Saturday, May 31, 2008

Probability: More Practice

Instead of starting a new unit, we worked on Probability. Mostly we went over the practice sheet we were given on Thursday. The questions we reviewed were one, two, six and nine.

Question One:

  • The probability of a randomly chosen car being defective is 1/3. Four cars are chosen randomly in order. Given that at least two cars are defective, what is the probability that the first car is defective?

The first thing we did, was figure out what the question was asking.
P(1st is defective I
at least two is defective)

Next we found the sample space. It's always easier if you think of it as a word, and then asking yourself how many times the letters in that word can be rearranged, like so:
DDGG -- meaning two of the cars are defective
DDDG -- three cars are defective
DDDD -- all four cars are defective
Then you can find the probability that the cars will be chosen that way, and that becomes your sample space, or denominator.

Next we looked at all the possibilities that the first car will be defective using slots, which, if you remember, was from combinatorics.

Put that numerator on top of the denominator that we found earlier, and you have your answer.

Question Two:

  • Two jars contain red and green marbles. Jar I contains 3 red and 2 green marbles. Jar II contains 4 red and 3 green marbles. A jar is picked at random and two marbles are picked out of that jar in order. If it is known that the first marble is red, what is the probability of the second marble being red?
You can start this problem by drawing a tree, and writing down the probability that each possibility will occur - if that makes sense. For example, if you look at the picture above, you'll see that the first branch is 'Jar I' followed by 'Jar II', the probability of choosing either jar is one half. The next branches are 'red' and 'green', and the probability of choosing red in 'Jar I' is 3/5, while the probability of choosing green is 2/5. So on and so forth. Keep in mind that the marbles are not replaced after you take it out, so the sample space is different everytime you draw.

In this question, what we're looking for is the probability of getting a red after drawing a red marble first. That becomes the numerator, and the denominator consists of all the possibilities, as shown above. So you plug in the proper numbers in their proper places, and you have that question solved.

Question Six:

  • Susan sees her friend, Tim, at his locker with a worried look on his face. She asks, "What's wrong?" Tim has to open his locker and change clothes within the next five minutes. However, he has forgotten the combination to his new lock. He knows that the lock requires three different numbers. He also remembers that all of the numbers are odd, and all of the numbers are divisible by seven. It takes 10 seconds to dial a locker combination and 1.5 minutes to change clothes. Is Tim likely to be ready for gym class on time?
I think you'll understand how we solved this problem just by looking at the above picture. The '1 min 30 sec', written in pink, is how long it takes Tim to change.

Question Nine:
  • Three identical boxes each contain two drawers. In one box, each drawer contains a gold coin. In another box, each drawer contains a silver coin. The remaining box has a silver coin in one drawer and a gold coin in the other. One drawer is opened and a gold coin is found. What is the probability that the other drawer in that box also contains a gold coin.
  • Michael claims that the probability is 1/3. Jessica claims it is 1/2. Raymond says the probability is 2/3. Explain how each person may have arrived at their answer. Who is correct? Justify your answer.

The probability of choosing the right box is 1/3, and the probability of choosing either a gold or silver coin is 1/2, since there's two drawers. Obviously if you get the box that only have two gold coins, you're not going to get a silver coin. But I'm sure we're all smart enough to realize that.

The probability of getting a second gold coin is shown by:
[P(IG)/(P(IIIG) + P(IG) + P(IG))] + [P(IG)/(P(IIIG) + P(IG) + P(IG))]

If we plug in the correct numbers, we have:
[(1/6)/((1/6) + (1/6) + (1/6))] + [(1/6)/((1/6) + (1/6) + (1/6))]
[(1/6)/(1/2)] + [(1/6)/(1/2)]
(1/3) + (1/3)

Therefore, Raymond is correct. Michael could've gotten his answer by thinking that there's only one box that contains two gold coins, and the probability of choosing that box is 1/3. Jessica could've simply thought, well there's two drawers, so there's a 1/2 chance of getting a gold coin.

So that's the end of that. There's also questions in the slides (9-18) that we could do for practice. In the afternoon class we had a pre-test that we're going to go over tomorrow. There isn't going to be a test for this unit because we don't have time. Soo.. we're starting a new unit next week.

And that's that. The next scribe will be Joseph because he asked me yesterday. Whoot. Good luck to everyone doing the English Provincial Exams tomorrow! Remember to pace yourself and.. uhh, I forgot what else Mr. V and Mrs. Strecker said.. but.. yes!

Thursday, May 29, 2008

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Reminder: Roxanne, Francis and Elven - Your D.E.V. Project is due FRIDAY, yeah that's right, FRIDAY by midnight.


If you haven't already, just give Mr.K a quick email to have him add you to the D.E.V. Blog. Do it now! 'Cause I know you want to.

On a side note: That dot on the Visitor map is getting awfully large. Makes us seem like a big target =( Lol, just kidding.

Stay Phi everyone,

Rence ~ Out

Workshop Period

Ohhh my goodness. I nearly forgot I was scribe today because of the basketball game XD. (Go lakers!)

So the title above describes todays class. Mr. K was away again but no worries, he'll be back tomorrow...i think. Anyways, we were given a worksheet that we worked on in our little groups for the whole class. I thought the sheet had some challenging question like 1 and 2. I hope we go over those tomorrow. Yeeaah it was a pretty uneventful period other than that.

So I'll get to the point. Next scriibe is NELSA. "Beautiful" right Nelsa?

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

AND NO, Unicorns aren't real. :) Sorry to burst your bubble guys. Lol.

Stay Phi everyone,

Rence ~ Out

Wednesday, May 28, 2008

DEV Work Period!

Well, today we had Ms. Gonzaga come back as our substitute since Mr. K was away today. Surprisingly, and thankfully, there was no quiz or anything (which some of us may have been expecting since we usually get one when there's a sub). But anywho, who cares about that right?! So then, basically, this class was dedicated to working on our DEV projects. The due dates for them are really close that's why. We were also told to work on Exercise 44 as well, and that's it! Same deal for the afternoon class! Okay, that's it!

Aaaannnnd!!! Here's something Jamie is probably expecting (inside-outside joke?!)...she's not scribe by the way, Joyce is, and no I did not mean harm to her by putting her name in red!

Well then, that's it for today!!!! Did I mention the next scribe? Well for those who don't care about reading scribe posts and just like scrolling RIGHT TO THE BOTTOM immediately, the scribe is Joyce.

Tuesday, May 27, 2008

Scribe List

Cycle 5


Hi I'm Justus


Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Today's Slides: May 27

Here they are ...

Combinations of Permutations and Probability and Combinations O_o

Why hello there!
Benofschool here. I never scribed for a while, I was beginning to think that you guys forgot about me =(. Well anyways time for a scribe. Mr.K was later than usual today but he wasn't late for school which was good. Today's class was a workshop. We were broke up into groups like every other workshop class. Unfortunately the slides aren't up. Questions were put up and we were off...

