♥SCRIBE 5/31♥

IT WAS A RAINY DAY…THE PERFECT DAY TO GET A COLD.

POD 5/31
Without using a calculator or your snowman sheet, find and exact value for sec (π /6)

It was absurd that no one remembered how to do it, so Mr. A helped us to remember by solving it.




Then we talked about what mistakes we made in the POD.

After that we continued to the bouncing ball problem:

A ball is dropped 100 feet and bounces straight. On each bounce the ball climbs half of the height of its previous flight. Assume that the ball bounces forever. How far will the ball travel?
We did this problem through steps.
1) List terms: 100+100+50+25...+...+....
2) Find if the problem is Arithmetic or Geometric?

Arithmetic
100+100+50+25...+...+.... ---------> NO
100-50 50-25 / /
50 25

We subtract the second number with the first one, the second one with the third one and the same with the other numbers. When we finish doing the subtractions we could see that the problem was not arithmetic because we got different numbers.

Geometric
100+100+50+25...+...+.... ----------> YES
50/10=1/2 25/50=1/2 We divide the second number with the first one, the second one with the third one and the same with the other numbers. When we finish doing the division we could see that the problem was geometric because we got the same number for the rate.


3) Find out if it is finite or infinite?
The problem is INFINITE
4) /r/ ?
r= ½
Sn= t1/1-r ∞ geometric series /r/ <>D= 100+limSn n-∞ D= 100+100/1-1/2 D= 100+200= 300ft D= 300 ftThen Mr.A showed Mac’s way to solve this problem. She divides this problem in two series. Up series Down series Snd= 100/1.1/2= 200 Snu= 50/1-1/2= 100 Snu + Snd= D D= 100+200 D=300ft It’s weird that the answer for 10 bounces is 299.8 ft… Mr. A then corrected a mistake that many people did on yesterday’s POD. He said that the r in 2πr stands for radians not radius. (this occurred in scribe 5/30) Later we changed the ball problem for that if it bounces to 2/3 of its previous height? Jaime solve this question Sn= t1/1-r R=2/3 T1= 100 Sn 100/-1-2/3=100/1/3 -----------> WRONG! 300+100= 400ft Mr.A fixed Jaime’s mistakes: T1= 2*100 2/3=400/3 Sn=400/3/1/3 Sn=400ft+100ft Sn=500ft Using MAC’s way: Sd=100/1-1/3=300 Sn= 2/3 100/1-2/3= 200 300+200=500ft When we finished with the ball problem, we went to the Achilles and Tortoise problem (www.mathacademy.com/pt/prime/articles/zeno_tort/) which is a geometric sequence.
10,1/1/10,1/100,,,,,1/10n or (1/10)n
Lim tn= Lim1/10= 0
n-∞ n-∞
t11-r^n/1-r
Sn = Distance covered= 10*1-(1/10)^n/1-1/10
10*1-(1/10)`n/9/10—100/91-1/10^n
Lim Sn = lim 100/9(1-(1/10)^n
n-∞ n-∞
Answer: 11/1/9
Homework...bring an umbrella
END OF CLASS, AND IT WAS STILL RAINING...
Next scribe is NK!!

Answer for #2 Mpooh

Now you find the tweflth term!!

First anser to Mpooh problem

The five terms of this sequence are 1/4, -1/8, 1/16, -1/32, and 1/64.

Limit Solutions

Below you will find the solutions to the limit problems listed in Raskolnikov's post from a couple of days ago. Try solving them yourselves first.

Ball Bouncing Problem

Trying to solve the ball problem:
Problem:
A ball is dropped 100 feet and bounces straight. On each bounce the ball climbs half of the height of its previous flight. Assume that the ball bounces forever. How far will the ball travel?

Approach:
First of all the question is asking for a distance of how far the ball will travel.
We know that it starts with a 100 feet bounce and every time bounces half of its previous flight.
We also know it bounces forever therefore we might think that the answer is infinity, but remember Sybil and Zeno´s Paradox.
Remember the sum of an infinite geometric Series.

Solution:
Now we can start by listing the first few terms.
Because we know it is going to be half of the previous flight, then:
N= 100, 100/2, (100/2)/2, … or N= 100, 50, 50/2 or simply n= 100,50,25…
But remember it is asking for a sum because how far will the ball travel asks for the sum of all the infinite distances.
So n= 100+ 100/2+…
We know now that t1 must equal 100 feet and r is .5 because the ball bounces 50% or let’s say half of its previous flight.
Then if we apply the Formula, Sum of an Infinite Geometric Series:
Sn = t1/1-r

Solve with plugging in:
Sn = 100/1-.5
=100/.5
=200 feet
Or else it says that the how far will the ball travel is no more or not even reach 200 feet

History and Famous Problems

I found this website and thought that it was really interesting. There are a lot of famous problems, some of which we have reviewed or studyied this year, and others that, at least, I have never heard of. It's interesting to see where the thoeries originated and it just gives a really nice explanation for each!!

http://mathforum.org/isaac/mathhist.html

MAY 30/07

Jaime's POD: Write a formula for the nth term of the given finite sequence

Everyone else's POD: Pier becomes possessed by the devil so that he can spin his head all the way around. Suppose that Pier's head is a perfect sphere and that he spins his head at a constant rate of 8 revolutions per minute. A small ant is perched on Pier's nose so that in 6 seconds the ant travels 60 cm. Find the radius of Pier's head.

***There are various ways of solving this kind of problem!!

Mr. A's HINTS:

1. Find the central angle that the ant sweeps.
2. Then use the arc lenght formula.

Isabella's Solution:

1. Find out how many seconds the ant travelled in 60 seconds, if it travelled 60 cm in In 6 seconds.

In 6 sec the ant travels 60 cm.

How many does he travel in 60 sec?
8rev/min. *8 revolutions per minute.
6sec-->60 cm.

