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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, November 30, 2005

Tangent Line Approximations and Newton's Method

I haven't exactly been helpful at all. I started something and I slowed down. I guess that's what happens sometimes. I don't know wether to say sorry or not. But here's one of the missing scribes.
Wednesday, November 30, 2005

We learned half of the last section of Chapter 4.
First area it covers was on Linear Approximations.
If we were given a function.
For example f(x)= square root(x).
And we were to find an approximation value of f(26).
What do we do then?
In this particular example, we find the closest perfect square.
Which happens to be 25.
What if we were given other kind of functions?
You try to find a coordinate that you know, and is close to the value you are finding.

Now we have a set of coordinates (25, 5).
Since 26 is close to 25, we can use a tangent line to find f(26).
Differentiate f(x).
f(x)= x^(1/2)f'(x)= 1/2(x)^(-1/2)
We then find the slope at our given coordinate.
f'(25)=1/2(25)^(-1/2)f'(25)=0.1 or 1/10
Now we have all the "ingredients" to put into our slope-point formula.
y-y1=m(x-x1)

y-5=0.1(x-25)

So basically, that equation up there is all we need to find f(26).
All we have to do now is solve for y when x=26.
In the end we get, y= 5.1 or f(26)=5.1

Let me sum up the steps of Linear Aprroximation.
1) Identify the function.
2) Identify what value you have to find.
3) Think of a set of KNOWN coordinate (x, y), close to the value you have to find.
4) Differentiate the function.
5) Find the slope of the tangent line at the KNOWN coordinate.
6) You have the slope and set of coordinate points, plug it into the slope-point formula.
7) Let x=the value you have to find, and solve for y.
8) *optional, but recommended. Do a check. You never know you might have made a mistake somewhere along the way.

That day we were also introcuded to Newton's Method.
Newton's Method is a method of finding zeroes of a function. You are given a function, an interval or point where the xero is close to. Zeroes of Tangent lines are then used to determine if its the zero of the function as well.
At the start we are given f(x) and (Xo, f(Xo)). From the given you can find f'(x) and the derivative at the given point.
y- f(Xo) = f'(Xo)(X-Xo) <----- slope-point formula
Since we are finding a root of the tangent line, we let y=0. From this equation we are trying the find the x value of the root. So we re-arrange the equation.
0- f(Xo) = f'(Xo)(X-Xo)
-f(Xo) / f'(Xo) = x - Xo
Xo - [f(Xo) / f'(Xo)] = x

x would then be the first x value.

When using Newton's Method it may take multiple times till you finally get close to a good approximation of where the zero of the function is. That's why doing this method by hand is very very very very painful. Thank You for our stupid calculators. In our calculators we have a program stored called "NEWT". It is used for this kind of method. We can also do this method manually, as showed during next class.

Hope this was some info. Hopefully is useful for someday reference. Sorry it was very very very very very very late.



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Newton's Method

One application of derivatives it to use the equation of a tangent line to approximate values close to a known value on a curve. i.e. finding √35 on the function f(x) = √x.

Another use of derivatives is to find a tangent line whose root is the same as the root of a more complicated funtion. Every approimation we make gets us closer to the actual root of the more complicated curve. This is known as Newton's Method and it's named after the man who first developed the technique, me! No, just kidding, Sir Isaac Newton developed the technique. Watch how it works here and here.

Here is a flash tutorial that explains the process we discussed in class today. You can watch Newton's Methis in action using this example. And then try these exercises to see how well you can apply what you've learned; detailed solutions are provided.



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Tuesday, November 29, 2005

Tangent Line Approximation

Stan the Clause - another one of the usual related rates problem Mr. K always gives every morning. As to what I did, I never preassumed that a wide set of concepts are to be considered in answering these kinds of problems. Claiming to that perception, I twisted my mind a little bit to come up with not "The Answer" but something just "close" to it. And to my dismay, I got way far from it. (laughs!)

A lot of mathematical ideas come into use when these kinds of questions are given:
- Find the rate at which the angle of elevation from the naughty boy's house to the flying can is changing in radians per second when the angle of elevation is pi over 12 -

To make life easier, visualize the whole problem. Try to illustrate. Know all the given facts or numbers. Then, give it your best shot! ;) It is just like solving an everyday math question, only it is more complex. Students might have to apply all the stuff they have learned throughout their mathlife. Regarding this problem, I was not able to get this one completely because I never thought this would include trigonometric concepts and all. I was stuck when I came to that point because I have slightly forgotten some stuff about it. So now I think I might have to work on refreshing myself with trigo.

