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Thursday, October 30, 2008

October.30th 2008

For today's class we continued Chapter 3.4 The fundamental Theorem of Calculus.
We get to read through the section and figure out the examples on our own.
Most of us did example 1 Determing cost from changing price rate.
Well the question and the solution is in the text book page 181.

Dr. Eviatar said that we will keep doing this chapter tomorrow.
Monday is the pretest, and Tuesday will be the test of chapter 3.

I'm done! The next scribe is Shelly

ps: Happy Halloween for tomorrow ^_^

Wednesday, October 29, 2008

Introduction to the Fundamental Theorem of Calculus

1. A particle is traveling along a straight line. Its position is given by S(t) = t^2 – 3t + 1. Find the change in position from t = 1 to t = 4.

∆y/∆x = [s(4)-s(1)]/(4-1) = [(4^2-3*4+1)-(1^2-3*1+1)]/(4-1) = 2

The question is asking for the change in position, which is the slope. Since time is the independent variable and position is the dependent variable, the derivative of the graph is the velocity, which is the slope.

2. Suppose a car is moving with non-decreasing speed according to the table below:
t (sec) 0246810
speed (ft/sec) 303640485460

a) What is an upper estimate for the distance traveled in the first 2 seconds?

36*2 = 72ft

b) Determine upper and lower estimates for the change in position for the first 10 seconds.

Upper estimate = 36*2+40*2+48*2+54*2+60*2 = 476
Lower estimate = 30*2+36*2+40*2+48*2+54*2 = 416

∫y’dx = change in y over the interval [a,b]

The definition of the Fundamental Theorem of Calculus is on Slide 6.


Chpater 3 Supplementary Problems
Chapter 3 Pre-Test and Test is postponed to next week.
Next scribe is Yinan.

Slides Oct 29

Here they are ... tomorrow we will work and TALK.

Ap Calc Oct29
View SlideShare presentation or Upload your own. (tags: apcalc2008 calculus)

Tuesday, October 28, 2008


Okay, so it wasn't a workshop class, and was kinda more like a quiz at the beginning of the class, but when everyone was done, it turned into something more like a workshop. So anyways, we were handed out sheets on definite integrals. Now, I can't explain what was on the whole sheet because we handed in the sheets.

First off, we were given a table of values, and we had to plot them on the given graph (I honestly don't remember those values.)

Next we were asked to find the lower and upper limits. Which were around 17.1 and 23.1?? ... Or something like that.

Then we were to estimate what the exact area was. I believe mine came up to around 19-something. Anyone else get the same thing?

Next we were to use different functions on our calculator to find the answer.
First question was to use the RIEMAN Sum program. I got a sum of 12.06.
When you use the program this is what you input:
NUMBER 10 <--- This is the number of subintervals

The next one, we were to use fnInt. The way this one works, is that you have to input the equation given to you in Y1.
Then input it like this fnInt(Y1, X, 0, Pi) *To put in Y1, press Vars, move right to Y-Vars and select Y1. Inputing this in simply as Y1 using Alpha will create an error.* For this one I also got a value of 12.07

The last one we were to use the 7th option under the Calc menu.
A graph would show up (because the equation is in Y1 already) and we were to use endpoints 0 and Pi. So you would choose X = 0 as the Left endpoint and use X = Pi as the right end point. After that, press enter and you should get a value around 12.07

And that's all we did today. So I guess the next scribe will be... Zeph!

Scribe List

Cycle 3

.:. J + ME .:.
le joséph

Not Paul

Hi I'm Justus

Quote of the Cycle ;

"Knowing is not enough; we must apply.

Willing is not enough; we must do."

-Bruce Lee

Rence ~ Out

Monday, October 27, 2008


I guess I ran out of creative titles for this scribe post, but it's a Monday. What can I do? Well, the majority of the time was spent dwelling on the little test we had. I'll put it lightly. MOTHER FATHER!!!! It was completely devastating...especially for my self expectations. It really brings me down you know?

Anyhow, I can't really re-explain the answers of the test considering that these tests were probably "recycled", so what I'll do is give reminders and notices for the important dates coming soon in class and maybe recap what I've learned in this chapter so far. So I won't have to guilt trip myself into scribing one more time. I'll...compensate for it. I guess that's the right terms for it.

Nothing really comes to mind except the likely test we will have on Friday this week. Happy halloween. That's what makes it so scary. Not the gore, the blood, or the costumes; it's the tests.

.:. J + ME .:.'s REVIEW OF CHAPTER 3 [I'm warning you..not to rely on this, it's probably not very informative. I'm just doing this to see if I understand this stuff or not. It's an effort, right? Wow.. whatever happened to my skills in English? I'm writing so informally. **sigh**]

This is the first part of the chapter which acts as a link between the concepts of the previous chapter 2 and preparation for what will be taught further into the unit.

