Infinite Geometric Series

OVERVIEW:

• Finding the Sum of Numbers in a Sequence
• Sigma Notation
• Infinite Geometric Series Formula
  

Finding the Sum of Numbers in a Sequence

(a) What is the sum of the integers from 1 to 5000?

Credit to Paul for being our Gauss today.

We imagine the sequence of all the numbers 1 to 5000 in the top row while the bottom row has all the numbers from 5000 to 1.

Example:
1, 2, 3, ... 5000
5000, 4999, 4998, ... 1

We find the sum of each column in our table and see that they're all 5001. Refer to slide 4 of June 3's slide for a better understanding of what I'm talking about.

We know that there are 5000 of these terms. We also know that if we want to find the sum of the integers between 1 to 5000, then we would have to multiply the number of terms by the total we have then divide by 2. We divided by 2 because there were two sequences we added.

Example: 5000 * 5001 / 2 = 12 502 500

Another way of solving this problem is to pull out the equation (found on June 3's slide #7) and plugging in the appropriate numbers, which Mr. K has a distaste for using this method because we're not really understanding the concept but just plugging numbers into an equation.

(b) What is the sum of all multiples of 7 between 1 & 5000?

The lowest multiple of 7 between 1 and 5000 is 7.
The highest multiple of 7 between 1 and 5000 is 4998.
There are 714 terms that are multiples of 7 between 1 and 5000.

We then imagine, again, that table, listing all the numbers in the sequence from 1 to 5000 (that are multiples of 7) in the top row and all the numbers in the sequence from 5000 to 1 (that are multiples of 7) in the bottom row, and find the sum of each column.

Example:
7, 14, 21, ..., 4998
4998, 4991, 4984, ..., 7

Add the top row and the bottom row, and you find that under each column in our imaginary table is 5005.

We multiplied the number of multiples of 7 between 1 and 5000 by the sum of the first and last terms of the sequence and divided that by 2 because we've counted the sequence twice and we get 1 786 785.

(c) What is the sum of all integers from 1 to 5000 inclusive that are not multiples of 7?

Well, we have the sum of all the integers from 1 to 5000. We have the sum of all the multiples of 7 between 1 and 5000. So if we find the difference of those two numbers, we find the sum of all the integers from 1 to 5000 inclusive that are not multiples of 7.

12 502 500 - 1 786 785 = 10 715 715

Sigma Notation

Left Question:

• The k = 1 means that you start with 1 and evaluate 3k.
• The number above the sigma symbol, 4, means to keep evaluating 3k until k = 4, then add all the terms.

Right Question:

• The k = 0 means that you start with 0 and evaluate 2^k.
• The number above the sigma symbol, 3, means to keep evaluating 2^k until k = 3, then add all the terms.

Then Mr. K says to find another sigma notation that is equivalent to the two questions on the slide. Of course, "mathematics is the science of patterns" is the common phrase that Mr. K always says, and yes, there're patterns here.

Also, a quick way of figuring out the answer to the sigma notation in question is to add the first and last term, then multiply by 2. Example: In the left question, [3(1) + 3(4)] * 2 = 30

Infinite Geometric Series Formula

The formula given in the above slide is something that we already derived from the previous class.

To answer the question "why is that the formula?" let's first use an example to help better our understanding of infinite geometric series. We make a square, cut it in half, and shade the half. We take the unshaded region of the square, cut that in half, and shade the half. Then we take another unshaded region of the square, cut that in half, and shade that half, and so on. So, firstly, we got 1/2 of the square shaded, then 1/2 + 1/4 shaded, then 1/2 + 1/4 + 1/8 shaded, then 1/2 + 1/4 + 1/8 + 1/16 shaded, and so on. If we continue this pattern and keep cutting and shading the square in half an infinite number of times, then we would've a shaded square.

1/2 + 1/4 + 1/8 + 1/16 + 1/32 ... = 1

The formula that we derived from the previous class was

Sn = t1*(1-r^infinity) / (1-r)

If we take r and raise it to the power of infinity, that would equal zero.
Example: r = 1/2

1/2 * 1/2 * 1/2 * 1/2 ... = 0

So, in the part of the formula where it says (1-r^infinity), that part of the formula is going to equal (1-0) = 1 anyways, and multiplying a number by 1 won't really have a visible effect. It's more efficient to just not include the (1-r^infinity) into the formula.

Plug in the numbers into the equation, BING! BANG! BOOM! you have the result. Have a nice day.


