BYE

Sir I just wanted to say goodbye, I couldnt personally say it in school I was in a rush, but I wanted you to know this is one of the, if not the best year I had. I hadnt learned so much in a long time. I hope you have a great life and keep changing peoples life, Good Luck and thanks for everything.
Jaime

Good Bye!

Dear Rumidog!

I know I talk in the name of everyone when I say you will be missed! Thank you very much for what you taught us this year, you were an excellent teacher, and I personally loved to go to your class. I wish you the best wherever you go in life, and don't forget to come to visit!

BYE!!!!

Larisa! (the "super" monitor)

Little Mistake

On problem #5, csc is 30 degrees, not 25 degrees. (francisco sevilla and martha alcocer)

Series Question!!

Can a series diverge with a finite sum? Its that I am tring to find out when the absolute value of r is greater than 1, the series diverges, but I do not know when does it goes to infinity or has a finite sum. Can someone help me..Thanx

Final Assignment -

Final Assignment
By: Martha L. Alcocer and Francisco Sevilla

1. lim ((9x^2) + 3x)/(4x - (5x^3))
(as x approaches infinity)

lim
(as x approaches infinity)


((9x^2) + 3x)/(4x - (5x^3)) x (1/x^3)/(1/x^3) =


((9/x^0)+(3/x^2))/((4/x^2)+5) = 0.





2.


















CSC 25° =


r/y = 1/(1/2) = 2.





3. Turn 75° to radians, and 1.2 radians to degrees.





75° x pie/180 = 1.31 radians.


1.2 x 180/pie = 68.76°





4. Find all 1st revolution angles for all angles[0,360]:


9sec o - 9 = 2


sec o = 11/4


(sec -1) sec o = 11/4 (sec -1)


o = sec -1 (11/4) = cos (4/11) = .99°


360° - 0.99° = 359.001°





5. Find the other five trig functions if cot 2/1 is on the 1st quadrant.





cot = 2/1 = x/y = 2 x^2+y^2 = r^2 = 4 + 1 = r^2 = squared root of 5.


tan = 1/2


sec = squared root of 5/2


cos = 2/squared root of 5


csc = squared root of 5/1


sin = 1/ squared root of 5





6. 10 - 3tanX = 8 + 2tanX





-3tanX = 8 + 2tanX - 10


-3tanX = 2tanX -2


-1tanX = -2


tanX = 2/1


(tan -1) x = 2tanX (tan -1)


x = 63.44





7. Mr. Alcantara told his students that they couldnt pass out of a certain sector of the trailer. If that sector has a radius of 5m and 1.2 radians. How long is the area of the sector in which the students can be?





s= (5)(1.2)


s= 6m


k= (1/2)(5)(6)


k= 15 m^2





8. Francisco is a very spoiled boy. His parents give him each christmas twice the presents they gave him the year before. His first year his parents gave him one present. Right now he is 17. How many presents has he received?





1,2,4,8,....(n=17)


s17= 1(1-(2^17))/(1-2)


s17=131,071 presents.





9. 3 x 4 + 4 x 5 + 5 x 6 + 6 x 7.... 30 x 31





tn= 3 + (n-1)(1) = 3 + n - 1 = 2 + n


tn= 4 + (n-1)(1) = 4 + n - 1 = n + 3


(n+2)(n+3)


n^2+3n+2n+ 6


n(n+1)(2n+1))/(6) + 5(n(n+1))/(2)


(28x29x57)/(6) + 5((28x29)/(2)) + 6.28


= 3654 + 2030 + 168 = 5852





10. 1/2 + 1/4 + 1/8 +....





Sn = t1/(1-r)


Sn = (1/2)/(1-0.5)


Sn = 1

Final Exam & Book Return

Hola, 11B.

Good luck tomorrow; it's easy.

Be sure to bring your textbook and your checkout sheet to the final. Otherwise you will have a hard time finding me. I will be leaving campus to grade your exams.