We discover that both integrals and derivatives have direct relationships with each other given that a derivative is a rate of change involving velocity and integrals involve determining the distance traveled, which is basically the area of the region below a function.

This chapter is easily explained remembering the idea of

distance = velocity x elapsed time.

This chapter focuses more on approximating the areas of rectangular regions below the functions and also parts of the regions above the function.

To find these areas, we take the limit of these approx. sums and find the area using the measurements of the intervals and the height of each interval that touches the function. The smaller the intervals are, the closer we are to being more exact in our approximates in area.

The name RIEMANN sums is just the name assigned to classify the method to find these sums.

Of course we have to take into account inscribed [below] and circumscribed [above] rectangles since, they do not fill in the spaces exactly.

Especially when looking for Riemann sums using something as obscure as a calculator, it makes it easier when we look for subintervals and choose midpoints [or left or right endpoints], so that it is easier to find what are known as RIGHT HAND SUMS [above fcn] and LEFT HAND SUMS [below fcn].

In this part of the chapter, a variation of the sigma notation is introduced, something that looks like a squiggly line. How fun to illustrate. But the subscripted and superscripted values are the coordinates of the main interval, represented by [a, b] in the examples. We are still finding the sum of course, and it is the sums within the vicinity of of the main interval given. There is an infinite amt of limits of the number of intervals between [a, b].

I'm a bit choppy on this part still, so I'm sorry. I don't want to mislead anyone.

But another thing worth noting is the distinguishable difference between MONOTONOUS fcns and NON-MONOTONOUS fcns.

Monotonous Functions: are functions that either increase or decrease, but never go both ways.

Non-monotonous Functions: These functions are capable of increasing and decreasing, resulting in an "unpredictable" function and thus, it's intervals that don't follow a specific pattern.

Well that's all from me tonight. Ummmmm. Who shall be dubbed the next scribe? Rence, I guess.

With great power comes great responsibility. - UNCLE BEN. haha wow. when will that ever get old?

Sunday, October 26, 2008

Definite Integrals

Hello Benofschool here. Wow it's been a while since I last scribed so here I go.

We started off the class talking about Riemann and his sums. The Riemann Sums was the term given to the sum of the different areas of the graph between each interval that we created during the last few classes with M(r)s. Karras and Dr. Eviatar. More info about Riemann and his sums can be found in the link on the first slide.

Now onto the Integrals. We were introduced to a new notation called the Definite Integral Notation. It is supposed to be a funky looking Sigma notation with a few twists. In the notation it shows the main interval [a,b] in which we are finding the sum, the function being used, and the size of the intervals between the closed larger interval [a,b]. So the definition of this notation involves limits where the number of intervals between [a,b] is to an infinite amount. Since it is human impossible to do this (Nothing is impossible if you BELIEVE) we try to get as close as possible.

Okay to work with this notation we locate the section of the function in which we are investigating. The definition of the notation is to get the sum of f(x) multiplied by the change in x up to the nth interval. If you remember from previous classes, the more intervals that we include in our calculations lead to a more accurate estimation of the integral of a function.

We looked at a couple of examples in the next few slides and continued on to a bit of something new. Monotonous Functions are functions that either increase or decrease, but never both. These functions give us the ability to determine the margin of error in our estimations. This is because in Non-Monotonous Functions the margin of error will be useless because of the change in the rates of change. Since Monotonous Functions either increase or decrease they will have a error that can be determined. This error shows how many intervals are required to find the Integral. Depending on the function some might need more than others.

Homework is Chapter 3.3 and do questions 1, 2, 4, 5, 7, 11, 17. But do enough that you understand it. No point in doing too much or too little.

The next scribe will be .:. J + ME .:.

Good night and do not let the Cimex lectularii masticate your epidermis, imbibing your blood.

Friday, October 24, 2008

Slides October 24th

Here they are ...

Ap Calc Oct24
View SlideShare presentation or Upload your own. (tags: definite math)

October 23rd: Return of the Definite Integral (sort of)

Starring: The Riemann sum
King Kong as Mysterious outline
and Parabolic Building as Opening Picture.

Sure to be a hit this fall, don't miss it!

Anyway back to serious business.

Today (or rather, yesterday if you want to be technical about it), we continued on yesterday's class about the definite integral. This time, instead of getting data from a table to find the mean of the upper and lower limits, we were provided with a function f(x) = x^2 and used that instead.

We'll start on slide 3:

Here we sketched the graph and tried to find the mean of the upper and lower limits.
We created two intervals: [1, 3/2] and [3/2, 2]. We tried to find the upper and lower limits, but give up.