In the remaining minutes of the period, we learned a new lesson that is the Tangent Line Approximation. This is basically applied in (as to what Mr. K used as an example) finding the value of a square root function with a non-perfect square number.

Given f(x) = √37 find the value

First, formulate its derivative equation - f ' (x) = 1 / (2 √x)
Then, obtain a perfect square number closest to the given number, in this example the closest is 36.
Substitute the x variables in the derivative equation by 36 and by doing that, the slope at that point will result. - f ' (x) = 1 / (2√36) = 1 / 12
Use this slope to form a derivative function for the given number 37 using the point-slope form. Y - Y1 = m (X - X1)

-> f(X) - 6 = 1 / 12 (37 - 36)

So, f (37) = 1 / 12 + 6 = 73 / 12

And using the calculator, the √37 is 6.08333 that is, almost as close to 73 / 12



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Monday, November 28, 2005

Derivative Acrostics

Kudos to Mrs. Armstrong for turning me on to the idea of acrostics in math.

Blogging Prompt
Your task is to create an acrostic "poem" that demonstrates an understanding of calculus related to any one of these concepts:

DERIVATIVE
POWER RULE
PRODUCT
QUOTIENT
CHAIN RULE
TANGENT LINE
NEWTON'S METHOD

As an extra challenge (worth an additional bonus mark) try to make a Double Acrostic, that is, each line should begin and end with a letter of the word you are working with.

Remember, this is a bit of a race. Your answers have to be posted to the blog in the comments to this post. If someone has already used a word or phrase in their acrostic you cannot use the same word or phrase. i.e. It gets harder to do the longer you wait. ;-)

Here is an example of an acrostic that Mrs. Armstrong wrote:

Always in 2 dimensions
Region between the boundaries
Entire surface is calculated
Answer is in units2

Be creative and have fun with this!!



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Sunday, November 27, 2005

Sunday Connected-Slide Funday!



ConSlide puzzles are a new type of sliding block puzzles invented by M. Oskar van Deventer... The pieces of a ConSlide puzzle move like regular sliding block puzzles, but some of the pieces have sections connected by bars of various heights. This means pieces can pass over and under one another as long as the bars and posts don't run into one another. The goal of each of these puzzles is to move the red block to the upper left corner. Play it here.

(Once again, thanks to Think Again!)



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Friday's class...

Last Friday we started our class by having our second review quiz in preparing for the Test in May. We're given 8 minutes to answer 4 multiple type questions. Then after that Mr. K showed us the rubric for the Prince of Calculand Project. He discuss it to us for a few minutes. After that Mr. K talked about the Editor's Initiative and he asks our thoughts about it.

After all the discussions we had about the project and the Editor's Initiative, Mr. K showed us and taught us how to answer this related problem that we have been working for a few days.

A spotlight at a school dance is fastened to a wall 8 m above the floor. A girl 1.75 m tall moves away from the wall at a speed of 0.75 m/s.
a. At what rate is the length of her shadow increasing?
b. At what speed is the tip of her shadow moving?

First in solving in this problem like other kind of problems is drawing a diagram on what is given to you.

Then next is analyze the drawing. In our diagram there are two similar triangles and based on that we could create a relationship between the two.
b/1.75= a+b/8 (a+b=c as shown in the diagram)

8b=1.75a+1.75b
6.25b=1.75a

Then find the derivative of that,
6.25 db/dt= 1.75da/dt
Since we know da/dt= 0.75 m/s, juss plug it in and we could figure out db/dt and justplug it in.

db/dt=1.75(0.75)/6.25
=0.21 m/s
So the length of the girl's shadow is increasing 0.21 m/s.

Then in order to know what speed is the tip of her shadow. We could use this relationship c=a+b and find the derivative of that,
dc/dt=da/dt+db/dt and plug in the values to get dc/dt
dc/dt=(0.75)+(0.21)
dc/dt=
0.96 m/s

The tip of the girl's shadow is moving 0.96 m/s.

That's all we did last Friday....The next scribe is Ara..I'm willing to accept any constructive criticism so don't be afraid to leave a comment. Thanks!




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Saturday, November 26, 2005

The Editor's Initiative

Instead of posting a pre-test reflective comment on your progress in this class you may undertake The Editors' Initiative. Here's how it works:

  Step 1: Scan through the previously posted Scribe Posts on the blog. Find one that has one or more errors.

