Almost Final Scribe Posteyness

Alrighty guys, as chosen by yinan I'm the scribe for todays class, which is great, because its a short scribe, and I have to go to work in about half an hour, so it works out perfect.

Anyways, todays class was a relatively simple one, as for reasons unknown to me, Mr. K wasn't there. Instead we had a sub, who's instructions to us were to work on the questions we had been given yesterday (I think? I wasn't all to clear about which questions he was talking about.) So the majority of the class did that. Myself, Lawrence, and Francis used that opportunity to do a bit more work on our DEV and go over some questions from the text that we had been struggling with.

Overall the class was mostly geared towards review and practice questions in preparation for our ever looming AP Calculus Exam (May 6th D:)

So I think thats everything, and its good too, because now I must get ready to go to work :] See everyone tomorrow!



SO theres one swear in it, I didnt notice it until just now. Theres also some mild drug references, although they're in good nature. Basically, make sure theres no young children around and it should be fine :]

Edit: Oh yeah, Kristina's scribbbbeeeeee Sorry for the lateness. I forgot ^^;

Belated BOB and Post Reflectionisms

So, as much as I'd like to say that I totally did this last night, I totally didn't. I 100% completely forgot, so here it is now, along with a special edition post test thought analysis kinda deal.

So Differential equations. For the most part, I found this unit quite easy conceptually. I understood most of what I was supposed to do and when and things, but I would always make stupid mistakes which cost me the whole question. I think this is mostly a matter of needing to do more questions so that I get out of the habit of overlooking those little bits. That said I don't think I did so well on the test D:

For the most part all I can really say about the test is I blanked. I was feeling pretty confident in the concepts and applications of them, and then as soon as I sat down to write the test, it was like someone opened a window in my brain, and whoosh, it all flew out. bleh, to say the least.

Anyways I'm looking forward to reviewing further, and covering all the loose ends I have with the course, as at this point in time it feels like I have alot ;P

But chyeah, Justus out for now.
Night all :]

Uggg spring break scribe :[

Okay guys here's my much delayed scribe from last Friday. I apoligize if it sounds funny, as I actually typed it during the class its written for, so most of it will be in present tense, although tried to keep it in past tense. SO basically, bear with me, and we'll get this thing under way.

Alas this is technically the last scribe of the course; however, it is not the last scribe of our meetings together. Although we may be finished the course today, Mr. K mentioned how he would like to spend some more time on this last little bit. The reason Mr. K gave for as to why he would like to drag out this last little piece comes from the fact that Differential Equations are, “The doorway to the rest of calculus.” Or something along those lines. Basically, he said you take Cal 2, 3, and then differential equations, and after that. You can take differential equations for the rest of your life.

After that little intro, we began to talk about our plans for April, and therefore the rest of our time together, now that the course is basically finished. The result of these discussions revealed that April will be the month we kick it into overdrive manic preparation mode. So Mr. K decided to help us out, and post pretty much infinity AP Calculus Exams from previous years so that we can a.) Work on exam style questions to prepare for this exam and b.) Prepare for the exam by beginning to recognize the patterns emergent in its construction. These practice exams can be found on the blog, in a little box on the right hand side, available for download as pdf's

An important thing to note, is that the solution/scoring guidelines are extremely terse/to the point. They also happen to include the main, “Mark docking errors” (ie. Which mistakes will cause you to lose marks.)

We also got into a discussion about how intense the AP exam is. Some interesting topics of discussion included;

-The fact that the exam takes approximately 5 years to create (in essence, the exam we will write was completed around 2003, 2004ish)
-You have to sign a contract/written oath that you will not divulge certain aspects of the exam (namely the multiple choice questions). If you fail to maintain this you can (and likely will) be sued.

After all that, we began the good stuff.

Solving Differential Equations Symbolically and Newton’s Law of Cooling

There are 6 types of differential equations you can do, and within each type you can have numerous orders (first second third, etc.)

