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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.

Thursday, December 22, 2005

a chapter to remember

so much for first impressions...

a moment of uncertainty...
an immediate response...
a stimulus reflexed instantaneously...

predicted... erroneously.
judged... superficially.

OPTIMIZATION
- a vague piece of powerful information
- distinctive from the rest
- yet, so interesting, I must admit

after an intensive work...
tremendous effort...
stupendous energy...
and humungous struggle...

it still hasn't paid off..
but it's worth infinite tries... afterall...

PATIENCE is an unbelievable virtue...

*keep that in mind*



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Wednesday, December 21, 2005

More Applications of the derivative

More Applications of the derivative

The first derivative test:
If c is a critical number and if f’ changes sign at x=c, then
- f has a local minimum at x=c if f’ is negative to the left of c and positive to the right of c;
- f has a local maximum at x=c if f’ is positive to the left of c and negative to the right of c.
A function f has
- a global maximum value f(c) at the input c if f(x) less than or equal f (c) for every x in the domain of f;
- a global minimum value f(c) at the input c if f(x) larger than or equal f(c) for every x in the domain of f.
The extreme Value Theorem
If a function f is continuous on a closed interval (a, b), then f has a global maximum and a global minimum value on this closed interval (a,b).

The Second Derivative Test
- If f’(c ) = 0 and f’’(c )>0 then f has a local minimum at c;
- If f’(c ) = 0 and f’’(c )<0 then f has a local maximun at c.

Find the vertical asymptotoes and horizontal asymptotes of f if f(x)=1/(x^2-1)
X^2-1 = 0
X=1 and x=-1 are vertical asympototes.

Lim(x-->+oo) =0
And
lim(x-->-00) =0
so, y=o is a horizontal asymptote.

Optimization problems
- read the problem. Identify the given quantities and those we must find
- sketch a diagram
- Using available information, express f as a function of just one variable
- Determine the domain of f and draw its graph
- Find the global extreme of f,
- Convert the result

Mean Value Theorem
If the function is continuous on the closed interval (a, b) and differentiable on the open interval (a,b), then there exists at least one number c in the open interval (a, b) such that
F’(c )= ( f(b)-f(a) )/ (b-a)
Ant derivatives:
Given the function f, determine a function F whose derivative is f. the function F is an antiderivative of f
Function -> antiderivative
X^n-------> (X^(n+1) / (n+1) ) +C
Cos X --->Sin X +C
Sin X----> -Cos X +C
(SecX)^2 -> tan X+C
e^X-------> e^X



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Antiderivative

An application of the Derivative
No one knows what C is
The family of curves are related to each other
Indefinite integral
Differentiate backwards
Equals the parent function
Rethink what each term is the derivative of
In each question remember to add C
Very often the power rule has to be reversed
Always check your answer by deriving it
These problem are different from the usual differentiating and solve for a max or min
In all likelihood, you'll forget to add C
Visualize the problem, draw and label a picture depicting the problem
Each problem is solvable!



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Tuesday, December 20, 2005

Infinity - No Limits


Introduced its symbol by Wallis, JohN
Named after the last greek letter some believed sO
Frankly others found it derived from a roman numeraL
It is not a real number unlike PI
Negative, it grows less than the number it was froM
If positive, it grows beyond, oui? OuI!
The quantity is endless and must be too big to counT
Yes! It does look like the number eight turned sidewayS



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Antiderivatives r us

Today in class, we did an optimization problem which took half of the class. The rest of class was spent on antiderivative problems. I'll show what the antiderivative problem looked like because it is something new that we haven't done before.

A train is traveling at 66 ft/sec when the brakes are applied. If the train decelerates at 8 ft/sec^2. How far did the train travel between the time the brakes were applied and when it stopped?