Question 1
We had to find the probability of getting an ace or a diamond from a deck of cards. Fairly simple. Francis went up to answer it. To get the answer we just add the probabilities of getting an Ace and a Diamond and subtract the probability of getting the Ace of Diamonds. We add because it is an "or" question and we subtract that single card because it was counted twice when we calculated the probability of getting an Ace or Diamond. So the answer was 4/13.

Question 2
The second question involved a venn-diagram and venn skills from grade 11 logic in pre-calculus. The question was what is the probability of picking a person who doesn't like Dr. Pepper or the diet version if 7 people liked Dr.Pepper, 11 people liked the diet version, and 3 people liked both. So first we have to create a two circle venn-diagram. Rule is start with the inside and work outwards. So we put in the middle which means that 3 people prefer both drinks. Then lets work with on side. But remember from grade 11 we have to subtract the number of objects in the center from the separate values. If we look at the first question, we didn't count the Ace of Diamonds because the probability of getting an Ace or a Diamond is not mutually exclusive as well as the Dr. Pepper situation. If one thing occurs, the other probability can still happen. So that means we have to subtract the number of people that liked both drinks from the separate preferences. We get the sum of those numbers (4 +3+8=15) and subtract from the total number of students in the class (25-15=10). That will be the numerator in the probability and the sample space would be the total number of students. So the answer is 2/5.

Question 3
The 3rd question is about picking dresses. There are 15 dresses: 6 green, 5 blue and 4 yellow dresses. We wanted to know what is the probability of getting exactly 2 green dresses from picking 6. This question involves both combinations and probability. So 6 choose 2 because they are indistinguishable. multiplied by 9 choose 4 for the remainder of the dresses divided by 15 choose 6 which is the ways that we can choose 6 dresses out of 15. So we get an answer of 37.8%.
Question 4
Another combination question. A couple has 4 children that are about to be born. What is the probability of getting at least 2 girls. So what we do is find the probability of getting 2 girls first then 3 girls and finally all girls. Then we just find the sum of those values and we get the answer. The answer is:

That were all of the questions we did in today's class. Remember DEV due dates are arriving in a future now. Check the calendar on the right side bar to check for your due dates. Tomorrow's double will be used as a DEV work day. So bring the stuff you want to bring to work on your DEVs. That is all I have to say about today's class. The next scribe will be kristina. Good Night and see you all tomorrow!!!

Costly Test

So yeah, the reason why I'm up so late was because I just got back from the Glow in the Dark concert... and I've completely lost my voice. Anyways, I found this really cool story so I thought I'd share it and hope that you get a laugh out of it too.

A professor was giving a big test one day to his students.

He handed out all of the tests and went back to his desk to wait. Once the test was over the students all handed the tests back in. The professor noticed that one of the students had attached a$100 bill to his test with a note saying, "A dollar per point."

The next class the professor handed the tests back out.

This student got his test back and $64 change.

Monday, May 26, 2008


and no ...
- being cagey does NOT mean you're inside a cage alot ...

So I was chosen for being the scribe for today. Let me tell you guys something. REVENGE GETS YOU NOWHERE. That's why I'm always going to steal Paul's seat from now on. Thanks for choosing me as scribe Paul.

- SLIDE 2, 3
- Identifying events / dice diagram

- SLIDE 5, 4
- Probabilities involving "AND" and "OR"
- Testing for independence

- SLIDE 6, 7, 8, 9
- A test for cancer

- SLIDE 10
- A test for industrial disease

- identifying events / dice diagram


- and so we started the class finishing off what we had last done in class on Friday.
- a list of events which the class had to determine whether it was dependent or independent. Mutually exclusive or not mutually exclusive. I'll take two examples.

b) one card - a red card or a king - is randomly drawn from a deck of cards.
- because it is only ONE event we know that it must be independent
- a red card can also be a king. Therefore it is not mutually exclusive

c) A class president and a class treasurer are randomly selected from a group of 16 students.
- because we can only have ONE president and ONE class treasurer, the event must be dependent. One student cannot be the president and class treasurer. Well in this case it can't. Therefore it is mutually exclusive



I had actually found this diagram on another blog of Mr. K's class.

This was done to answer the question:
e) Rolling two dice and getting an even sum or a double
- It is independent because the first roll does not affect the chances of the probability for the next roll.
- As you can see on the diagram, you are able to roll a double AND get an even sum. There fore it is not mutually exclusive.

- Probabilities involving "AND" or "OR"
- Testing for independence


here is the slide ...


We compared the result of the probability of getting the flu shot and getting the flu. (0.10)
To the result of the probability of the seniors getting the flu (0.15)

There fore the event is dependent if you get the flu shot or not.

- A test for cancer


- The given question, information and possibilities.

0.5% of 1,000,000 = 1,000,000 x 0.005 = 5000
To find who does not have cancer you would subtract 5000 from 1,000,000
98% of the time, the test will be positive.
The amount of people who have cancer and will test positive would be ..
5000 x .98 = 4900
The amount of people who have cancer and will test negative would be ..
5000 x .02 = 100
982% of the time, the test will be negative.
The amount of people who don't have cancer and will test positive would be ..
995,000 x .02 = 19,900
The amount of people who don't have cancer and will test negative would be ..
995,000 x .98 = 975100
Adding it up the amount of people who tested positive for cancer
4,900 + 19,900 = 24,800
When only 4,900 actually do have cancer.

The probability that a person who actually does have cancer and tests positive is the amount of people who do have cancer divided by the total amount.

4,900 / 24,800 = 19.75%

Here are the tree diagrams to see how this all works out.

- A test for industrial disease


The probability of a person who tested positive and actually does have the industrial disease is much similar to what was done in SECTION C.

The probability of people with industrial disease and tested positive.
The probability of people with industrial disease and tested positive,
The probability of people without industrial disease and tested positive.

^ does that make sense?!

In other words, and in this case ..

0.0099 / 0.0099 + 0.0099 = 0.5%

because there is still a great chance that even if you do NOT have cancer, the result may still test POSITIVE

- not intentionally trying to scare any of you, or myself.
- not intentionally trying to offend anyone who is reading this.
- It's just a joke! Humor in OUR math class is NO laugh matter.

That's it, That's all ladies and 'gents! Have yourself a great day. Go out there and commit random acts of kindness! Cheers!

The next scribe will be ...benofschool

Today's Slides: May 26

Here they are ...

Sunday, May 25, 2008

Types of Probability and Exclusivity

So, this is a double scribe post, which covers the lessons we did on May 22 and May 23 (Thursday and Friday respectively).

So lets get started, shall we? First off is our lesson from May 22...


So last Thursday, we talked about the two types of probability. There are dependent and independent probabilities, and thats what we'll explore here.

Slide Two:
The first thing we did was make a tree, which we've all done before. This tree displays all the possible outcomes of flipping two coins. Easy right? We know probability of each outcome is 1/4 because there are four possible outcomes, and each one is one of those four.

Slide Three:
Moving forward, we have "an entirely different question that is not at all similar to the one we
just did." Please note the quotation marks. Because, this question is very similar to the one we just did. Instead of heads and tails, there are reds and blues. But this is similar but not the same. Why? Because unlike our previous probability question, this question deals with a dependent probability, whereas our previous one dealt with an independent probability. Remember, this lesson was supposed to be about dependent and independent probabilities?