END NOTES

• Start on Exercise 47: Infinite Geometric Series. You should have finished all the questions in the exercise book up to 47.
  
• Tomorrow, we will finish off the lesson the infinite geometric series.
  
• Hopefully, we will be on task and finish the course tomorrow, and do a review on Friday.
  
• Next scribe is Francis.

Geometric Sequences

WARNING: The geometric sequences unit does not have any new material; rather the unit gives a new perspective to what we already know from previous learning.

OVERVIEW:

• What is a sequence?
  
• Recursive definition vs. implicit definition
• Common difference vs. common ratio
  
• Determining any term (the nth term) in an arithmetic sequence vs. determining any term (the nth term) in a geometric sequence
  

SLIDE 2
An example of an arithmetic sequence is the first sequence of numbers found on SLIDE 2. Examples of geometric sequences is the second and the third sequence of numbers. Note that arithmetic sequences and geometric sequences aren't the only sequences that exists, such as the Fibonacci sequence, but the main scope of the unit focuses on geometric sequences.

We find the next three terms by determining a rule for each sequence of numbers.

1. For the first sequence of numbers (4, 7, 10, 13...), we see that 3 is added to any term to get the next term.
2. For the second sequence of numbers (3, 6, 12, 24...), we see that 2 is multiplied to any term to get the next term.
3. For the third sequence of numbers (32, 16, 8, 4...), we see that 1/2 is multiplied to any term to get the next term. (Remember, in grade 12, we multiply by 1/2 instead of dividing by 2; it's to make our calculations easier to do.)
   
4. For the fourth sequence of numbers (1, 1, 2, 3...), we see that we have to add the previous two terms to get the next term.

SLIDE 3
Let's take a closer look at the first sequence of numbers (4, 7, 10, 13...).

How did we determine that the rule is to add 3 to any term to get the next term? We find that out by determining that the common difference is 3 (green font). We then find out that if we're given any term, n, then 3n+1 is the equation we use to get what n equals. For example,

3n+1 = 3(1)+1 = 4.

So if the rank (n) is 1, then its term is 4. If the rank is 2, then its term is 7. If the rank is 7, then its term is 22. This is expressed with the equation 3n+1.

The rank represents the term. For example, in this sequence, the rank of the first term is 1, the rank of the second term is 2, the rank of the third term is 3, the rank of the nth term is 3n+1, and so on.

Graph 3n+1, and we find that it is the equation of a line. Also, we can determine the 0th term by plugging 0 into the equation.

3n+1 = 3(0)+1 = 1.

For the zeroth term, the output 1. It is also the y-intercept of the graph.

Term 1 (t1), or in some textbooks it's called a, equals 4. The common difference, d, is 3.

The recursive definition, in this sequence of numbers, is to take a term and add 3 continuously.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3n+1.


SLIDE 4

Let's take a closer look at the second sequence of numbers (3, 6, 12, 24...).

How did we determine that the rule is to multiply 2 to any term to get the next term? Firstly, there isn't a common difference in this case but a common ratio. The common ratio is 2. We then find out that if we're given any term, n, then (3/2)(2^n) is the equation we use to get what n equals.

tn = (3/2)(2^n) = [3 * (1/2) * 2^n] = [3 * 2^(n-1)]

(Refer to the bottom-right corner of SLIDE 4.)

So if the rank (n) is 1, then its term is 3. Similarly, if the rank is 2, then its term is 6. If the rank is 7, then its term is 192. Etcetera.

Graph [3 * 2^(n-1)], and we find that it is the equation of an exponential function. Also, we can determine the 0th term by plugging 0 into the equation.

tn = [3 * 2^(n-1)]
t(0) = [3 * 2^(0-1] = 3/2

For the zeroth term, the output 3/2. It is also the y-intercept of the graph.

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 2.

The implicit definition, in this sequence of numbers, is the equation of the sequence, tn = 3*2^(n-1).


SLIDE 5
Let's take a closer look at the third sequence of numbers (32, 16, 8, 4...).

The recursive definition, in this sequence of numbers, is to take a term and multiply it continuously by 1/2.

The implicit definition, in this sequence of numbers, is the equation of the sequence.


VOCABULARY/SUMMARY is found on SLIDES 6 to 9.


HOMEWORK

• Exercise 45: Geometric Sequences
  

Next Scribe is Nelsa.

BOB Version 6: Conics

Haha I like Ben's creativity... Unfortunately, time isn't a luxury for me.