  Step 2: Discuss the error(s) and what you think the correction(s) should be with me. If I agree with your editorial proposal go to Step 3.

  Step 3: Discuss the editorial change with the author of the post. The author will chose to proceed in one of the following two ways.


3a3b
The Editor is briefly allowed administrative privileges on the blog. They will edit the post to make any necessary corrections. They then sign the post at the bottom:
Edited by: [name] on [date]
The author will edit the post in consultation with the editor who will vet the author's changes until they are correct. The author then signs the post at the bottom:
Consulted editor [editor's name] on [date]

Students may chose to make more than one edit. Each additional edit will earn them a bonus mark on the next test. Your mark on the previous test determines the maximum number of edits/bonus marks available to you.

Mark on Last Test / Max Edits Allowed
> 90 / 1

80-89 / 2 (1 bonus mark)

70-79 / 3 (2 bonus marks)

60-69 / 4 (3 bonus marks)

50-59 / 5 (4 bonus marks)

40-49 / 6 (5 bonus marks)

30-39 / 7 (6 bonus marks)

20-29 / 8 (7 bonus marks)

10-19 / 9 (8 bonus marks)

0-9 / 10 (9 bonus marks)

You may also assume the role of Content Consultant to earn marks as outlined above. Here's how it works:

  Step 1: Scan through the previously posted Scribe Posts on the blog. Find one that doesn't provide enough detail or leaves out too much information. Decide what additional content should be added.

  Step 2: Discuss the new content you think should be added with me. If I agree with your editorial proposal go to Step 3.

  Step 3: Discuss the editorial change with the author of the post. Together, you will chose to proceed in one of the following two ways.


3a3b
The Content Consultant will add a new post to the blog inserted at the appropriate time and date. They then sign the post at the bottom: Additional Content by: [name] on [date]The author will edit the post to include the additional content provided by the consultant. Additional content will appear under a heading "Additional Content". The author then signs the post at the bottom: Additional Content Provided by [consultant's name] on [date]

Students may chose to make several additional content contributions for bonus marks according to the table above.

As we discussed in class, you can edit your own scribe posts only if the whole class agrees with your proposal and we are convinced that you tried to do your best work the first time around. When it's your turn to be scribe try to write a post that is so excellent no will be able to edit it. ;-)



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Friday, November 25, 2005

Mr. Dixon's on eBay - How Trusted is He?

Mr. Dixon is in my classroom right now and he has a question he needs your help with.

His trust rating on eBay is 99.2%. Out of 122 transactions he's had, 121 people said he treated them well and he can be trusted.

Mr. Dixon wants to know when will his "trust rating" reach 99.3% and he needs your help.

(1) Can you write an equation he can use to solve this problem?
(2) Can you solve the equation and tell Mr. Dixon how many more favourable exchanges he needs to have to raise his trust rating to 99.3%.

All of my classes are working on this. The first student in each class who solves Mr. Dixon's problem correctly gets a chocolate bar. ;-)



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Wednesday, November 23, 2005

More Related Rates

Today in class, we attempted a problem in separate groups. It was called Scuba Steve's Slick Bacteria. The problem goes like this:

While on vacation in Mexico, a tanker accident has spilled oil in the Gulf of Mexico. Scuba Steve volunteered to help by spreading oil-eating bacteria which is gobbling 7 ft^3 per hour. the oil slick has the form of a circular cylinder. When the radius of the cylinder is 250 feet, the thickness of the slick is 0.02 feet and decreasing at a rate of 0.002 ft/hr.

1. At what rate is the area of the circular top of the slick changing at this time?

2. Is the area of the slick increasing or decreasing?

To solve this problem you have to identify what the problem is giving you in terms of information and find out what it is asking of you. The problem talks about a cylinder and the thickness of it, so one must realize that the problem will require the use of the volume formula for a cylinder.

V = πr^2 h

dV/dt = -7 ft^3/hr dh/dt = -0.002 ft/hr r = 250 ft h = 0.02 dA/dt = ?

Because of the two variables r and h, you have to use the Product Rule.

dV/dt = π(r^2)(dh/dt) + 2πr(dr/dt)h

You solve for dr/dt, by using substitution.

-7 = π(250^2)(-0.002) + 2π(250)(dr/dt)(0.02)
-7 = -125π + 10π(dr/dt)
dr/dt = (-7 + 125π)/(10π)
dr/dt = 12.2772

Now to solve the area question, we use the formula for the area of a circle.