Separable differential equations are equations where it is possible to separate the variables. These are the type of differential equations we're concerned with in this course.

The reason we like the liebniz notation is because you can use it in separable equations, something you can’t do with f’(x) notation.

So now heres an example of this :]






- To start, your gonna rewrite dy/dx = ky as dy = ky dx, as seen in the slide.
- From there your gonna continue by getting all the y's on one side and the x's on the other by multiplying both sides by 1/y
- Next your going to take the integral of both sides which brings you to the next line on the slide.
- Now because we have Two C's on either side of the equation, we'll just combine them into one, since we know that they're constants.
- To continue we take the ln of both sides so that we might find y. This gives us y = e^(kx) * e^c
- The next thing we do is let e^c = A or something like that, which gives us our final solution of y = Ae^(kx)

After that problem, we got to the fabled, "Newtons Law of Cooling Problem"

Newtons Law of Cooling problem.

According to Newtons law of cooling a hot object cools at a rate proportional to the difference between its own temperature and that of its environment. If a roast at room temperature, 68 degrees F, is put into a 20 degree F freezer, and if after 2 hours the temperature of the roast is 40 degrees F

a.) What is the temperature of the roast after 5 hours
b.) How long will it take for the temperature of the roast to fall to 21 degrees F?


Looking at the question we find that we know there is a relationship between temperature and time, but we don’t know the parent function of this relationship. Instead we know the derivative for this relationship (temperature with respect to time.)



So just like with most problems we've solved, we begin by lying out what we know, and what we have to work with. Following the strategy we used before, we begin by getting all our x's on one side and our y's on the other (or in this case, the T's on one and the t's on the other.) We then take the integral of both sides, and fill in everything we know.



T(0) = 68 and T(2) = 40.

By plugging in the value of T at 0 we can find the value of A. This allows us to begin the process of finding the value of k, the point in the first place. Some more plugging in and algebraic massage, and we find the value of k to be

this.



Part B. of the question is solved mostly by the tried and true plug and chug. Some of this process is started on the next slide, where the empty box, is used to represent the value of k.

Okay guys, I think that's everything, hopefully I didnt muck to much up. Sorry for the late post, I had lots of work to do this break. So much so it didn't really feel like a break. The scribe for monday I pick to be Lawrence, if lawrence already did it, kristina, and if she did it I'll pick on monday, cause the last scribe list is olddddd.

Alrighty, night guys, see everyone tomorrow :] :[

Applications of Integrals Bobbbbb

Hey guys, I guess its that time of unit again. To talk about all the little finicky bits and the easy bits and the ones you liked and didn't like and all that stuff.

So since I'm super tired, I'm kinda gonna cut to the chase, so I can get all my z's haha.

SO overall I found the unit wasn't to bad. I understand the concept of taking a slice of whatever your looking at, analyzing it, and applying what you understand about that slice to the rest of the solid.

I'm finding the hardest part to generally be figuring out what the solid looks like and what exactly the question is asking. I also tend to forget some bits of the question (like when dealing with washer problems, I sometimes forget to subtract the inner circle :P)

Anyways the final verdict, this unit wasn't that terrible, although it wasn't my favourite either.

K, well I'm off to sleep, good luck tomorrow everyone!

~Ciao

Boston Bay problemmmsss

Okay so I was all ready to do this scribe post, and then I discovered, I left all my notes at school :[ SO basically all my diagrams are not here with me (atm) and I have to solve alot of the more finicky bits of the problems myself. Yay me! :[

Anyways enough griping, what we did today. To start, we had another little quiz. I'm not really quite fond of these things, although I suppose they're good tools for both Mr.k's and our own assessment of where each individual student is/how well they're doing in the course. As usual we had four questions, and this happened to be one of the no calculator varieties.

So onward with the first question.