Anyways, this unit was nice and short, even though the unit had an emphasis on graphing. Once I was able to visualize the graph in my mind's eye--the expression that Mr.K likes using--graphing the graph made things a lot easier, including the origami portions of the unit as well.

Along with remembering the terms, remembering the equations of the circle, ellipse, parabola, and hyperbola is a bit of a doozy, but mathematics is the science of patterns, and contrasting and comparing the formulas between the horizontal and vertical orientations really helped.

I really liked the shward fiasco, the horrible dubbing by DJ K, and his whole samurai act. We may have been fooling around excessively during that class--like what Benz says--but that's what gives this unit its uniqueness. (And maybe there's going to be more unique classes like that when we enter AP Calculus.)

There were some questions left unanswered though, like how golf is evidence that aliens visited our planet Earth and how the focus of the circle is infinity, but if time really isn't a luxury for us, then I guess it's understandable, seeing that THE PROVINCIAL EXAM IS ONLY 14 SCHOOL DAYS AWAY.

If I were to rate this unit, I would give it 5 stars, for being nice, short, unique, and fun.

P.S. Vote ZEPH for V.P.

BOB Version 5: Combinatorics

I didn't like this unit. This unit wasn't very orderly like logarithms where there was sort of like a "process" to answer a question. Studying this unit felt studying like something that's abstract, but nonetheless, this is math. Although there wasn't any graphing to do...

I was a bit slow in this unit at first (and yeah I'll admit I struggled a bit), but I managed to learn the basics of permutations and combinations. The poker combinations took a while to wrap my head around, so I may still not be comfortable doing those types of questions, although being scribe for the class on the poker combination class really helped.

On the other hand, learning about phi, the Fibonacci sequence, and Pascal's triangle was... is eye-opening a good word?

Well, I hope I do at least satisfactory on the test today. Good luck.

Poker Combinations

OVERVIEW:

• SLIDES 2 to 6 are a review the basics of combinations and permutations
• SLIDES 7 to 16 deal with poker combinations

SLIDE 2

In how many ways can 8 books be arranged on a shelf if 3 particular books must be together?

There are 8 books. We grouped 3 of the books together. (If you imagine that the 3 books are together in a bag, this might be helpful to you.) So now we have 6 objects. We can shuffle the 6 objects 6! ways for a different arrangement. We can also shuffle the 3 books in the bag 3! ways for a different arrangement.

Answer: 6!3! = 4320 ways

SLIDE 3-4

There are 10 football teams in a certain conference. How many games must be played if each team is to play every other team just once?

SLIDE 3 is wrong. SLIDE 4 is correct.

To understand this problem, let's simplify the problem to 3 football teams. Let's imagine Mr.K's team, Rence's team, and AnhThi's team are together in a conference, and they're to play each other in a game just once. Mr.K versus Rence in one game, and Mr.K versus AnhThi in another game. So far, there are 2 games played. Rence versus AnhThi in a game, so the maximum number of games that can be played between the 3 teams is 2+1 = 3. Here we see a pattern that we're adding. WE'RE NOT USING FACTORIAL. The student who did SLIDE 3 made that mistake and got the wrong answer.

So if there are 10 teams that are to play against each other and they're to play each other in a game just once, then the first team can play any of the 9 other teams, the second team can play any of the 8 other teams, the third team can play any of the 7 other teams, the fourth team can play any of the 6 other team, etc. And so, the maximum possible number of total games played in a game is:

Answer: 9+8+7+6+5+4+3+2+1 = 45.
---

Another way to solve the problem is to see that there are 10 games and 2 teams are chosen to play against each other. This can be expressed as:

Answer: 10C2 = 10!/(8!2!) = 45

SLIDE 5

There are 9 chairs in a row. In how many ways can 4 students be seated in consecutive chairs? (Hint: First find the number of ways of choosing 4 consecutive chairs.)

First, we counted the number of ways of choosing 4 consecutive chairs. By moving the red "container" (that holds only 4 chairs) at a time to the right, we find that there are 6 ways to get 4 consecutive chairs.

Now, we ask ourselves, how many ways can 4 students sit in a row of 4 chairs? Answer is 4!.

6*4! = 144

SLIDE 6

Seven people reach a fork in a road. In how many ways can they continue their walk so that 4 go one way and 3 the other?

