## Wednesday, November 09, 2005

### More Tutorials that you guys liked =D

Remember I said that there were other tutorials on the site, but we have no use of them yet? Well if you do remember, HERE THEY ARE! We can finally use them. If you don't remember or you haven't seen the tutorials that I had up on here for the limits and discontinuity sections, then you're mission out on how great those and these tutorials are. I suggest you to check out the site Calculus Help. The site is made by one of the coolest study guides author Mike Kelley. He puts the fun into learning in books.

Chapter Two: Finding Derivatives
Lesson 1: The Difference Quotient
Lesson 2: The Power Rule
Lesson 3: The Product Rule
Lesson 4: The Quotient Rule
Lesson 5: The Chain Rule

Yes, some of those rules we haven't gone over yet. But then again, I just wanted to put them up, so that way when we do go over it, it'll be right here on the site for you to go over with again. Have fun!

P.S. Did you know students in the U.S. could just sign up to take any AP exam from the collegeboard (the board where our AP exam is coming from), and teach themselves the course with a help of a book, textbook or any kind of resources? All they have to do is pay a fee. Less than \$100 dollars I think. Imagine that.

The books from that guy who created those tutorials puts a bit of a sense of humour into his teaching. He really relates his teaching to everyday stuff. You'll know what I mean if you've watched the tutorials for Limits and Discontinuity. Amazing. I should purchase one of his books through the net sometime after Christmas, "Peterson's AP Calculus AB and BC." Lol. We'll see.

Later dayz.

## Sunday, November 06, 2005

### An Integral

The end of chapter three. That went by pretty fast if you ask me. A short chapter, but a concept somewhat hard to grasp, like derivatives.

So let me start of summarizing the chapter from the beginning.

Concept of finding slope of tangent lines at a point leads us from the:

Parent Function -- > First Derivative -- > Second Derivative

That was last chapter. Chapter 3 leads us from:

Second Derivative -- > First Derivative -- > Parent Function

This time it is the concept of finding area between the function and the x-axis.
In other subject areas, we have always known that area is a scale. So is speed, and mass. Unlike velocity, acceleration or displacement, those are vectors. So basically, we have not heard of an area that is negative. But within the concept that you are learning, an integral, is the area between the function and the x-axis, which can be a negative value. Why? Because technically an integral is not an area, its just the method.

So what is an INTEGRAL?

A number. =D But to be able to say that you have to understand the whole concept of it, but in the end its just a number.

What is the concept?
An integral is an infinite sum of rectangles within an interval. The height is given by the output value of the function. The width is a very small number, close to 0, but not equal to 0.
{ formally: It is the total change in the parent function over the given interval. }

** If the height is given by the output value of the function, and most functions go below the x-axis, the output may be negative. As for the width, will be positive. Therefore, the signed area given may be negative.

=D Gets so far? If so, please continue on reading. If not, try reading it again, but more slowly this time, and maybe try to visualize it in your head.

Riemann Sums.
We learned 4 types of Riemann sums, and I’ll go over each sum with you.
Let’s use x^2 as an example.

Let’s pretend we have a monotonic increasing function.
For x^2, it is monotonic increasing on the interval (0, infinity). But to make explaining easier, I’ll be using the interval [0,5] and with 5 subintervals.

The following table of values is resulted.
X Y
0 0
1 1
2 4
3 9
4 16
5 25

Left Hand Sum:
This sum is calculated using the left value per interval. Since we have a width of 1, we’ll factor it out to make calculations easier.
LHS = 1 ( 1+4+9+16)
= 30
So therefore, the interval [0,5], LHS is equal to 30.

Right Hand Sum:
This sum is calculated using the right value per interval. Since we have a width of 1, we’ll factor it out to make calculations easier.
RHS = 1 ( 4+9+16+25)
= 54
So therefore, the interval [0,5], RHS is equal to 54.

Trapezoid Sum:
If you recall, the formula for finding an area of a trapezoid is =
{b(short side + long side)} / 2
So therefore, this sum is calculated using the average of the left and right hand sum. Since we already multiplied the base or width within the calculations of right or left sum, we could skip that step. Then divide by 2 to get the trapezoid sum.
Trapezoid Sum = (30+54) / 2
= 42
So therefore, the interval [0,5], trapezoid sum is equal to 42.