So in this question, your given the graph of f(x) and asked to find some information (specifically, which of the 5 points f'(x) < 0 and f''(x)> 0. This for the most part is testing to see if you know what the first and second derivative tell you about the parent function (namely that the first derivative tells you weather the parent function is increasing or decreasing, and the second tells concavity.) Knowing this we find that solving the problem is a simple matter of looking at the concavity around the point, and if at that point on the parent function, it is increasing or decreasing.

Second question



This question 100% wont be on the exam

Why?

Because it's way to simple to be on the exam.

quote endquote of the conversation involving this slide. Pretty much, you just do the red in the picture, take the value of f(x) at 4, subtract is from the value of f(x) at 1, then put that over 4 - 1, then voila. answer. (in this case 4/3)

Neeexxxttttt



This question was also somewhat easy. If you read through the question properly you'll find it's just a direct application of the chain rule. Simply follow the values on the chart, plug em in, and away you go.

FINAL LAP!

dundun dundundun dundeedee!



SO, this one turned out to be a related rates sort of question. The first step here was to write the formula for the volume of a sphere, then differentiate that. Then, you simply begin plugging in some values (namely, a value smaller then 1 and one bigger then one) What you find is that with values of r smaller then 1, the sphere decreases in size, and that with values of larger then 1, the sphere increases in size. Turns out, this is one of the answers. yippee :]

Okay so this is where it gets kinda foggy, so please bear with me guys D:



So this question here, is basically a continuation of the stuff we were doing yesterday, with the funky density questions and the latter rho (whoa Iknorite?). Benchmen showed us the light with this one, so props to him for that goodness.

So in part a.) we need to write a function which will give us the number of cars. We know that the function given describes the traffic in terms of cars/km. To find cars we need to multiply this my Delta km (change in km.) Now as bench said we're actually trying to get really small changes to Delta km ends up ad just dx. (As Shown on the slide)

After all that we put it back together, and find that to find the number of cars, you just do this. (its kinda hard to explain in words so I just cropped it xD)



The next bit of the question (aka. part B.) goes something like this.



Quite simply, all you do is evaluate over the interval 0 to 30 (since you have 30km of road your working with.)

Alrighty guys, its not time for the post titled, BOSTON BAY PROBLEMMMM

dum dun dundundundun.



Now, instead of tackling this whole thing at once, lets start with just part a (or what we started of it anyways before the class ended D:)



So, from what I can tell by looking at the blog, we got as far as setting up the units, (aka, out delta whatever, that becomes the dx.) I'm pretty sure it was around this time the bell rang, and we got our homework and things.

Okay everyone hopefully that wasn't a nightmare to follow, if it was, I apoligize, thats what I get for forgetting my books -_-;

So ja, the next scribe is Paul because I havent picked him in awhile lol.

Ciaooooo

edit: Btw, homework is 8.5 6-10 and the rest of the boston bay problem :]

edit edit: Soo I couldn't leave without some goodness haha

Bobs, antiderivatives, and integrals oh my!

So this is my bob post for this unit. How did this unit make me feel? well for the most part I definitely understood more of it then just about any unit we've done so far, the mechanical aspects of it make it appealing to me on so many levels ;p The substitution stuff was interesting, I found that to probably be some of the easier stuff, along with the anti-derivatives of the arc-trig functions.

The harder stuff would mostly be everything I wasn't in class to learn haha (Namely, integration by parts, and the stuff from tuesday) although it wasn't really all that bad. Overall I can safely say that I enjoyed this unit.

So I think thats everything, I'm off to do a bit of last minute studying before I hit the z's.

Night, and good luck everyone :]

Omgosh fast scribelists and approximations.

Okay guys, I'm here again, scribing, which makes no sense to me cause like, I scribed only a few weeks ago it feels like. Anyways todays topic, beside omgosh lightning fast scribe posts, is Approximation, or more specifically, how approximate is approximate?



Now the real location of that campground is somewhere around the big purple dot and letters, NOT near the big purple x and the lies that google tells us. According to google, approximate locations means somewhere within about 40 miles which I can safely say is a bit much.