*In 6 sec., the head spins 60 cm.

x --> 60 sec. *If the ant travels 60 cm in 6 seconds, how many cm does he travel in 6 sec?
60 cm.<--6sec.

2. Solve for x

x=3600cm(sec.)/6sec. *The seconds cancel out
x=600 cm. * There are 600 cm. in 60 seconds or also 8 revolutions.
8 rev.=600 cm.

Mr. A reminded Isabella of writing the conversion because it was easier to understand were the x comes from.

Conversion:

x/60cm=

60 sec./6sec.

* This also means that how many cm there are in in 6o sec, if there 60 cm in 6 sec.

3. Find out how many cm the ant travels in 1 rev. if in 8 it travels 600 cm.

600 cm./8 rev.= 75cm.
75cm= 1 rev.

4. Then use the formula for circumference to find the radius.

2пr=75
75 cm/2п=r

r=11.94 cm.

***You can find another solution for this problem in the 11A Blog.

MAC's solution:
1. Find how many rev./sec. if if 1 min. there are 8 rev.
8 rev./min. x 1 min./60 sec.= .133 rev./sec

*Conversion of rev per min times 1 min over 60 seconds.

2. Find how many revolutions in 6 seconds.
.133rev./sec x 6 sec= .8 rev

* By multiplying the revolutions per seconds times the t 6 seconds

3. Find the distance in 1 rev if in 8 rev, the ant travelled 60 cm. Use the circumference formula.
8 rev./60 cm.= 1 rev./x *x stands for the circumference of the head

x=75 cm.

4. Solve with arc lenght formula for the radius.
75 cm.= 2пr

r= 11.94cm.
Rumidog's solution:
1. Solve for x using the circumference of a circle and its degrees.
8 rev./min.
8 ·360º/min= 16П/min

1 rev= 2Пr

8 rev. = x

* 6 sec. is 1/10 of 1 min.

6 sec.= 1/10 min.
***Try and understand all methods of solving the problem.
Try that at least one of the makes sense to you, if not then visit REMEDIALS!!

Sybil the Rat:
A mathematical answer:

**tn= fraction of room to cross

***When in doubt of how to solve or approach a question:
1. List the first few terms.
For Sybil:

1/2, 1/4, 1/8, 1/16...

2. Then write an explicit formula:

***What happens to n as n reaches infinity?

Tn approaches zero
>>>Support with a graph.

Provide a graphical answer:
a) Must label both axes
b) Show points not lines

c) Interpret graph

11A Graph: As you can see in the graph below, the limit seems to approach zero, but Mr. A emphasized that to prove limits with graphs some points have to be followed. In the graph, tn values approach zero as n values approach infinity.

Algebraic answer:
Jorge said limit is zero.

Book:

If the absolute value of r is less than 1, then the limit of r to the nth term, as n approaches infinity, is zero.

If you keep multiplying always by the same number, this is what happens when a numer is squared, the denominator will always increase, but because it is being divided by 1, then the answer is reaching zero. n/n^2+=0

Sn=total amount of room crossed.

For Sybil to croos the room, then Sn must be less than or equal to 1.

**Write as a series:

Since we are talking of adding all those infinte numbers that Sybil has to walk to cross the room, then the problem can be solved as a series.

1) List 1 few terms: 1/2, 1/4, 1/8, 1/16...

2) Find S1,S2,S3=
S1= 1/2
S2= 3/4 ( This comes from adding up the first and second term)
S3= 7/8 (The sum of the first 3 terms)
S4= 15/16

*Some people say that the limit is reaching one.

*Others like jaime say that the limit will never be zero because as the numerator and denominator increase it looks like the numerator will never equal the denomiantor, rather pass the denominator.

Algebraic form:

S is a Geometric Series where

r= 1/2

Then use the Geometric Series Formula

Sn=Distance covered=

**In the picture we used the Geo. Series formula and plugged in
to find Sn.
t1= 1/2

r=1/2

Sn=t1(1-r^n)/1-r **GEOMETRIC SERIES FORMULA**
Sn=1-(1/2^n) *We can take the limit of both sides of the equation.

In the picture above in the first step we are taking the limit of both sides of the equation after we plugged in values of t1 and r to find Sn. Then we remember that 1/2 squared or to the nth term equals zero. So 1/2 to the nth term is 0 which leaves the limit of 1-0, which also equals the limit of 1. Finally it appears as the a approaches infinity, the limit of Sn is 1.

The 1 is exactly the limit of the Sybil problem. Which means the limit the rat has when crossing the room, or rathe rthe limit is simply the lenght of one trailer.

Sum of an Infinite Geometric Series
*Here the first step is to plug in r and t1 and show that 1/2 to the nth term cancels out. The theorem says that for any number in between -1 and 1 equals zero.This is calles the sum of an infinte geometric series. With this problem you can approach and solve many problems like the Tortiose Paradox ans Sybil the rat.

New formula: Sn=t1/1-r
Last 3 min. of class:

A ball is dropped 100 feet and bounces straight. On each bounce the ball of its previous height. Assume that the ball bounces forever. How far will the ball travel?

*****HOMEWORK:READ PG.500-503, PG.502, #1-19 ODD
NEXT SCRIBE: ELI OTOYA

Mpooh Has Got Problems

Pooh says:
Hello, Since I am having some difficulties with Geometric Series problem, I thought It was kind of nice to post some problems, some which are very challenging, to those interested. The answers I did not post, but if anyone wants them, they can try the questions and are invited to ask me for the answers. Feel free and try these problem, post comments with your answers!!

Ready?

There are Sequence and Series problems! GOOD LUCK!

Find all of the terms of the finite sequence?

1a.


; 1 < n < 5

Find the first five terms and the twelfth term of the infinite sequence.
2a.


Write a formula for the nth term of given the infinite sequence.

3a.