Solution:

First you have to start with the acceleration function because that is what the problem gives you.

a (t) = -8
Now find the antiderivative.
v (t) = -8t +C
Solve for when t=0 and the velocity equals 66
v (t) = -8(0) +C = 66
C = 66
v (t) = -8t +66
Now find the antiderivative of the velocity function to find the position function.
s(t) = -4t^2 +66t +C
When t =0, the position is 0, so solve for t=0
s(0) = -4(0)^2 +66(0) +C = 0
C = 0
s(t) = -4t^2 + 66t
Solve for t when the function v(t) = 0
0 = -8t +66
0 = -2(4t -33)
t = 33/4 = 8.25
Now evaluate the position function for when t = 8.25
s(8.25) = -4(8.25)^2 +66(8.25)
s(8.25) = -272.25 + 544.5
= 272.25 ft

The next scribe is Chris



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Monday, December 19, 2005

Yule Time


Need not to worry, I got this day covered. Prince told me last Sunday that Im the scribe. Ok? good! To start off today's class we had a group work about optimization. It is about Topher and Sully. The question goes like this:


Topher and Sully decided to go cut down a Christmas tree together, so they ventured out into the snowy woods. After wandering for close to three hours, the two found the perfect tree but also decided that they were lost. Luckily Sully found a map nailed to a tree which showed that they were four miles due north of a point on a perfectly straight road. Six miles east of that point was the park that the two called home. If Topher and Sully can walk and drag the tree at the rate of two miles per hour through the snow, and three miles per hour on the road, what is the minimum amount of time that they would need to get home?

Constraint
T=d/r

Optimization
T(x) = ((x^2+16)^1/2 ) / 2 +(6-x)/3
T(x) = 1/2 (x^2 +16)^1/2 +2 - x/3
T ' (x) = (1/2)(1/2)(x^2+16)^-1/2 (2x) - 1/3
T ' (x) = (3x - 2(x^2 +16)^1/2)/ (6(x^2+16)^1/2

After finding the derivative we find where there are zeros or where the derivative is undefined. The denominator will never equal zero so we just focus on the numerator.

3x - 2(x^2+16)^1/2 = 0
3x= 2(x^2+16)^1/2
(3x)^2= (2(x^2+16)^1/2)^2
9x^2 = 4(x^2 +16)
4x^2 + 64 -9x^2 = 0
(5^.5x-8)(5^.5x+8) = 0

therefore x = 8 / (5^1/2)
then we do the 1st derivative test
- +
T'<---------------8 / (5^1/2)------------->


by the first derivative test 8 / (5^1/2) is a min

then we plug in the root to the parent function to get 3.894
therefore they need three hours to minimize the time they take to walk


After the group work Mr. K gave us derivative functions which we need to find its parent function.

F(x).................f(x)

x^3.................x^4/4 + C
-cos(x)............-sinx + C
sinx..................-cosx + C
sec^2x............. tanx + C
1/x....................lnx + C
e^x....................e^x + C
2^x...................2^x/ln2 + C
x^1/2................2/3x^3/2 + C

Well thats the class...

Remember
This December,
That love weighs more than gold

the next scribe is steve




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Applications of Derivatives Acrostics

Here's the new set of acrostics for you.

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

DERIVATIVE
OPTIMIZATION
ANTIDERIVATIVE
MEAN VALUE
INFINITY
APPLICATION

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, December 18, 2005

Sunday 3x the Funday!

A triple header this weekend.

Mr. Zhong Kui will make you laugh. I think his "problems" are the easiest ones to solve.

Rat is another "escape" puzzle. Every time you do something wrong he squeaks.

No. 5 is a set of three puzzle/adventures to get a little boy out of trouble.

Have Fun!



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Thursday, December 15, 2005

Antiderivative

In today’s class, we went to @_@group work.
Will’s pet Bob~^.^~
Will the wallaby has been looking for the perfect pet for a very long time and he has finally found it – Bob the boa constrictor. Now Willy has to make a close, rectangular cage for Bob but it has to be 4000 cubic feet in volume and the length has to be 5times that of the height of the cage. The material to make the cage costs $0.25 per square foot.
1) What are the dimensions of the least expensive cage?
2) How much does it cost?
Given V=4000
L=5h
V =L * W * h
4000 = 5h w h
4000 = 5h^2 w
W= 800/(h^2)

Closed rectangular cage
S. A = 2 L h + 2 L h + 2 h w
S.A = 2 (5h) w + 2(5h) h + 2 h w
S.A = 10 h w + 10 h^2 + 2 h w
S.A = 12 h w +10 h^2