So the main thing to take away from this, was that when dealing with things as approximations you need to know how good the approximations are. Moving on.



You should recognize these as the questions homework from the other day. Mr. K kindly went through the solution of one lightning fast, while simultaneously commenting on how the other should also work out to be an arctan. The actual solution process shown there is somewhat straightforward. Bring the 100 out of the brackets so you get 1 + something, then make the x²/100 a square by turning it into (x/10)². From there apply substitution, and voila. The second requires some completing the square to solve, nothing really too complicated, just good practice.



K so the next topic, is shown here on the slide. How do you find an integral when you can't find an anti derivative? On the slide you see us trying some methods we know to integrate e^x2. We find substitution doesn’t work because, say you choose x squared as your u, you can get rid of the 2 once you find du but you cant get rid of the x. Also, integration by parts doesn’t work because your f’ is e to the x squared which we're trying to integrate in the first place, and conversely, don’t know the derivative of.

Here's some notes about the above slide that I took in class.

- e to the x squared is u shaped, but not a parabola. The parent function of this is similar to x cubed, but is not x cubed because things happen much faster. Everything is very abrupt.

- Finding the definite integral (I think?) of this function on your calculator gives an approx value of 1.4627

Next slide(s)!


Link: www.tinyurl.com/bu9kzs


Link: www.tinyurl.com/c4mwmo


link: www.tinyurl.com/Cojrp9


Okay, so all this might be a bit overwhelming, and confusing, but all we were really doing here, was comparing the accuracy of these various different approximation methods. As you can see, we compared normal old boring left and right hand sums, moderately exciting trapezoidal sums, and superb midpoint or tangent sums. Overall there were a couple main things to take away from the comparisons.

When trying to figure out if your estimate is and underestimate or an overestimate what it is depends on;
-The tonicity of the function
-If it is concave up or concave down
And
-If you are taking a right hand sum or a left hand sum.

But what’s the difference between a value of 3 and a value of 300?
-How concave the graph is.

And

The first derivative test deals with increase and decrease, while the second derivative test deals with concavity.

The main thing to take away from this scribe (and therefore this lesson) can be seen in this quote taken from a very knowledgeable person.

"When you do an approximation its important to say how approximate is approximate. Being off by 40 miles is not cool. In other words, you need to know how accurate it is. "

-Mr. K

Haha. I think he said that anyways. I might have translated it into my own words or something D:

J + Me Is gonna be the next scribe, and your homework is the homework from yesterday if you didnt do it, and the stuff on the last slide. To do the stuff with Trapezoid sums, use the riesum program on your calculator to figure out the right and left hand sums, then add those two, and divide by two. For the midpoint sum, Set your X Choice value in the Riesum program to .5 for the middle :]

Hopefully that wasnt too fast, and hopefully it all made sense.

Justus saying ciao, and goodnight.

:]

Techniques and Substitution

So, I know this is long overdue, and I'm sorry guys I've been ridiculous busy. Anyways it's not monday yet, so technically, it's not late but yeah ;p

Anyways here we go with the blog for jan 23rds class on substitution! :]

Alrighty, so in our quest to be able to anti differentiate effectively we've found that its much easier to go forwards, then it is to go backwards, because the nature of the original problem isn't always apparent from the differentiated version (ie, the result, isn't always indicative of what the beginning was like.)



Now the whole idea of substitution kind of ties into running the chain rule in reverse thing (as seen in the above image.) In the example function to the right of the definition box thinger, the inner function is shown to be

x² + 3.

Well using substitution we just change this out for something slightly simpler. Sayyy, u? Sounds good to me. So now that we've got that out of the way, we need to actually differentiate the value of u (which we know, by the instructional box, to be g(x). In this example, we end up with

u = x² + 3
du = 2x dx

Now, what we want to do, is isolate xdx. To do this, we divide du by 2

du/2 = xdx

So after thats said and done, we end up with





Which can then be converted into





THUS we are on the next step in the "down, over up" process, that being anti differentiating integral symbol F'(u) du (aka, anti differentiate what we have above.) Doing this gets us




Continuing along with the process, we now have to substitute the value of u back in, in order to get our final answer.



