Monday, November 5, 2007

Monday's Scribe (this is monday right?)

Hi, I'm tired and might make a few mistakes to bear with me please and if you something wrong let me know by email or facebook.


The teach gave us this question. We had to try to find the derivative of f(x) g(x).





multiplying the derivatives didn't give us the right answer. so we had to find a new way of find derivatives. We learned of 2 ways, that revolve around this formula.



This next slide is the classic formula. The basis of this formula is that you add and subtract same number so its like adding a zero.



After you add the "zero" you want to write out the whole equation.

You want to single out f(x+h) and g(x)

You the want to find n as it goes to zero for each part of the equation. And as n goes to Zero the equation g(x+h)- g(x) is the same as g'(x). Rewriting the equation by putting in g'(x) and f'(x) gives you your formula.





Sorry if its not that clear. Now the second way of doing it is the non traditional proof. Basically you can think of it as two rectangles. And all your doing is finding the difference between them. Take the differences, add them up and divide all that by h as h goes to zero. Remember that since h will be zero anything times to h will also be zero so therefore g'(x) f'(x) times h will be zero.












This is the whole procedure, have fun with it.

Now we had to do this all over again except this time with dividing.
As you can see we once again need a new formula.but this time there is only one.









There is the the method of doing that and to end class we did a question(below).


I know this is not the best, but its all you will get for now. :) The next scribe will be........Robert? i can't tell who's done it or not by the scribe list.
Oh yea before I forget here is one of the sites the teach used in class. I managed to copy it down.
And lets not forget the little ditty we learned.
low de high minus high de low, all over low low :) have fun with it.

Today's Slides: November 5

Here they are ...



Sunday, November 4, 2007

e is special (=

Hi everyone! The title says what the majority of this class was about, and I'll detail how this is true further into this scribe post. Now I know that this scribe is going to be a doozy, but not only for those in seek of amelioration or those that did not partake in Friday's class, but the entire class will benefit form the beginning portion of this scribe. And just as a side note, I would have attempted to create a visually stunning scribe, as I vowed to myself that I would, but I barely have access to the computer or even any free time during this weekend since it's rather busy here. Well, I think it's time to continue on with Friday's scribe!

AP Calculus marks:
Before class began, I came into class approximately 10-15 minutes early to talk with Mr. K since I had finished my fourth period biology test early. As we talked about AP classes for a short period, he quickly delved into a topic that he would have wanted to inform the class about on Friday, but he felt that the lesson for the class was too significant to give up any time for such a topic, so I'll project to the class what he informed me of (even though he will go over it quickly on Monday).

The issue regarded our marks for AP calculus for this year, as many of us in the class probably have pondered throughout the few recent weeks. This concerns how we will be marked in this class.

As he has told us before, he will be taking the raw score of our marks from within the class, but he has not told us before about how the marks will be manipulated to better represent our progress as best as possible. He asked other AP teachers how they have achieved such appropriate representation, but he found them to be rather arbitrary or completely asinine. Examples of such include the utilization of multipliers, or in one case using the square root of the class average as a multiplier. But Mr. K's method is rather different, and if my explanation does not suffice, I do believe that Mr. K will outline his method once again on Monday to the whole class.

Class marks are often illustrated as a normal distribution, where quantities are represented by a distribution that indicates many quantities to be centralized around the mean (average) value, but become more infrequent as they approach the extrema. In other words, there will only be a few really low or really high marks, but there will be more values in a certain area as marks become more stabilized towards the average. This can be shown by the graph below (also known as a 'bell curve')
Mr. K has compiled our pre-calculus 40 marks, and graphed them as a normal distribution curve (like the one above). He calculated the mean value of the distribution, otherwise known as our class average, and also found the first and second standard deviations (in a normal distribution, standard deviations show which area of values a quantity resides or is spread out relative to the mean). Naturally, most of our marks would go down from our pre-calculus marks, so the distribution of our marks in AP calculus is expected to appear shifted to have a lower mean and a lower centralized curve. While our pre-calculus class average might be around 70-80, our AP calculus class average might even be around 60-70, which does not fairly represent our potential as students in mathematics. An example of such curves can be seen below, where the graph on the right (the solid blue graph) can be an example pre-calculus distribution, whereas the slightly transparent red curve on the left depicts an example AP calculus distribution of our raw marks.
Mr. K plans to apply a translational transformation to the AP calculus curve to nearly match that of the pre-calculus class. In this way, the class' overall achievement in the course can more fairly be represented relative to how we performed in calculus relative to the mean after it has been shifted. That's mostly everything, but that is not nearly the end of this scribe post.