But Paul, you say, what the heck are dependent and independent probabilities anyway? Im confused!

Well, to put it simply, independent probabilities are probabilities that are affected by the steps before them. Why don't you think about it in context, like this:

If you flip a coin, and then you flip another entirely different coin, does the chance of you getting tails on the second coin change because you flipped the previous coin?

Answer? Of course not, excluding far-out possibilities such as your hand got so tired flipping the previous coin that you put less energy into flipping your second coin and gave it 0.001% better chance of landing on tails. This is an example of an independent probability, where both probabilities are, well, they're independent of each other. Makes sense, right?

However, with our second question, our probabilities are dependent. Why? Because when you choose a marble (wearing a blindfold, earmuffs, and nose plugs so you cannot possibly see, hear or smell which marble you're picking), you also
remove that marble, thus changing the probability of you getting the same colour marble when you draw a second one. This is all displayed in the slide two tree, where if you drew a red marble and removed that, your chance of drawing a red marble again the second time is 2/5, whereas if you drew a blue marble the first time, you have a 3/5 chance of getting a red marble when you draw a second time. See? They're different, which is exactly why this is a dependent variable.

And that is basically the gist of this entire lesson, so I'll just summarize the rest of the slides...

Slide Four:
Okay, something changed in this question, and I'll give you a cookie if you can spot it.

Alright, found it? Here, have a cookie. You clearly noticed that our question changed in that the marble is no longer simply discarded (or "thrown out the window" as Mr. K would say [hey, I rhymed]). Instead of removing a marble when we draw it, we just put it back. Because we put the marble back, our probability doesnt change, and it becomes independent. Now that its independent, all you'd have to do is change those R's to H's and B's to T's, and you've got your original heads and tails question! Magic right?

Which brings up another point Mr. K mentioned. A lot of the time, your question boils down to stuff much like the heads and tails question, just wrapped up in a pretty little. Remember that (say it to yourself five times, or something), and you'll find things will be a lot easier.

Slides Five & Six:
Just a rehash of what I already wrote, basically definitions of what dependent and independent variables are (in simpler, less rambling and rant-y [is that a word?] terms).

Slide Seven:
A few questions, dealing with figuring out whether or not the question is describing an independent or dependent probability. Simple, right? I'll explain why quickly.

a) Independent because the outcome of the coin toss doesn't affect the outcome of the dice roll (again, this is factoring out such insane possibilities and how your possibility might be changed if you did both at once, or something along those lines).

b) Dependent because when you draw the first card, you remove it from the deck and change the amount of cards in the deck. So when you draw your second card, your deck is one card smaller than it would be had you not removed the first card, and you can no longer draw that particular card. Thus, the first outcome affects the second outcome.

c) Independent because now your deck is static and the probabilities always remain the same (excluding time travel and such nonsense).

Slide Eight:
Slide eight takes our second slide's question one step further by adding another coin flip. Nothing terribly new here, it just makes the chances of getting a particular combination smaller. I won't bother explaining this in great detail, but I'd like to point out that the final probability of an outcome (say, HHH, or flipping three heads in a row) is equal to the product of its previous steps probabilities (in other words, 1/2 * 1/2 * 1/2 = 1/8, not a coincidence), where each coin flip had a 1/2 chance of landing Heads.

Slide Nine:
Now to follow our trend of making small changes to our previous questions, we take our old question and put a fresh splash of paint on it and volia, we get something entirely new. Fortunately for me, all I have to do is take our old solution, paste it on to our new slide and say Boy = H, Girl = T. But wait, our question says, whats the probability of mom and pop getting exactly two girls and one boy. So what do we do? We find all the answers that have two girls (T's) and one boy (H's). Since there are three possible outcomes out of 8 that have two T's and one H, we determine that the chance of getting two girls and one boy exactly is 3/8.

Slide Ten:
Ignore this, this was my poor and not at all though out attempt at solving the question, and it is completely wrong.

Slide Eleven:
Our final slide, with the correct solution, courtesy of Kristina. Another variation on our girl/boy question, except with a different amount of steps and different conditions. So for the first draw, if we draw a blue marble first, our bag with 3 red and 3 blue becomes a bag with 4 red and 2 blue, and vice versa. So when you draw a second marble, you either have a 1/6 or 2/6 chance of drawing a blue marble the second time depending on the colour of the first marble you drew.

And that concludes our lesson on the Types of Probabilities.


And without delay, I give your my scribe post for our May 23rd (Friday) class on Mutual Exclusive Events.

Slide One:
Contains a fox that is entirely unrelated to probability. However, I would like to dub him (or her) the Probability Fox, just for fun.

Slide Two:
A simple slide with a simple question. If you have 56 listed and 144 unlisted phone numbers, you have a total of 200 phone numbers. So if you want the probability of choosing a listed phone number from those 200 total numbers, you have a 56/200 chance, or a 28/100 or 7/25, or .28, or 28% chance.

Slide Three:
Okay, so we have a horse named Gallant Fox (makes total sense, right?) and another horse named Nashau. Gallant Fox runs a race and has a 2/5 chance of winning. Nashau runs an entirely separate race and has a 1/3 chance of winning. What is the probability that:

a) Both Nashau and Gallant Fox win their respective races. That's easy, we simply multiply their probabilities together and get 2/15. Remember earlier how our final probability was the product of all the probabilities of the steps before it? This is just like that, where our first step is Gallant Fox wins (2/5) or he loses (3/5), and then our second step is Nashau wins (1/3) or he loses (2/3). Since we want the path that has both Nashau and Gallant Fox winning, we want to multiply 1/3 by 2/5, which gives us our final probability for that outcome.

b) So this is just a in reverse, and instead of taking the winning path, we take the losing path. So 2/3 * 3/5 = 6/15.

c) Well, there are two ways we could solve this question. We could find the probabilities for all the routes where one of the horses wins their race and then add those together. But thats just long and tedious, and considering what we already know, we have a much simpler solution at our doorstep. The question asks us what the probability is that one of the horses wins, so as long as they both don't lose, we're good right? Wait, don't we already know the probability of them both losing? Well... if we know that, cant we just take all the possibilities and remove the ones we don't want to get the ones we do want? Something like this... 15/15 is our total possibilities, but 6/15 of them are ones that have both our horses losing their races. So we just subtract 6/15 from 15/15 and... our answer is 9/15.

Slide Four:
This slide was probably the most confusing question we had encountered in a while, and it took a while to figure out. But basically, it goes like this:

Chad wants to meet his girlfriend in either the Library or the Lounge. If he goes to the Lounge, he has a 1/3 chance of meeting her (apparently Chad has poor arranging skills as he still has no idea where they're meeting despite the fact that this has all been pre-arranged, but I digress), but if he goes to the Library, he has a 2/9 chance of meeting her.

a) What is the total probability he'll actually meet up with her, either in the Library or the Lounge. Because its an "or" probability, we add the probabilities together (the Library OR the Lounge), and get 5/9 chance they actually meet.

b) Now this is where it kinda gets confusing. The probability of them not meeting is 2/3 * 7/9? Wait, just were adding just a moment ago, why did we suddenly switch to multiplication? Now we're dealing with an "and" probability, where she is not in the Library AND not in the Lounge. Thus, we get our 14/27 chance they don't meet.