And we're done! Now that might not seem like to much to do, but it's VERY VERY important to follow the process, and have good bookkeeping. IE. Keep track of your substitutions, and where everything is, because it's very easy to get lost and things.

Now on the next slide, is another example of substitution in action.



However, what in these two examples, all the "pieces" are these so to speak. But, what do you do, if they aren't? D:



In the example above, you see that if x is equal to -7, some problems arise. However, as seen in the slide, we may get around this problem via some algebraic massage.

Since u = x + 7

Then you can say that

u - 7 = x

and

du = dx

From there its all a matter of following the process like we have been for every other question, albeit, with slightly more algebraic massage :D

Alright guys, I think that about sums everything up. Hopefully it all made sense D:

The next scribe shall beee (checks list) J + Me, since I've hooked her up with a bunch of new tunes, she needs a good excuse to sit on her laptop listening to them, and a scribe post is just that excuse I think! :]

Now I'm off, to finish our video D:

Ciao!

Integrals Bob

OMGosh so much blogging in one night haha.

Anyways, this unit. Not to bad I must say. I kinda like this stuff, although it is a bit confusing at times. Overall, not my strongest unit (the first integrals unit gets that award), although definitely not my weakest (since it IS a cousin of the other one)

Easiest bit: The mechanical process of taking the inner function and replacing things and mixing it around. It's mah fav :3

Hardest bit: Remembering that the inner function is the derivative of the accumulation function, and the parent function is found by anti-differentiating the inner function D: Mind twisting stuff there.

Overall: I'm ready for round 2! :p

Anyways, I need to go to bed, so I can have a decent sleep in time for tomorrows test D:

night guys, and good luck!

Pre-Testing Revised, David vs. Goliath Stories, and Area

Okay, so I'm the scribe today amirite? Sounds good to me. Before I begin I'd like to explain the David vs. Goliath story thinger.

Basically, today at my basketball game, we beat Oak Park by about 4 points. Now this might not mean anything to you, but it does to us (the basketball team.) On paper, Oak Park should destroy just about any team in manitoba. Hence the DvG story.

Anyways onward with the math. Today we had a revised version of the pre-test, which I liked much better. Coming out of that pre-test I felt much more prepared for the test tomorrow then I usually do, which is a good thing! :p Basically, what we changed was, instead of working by ourselves for a large portion of the pre-test, we worked on it independently for about 20 mins. We then went (read: Were placed) into small groups, and treated it somewhat like a workshop. After about 10 minutes of that, we spent the rest of the class going over the answers. It's those answers/ After pre-test discussions I'll be going over now ;p

Question 1.
















What the writing means:

Essentially, how the problem is set-up means that F(x) = the stuff in the slide written in black.

Now we know that the derivative is the underlying function. It is from this notion that we end up with the next bit, (starting with F'(x))

What see you there in blue is the anti-derivative of F'(x) and therefore is F(x).

From there it's simply a matter of finding the values of those functions at (2) (the bit you see in the green there.) One thing to remember for this last step here, is that the integral of any function from x to x, is 0.

Answer: E

Question 2.

















Alright, as you can probably guess by all the blue circles the main idea here is that the accumulation function is actually a COMPOSITE OF FUNCTIONS. This means, that you need to use chain rule when differentiating.

SO the procedure here is fairly standard, take the derivative of the outer at the inner, multiplied by the derivative of the inner. The result is the green stuff below. Then evaluate this function at -1 and you have your answer!

Answer: E

Question 3.

















Okay, lets see if I remember this. Basically, the first thing to do, was sketch the graph (which is usually a good move.) Next you need to do some logistics work, by looking at the actual integral itself. You should notice that the inner function f(x) is all that there really is, and this is given in the question, so you don't need to do any guess work there. Next, start going through the options.