1. Write a formula for the nth term of the geometric sequence 7, 28, 112, 448, .... Do not use a recursive formula.

2. Write a formula for the nth term of the geometric sequence 16, - 4, 1, -1/4, .... Do not use a recursive formula.

3. Find the first term of a geometric sequence with a fifth term of 32 and a common ratio of -2.

4. Find the common ratio for a geometric sequence with a first term of 3/4 and a third term of 27/16.

5. Find the sum of the finite geometric series 3 - 6 + 12 - 24 + 48 - 96.

6. Write a formula for the nth term of the given geometric sequence. Do not use a recursive formula.
 125, 25, 5, 1, ...
 4, -12, 36, -108, ...

7. Find the sum of the given finite geometric series.
2 + 14 + 98 + 686 + 4802 + 33614 + 235298

Zeno’s Paradox of the Tortoise and Achilles

Hello,
I was doing a little research on the Tortoise Paradox and found out the following:
When solving this kind of problems it is better not to get a headache over them, but try to solve them the easiest way through logic.
First of all we are trying to find out how many miles does it take to Achilles to catch up the Tortiose? But let us see it this way better: How many miles would it take Achilles to finish the race if the race is one mile?
If we already know that every time he reaches half of the distance he has to run, he has 1/4 of the distance of that half he has left, and then as he continues he has 1/8 of the distance, by then running half of the 1/8 he has 1/16, 1/32 and so on. This creates infinte numbers that derive from a finite distance. The finite distance is 1 mile and is practically divided into small and infite numbers. What seems most logic is that Achilles' limit is probably an infinite number, but in fact it is not. On the other hand it is a finite sum. The distance Achilles will spend travelling 1 mile is exactly 1 mile.
Ok, to further understand why this is. First we realize that if we divide the finite distance (1 mile), it divides into infinite small distances, (1/2, 1/4...), then by logic if we add all those infinte numbers it equals us the finite distance we started with, or to say 1 mile!
When an infinite sum like the one in the Tortiose' paradox, also known as an infinite series, adds up to a finite number, the series is summable.
In Zeno's Paradox, Achilles gives the Tortiose a 10-meter head start, but as Achilles catches up with that head start, the tortiose has already walked 1 meter more. From that meter on, Achilles is always trying to catch up with the distance and every time the Tortiose is creating a new meter of difference between them. Therefore it is a finite number creating an infinite number of distances and to find the distance where they meet is equal to the one they started with. 1 meter.

Scribe Post May 28

Scribe Post

May 28 of 2007

The class started with the following POD: if the distance from Larissa’s wrist to the head of a fly swatter is 50 cm, and Larissa swung the swatter through and angles of 52 degrees, then how far did the head of the fly swatter travel.


The class had some time to solve the problem and Jaime put his answers on the board.

First he converted the angle into radians (he converted into radians to use the following formula)

52 degrees x (π radians/180 degrees)
≈ .907 radians

Then he used the formula we had already learned for arcs:
S=Rθ
S≈50 cm x .907 radians
S≈45.38 cm

He also showed how to solve the problem with a proportion instead of the formula

52°/360° = S/2πr

52 degrees are proportional to the arc as 360 degrees are proportional to the entire circumference.

Both methods result in the same answer.

After the POD Mr. Alcantara read us a story from the following webpage www.mathacademy.com/pt/prime/articles/zeno_tort/

It talked about one of Zeno’s paradox about a Tortoise and Achilles, which resembled the “ratty problem” we had already discussed regarding Sybil the rat.
The Tortoise tricked Achilles into conceding a race as she clamed it was impossible for Achilles to pass her after he had given her a head start, considering he will always have to cover some distance to catch her, no matter how small.

We discussed the problem in the class and were confused by the fact that in real life we had already seen people winning a race even after giving a head start.

Mr. Alcantara then proposed three choices for the class:
We could continue to think about the problem and try to solve it.
We could practice limits with the following excercises

Lim of (5n^2/(9n^5+5n)) as n approaches infinity

Lim of (5n^5/(9n^2-8n)) as n approaches negative infinity

Lim of ((8n^2-3)/(2n^2+6n-1)) as n approaches zero

Lim of (sin n/ n) as n approaches negative infinity

Lim of (sin n/ n) as n approaches zero

Lim of ((10-6n^2)/(2n-5)^2) as n approaches infinity

Or we could correct the quizzes he had handed out (the students whose quiz grade was below a 60 can hand in their corrected quizzes tomorrow and earn 5 points)

On the board Larissa, Nancy, Isabella and Jaime did problems 4,5 and 8 of the quiz.

Problem 4

Problem started off with 3 kernels of popcorn, for each second that passed there would be 1.8 times as more kernels than before. We had to find the number of kernels that would have popped by second 15.

3x1.8=5.4
5.4x1.8=9.72

9.72/5.4=1.8=r

S15= 3 ((1-1.8^15)/(1-1.8))
=25,296.152 kernels popped after 15 seconds

Problem 5

The following problem was about an arithmetic series. It asked how many donuts a bakery would sell after 2 weeks if it sold 20 units on the first day and each day sold 12 units more than it did the day before.

D=12

First we had to find T14 to consequently find S14

T14=T1+ (13)(12)
T14=20+156
T14=176 donuts

S14=(14/2) * (20+176)
S14=7*196
=1372 donuts

Problem 8

Problem 8 asked how much interest would be earned after 10 years if there was 4% interest compounded annually and the initial amount was $1000.

Isabella’s solution

1000*.04=40
1000+40=1040

1040*.04=41.6
1040+41.6=1081.6

Sum of Interest

40+41.6+…

41.6/40=1.04=r

S10=40 ((1-1.04^10)/(1-1.04))
=$480.244 total interest earned in ten years.

Jaime’s solution

At=A(1+r)^t
A10=1000(1+.04)^10
A10=1480.244
1480.244-1000=$480.244


After solving the problems, and with no new homework besides correcting the quizzes, the class was dismissed.

I choose Maria Carolina to be the following scribe.

Sir I was not able to post this yesterday because as you know I don’t have a computer but I used one of the computers in the library to do it today.

