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...

TYPES OF PROBABILITY: 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.

EXCLUSIVITY IN PROBABILITIES: MAY 23

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.

No, the next scribe will be SOMEONE WHO HASNT BLOGGED YET, WHY ISNT THE SCRIBE LIST UP TO DATE WHAAAAAARGHADSADJKSAHDJKAHSJDL.

>:(

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.

Permutations of Non-Distinguishable Objects and Circular Permutations

Okay, so its that time again, time for a class blog. I doubt this one will be as long or epic as say, a blog by Justus or Francis (seriously, just make a book or something), but hopefully it will be just as informative.

Today's topic was "Permutations of Non-Distinguishable Objects and Circular Permutations."

Whoa, that's pretty long. Someone, create Tinytopic.com, quick!

So does everyone remember what a permutation is?

(As defined by Mr. K)
Permutation - An ordered arrangement of objects without repetition.

Or (as defined by me)
Permutation - A set of objects where the order matters.

A permutation is not to be confused with a combination as we discussed in the last class. In short, a combination is a set of objects where the order does not matter. A permutation is a set of objects where the order does matter. That's why a "combination lock", should technically be a "permutation lock."

Remember that? Yeah, you better, because it's important for our formula, the aptly named "Pick" formula, which goes like this:

nPr = n!/(n-r)

(n and r are subscript, they are not being multiplied)

Where:

• n is the number of objects to "pick" from
• r is the number of objects to arrange

And when we use this formula, nPr is read as "n 'pick' r."

3P₄ is read as "three pick four."

What's totally awesome though is that this function can be accessed through your calculator!
So it goes something like this...

• Enter your "n" value first
• Press the [MATH] button
• Press the [<] button
• Press [2]
• Enter your "r" value
• Press [ENTER]

Isn't technology wonderful?

So after that, we talked a little bit more, but since there's not much to the "Pick" formula, we went on to talk about really huge permutations.

TinyURL.com is a site that takes immensely long website URLs and turns it into a short and sweet (although probably not so easy to remember) URL. So http://www.internetisseriousbusiness.com/index.html would turn into something like http://www.tinyurl.com/9aidso. Thus, it generates a unique (this means it's a permutation and not a combination) url every time someone wants to make a TinyURL.

So, Mr. K brought up a good point. Just how many URLs can TinyURL generate, and how long would it take to generate all those URLs?

Well, lets think about this for a second, and look at the example url TinyURL generated for us.

http://www.tinyurl.com/4kaqlv

Well, the fact that there is a number and letters there tells us, the value of a slot can be either a number (0 to 9) or a letter (a to z). Lets assume for now that:

a) The number can be in any slot, but there can only be 1 number
b) The letters are not case sensitive, so we can assume they'll always be upper or lower case (if it was a combination, we'd have to count upper and lower case a's as two seperate objects, but more on that later).
c) Letters can repeat (so your string could look like "1aaaaa"). This means you cannot use the "Pick" formula!

Why cant you use the pick formula?
Because the pick formula calculates the number of permutations where once an object or character is placed in the set, it cannot be used again. This is obviously not the case here.

So, we've already determined that a slot can have one of 36 values. 10 of those values are for each of the 10 numbers, and 26 of those values are for each of the 26 letters.

And we have 6 slots. So for every slot, we have 36 possible values. Thus our number of possible permutations is 36*36*36*36*36*36 or36^6 .

Which equals:

36^6 = 2 176 782 336

So thats two billion one hundred seventy six million seven hundred eighty-two thousand three hundred thirty six. (Say that 10 times fast!) And remember, a billion is big. Godzilla big.

Infact, its so big, we determined that it would take ~32 (approximately thirty-two) years to give Lawrence a billion dollars if we gave him 1 loonie (1 dollar) per second consistently. Times that by 2.176782336 and you have approximately how long it would take to generate every possible URL if one URL were generated every second consistently:

2.176782336 * 32 = ~69 years. The one on the slides says 64 years, because there we simply multiplied 32 by 2 (which is how many whole billions we have). I counted the fraction that would have been generated all the way up till the 2 176 782 336th day. However, since the 32 itself is an approximation, this is not completely accurate (hence the tilde [~], which means approxmiately).

And thats just if we're not counting capital letters!

If we did count capital letters, our number would be even huger! Because now our number of possible values in each slot is now larger. Instead of just "t", we now have "T" and "t". Thus, we don't just have 36^6 possible values, no. Now we have 62 possible values, because we have 52 possible letters and 10 possible numbers.

So how many permutations is that? Thats 62^6. Which equals...