Option I : F(1) = 6.5

This is confirmed by counting up the squares. If you did it right you should end up with 6.5.

Option II : F'(1) = 3

The thing to remember here, is that the derivative of F(x) is f(x) which is shown in the graph. Therefore, all you need to do to confirm this, is go over to 1, and read the according value on the y axis.

Option III : F''(1) = 1

To check this one, you need to remember a continuation on the first idea from option 2. As said before, the derivative of F is f(x), so the next step would be to look at the second derivative of F. It just so happens that the second derivative of F is the first derivative of f(x). Looking at the function for f(x) and taking the derivative of it's two pieces you get f'(x) = 0 and f'(x) = 1. It just so happens, that f'(x) = 1 is just what we were looking for. Thus, myth confirmed!

Answer: E

Question 4.







This one was easy enough, a fact confirmed by the lack of needing an explanation.

Basically, use your calculator for this one. Plot the graphs, find the interval in which they overlap. 2nd calc > 7 select your lower and upper bounds. Repeat for the other graph, and subtract the values. If that all went according to plan you should have something like, 1.66666666666666666666666666666666666666666666666666667, or 4/3.

Answer : C

Question 5.



















So, this problem is a simple definite integral.

Start by anti-differentiating, to end up with the first bit of blue seen in the slide.

After you anti-differentiate, substitute in the values seen on the definite integral (-2 and 2), and subtract. Its all simple algebra from there.

If it all went well, you should get k = 4, which happens to be answer d.

Answer : D

Question Free Response

















Part a.

So, you need to find the value of the accumulation function, G, on the interval from - 4 to 4 at -4. This should be simple enough, since we know that if the interval of an integral is from x to x, then the integral is 0.

moving on.

Part. b

This is a bit trickier, although not much moreso. G'(-1) is the same thing as asking for the value of the inner function (f in this case) at that point. Since we have the graph of f right here, lets do that shall we? Simple look on the graph to discover that f(-1) = 2 so therefore, G'(-1) also is 2.

Part. c

on the graph you may notice that it on the intervals (-4 , -3) U (-1, 2) the graph of f is decreasing, and therefore the first derivative of G is decreasing on those intervals as well. So, by taking the derivative of f(x), you get the second derivative of G, and can go ahead and apply the second derivative test. (I think D:)

You also have to justify why this is true (as in write a short little ditty about why its decreasing on those intervals and things.)

Sorry if this doesnt make sense, I kinda need to clarify some things here D:

Part. d

This bit is also somewhat hazy for me, but what I have in my notes here, is that you need to use extreme value theorem, and examine both the endpoints, and where the derivative is zero.

I also recall hearing that if you dont examine the endpoints (read: forget them)


Okay guys, I think that about wraps it up for todays scribe post. Since it's a new cycle I think I will pick francis to be the next scribe mmkay? Hopefully you kinda understood what I was talking about here. I apoligize if you didn't. Any comments/critique would be appreciated!

~Justus out.

Team FLJ Timeline

This is FLJ's Timeline as of today. Timeline is subject to change.

TIMELINE:

JANUARY 9 - Record Copied Examples of Questions for Project Use -Complete-

• Six examples are to be copied from Calculus textbook to be used as models for questions to be created.
  

JANUARY 23 - Create Six Questions from Copied Questions for Project Use

• From the six examples taken from the book, they will then be used to create questions that are going to be displayed in the project.
  

FEBRUARY 13 - Solve and Annotate the Six Created Questions

• The six created questions are to be solved, corrected, edited and annotated as to be properly displayed in the project
  

MARCH 6 - Begin Filming Gameplay

• Gameplay filming will begin via Xbox 360 and Halo 3. Filming will begin as soon as a HDPVR is acquired. Many long nights of filming will occur.
  