Scribe Post May 28

Scribe Post

May 28 of 2007

The class started with the following POD: if the distance from Larissa’s wrist to the head of a fly swatter is 50 cm, and Larissa swung the swatter through and angles of 52 degrees, then how far did the head of the fly swatter travel.


The class had some time to solve the problem and Jaime put his answers on the board.

First he converted the angle into radians (he converted into radians to use the following formula)

52 degrees x (π radians/180 degrees)
≈ .907 radians

Then he used the formula we had already learned for arcs:
S=Rθ
S≈50 cm x .907 radians
S≈45.38 cm

He also showed how to solve the problem with a proportion instead of the formula

52°/360° = S/2πr

52 degrees are proportional to the arc as 360 degrees are proportional to the entire circumference.

Both methods result in the same answer.

After the POD Mr. Alcantara read us a story from the following webpage www.mathacademy.com/pt/prime/articles/zeno_tort/

It talked about one of Zeno’s paradox about a Tortoise and Achilles, which resembled the “ratty problem” we had already discussed regarding Sybil the rat.
The Tortoise tricked Achilles into conceding a race as she clamed it was impossible for Achilles to pass her after he had given her a head start, considering he will always have to cover some distance to catch her, no matter how small.

We discussed the problem in the class and were confused by the fact that in real life we had already seen people winning a race even after giving a head start.

Mr. Alcantara then proposed three choices for the class:
We could continue to think about the problem and try to solve it.
We could practice limits with the following excercises

Lim of (5n^2/(9n^5+5n)) as n approaches infinity

Lim of (5n^5/(9n^2-8n)) as n approaches negative infinity

Lim of ((8n^2-3)/(2n^2+6n-1)) as n approaches zero

Lim of (sin n/ n) as n approaches negative infinity

Lim of (sin n/ n) as n approaches zero

Lim of ((10-6n^2)/(2n-5)^2) as n approaches infinity

Or we could correct the quizzes he had handed out (the students whose quiz grade was below a 60 can hand in their corrected quizzes tomorrow and earn 5 points)

On the board Larissa, Nancy, Isabella and Jaime did problems 4,5 and 8 of the quiz.

Problem 4

Problem started off with 3 kernels of popcorn, for each second that passed there would be 1.8 times as more kernels than before. We had to find the number of kernels that would have popped by second 15.

3x1.8=5.4
5.4x1.8=9.72

9.72/5.4=1.8=r

S15= 3 ((1-1.8^15)/(1-1.8))
=25,296.152 kernels popped after 15 seconds

Problem 5

The following problem was about an arithmetic series. It asked how many donuts a bakery would sell after 2 weeks if it sold 20 units on the first day and each day sold 12 units more than it did the day before.

D=12

First we had to find T14 to consequently find S14

T14=T1+ (13)(12)
T14=20+156
T14=176 donuts

S14=(14/2) * (20+176)
S14=7*196
=1372 donuts

Problem 8

Problem 8 asked how much interest would be earned after 10 years if there was 4% interest compounded annually and the initial amount was $1000.

Isabella’s solution

1000*.04=40
1000+40=1040

1040*.04=41.6
1040+41.6=1081.6

Sum of Interest

40+41.6+…

41.6/40=1.04=r

S10=40 ((1-1.04^10)/(1-1.04))
=$480.244 total interest earned in ten years.

Jaime’s solution

At=A(1+r)^t
A10=1000(1+.04)^10
A10=1480.244
1480.244-1000=$480.244


After solving the problems, and with no new homework besides correcting the quizzes, the class was dismissed.

I choose Maria Carolina to be the following scribe.

Sir I was not able to post this yesterday because as you know I don’t have a computer but I used one of the computers in the library to do it today.

Scribe Post May 28

Scribe Post

May 28 of 2007

The class started with the following POD: if the distance from Larissa’s wrist to the head of a fly swatter is 50 cm, and Larissa swung the swatter through and angles of 52 degrees, then how far did the head of the fly swatter travel.


The class had some time to solve the problem and Jaime put his answers on the board.

First he converted the angle into radians (he converted into radians to use the following formula)

52 degrees x (π radians/180 degrees)
≈ .907 radians

Then he used the formula we had already learned for arcs:
S=Rθ
S≈50 cm x .907 radians
S≈45.38 cm

He also showed how to solve the problem with a proportion instead of the formula

52°/360° = S/2πr

52 degrees are proportional to the arc as 360 degrees are proportional to the entire circumference.

Both methods result in the same answer.

After the POD Mr. Alcantara read us a story from the following webpage www.mathacademy.com/pt/prime/articles/zeno_tort/

It talked about one of Zeno’s paradox about a Tortoise and Achilles, which resembled the “ratty problem” we had already discussed regarding Sybil the rat.
The Tortoise tricked Achilles into conceding a race as she clamed it was impossible for Achilles to pass her after he had given her a head start, considering he will always have to cover some distance to catch her, no matter how small.

We discussed the problem in the class and were confused by the fact that in real life we had already seen people winning a race even after giving a head start.

Mr. Alcantara then proposed three choices for the class:
We could continue to think about the problem and try to solve it.
We could practice limits with the following excercises

Lim of (5n^2/(9n^5+5n)) as n approaches infinity

Lim of (5n^5/(9n^2-8n)) as n approaches negative infinity

Lim of ((8n^2-3)/(2n^2+6n-1)) as n approaches zero

Lim of (sin n/ n) as n approaches negative infinity

Lim of (sin n/ n) as n approaches zero

Lim of ((10-6n^2)/(2n-5)^2) as n approaches infinity

Or we could correct the quizzes he had handed out (the students whose quiz grade was below a 60 can hand in their corrected quizzes tomorrow and earn 5 points)

On the board Larissa, Nancy, Isabella and Jaime did problems 4,5 and 8 of the quiz.