On slide 4, we change the equation to f(x) = x +1 to make things easier.

So we find the lower limit using the Riemann sum (described in detail on the next slide) and by using some basic algebra.  The answer is the same because the function is linear, but obviously the whole solving by subtracting rectangles wont work so well with a wavy and unpredictable function.

I'm going to try to explain the algebraic thing, but it'll likely be unclear and using general terms.

Since f(x) = x +1, we know its linear, meaning its a straight line. No bumps, curves or waves. Because we want to find the area beneath the function, we can simply get the values by using an interval, in this case [0,4]. So we make a rectangle from x = 0 to x=4 and it goes as high as the function does at this interval (5). So we have a length and width, and thus an area (20). Now, because we want the lower limit, we dont want this extra space thats above the function (because the function is not rectangular, rather it is trapezoidal). So we subtract the extra space by finding the length and width of it (4) and dividing that area by two (because we want to keep half of it). So we get 20 (the whole area) - 16 (the area around the function on the interval [0,4]) / 2 (because we want the half of it that is still under the function) = 12

Which is the same result as we get when we used the Riemann sum.

On slide 6 we tried some more with the f(x) =  x^2 problem. I believe here we compared the accuracy of using smaller intervals vs larger intervals. 

Slide 7 is just talking about intervals in formulas. Here delta t represents the difference between two values on an interval while delta v is the output difference related to delta t.

Slide 8 is an ad.

Well, sorry this post is so late and mostly likely incomprehensible. The next scribe is benofschool.


And because we all love this video now, I give you:

Tom Lehrer!

This is a great song to write blog posts to. It's so catchy.

Thursday, October 23, 2008

Wednesday, October 22, 2008

Scribe List

Cycle 3

.:. J + ME .:.
le joséph

Not Paul

Hi I'm Justus

Note: Yes, I'm aware they are no longer in the class. I just have a case of nostalgia. Haha 8)

Quote of the Cycle ;

"Knowing is not enough; we must apply.

Willing is not enough; we must do."

-Bruce Lee

Rence ~ Out

October 22. The Definite Integral

I guess I'm scribe again since I've been caught by Mr. K. That's alright though, no big deal. At least he's still around. :D

Today we started off my watching a youtube video entitled "Lobachevsky - Tom Lehrer" The video is about a mathematician named Lobachevsky who supposedly plagiarized from a fellow mathematician named Gauss.

Well back to the subject. We started out by graphing some values.

The red bars represent the velocity or speed at the given time intervals as it changes gradually. The orange and green bars represent pretty much another graph, where the change in velocity was instant. Whether the change in speed was gradual or instant, we can't determine so. The curve in between the orange and green is the distance travelled, and this is what we will be trying to find.

We then try to find the lower limit by finding area of the lower limits (found in red bars).

Lower limit = sum of area of least possible #'s.

We multiply each value by the time interval (1).
Lower limit = 1.4(1) + 2.7(1) + 3.5(1) + 4.5(1) + 5(1) = 17.1 ft.
Don't add the last value, because this is the upper limit and it can't get any higher than this (t=5).
Now we calculate the upper limit, which includes the orange and green bars.
Upper limit = 2.7(1) + 3.5(1) + 4.5(1) + 5(1) + 5.7(1) = 21.4 ft.
Now estimate the distance by taking the average of these two areas. (17.1 + 21.4)/2 = 19.3 ft.
If we're talking about limits, the range would be 17.1 ft. to 21.4 ft.
At this point we find the ambivalent area (area of uncertainty) which is pretty much the orange-green bars. Ambivalent area = 1.3 + 0.8 + 1 + 0.5+ 0.2 = 4.3 ft.
We have an estimated area of 4.3 ft. under the curve which doesn't exactly match the previous results we estimated.
Now we decrease the time intervals to be more precise, and to see what happens.

Let's find the lower limit, or left edge of this graph. Left edge = 0.7 + 1.35 + 1.5 + 1.75 + 2.05 + 2.25 + 2.4 + 2.5 + 2.7 = 18.2 ft.

Now we find the upper edge which is pretty much the same, but we exclude the number of least value and include the number of most value. It should equal 20.35 ft.

The distance will be between 18.2 ft. and 20.35 ft.
The mean distance is (18.2 + 20.35) / 2 = 19.3 ft.

The average distance remains the same as our last graph.

Therefore by decreasing the size of the time intervals we have brought both the upper and lower limits' ranges lower.
At 1 sec. interval range = 17.1 ft. and 21.4 ft.
At 0.5 sec. interval range = 18.2 ft. and 20.35 ft.
The mean stayed the same at 19.3 ft.