There are 7 people. These are your slots.
_ _ _ _ _ _ _

Each person (represented by a slot) either goes left or right because there's a fork in the road, but there must be 4 that go in one direction, while the other 3 go the other direction.
_ _ _ _ _ _ _
L L L L R R R

Remember, the formula for permutations of non-distinguishable objects says:

n! / (k1! k2! k3!)

n is the number of objects that contain k1, k2, k3..., which are non-distinguishable objects.

Answer: 7!/(4!3!) = 35
---

Another way of solving the problem is to choose 4 people from the 7 to go in one direction, while you choose 3 people to go in the other direction, which is expressed like so:

7C4 * 3C3 = 35
---


The rest of today's lesson is the "juice" of today's lesson.

POKER COMBINATIONS

Given a standard deck of 52 cards, how many ways are there to draw 5 cards to obtain each hand? (SLIDE 7)

In other words, "how many different 5 card poker hands are possible?"

Well, there are 52 cards and we're choosing 5 from them. This is expressed as:
52C5 = 2 598 960 ways.


(a) Royal Flush [ace, king, queen, jack, ten in the same suit]

There is only one way to get the sequence: ace, king, queen, jack, and ten if they're all in the same suit. This is expressed as:
4C1 = 4.


(b) Straight Flush [five cards in sequence and of the same suit, but not ace, king, queen, jack, ten] (SLIDE 8)

* This problem is similar to the problem with the red container and chairs.
* The black work on SLIDE 8 is wrong, by the way.

Here are your "slots."

A 2 3 4 5 6 7 8 9 10 J Q K

Because of my limitations of only describing things to you and not by means of animation, please bear with me.

Imagine the container again and this time the container can hold up to five slots. Starting from [A, 2, 3, 4, 5], we move our container to the right and count how many ways the container can hold consecutive slots. We count that there are 9 ways to have five different slots consecutively in the container.

By definition, a 'straight flush' must be of the same suit, so we can only use one out of the four suits. This can be expressed as 4C1, "4 choose 1". (From the four suits, only one of them is being chosen.)

Answer: 9 * 4C1 = 36


(c) Four of a kind [four cards of one face value and one other card] (SLIDE 9)

The first four cards...
There are 13 different face values per suit, and 1 face value is being chosen of that thirteen; this is expressed as 13C1. In total, there are 4 face values of each suit, and any of those 4 face values of each suit can be chosen; this is expressed as 4C4.

The fifth card...
The fifth card can't have the same face value of the first four cards, which leaves us with 12 different face values remaining. Of those 12 face values, we just want 1 of them; this is expressed as 12C1. There are 4 suits, only 1 will be chosen; this is expressed as 4C1.

Answer: 13C1 * 4C4 * 12C1*4C1 = 48


(d) Full house [3 cards of one face value and 2 cards of another face value] (SLIDE 10)

"3 cards of one face value"
There are 13 different face values, and only 1 of the face values is being chosen; this is expressed as 13C1. There are 4 different suites, and 3 of one face value is being chosen; this is expressed as 4C3. Now we have our three cards of one face value.

"2 cards of another face value"
Since, by definition, a full house includes 3 cards of one face value plus 2 cards of ANOTHER face value, we can only choose from the remaining 12 face values that hasn't yet been chosen by the first 3 cards. Of the remaining 12 face values, only 1 is being chosen. There are 4 different suits, and only 2 of one face value are being chosen; this is expressed as 4C2. Now we have our 2 cards of another face value.

Answer: 13P2 * 4C3 * 4C2 = 3744.


HOMEWORK

• You should be at Exercise 35.
• Slides 11 - 16.

NEXT SCRIBE
By the process of elimination, Eleven is scribe.

BOB Version 4: Exponents and Logarithms

Dear BOB,

I'm not comfortable with the graphing section of this unit.

Yes BOB, that's my first sentence of this post. Yes, I'm breaking the traditional rules of paragraph-writing by moving a detail up to the front line, where my introduction should be, so that my point stands out definite, clear, and observable. This just means I have to put time in to graphing exponential and logarithmic functions because, as the saying goes, practice makes perfect.

Well if reading that first statement gave you an unhappy impression, well guess what? As of now, I haven't done Exercise 27 or looked at the stuff that our class did while I was away at my mini-u and conferences, honestly speaking, which means I have to put a bit more emphasis in that section of this unit when studying.