Slide Five:
This slide is a little demonstration on how Chad cannot go to the Lounge AND the Library. Thus, going to the Library or going to the Lounge are mutually exclusive events, which is what this lesson is all about.

Slides Six & Seven:
These slides give you the basic definition of Mututally Exclusive Events, and some examples. These are pretty self-explanatory.

Basically, if you have one event that makes another impossible, then they are mutually exclusive. You cannot turn left and right at the same time, you cannot be facing North and South at the same time. Facing North is mutually exclusive to facing South.

Slide Eight:
So we want the probability of drawing a King or a Spade in a single draw from a pack of 52 playing cards.

Our first event is to draw a King. Our second event is to draw a Spade. Drawing a king has 4/52 chances, and drawing a Spade has 13/52 chances. However, because of the one card that is both a King and a Spade, we must remove that card (1/52). So our formula:

Probability(Event A or Event B) = Probability(Event A) + Probability(Event B) - Probability(Event A and Event B)

Means this:

The probability of Event A or Event B is equal to Event A's probability plus Event B's probability subtract any Probabilities that fulfill both Event A and Event B.

And inputting our data, we get the Probability of drawing either a king or a spade as 16/52.

Slide Nine:
And finally, a couple practice questions. Drag'n Drop Baby!

a) Independent because the marble is returned, mutually exclusive because you cannot draw a red and blue marble at the same time.

b) Independent because there is only one step, not mutually exclusive because it is possible to draw a red king.

c) Dependent because once a president or treasurer is selected, the number of people in the selection pool changes and thus so do the probabilities for the next selection. Mutually exclusive because the president cannot be the treasurer at the same time (they are removed from the selection pool once they have been chosen).

d) Independent because there is only one step, mutually exclusive because there is no card that is a red king and black queen.

e) Independent because the second step does not change the probability of the second, mutually exclusive because its not possible to get an even and odd number at the same time.

And that concludes this super long double buttery extra flavoured scribe post, again posted in the middle of the night and written in the dark. I'll see you all in a few hours. Please feel free to point out any mistakes, I'm fairly sure I made a few.

And because he asked so nicely, my next scribe will not be Thi.



I'll just tell you guys in the morning since nobody will read this until then anyway. Good night, and have a pleasant tomorrow.

P.S. That Google translate thing works, atleast, I think so (I cannot read Japanese and verify this). However, from this I have learned that "Eleven" is "11" in Japanese.

Thursday, May 22, 2008

Today's Slides: May 22

Here they are ...


Once again, it’s been days since I’ve gotten access to the blog. But I still manage to get things done. Conics—the in-depth look at the aspects of a cone. It sounds like I’m writing a feature article on it or something. I sort of enjoyed this unit, it had its moments. It was pretty reminiscent of the days when I was master paper folder in origami club. By the end of emphasizing each shape that evolved from a cone, I simply wondered, “what happened to just making cranes?” I also commend Mr. K for his innovative [yet deadly] ways of disarticulating a cone with a shward.

This may have been a short unit but I really have to admit that I learned a lot. I didn’t know that so many shapes could come from a cone, or two cones to be precise. A “hypercone”, just kidding… The proper term of course would be the double-napped cone that gives way to a hyperbola, a parabola on steroids. Sarcasm, my native tongue.

The class learned each of the formulas and also learned how to graph the formulas. Call me juvenile, but I still have trouble with the grade 12 version of the parabola. I think the 4p always throws me off. I also feel like using the other formula from grade 11 though. But using that idea again should be avoided since thinking would not be lateral. This is why practice makes perfect. I just need to do a little adapting. Otherwise, I’d think I’m ready for the test today. Just keep my hocus pocus my focus, look it’s a locus.

BOB: Conics

WOAAAHH!! I almost forgot to do my BOB again! Sorry this is really really late XD

Okay, this unit was one of my favorite units. I actually enjoyed graphing for this unit and I thought that the concepts were very simple. It was a good break from the Counting unit, which I didn't really like much. I also enjoyed doing all that folding during class, such easy homework! This unit was also full of laughs, we really couldn't seem to stay serious most of the time. As for the parts that I found the hardest for this unit, I would probably have to say that the hyperbola and ellipse were definitely the most difficult, although not by much. I got confused at first with how to find the foci and the asymptotes (hyperbola) but then it all made sense after some explaining. I also remember messing up graphing my vertical hyperbolas since I switched around the transverse and conjugate axis. All in all, tis was a good unit, now I got to get ready for school! Good luck on the test guys :P

Wednesday, May 21, 2008

BOB on Conics

Hi guys this is Richard and here is my bob on the unit of conics.

This unit was not as hard as the other units there were some parts that i got really easy like the ellipse. I probably found that this topic was easiest because i had to scribe for it or it is just the simpler out of the conic shapes. Another thing that was cool was that we got to do a lot of fun folding. That led to the geometry of the conic shapes.

There were also topics that i found difficult for example the parabola. I think that the main reason i don't understand it is because i was not present for that Pre cal class. I almost forgot that the graphing of the parabola is also hard. i also felt the same was that Francis felt on that Pre test. I was also puzzled by the transforming parabola question . Well that's just me ah ha

Richard signing off.

Conics Bob

Ahhh Conics. Too bad Mr.K didn't show us any more of his samurai skills with this metre-stick-katana.

To the point. I will admit, like every section in this course, I had troubles in one way or another. The origami was cool, as it went with the whole Japanese theme Mr.K was trying to capture (TOYOTA-HONDA-MITSUBISHI-MAZDA-YOKOHAMA-SONY-etc...) Like every other type of graphing, I don't like to graph. Granted it does give me a better look at whats happening and I can tell what's happening, but like everything else in life, just because you know how to do it doesn't mean you're supposed to like it. Like graphing.

I liked how parabolas, circles, ellipses and hyperbolas had their distinct way to tell each other apart without having to think hard but rather just look at what's switching to make it graph that way. But, that's probably about it. The math was pretty simple once I took the time to sit down and study it all, like I am tonight.

I didn't like word problems however, how they make it seem like it's supposed to be something it's not, but yeah, I know, I'm just over complicating it.

I will be deep in study for a while because well, exams are closing in surprisingly fast. OH and don't forget about your DEV projects.