That was all we did in class, our homework is Exercise 3.1 questions 1, 3, 6, 7 and 9. Hope you enjoyed, if any issues, be sure to tell me and I'll be sure to correct them. The next scribe will be well, not Richard, but Not Paul.

Tuesday, October 21, 2008

October 21, 2008

Wow, I'm scribe again! Haha, thanks Lawrence.
Today in class we had our test on Derivative Functions, it was quite tough, because there were questions in there that we hadn`t reviewed before, but next unit should be easier.
That is all we did in class today, now you know why I wanted to be scribe.
The next scribe will be Joseph!

Monday, October 20, 2008

Scribe List

Cycle 3

.:. J + ME .:.

Not Paul

Hi I'm Justus

Quote of the Cycle ;

"Knowing is not enough; we must apply.

Willing is not enough; we must do."

-Bruce Lee

Rence ~ Out

The Pre-Test

Hope you've all been studying, because I know for a fact that some people were as confused as I was on that pre-test.

Today we just had a pre-test and went over the answers. We were all in class, so I don't think I have to go over the answers in detail.

1) A
b)f(x) = -2.25 (x-5.2) + 8.8

So by request, Francis asked to be scribe, for the same reasons everyone would want to be.
Therefore, Francis will be tomorrow's scribe.

Slides October 20

Here they are. Good luck tomorrow!

Ap Calc Oct 20
View SlideShare presentation or Upload your own. (tags: apcalc2008 calculus)

Sunday, October 19, 2008

Scribe List

Cycle 2
.:. J + ME .:.

Not Paul

Hi I'm Justus

Quote of the Cycle ;

"There is no such thing as good or evil...

...But thinking it makes it so."


Rence ~ Out

Friday, October 17, 2008

Friday's class of a couple Theroms.

Today in class we learnt about a couple new theorems. The first one is the "Intermediate Value Theorem". This theorem pretty much states that according to slide 1, found on the graph, that if y= f(x) is continuous on an interval [a, b] and y = k in between points y=f(b) and y=f(a) then there is an x-value of c where f(c) = k. If a discontinuity is found, remove it. I believe that what Mrs. E means is if there is a discontinuity between the chosen interval of [a,b] then chose a new interval by moving it to a place on the graph where there is no discontinuity. Confusing? Yeah, just look at the graph and read the notes a couple times. That's what I did.

We also looked at how to find zeros on a continuous function without our calculators. We were given an example on slide 2. f(x) = x3-3x+1. To find the zero of this function, we chose points [0,1] for this example. We then found out where the x-values were either positive or negative. On the example the point 0 was input into the function and equaled to 1. Which was positive ( f(0) = 0-3(0) +1 = 1). At the point of 1, the function was a equal to a negative number ( f(1) = 1 -3(1) + 1 = -1). Now we know that the zero of the function will be in between those two points. Now we want to hone in on a smaller region, so we should cut that interval of 0 to 1, in half. So we find the f(1/2) and see if it's positive or negative, so we know where to look next. The function of f(1/2) turned out to be negative, as found on the slide. We then know to look between points [0 and 1/2] because the zero will be in between a negative and positive value. It's best to divide the section by half to cover greater amount of space between the points. The next point we used was 1/4. We input this into the function and found it to be positive. Now we can look between points [1/4 and 1/2]. We then used the point of 3/8 in our function, which is in between our given points. It turned out this was negative, so now we know our zero is between 1/4 and 3/8 which is small enough of an interval where our zero is found, so we stopped there.

The next theorem we used was entitled the "Extreme Value Theorem". This states that on a continuous function, it will always have a maximum and minimum and will be an extremum value from where all points on the function will either larger or smaller depending if its the maximum or minimum extremum. Lets say that on a continuous function on interval [a,b] there exists numbers c and d where all x-values in [a,b] f(c) ≤ f(x) ≤ f(d). d will be the maximum and all x-values will be less that or equal to d. c is the minimum and all x-values will be greater that c or equal.

That's all we did in today's Calculus class. If anything is wrong, talk to me and I'll be sure to change it. The next scribe will be Lawrence!

Slides October 17th

Here they are ...

Apcal Oct17
View SlideShare presentation or Upload your own. (tags: math calculus)

Thursday, October 16, 2008

Continuity, and some more Limits

Alrighty guys, my name is Justus and I will be your scribe for today. Lets get started.

This image is basically a modified version of the first slide. Now this example was used to explain the concept of continuity. If you look at the image, I've drawn 3 sets of lines, the magenta (purpley) ones, the maroony ones, and the aqua ones. If you try to travel from one aqua line, to the other, you can do it easily, and without any kind of sudden changes. Therefore the graph is said to be continuous along that interval. The same goes for the magenta. However, when trying to travel from one maroon line to another, difficulty arises. At x = 0, there is a sudden change in direction, the line is no longer continuous, and there is now said to be a discontinuity. It is called a discontinuity because there is no way to get from one maroon line to the other in a continuous fashion. The next point we got into, were 3 defining rules of continuity, which I shall type out
here for everyone to see.