Cost =1/4 surface area
Cost = 1/4 (12 h w + 10 h^2)
C (h) = 3h (800/h^2) + 5/2 h^2
C (h) = 24000 h^-1 + 5/2 h^2
C’ (h) = -24000 h^-2 + 5h
= -5 h^ -2 (480-h^3)
- - +
--------------1--------------------1---------------------
2 roots: 0 3root480

By the first derivative test, cost is Mimi zed when h = 3 root 480 feet

L= 5 h= 39.15 feet
W = 800/h^2 = 13.04 feet
Cost = 3h (800/h^2) + 5/2 h^2 = $ 331.05


Antiderivatives
Given the function f, determine a function F whose derivative is f. That is, F’ = f. If F’= f, then the function F is called an antiderivative of f.

A function f has an antiderivative it will have a whole family of them and each member of the family can be obtained from one of them by the addition of an appropriate constant.

If f’(X) = 2x, find different possibilities for f, the parent function.
1) f (x) = x^2
2) f (x) = x^2 + e
3) f (x) = x^2 + 10^100
4) f(x) = x^2 + 1 …

Next scrible Prince. +_+



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Wednesday, December 14, 2005

M V T (Mean Value Theorem)

Now that the plague of Related Rates problems has ended (i don't know with the others but it was for me... laughs!) , a fresh new topic has emerged from its shell for us students to focus on for the next couple of days. I found Optimization problems much easier than related rates... or is it too early to judge?

This morning was the start of the Solving sessions in groups with the new topic Optimization. We had to work on Scuba Steve's Shark Cages. Given the total perimeter of the rectangle 450 feet, we had to divide the whole area into two equal parts so that the sharks wouldn't kill each other. The question is, what is the maximum area of each section of the cage that Scuba Steve can build?

First, know your constraint and optimization equations. From the constraint, solve for one variable, whichever is easier, the length (L) or the width (W). Use this to solve for the Area of the rectangle. Then, get the derivative of that function. After that, do the first derivative test and there you have it, a good start on Optimization problems. ;)

We also started on a new one today, the M V T. Most Valuable Tlayer? (laughs!) Nope, it's the MEAN VALUE THEOREM. M V T states that if a function is continuous on the closed interval [a , b] and differentiable on (a , b), then there exists at least one number c in the open interval (a , b) such that:

f ' ( c ) = f ( b ) - f ( a )
b - a

So given a function and its interval, get the values for intervals a and b. Then use the MVT to get the slope of that interval. Next, get the derivative of that function and equate it to its slope. So at X = c , the slope at c is then equal to the slope of the secant line in the interval a , b.

Sounds easy? Let's just wait and see. :D
Meanwhile, the next scribe is Xun. ;)





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Tuesday, December 13, 2005

OPTIm-EYES-a-SHUN

Is it just me or just the word Optimization tells us to really use our eyes, OPTI, EYES. So therefore we should use more PICTURES. Shun, means to avoid. How about avoid just reading the problem. Don't just read it, do more with it.

=D

Hello to everyone.

Today’s class we started off with going over questions from homework. Exercises 5.4: 1-9. The problems Mr. K went over were #4, #5, and #9. Hope no one else had any problems with the homework.

TIPS for OPTIMIzation:
(optimi-sm: to look on the best side of things)
1) Remember your geometric formulas.
2) Math is the science of patterns. Always look for patterns.
3) If you ever get stuck with your geometric figure, it’s never wrong to just drop a Cartesian plane, it might even help you see something you didn’t see before.

After going over the three questions from homework. Mr. K put up three optimization problems on the board. The first one being the classic optimization problem.

1) A sheet of cardboard 3 ft by 4 ft will be made into a box by cutting equal size square from each corner and folding up the edges. What will be the dimensions of the box with largest volume?



In each optimization problem, there will be constraint equation we will be working with. The classic optimization problem up there, the constraints were the dimensions of the box and what they did with it (cutting equal size square from each corner). Some will give you a function as a constraint. Read the problems carefully. Write down what is given to you and draw a picture to understand of what is being asked of you to find.