Midpoint Sum:
Midpoint sum is found using the midpoint between each subinterval, and taking its output as the height, and width would still be the chosen width from the start.
In our example, lets first generate a table.
X Y
0 0
0.5 0.25
1 1
1.5 2.25
2 4
2.5 6.25
3 9
3.5 12.25
4 16
4.5 20.25
5 25
So in our calculations, we would be using the output we calculated from each midpoint. And our width would still be 1.
Midpoint = 1 ( 0.25 + 2.25 + 6.25 + 12.25 + 20.25)
= 41.25
So therefore, the interval [0,5], the midpoint sum is equal to 41.25.

As for the programs we have in our calculators. Once you know what each program does, and what each is actually calculating for you. You will want to use the most efficient program to get your answer faster.
1) I believe RIESUM is the program on the T1`s that just calculate the RHS, LHS, and the Midpoint sum without showing graphs.
2) RIESUM is more efficient than RSUM which shows the graphs as well.

There are also built in functions in our calculators that calculate the exact value of the integral.
1) Math 9 is the command called fnInt. You use it on the home screen and going to the math menu and it is number 9 on the list. FnInt(funtion or Y1 , X, lower point, upper point)
For example, going back to x^2, from the interval 0 to 1:
Command would be as follows: FnInt(X^2, X, 0, 1)
When you press enter it should show you the value = 1/3 or 0.3333...
2) There is also the command where you graph the function you want to integrate, then go 2nd Calc menu and use option 7. Option 7 is the command the integral of f(x) dx. You choose your lower bound, then your upper bound. This command should shade the area within the graph and the x-axis from your stated lower and upper bound. It should then show you the exact value after shading.

Well sorry this is late guys. I finally got a new job, and I`ve been working
for the past weekend plus English homework *sigh*. Well hope you guys the best
of luck on the test. Hope everyone does good on the test. Did I mention we have
a very very very very very very very very SMALL class? If I haven`t , I shall.
We have a very very very very very very very very SMALL class. Lol. That`s all
for this unit. To the next unit. I think I like the next unit. *wink* *wink*
Till next time.

## Sunday, October 30, 2005

### Inversing the Derivative

So we learned how to find the rate of change, but now we are being taught to go backwards, finding the parent function from the first derivative. Or finding the first derivative from the second derivative.

Strange eh? In life we are always told to focus on now and the future, never to worry about the past. Now all of a sudden we are taught to go back. But then again it is calculus :P

Alright Ladies and Gentlemen. Have you read Ara’s scribe about Thursday’s lesson back at the AP Calculus blog? If you haven’t shame on you! Now go read it.

Pretty much this scribe will be about Thursday and Friday’s lesson. The beginning of Chapter 3 or shall I say half of Chapter 3. If you look forward on the textbook I think there are only 4 sections in this chapter. It will go by very fast. I think so anyways. So when the chapter is done, I’ll sum up the whole chapter again.

Here we go:

Velocity = change in distance / change in time

We were taught to find velocity that way. But now we want to find out the change in distance over the given change in time, how do we find it?

Change in distance = velocity x change in time

Drawing this image might help you when you’re trying to figure out either variables. I like saying “Very dumb things” in other words “v=d/t”.

So graphically what exactly are you doing? Given a velocity graph, How do you find the change in distance?

You take the area under the graph, or maybe I should be putting, the area occupied between the function and the x-axis. Why? Because someday we will be dealing with functions that have certain intervals that are beneath the x-axis. Why do I say change in distance? Because of intervals beneath the x-axis, when we do calculate them, you’ll see why it’s a “change” in distance.

Within the two days: We were taught about Riemann Sums.

The function being showed is monotonic decreasing, in other words in the shown diagram the function is ONLY decreasing. If it was said to be monotonic increasing, it’ll ONLY be increasing. Many functions are not monotonic. But if you look at it in segments, in certain interval, they can be monotonic.

