had known that before, I've probably aced some of my past tests... or maybe not... who am I kiddin'? hahaha!!! But honestly, that's like so amazing. After practicing on some of those exercises I came up to the point where I said,

The Calculator's not that dumb after all. (laughs)

### Blogging on Blogging on Integrals

When it comes down to integrals, they are kind of confusing. The fact that integrals and derivatives are inverses confuses me. I think the the hardest part of this unit was composite functions involving integrals. Finding the derivative of an accumulation function requires the chain rule.

a=0
b=2x
f(x)= f(t)= tÂ²

You take f(t)substitutee 2x for t and derive and multiply that by the derivative of 2x.

F' (x)= (2x)(2)
F'(x)= 4x

I think that the confusing part of it is remembering that the upper limit or 2x in the previous example, is the inner function and that the outer function is f(t). Having clarity on this makes such a difference when solving for the derivative of F(x).

## Sunday, January 15, 2006

### Sunday Slither!

Thanks go out to Mrs. Armstrong for pointing to today's game from Think Again!.

The game, from Japan, is called Slither Link.

1. Connect adjacent dots with vertical or horizontal lines.

2. A single loop is formed with no crossing or branches.

3. Each number indicates how many lines surround it, while empty cells may be surrounded by any number of lines.

Play here.

## Saturday, January 14, 2006

### Composite Functions and Integrals

Friday, we learned about accumulation functions that are a composite of functions. when finding the derivative of a composite of functions involving integrals, is not as difficult as it may seem.

a= 3
b= x^2
f(x)= √(t^6 +1)
dy/dx = (√((xÂ²)^6 +1)(2x)
The inner function is xÂ² and the outer function is the square root function.
dy/dx = 2x√(x^12 + 1)

Another part of the class was about the area between the curve and the x axis. Integral means signed area, and Area is not the Integral. Due to the fact that many graphs have roots and an area below the x axis as well as above it, to compensatete for this, you have to take the absolute value of the function.
abs(f(x))
If you have two functions, the area between them involvesolvingng for the intersection points, then finding the integral of the function that is higher and then subtract the integral of the lower function.
The next scribe is Sarah.

## Thursday, January 12, 2006

Well, do I really have to say this???Its been said over and over, well at least for my blogs... If I asked you what we did today, I'm pretty sure you know what we did... :D right?right? Yes I know what you're thinking "well get on with it already..." heheheh well ok, after saying this a million times, here I go again, to start the class, we had to do questions regarding yesterday's class. We are again faced with problems specifically made so that each problem will have confusing parts so we learn better. The first and second problem was about what we learned yesterday. We had an accumulation function and we needed to complete the table of values for the first problem, sketch the graph of that function, name critical numbers of that function and determine the intervals where that function is increasing. All of this is not new to us because we did this before and all of it is just a review so I will not bother discussing it.

What I would like to talk about is the second part of the fundamental theorem of calculus which states that:

If F(t) is continuous and A(x) is defined as an accumulation function:

the A'(x) = f(x)

To be able to remember this we have to go back to our roots. We learn addition first then subtraction. And we know that they are inverses of each other. Just like addition and subtraction, Differentiation and integration are inverses of each other. Even though its not straight forward because getting the slope of tangent line and calculating bunch of little squares are not really making sense but the because of the first part of the theorem we know it is true. Because we know that the first theorem is true then we understand that if we get the derivative of an integral then the function is just the underlying function; that simple.

## Wednesday, January 11, 2006

### Integral or Derivative?

Todays class(and yesterdays class for the three of us not writing our english) started with Mr. K asking us to solve a lot of integrals, but geometrically, not using the fundamental theorem. The integrals were all on the function f(t)=t with the interval between 0 and a series of different values of b. That was all fine until we came upon integrals with a negative value for b. However Mr. K explained that we solve them exactly how we solve any other integral.
For example: on the interval 0 to 1, the integral is solved F(1) - F(0)
on the interval 0 to -1, the integral is solved F(-1) - F(0)

After that, Mr. K asked us to complete a table of values with values ranging from -4 to 4 for the function

We were then asked to graph it, the graph itself turned out to be the same as the function f(x)=(1/2)x². The function f(x)=(1/2)x² is also the anti-derivative of the identity function(f(x)=x). The lesson is a demonstration that integration and derivation are inverses of each other, just in the way that addition and subtraction are.
The function A(x) is an example of an accumulation function. It is called an accumulation function because as x progresses, the function accumulates area.

The next scribe will be..... Jayson

## Saturday, January 07, 2006

### Just For Fun (or Getting Ready to Think Again!)

I found this "test" over at Teaching and Developing Online. Try it out .... just for fun. ;-)

Below are four (4) questions and a bonus question. You have to answer them instantly. You can't take your time, answer all of them immediately.

OK?

Let's find out just how clever you really are. No looking at the answers in advance.

First Question:

You are participating in a race. You overtake the second person.

What position are you in?

Answer:If you answered that you are first, then you are absolutely wrong!

If you overtake the second person and you take his place, you are second!

Try not to mess up in the next question.

To answer the second question, don't take as much time as you took for the first question.

Second Question:

If you overtake the last person, then you are...?

Answer:If you answered that you are second to last, then you are wrong again. Tell me, how can you overtake the LAST Person?

You're not very good at this! Are you?

Third Question:

Very tricky math! Note: This must be done in your head only.

Do NOT use paper and pencil or a calculator. Try it.

What is the total?

Did you get 5000?

The correct answer is actually 4100.

Don't believe it? Check with your calculator!

Today is definitely not your day.

Maybe you will get the last question right?

Fourth Question:

Mary's father has five daughters: 1. Nana, 2. Nene, 3. Nini, 4. Nono.
What is the name of the fifth daughter?

NO! Of course not.

Her name is Mary!

Okay, now the bonus round:

There is a mute person who wants to buy a toothbrush. By imitating the action of brushing one's teeth he successfully expresses himself to the shopkeeper and the purchase is done.

Now if there is a blind man who wishes to buy a pair of sunglasses, how should he express himself?

He just has to open his mouth and ask, so simple.

So simple it is ... ;-)

## Wednesday, January 04, 2006

### A Comment From a Former Student

I know many of you don't read the comments left on each other's posts — you really should — sometimes a real gem is hidden there; like this comment from a former student. I think he wanted all of you to read it so here it is:

As far as I know I have been through pre cal and calculus in post secondary school and it's different. Mr K., as strict or fast paced as he may seem, is teaching all of you very well as he is preparing you for the future. I know he has been a great help in my future studies. He has been able to explain things in different ways unlike my new profs who use the book/text only as a reference and do things like completing squares to find roots; which wasn't taught to me in university but was taught in high school and has made life easier.

Also as much as you may think it is stupid to remember the cosine song I wish I could remember it as it would have gained me an extra 10% in post secondary. Also don't ever forget that a log is an exponent. It may seem stupid or useless but it helped me get that extra 10% needed to get a B+ in math. When you are in doubt a log is still an exponent.

This is written from a past student of Mr. K. and he is doing his best and succeeding in preparing you for post secondary. (I wish I would have tried harder). Also, don't mind my English as that wasn't my focus so it may be poor.

Good luck and assignments vs final mark are very close together.

Happy New Year! ;-)