A function is continuous at x = a if;

1.) lim f(x) exists

2.) f is defined at a

3.) lim f(x) = f(a)

So after going over these rules, we got some examples on the board, so we might get a better understanding of what each means.

For the first we came up with the following

1.) Example: lim f(x) = 1/x
Counterexample: (look at slide)

So as seen in this slide image here, the example graph produces a nice, rule following reciprocal graph (reciprocal of the graph f(x) = x), whereas the counterexample gives a piecewise function, which has that huge gap/jump/aka. Discontinuity at x = 10 approximately. Alright, moving on.

2.) Example: f(x) = x/2 at x = 1
Counterexample: f(x) = 1/x at x = 0

In the second example there, it is pretty obvious to see what the problem is. In the counter example, the function is NOT defined at x = 0 because you are then dividing by zero, and we all know how much fun that is. In the example however, all is good and well, and f is defined at a, as required to be continuous.

3.) Example: lim 1/x = 1 = f(1)
Counterexample: (check slide)

Okay, so the counterexample in this one, might seem a little confusing but I'll try to explain it as best as I can. Basically, its saying that f(x) = x2 when x does not equal 2; When x DOES equal 2, then f(x) = 5. Thus you get the graph with the weird point floating in the air. This point also leads to that magical thing called a discontinuity again.

Now it was at this time Dr. E said to checkout Pg's 140-141.

The final little tidbit in the lesson involved continuity on an interval, which basically says that if function is continuous from points a, and b, it is continuous on that open interval a to b. This is shown in more detail on the following slide

The lim f(x) = f(a) basically means that starting above a and moving towards it, the function
is continuous. The opposite is true for the lim f(x) = f(b) which says that starting below b and x->b-
moving towards it the function is continuous.

Now thats about it for my scribe post for today. I know we did a couple other things, (two slides), but I forgot my textbook at the school, and I didn't really understand them at all, so if someone could help out with that (ie by talking about it a little bit in their post, that would help me tremendously, and I could then go and fix mine.) Either way, like I said, I will need to get my text to help me explain those two slides.

So I think thats everything, which only leaves choosing the next scribe. I've decided it will be francis mkay. kay.

OH And I almost forget, theres alittle note about pg 143 here in my notes, which means it's probably a good idea to check that page out. As far as I remember there weren't any questions assigned for homework for tonight though.

Anyways I'm off, because I have lots of other things to do now. Hopefully my post made enough sense, and I apoligize for any discontinuities in it; feel free to point those out btw, so I may fix them.


Slides October 16th

Here they are ....

Slides October 15th

Here they are ....

Ap Calc Oct15
View SlideShare presentation or Upload your own. (tags: apcalc2008 calculus)

Slides October 14th

Here they are ....

Ap Calc Oct14
View SlideShare presentation or Upload your own. (tags: apcalc2 calculus)

Wednesday, October 15, 2008

Cycle 2
.:. J + ME .:.

Not Paul

Hi I'm Justus

Quote of the Cycle ;

"There is no such thing as good or evil...

...But thinking it makes it so."


Rence ~ Out
Hey bloggers! Its me, Shelly, here to blog again for this cycle, including my little ROFL corner. To start things off we have received handouts of the equation that we were working on Friday and had finished yesterday, but there was a problem, some people were a bit confused due to the long 4 day intermission we had and us continuing off were we had left off in the middle of the equation we were working on. So to clear things up we were given hand out that shows us how to solve the equation.

Hand out
page 1
page 2
page 3
page 4

As we head into something new, this handout of Algebraic Limit Rules is handy (which was given to us on Friday)

Limits and Functions
Suppose that limits

exist and let "c" be a constant

CMR (Constant multiple rule)
The limit of a constant times a function is the constant times the limit of the function

lim c f(x) = c lim f(x)

SR (Sum of Function Rule)
The limit of a sum is the sum of the limit

lim (f(x)+g(x)) = lim f(x) + lim g(x)

DR (Difference of Function Rule)
The limit of a difference is the difference of the limits

lim (f(x) - g(x)) - lim f(x) - lim g(x)

PR & QR (Product of Function and Quotient of Functions Rule)

PR: lim (f(x)g(x)) = (lim f(x))(lim g(x)) QR: lim (f(x)/g(x)) = lim f(x)/lim g(x)

Example 1:

Example 2:

Example 3:


The limits are the same but the function itself is different, for
f(x) = (x2/x-1 - 1/x-1) there is a discontinuity at x = 1

For more examples similar to these one look at example 2 and 4 in the Algebraic Limit Rules booklet.
Example 2
Example 4

Homework: Exercise 2.7
1-9 (odd)
14-27 (odd)
and 24

And now it is time for the ROFL corner!
Since the last riddle was a tad difficult, I'll give you guys an easier one, which means that you won't need a hint for this one. ;D

How many times can you take 5 out of 25?