Rence ~ Out

BOB on Conics

This unit was quite short, but I'm not celebrating. Personally, I need a bit more time with this unit, only because I don't think we went far enough with transformations that can happen with each conic shape. On the pre-test I saw that one question where the graph of a hyperbola was stretched 4 units, and moved down 1 unit, I was sitting there staring at the page for like 5 minutes, forgot all about those transformations. Plus, I don't think I can remember all those equations such as the equation of a circle, ellipse, hyperbola and the new parabola equation. Hopefully we'll get a formula sheet, or some other type of aid. On top of that, I don't fully understand the total anatomy of all these conic shapes, such as the major, minor, semi-major, semi-minor, conjugate, and transverse axis', and some, I suppose. I still don't exactly know about graphing these guys either. Other than that, I really enjoyed folding the paper, being interactive in a class is always a plus.

Good luck on the test everyone.

Until next time,

Bobbing on Shwards and Things

So, the infamous conic sections unit was completed late last week, and this would be a bob for said unit.

Where to begin? the good? the bad? the ugly? Lets start with the ugly

THE UGLY *dun dun dun*

-GRAPHS! Yes I'm terrible with graphs but I am getting better at them, thanks to a certain "perspective." More on that later though.

The Bad

Not to much here this time, thankfully.

-One of the most difficult parts of this unit is how most of the equations, have another equation which is nearly identical. This made memorizing/remembering which was which difficult at times.
-Going from standard to general form, and remembering which is which was often a pain.

The Good

-Short unit
-Easily understood by used of "visual" tricks (what I meant in the ugly section. By thinking of what the graph of some equation looks like, the question is often made much easier.
-Overall the concepts in this unit, and the difficultly level of the questions was not very demanding.

So yeah, there is it, my small and not very impressive bob post. :P Now I'm back to studying haha.


An Introduction to Probability

Alright young fellers, american idol is almost on so I'll try to finish this before it starts but if I don't..I'm sure I can do both! (:

Slide One
The first thing Mr. K asked us today was, "What is the probability of the sky falling?" (chicken little yay!) Each of us were given about a few minutes to write down our answer and we all shared our answers to each other. *refer to the first slide* Most of us came up with different numbers but some of us had the same answer as others. Most of us just chose a random number, like me, but some of us also took some time to why they choose that number. After this dicussion, he then asked us what the numerator and the denominator represented. The numerator is the number of chances of the sky actually falling while the denominator is the set of all possible outcomes.

Slide Two
The next slide explains the defintion of an inderterminate form. So if we take any number such as 6 and multiply it by one then divide by one your answer would be 6. And if you take the 6 again and multiply it by 1/10 then divide by 1/10, you still get the answer 6. Continuing the pattern, you'd eventually reach zero. So you would multiply 6 by zero then divide by zero. Therfore it is in an inderterminate form. If you recall on slide one, 0/0 and 0/∞ are examples of indeterminates as well as ∞/∞ and ∞/0. 1/0 is not in an indeterminate form, it is however, an undefined value.

Slide Three
The next slide includes various vocabulary words which we should know for this unit as well as different examples to explain the meaning of each words.

Slide Four
Calculating the Probability of Event A
This slide includes the formula P(A) = n(A)/n
P(A) - Denotes the probablity of A
n(A) - number of favourable outcomes (number of accourences of A)
n - sample space (the total number of possible outcomes
Probability can not only be expressed as a fraction but also as a ratio, decimal and a percent. However we will not be looking at it as a ratio in this unit.
IMPORTANT TO REMEMBER: Probability of any event is always a number between 0 or 1. This means that if your answer is bigger than one or it is a negative, then your answer must be wrong.

Slide Five
The fifth slide is also a continuation of the different vocabulary. In this specific slide we dicussed mainly about the complimentary event. The example he asked was, if there was a sudden death single tennis tournament and there were 54 players, how many games would be played if there could only be one winner? The easiest way of coming up with the answer is asking yourself, "how many losers would there be" The answer is 53 because only one player could win. Another example is using a coin or a die. Basically, you're looking for the opposite. So if P(E) = a then P(E1) = 1-a.

Slide Six
In the next slide, we had to determine the sample space when a fair die is rolled once. Since it is possible to roll either 1, 2, 3, 4, 5 or 6, the answer to the question is {1, 2, 3, 4, 5, 6} *note: if you are asked to determine the sample space you must use brace brackets { }*
The next question was to determine the sample space for rolling a six sided die and flipping a coin. You could either use a tree diagram or a chart to find the answer. In this case we first used the tree diagram to find the answer. The first two branches consited of Heads and Tails and each had 6 branches to indicate the numbers on the die. Each number could either get H1 to H6 or T1 to T6 so the answer was 12.

Slide 7
We then had to determine the probability of rolling a 2 when rolling a fair die. The answer is 1/6 because there's only once chance of getting a 2 out of 6 different outcomes. We write this as P(2) = 1/6
In the next question it asked to determine the probability of getting a head and an odd number when rolling a die and flipping a coin. In this case we used the previous chart we made on slide 6 to find the answer. We then found that there are 3 possible chances out of 12 possible outcomes which also redues to 1/4. *note: remember to always reduce your fractions.

I believe I covered most of what we learned and talked about in class, so that means..I'm finished! (: This scribe actually took longer than I clearly expected considering I was running back and forth to my room and my sister's to watch American Idol. But enough of my life..toodles everyone, American Idol isn't over yet!

BTW..Homework is on slides 8-10 and don't forget to study for tomorrow's test! Goodluck and Ciao! OH and.. Next scribe is... "what it do PAUL wall" (:

BOB For Conics

This was a short unit that I was hardly there for because of the music trip to Moosejaw, which was fun, buut now it's time to get back to school and all that piled up work that I have to catch up on. Whooo.

Missing classes for a long period of time has never been a good thing, especially if it's your weakest, and most especially if it's pre-calculus. But some things can't be helped, I think. Well, either way I get in trouble soo..

The concepts were all there for me. Probably the hardest part of this unit was breaking down the question and figuring out which numbers I could use to make an equation. Of course graphing was hard (when is it not hard), but after you have that image, it makes everything so much easier. Everything could've probably been more straight forward if I was actually there (I'm not much of a self-learner when it comes to math) and that's just my fault I guess. Honestly, I'm not looking forward to this test, but I don't know, I think I've done what I can. Hope for the best I guess.

Hahah, as if I'm the very last one to BOB (I think so anyways), but better late than never. Good luck everyone.

Today's Slides: May 21

Here they are ...

Tuesday, May 20, 2008

BOB: Conics

Time to Bob! (:

I'm quite happy that this unit was pretty short but I'm not going lie because truth be told, I did have some trouble in a few areas. The most trouble I had in this unit was graphing the parabola and I have no idea why I find that difficult compared to the others. The circle, ellipse and the hyperbola were probably the easiest because it was sort of similar to each other. I suppose I had trouble with the parabola because the formula seemed much different compared to others because it included 4p. Hopefully, with a bit more practice I'll feel much more comfortable graphing them. (:

The best part of the unit was folding the paper and visualizing the geometry in each conic section. I would probably have a harder time understanding the whole unit if Mr. K just explained each conic section without folding the paper. Though it took a lot of time and made me quite frustrated beacause the paper wouldn't fold the way I wanted it to, it helped me a lot to understand the anatomy of each conic.