For homework today is the rest of 5.4: 9-18. Hope everyone does ok with the rest of the problems.

The homework question for today is:

A cylindrical can is to hold 20 pi meters cubed. The material for the top and bottom costs $10 per meter square and material for the side costs $8 per meter square. Find the radius and height of the most economic can.

Good luck.




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Sunday, December 11, 2005

Sunday Jumping Funday!



'The goal of the puzzle is to switch the the pegs on the left with the pegs on the right by moving one peg at a time.

Move pegs by clicking and dragging them to open slots. A peg may only be moved to an open slot directly in front of it or by jumping over a peg to an open slot on the other side of it. You may not move backwards. The game ends when you win or get stuck.'

Play the game here. Can you win the 8 peg game? ;-)

(Thanks again to Think Again!)



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Thursday, December 08, 2005

Not much happening

In todays class, we didn't learn anything new really until closer to the end of the class because Mr. K wanted to keep the talking to a minimum in todays class. So, he gave us a few questions to do, and he gave us a good deal of the class to do them as well. After wrapping up the questions and going over them with us, he began to talk a little bit about more of the applications of the dirivative. We didn't really get very far into it so I do suppose that he will continue this conversation in class tomorow. I wish that I had more material to write on but alas, perhaps next time.



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Monday, December 05, 2005

C.O.O.L.


You know what's cool? Here it is, I never knew Mr. K never cared what are marks were. It was something I never thought he'd say. I understood what he was saying that our marks will only get us IN university and after that well it doesnt matter anymore.What matters the most is that if you have a deepp enough foundation to move forward. Once we're in there if we learned nothing then there is only one way for us, it is downwards. In other words, marks are not important it is better to have a low mark with a high understanding of the concepts rather than to have a high mark but really just know how to do it not why we did the problems.



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Convey and Describe

Our class today was not about learning new stuff, it was about understanding the deeper concepts of what we learned before. Basically it was a day for enhancing our knowledge about the language of calculus.

We learned about the Extreme Value Theorem which states that if a function is continuous and is in a closed interval it has to( Mr. K repeatedly said it has to meaning its important) have a global max and min either in the critical points or the endpoints.


We also learned the 1st Derivative Test which is used to determine if a root of the derivative is a max or a min in the parent function. If the derivative is negative on the left and positive on the right of the root the then it is a local min if the left is positive and right is negative then it is a local max.

Critical points are on the parent function( max or min or where it is undefined)

Critical numbers are the zeros or root in the derivative function

The first example we had is to find the critical points of the given function.
f(x) = x^2 -4x +5
f'(x) = 2x-4
= 2(x-2)

the critical number is x = 2

to get the critical point we get the value of 2 in the parent function. Which is 1
therefore the critical point is (2,1)

we describe this by saying that by the derivative test that there is a min at 2 and the value is 1

Again all we talked about today wasn't something new. All we are learning now are concepts of what we discussed before. The point is we need to know how to convey and describe what is being asked and not just push some numbers in the paper.

The next scribe is.....
CJ



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wild crazy stuff!

In this chapter I have experienced one of the easiest and one of the hardest topics ever that I have encountered in Calculus.

Getting the derivatives of polynomials are not much of a tedious task, although a lot of times, I miss seeing my mistakes so I hardly perfect them. :) Memorizing the derivatives of such trigonometric functions is quite tough but it does not really blow my mind. Chain rule? well... I still forget steps when doing it, but at least I know how to do it.

But if I see a number with a long paragraph beside it... my heart beats faster.. then I FREAK OUT! Yes Mr. K, YES! I am still lost when it comes to related rates. I tried answering 4.6 the best way that I can but I am definitely, positively and undoubtedly sure that only 1 out of 10 numbers will I get correctly. Not that I am being pessimistic, it's just that, for me, IT IS REALLY, REALLY HARD. But I'm still trying, even until now. I keep on reading and reading until the whole problem finally gets stuck in my head.

So, that's about it. I hope we all pass guys...

(...Lord please enlighten our wonderful teacher, that he may be easy on us tomorrow in our test...)