Problem 4

Problem started off with 3 kernels of popcorn, for each second that passed there would be 1.8 times as more kernels than before. We had to find the number of kernels that would have popped by second 15.

3x1.8=5.4
5.4x1.8=9.72

9.72/5.4=1.8=r

S15= 3 ((1-1.8^15)/(1-1.8))
=25,296.152 kernels popped after 15 seconds

Problem 5

The following problem was about an arithmetic series. It asked how many donuts a bakery would sell after 2 weeks if it sold 20 units on the first day and each day sold 12 units more than it did the day before.

D=12

First we had to find T14 to consequently find S14

T14=T1+ (13)(12)
T14=20+156
T14=176 donuts

S14=(14/2) * (20+176)
S14=7*196
=1372 donuts

Problem 8

Problem 8 asked how much interest would be earned after 10 years if there was 4% interest compounded annually and the initial amount was $1000.

Isabella’s solution

1000*.04=40
1000+40=1040

1040*.04=41.6
1040+41.6=1081.6

Sum of Interest

40+41.6+…

41.6/40=1.04=r

S10=40 ((1-1.04^10)/(1-1.04))
=$480.244 total interest earned in ten years.

Jaime’s solution

At=A(1+r)^t
A10=1000(1+.04)^10
A10=1480.244
1480.244-1000=$480.244


After solving the problems, and with no new homework besides correcting the quizzes, the class was dismissed.

I choose Maria Carolina to be the following scribe.

Sir I was not able to post this yesterday because as you know I don’t have a computer but I used one of the computers in the library to do it today.

Practice limits

This is a cool page that will help you if you have any dificultices with limits .It has example problems, practice problems and quizzes .We hope it will help you to understand better limits.
mpooh and Eli

http://archives.math.utk.edu/visual.calculus/1/limits.7/

Next scribe

The next scribe is...................




Mascaaa hmmm belllaaa...

(Isabella Martinez :D:D:D)

;)

Limits scribe (Georgette Manzur, Martha Alcocer)

Today's class, as usual, started with the following P.O.D:


*Use a graph to find the limit.

Lim 6n^2 / 3n^2 + 7n(as n approaches infinity)

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Mr. Alcantara asked "in which quadrant should we focus in order to find the answer easily", and Alejandro Covo answered that we should look the 2nd and 3rd quadrants, which was not the correct answer. Why? Because those quadrants are negative, therefore the line doesnt approach infinity.
After we graphed the function in our calculators, we all noticed that it did not reach two, so we concluded that the answer should be around that number.
Because with the graph you are not able to prove the exact answer, Mr. Alcantara teached us a new limit trick; manipulating the equation algebraically.
Mr. Alcantara first showed us the process with the P.O.D. problem:

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As you can see, after you have worked out the equation algebraically, you take the limit of the answer, which means in fact the same, but is actually much more simpler.
At that time, as shown in the bubble share slides, you're left with Lim (6)/(3 + 7/n) (as n approaches infinity). Here you can notice that as n (the only variable left) gets really big, the number is getting each time smaller because n is placed in the denominator. So because of this you can see that that whole number (7/n) will finally approach zero.
A better example of this can be seen in the P.O.D we did on Friday, which was to find lim 1/n (as n approaches infinity). By this we learned that when a variable is placed in the denominator, as that variable gets bigger the whole number gets smaller and approaches zero.
After this, Mr. Alcantara gave us another equation in order to practice and understand the new limit trick better.
Find:
Lim 6n^7+3n-5/12n^7+7n^5 (as n approaches negative infinity).
After trying by ourselves, we solved the process all together in the board:
Lim (6n^7+3n-5/12n^7+7n^5) x (1/n^7)/(1/n^7)
(6n^7/n^7+3n^2/n^7-5/n^7)/(12+7n^5/n^7)
(6+3/n^5-5/n^7)/(12+7/n^2)
At this time you can cross terms that go to zero, leaving the equation like this:
Lim 6/12 (as n approaches negative infinity) = lim 1/2

In this equation, we also solved using the fact that if n is in the denominator, the whole number will eventually become zero as n gets bigger, so that is why we crossed out all n's.
So, the answer was: 1/2
Remember, the analytical trick is better to use because you get an exact answer.
After this, Mr. Alcantara gave us several steps in order to understand the trick better. Those are as follows:
1. Only applies when x approaches positive or negative infinity.
2. You have to multiply the top and the bottom by the reciprocal (divide) of the variable raised to the highest power in the denominator.
3. Distribute.
4. Eliminate terms that go to zero.
5. Reduce the fraction.
Then he gave us three other equations to practice.
We did the first one together using the trick.
Find: Lim - ((7n^3)/(n^2-2n)) (as n approaches infinity)

First, as Mr. Alcantara said, you have to multiply the top and bottom by the reciprocal of the variable raised to the highest power in the denominator.
(DONT FORGET THAT THIS TRICK ONLY WORKS WHEN X APPROACHES INFINITY OR NEGATIVE INFINITY)

Lim -((7n^3)/(n^2-2n)) x (1/n^2)/(1/n^2)

Then you distribute:
- (7n^3)/(n^2) / (n^2/n^2)-(2n/n^2)

This equals - (7n/1-(2/n)), where you cross out 2/n because that term eventually goes to zero.

You are finally left with Lim - (7n/1) which practically equals negative infinity over one, which is the same as negative infinity. So the answer is D.N.E. (DOES NOT EXIST).

NEXT SCRIBE

The next scribe is a couple.................................23462352...............................?????????????? M.....A...rtha and Georgette.........!!!

... a little mistake...

Fibonacci sequence: 1,1,2,3,5,8,13,21,34...

Scribe Post 05/18/07 Francisco and Miguel

(In this period, we continued talking about limits)
This period, we started as usual with a POD:

tn= 1/n find:

A.- Lim tn=
n --> oo

B.- Lim tn=
n --> -oo

C.- Lim tn=
n --> 0

How to do it:

On the firt problem, n is aproaching to infinity, so we have
to make a table of values with positive integers, for example:

n 1/2 1 2 3 4 5
tn 2 1 1/2 1/3 1/4 1/5

With this table of values, we can do the graph to check and prove the answer.