Once again the idea of “many” is brought up in this chapter. In the past chapter we used “many” in the number of zeroes after the decimal. As “h” gets infinitely close to 0. Now in this chapter, we need to have “many” subintervals. The more we have, the smaller the width of a subinterval, and the smaller it is, the more the rectangles become looking like the area under the graph. We have seen that with our calculators.

Right Hand Sums and Left Hand Sums doesn’t give the exact change in distance, but it gives you a range of where it may be. Finding the average of the two sums or in other words taking the trapezoid sum gives you a much better estimate. We were also taught about midpoint sums, using the value of t in between the subinterval.

I know this isn’t the best explanation I’ve given at all. It’s rather a rushed explanation. So sorry. But really you guys should read Ara’s scribe. She did a good job. Till sometime this week. I’ll summarize the whole chapter for you, and hope it helps for the chapter test. Till then. L8r days.

## Thursday, October 20, 2005

### Finishing up Derivatives

So in class you guys told me how you liked those tutorials. You liked the fact that there was someone talking and that it used movement in the pictures. So I went home, after dealing with my first priorities (*ahem*babysitting*ahem*), I went ahead and googled some flash tutorials. I came up with none. The site where those flash tutorials came from does have a chapter 2, but we have no use for it YET. The ones we really need still says (COMING SOON!). I guess we’ll see it someday, and unfortunately that someday isn’t today. But, the first tutorial in chapter 2 is somewhat useful if you want a more clearer picture of secant lines and tangent lines. It’s on “The Difference Quotient”. You guys can take a look at that.

So I guess pictures will have to do for now. Sorry guys. I hope it helps though. Even though it’s boring.

Parent Function -> First Derivative -> Second Derivative

So looking back, first derivative. The first derivative tells us three things about the parent function. For the following graphs consider the lime green graph the parent function and the red graph the first derivative.

1) Where the parent function is increasing and decreasing.
-------- > How?
Where the first derivative is positively valued (above the x-axis), the parent function is increasing (slope of tangent lines are positive). Where the first derivative is negatively valued (under the x-axis), the parent function is decreasing (slope of tangent lines are negative).

2)Where the parent function has a minimum.
-------- > How?
Where the first derivative has a root (equal to 0), the parent function has a CANDIDATE for a minimum OR a maximum.
-------- > How is the candidate a minimum NOT a maximum?
When it’s negatively valued (under the x-axis) on the left side of the root, and positively valued (above the x-axis) on the right side of the root, it is a minimum.

3) Where the parent function has a maximum.
-------- > How?

Where the first derivative has a root (equal to 0), the parent function has a CANDIDATE for a maximum OR a minimum.
-------- > How is the candidate a maximum NOT a minimum?
When it’s positively valued (above the x-axis) on the left side of the root, and negatively valued (under the x-axis) on the right side of the root, it is a maximum.

So now the second derivative. The derivative of the first derivative. Confusing rights? Hmm not necessarily. If you look at this calculus stuff and not knowing what question you have in mind, maybe then it’ll confuse you. But if you know what you are being asked, you’ll know which part of the “BIG” picture to look at. Let’s get something straight first.

Behind the first derivative and the second derivative is some PARENT function. It’s what you first started out with.

The first derivative tells us where the parent function is increasing or decreasing and where it has local minima and local maxima.

The second derivative tells us where the parent function is concave up or concave down and where it has a points of inflection.

Remember the second derivative is the derivative of the first derivative? Meaning the second derivative also tells us where the first derivative is increasing or decreasing and where it has local minima and maxima.

Now onto the how can you tells.

The Second Derivative

The second derivative also tells us three things about the parent function. (The next following graphs, red graph is the parent function, the lime green graph is the first derivative, and the blue graph is the second derivative.) They are:

1) Where the parent function has a point of inflection.
-------- > How?
Where the second derivative has a root, the parent function has a point of inflection.

2) Where the parent function is concave up.
-------- > How?
Where the second derivative is positively valued (above the x-axis), the parent function is concave up.
3) Where the parent function is concave down.
-------- > How?
Where the second derivative is negatively valued (under the x-axis), the parent function is concave down.