- just kiddin'! I know that the harder the problems we take, the better knowledge we gain. :D

GOODLUCK everyone! ;D



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Blogging On Differentiation rule

In this Chapter, I found the most difficult stuff was relates rates problems.
We need to read the question and find a formula, or even probably two relate to the information that question told us. We need to know everything that might relate to the question, and solve the problem.
Exp: Water is flowing into a cone-shaped tank at the rate of 5 cubic inches per second. If the cone has an attitude of 4 inches and a base radius of 3 inches, how fast is the water level rising when the water is 2 inches deep?
The problem is given us the rate of change volume, dv/dt =5, r = 3, and asked to determine the rate change of height when h is 2, dh/dt when h=2.
We know the formula for conical volume is: V = 1/3 pai (r)^2 h
By similar triangle we have r/3=h/4,
So, r=3/4 h
V=1/3 pai r^2 h
V=1/3 pai (3/4h)^2 h
V= 1/3 pai 9/16 h^2 h
V= 3/16 pai h^3
Defferentiate implicitly:
dv/dt = 3/16*pai* 3h^2 (dh/dt)
5 = 3/16 pai 3(2)^2 (dh/dt)
20/9pai = dh/dt
The water level is rising at a rate of 20/9pai inches per second when the water is 2 inches deep.



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Blogging on Related Rates

Instead of reflecting, I'm going to list some information that might help you understand differentiating functions.

Constant Function f (x) = c f'(x) = 0
Constant Multiple Rule d/dx [c*f(x)] = c*f'(x)
Sum and Difference Rule d/dx[f(x) + g(x)] = f'(x) + g'(x) you can replace the + signs with a - sign
Power Rule d/dx(x^n) = nx^(n-1)
Derivative of e^x d/dx(e^x) = e^x
Exponential Function d/dx(a^x) = (ln a)*a^x
Product Rule d/dx[f (x)*g (x)] = f (x)*g'(x) + f'(x)*g (x)
Quotient Rule d/dx [f(x)/g(x)] = [g(x)*f'(x) - f(x)*g'(x)]/[g(x)]^2
Derivatives of Trigonometric Functions
d/dx (sin x) = cos x
d/dx (cos x) = -sin x
d/dx (tan x) = sec^2 x
d/dx (csc x) = -csc x cot x
d/dx (sec x) = sec x tan x
d/dx (cot x) = -csc^2 x
The Chain Rule [fog]' (x) = f'(g(x))*g'(x)
d/dx (ln x) = 1/x
Local linearization of f at x = a f (x) ~ f(a) + f'(a)(x-a)

All of these derivative rules are probably going to be on the test, so it is a good idea to learn how to use them. I excluded related rates because that depends on the question, and there is no definite formula for all related rates problems. These formulas and derivatives cover most of the unit and they are good to know.



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Saturday, December 03, 2005

Cubeoban Sunday



The objective of Cubeoban is to push/pull all the blocks to their corresponding lights. Do this by clicking on the blocks and drag them in the direction you want to push them. Play it here.

Level 1 was so easy that even I could do it. Level 2 (the image), started my thinking.

(Thanks again to Think Again!)



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Friday's Pre-test

Like other unit before the test we had a Pre-test, in order for us to get ready for the test on Tuesday. The test is consist of 5 practice multiple type of questions and one Open response question. In today's pre-test we got this question:

The sides of this rectangle increase in such a way that dz/dt=1 and dx/dt=3dy/dt. At the instant when x=4 and y=3, what is the value of dx/dt?


First we create a relationship in about this problem. and we got this z^2=x^2+y^2
then we find the derivative of that relationship.
2z dz/dt=2x dx/dt + 2y dy/dt. The two are reduces z dz/dt= x dx/dt + y dy/dt.
By the pythagorean theorem we found that z=5. So after that we just plug in the values.
(5)(1)=4(3)dy/dt+ 3 dy/dt
=15 dy/dt / 5
= 1/3 dy/dt
Since dx/dt= 3 dy/dt we got the value of dx/dt= 1.



Now our pre-test was done all we have to do is study for our test on Tuesday. Good Luck guys.... The next scribe is Steve....



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