Exponential function differentiation:

It's finally time for another one of those great "ta-da" class' that makes the teacher want to present the board like their advertising a bottle of shampoo and proclaim "isn't that cool? ehh? ehh?" First off, we established the fact that we began to determine rules for differentiating (finding the derivative of a function) basic functions, but we have not encountered differentiation of exponential equations. What a beautiful segue into the inevitable lesson for the day; learning the process and foundation for differentiating exponential functions.

Mr. K announced that we would need our calculators for this lesson, and that we want to be able to determine the derivative of any exponential function. To do this, we input the following equations into our calculators:
y1 = Ax <-- The parent exponential function

y2 = [A(x + h) - Ax]/h <-- The derivative of the parent function using the limit definition.
We then defined the function by setting A = 2 (using the [sto>] and [alpha] keys), and since we want h to approach 0 but not equal zero (as that would cause an error in the calculator) we need to use a value extremely close to 0, which in this case was 0.001. So far we had A = 2 and h = 0.001 set in our calculators, and before we graphed them we turned off y2 momentarily, so that we could fully focus on analyzing what's happening in y1 and to the values of y1 and y2 in a table. Here is what we viewed as we graphed the seemingly simple function Ax:
We then deduced that we have to try and look at it algebraically, but first Mr. K digressed and we began to analyze the table values for y1 and y2 in [2nd][graph]. From this, the class noticed that there was a doubling effect happening for each subsequent value of f'(x), starting from 0.69317. Now, the relationship between the parent function f(x) and it's derivative f'(x) is shown in this table for when A=2, in which case there is a multiplier of 0.69317 (as can be seen when x=0) Mr. K then told us to look at the two graphs (turn on y2) and try to identify how y1 is related to y2 graphically. Graeme suggested that the derivative appears to be a translation of the parent function, but by referring back to the table of values it became apparent that this was not true. Instead, Mr. K asked if there was any way to determine this multiplier, which was determined in a similar fashion as calculating the common ratio of a geometric sequence (since geometric sequences are exponential in essence). We then added another function into our list of functions in our calculator; y3 = y2 / y1. This allowed us to determine the fact that each respective y value is multiplied by a common factor of 0.69317 (as we determined before) and that this factor is equivalent to f(0). This meaning that the value of the derivative at x=0 is in fact the multiplier. Followed this was a table we set up ourselves to show the multiplier values (c) for specific bases.
Base ------ c
2 ---------- 0.693
3 ---------- 1.099
pi --------- 1.145
5 ---------- 1.609

We can easily determine that by looking at our data tables, each output of the parent function is simply multiplied by the multiplier that is given by when f'(x)=0. This can be shown symbolically by:
f'(x) = c * bx
But now we encountered a question that probably riddled many of the students in the class. How can we find a perfect base number such that c will be equal to 1? Well, we can see that 3 is slightly too large a base to give a multiplier of 1, but 2 is quite a bit smaller to do so as well. We played around with numbers around 2.7 and determined that the number that gives a multiplier is 2.718281828459, otherwise known as e. e is the perfect number that we were searching for, and once we adopted it's value as A, we received a multiplier of 1.0001. It was not exactly equal to one since when we provided a value for h close to 0, we only approximated 0 thereby giving a selected margin of accruacy or a chosen error value accompanying our calculations.
We also determined that e is the perfect number through the use of algebra:
We solved for this equation to arrive at e by inserting the graph as y4 in our calculators, and solving for the root of the equation or by viewing the table of values when x=0 to see if it in fact arrives at a multipler of 1. We then learned that we could then say two things about e:
- the perfect base for e x retains a tangent line of slope 1 at x = 0
- e = lim(x->0) (1+x) ^ (1/x)

Now comes an extremely important constituent of the rule for differentiation exponential equations. That is to find a rule, or a true relationship between the different bases and their respective values of c. Our main goal at this point was to determine a rule that would determine what c is for absolutely any base. To further develop the aforementioned base / multiplier table, we added values for bases 6 and 7.
Base ------
c
2 ---------- 0.693
3 ---------- 1.099
pi --------- 1.145
5 ---------- 1.609
6 ---------- 1.791
7 ---------- 1.946