But if you're feeling so depressed, don't worry about it! On the brighter side, I can handle the basics of this unit and the stuff we previously learned near the beginning of this unit, i.e., solving exponential and logarithmic equations and using logarithmic theorems. PLUS, I STILL REMEMBER THAT A LOGARITHM IS AN EXPONENT! (Even though I'm admitting to Special K that he was right--that I did forget that important fact from time to time.)

I hope we can come to an understanding in precal. Let's hope that in the end we all end up as happy smiling people infecting the world with our angelic smiles.

That is all for now. Good day.

Sincerely yours,

zeph

The Natural Logarithm (Continued)

OUTLINE:

1st Session

• Properties of exponential functions
• Properties of logarithmic functions
• Change of base formula

2nd Session

• Simple interest versus compound interest
• Numeracy: how big is a million, billion, trillion, googol, and googolplex?
• e, ln, the exponential function, and the natural logarithm
• Solve equations using ln

PROPERTIES OF EXPONENTIAL AND LOGARITHMIC FUNCTIONS:
A SUMMARY

Refer to SLIDES 2 and 3, for exponential functions; 4 and 6, for logarithmic functions.


Definitions for the exponential and logarithmic functions.

Domain: All permissible x-values of a function on a graph.

Range: All permissible y-values of a function on a graph.

y-intercept: The y-coordinate when x = 0.

Increasing: Function with a graph that goes up as it's followed from left to right. (e.g. exponential growth)

Decreasing: Function with a graph that goes down as it's followed from left to right. (e.g. exponential decay)

Concavity:
Concave up - Function is concave up if the graph is facing up the y-axis, like a cup having its opening facing up.
Concave down - Function is concave down if the graph is facing down the y-axis, like a bell having its opening facing down.

Asymptote: A line in which the function on a graph approaches infinitesimally closely but never meets.

CHANGE OF BASE FORMULA

This is a review of a previous class.

Refer to SLIDE 5 for a better visual of the proof of the change of base formula (how we derived the formula using an example and the definition of a logarithm), examples of the change of base formula, and the graph of the change of base formula in action.


log_2(x) = y

1. By definition, a logarithm can be expressed as this:
2^y = x

2. Find the log of both sides.
log(2^y) = log(x)

3. Use power law.
ylog2 = log(x)

4. Solve for y.
y = (log(x))/(log(2))
---

This is an example using the change of base formula:
7^y = x
y = (log(x))/(log(7))

NUMERACY:
HOW BIG IS A MILLION, A BILLION, A TRILLION, GOOGOL, AND GOOGLEPLEX?

1 million seconds = 12 days
1 billion seconds = 32 years
1 trillion seconds = 32 000 years

Note: In some countries, such as Canada and the United States, a billion is a 1 followed by 9 zeros, and in Great Britain by 1 followed by 12 zeros. The above billion is the Canadian billion. Also, the above calculations are approximations.

We watched the One Billion is Big video:
http://numeracy.wikispaces.com/Online+Numeracy+Videos

googol = 10^100
googolplex = 10^googol = 10^(10^100)

We googled googol and realized that googol is a number that's greater than the number of the atoms in the observable universe.
http://en.wikipedia.org/wiki/Googol

e, ln, THE EXPONENTIAL FUNCTION, AND THE NATURAL LOGARITHM

Refer to SLIDE 8 for the calculation of e to 1 000 000 digits.

---

Pronunciation of ln (the log_e) is found on SLIDE 9 as well.
---

Refer to SLIDE 11 for the concept of ln(e) = 1.

ln(e) = 1

The ln_e, or "ln of e", is the "log of base e of e, " which is 1.

This is similar to log_2 (2) as being "the log of base 2 of 2 is 1."
---

If we ever talk about THE exponential function or THE natural logarithm, then refer to SLIDE 9.

f(x) = e^x
...is the exponential function.

ln(x) = log_e(x)
...is the natural logarithm.

SIMPLE INTEREST VERSUS COMPOUND INTEREST

Simple interest is when you get interest from the original principal payment.
For example, Mr.K invests $100 and has a 10% simple interest annually.
In Year A, he would receive 10% of the original payment of $100, plus the $100, which is $110.
In Year B, he would receive 10% of the $100 again, plus the original payment of $100, which is another $110.

Compound interest is when you get interest from the previous payment you received.
For example, Mr.K invests $100 and has a 10% simple interest annually.
In Year A, he would receive 10% of $100, plus the $100, which is $110.
In Year B, he would receive 10% of the $110, plus the original principal payment of $100
---

Refer to SLIDE 7 for the calculations using the formula for the compound interest.