And if that one is too easy try this one! (Sorry!! I don't have a hint for this one!)

What belongs to you, but is mostly used by others?

Have fun!

Finally the next scribe will be...... Hi I'm Justice!

Tuesday, October 14, 2008

Continuing From Last Class.

Well first we started off on the wrong track because we all forgot where we left off we then realized how did we get those numbers. we came to the conclusion that we had skipped a few steps will here is the notes but in the right order. i think.

First you use the critical points that you found and plug the values into the second derivative test and you will find the local max and min.

Next you find all possible inflection points by making the second derivative equal to zero and solve for x.

After Figuring all that stuff out you solve for the original function and use the critical points that we have found for the x values.

Now you need to find out if it is concave up or concave down.
To find the that out you need to choose a value from the ranges and plug it in to the second derivative formula which is f ''(x) = 4 - 12x

Next you solve for f(x) but this time you use the infliction points that we found for the x value.

This next picture is a picture of the second derivative i think. it shows when the graph is concave up and concave down.

Well thats all for today tune in next time. Next scribe post will be featuring Shelly.

Rguy out.

Friday, October 10, 2008


Hello blog. It's me Jamie again with today's scribe post or yesterday's really. I'm starting it up again, after it abruptly stopped for that day. At the moment the class is basically at the peak of chapter 2. It's still a bit vague for me but having class with Mrs. Karras and her famous notations like "fcn" helps.

First we spent the first seven minutes or so settling in and having a few warm up discussions, relating to math of course. Most of it. Time is golden, right? We were deciding amongst ourselves whether we were audio/visual learners who absolutely adore diagrams [and possibly lectures] and those theoretic learners who love...well, theorems and of course, there are the extraordinary who are both. But let's get down to business.

We started the class reviewing what we should know... and I guess emphasizing that knowledge on derivatives. We began with the FIRST DERIVATIVE TEST, although it was not called that in the text. Of course, this is one way to find the relative maxima or minima of a function, without being too tedious.

Like on the slides, there are the notes describing the function itself. But to carve this inside your head, I guess, I'll repeat. If this derivative test is to be used, it follows these conditions:
  • f is a continuous function located on a closed interval [a, b]
  • is differentiable on an open interval (a, b)
Along with those two conditions, are indicators of which direction the intervals would be, which are:
  1. If f '(x) is posve for the x values in the open interval, then '(x) is INCREASING on the closed interval, [a, b]
  2. If '(x) is negve for the x values in the open interval, then '(x) is DECREASING on the closed interval.
Secondly, to make sure we were all on the same train of thought, we defined EXTREMA. This the term describing either the local minimum or maximum of a differentiable fcn. 

Extrema occur in a fcn where f is on an interval (I) and c (a critical point; more info on critical points here) is a number in I. They also follow these conditions:
  • f (c) is the max value of the function on the interval given that '(x) is less than or equal to (c) or vice versa, which is shown in the next pt. [LOCAL MAX]
  • (c) is the min value of the fcn on the interval given that '(x) is greater than or equal to (c). [LOCAL MIN]
But if (c) is constant, then the function is BOTH a max and min value for EVERY real number c, of course considering that not every function has both a minimum and/or a maximum.

[more on...] CRITICAL NUMBERS 
This next section was a further explanation of critical numbers. But basically, if a function has a local extrema then either f '(c) equals zero or is undefined.


This was one moment in class where I was quite intrigued. It was one of those moments where you learn a concept through a different perspective, in this case, I learned a new way to describe what concavity is. Concave UP was when the function was curving upwards above the tangent line and concave DOWN was if it was curving downwards below the tangent line.

**NOTE VOCABULARY: INFLECTION POINT [click on the word for a link]


This occurs when a function is differentiable on an open interval containing "c" and f '(c) = 0. Once again, this indicates where there is a local extrema. If f "(c) is negative, then f has a local MAX at "c". If f "(c) is positive, then f has a local MINIMUM at "c".

With taking notes aside, we were given a warm up question that applied some of the things we've learned so far:
If f (x) = 12 + 2x2 - x4 , use the 2nd derivative test to find the local max and min for f . Discuss concavity, find points of inflection and sketch the graph.