A = πr^2

dA/dt = 2πr(dr/dt)

Now use substitution to solve for dA/dt.

dA/dt = 2π(250)(12.2772)
dA/dt = 19284.9807 ft^2/hr

If you see the word rate in a question, you have to use the derivative function.

For the Prince of Calculand project, the questions are at http://calculandquest.blogspot.com and the solutions are at http://calculandans.blogspot.com

The next scribe is Prince.



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Tuesday, November 22, 2005

Related Rates Homework

Read through this tutorial on related rates problems. It's a lot to read but it's worth your time. The beginnig will take you through the general steps to take in solving related rates problems, then threre are several examples that have animations to show you what is changing in the problem and how.

Here is another tutorial with animations and a problem set. You can do as many questions from the problem set as you like. But your assignment is to do the first 5 questions here. Answers are provided so see if you can solve them without looking at the answers. ;-)



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Related rates of change

In today’s class, Mr. K told us about the exam. Woo... Which we know it will be really hard. And Mr. K gave us a quiz. It had 4 multiple questions on that sheet. We got only 8 minutes to do with yet, which every question you only gave two minutes to do with it. And we were not allowed to use our calculator. We need to do a little practice to get familiar with those questions which will appear on our final exam.
I can image how hard the exam will be, we need to do a lot of exercise to be comfortable when we white that exam which will be on May.

There were three questions on the white board. It was about related rates of change.
Air is blown into a spherical balloon at a rate of 12In^3/ sec. How fast is the radius increasing at the instant the diameter is 1 foot.

From the question that we know the rate of change volume dV/dt=12
1 foot is equal to 12 inch. So, the radius is 6 inch.
And the question is to find the rate of change radius dr/dt.
Volume of Sphere formula: V=4/3 pai r^3
Up to here, we know to fine the rate of change; we need to find the derivative of the function.
According to the Product and chain role:
dV/dt = 4/3pai* 3 r^2 (dr/dt)
dV/dt = 4 pai r^2 dr/dt
12 = 4 pai 6^2 dr/dt
12=dr/dt

Two questions left, we didn’t have enough time to finish it. And the homework will be on the Bolg tonight.



Nest Scribe is Steve:P



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Monday, November 21, 2005

Focus


Today, again, like all the other days, we had to do problems on the board. These problems were like reviews of last friday's class on how to differentiate functions implicitly. Again we differentiate functions implicitly when the function is not in the y=f(x). We are reminded not to make little mistake like negative signs and forgetting to use the chain rule when differentiating y.

We had two questions on the board. First was about differentiating and finding the equation of the of the tangent line and next was about the related rates of change. For the related rates of change we apply everything we learned from before. For these kind of questions, we are not given a formula to use. We need to analyze the question to find out what we need to do. For example

***** All edges of a cube are expanding at a rate of 3 cm/min. how fast is the surface area changing when each edge is

a) 1 cm.

Surface Area = 6x^2 <----------------- this formula of surface area is static ( no changes) to make it dynamic, we have to take the derivative of this formula with respect to time. d S.A./dt = 12x(dx/dt) next we found out that 3 cm/min is the rate of how much the sides are changing. thus it equals to (dx/dt) therefore we just substitute what is given to our derived formula x=1 (dx/dt) = 3 to find the rate of change of the surface area.

d S.A./dt = 12(1)(3) = 36 <--------------- at that instant the rate of change of the surface area is 36


To conclude, there are a lot of other problems like this which uses other formulas. The key is analyze what is being given to you and understand what you need to do.


Thats it... the next scribe is xun :D



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Forget me not

"I write down everything I want to remember. That way, instead of spending a lot of time trying to remember what it is I wrote down, I spend the time looking for the paper I wrote it down on."
Beryl Pfizer

We started the day by going into groups to answer the given question. The question was about graphs of different function and we had to determine which function is parent function, the derivative and the second derivative. After figuring out which is which, we had to explain why we made the decision. After the group discussion, Mr. K gave us problem in the board that asked us to differentiate equations. We learned how to differentiate implicitly. We know that we should use this form when functions are not in the form y=f(x).

ex.)
x^2 + y^2 = 4 <-----------------differentiate using the tools we learned before
2x+2yy' = 0 <--------------- chain rule for differentiating y because there might be an inner function, we just don't know what it is.
yy'/y = -x/y
y'=-x/y

To conclude, differentiating implicitly is used when the function is not in the form y=f(x). We have to remember to use the chain rule when differentiating the y. Also remember not to make little mistake like the negative signs.