5.680023558 x 10^10
Which equals to 56 800 235 580, or fifty-six billion, eight hundred million, two hundred thirty-five thousand five hundred eighty. Permutations.

So, 56 * 32? Thats ~1792. That's how many years it would take to generate all those permutations if one were generated every second!

But enough about that. Lets look at distinguishable objects versus non-distinguishable objects.

(Im going to speed up here because it's now 2am and I'm getting sleepy)

So on slide 7, we have a little example with the word book. How many ways can you rearrange the word book? How many unique permutations are there.

Well, we figure that out easy: 4! (the number of slots) divided by the number or repeating letters in this case, that letter is "o" and it repeats twice). Thus, our solution is 4!/2!, which is 12.

But then there a plot twist. What happens if one of the "o"'s is red, and is counted as a unique letter? Then we have even more permutations! Because now, instead of having only 3 possible values for a slot (where one slot has the same value as another slot due to one of the values repeating), we have 4 possible values. Thus, we have 4x3x2x1 = 24 possible permutations. But if you turn that red "o" back into a regular "o", some of your permutations become repeats. Thus, the number of permutations is halved, and you get your original answer of 12 permutations. Wow!

After being introduced that concept, we do a little practice with some other words. Then its on the the main course. Circular permutations.

Circular permutations looks wacky and complicated, and it takes a while to wrap your head around it, but really its just an expansion of the concept of distinguishable objects. The basic idea here is that when you have a circle, the first point you put on it serves as your reference point, and you work from that point on. Because your reference point is wherever you want it to be, your perspective as to the rotation of the table can be anything you want. Thus, with circular permutations, your duplicates stem from the fact that some permutations are in fact existing permutations that have just been rotated.

Alright, I give up now. This scribe post is incomplete, and I apologise to everyone that I'm posting so late. I'll expand and finish this post properly later today and make sure it's done as it should be, but for now I cant think when I'm half asleep.

Tomorrow's scribe (or should I say, today's scribe) is now Nelsa.

Also, a million points and a PHD to someone who can tell me the adjective of "bracelet." Seriously, a braceular table? I don't think so. I wonder if that's even an actual legitimate shape, or is it just a concept...

Hi my names BOB, and here about Identities...

So this is my BOB for our Identities unit.

At first, I didnt really like identities. The whole concept of "proving" instead of "solving" bothered me and I really didnt get it or like it at first. It took a little while and alot of practice before I was able to do it.

In general I think the identities stuff was pretty great, kinda easy. The hardest part was and still kinda is solving identities "elegantly." Its pretty confusing sometimes and I can never seem to remember some trig identities right. Since this was a really short unit and we didnt cover a wide array of different stuff, it was much easier to swallow, especially with all the practises we did. The sine dance helped alot, and was great fun.

I think im fairly well prepared for tomorrow, although I'll be sure to review notes before I go to bed and remind myself that cosa != a. Always making that mistake...



Anyway, good night everybody and good luck on the test!

Trigonometric Identities: Double Identites

The slides:

Link to the slides post.
Link to the slides page.

Hey guys, its Paul here posting my scribe for our class on Double Identities on Tuesday (just in time amirite?).

So we started off the class with a review of the previous class on slide 2. The basic concept here is that the distance between Q and P is equal to cos(a-b).

In slide 3, we used this idea to find the Sum Identity with Even and Odd functions since we already have the Difference Identity.

Sum Identity: cos(a+b) = cosacosb - sinasinb
Difference Identity: cos(a-b) = cosacosb + sinasinb

Cosine is an even function, sine and tangent are odd.

We use the same concept for finding the Sum and Difference identities of the distance between R and Q on slides 4 and 5.

Then we take a break from our "regularly scheduled programming" (I wonder if that counts as one of Mr.K's catchphrases that we should put on the board) with a short and simple quiz on slides 7, 8 and 9.

Slide 7: Simplify the expressions. Pretty self explanatory, use identities to find the simplest expression.

Slide 8: Prove the identities. Again, stuff we've done before.

Slide 9: Find the exact value of sin(5pi/12). This one took a little more work, but it was actually easy because the values a = pi/6 and b = pi/4 were given to us. As Mr.K explained, on tests we will simply be given a value to find with a formula, which means we'll have to find a and b on our own. Since they're given to us, we can just plug the values into the formula and get our answer.

As we return to our regularly scheduled programming on slide 10, we are given and identity to prove. The Sum Identity for tangent, to be precise.

The solution is pretty long and looks complicated but it basically follows these steps:

tan(a+b) = sin(a+b)/cos(a+b)
And since we already know what sin(a+b) and cos(a+b) equal to, we get a really long equation (one I won't bother to type since its right there in the slides). Once you have your really long equation, most of it simplifies or reduces to the proper solution of tana+tanb/1-tanatanb.