To calculate compound interest, use the formula,

A = P(1+r/n)^(tn)

whereas A is the amount, P is the principal (the money invested), r is the rate expressed as a decimal in the equation and not as a percent, n is the number of times taken, and t is the time (such as the number of years).

Note: Annually means once a year. Biannually means twice a year.
---

Refer to SLIDE 10 for the calculations using the formula for the compound interest.

Another way of calculating the compound percent is to use the formula,

A = Pe^(rt)

whereas A is the amount, P is the principal, e is e--as in ln (the log of base e), r is the rate, and t is time.

SOLVING EQUATIONS USING LN

Refer to SLIDE 12 for solving a sample equation using ln.

Solve.

19^(x-5) = 3^(x+2)

1. Find the ln, which is the log of base e, of both sides.
ln19^(x-5) = ln3^(x+2)

2. Using the definition of a logarithm, we can rewrite the equation as...
(x-5)(ln19) = (x+2)(ln3)

3. Like solving for a simple jr. high math equation, we apply the same rules here using distribution.
xln19 - 5ln19 = xln3 + 2ln3

4. Like solving for a simple jr. high math equation, we check to see if we can collect like terms, but we can't, but we can put the variable x to the left side of the equation, thus making us closer to isolating x.
xln19 - xln3 = 5ln19 + 2ln3

5. Factor out the x.
x(ln19 - ln3) = 5ln19 + 2ln3

6. Solve for x.
x = (ln19 - ln3)/(5ln19 + 2ln3)

The second question on SLIDE 12 wasn't touched upon, but that can be done for homework, if you want, as preparation for tomorrow's lesson.


HOMEWORK:
Exercise 26: Natural Logarithms

Next scribe is nelsa.

Board of Trig

BOARD OF TRIG V 2.0


I originally used my bulletin board as my photo to show trigonometry, but I thought it was too contrived, so I adjusted the photo instead.

http://flickr.com/photos/24464847@N04/2391625670/


BOARD OF TRIG V 1.0

BOB Version 3: Trigonometric Identities

I found this unit less daunting than the other previous units we've done so far. (This is because we don't have to memorize something huge that's like worth 1 megabyte of our memory, such as the unit circle, or get anxious over a graphing portion of a unit--because there is no graphing portion in this unit.) Comparing topics in this unit, I liked the "prove the following identity" part because it seems to be the easiest topic to absorb. Remembering how to do the sine dance (which aided us in remembering how to remember the 'dancing equations'--that's what I call them) helped out A LOT, and so, I found doing those questions without difficulty.

As of now, I have to go over the slides for the proofs of sum and differences because I haven't yet looked at them! I'm not too comfortable with questions where the questions tell what alpha and beta are, then we would have to draw a circle to find out quadrants they're in, then use the dancing equations to determine where a point is in which quadrant--because right now I'm having trouble wrapping my head around the 'setting up' part where we have to draw the circle to see what angle alpha, beta, a, or b are in. If you don't know what I'm talking about, then refer to March 20's slide #4 and 5, please and thank you! But now, I've just recently had a moment of clarity and doing those kinds of questions make sense to me.

Also, one of the questions in our homework that was assigned, EXERCISE 16: SUM AND DIFFERENCE IDENTITIES I, #9, it tells us to graph sin (t + 3pi/2) = -cos t, in which I have no idea how to graph that. And there maybe more questions that I'm not comfortable in doing because I haven't looked at them yet.

The test on trig. identities is set for Friday, and I'm sure I can prepare myself before that day. So 'til then, may we all grok in fullness!

P.S. Is it just me or did Mr.K not posted up today's slides? I was planning on using those slides to study off of because I was going to look over those slides thinking it would be more convenient for me to study off of since I am on the computer now typing up my blog post.

Pythagorean Identities

OUTLINE:
* Discussed a case study, a blogging issue, that is of recent concern in Australia.
* Continued on solving the introductory expressions on trigonometric identites.
* Discovered the concept of corrolories and how it they can make 'algebraically massaging' an expression much easier.


CASE STUDY: AL UPTON AND THE MINILEGENDS ARE SHUT DOWN

The day started off with an appealing discussion on a man named Mr. Al Upton, a sixth grade teacher in Australia, whose blog for his students was recently shut down by order of the Ministry of Education. The reasons behind the shutting down of his educational blog is not specific, but this decision is believed to contradict Article 13 of the Convention on the Rights of the Child in which the child is allowed to use any type of media for her or his education, basically. This alarming issue was discussed by Mr. Kuropatwa to his class as a warning that our blogs could be shut down as well. Mr. K encouraged us to comment on Upton's blog so as to raise the concern even further that Canadian children (us) are concerned about this as well and that children should be able to use blogs for educational purposes.