Remember, some of you must post your final assignment by 5:00 pm.


See you at 7:15

Blog Final Problems

3.Simplify and find the sum.

10
∑ 2+(n-1)5
n=1


10
∑5n - ∑3
n=1

10
5∑n - ∑3
n=1


5(11)(5) - 30


275 - 30 = 45


5. Find sin of a 30 degree angle.

Y= 1, X= √3, R = 2

Sin= Y/r, so sin= 1/2


6. Convert 45 degrees in radians.

450 degrees * л/180 degrees = 2.5л or 5л/2



7. Find all first revolution angles if cot= 3/5

cot=x/y, tan=y/x

tan= 5/3, then you get its inverse.

so the angle= tan-1 (5/3)

angle= 59.03, 239.04



8.Find the recusrive and explcit definition for the sequence 2,4,6,8,10. Then find T20.

2,4,6,8,10

2 2 2 2

t1=2, d=2

Explicit= T1 + (n-1)d

= 2+ (n-1)d

= 2n



Recursive= T1=2

Tn=T(n-1) +2



T20 = 2(20)

t20 = 40



9. Find the sum of 2,1,0.5,0.25,0.125

1/2 = 0.5, 0.5/1 = 0/5

Rate = 0.5

Sn=t1/1-r, Sn= 2/0.5

Sum =4



10.Find 5 trig functions if cot=3/4 in the 1st quadrant.

Cot=x/y, so y=4 and x=3

Using pythagorean theorem, find r.

3^2+4^2=r^2

9+16=r^2

25=r^2

√25=r

5=r
sin=y/r or 4/5

cos=x/r or 3/5

tan=y/x or 4/3

sec=r/x or 5/3

csc=r/y or 5/4

11. Solve for 2sin -9 =3

2sin = 12

sin = 6

Since sin = y/r, and y cant be greater than the radius, this is undefined.

12.A circle has a radius of 30cm. Find the sector of a 30 degree angle.
s=r * angle

convert the angle in radians

30 degree * π/180 degrees = π/6

π/6 * 30cm = 15.7 cm

14. Find the sum of 1 + 3 + 5+ 7+ 9+ ..t15 using sigma notation.

difference= 2
t1=1
Explicit formula = 1 + (n-1)2
= 2n-1

8
∑ 2n-1
n=1

8
∑2n - ∑1
n=1

8
2∑n - 8
n=1

72-8 = 64

Assignment

Juliana Lecompte and Sofia Schuster

Assignment

1.
Lim 2x­2 + x + 3 / 5x3 + x2 + 2x – 4
n to ∞

Lim 2x­2 + x + 3 / 5x3 + x2 + 2x – 4 * 1/x3 /1/x3
n to ∞

Lim 2/x + 1/x2 + 3/x3/ 5 + 1/x + 2/x2 – 4/x3
n to ∞

Lim 0/5 = 0
n to ∞

2.
Finite Sum

10+5+2.5+1.25
R= .5
S∞ = T1/ 1-R S∞ = 10/.5 = 20

Not Finite Sum

1+2+4+8+16
R=2
S∞ = 1(1-2n/1-R) S∞= 1(1- ∞/-1) Turns positive so: S∞ = ∞ = DNE

3.
Simplify and Evaluate

6
∑ 4+ (n-1)4
n=1

6 6 6
∑ 4 + 4 ∑n - ∑1
n=1 n=1 n=1

24+ 4[3(5) – 6]
24+ 4 (15-6)
24+4(9) = 60

5.
Find the value of any trig function

√ 2

1

1


Sin 45 = 1/ √2 * √2/√2= √2/ 2

6.
Convert

150 to radians π

150 * π/ 180 = 2.68

5 radians to Degrees

5* 180/π = 286.5

7.
Find first rev angles

Cos Ө = 4/5
Cos -1= 36.90
360 – 36.9 = 323.10

Cos it’s positive in the 1st and 4th quadrants

8.
Recursive and Explicit definition

5, 9, 13, 17
D= 4
Recursive def = Start in 5 and add 4 to each term it gives you.
T1= 5
Tn = (Tn-1)+4
T5 = (T5-1) +4
T5 = T4 + 4
T5 = 17 + 4
T5= 21