Just Remember to not to Forget!!



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Sunday, November 20, 2005

Fourth Powers


- I hear you have extraordinary powers!

- Not true! What I said to the press was that I have fourth powers at home.

- What do you mean?

- 1, 16, 81, .... That kind of stuff.

- But these are square numbers!?

- Yes, but they are more than that. They are square numbers squared.

- I see. And you have these at home?

- Some of them. Not all of them.

- And what do you do with them?

- Play with them, of course, what else?

- What kind of games?

- This morning I added them.

- That must have been great fun!

- That's what I said to the press, but they didn't believe me.

- How did the game go?

- Not very well. I tried to pick less than nineteen of them, one several times if
neccessary, to get any number in the universe.

- How did you get 20?

- 20 was easy. 1 + 1 + 1 + 16 = 20. I only had to use five of them.

- Which number gave you problems?

- I found a number less than 500 that required nineteen fourth powers.

- So you lost the game?

- Yes, and I can't even remember which number it was.



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Friday, November 18, 2005

Differentiating and a little bit of manipulating

Well, I guess I'm up. Today in class we worked a bit with some problems using our bank of derivative rules and the chain rule. The questions mostly involved differentiating several functions, and also served to refresh our memories on how the first and second derivatives apply to the parent function. Closer to the end of class today, Mr. K began to tell us how we could take some equations such as the equation of a circle, and manipulate the domain to make it into a function.

For example: if we look at the circular funtion x²+y²=4
On its own, it isnt a function because it fails the vertical line test.
We could break the function the function up into:
f(x)={(4-x²)^½}
{-(4-x²)^½}
Afterwards, we could then manipulate the domains of both of the functions parts. For example:
f(x)={(4-x²)^½; -2 ≤ x ≤ 0}
{-(4-x²)^½; 0 < x ≤ 2}

After the manipulations the function passes the vertical line test and the graph looks like this


Class ended off there, but Mr. K is going to continue this discussion tommorow. See you all tommorow. Scribe for tommorow is Jayson.



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Wednesday, November 16, 2005

What Happened in Class today....

First, like usual we answer some questions on the board. We're asked to differentiate different kinds of functions. On those questions we're able to differentiate functions using the power rule, the quotient rule, the product rule. Also, today we found out that eventhough we're finish with the calculus sometimes it's better to simplify it algebraically because most likely in a multiple choice question we would get the simplified answer in one of our choices.
Next, thing we learned and was added to our bank of derivatives was the chain rule:
The Chain Rule: [ f 0 g]' (x)= f'(g(x))*g'(x)
That's the Chain Rule, Mr. K didn't prove it to us yet. However checkout this site, this would show us the proof of the chain rule. Also, check this one too, this would also show us the proof of the chain rule.
Lastly, Mr. K showed us what's the derivative of ln x. d/dx lnx=?
e^lnx=x
e^ln x (d/dx ln x) = 1
x (d/dx ln x) = 1

d/dx ln x = 1/x

Basically, that's all we did today..... The next scribe is CJ.....




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Monday, November 14, 2005

Bank of Derivatives (so far...)

Our rules SO FAR...

Any constant: d/dx C = 0

Power Rule: d/dx cx^n = ncx^(n-1)

Product Rule: d/dx f(x)g(x) = f(x)g’(x) + f’(x)g(x)

Quotient Rule: *lo di hi minus hi di lo all over lo lo*
In mathematical terms….
d/dx f(x) / g(x) = [ g(x)f’(x) - f(x)g’(x) ] / [ g(x) ] ^2

There were also the derivatives of the six trig functions that Mr. K proved in the end of class today.

d/dx sin x = cos x--------------- d/dx csc x = - csc x cot x

d/dx cos x = -sin x-------------- d/dx sec x = sec x tan x

d/dx tan x = sec^2 x----------- d/dx cot x = - csc^2 x


See any patterns?
Look at the functions that starts with “C”, their derivatives are all negative.
The function tan x and cot x, their derivatives are squared.


So today we started off our class with a problem. Boys vs. Girls. Who won the race between Willy the wallaby and Roo the kangaroo?

A verytrickyy tricky question. Yet very funny in some sense. We need more practice on it. And we shall get another one tomorrow =)

For the rest of the class, we filled out our bank of derivatives with a couple of new ones. It’s growing larger everyday (except weekends).