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So, with that we can know:
Lim tn= 0
In In this problem, the answer is 0, beacuse n will never pass 0.

On the next problem, the metod to solve it is very similar, the difference is that on here,
n is approaching to -oo ( negative infinity). So the table of values have to be like this:

n -1/2 -1 -2 -3 -4
tn -2 -1 -1/2 -1/3 -1

With this table, we can see a graph like this:

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So with this graph, we can conclude that:
Lim tn= 0
n --> -oo

This answer is beacuse the n never going to pass 0, so the limit is 0

The last problem was a littlebit different than the previous problems,
the difference is that in this problem, n is approaching 0, so the table of values
should be similar than this one:

n .01 .001 .0001 .00001
tn 100 1000 10000 100000

With this values, the graph should be:

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Lim tn= DNE
n --> 0

This answer, is beacuse n will never reach 0 on one side and neither the other side.

Very Important Note: When we finished the POD, mr. alcantara told us that some one in
AFRICA is watching our blog (.. very interesting...)

Later, we started doing some exercices about limits.

* Solve without using calculator:

1.- tn= cos (n) Lim tn=
n --> 00

To solve this problem, we have to know the shape of a cosine graph:

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Knowing the shape of the graph, we can see that it doesnt have limits, so we have to answer:
DNE that means: The Limit Does Not Exist

tn = Cos (n) Lim tn= DNE
n --> oo



2.- tn= Ln (n) Lim tn=
n --> oo


To solve this problem, we have to know the shape of Lnx, so we can know it easily if we know the shape of e^x, because Ln is the inverse function of e^x, so we can know the next graph:

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With this graph, we can conclude that:

tn= Ln (n) Lim tn= oo
n --> oo


Nearly of the end of the period, Mr. Alcantara puted us another problem:

* Find the limit ( accurate to 3 decimals) of the sequence formed by taking the ratio
of successive terms of the fibonacci sequence:

Lim Fn/Fn-1=
n --> 00

To answer this problems, firts we should know the Fibonacci sequence, that is:
1,1,2,3,5,8,12,21,31, ...

So we have to make a table of values, for example:
n 2 3 4 5 6
tn 1 2 3/2 5/3 8/5

With this values, we can do this graph:

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As we can see on the graph, the integers are not passing 2 and 1, but as Mr. Alcantara told us,
the numbers are going to stay between a specific number, and that number is:
1.618

So the answer of the problem is:

Lim Fn/Fn-1= phi
n--> 00


When we finished this problem, the period was OVER.....

Remember: Quiz Wednesday, about : Sequences, Series, Limits.

Sum of First Positive Squared Numbers

I have been practicing with the difference analysis technique because it seems to me a very good tool, which has many benefits. We practiced very little of the technique in class, so I asked Antonio to help me, and put me an example. He told me that the problem got harder depending on te quantity of levels, so he told me to try a 3rd level problem.

My goal was to reach a formula for the sum of the first positive squared integers.

Here is my work:



First I listed the first terms

1, 4, 9, 16, 25 - But since what I want is a formula for the sum, I created a sequence with the sum of all previous terms.

1, 5 (1+4), 14 (1+4+9), 30 (1+4+9+16), 55 (1+4+9+16+25).......



Now to find the level, I have to reach the common difference.

1, 5, 14, 30, 55..

Ist 4, 9, 16, 25

2nd 5, 7, 9

3rd 2, 2

(Now I can assure the formula is raised to the 3rd degree)





I identified the values so when x = 1, y = 1.. this means that n = x and Tn = y.

x = 1, y = 1

x = 2, y = 5

x = 3, y = 14

x = 4, y = 30



Next I plugged in the values in the formula ax^3 + bx^2 + cx + d = y, and then solve for the variables by subtraction (which is pure algebra)



a(1)^3 + b(1)^2 + c(1) + d = 1, so, a + b + c + d = 1, when x = 1

a(2)^3 + b(2)^2 + c(2) + d = 5, so 8a + 4b + 2c +d = 5, when x = 2

a(3)^3 + b(3)^2 + c(3) + d = 14, so 27a + 9b + 3c + d = 14, when x = 3

a(4)^3 + b(4)^2 + c(4) + d = 30, so, 64a + 16b + 4c + d = 30, when x = 4



Now, by subtraction solve for the variables.

8a + 4b + 2c + d = 5, minus

a + b + c + d = 1, which equals

7a + 3b + c = 4 ( We will use this later and call this formula 1 or f1, to return quicker)



Then repeat the same process with different formulas

27a + 9b + 3c + d = 14 minus

8a + 4b + 2c + d = 5, which equals

19a + 5b + c = 9, hang on there keep this one to, and call it f2



Now we can use this to formulas to eliminate one variable.

19a + 5b + c = 9 minus

7a + 3b + c = 4 which equals

12a + 2b = 5, f3



Repeat again

64a +16b + 4c +d = 30 minus

27a + 9b + 3c + d = 14 equals

37a + 7b + c = 16, f4



Now subtract f2 from f4 to eliminate c once again.

37a + 7b + c = 16 minus

19a + 5b + c = 9 equals

18a + 2b = 7, this is f5



Okay were almost finished, now subtract f3 from f5

18a + 2b = 7, minus

12a + 2b = 5 equals

6a = 2, Finally we reach our first variable a = 1/3



Now just plug in a in f5 to solve for b,

18(1/3) + 2b = 7

6 + 2b = 7

2b = 1, now we also have b = 1/2



With a and b now we can solve for c plugging in f4

37(1/3) + 7(1/2) + c = 16

(37/3) + (7/2) + c = 16 so since the Least common multiple is 6 we multiply the denominators to get 6, then we are able to subtract.