Now on to today's lesson.


INTRODUCTORY EXPRESSIONS OF TRIGONOMETRIC IDENTITIES

Refer to SLIDES 4 to 7.

We 'algebraically massaged' (sec t)/(cos t) - (sec t)(cos t) and (1 + tan^2 B)/ tan^2 B. But that work is now lost and Mr.K had to recreate them as that is mentioned in his post.


PYTHAGOREAN IDENTITIES

Refer to SLIDE 8.

One can find that any triangle on the circle that the equation...

sin^2 X + cos^2 X = 1

...with the case of degenerate triangles at the corner points on the unit circle (i.e., 0, pi/2, pi, 3pi/2, and 2pi in the domain of [0,2pi]) in which the 'triangle' has an adjacent or opposite side equal to zero, therefore, "missing a hypotenuse" and the 'triangle' can be seen as "degenerate," but still abides to the "sin^2 X + cos^2 X = 1" rule.


DERIVING COROLLARIES FROM sin^2 X + cos^2 X = 1.

Refer to SLIDE 9.

If sin^2 X + cos^2 X = 1, then the following must also be true...

sin^2 X = 1 - cos^2 X
cos^2 X = 1 - sin^2 X

...because they are algebraically the same.

So anytime we can see sin^2 X, cos^2 X, 1 - cos^2 X, and 1 - sin^2 X in doing trigonometric identities questions, we can substitute those in.

For example, refer to SLIDE 10.

From our previous class, we 'algebraically massaged' (1 - cos^2 A)/cos^2 A. We substituted/replaced the 1 - cos^2 A with a sin^2 A and found out that the equation can further be simplified to (sin^2 A)/(cos^2 A) or tan^2 A.

Refer to SLIDE 11.

And if we use that same equation, sin^2 X + cos^2 X = 1, and divided that whole equation by sin^2 X or cos^2 X, we can derive more corrolaries that are algebraically the same as its siblings.


SUMMARY
* Mr. Al Upton and the miniLegends' blog was shut down by the Ministry of Education in Australia because the ministry felt the blog violated some laws. This can happen to us!
* From sin^2 X + cos^2 X = 1, we can derive many more equations--the corrolaries--and use them in 'algebraically massaging' trigonometric identities questions.


HOMEWORK:
Exercise 14: Trigonometric Identities I

Next scribe is Joyce.

Nothing Less Than The Best....Group

SLIDE 18

On a typical day at an ocean port, the water has a maximum depth of 20 m at 8:00 a.m. The minimum depth of 8 m occurs 6.2 hours later. Assume that the relation between the depth of the water and time is a sinusoidal function.

Let's draw a graph!


a) What is the period of the function?

From the information we have been given...
* We can set the 8am as t = 0 hours.
* A maximum value is when t = 0 hours and when d = 20.
* A minimum value is when t = 6.2 and d = 8.

We can see from the graph that the period is 12.4 hours.

b) Write an equation for the depth of the water at any time, t hours.

cosine equation's parameters...
A = 6
B = (2pi) / 12.4 = pi/6.2
C = 0
D = 14

To get A, amplitude, calculate the distance from the sinusoidal axis to a maximum value or minimum value.
sinusoidal axis = (20+8)/2 = 14
amplitude = 14-8 = 6

B = pi/period = pi/6.2

C, the phase shift, is 0.

D is the sinusoidal axis, 14.

D(t) = 6cos [(pi/6.2)t] + 14


c) Determine the depth of the water at 10:00 a.m.

10:00 am = 2 hrs from when t = 0 or 8:00 am. Plug in the 2 as t into the equation to get the answer.

D(2) = 6 cos [(pi/6.4)2] + 14 = 17.1738 metres

d) Determine one time when the water is 10 m deep.

The wave is 10 metres deep, so the qestion is asking for what the time is when D = 10. Plug in 10 as D, then solve for t.

10 = 6cos[(pi/6.2)t] +14
-4 = 6cos[(pi/6.2)t]
-4/6 = cos[(pi/6.2)t]
arc cos(-4/6) = (pi/6.2)t
2.3005 = (pi/6.2)t
2.3005/(pi/6.2) = t
2.3005 x 6.2/pi = t
14.2632/pi = t
t = 4.5401

Convert the 4.5401 into "actual time" because we use hours:minutes:seconds to show time, so...

