Bob

Hi, here I'm 7:50 in the morning to do the bob.
This unit is mostly about xolume, and who u can see the washer.
We looked volume by shell and slicing.
We learned a new way to find average value of a set of numbers.
This umit covers a lot of things that is kind of hard.
I try mu best to usr the rest of the morning to study and do good on the test.
Well, see u!!!
Hope every one do good on the tesr.
BYE!!!!!!!!!!!

Pre-Test: Definite Integrals

Today we had our pre-test for the unit Definite Integrals. Since Dr. Eviatar is currently smart board handicapped and cannot post any slides, I shall steal the one from last year's blog :D

AP Calculus Slides October 30, 2007

View SlideShare presentation or Upload your own. (tags: apcalc07 definiteintegral)

Anywho, the first two questions are basically the same type of question. All you have to do is input the functions given into your calculator and then use fnInt to get the answers. Easy peasy.

The next question is basically just thinking each of those answers through and if you could understand what each are trying to say, then you'd find that the only true answer is D.

Question 4 is a bit trickier. First you have to make the function so that you could write it into the definite integral form. After constructing the function from the info given [(5t+4)-.5t²], you then make your definite integral and add 10. The reason you add 10 is because at t = 0, the tank should contain 10 gallons of water. From there, you just solve and voila, you should get the answer on the slide.

Number 5 is just a simple analysis of the graph given.

Part A for the last question shouldn't be too hard if you had been keeping up with your homework. The reason why the interval shown is from 0 to 4 is because it takes 4 years to get to 1996 from 1992.

Part B just deals with finding the values for f(t) for the first 4 intervals. You just start with f(0) and go up to f(3) then add them all together. d(-_-d)

Part C was just asking to basically write a story about how the oil consumption was changing during the interval given in part B.

Alright that's it. The next scribe is not Justus, but Joyce. Good luck on the test tomorrow everyone and a reminder to the scribe that you don't scribe on test days! =D

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 ^_^

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

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

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

http://apcalc07.blogspot.com/2007/10/todays-slides-october-25.html

http://apcalc07.blogspot.com/2007/10/thursdays-post.html

END NOTES:
3.4

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

Workshop

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:
LEFT: 0
RIGHT: Pi
X CHOICE: 0
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!

TEST RESULTS/REVIEW

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.

IMPORTANT DATES:
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**]

3.1 CALCULATING DISTANCE TRAVELED
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.

3.2 CALCULATING AREAS: RIEMANN SUMS
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].

3.3 DEFINITE INTEGRALS
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?

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.

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.

Orange.

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.

Slides October 23rd

Here they are ....

Ap Calc Oct23

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

Slides October 22nd

Here they are ... watch out for Mr. K, LOL.

Ap Calc Oct22

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