So that pretty much conclude my post on derivatives. I'm not sure if I should post up more on limits and continuity. Tell me if I should. So the pre-test is on Monday and the test is on Tuesday. Don't forget to do your test blogs guys. Wouldn't want to lose a mark on that. I'm thinking of posting up another blog tomorrow. I'm thinking of applying the stuff on derivatives in 2-3 questions. If you have any questions in mind, put in your comments, or just tell me in the shoutbox. Hope those pictures helped and you understand the written stuff up there. Now just apply it to questions. Till tomorrow.

P.S. October 20, 2004 ---> The day I transferred into Calculus last
year. Yups I was 2 chapters late when I got into Calculus. I didn't understand
what the whole derivative stuff were. Much less the pre-cal stuff. I was also
taking Pre-Cal 40 at that time with Ms. Antymniuk. But you know what? I didn`t
let that stop me. Mr. K explained the concepts to me, even though it was
somewhat brief. That whole week, I made notes from the textbook. I learned half
of Pre-Cal 40S from the Cal Textbook in 2 days. The Calculus stuff it took me a
longer time to analyze it all. But I understood LITTLE parts of it. I wrote
the Chapter 2 test the same time the class wrote it. What I got in it, I was
proud of. 63%. You might ask why I would be proud of such a mark?
Considering I started the class when there were already in Chapter 2.6, and
nearing a test. The fact that I ACTUALLY passed, was good enough for me at that
time.

Unlike Sysiphus we are NOT set up TO FAIL, we are set up TO SUCCEED.
It all depends on the effort you put in, and how hard you try.

## Tuesday, October 18, 2005

### Flash Tutorials for 2.7 and 2.8

I did happen to say I’ll be doing weekly summaries in this blog. But since the whole chapter is almost over, I kind of thought otherwise. I’ll post up another blog in the following days but for now I’m posting about the last two sections in the chapter.

Chapter 2.7 deals with limits and Chapter 2.8 deals with continuity. For the next few days you probably have those questions in your head and don’t want to voice it out. When I first started out hearing about limits and continuity, I didn’t understand it completely, and I admit it, I still don’t. So searching on the internet for possible good explanations was my first or second resort. (Yes, shame on me, my last resort is the teacher.) If you look on the right side of this blog, there are the links. Some of those links are useful, for practice questions, for good explanations, and for good definitions. I ask of you to take advantage of those links, they’re there for your use.

Remember your VARK results? If you’re multimodal, or you have one-two-three preferences, the link that I’m about to direct you to is a site where there are available flash tutorials on Limits and Continuity. They’re like Mr. K teaching. They use voice, text, math pictures, picture analogies, and straight out definitions. There are six flash tutorials on the site. Below are what the flash tutorials' titles are:

Chapter One: Limits and Continuity

Lesson 1: What is a Limit?
Lesson 2: When Does a Limit Exist?
Lesson 3: How do you evaluate limits?
Lesson 4: Limits and Infinity
Lesson 5: Continuity
Lesson 6: The Intermediate Value Theorem

Hope you guys like the tutorials, and have a better understanding about limits and continuity. Till my next summary, which might also be a Test summary. Or maybe I'll seperate those two. Eh, who knows what you'll see in this blog. For now, later days.

## Sunday, October 16, 2005

### The Derivative Function

What we know so far…

Vocabulary Wise

Tangent line - A line that touches the curve at one point, in result of an infinite numbers of secant lines.
Instantaneous velocity - Taking average velocity in smaller and smaller intervals at a point. Having a very small change in x, close to zero, but never equalling zero, gives you the instantaneous velocity. The rate of change at that instant.
Derivative - a slope of a tangent line, or to be exact, it is the total change in the parent function over that interval.
Critical numbers - Is the root(s) on the derivative function or where the derivative function is undefined and is showed by using the x-value.
Critical point - Is the relative maximum and minimum points of the parent function and is shown as a coordinate of points.
Inflection point - Where the function changes concavity (u-shaped or umbrella shaped). (We’ll go in greater depth about inflection points and what it is related to in the following week ;-) )

With the concepts

Consider the lime green graph the parent function and the red graph its derivative function.