Under deep analysis of this given data, it became apparent that there was an underlying logarithmic connection. We attempted to model a natural logarithm regression, and it turned out that it was an exact fit. We even entered the tested base values into the natural logarithm, and the respective multipliers determined previously were a match. Now since the natural log of the base, (ln(a)), gave us the multiplier, c, then:This ultimately gave us the final key to determining how to differentiate an exponential function. The following equations represent the rules for determining the derivative of an exponential function:



We then concluded the class with three sample calculations by applying not only the newly acquired exponential differentiation rule, but by using a combination of other differentiation rules we have learned previously.

So, I think that this scribe pretty much states everything that happened in class. What's really weird though, is that my computer can't seem to load the blog at all. It has been loading for more than an hour already, and refreshing won't change anything. Well, before I leave this post to head off to bed for a good night's sleep, I must say don't forget to add tags to your del.ici.ous accounts people because I think we're far behind our tagging priorities. I hope I helped at least someone with this post, as I put a lot of effort into it (as for any of my posts) but there's always room for improvement along with anything, so if there are any questions, comments or concerns pertaining to this post, please feel free! Until next time, I'm MrSiwWy. Have a great weekend everyone!
Oh yea, the next scribe will be...Ethan
Cya in class everyone!
Dixi~

Thursday, November 1, 2007

The scribe for Wednesday.

Well, today was a rather small class. 4 of the 11 were missing. Due to other classes needing students for events, or parents keeping the kids home, afraid of the hoax shooting. Anyways, Happy Halloween.

I also appoligize for posting this one day late. I was fairly busy yesterday. And I wasn't even supposed to scribe. Oh wells.

Let's start with:

Graeme's Selected Word of the Day

Platitude :
a trite or common remark delivered solemnly.

Alright. Today, we like food, because, we got an awesome food line-up. It starts with the Potatoes.

But first, let's talk about how great Yogurt is. And the spell check for blogger is saying that Yogurt is the correct spelling, and we have it spelled Yogourt. Whatever. Mr. K loves his yogourt.

And now on to the Potatoes. They taste okay on their own.

Mr. K begins the class with, how we find derivatives at certain points on the graph of f. We do a few practice questions, and he says "There's gotta be a better way. A faster better way. We don't want only the derivative of just one point. We want the derivative function!"

Slides are here:

http://apcalc07.blogspot.com/2007/10/todays-slides-october-31.html

That's what Slide 2 says. We did practice questions, and now we're going to learn rules. As Mr. K says, "Mathematics is the science of patterns."

Slide 3

When we are given a linear function, to find the derivative of that, whatever the slope of the linear function is, is the derivative, given as a constant.

m = slope
so m = derivative

Slide 4

When the given function is a constant, with no variables, the slope on the constant function is 0. Therefore, the derivative is 0

Slide 5

When the given function is being multiplied, you can factor constant out, then apply the rules that you already know.

Ah, and since the potatoes are now served, we shouldn't eat them yet. We gotta have the, MEAT

Slide 6

When given 2 separate functions, and then asked to find the sum of their derivatives, you can find the derivative of each function individually and then add them together after you do the algebra.

Slide 7

Same thing as 6, except it's a differences, and you should subtract.

Okay, that's some pretty good smellin' meat. Let's add on the gravy to the potatoes, and we can enjoy the meal.

Slide 8

Whatever n is equal to, you multiply it by the coefficient then subtract one off of the power.

Slide 9 & 10

Continuation of the proofs and an example.

Alright, and that's the end of it.

Homework for tonight is exercise 4.1 all odd, 18 and 26.

Next scribe for Friday will be Chris!

No scribe for Thursday, because of the test. (this is treated as Wednesday)

Bob 2

My oh my!! I almost forgot about this. Well then, this unit was pretty short. It was a little challenging. I can definitely say that I don't like it being last period everyday. However, I'm managing to make my way through. This unit isn't too hard to grasp though. So good luck to everyone on the test! (:

BOB

Well, its just that time again. Another test has arrived; definite integrals. There isn't much for me to say other than that this unit was a breeze for me and so i shouldn't run into much problems on the test. Can't believe i almost forgot to do my reflection.

Goodluck everyone.