That was pretty much what we did today. I have three words before I who's you whos the next scribe and that is...

MEMORIZE MEMORIZE MEMORIZE!

Next Scribe is Prince. Sorry Prince. Hope you're feeling better by tomorrow. I think your the last one. =S I'm not sure.



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Sunday, November 13, 2005

The View From Sunday

This week's puzzle comes from Think Again! again. ;-)

The Other View


With nine pieces as the one shown above, a 3x3x3 cube was made. Below you can see the cube viewed from South, East, and North. What does it look like seen from the West?



Say .... uh .... you've already finished your homework right? The stuff three posts below?



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Saturday, November 12, 2005

Visitors From Alabama

This was left as a comment to Jayson in his pre-test reflective post. I thought I'd bring it out here for all of you to see .... our class has just grown. ;-)

I just want you to know that I am an AP Calculus AB teacher in Birmingham, Alabama, and I was just introduced to your class' blogs yesterday during a workshop. I think that this is a really neat and enriching idea to do in a mathematics class. I am going to try to learn how to get this set up for my classes, but until someone has enough time and energy to train me, I am going to require that my students visit, learn from, and comment on your class blogs. I really respect this particular blog of yours, and I will read it first thing Monday morning to all of my classes. I hope that you all and Mr. K keep up the incredible work!!!
Mrs. B from Alabama



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del.icio.us Guide

You can download a pdf guide to del.icio.us right here.

It is 12 pages long. Flip to the section where you feel you most need help. If you like to work off a hard copy you can print it up. If you don't have a printer at home you can print it up at school. If enough of you want to have the hard copy let me know in class and I'll have a bunch of them copied for you. ;-)

Don't forget to do your homework in the post below this one!



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Friday, November 11, 2005

Product and Quotient Rule Homework

As promised in class, your homework this weekend is online.

We'll begin with this combination quiz and review of the product and quotient rules. The page is a little long but make certain you work through ALL the examples and exercises.

Sarah continues to do excellent work on her blog. Part of your homework this weekend is to visit her blog and work through the first 4 movie links. You can also watch the Chain Rule movie to get a preview of what we're doing next week.

Then you can take this quiz. Actually there are two quizzes on this page, you only have to do quiz 1 questions number 1 through 6. The other questions involve derivatives of the natural logarithm function and composites of functions. We'll be covering these topics next week -- so you can come back to this quiz then.

Lastly, take this quiz on the product and quotient rules. There are twelve questions -- you have to do the first eleven. If you watch the video over at Sarah's blog on the chain rule you'll be able to do the twelfth question as well. You'll come across some notation you're not familiar with in this quiz in 1 or 2 questions. You can click on the answer link to learn how it's used and what it means.

Have a great weekend!



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Product Role & Quotient Role

Product Role and Quotient Role ~^.^~

In today’s class, we learned another way of how to get derivative function.
We used the definition of derivative and a rectangle diagram proved the product role.
Product role:
If the function is F(x) =f(x)g(x)
Then the derivative of the function F’(x)= f(x)g'(x)+f'(x)g(x)

Example *+_+* : H(x) =X^2X^3
According to product role:
H’(x) =(x) ^2(3x^2) + 2x(x^3)
H’(x) =3x^4+2x^4
H’(x) =5x^4

Quotient Role:
If the function is F(x)=f(x)/g(x)
Then the derivative of the function F'(x)=( g(x)f'(x)-f(x)g'(x) )/( g(x) )^2


And next Scrible is Sarah



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Wednesday, November 09, 2005

ok, my turn

i never knew i was the scribe until I saw everyone's head move 90 degrees to the right, eyes quite astonished, looking straight at me, wishing to inform
" you are Ara... yes you are. "
And for all the days that i usually checked the blog... ( s i g h ) .... hey, i just missed it by a blink of an eye! ( l a u g h s )

anyway...
Today was a continuation of getting the derivative of a function particulary the EXPONENTIAL FUNCTION.

A certain was given for us to find its derivative. To find its derivative we can use the definition of the derivative:







Well, we did not really do it this way, but it's fundamentally the same. We did it in another perspective, which is using the calculator, and i can't put all those things here so i did it this way.

So basically, the rule to get the derivative of an exponential function is:

The derivative is simply the original function times the derivative of exponent times the natural log of the base.



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Tuesday, November 08, 2005

Suddenly Not So Ugly

Today in class, we started the fourth unit, and yet again the derivative function shows itself once again. We learned how to convert a parent function into a derivative function without using the Definition of the Derivative Function.