Parent function = f
Derivative function = f’

Where f has relative maximums and minimums, f’ has roots or is equal to 0.

Where f is increasing (positive tangent slopes), f’ is positively valued (above the x-axis).
Where f is decreasing (negative tangent slopes), f’ is negatively valued (below the x-axis).

Where f has a relative maximum, f’ is a root, its positively valued on its left, and negatively valued on its right.
Where f has a relative minimum, f’ is a root, its negatively valued on its left, and positively valued on its right.

Looking at a line analysis like the lines under the graph could help. Remember to always label what the line analysis represents, the parent function (f), the first derivative (f’), or (soon we’ll learn) the second derivative (f’’).

Another thing we haven’t learned in great depth yet, is where f changes concavity, where it has an inflection point, f’ has a critical point.

Overall

The first derivative tells you whether the parent function is increasing or decreasing, if it has a max or min, and where it has an inflection point.

Parent Function --> First Derivative --> Second Derivative

Looking back to Physics:

Displacement --> Velocity --> Acceleration

Applying that to Calculus:
Velocity is the first derivative of displacement, acceleration is the second derivative of displacement, and velocity is also the first derivative of acceleration.

So now that we know how to look at the derivative function graph and we know how to reflect it back to the parent function graph. Let’s do some algebra.

Calculating the Derivative

To find the average velocity, we take the slope of a secant line, which could be showed in three ways.

Taking a look at the parent function x^3 - 3x + 3. Consider the function the displacement of a particle per second.

If we were asked to find the average velocity between t=-1 and t=1, we would do the following:

Therefore, the average velocity between t--1 and t=1 is 2 units/s.

Consider the following table of values. If we are asked to find the instantaneous velocity at x=0.5, we would do any of the 2 following steps:
1) Using a one sided difference quotient. But then we know we are either going to be over or under. 2 )SO, to be precise we use the symmetric difference quotient. We go 0.1 units to the left of the point, and 0.1 units to the right of the point and take the average of both answers.

Using the definition of the derivative, we could use it to find the derivative of any given function. Consider the following function: x^2.

There are so many ways to find the derivative. Mr. K gave us a list of the ways we could find the derivative:

1) If we were given the graph, we could draw a tangent line, and find the slope using where the line intersects whole number coordinates.
2) Given the function, we could use our graphing calculator and zooming in at the point, till what we see on our screen is a straight line. Then using the slope program installed in our calculators.
3) We can use the definition of a derivative and work out the algebra.
4) Using our calculator, we can draw the tangent line. The given formula of the tangent line would be in slope-y-intercept form.
5) Given a table of values, we could use the symmetric difference quotient.
6) Using the stored function in Y0 in our calculators.
7) Using the command Nderiv (on T1’s) on our calculators.

We’re only on chapter 2 of our textbook and only a little over halfway through the chapter, believe me there are more ways. More ways to make the whole “BIG” idea a lot easier.

Well hope that helped as this week‘s scribe. In the following week, I’ll try and find out how to get everyone to make their post in here. I’m still new on using this whole blogger thing. Hope you guys like the idea and take part in it as well. I’m here to help you guys out, we’ll go through the whole Calculus course together.

## Friday, October 14, 2005

### A Start

To my classmates who just happen to go across this page over the time its under construction I'll give a little introduction to what this blog would be basically about.

I just thought of it as a great idea to do something like this. I've noticed even before the first test, some already had second thoughts of dropping the class. Other than the fact that there's an advantage of taking the course, with it also comes a disadvantage. We know not everything is fair. So I thought I could try making it a bit easier? If possible a lot easier. Things will get tough, but as a class, maybe it wouldn't be so hard. Did I ever mention that this year's class is very small compared to last year's? Well since its small, it means we can all help each other out more.

Some of you probably already know. I had taken the course last year already. To make things a bit easier, I thought of starting this blog. It's a place where we can all talk about what we don't understand, what we need a little more work on, or what needs to be explained a little more. Along with that instead of a daily scribe, I'll try and post up a weekly summary of what we learned so far in the past chapter.

To those in favor of starting this blog, post "aye!"

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