Explicit def
Tn = 5 + (n-1)4
T6 = 5 + (6-1) 4
T6 =5 + 5(4)
T6 = 25

9.
Find the sum of finite and infinite geometric and arithmetic series.

Arithmetic Finite
1+2+3+4+…+20
S20 = 20/2 (1+20)
S20 = 10(21)
S20= 210

Geo infinite
20 + 5+1.25
S∞ = T1/ 1-R
S∞ = 20/.75= 26.7

10.
Tan Ө = -5/4 quadrant # 2

52 + 42 = R2
√41= R
R= 6.5
Cot Ө = 4/5
Cos Ө = -4/6.5
Sin Ө= 5/6.5
Csc Ө = -6.5/5
Sec Ө= 6.5/4

14.
8 + 10 + 12 +14 +…+22
n
∑8 +(n-1)2
n=1

(22-8/2)+1 = n
n= 8

8 8 8
∑8 + 2∑n - ∑1
n=1 n=1 n=1

64+2[4(9)-8]
64+ 56= 120

Final Assignment

By: Larisa Jasbon and Kristina Wick

In order to see the larger image, click on the BubbleShare. If anything is not clear, please, let us know.

Problems:

This album is powered by BubbleShare - Add to my blog

Answers:

This album is powered by BubbleShare - Add to my blog

We hope that these problems are a helpful study tool for the final exam. If you have any questions, we are willing to help you in any way that we can.

Sayonara!!!

Radians and Degrees

Conversion between radians and degrees!!

Since, and



we have the following conversion rules.

To convert from degrees to radians, multiply degrees by



To convert from radians to degrees, multiply radians by

Radian measure!!

Radian Measure
Converting from radians to degrees

IN THE RADIAN SYSTEM of angular measurement, the measure of one revolution is 2π.



Half a circle, then, is π. And, most important, each right angle is half of π: π/2.
Three right angles will be: 3· π/2= 3π/2.
Five right angles will be 5π/2. And so on.

Radians into Degrees
Students should have a clear picture of this:




π/4 is half of π/2, so it equals a 45°.
π/4 is one fourth of π.
π/3 is one third. Which equals 180° ÷ 3 = 60°.
π/6 is one sixth: 180° ÷ 6 = 30°.


5π/4
= 5 · 180/4
= 5· 45°
= 225°.

2π/3 is a third of 2π. So 360° ÷ 3 = 120°.

***To convert from radians to degrees, you convert the pie π into degrees : 180° and solve for the angle.
If it is π/4 angle, you divide 180 into 4 which is 45. Therefore π divided 4 times equals a 45°.

Solve:
Convert from radian measures into degrees:
a) π/8
b) 2π/5
c) 7π/4
d)9π/2
e)4π/3
f)5π/6
g)7π/9

ANSWERS IN COMMENTS!!

Arc length

>Practice Problems

a) At a central angle of 2.35 radians, what ratio has the arc to the radius?
b) In which quadrant of the circle does 2.35 radians fall?
c) If the radius is 10 cm, and the central angle is 2.35 radians, then how long is the arc?

*****ANSWERS ARE IN COMMENTS!!**********


Problem 1
a) At a central angle of π /5, approximately what ratio has the arc to the radius? Take π=3.
b) If the radius is 15 cm, approximately how long is the arc?
Problem 2
In a circle whose radius is 4 cm, find the arc length intercepted by each of these angles. Again, take π = 3.

Trigonometry- Arc Length

The definition of a radian meadure
An angle of a 1 radian
Examples!!



Let the letter s (for space) symbolize the length of an arc, which is called arc length.
Now the circumference of a circle is an arc length.
The ratio of the circumference to the diameter is the basis of radian measure.
That ratio is the definition of π.