Overall, this unit was pretty straight forward and hopefully I'll do better on the test on Thursday than the pretest we had today. *cross fingers and toes* Goodluck to everyone! Ciao! (:
PS. school's almost over! yay!!

Conic Workshops and Pre-Tests

This will be basic and short as I don't really have any time to spare.

Our first class consisted of
  • A workshop on Conic word problems

Our Second class consisted of a Pre-Test.

Our first class was pretty straight forward and Mr. K just put us in groups to solve a number of word problems. When we started on the first one. Some of us had an answer, but we didn't put it up, whether it was right or wrong. Mr. K pointed out that we were afraid to get out our Ideas because most of us, if not, all, are afraid of getting the wrong answer, and we'll be "bad people". He also pointed out that we shouldn't be scared in getting our idea's out, because it wouldn't make us "bad people" and if everyone was always right, we wouldn't need to attend the class. Because how else do you learn? By making mistakes. So take that into consideration.

Anyways, we worked on a few questions but we sometimes made it a little complicated by looking at it wrong. Remember Mr.K's block of wood. Look at it in different perspectives. I won't really go into detail, and you can more or less pick your parts and pieces from the slides.

In the afternoon class, we had a Pre-Test, and again, the answers are posted up on the slides. Our test is on THURSDAY. Do recall that Mark has kindly posted Links for us to brush up on our Conics. I know I'm gonna hit those links up later, so you should too :) Mr. K also has practice exam's and exam information in his LINKS section (because you know, the exam is in a bout two weeks roughly) Sorry this is short and not into detail but I have to go. Kthxbye!

Oh right.. Scribe.. The next scribe shall be... ROXANNE :)

Today's Slides: May 20

Here they are ...


Here is my BOB for the unit Conic Sections.

What more can I say than .. "I've missed too many classes" ..
Although I've missed probably more than half the classes for this unit, it really seems like a fun unit. It seems like a unit that involves lots of algebra and equation handling. Anyways I'm hoping to participate in this last couple of days in this math class. Good luck to everyone! Especially me. I've read the blog, but it might not be enough. Hope I can keep up.

- This is Elven, cheers.

Monday, May 19, 2008

Conics quizzes

Well, since i heard that the test for this unit is coming very soon I decided to compile a bunch of links that will help all of us to study, so without further ado here it is:

Link 1
Link 2
Link 3
Link 4
Link 5
Link 6
Link 7
Link 8

I hope that those links will be very useful to all of us on the upcoming test and especially the provincial exams which is only 20 days away. Leave a feedback if any of those links doesn't work.


Scribe List

Cycle 5


Hi I'm Justus


* Note; Rence is crossed off as he is scribe today [May 20th].

Quote of the Cycle ;

"We say we love flowers, yet we pluck them. We say we love trees, yet we cut them down. And people still wonder why some are afraid when told they are loved."

Stay Phi everyone,

Rence ~ Out

Bob for Conics

*Sigh* What a nice time to have a long weekend. Now I can bob early.

Anyways, this unit was really fun! Not to mention it was also short and simple. What really helped make me understand this unit more was the folding exercises with the ellipse, parabola, and hyperbola. Yes it really did help I'm not just making my bob extra fluffy this time XD. Folding was great for homework assignments too! I usually have trouble with anything that deals with graphing but not in this case. I found it really simple and *coughs* I enjoyed doing them. But what I'm worried about is mixing up all the standard formulas for the ellipse and hyperbolas because they're so similar. Other than that there wasn't anything that troubled me surprisingly and I'm really confident with this unit! Hopefully I'll still feel the same way about this unit after we get through word problems.

Sunday, May 18, 2008

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.

Conically BOBING

The Conics Unit was pretty simple. The thing that really helped me in this unit was actually visualizing the geometry of each conic section. I started to see how each formula was related to the sections. Overall I'm doing fine so far in the unit. Just like what happened during the quiz. The circles and ellipses were easier to work with than the parabolas. I guess it was because that the
circle formula didn't change at all and that the ellipse formula
was very similar to the circle formula. The parabola
formula was a bit of a change. The a became 4p
and there are 2 binomials now. But it is like
what they all say, "Practice makes
perfect!" The Hyperbola
is still pretty new to
me but what I
that it
used a lot
of our older
analytic geometry
skills to graph the section.
It used the linear graphing skills
as well as many others. Now for my
thoughts about the class. I'm sorry if I am
being a bit mean but I would like it if we stayed
on task a bit more. It just seems like we were fooling
around quite a bit excessively, especially during the second
period. We would get a lot more done if we stayed on task. I don't
mind if we do have a little fun. But it would be great if we toned it down
just a tad. That's all I really have to say. I'm looking forward to the DEVs and
the exam. I'm not really nervous about it at all. As you can see I'm having some fun with my BOB. I am also looking forward to the word problems with conic sections as well as the test that is coming up. So until next time good luck on the test and YAY Calculus is going to work out. Calculus class of the future. Bye and TABBERNACK

Saturday, May 17, 2008

WORKING WITH A(N) HYPERBOLA - CONCLUSION [beware shameless advertising.]

Sorry the post is late, but I guess this is why I needed the weekend. Thank you Justus. Nice doodle there, by the way. I don’t think I can incorporate images of the slides in my post, but I’ll still try to use some form of an outline and I’ll also parody a certain [we]blogging method. Yes, I said PARODY. I thank Richard for this.

On Friday’s class, our goal was to conclude our lesson about the hyperbola. We leaed to: rn
  • Write the standard forms of a hyperbola [horizontal and vertical]
  • Find the similarities and differences between vertical and horizontal hyperbolas
  • Find the Pythagorean theorem property in hyperbolas
  • Finally, apply everything we’ve learned by graphing and finding the coordinates required in every hyperbola.
  • As well as that, we also had a sort of English epiphany, but we’ll discuss that later.


On Thursday, the standard formulas of a hyperbola were:
HORIZONTAL: [[(x – h)2 / a2] - [(y - h)2 / b2]]
VERTICAL: [[(y – h)2 / a2] - [(x - h)2 / b2]]
As soon as the class first caught a glimpse of the equations, we wondered why it was the same as the ellipse, but as we examined closely, there were differences as well as there were similarities between them.



  • they have the same terms
  • they both equal 1
  • both have the same denominators
  • both have square binomials
  • (h, k) indicates centre


  • numerators have different orientation
  • numerators have opposite signs

We also emphasized the differences between the ellipse and the hyperbola by cloning the slides from the ellipse and altered them. We rewrote the formulas and redrew the diagrams, interactively of course. There was even a point when Mr. K was going to redraw a diagram that was on the other page. For the sake of elegancy that rubbed off on the class, they argued for him to clone it, cut it, paste it and shrink it…[that’s sounded like I was singing a new version of “Technologic” or something haha DAFT PUNK man. All the way.] The point is “Just Shrink It”, man. That should be the new Nike slogan. Jeebers, I shouldn’t call out on Mr. K, it’s not even June yet. By the way, it was deemed necessary to rotate the “O” in the graph.