If the parent function is a constant function, for example: f (x) = 2
The derivative function is automatically zero (0), f ' (x) = 0

If the parent function is a linear function, for example: f (x) = 2x
The derivative function is the slope of the line (m), f ' (x) = 2

If the parent function is a power, for example: f(x) = x^3
To convert from the parent function to the derivative function by taking the exponent and make it the coefficient, then subtract 1 from the exponent to get the derivative function.

f (x) = x^n
f ' (x) = n(x^(n-1))

Now with these short cuts the derivative function is not as ugly as it once was.

The next scribe is Ara.



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Monday, November 07, 2005

Better late than never

Better late than never is often used as a polite way to respond when a person says "sorry" for being late. :D

Last thursday we did the pre-test about definite integral. As usual we had about 30 min. to do the test and after we when to groups and discused what the test was about. In groups we figured out decided what the answers should be and handed one paper with all our names on it.

Another things was Sarah and Chris went to Ms. Armstrong's class to discuss what we are doing here at blogger and how it influence us in ways other resources can't.

Again sorry for being late and the next scribe is...
Steve



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A del.icio.us Idea

I recently received this email from a student in another class:

Hey Mr. K.

This is one of the websites I was looking at that had simplifying radicals..

http://regentsprep.org/Regents/math/radicals/pracRad.htm

I found a few that I thought were good just by typing "radicals" in google, they really helped me out.

See you Monday,
************


Students often find more, and better, sites than I do. You're better websurfers than I am. ;-) That got me thinking .....

I spend a lot of time looking for good websites that help us learn in this class. But what if we all spent a little time doing that? What if there was an easy way for us to both save our bookmarks (without cluttering up our favourites list) and share them with the whole class with the click of a single button? And what if we could access those bookmarks not just from home, but from any computer in the world? Hmmm .....

Well, there is an easy way to do that! Instead of saving bookmarks on your home computer sign up for a free account at a site called del.icio.us. You can then access them from any computer in the world. You can easily install a little button/bookmark that allows you to save any webpage you're looking at without interupting your surfing. Tag it using this tag:

apcalc

Now we can all get each others bookmarks with the click of a single button in our del.icio.us accounts.

I'll go one better than that. If you all jump in on this idea, I'll write a post on our blog (with a permanent link in the sidebar) that will load the 10 most recently saved links automatically as you find them. I'll also include a link to the archive that you can browse at your leisure.

If this interests you (and I can't imagine how it doesn't) read this tutorial on how to get started with del.icio.us. You might also be interested in watching this screencast that illustrates just how powerful this web tool is.

When you've signed up for a del.icio.us account (register here) leave a comment on this post telling me so. When I see some action here I'll blog that self-updating post. ;-)



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Why Make It Harder For Ourselves?

Well the question is why? Why are we in this class? Why do we make ourselves suffer through all the hardships, the confusions, the frustration of learning AP calc? What's in it for us? Why do we keep on climbing the mountain even though our rock is always rolling down from the mountain? Well you know why, because we are not designed to fail. We strive to move on because we know that there is light at the end of the tunnel. We know that there is some reward to all this. We believe that all our hardships will be worth it after a while. Also because we have people who are willing to help us. We have our classmates and Mr. K to assist us in our journey towards success. The truth is I don't do my homework and I really don't understand why do have to do all this extra things like the project or the scribe and/or the blog stuff. I ask myself what if we just focused on math and just do it you know, no extra stuff that might get us out of focus on what we are learning. Despite my ignorance, I realized something, those things are there to help us not to give us a hard time. It is hard because we make it hard. Like for me, I left doing my homework in the last minute that's why its hard. Ask for help and give input to Mr. K so he knows what we need. We can say that it is a blessing to have enormous amount of resources that we know is there to help us like Sarah's blog and Mr. K. The blog and scribe activity just helps us grow. It makes us responsible to do what needs to be done.

The point of my blog is this that we make it harder because its good if its hard. That is the time that we know that there is something new to learn. Hardships makes us stronger and it builds our character. One thing is for certain in our class failing is not an option. We are here to succeed!



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Sunday, November 06, 2005

THis is so DRAINING!

Chapter 3 really twisted my mind. I read the whole thing for more than a hundred times already and I still can't quite get them. I was quite lucky though because it only had 4 subchapters or subunits, but then...

all those four KNOCKED ME OUT! They were brain - DRAINING!