Use the power law to differentiate the first and second derivative. The power rule is also known as: f ' (x) = nxn - 1 
f ' (x) = 4x - 4x3  first deriv
f " (x) = 4 - 12x2 = 4x (1 - x)(1 + x) second deriv

To find critical numbers, we must know that critical numbers exist when f '(x) [or the 2nd deriv] = 0, meaning that the critical numbers were {-1, 0, 1}.

Then to continue this problem, finding the inflected points makes finding the concavity easier since we can find whether the function increases or decreases by separating the intervals accordingly by using the critical numbers and plugging numbers that were in b/w intervals into the first derivative. But then the bell rang and we are to conclude this problem on Tuesday. Also, Mrs. Karras will be back to teach us about limits. 

Later that week though, just a reminder, when Dr. Eviatar comes back, we will have our pre-test for this unit and the test the following day. My post today wasn't quite exciting, but at least there was no major advertizing... [or is there?] But anyhow... I'm famished-- so are my fingers typing, so I'm done for tonight. The next scribe derived from randomosity is Richard.

Slides October 9th

Here they are ....

Tuesday, October 7, 2008

Scribe List

Cycle 2
.:. J + ME .:.

Not Paul

Hi I'm Justus

The class just seems to be getting smaller and smaller :(

Quote of the Cycle ;

"There is no such thing as good or evil...

...But thinking it makes it so."


Rence ~ Out

Slides October 7th

FWIW. Sorry for the confusion. Together, we will defeat this book AND learn calculus (and have fun, I hope).

Monday, October 6, 2008


Hi, it's me YiNan! I just found out that I'm the scribe for today!!
Well, we did section 2.5 Critical Numbers; Relative Maximum and Minimum Points.

We start off with the local maximum and the local minimum.

A local minimum, also called a relative minimum, is a minimum within some neighborhood that need not be (but may be) a global minimum.

A local maximum, also called a relative maximum, is a maximum within some neighborhood that need not be (but may be) a global maximum.

Global minimum is the smallest overall value of a set, function, etc., over its entire range. It is impossible to construct an algorithm that will find a global minimum for an arbitrary function.

Global Maximum is the largest overall value of a set, function, etc., over its entire range. It is impossible to construct an algorithm that will find a global maximum for an arbitrary function.

As x^2 is increasing, 2x is also increasing.

Critical Point is where f'(c)+0 or does not exist.

Then we did Example 1 on page 115.

Homework for today is 1,7,9,11,15,19,22,23,25 on Exercises 2.5

I'm done! next scribe is Yi Cong ( good luck! ^_^)

Slides October 6th

Here they are ...

Sunday, October 5, 2008

Me and My TI-83.

In Fridays class, we explored the different ways of finding derivatives with our handy calculator.

On slide 2, we used the change in y / change in x to find the derivative. That method of course is where we find the slope of two points that are very close to the point we want which is where x =2.

On slide 3, the method we used is the quotient one. All we did with this question was to plug in 3 in x and chose 0.01 as our interval. From there we simplified and got the answer of 0.33. This is my favourite method for some reason. We also saw that we'd get the same answer with using the change in y / change in x way. After plugging in numbers like 2, 3, and 4 we noticed there was a pattern. The derivative for x=2 was .5, x=3 was .33, and x=4 was .25. Therefore, the derivative for ln(x) was 1/x.

From what we learned from Mrs. Karras, not all points are differentiable. The points that aren't are where the graph is....pointy! Like on the slide, the absolute graph is not differentiable at x=0. This is because the tangent slopes are -1 from the one side and 1 on the other side.

The other methods we learned were the dy/dx thing in our calculator and the nderive function in our calcs.

That's pretty muuuch it. *looks at scribe list.* The next scribe will be Yi Nan i suppose. Have fun scribing tomorrow. Now is time for me to enjoy the last hours of my weekend :(

Friday, October 3, 2008

Slides October 3rd

Here they are ... one of these days I'll make pretty slides like Mr. K. For now, I'm glad to be putting them on the blog on a regular basis!

Thursday, October 2, 2008

The Power Rule...

Okay this is Benofschool and going for another attempt to explain the Power Rule...
The following image shows the equation work that was used to derive the rule. I will explain each line after the image.

Line 1: Okay lets find the derivative of x to the power of n at x

Line 2: So we just throw it into the derivative formula like usual

Line 3: We have to expand the binomial in the brackets. There are 3 things to realize. First the first term will be x to the power of n regardless to what n is. Secondly, the next term will always have n as the coefficient and n-1 as the exponent and also multiplied by one h. Finally, every other term will have a unknown coefficient (this coefficient will become irrelevant and you will see later) and the exponent of h will increase by 1. So in the third term the power of h will be 2 and in the fourth it is 3 and so on

Line 4: Now we can get rid of x to the power of n because it is subtracted to zero by the other term.

