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