4.5401 hrs + 8am = 12.5401 hrs

Obviously, its not efficient to say .5401 hrs so we convert that to minutes.

0.5401 x 60 = 32.406 min.

12:32:24pm

We can round that to 12:30pm.

BOB Version 2: Transformations

At first, I didn't realize until later that our unit, Transformations, placed a huge emphasis on graphing. From graphing square roots to graphing absolute values to graphing reciprocals to graphing inverses and the like, with an added mix of stretching and compressing the various graphs, I spent hours practicing the art of graphing.

After figuring out how to graph absolute values, I found it easy compared to graphing reciprocals because of the number of things I had to do and remember, but nonetheless, I had acquired the skills needed to do those...I hope. I found the reciprocal functions to be sort of cool though, like, they look like people if you graphed them a certain way... E.g., f(x) = 1/(x^2-4) (Also, the graph on one of March 3rd or 4th's slides looks like a DNA strand unwinding...)

Reminiscing on my progress through this course, I remembered the "STRETCHES BEFORE TRANSLATIONS" rule (as being bold and caps locked by Mr. K on his SMARTBoard during one of his lessons), which eliminated a lot of confusion related to the decision-making part of which parameters of a translated, compressed/stretched function should be thought out first. When I finally understood that rule, and with the help of Mr. K, I felt that this should be included into my BOB and that it should also be merited as one of my EUREKA! moments.

Although I'm not yet comfortable in writing the test as of yet (but I will be later), I feel the need to look over and redo more of my questions in the exercises (because practice makes perfect), as symmetry and reflections are a bit vague in my memory...

BOB Version 1: Circular Functions

Nope, not done blogging for today. Just doing my BOB…

And so, Circular Functions, the unit, put a lot of emphasis on the unit circle. Memorizing (no, not memorize, but remember!) the values using mnemonics was the muddiest thing that I didn’t like about this unit. But remembering the unit circle did help me out as we progress through the course. Also just as equally as muddy was the group work assignment, The Unit Circle Triangle. I didn’t like that assignment ‘cause of its “challengingness.”

I remember in gr.11 we were studying how to graph trigonometric functions and Mr.K exposed us to the equation f(x) = AsinB(x-C) + D. He taught us what A, B, and D did to the sine graph but told us that we’ll learn what B does in gr.12 precal. Now, after over a year of waiting, I found out what B meant and that became my moment of clarity. B stretches/shrinks the graph horizontally.

I also liked how Mr.K did his 1st class on the 1st day of school because I remembered it as inspiring.

I also liked the helpful advice Mr.K said to one of his students: Remember folks, Mr.K says, “If you don’t memorized the unit circle, you will fail.”

By the way, does anyone know how to do Exercise 2 #15?

If f (x) = 2x + 3, find k so that f(k+2) = k + f(k).

Answer key says:

2(k+2) + 3 = k + 2k + 3
2k + 7 = 3k + 3
4 = k

...which doesn't make sense to me.

Also, one of our questions in our homework asked us to graph the tangent function. We weren't taught that.

Wrapping Up and Concluding Circular Functions

3 assignments, 2 periods, 1 class. This is PC40SW08 (Pre-Calculus 40S Winter 2008).

As of now, Mr.K is still out-of-province because he was selected by The Council at the Conference last week to help write a textbook, but will return from his quest hopefully by Monday. And so, we had the same substitute as previous class'.

Period 1:

Our first assignment was Quiz 2 on Circular Functions in which we were not allowed to use our calculator. It was not a hand-in because the substitute says "according to the instructions Mr.K gave me, it did not say to hand anything in." So we assumed that we are not to hand that in. The class worked quietly during that quiz.

After finishing Quiz 2 on Circular Functions, we were assigned Graphing Trigonometric Functions. We were also told not to use a calculator when doing that assignment, but were allowed to use the calculator to check our answers. That is for homework, folks.

Period 5:

In our afternoon session, we were told to continue working on this morning's pre-calculus assignments and that Exercise 7: Translations #11-20 is also for homework.

HOMEWORK:
* Graphing Trigonometric Functions
* Exercise 7: Translations #11-20
* Circular Functions Test on Tuesday 26 or Wednesday 27

Next scribe is AnhThi.


That is all.