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.