π = C/D
C = Circumference
D = Diameter

Since D = 2r, then

π = C/2r
or,
2π = C/r




That ratio of the circumference of a circle C to the radius r -- 2π -- is called the radian measure of 1 revolution, which are four right angles at the center. The circumference subtends those four right angles.




Radian measure = θ s /r

Thus the radian measure is based on ratios -- numbers -- that are actually found in the circle. The radian measure is a real number that indicates the ratio of a curved line to a straight, of an arc to the radius. For, the ratio of s to r does determine a unique central angle θ.




EXAMPLE:
In a circle whose radius is 10 cm, a central angle θ intercepts an arc of 8 cm.



a)What is the radian measure of that angle?
b) what is the arc length if the radius is 5 cm?

Answers
a)
θ = s /r
= 8/ 10
= .8
b)For a given central angle, the ratio of arc to radius is the same. 5 is half of 10. Therefore the arc length will be half of 8: 4cm.
An angle of 1 radian
Note that an angle of 1 radian is a central angle whose subtending arc is equal in length to the radius.

question!

Sir, does the hard copy of the questions due on Tuesday has to be typed or handwritten?

TRIG functions!!

Cosine Overview

A type of trig. function
Definition 1 is the simplest and most intuitive definition of the cosine function. It basically says that, on a right triangle, the following measurements are related:

the length of the triangle's hypotenuse
the length of one of the other sides
the measurement of the angle (q) adjacent to that other side






Graph of Cosine and Sine




Fundamental Trigonometric Identity

sin^2t + cos^2t = 1

Study ..Final Exam

http://en.wikibooks.org/wiki/Trigonometry/Solving_Trigonometric_Equations

http://www.analyzemath.com/Trigonometry.html

Study..Final exam

Final Topic List & Assignment

Below you will find a list of skills that you will need for the final exam.

For those of you who did not write two scribe posts, you have the following assignment:

• Choose 10 of the 14 topics
• Write an example problem for each type of problem in the list.
• Number the problems with the numbers of the problem types found on the study list below.
• Solve your problems.
• Turn in a hard copy of your problems and solutions at the beginning of class on Tuesday.
• Post your problems and solutions to the blog by 5:00 pm Tuesday ( I will check at 5:00)

You can do this assignment with a partner or by yourself.

Look back in your notes, text, and quizzes for ideas. Be sure that you create your own problems. Each group must have different problems. Suspected copying from the Internet, from other students, or elsewhere, will be rewarded with a zero. Also, such an approach would not be good preparation for your final exam.

Remember, this assignment is designed to help you and the rest of your class. If you do a good job on this, then the final should go quite smoothly for all.

Trig. Graphs!!

Study for Trig. Graphs:

Sec Graph!!


A graph of sec(x). sec(x) is defined as 1/cos (x)

Cos Graph!!


A graph of csc(x). csc(x) is defined as 1/ sin(x)

Sine Graph!!


Careful analysis of this graph will show that the graph corresponds to the unit circle. X is essentially the degree measure(in radians), while Y is the value of the sine function.

Cos Graph!!


As with the sine function, analysis of the cosine function will show that the graph corresponds to the unit circle. One of the most important differences between the sine and cosine functions is that sine is an odd function while cosine is an even function.

Sine and cosine are periodic functions; that is, the above is repeated for preceding and following intervals with length 2π.

Cot Graph!!


A graph of cot(x). cot(x) is defined as 1/tan (x) or cos(x)/sin(x)

Tan graph!!


A graph of tan(x). tan(x) is defined as sin (x)/cos (x)

Review

STUDY FOR FINALS!!