Line 5: Now we factor out an h. Notice that the first term in Line 4 no long has an h variable in the term and every other term afterward still has an h variable remaining. This is the key to seeing the rule

Line 6: Now the h reduces and you are left with what was in the brackets in Line 5.

Line 7: Now we apply the limit and substitute all the h values with zero since there is no h in the brackets anymore. This will cause all terms containing the h variable to become zero. Since the first term does not contain an h variable, it won't be affected by the limit.

Line 8: This leaves us with the rule because all of the other terms are turned into zero thus not affecting the first term at all.

I hope this helped. If there is any more help required, please talk to Dr. Eviatar or myself for help...

Okay well I tried =P

The Derivative Function (Cont'd)

  • Alternative Notations for the Derivative
  • y = nx^(n-1)
  • Derive f(x) = x^3
  • Derive y = -1/x
We finally finished our lessons with the topics covered in 2.3 of our Calculus: Concepts and Calculators textbook. According to yesterday's substitute's substitute, the textbook we use is the worst textbook she has ever seen in her life because it's convoluted and because its explanations can be explained better too, but that is why we have the BLOG, our online interactive textbook to help better our understanding.

Alternative Notations for the Derivative

is the symbol used by Lagrange. E.g. If f(x) = x^2, its derivative is f'(x) = 2x.
dy/dx is the symbol used by Leibniz. E.g. If f(x) = x^2, its derivative is dy/dx = 2x.

y = nx^(n-1)

Then the Benchmeister took over the class and starting teaching us. He taught us a shortcut in finding the formula of a derivative by using a formula. We were so amazed at his genius work that probably only Dr. Eviatar and Bench can understand it, because I know I didn't. Mr. K says, "If you don't understand something in class, you can bet that six other people in that class probably didn't get it as well." So I'm counting on other people to fill in the blanks that I've left in this blog post.
Also, the formula that Bench gave us is something to be taught some time in the future. Anyways, we ended up getting the formula, y = nx^(n-1).

Derive f(x) = x^3.

Derive y = -1/x.
Using the y = nx^(n-1) formula,
y = nx^(n-1)
y = (-1)x^[(-1)-1]
y = -x^-2
y = -1/x^2


  • Homework: 2.3 The Derivative Function #1, 5, 7, 9, 11, 12, 21, 27
  • Note: The principal says we won't get our laptops until he talks to Mr. K (assuming that Mr. K gets back in action at our school). Well, guess what? WE ARE THE RADICAL CALCULUS GROUP. WE WANT OUR LAPTOPS, AND WE WANT OUR LAPTOPS NOW!
  • Next scribe is Joyce.

Wednesday, October 1, 2008


Hey guise, its me Not Paul making my blog post for today because I'm Not* Scribe.
*Truthiness may vary

Anyway, onto actual Mathematics, those darn derivative deriving derivates and what in the heck a differentiable function is. I cannot however tell you with absolute certainty what a differential is, but I have a good idea we're going to find out soon enough.

The basic definition of a differentiable is a function where the derivative can be found for any point of a defined segment of that function.

Which is why f(x) = x^2 is differentiable, because we can find the derivative for every point in the function or any portion of the function.

If you get a function with "kinks" or very pointy points when you graph it, it won't have derivatives at those pointy points and thus is not differentiable for the entire function.

However, if you define a segment of that function that doesn't have pointy points then you can say that the function is differentiable for this segment of the function.

The teacher's definition is as follows:

"A function (fcn) is said to be differentiable at variable a if at variable a f '(a) exists. It is called differentiable on an interval if it is differentiable for every number in the interval."

Here's the example from yesterday that we did today [Ex 1] (which we also did yesterday), and another example we did [Ex 2]. Both of these are differentiable, because the derivative f '(x) exists at x = 0. (in Example 1 replace variable a with x)

Ex 1)

(Click to view larger)

Ex 2)

(Click to view larger)

This is an example where f '(0) doesn't exist, therefore it is not differentiable.

Ex 3)

(Click to view larger)

As shown in the image, the function f(x) = |x| is not differentiable because when x = 0, its derivative does not exist/cannot be determined tangentially. If graphed, y = |x| would show it has a "kink" or pointy point at x=0.

At the end of class we started talking about the different people who invented Calculus, and importantly Joseph Louis Lagrange's contribution of f '(x) = dy/dx, which will come up later in the course I'm sure of it, because of this Wikipedia article.

The next scribe shall be...

Zeff (zeph)

Sorry this post is uncredibly late, you can blame Wired and NeoGAF for their unfairly up-to-date news. I started working on this post at 9pm and finished 4 hours later. Of course atleast one of those hours was spent trying to figure out OpenOffice...