MARCH 31 - Complete all Gameplay Film

• It is expected that by this date, filming will be completed
  

APRIL 1 - Begin Editing Gameplay Film

• Post-production of the project will begin. Special effects, voice acting and video editing will take place. Many long nights will again, also occur.

APRIL 13 - Begin DEV Blog Construction

• A separate blog will be created for the DEV to be published to. Will consist of posts for Solved questions, Reflections, Sources and Credits.

MAY 1 - DEV Revision

• Prior to "handing the project in", the blog will be revised that it is in correct layout, questions will be revised for errors, and links and videos will be tested for optimal viewing.
  

MAY 3 - 100% DEV Project Completion

• Film will be completed and fully uploaded. Link to DEV blog will be revealed on the Expert Voices blog. Digital copies will be created and will be published via DEV blog. Project will be "handed-in" for evaluation.

This project is a sequel and will most likely require you to view the previous project : MITSOP.

This Project is Rated EG for Educational Guidance

This film contains scenes of Intense Math, Frightening Equations, and the Use of Patterns

Bobbing for 2009

So, it seems just by glancing over everyone's bobs, that the break really took a toll on their "remembrance" of the material we did prior to the break. I hate to admit it, but I'm pretty much in the same boat.

For me, it seems that the pre-test, was basically a bunch of questions, that I knew what to use to figure them out, but couldn't remember how to do it at all. Like I'd see one, and be like, "Oh! you need to use mean -value theorem to solve this. Wait. How do you do that again?"

As you could probably imagine that became terribly frustrating quite quickly.

In terms of the unit itself, I found it somewhat high on the difficulty scale, not really super mega hard, but no where near as easy as integrals. The easiest part was definitely the bit with taking the limit of a rational function as it approaches infinity, where half the terms like disappear and things. That was a fun bit! :D

Besides that though, I'm not enjoying these units. :[ Difficulty wise anyways ;p

Hmm, so thats all I can really think of for this time, so I'll see everyone tomorrow. Good luck!

ciao

(Di)Lemma's, Proofs, and Other Things.

By now it should be no mystery to everyone that I, (Hi I'm Justus, glad to meet you) am the scribe for today. Now I regret to inform you that this scribe post might now be wholly as spectacular or colorful as the one I last left you with, and this is due to the fact that well, I just don't have that much time on my hands today ;p As a result you get the scribe post you may or may not be about to read. I apologize in advance for any trouble this may cause.

So without further ado, onward with the show!

Where to begin...where to begin...

This is indeed a hard question, since we really didn't start with anything in particular, but instead jumped all over the place across a few topics we all felt would be good to talk about. Well I'll just start with what I have first in my notes here.

All right, so to kick things off we started with some more proofs (Or rather, logical explanations) for the differentiation rules (Which Jamie so graciously typed up for us in a lovely format.)

The first (according to my notes) was;



Now Mr.K explained to us that it is quite possible to go through the algebraic explanation of this differentiation rule, but also said that it is much simpler, so just remember the graphs of sine(x) and cosine(x)


Sine Graph



Cosine Graph



Looking at the sine graph as the graph of some function we want to differentiate, we may pick out the important points/bits of information found within the graph itself. By doing a quick glance you can find the local minimums and maximums, as well as the inflection points for the graph. These will be used to construct the derivative graph of sine(x).

Our experience with the graphs of functions and the graphs of their derivatives tells us that the red points are the local minima/maxima and that the blue points are inflection points. We know that because the red dots are local extremum, they will become roots for the derivative graph of f. We also know that the blue point(s) is a maximum because the slope at the inflection point is positive.




At this stage it should be easy to see, that connecting these dots gives you a derivative graph identical to cos(x). Therefore;



We also dealt with the derivative of cos(x) being -Sin(x) following basically the exact same steps. Those can be seen in the following image.