Graph of Cubic function, where y=x^3
Polynomial Functions
Basic definitions:
1. When numbers are added or subtracted, they are called terms.
When numbers are multiplied, they are called factors.
2. A variable is a symbol that takes on values. A value is a number.
3. A constant is a symbol that has a single value.
4. A monomial in x is a single term of the form axn, where a is a real number and n is a whole number.
The whole numbers, recall, are the non-negative integers: 0, 1, 2, 3, 4, etc.
5. A polynomial in x is a sum of monomials in x.
6. The degree of a term is the sum of the exponents of all the variables in that term.
7. The leading term of a polynomial is the term of highest degree.
8. The leading coefficient of a polynomial is the coefficient of the leading term.
9. The degree of a polynomial is the degree of the leading term.
10.. The constant term of a polynomial is the term of degree 0; it is the term in which the variable does not appear.

Scribe 6/6

Today as usual we started with the POD.

Pod 6/6: Find lim 6x^4+ 3x^2/12x^3- 8+x^2. Show your work. No graphing calculator.
x:-∞

In order to solve this problem we used the trick given for limits(remember not all limit problems apply):
**Trick: multiply top and bottom of the fraction bythe reciprocal of the variable raised tothe highest power in the denominator.

lim 6x^4+3x^2/12x^3-8+x^2 * 1/x^3/1/x^3
x:-∞
lim (6x+3/x)/(12-8/x+4/x) = 6x/12 =
x:-∞
The limit does not exist because the number will increase negatively as the x gets bigger!
DNE :-∞

With the last property we got yesterday we can solve the extra credit problem of the last test:

A brick pyramid is built in square layers such that smaller layers are stacked on top of larger layers. The lowest layer is made with 100^2 bricks. The next layer with 99^2 bricks, the next with 98^2 bricks, etc...
How many bricks are in the pyramid if the top layer is made of 40^2 bricks?

b =number of bricks

This album is powered by BubbleShare - Add to my blog

After solving the previous problem we were asked to solve the next problem using the poperties previously given:

This album is powered by BubbleShare - Add to my blog


Practicing with another problems:

Given that a 4-square court has a total of 5 squares.
a. How many squares would a 9 square court have? 14 squares
b.How many would a 25 square court have? 55 squares
c.Use a summation formula to find the total number of squares in a 100 square court.
10 sides

This album is powered by BubbleShare - Add to my blog

Practice......
Problem # 26 from the Homework
Express the series in Sigma notation

27-9+3-1+1/3-1/9

This album is powered by BubbleShare - Add to my blog



Now that we have expressed the series in sigma notation we are going to practice gettingthe solution of the sum:


Express the series in sigma notation and then use sigma properties to find the sum:
3*2+3*3+3*4+3*5+.....+3*18

This album is powered by BubbleShare - Add to my blog


At the end of the class Mr A gave us the last property:

5th property:
The constant never changes so you just have to multiply it by the number of terms.

This album is powered by BubbleShare - Add to my blog




Homework pg 509 # 25,27,33,35


Tomorrow´s scribe would be done by Allison

Basic Graphs

Basic Graphs

Here are the graphs of a parabola, where y=x^2 and square root function.

Also the graphs of y=x and absolute value, y = lxl.

Logarithms

STUDY!!

A LOGARITHM is an exponent. It is the expontent to which the base must be raised to produce a given number.
For example, since
2^3 = 8
then 3 is called the logarithm of 8 with base 2.
3 = log2 (8)
3 is the exponent to which 2 must be raised to produce 8.

Definition:
logb (x) = n means b^n = x.
That base with the exponent produces x.
**With any base, the logarithm of 1 is 0.

Practice:
Write in exponential form: log2 (32) = 5

Write in logarithmic form: 4^−2 = 1/16

Evaluate log8 (1).

Evaluate log5 (5).

Laws of Logarithms

TO STUDY FOR FINALS:

3 Laws of Logarithms:

1. logbxy = logbx + logby
"The logarithm of a product is equal to the sumof the logarithms of each factor."

2. logbxy= logbx − logby
"The logarithm of a quotient is equal to the logarithm of the numeratorminus the logarithm of the denominator."

3. logb x n = n logbx
"The logarithm of a power of x is equal to the exponent of that power times the logarithm of x."