I don't know about the others but I'm having a SUPER hard time! Whenever I am in class, I seem to get it. But whenever I go home and try to answer the assignments, I seem NOT to get it! I know it's basically

DISTANCE = VELOCITY * TIME

but the whole concept is way complex. That is the FUNDAMENTAL THEOREM OF CALCULUS : PART II.

In learning this, I feel like I'm already in post secondary education, which is good, only, it is SO HARD. I'm trying. I'm really, really trying. But I seem not to get any closer. To be able to answer the problems tomorrow I think I might be needing a copy of all the lessons because I'm gonna be lost without them. I'm not sure about tomorrow's test but I'm still positive. I still have hopes. :)

I think now I need to look over my notes again for the test. :D

so GOODLUCK people. ;)




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Perseverance

"The difference between the impossible and the possible lies in a man's determination." At first, I thought its impossible for me to pass this AP Calculus course and understand all of the stuff about this subject. I thought everything in this course is hard to understand and confusing. I admit that I think I fail on those previous tests and I'm ready to give up. However, I told myself what would I get if I quit and not try to go to class. What I'm learning in this course is an opportunity that I need to take a risk. This might be the first class that I would fail into but it wont stop me to learn. It would just makes me persevere more and try my best as hard as I could in order to succeed in this course. I believe in my capabilities and if I put all my effort I would probably will pass this course. Like what Mr. K told us sometimes we found ourselves like Sisyphus struggling to achieve its goal but unlike Sisyphus we could succeed. Also, I want to point out doing our homework it really helps.
By the way, I need to remember this formula for tomorrow's test,
That's the Fundamental Theorem of Calculus.



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It's Sunday Again!


image

In the game of Kayles a number of pins (or coins) are arranged in separate rows. A legal move consists of knocking down either one pin or two adjacent pins from the same row. This may break up the row into two smaller rows. Whichever player knocks down the last pin wins. What's the winning strategy for the game of Kayles?


Play here. Discuss your winning strategies in the comments to this post .... and have fun doing it!




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Blogging on Blogging on Integrals

When you look at an Integral (signed area between the curve and the x axis), You then have to look at the Fundamental Theorem of Calculus. If you find the Integral of the derivative function over a specific interval [a, b], then you have found the change in position or Total Change in the parent function.



I find that referring to this helps when solving problems with Integrals. This Integral stuff is fairly easy for the most part, as long as you the difference between Left Hand Side and Right Hand Side. The only confusing part of this unit was learning to use the "big S" with f(x) and dx and remembering to subtract f (a) from f (b).

This was the easiest unit we will probably have to do in calculus.
This is my blog for the Integral unit.



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Wednesday, November 02, 2005

Scribe


=^-^=
What did we do today? --_--# Sorry, I don't really know what should I say about this scribe. All we did in today's class was the online review questions for chapter 3 - the Definite Integral. http://www.wsd1.org/dmci/kuropatwa/calweb/ch3/q01.htm ^O^


~^.^~ A very good challange online review questions, I guess everyone learned something from those questions. ;P

That's all. And nest scribe is Jayson.>0<



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Tuesday, November 01, 2005

The Fundamental Theorem of Calculus

In class today we went on another road trip, or rather, Xun and Ara went on another road trip. Though it might have been a good idea to get that odometer fixed before the trip though. :P

The highlight of todays class was:
The Fundamental Theorem of Calculus

We were only taught the second part of the theorem though, we will be taught the first part later. I don't know about you guys though but I don't like seeing the second part of anything before I see the first, especially when it comes to movies. Mr. K has his reasons for teaching it to us this way though.

Now although it must have been a rough trip having to write out the values every 2 seconds, they persevered and brought us back the table of values which was the basis of todays class. That aside, onto the theorem.

The theorem is




That meaning that in the interval [a,b], the function f(x) multiplied by dx(meaning an extremely small change in x) is equal to the change in x in the interval.
A simpler way of wording that would be that finding the integral over a specified interval gives the total change in the the parent function.

I feel bad for Xun having to go after me, because she was the person I had to pick the last two times, because I'm pretty sure that there is nobody left after me. See you all in class.



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Another Feather in Your Cap

About a month ago I was contacted by Meg Gwaltney who works for Stein Communications, an education marketing firm in the US. She had stumbled across our blog and was impressed with our work. So impressed she wrote an article about us that is distributed across the United States. You can read it here.

You continue to make an impact. Have you decided how you're going to move this to the next level?



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