Anyhow, we then realized that a2 [pronunciation: “ay squared”] could be smaller than b2 [bee squared] after seeing the animations that proved so. These diagrams revealed what would happen if there were variations in the lengths of a [semi-transverse] and b [semi-conjugate]. There were also additional animations for the variation of points h [ey-ch] and k [kay]. “h” moves center and hyperbolas horizontally because it is an “x” coordinate. “k” moves center and hyperbolas vertically since it is a “y” coordinate.

The class then asked themselves this question: if a2 can be greater than b2, how will a vertical graph be distinguished from a horizontal graph? The answer lies in this statement:

If the “x” coordinate is positive and the “y” coordinate is negative [y being subtracted from x], then the hyperbolas are HORIZONTAL. If the “y” coordinate happens to be positive and the “y” coordinate negative [x subtracted from y], then graphically, the hyperbola is VERTICAL.


This is where the English lesson came in. Mr. K insisted that it was “an” hyperbola since using an in a consonant was true in “an hour”. We realized that the English language is at times illogical and contradictory. If all of the rules were followed, we would either choose to have the silent letters or disregard them in the words: “talk”, “milk” and “walk”. Imagine what the Hulk would sound like. OMG Edward Norton as the new hulk. Unexpected, but I'm okay with it. By the way, you don't want to make Mr. K mad. He'll turn green [environmentally friendly?]

I'm such a subliminal message. But let’s not derail the train of thought here. The formation of a(n) hyperbola on a graph depends on a rectangular box, which forms a right angle triangle inside. The legs of the right triangle are composed of:

semi-conjugate + semi-transverse = c [the hypotenuse also equal to the length of the center (O) to a foci point.]

simply, this is where the Pythagorean [pie-tha-gore-eeiy-an] theorem is applied; a2 + b2 = c2


Then we spent the remainder of the class solving a series of hyperbolas in two different perspectives: graphically and symbolically.

To start we find the standard form by multiplying by (1/9) and (1/25) on each side
leaving us with [(x2 / 9) - (y2 / 25)] = 1

From this equation, we can derive the rest:

Center (0, 0) because nothing is being added to either of the x or y coordinates.

TRANSVERSE AXIS is found by finding the square root of 9 (a2) and multiplying by 2 (2a), in this case 2a = 6

CONJUGATE AXIS is found by finding the square root of 25 (b2) which is 5 (b) and then multiplying by 2. 2b = 10

VERTICES are the distances from the center to the vertex of both parabolas in the hyperbola. It helps if you know the length of the semi-transverse axis the coordinates of the vertices are (3, 0) and (-3, 0)

To find the FOCI, we start by recalling that they are the same distance away from the center as the hypotenuse of the right angle triangle formed by the semi-conjugate and semi-transverse axis serving as the legs of the triangle. Therefore we use a2 + b2 = c2.
In this case, 9 + 25 = c2
c2 = 34

Find the root of that and add/subtract from the x coordinate of the centre because this is a horizontal hyperbola, resulting in (root 34, 0) and (-root 34, 0)

Finally the asymptotes that help graph the two parabolas in a(n) hyperbola, it is important to recall grade 10 pre-cal where we spent time learning about oblique lines and how we find the slope. y = mx + b or simply "rise over run" and use a corner of the "box" made, meaning that we use the two legs of the triangle which happen to be the length of the semi-transverse axis and semi-conjugate axis. Rise 5, run 3 in both directions.
ASYMPTOTE = (5/3) and -(5/3)

Sorry if the post is messed up, but the library computers are timed... As for the next scribe.. I can't doodle, but nevertheless I dub with my very own schward the next scribe Rence.

PS. [post scribe, hahah] I know I said I was a hopeless apolitic when it comes to school elections and that I'd rather not vote. But I still encourage all of you to vote JOSEPH vice-prez.

Thursday, May 15, 2008

Hyperbolae Unveiled...Sorta

Alright guys, no shenanigans this time, I'm tired, just got off work, and would like to sleep. So I apologize if I'm a little bit frank, and a little less funny an animated in my scribe post compared to my usual ones. Hopefully the quality is still there though, if not, I'll fix it up tomorrow or w/e. Alrighty, here we go.

*note* links added under appropriate images for full sized goodness. Like I mean, it wont matter on the smartboard, since that things huge, but at home I doubt everyone has like 70 inch moniters, so ja. Thats why there there. Why arent they hotlinked or w/e its called? Cause its 1:24am haha.

Morning Class

So this mornings class wasn't much of anything special, which is kinda different from the norm (kinda paradoxical I know.) Anyways we started going through Mr. K's slides, and we solved the little equation for one of the friendly Ellipses which can be seen on the following slide.

As always, fully more large sized versions of these slides can be found in the slides themselves. Which Mr.K posts quite consistently :P

Moving on. After that we filled out a health survey. That basically took up the rest of the morning class. However, we did have a moment to do a quiz, which was marked the following afternoon class. If you were not there *cough cough* then talk to Mr.K, see if you can write it. The solutions are on the slides. If you were here, then you don't really need an explanation right? Cool.

Moving on.

Afternoon Class

The afternoon class was slightly more work oriented, and was not interrupted by some unnecessary forms/paperwork.

We began by correcting the quiz thinger from the first period.

Some usefull things to take out of that include the following.

How to distinguish Various conics from their equations.

-If the Equation has an x2 term but no y2 term it is a Horizontal Parabola
-If the Equation has an y2 term but no x2 term it is a Vertical Parabola
-If the Equation has an y2 AND an x2 term, with coefficients that are THE SAME, it is a Circle.
-If the Equation has an y2 AND an x2 term, with coefficients that are DIFFERENT, it is an Ellipse.

Remember those :)

Next we took out our previous nights homework (that being measuring, and finding the difference, of the lengths from one point on a branch of the hyperbola, to both foci. That will make more sense when the image comes up in a moment.)

The results of taking up that homework are shown as follows.

Okay, I'm going to try to explain this as best as I can, but like I said before, I just got off work, I'm crazy tired, not running on anywhere near enough sleep and a host of other not so great things. So if I phail (yes thats a "ph" fail. That doesn't mean an acidic fail but rather, means a really bad fail for all you en zero zero be's out there.)

So, the homework was to pick a point on one of the branches of the hyperbola, and measure that distance from the point, to both foci (that is to say, measure from say, point P, to F1, and then from point P to F2). We then too those values, and found the difference, (PF1-PF2). Voila! homework done. By now it should have come as no surprise that the result were values which should have been basically the same.

Once we did that, things started to get tricky. The next thing we did, was find the Vertices (sp?) of the hyperbola. To do this, you take your ruler, and line it up so it goes through both focus points. The POINTS at which the line made by the ruler intersects with the curve of the hyperbola make up it's vertices.

Okay, so if you followed my last paragraph's directions correctly, you should end up with two points at what are the vertices of the hyperbola. The next step is to connect those two points with your handy dandy ultra straight ruler. This creates what we called the TRANSVERSE AXIS. This axis can be described as the line which connects the two vertices of a hyperbola, and is seen AND labeled in green on the slide crop up there. Got it? I hope so. Moving along.