Problems to solve with Log:

a) log ab/c
b) ln (sin²x ln x)
c) ln (a2x − 1 b5x + 1 )´

Scribe 6/5

Mr A started our class as usual, with a POD.

POD 6/5:
Find, in radians, all 1st revolution angles, where cot = - √3.
(NO CALCULATORS OR SNOWMAN SHEET)

Mr A´started us off by saying how he would have solved this problem. He told us how he would ignore the √3 and focus on the negative sign. The tan / cot angles are negative in both the second and fourth quadrants.

This album is powered by BubbleShare - Add to my blog


*****REMINDER: In the final exam 1/3 of the quiz will be without a calculator
2/3 of the quiz will be with a calculator

Going back to what we were doing yesterday...
We left a problem without solution:

This album is powered by BubbleShare - Add to my blog



Now that we have solved this problem, we are ready to prove the first properties we were given yesterday, by solving an exercise we have done before:

3+24+1.92+1.536+... FIND the sum of the terms 5 - 18

Answer: S5-18≈ 5.87


Find:

18
∑3*0.8^k-1
k:5

Explicit formula T1*r^k-1
(Geometric Sequence)

1st property (for further info. check Daniel's scribe)
18 18
∑3*0.8^k-1 = 3∑0.8^k-1
k:5 k:5



By using the 2nd property we will brake the equation into 2 geometric series eliminating the first four terms giving us as a result the sum of the terms 5 - 18

18 18 4
3∑0.8^k-1 = 3[(∑0.8^k-1) -(∑0.8^k-1)]
k:5 k:1 k:1

Use the Geometric series formula to evaluate:
Geometric Series formula Sn: T1(1-r^n/1-r)
18 4
3[(∑0.8^k-1) -(∑0.8^k-1)]
k:1 k:1


T1= 1 (FACTOR THE 3 OUT)
r= .8

3[1*(1-.8^18/.2)-1*(1-.8^4/.2)] = 15 (-.8^18+ .8^4) ≈ 5.87

5.87= 5.87

Mr A proved us that solving this series problem with sigma notation does work, and told us that it would be more useful to solve more complex exercises.

Solving another problem:
Find:

20 20 9
∑2n = 2[∑n-∑n]
n:10 n:1n:1

2[(20/2(21))-(9/2(10))]
2(10(21)-(45))
2(210-45)
=330

TIP: List first terms if you doubt which type of series you are being asked


Practicing with another problem...

30 30 30
∑3p-∑9p = ∑-6p
p:1 p:1 p:1

30
=-6∑p
p:1

=-6(15)(31)

=-2790


MORE PROPERTIES

3.The sum of the 1st n positive integers.

n
∑+=n(n+1)/2
i:1


4. The sum of the squares of the 1st n + integers. (For further info. look at Jaime´s post)
n
∑i^2=n(n+1)(2n+1)/6
i:1


Proving property #4
n
∑i^2=n(n+1)(2n+1)/6 [expand and add]
i:1

4(5)(9)/6 = 180/6 = 30

With property #4 we can finish yesterday´s problem:

find

12 12
∑5k-k^2 - 390 - ∑k^2
k:1 k:1

n(2n-1)(n+1)/6

= 390- [12(25)13/6]
=390-650
=-260

390(12*13*25)/6
= -260



**** LAST REMEDIAL AT 6:15 ****

Today's Scribe By: daniel g and Jorge V.

04/06/2007 Scribe Post

As usual the class started with a POD

Today’s was:
If the angle is in the 4th quadrant, and its cosine is 4/5, find the values of cosecant and tangent of the angle.
First we had to find y:
We used the Pythagorean Theorem (x^2+y^2=y^2).
4^2+y^2=5^2
16+y^2=25
16+y^2-16=25-16
y^2=9
We then got the square root of each side which ultimately crossed out in one side and in the other side it gave 3, so y equals 3.
We then needed to get the cosecant of the angle. Since we know that cosecant is the inverse of sin, which is y/r, it meant that cosecant must be r/y.
That made the cosecant = -5/3
Then we got tangent. Tangent equals x/r, so the answer was -3/4.
As you can see both answers were negative. This happens because since in the 4th quadrant all y’s are negative and all x’s are positive the division of both will result a negative answer.