(74/6) + (21/6) + c = 16

16 - (95/6) = c

(96/6) - (95/6) = c

So, finally we have c which equals 1/6 = c

We only have one last variable to find which is d.

So, we repeat the same process. Now we can just plug in a + b + c + d + = 1, rememeber this is the first formula we used.

1/3 + 1/2 + 1/6 + d = 1, so we repeat the LCM rule and we simplify to

2/6 + 3/6 + 1/6 + d = 1

6/6 + d = 1

1 + d = 1

d = 0



We finally have all our variables defined, that was the long part so,

a = 1/3

b = 1/2

c = 1/6

d = 0



We plug all this in the normal formula ax^3 + bx^2 + cx + d = y

Also, lets replace x for n so that we now we are talking about sequences and series.

(1/3)n^3 + (1/2)n^2 + (1/6)n = Sn, where Sn is the sum to the nth term of all positive integers squared.

So lets proove if it works.

Lets try to find the sum to 6th term

S6 = (1/3)6^3 + (1/2)6^2 + (1/6)6

S6 = 72 + 18 + 1

S6 = 91



1+4+9+16+25+36 = 91, so it does work



I really think this is a very important tool for seqeunces and series, beacuse we have only been taught to find formulas for geometric and arithmetic series, as you can see this is neither, this is called a complex sequence. I think you should try this and get good at this, it can save you many times.



Example problem, try maybe an easier one to practice try to find the sum for the sequence for first positive odd integers. In another post I did it for the terms, now you do it for the sum.

The terms are:
1 , 4, 9, 16, 25....

nooo

nooooooooooooooooooooooooooooooooooo ..... georgieeee.... pk...???

the next scribes

the nexts scribes are...................................................................................................................................................
............................................................................................................................
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................................................................................................................................???????????
?????????????????????????????????????......................................................
.................................................................FRANCISCO AND MIGUEL!!!!!!!!!!

Scribe Post 05/17/07

POD 5/17/07

The east section of a sports stadium has 30 rows of seats. The tenth row has 100 seats and every row has 3 more seats than the row below it. How many seats are in the east section of the stadium?

Melissa answered:

10th row=100 seats d=3 amount=30 rows t30=?

Find t1
10-1=9 = 9jumps
9 * 3(distance) = 27
100-27=73
t1=73

2. tn=t1+(n-1)d
t30=73+(30-1)3
t30=160

Sn=n/2(t1=tn)
S30=30/2(73+160)
S30=15(233)
S30=3,495 seats

Different way for finding t1

t1= t10 (n-1)-3
t1=100 = (10-1)-3
t1=100-27
t1=73



tn= (.99)^n= whose limit is 0.
(in our graphing calculators)
Graph y=.99^x
Adjust window: Xmin=0, Xmax=500, Ymax=1

*We used this procedure to prove that the limit of the function is 0. This way we acquire a clearer view.

The graph shows that the sequence will never reach or cross the line Y=0.

*The limit is always a Y value(tn for a sequence)
*The limit is a boundary on the range.


Practice: find the Lim tn as n approaches infinity for the following sequences.


*Two acceptable strategies
1. list the 1st few terms
2.THINK


Example 1:
tn=(-n)^2
Answer: the limit of tn as tn approaches infinity = infinity.

First few terms: 1, 4, 9, 16 …. 10,000
The numbers keep getting bigger, therefore there is no limit.

Example 2:
tn=(-1)^n-1(n/10^n)
Answer: The limit of tn as n approaches infinity = 0.

First few terms: 1/10, -1/50, 3/1000, -4/10,000….
While n is approaching infinity, tn approaches 0, however never gets there.


Example 3:
tn=(-1)^n-1*n
n+1
Answer: DNE

First few terms: ½, -2/3, ¾, -4/5…
There is no limit because it isn’t reaching any number, it is isolating between two different numbers.


Sir we had problems with the camera, therefore we werent able to post the graphs. We will post the graphs tomorrow with their explanations, as a comment to the scribe.


Daniel gilchrist and Jorge Vergara

Scribe Post May 16, 2007

May 16, 2007
As usual we started the class with the problem of the day.

POD 5/16
My son Julian was swinging on his swing and I was pushing him so that the arc lenght of his swing was 3 meters. After I stopped pushing him the arc lenght of each swing decreased 1/6 of the previous distance. How far does Julian travel during 10 swings after I stop pushing him?

Nobody in the class was able to answer the problem so Mr. A answered the problem.

There are two major things you have to understand in this problem.

1. Is the sequence geometric or arithmetic.
   
2. What to do with the 1/6.
   

Mr. A then gave us a tip, "Always try to list the few terms first, to have a btter view of the problem."
So since we know T1= 3, and that 1/6 of 3 is 0.5, we can assume that T2 = 2.5. Doing this step many times would be very long so there is an easier way.

Instead of multiplying by 1/6 and subtracting from the previous term, just multiply the term by 5/6.

Why?

By multiplying the # by 5/6 you are throwing awy the 1/6 we dont want and conserving the other 5/6 we need. Then by multiplying 2.5 x 5/6 = 2.0833, we get the first three terms.

3, 2.5, 2.0833
(Now its clearer to see that the sequence is not arithmetic, but geometric)

Now, we only have to plug in the Geometric series formula.
Sn = T1(1-r^n)/(1-r)
S10 = 3(1-(5/6)^10)/(1-(5/6))
S10= Approx 15.09 meters

Now that we were finally finished with the POD, we started with limits.

Remember that a limit is a boundary, and that this statement means "What happens to Tn as n gets really big?"

So we had an example to solve:
Question: Find the limit of Tn as n goes towards infinity if the explicit formula is Tn= (0.99)^n

Mr. A, then showed us that there are four was to approach this answer.