Moving along nicely then :]

The next topic of discussion, is kind of a twisty windy, jump off into several lemmas, resulting in an organization dilemma, and me having to sort out some thoughts kinda topic. For the most part, it involves this;



and the question, how would you find the derivative of that logarithm? This is indeed a tough question, as we didn't (at that point in time) have any differentiation rules for logarithms of any base (which in reality, is what we were trying to find at the time.) However, by using the change of base law, and some knowledge about differentiation rules with did know at the time, Mr.K showed us the way.

LEMMA TIME

(something about lemmas imagery here)

Change of Base Law, how it works, and what it looks like on the inside.

So, say you have a logarithm base "a", that equals "n", that you would like to change to base "c."



You must first start by re-writing the logarithm as a power (In this case, a to the n equals b.)



The next step is to take the log of both sides giving you the following;



After you have completed that, you would use the power law to bring down the "n" exponent on the left side leaving you with something you can work with.



After that, all you have to do, is isolate N. Looking back at the beginning we see that in the very definition of what N was, lies the logarithm we wanted to change the base of. What we're left with, is N (aka. our logarithm) equals a new logarithm, now with a different (hopefully more useful) base.



I hope that made sense ^_^;, just in case, here's all the steps together, so you can see it as one fluid thing, instead of a bunch of seperate steps.

1.

2.

3.

4.

5.

Back To Our Regularly Scheduled Programming.

So we have this logarithmic function



which we would like to find the derivative of. The first step would be to apply our newly a wholly understood change of base law to end up with this;



From here, we will pull out 1/ln2 so that we have;



the reason for this lies in the fact that although ln2 is a logarithm, it is still just a number, and therefore, a constant. By pulling it out of the whole thing, we get the nice and easy to differentiate, lnx by way of the constant multiple rule. For those of you who may be having trouble remembering exactly what that is, it basically means that;

"The derivative of a constant times a function is the constant times the derivative of the function."

As a result of this differentiation rule, all that remains is to find the derivative of lnx which we know to be 1/x

*Note* We did a Lemma for the differentiation rule for lnx but when I looked at my notes I couldnt really makes heads or tails of it. I'll need to check it over with Mr.K and get that up here once I understand it. Sorry guys D:

The result, is the derivative, which in this case, happens to be



Now if we think back to the beginning, the primary goal was to see if we could find a rule for all logarithmic functions, and to this end I believe we achieved our goal. By simply substituting a variable (lets say a) into the spot of 2, you get the rule for differentiating logarithms.



With that stuff out of the way we moved onto the proof for d/dx of tan x, which happens to go something like this.




After converting tan into something more friendly, you simply apply the quotient rule.







Tadaa! If you wanted to find the derivative of say Cotangent, you would do the exact same thing. Turn cotan into Cos x over sin x, and apply the quotient rule from there.

Another neat little tidbit of information for you. According to Mr.K, the derivative of all "co" trig functions is negative. Just thought you'd like to know ;p

Now, after alllll that stuff behind us, we got back to antiderivatives. Remember that an antiderivative lives on the idea that anything that can be done, can be undone. However, with derivatives, because of how we figure them out, there is a small kink. Any constant in a parent function, is lost when you take the derivative of that function. Essentially, you may know a functions derivative, but you may NOT know its vertical positioning by the same method. In effect, when you differentiate a funtion, you are finding a family of functions, not just one. Now there is a way to figure out what the constant was, if you have a point from the graph you'd like to single out. These are called intial value problems. Lets see an example.




Because we know an antiderivative gets rid of the constant, we need to add that back in. We do so with the value "c", seen above.

Know this is called the inital value theorem because we have a value at the beginning (initial) of the problem. In this case its the value of f at 1



From this point we substitute the initial value into the antiderivative and solve for c.





And thats really all there is to it :] In case that didnt make much sense, here's the slide that explains it better then me I think D:



Okay guys, I think that about wraps it all up for tonight. Not gonna go into a big spiel this time, its late, and I'm tired, and I have basketball tomorrow, and the day after, and the day after lol.

SO with that said, lawrence shall be scribe for the next day (which is technically today.) Alright? alright. Night all :]