On Friday we left on this problem:

Find the sum of this infinite series: 3+2.4+1.92+1.536+…
Today we wrote it more efficiently like this:
Σ 3(.8) ^ (k-1)
The expanded form for this formula is what we already know as:
3(.8) ^(1-1) + 3(.8) ^(2-1) + 3(.8) ^(3-1) + …
The Σ stands for Sigma Notation.

Sigma notation: a fancy but efficient way to write complicated addition problems.
Vocabulary for Sigma Notation:
Sigma: Greek letter- means add the following.
Summand: the expression that is being added. n the last example: (3(.8)^(k-1))
Index: the variable that is changing. Example: k.
Limits of summation: upper and lower bound of the index (1,∞).

Example:

For the following:
Σ-2k its limit bounds are (1, 6).

Write the expanded form.
Evaluate the expression.
Identify the summand.

Solution:
-2(1) + -2(2) + -2(3) + … + -2(6)
-2 + -4 + -6 + -8 + -10 + -12 = -42
the summand is -2k

Shortcut
Since the series changes arithmetically because every time there is being added -2 more we can use the arithmetic series formula as a shortcut.
S6 = 6/2 (-2-12)
S6 = -42

We then started studying the properties of the Sums

This are:
When the limit is (1,n) Σcai it is the same as cΣai.
What is being done here is that a constant of multiplication is being removed where ai is some expressions dependent on i.

Example: find the sum of the 1st 30 multiples of 7.
If the limit of Σ7k is (1,30).
It’s the same as: 7Σk To solve this we can use Gauss’ technique: [n/2(n+1)]
7 [30/2 (30+1)]
7*15*31= 3255

Rule # 2 states that:
ΣAk+Bk = ΣAk + ΣBk and ΣAk-Bk = ΣAk - ΣBk
This means that the sum of a sum is the sum of the sums.
And the sum of a difference is the difference of the sums.

Class was over and we finished there.
There is a new homework which is:
Read 506-507 and do in page 508 # 1-15 odd.

Tomorrow’s scribe since we get to choose would be ………………………………………………………..........................
Alejandro Covo

Scribe June 1

Well, first of all I want to wish good luck to all the students running for an officer position at STUCO. This is not relevant to Pre-Calc, but still, our classmates need to know we support them!

As always, I was late to class and missed the POD. However, Eli copy it for me. So please if there's any mistake tell me!!!.. jajajaja
POD
When Frieda was born her parents gave her 2000 dollars. On each subsequent birthday, they gave her 4/5 of the amount they gave her the year before. On Frieda's 18th birthday, what is the total amount since birth that Frieda would have received?
-So, first of all this is a geometric series! It ask for the TOTAL AMOUNT, not how much will she get on the 18th birthday.
The rate is = 4/5
t1 is = 2000
and we need to find S18
STEPS
1) Get the second and third term to see if the final answer is correct.
2000(4/5)=1600
2000+1600..+.......+.........
2) To get our final answer we use the formula Sn= t1((1-r^n)/(1-r))
So, PLUG IN the given numbers!
S18=2000((1-(4/5)^19)/(1-(4/5))
*WHY 19??? well, there are 18 birthdays + the actual day she was born so it's the 19th number!
3) Get the answer, which is aprox $9855.88

Now, changing the question a little bit:
Assume that Frieda and her parents live forever (inmortals, yeah right), how much money will Frieda collects?
If you remember, on Wednesday, Mr. Rumidog gave us a formula to solve this type of problems. The formula is S=(t1/(1-r)).
So again, just PLUG IN!
S=(2000/(1-(4/5)))
S=$10,000
Which means that our limit is $10,000, when n goes to infinite.