And since Math is the science of patterns, you'll probably know what we're going to do next. Next, we find the Difference Identity of tan (tan(a-b)) since we have the Sum Identity. We do this on Slide 11 in an equally long but rewarding process.

In summary:
tan(a+b) = tana+tanb/1-tanatanb
tan(a-b) = tana-tanb/1+tanatanb

On Slides 12 and 13 we learn about Double Angle Identities (which is different from a double identity). Here we apply some old stuff to a new problem, and turn something like:

sin(2theta)
Into:
sin(theta + theta)

Which would look more familiar as say... sin(a+b)?
Since its exact same thing (so long as b = a), we can rewrite sin(2theta) as:
(sintheta)(costheta) + (costheta)(sintheta)
And then simplify it so it looks like
2sintheta(costheta)

We then apply this concept to find the Double Angle Identities of cos and tan.

sin(2theta) = 2sintheta(costheta)
cos(2theta) = cos^2(theta) - sin^2(theta), 1-2sin^2(theta), 2cos^2(theta) -1
tan(2theta) = 2tantheta/1-tan^2(theta)

And that basically sums up what we did that day in class.

Now I'd like to add a small note to the sine dance to make provisions for the tan identities, which is obviously a pain to figure out via sin(a+b) and cos(a+b). Therefore, I propose a tan dance to help us remember the tan identities.

Since this is all in terms of tangent, we do the dance in variable order, something like this:

Alpha, Beta, Divide, 1, Bust-a-move, Alphabeta

or in stick man form...

Im just putting it out there. Seems easier than solving sin(a+b)/cos(a+b) every time you want to do a tan identity to me.

And that concludes my painfully brief scribe post about Double Identities. Hope you guys get a good nights sleep and enjoy our last day before SPRING BREAK.

Yeah man.

Proving Identities with cos and sin

Hey everybody, this is Paul, posting the scribe post for our Thursday class which was on the 20th. Sorry this is so late, its been a hectic week.

Note: This is the symbol for theta in my blog post: Θ.

Okay, so Thursday we continued the topic of using identities with sin and cos and expanded it so we could prove stuff like sin(pi/6+Θ) + sin(pi/6-Θ) = cosΘ. This is demonstrated in our 2nd slide.

At first the solution confused me, but I realized the solution works by using the formula we used previously to convert a formula that looks like:
sin(α+β) + sin(α-β) = cosβ

And use formulas we already know to convert it to:

(sinαcosβ + cosαsinβ) + (sinαcosβ - cosαsinβ)

And then reduce to get cosΘ.

In slide 3, we were given a question that asked us to find the cosine of alpha (cosα) and the sin of beta (sinβ). Initially, we got the wrong final answer because we made the mistake of saying that α = -3/5, which is WRONG.

The truth is we never get the value of alpha, we only get the value of the cosine of alpha.

Once we realized our error, we solved the problem properly by using the formula cos(α+β) = cosαcosβ - sinα+sinβ, giving us the correct answer of -33/45. The proper solution is on slide 4.

After that, we were given the same question except we were told to solve for sin(α+β). This takes place on slide 5. The solution is pretty straightforward since you just change the formula but keep all the values you already found in the previous question.

Mr.K then started talking about his "clever idea." He explained to us that the R, Q, and P can be defined in terms of sin and cos on slide 6. He then started to talk about the values of cos and sin in the rotated triangle, but didn't finish because the class ended. I guess we'll find out how clever his idea is on Monday.

Slide 7 shows us how to rotate a point 90 degrees in terms of coordinates. If your coordinates are (x,y) unrotated, then when you rotate it 90 degrees your coordinates will be (-y,x) (such that x = -y and y=x). For 180 degrees, your outcome would be (-x,-y).

And the sums up my scribe post. Sorry it took so long to be posted, especially since its rather short. I didn't manage to get around to doing it until now.

And since nobody's told me they want to be scribe, I will randomly choose a name from a hat (in my mind).

The lucky winner is... ZEPH.

And colour me surprised, the name's not in pink. (gasp)
Somehow it still stands out from the rest of the post.

Good night and farewell.

P.S. And Mr.K, I think the scribe list needs some updating?

BOBbing for.... knowledge

Hello, this is Paul, posting his Blogging on Blogging post.