1. Plug in large values for n, this way we can analyze what is Tn getting closer to.


2. List terms, as Mr. A said this helps us get a better view.


3. Repeated multiplication of a number less than 1, and one bigger.


4. 99% or as Isabella said 99/100. Why? 99/100 x 99/100 x 99/100, shows that the denominator is always bigger, therefore the numerator will get smaller.

This are basic ideas to reach the limit of the equation. We got to the conlusion that the equation tries to get closer to 0, even thought it yet never reaches 0.

So, the limit of Tn as n approaches infinity is 0.

At last in class, Mr. A taught us a second perspective of how to analyze limits: GRAPHS.

As an example we used a complex sequence.

1/2, 3/4, 7/8, 15/16

Still, to graph this is the GC there are several steps.

• Under mode change to seguqence and to dot.


• Go to Y and assign the following values.


• nMin= 1


• u(n) or equation= (2^n-1)/2^n


• u(nMin)Which is the first term = 0.5


• Then go to ZStandard in Zoom.


• Go to Window: Ymin= o and Y max= 2


• Graph

This graph can also be made as a simple function:

F(x)= (2^n-1)/(2^n)

CLASS ENDS!

There is a new HW - Read pg. 493-496. pg 496 #1-25 odd

Remember there is a test next week.

The next scribes are DANIEL GILCHRIST and JORGE VERGARA!!!!!

Sybil Problem

What I think...
Well thinking about todays class and what we learned. I guess in a way I can relate Sybil to limits. In one hand Sybil, mathematically, can never get to the other side of the room because it would always have left have of what it has remaining to go. In other words every time Sybil moves forward, although it has achieved some space, it still has half of what is left of the room to go. He will always have room to go. That is because limits can never be exceeded nor achieved. In todays class we were said to think of limits as boundaries that may or may not reach it. For me the Sybil question has a boundary of infinity. It has no limit. DNE. I dont know if this is right, I am just stating out my ideas.

May 15 Scribe

May 15/2007

Pooh and Alli say:

• Tuesday started out with a POD, as usual.

The POD was:

t3= -7
t6= -2401
Find t10.

• After a few minutes of work, Jaime went up to the board.

His solution:

1. Find out how many jumps
T6 - T3 = 3 jumps *From t3 to t6 there are 3 jumps

2.Solve for r,
Set r to the third power because there are three jumps
Then solve by multiplying the third term with r cubed which equals the sixth term.

-7 x r3 = -2401
-7 -7
r3 = 343
r = 7

3. Find out how many jumps are between t3 and t1.
T3 - T1 = 2 jumps

4. The third term divided by r equals t2.
-7/7 = -1 = t2
With t2 you divide it by r which is 7 and equals t1.
-1/7 = t1

5. Apply Geometric Sequence Formula

Tn= t1 x r^(n-1)
T10 = -1/7 x 7^9 Plug in 10 in tn because we are looking for the tenth term
T10 = -5,764,801

OR

• As Rumidog said, another way of solving the problem, is by assuming that T3 is T1, so that T10 becomes T8.
  

T8 = -7 x7^9
T8 = -5,764,801

After a while of struggling with the POD and getting some sense out of Jaime’s work most of the class understood the process. You do not need to know the formula to solve the problem, you need to understand how a sequence works.

• Definitions:
  Finite Sequence: problems we have already solved.
  Infinite Sequence: a sequence that goes on forever.
  May or may not have finite limits.

• Limits
  Limit: Think of it as a boundary. May or may not reach it, but can never exceed it.

• Bouncing Ball Sequence:
  The bouncing ball sequence had a limit:
  100, 50, 25, 12.5…

¨The limit of tn as n approaches infinity is zero¨

Lim (tn) = 0
n-->∞

Lim (tn)= ? What happens to n as it gets really big?
n-->∞

The sequence, Tn=n
Write out the first terms of the sequence? Write the limit if possible.
Sarah says the first four terms of her sequence
1, 2, 3, 4

****For a LIMIT to exist it must be a REAL #!!!***

The sequence, Tn=n
The limit Does Not Exist DNE

• To explain why a Limit does not exist, write the following:
  

Lim(tn) = ∞
n-->∞

• New Sequence
  -1, -4, -9, -16,…
  

What is the Limit?
Lim(tn)= - ∞ DNE
n--> ∞
Lim (tn) = 0
n-->∞

****Infinity and negative infinity are not the only two ways a limit can not exist.****

New Problem:
Solve. Find Limit.
The grains of salt on an infinite chess board?
1,2,4,8

Answer- DNE

Lim(tn)= 8
n->8

• New Sequence-
  ½, ¾, 7/8, 15/16…
  Write Limit.
  
  How to solve:
  One way is to put them in decimals.
  In this sequence, numerator can not be greater than the denominator.

NOTE:
Test is next week: it will have- sequences, series and limits
Remedial Tomorrow

NEXT SCRIBE IS JAIME!!

WORD PROBLEMS

POOH SAYS:

Arithmetic

• Carlos goes to the doctor and the doctor tells him he has a growth problem. Carlos is sent a prescription of injecting 10 ml of medicine each week for 2 months. After the two months, Carlos has to increase his dose every week injecting 13.5 ml more than the day before of what he is already injecting. How much ml of Medicine has Carlos injected after 3 months?
  Making the assumption that every month has 30 days.

• Mario loves to read. His grandma told him that if every month he had a good grade, she will give him a new book for his collection. Mario has already read 3 books and used them to start his book collection. Fortunately Mario is a good student and every month he is working better to improve. As a result his grades are getting higher and his grandma decided that because he is improving so much she is going to give him 2 books more than the last one, every time a new report card is given to the students. How many books will Mario have in his collection after 3 years?
  Assume there are 4 report cards per year and Mario is always improving his grades.

Geometric

• Carla wants to buy a new apartment. The problem is that she hasn’t decided therefore she decides to start the first 6 months with 2 apartments. After a year she raised her money and now has 4 apartments. A year and a half later, Carla has 8 apartments and so on. How many apartments will Carla have after 12 years and six months?