Now, change the problem again…
What if Frieda’s parents gave her (5/4) as much as every year for 18 years? Forever?
-18th birthday:
S18=2000((1-(5/4)^19)/(1-(5/4))
S18= $547,111.51
-Forever:
S=2000/(1-(5/4))
S= -$8,000
*WRONGGGGGG!!!!!!!!!!!!!!!!!!
The “Sum of an Infinite Geometric Series” states that: If r≥ 1, tn ≠ 0. This means that the equation we just use can only be used if r is ≤ 1. *5/4 is ≥ 1*
The correct way to answer the question is the following:
Sn=t1((1-(r)^n)/(1-r))
So, how she is earning for ever, we may say that r^n is = .
Answer: DNE!
*Isa then stated that if the rate is greater than 1 then the answer will be INFINITE.

Finally, after the LONGEST POD EVER, Mr. Rumidog recommended to visit B.Bustillo’s everything site.

HEY! We are also going to have a QUIZ on FRIDAY!!!!
The topics on the quiz will be:
· Limits
· Infinite series
· And a new topic that he is going to teach on Monday called “The Sum of an Infinite Series”
To practice for the quiz there is HOMEWORK! Pg: 503 #23-31 odd and 35


So, going back to the Achilles’ problem, we may now use our formula.
So the series will be 10+1+(1/10)+(1/100)
Formula: S=(t1/(1-r))
r=1/10
t1=10
So, PLUG THEM IN!
S=(10/(1-(1/10)))
S=(10/(9/10))S=100/9S= 11 1/9 m

Now, a practice problem:
Find the sum of the following series:
3+2.4+1.92+1.536+…
r= 2.4/3 = .8
Sn=(3/(1-.8)) = 15
Dark Angel said: The rate is less than the absolute value of 1!!

New task:Create your own problem!!1. Choose some real # for t1, NOT 02. Choose some R such that r<1 and r≠03. List the first 4 terms of your series4. Find S10 and S 5. Exchange just the terms6. Solve each others problems7. Check with the author

So, Dark Angel and I did the task together… this is our problem:1. T1= 112. R= .13. 1+1.1+.114. S10= 11((1-.1^10)/(1-.1)) =12.2
S =(11/1.01) = 12.2

After solving these problems, Kris asked about Infinite Arithmetic Series.
The Arithmetic Infinite Series consists on adding an infinite # of numbers. (Sn= t1+t2+t3+…+tn)
A practice problem:
t1= 5d= 8(5+13+21+…)
or
d= -8
(5+ (-3) +(-11) +(-19)…)
Sequence: Start at some number (t1) and repeatedly add some # d (not 0).
D>0, t1 increases without bound. (Sn=? = ) DNE
D<0, t1 decreases without bound (Sn=? = - ) DNE
Infinite arithmetic series cannot have a finite sum.

Next Scribe:
Daniel + Jorge = Monday!!!

War and Peace

Last night, I sat down in my hammock and started reading pg. 985 of Leo Tolstoy's War and Peace. This is the classic novel of Russian life at the time of Napolean's empire. I recommend reading the entire text sometime in your life when you have the time to tackle its 1455 pages.



You will probably understand my amazement when I tell you that Tolstoy started Book 11, Chapter I by discussing Zeno's paradox of The Tortoise and Achilles!!!!!!!



He then goes on to discuss the implications of calculus on the study of history. Some of you might find these three pages quite interesting. You can read this short chapter here.



I would be interested to hear your thoughts on this as, I imagine, would Mr. Janke.

Rain as Pathogen

Rain DOES NOT cause infections!

Infections are caused by microorganisms such as viruses, bacteria, protozoa, yeasts, and AC Milan players.

Doctors prescribe antiBIOtics not antiLLUVIAtics.

Despite the best scientific evidence, if you still fear the cleansing effects of rain, please purchase and carry an umbrella. Another option is to lift and carry a smaller student over your head.

♥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