For this unit, Transformations, we covered a wide variety of topics that relate to graphing. Most of these were pretty simple, like DABC and flips, shifts and stretches. However, for me atleast, I find the questions involving inverses to be confusing most of the time despite how the answer is fairly simple. Also, Im still not really getting how to graph the 1/f(x) questions, although I understand what is being said I can't seem to apply it myself. The question we did in place of the test also helped me alot in understanding how to solve "asin[b(x-c)]+d" questions, which I did not fully know how to solve before. Now I can solve one with a calculator and some paper, yay. Thanks Mr.K.

So in summary, difficult things are:

Graphing 1/f(x) functions. I think my problem here is that Im still looking at the graph as points and then trying to apply the "1/f(x)" (which was our approach for graphing "asin[b(x-c)]+d"), which means I end up with points but not a proper graph. I should definitely review the scribe post for that class before the test.

Inverses are just a little confusing, but I do understand what they do and mean. Probably the hardest part for me is graphing one because Im not totally sure about how to, but a look over that scribe post will probably help.

Other than that I think I understand this unit, or I understand well enough to pass this test (hopefully).



I also would like to apologize to Lawrence and Nelsa that I did not post our group's solution in time. I unfortunately did not anticipate that my alarm clock would not wake me up. Due to my alarm clock being off, I did not wake up until nearly 3pm. Much thanks to Lawrence for posting our solution.

So good luck guys, and don't forget to bring delicious pie!

Group Second to None

Okay, late I know, but unfortunately our group didn't post in time, So I'll try to do this quick fast.

Anyways, this is our solution.

SLIDE 17:

This equation gives the depth of the water, h meters, at an ocean port at any time, t hours during a certain day

h(t) = 2.5 sin[2pi(t - 1.5)/12.4] + 4.3

A) Explain the significance of each number in the equation

I) 2.5 - This is parameter A, which will determine the amplitude of the function


II) 12.4 is the Period of the function, which is obtained from parameter B - 2pi/12.4

III) 1.5 - This is parameter C, which is the phase shift, in which this function, the graph shifts to the right 1.5 units/hours.

IV)4.3 - This is parameter D, which is the sinusoidal axis, which shifts the sinusoidal axis up 4.3 metres..



B) What is the minimum depth of the water? When does it occur?




Now, We can go backwards and use that point, but since we can use whichever point, we decided to use the next one, which minimum depth, 1.8 metres, occurs at 10.8 hours.



C) Determine the depth of the water at 9:30 am.

So then we just plug it in.

h(9.5) = 2.5sin[2pi(9.5 - 1.5)/12.4] + 4.3
*t is 9.5 because time ~ 9:30 AM ~ is converted to 9.5 because :30 minutes is .5 hours.*

in which we get... 2.323 metres.


D) Determine one time when the water is 4.0 metres deep.

So then we go something like...

4 = 2.5sin[2pi(t-1.5)/12.4] + 4.3



First we subtracted 4.3 from both sides like so.

-0.3 = 2.5sin[2pi(t-1.5)/12.4]


Then we added the phase shift to both sides to get...

1.2 = 2.5 sin[2pi(t)/12.4]


Then we divided parameter A out.

0.48 = sin[2pi(t)/12.4)


We then divided 2pi/12.4 in which .48 would be multiplied by the reciprocal of 2pi/12.4, and moved sine to the other side to change it to ARCSine to isolate t.

ARCSine(0.9473) = t

Which would equal to --> 1.2447 hours.


To make it efficient, we'll multiply the .2247 by 60 so


Depth of water of 4 metres occurs at 1:13:48 am.



Again, sorry for posting so late. Original personnel that was to post, did not post. Feel free to comment, as we are supposed to.

Odd and Even Functions

Hello, fellow classmates. My name is Paul, and I will be your scribe for today, and I will be writing about "odd" and "even" functions in an overview of Friday's class.



So our class today started with a review of yesterday's class (slide 2), which basically stated that:

In the formula y = af[b(x-c)]+d:
- If a > 1, the graph is stretched vertically
- If 0 < |a| <> 1, the graph is compressed horizontally
- The y-coordinates of f are multiplied by a
- If b > 1, the graph is compressed horizontally ("speeds up")
- If 0 < |b| < 1, the graph is stretched horizontally ("slows down")
- The x-coordinates of f are multiplied by (1/b)

Following the review, we spent some time practicing graphing functions, which can be seen on slide 3.

REMEMBER!
Stretches before translations!

Here's a quick rundown of the formula and what each variable does:

y = af[b(x-c)]+d
a -> Vertical stretch/compression
d -> Vertical shift
b -> Horizontal stretch/compression
c -> Horizontal shift

Note: Remember that you have to stretch the graph first, but it doesn't have to be both stretches first (abcd/abdc)! You can stretch the x axis and then shift it before stretching the y axis(adbc), the important part is that the stretch for each axis must happen before the shift for its respective axis. This was discussed on slide 5, where we wrote the possible orders you can apply transformations.