Next, we bisected this transverse axis.

"How do we do that Mr.k?"
"Well I'll show you."

Thats a rough translation of what probably everyone was thinking. Except the Ogre 1 and 2 of our class. They knew the answer already, (Outside joke.)

So to bisect the transverse axis, we folded our papers for one more time. This time, we folded one focus point, directly on top of the other. Thus we obtained a crease which defined not only the bisector, but the perpendicular Bisector of the Transverse Axis. We see this as a vertical red line on the image above. Lets step things up a notch (I know Mr. K did.)

So, do you remember the Transverse axis? and how we just cut it in two? Were going to take this line, and modify it a little bit. Using our rulers again, we measured the line from one of the FOCI (in this case F2) to that vertical line (which we now call the CONJUGATE AXIS.) This gave us some length.

Now this next bit is tricky, we took the length we measured from the PREVIOUS STEP (that being from the conjugate axis, to Focal point F2) and moved it so that one end of it was on the (whatever the singular of vertices is. Verticii?) of the nearest branch, and so that the other end of the length lies on the Conjugate axis. Since I bet your saying "what the frig" I made a quick little paint mockup to get the point across, or rather, to help you "see" what I mean.

Okay, so as I wrote on the picture, the light green is the length Conjugate Axis to F2. Then by moving it like I mentioned before, we get the makings of a triangle.

Step 3. Measure the distance from the point where the light green line meets the conjugate axis, to where the dark green axis meets the conjugate axis. This we can call length b for explanations sake. The next step is to take this length, and go from the intersection of the Transverse axis and the conjugate axis, (aka the cross in the centre) and "b" units down. I'll show you what I mean again.

So, as I wrote again, the light blue, and dark blue lines are the ones I was talking about before, and they are of equal length. Thus we get those points G1 and G2. This brings us to our next step.

By drawing perfectly 90 degree lines from the vertices (going through them, vertically) and from the end points of the light and dark blue lines (going through points G1 and G2, horizontally) we can construct a box of sorts. This is seen once again on the slides.

So you see the box right? The black dotted lines there are actually it :P So yeah after that we took our ruler again, and drew that orange cross that goes far off the page kinda. We drew that cross by going from one corner of the box to the other (and beyond.) After we had done this Mr.K revealed to us that those orange lines, are actually asymptotes for the graph. In other words, neither branch of the hyperbola will actually ever touch those orange crossey lines.

The final slide here, is the equation of the hyperbola, along with some similarities, and differences, which we came up with together, to help us remember the equation, and keep it distinct from the others.

To Conclude: The basic parts within the anatomy of a hyperbola
- Transverse Axis.
- Conjugate Axis
- Focus Points
- The Asymptotes
- And finally, you should know the equation, which is shown in the above slide image.

Alrighty, I think that about wraps this blog post up. I'm not entirely sure, because at 1:00am I kinda start to get a bit loopy and my thoughts aren't so coherent (thats probably evident based on how my writing progressed.) If you guys have any questions or whatever lemme know. I'll try to answer them, and post em up. Or if you find any errors, or know something I dont/forgot to add, lemme know so I can add it.

Well I'm off, but of course, not without letting you know who the next scribe poster is :)

Remember, vote for Zeph for VP.

K? Kthxbai

Justus- FINALLY out.

Today's Slides: May 15

Here they are ...

Wednesday, May 14, 2008


Today's class we didn't do a whole lot of anything other then hyperbolas. We recapped on the equation of an ellipse, refer to last class: Richard's scribe post.

We learned about how to put subscripts and super scripts, that are used to right those tiny numbers, or letters, that are used to show various things, such as an exponent on a power
Example: 21 - The 1 in this equation would be the super scripted number.
Use: This is given with <.sup> and end with <./sup> Get rid of the periods.I put them in so it wouldn't actually work.
Example2: log2 - The 2 in this log would be the subscripted number.
Use: This is given with <.sub> and end with <./sub> Get rid of the periods. I put them in so it wouldn't actually work.
This is a pretty useful html code to use. Remember you can only use these codes in the blog if you are in the "Edit Html" tab, shown beside the "Compose" tab at the top right, just above the various posting tools (spell check, insert picture, etc.).

We were given 1 choice of 3 different questions that we were asked to solve. We started by evaluating this equation: 16x2 + 9y2 = 144

The first part of the question was to change it to standard form, which was quite easy. To do this, you would have to reduce the coefficients so there wouldn't be any on the x or y values. This could be done by multiplying each side by (1/(16)9). By doing this, it would reduce the coefficients, giving us the standard form of the equation: (x2)/9 + (y2)/16 = 1.
For the 2nd part of the question, we were asked to find the centre, major axis, minor axis, vertices and the focii points.

The centre being (h,k) but because there is no h and k values in this equation, the centre is at (0,0).

Major Axis:
the major axis being 8, because 16 = b2, and b is the length of the semi-major axis from the centre to one vertices, so square root 16, to get a semi-major length, which is 4, then multiply that by 2, to get the other semi-major axis, which gives us the major axis(two semi-majors = 1 major), which is 8.

Same idea with the minor axis, but using the a2 value, which is 9, square rooted giving us 3, which is the semi-minor axis, then multiplying by 2, to get the minor axis, which is 6. the vertices would be the endpoint of the semi-minor axes, and semi-major axes, starting at the centre (0,0).

The vertices of the minor-axis spanning a value of 6, from the centre and outwards horizontally, because it's a vertical ellipse, due to the fact that b2 is below the y-value (need more info?, refer to Richard's post). So if a value of 6 was spread outward horizontally, from the centre (0,0), it would give the vertices a value of (3,0) and (-3,0). The other vertices would come from the value, of 8, spanning outward vertically from the centre (0,0) giving the other vertices points at (0,4) and (0,-4).

The 3rd part of the question was to graph it: (I'll edit this in later, when the slides come out, so you can see the actual graph)

Finally, is the hyperbola, the last section of todays class, we made these hyperbola by folding a piece of paper. It started off by having a circle that was off centre on the paper, and the circle had a given centre point. We were asked to make a point about 2 cm off, out of the circle. If you were to draw a line perpendicular to the edge of the paper in landscape view that bisected this centre point, the point out of the circle should be close to touching this line. We were then asked to draw atleast 25 dots with 5 of the dots bunched in the area on the edge of the circle closest to the outside point. These points should be touching the edge of the circle. We then folded all these points, onto the points outside of the circle. This gave us an outline of a hyperbola, with the centre point of the circle, and the outside point the focus points of the hyperbola. Voila a hyperbola.

For homework, we were asked to pick a point on the left branch of the hyperbola and 2 points on the right branch of the hyperbola, after that, we measure the distance from this point to both of the focal points, then find the difference. What do you get? Find out.

That was pretty much everything we did for today's pre-cal class, and now I feel like a world-class math shaped origami teacher. I''m sure everyone feels this way, after we folded 3 different different shapes. I can't complain though, I found it quite fun. I propose the next scribe will be: eeeny meeny miineyy, Justus. Surprised?.. I thought so.

Until next time,