After the practice with drawing graphs, we were introduced to images, which is the graph after it has been shifted/stretched (slides 4 and 6).

Also, when talking about an image, the "formula" is:

(image) is the image of (original coordinates) under (function).

So an example would be:

(0,7) is the image of (-2,-2) under the function y=-3f[1/2(x-4)]+1



We then proceeded to learn all about the online bookmark service and website Del.icio.us.
I won't really go into detail about this, since it's pretty self-explanatory. However in the interest of thoroughness, I've made a quick image here:



And with that, we proceeded to our next topic of reflections (slide 7):

- Basically, a vertical reflection is a reflection along the x-axis which occurs when the y-coordinates of any function f(x) are multiplied by (-1).
- Conversely, a horizontal reflection is a reflection along the y-axis which occurs when the x-coordinates of any function f(x) are multiplied by (-1).

Slide 8 contains a picture of a reflected sine wave, neato.



Inverses are also introduced on slide 7:

As stated by the slide, the inverse of f(x) is f^-1(x) [f to the power (-1) then/applied to (x)]. Which looks like this (image follows):

Thank you, Paint, for that excellent demonstration.

And in BIG SCARY LETTERS, I WILL MAKE SURE YOU DO NOT FORGET THAT f^-1(x) IS NOT EQUAL TO [(1/f(x)]!!!


The truth is that (can you handle this?)... [f(x)]^-1 (note how the power is on the outside), is the one that is equal to [1/f(x)].

So in summary:

[f(x)]^-1 = [1/f(x)] √ Correct
f^-1(x) = [1/f(x)] X Wrong

Its also pronounced "eff inverse." And finally, f^-1(x) undoes what f(x) does.



Lastly, we started talking about "odd" and "even" functions (slides 10 and 9 respectively):

A function is "even" if it is symmetrical about the y-axis, which occurs only when f(x) = f(-x). If you look, a cosine curve is a nice example of an "even" function.

A function is "odd" if it is symmetrical about the origin, which occurs only when f(-x) = -f(x) (the negative of the whole function). If you look, a sine curve is a nice example of an "odd" function.



And that concludes my scribe post summarizing the topics we covered during our class last Friday. Now about the scribe...

Since some people actually want to be scribe (unlike myself who was hoping to be scribe for pi day), I hereby declare that benofschool has the honour of being the next scribe for Monday's class.

And don't forget to get a Delicious (mmm, delicious) account if you haven't already! Here's a direct link for you who are too lazy to type it (shame on you), or like convenient things.

Del.icio.us

So I bid you adieu and offer my sincerest apologies for the lateness of my post. Unfortunately, I had work Sunday and was busy the majority of Saturday. I did infact start this post on Friday, but did not manage to finish it until today. I am highly aware of the time I am posting this at, especially considering I had resolved to post it first thing on Friday.

Good night, and I'll see you tomorrow.

P.S. I would like to offer the following advice to the following scribes:

Blogger's draft system is not without its flaws, and I have had to rewrite the second paragraph (involving the overview of the a and b variables) numerous times because the draft system confuses my use of the arrow signs (<>) as parts of html tags and deletes everything inbetween. Thus, if you think you're not going to finish your post, just save yourself some hassle and put it in a notepad document instead of using the draft system.

Hey everybody, Paul here, doing my scribe post (since I got "voluntold" to). I hope 8pm isnt too late for a scribe post, but I guess you guys don't really need it all that soon anyway as we didn't do a lesson.

So yeah, today in Pre-Calculus 40S we had a substitute (Mr. Rekrus?), and he gave us an assignment per Kuropatwa's orders. The assignment was on circular functions and it went something like this:

You have a unit circle, and on this circle you have a point P, which is along the circumfrence of the circle and in Quadrant I. You also have point A, which is at the center of the circle (thus making it the origin), and a point B, which is positive along the X axis. If triangle APB is a right angle triangle, find the value of P that creates the largest area of the triangle.

Since this was a group assignment, the class was split into three groups. There was also supposed to be a time limit of 20 minutes, but the question utterly stumped the groups for most of the class, so the time limit was disregarded. Some groups finished the question before the end of class, some spent the entire class trying to solve it and some groups made little paper cranes.

And that was what we did today in Pre-Calculus.

Today's homework is Exercise 6, Questions 1-20. I suggest you atleast try them since we do have a test coming up.

As for the next scribe, I have chosen Zeph since he asked so nicely.

Good night and farewell.