Scribe Post: 70

Hey everyone, sorry for the later post, but it's up like I said it'd be =]. Didn't know I was scribe for this class, but here's what best I could explain it from off the top of my head.

Note: for the following problems... [ ^b∫_a ] should be read as "the intergral over the interval from a to b" or as pictured like this...
1. Find g'(x):

a) g(x) = [^x∫₀] sin(t) dt
g'(x) = sinx
- We get this answer because we know and committed to memory as to what the antiderivative functions are for trig. functions which we learned are also 'transcendental' functions (don't mind the big word, isn't much useful) and we get -cosx. Then we use the Fundamental Theorum of Calculus Part 1... [ ^b∫_a ] f'(t) = F(b) - F(a) because we're trying to find the definite intergral because when it is over an interval, we're looking for a number. And when there is no interval, then it turns out to be an indefinite integral.
- Using the FTC-part 1, we get g(x) = .
- Simplified g(x) = -cosx + 1.
- Then we simply differentiate g(x) for g'(x)

From this problem, we find out that the derivative of the intergral function, is the same as the underlying function of the intergral. This is true for all functions and is interpreted through the FTC-part 2...
The reason why the variable changes is because, we are only looking and changing x as the other variable t only changes as x does.

b) g(x) = [^x∫₀] √(1+t³) dt
g'(x) = √(1+x³)
- Using the FTC-part2. (See the pattern?)

c) g(x) = [^3x²∫₀] sin(t) dt
g'(x) = sin(3x²) · 6x
- The difference about this question to the first is the variable x has changed.
- We interpret this as a composite of functions, and then use the chain rule as we normally do to differentiate these functions...
g'(x) = f'(g(x)) · g'(x)

d) g(x) = [^sinx∫₀] √(1+t³) dt
g'(x) = √(1+sin³x) · -cosx
- Using FTC-part 2 and the chain rule law.

2. Graph of f


g(x) = [^x∫₀] f(t) dt

a) Find g(1), g'(1), g''(1)
g(1) = 1.5
g'(1) = 0
g''(1) = -3
- g(1) is found by finding the sign area under the curve from 0 to 1, (f(1)·1)/2 'because it's a triangle.
- g'(1) is found by reading straight off from the graph, as the FTC-part 2 proves that a derivative of an intergrated function is just its underlying function.
- g''(1) is found by finding the slope of the tangent line at g'(1)

b) For what values of x on the interval (-2,2) is g increasing? Explain your reasoning.
(-1, 1)
Since g'(x) = f(t), we just look at where ever it is positively valued on the graph.

c) For what values of x, on the interval (-2,2) is g concave down?
(0, 2)
Since g'(x) = f(t), we just look at where ever its slope is negatively valued on the graph.

d) Sketch the graph of g

This is just one of the possible graphs for g as it is interpreted as a parent function derived from its derivative.
The graph of g'(x)=f(t), and the derivative only tells you the shape of the parent graph not where it is vertically. "If you were given a point on the graph, you would be able to position it, but if not, there would be infinitely many answers. (e's correction)" I got this answer by looking at the derivative and second derivative for increasing/decreasing and concavity characteristics. Aswell as integrating the function from -2 to 2, to find the total change for more accuracy. Using different intervals when intergrating will give you the same shape of the graph but it will be positioned different vertically.


Take care everyone!

Next scribe will be Charlene Linger, yay for the weekend.

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What did we do in class today?

For those of us who were writing the ELA provincial exam yesterday, today in class we buddied up with the students who were present in class and went over the questions of functions that required the use of integrals to solve. We pretty much did this for the entire class because some of us (me) didn't understand why the graph looked the way it did. Eventually I had one of those " lightbulb goes on moment " and figured it out.

Mr. K reviewed yesterdays class and explained everything and stressed the importance of understanding the domain of the functions. Which Suzanne has nicely explained in her scribe post (the one before this one).

Referring to the prior question in Suzanne's post, we learned about the domain of g(x):



We also have this question qhich we did not complete but will probably go over tomorrow in class:

A) What is the domain of g?
B) For what values of x does g'(x) = -1
C) Sketch the graph of g over it's entire domain

h m m . . . I sort of wished I had some math jokes but I don't =(

H o m e w o r k : Section 6.3 all the odd numbers

N e x t s c r i b e : manny

Scribe Post

Accumulation Functions Continued

For all of you guys who wrote the provincial English exams today, I hope you did well. We covered a lot in class today so I STRONGLY RECOMMEND YOU READ THIS.
As you probably remember, yesterday we looked at the graph of a function f(t). At the end of class we were given five integrals, and were asked to draw the corresponding graphs. We spent the class going over the answers to these.

First of all, here is the graph f(t) and the given information, which the whole class was based around.




The graph of y=f(t): Is defined on the interval [0,4]
Has odd symmetry around the point (2,0)
On the interval [0,2], the graph is symmetric with respect to the line t=1

In yesterday’s class, we examined the graphs created by examining integrals on intervals, all using the function f(t). We discovered that f(t) was the derivative of all the different integrals we looked at. In other words, the graphs we created were all parent functions of f(t). In today’s class, we did the same thing, except that we had to do different transformations on the graph of f(t) and THEN find the graph of the parent function using the new graph.

I’m going to show the transformed graph for each question, followed by the graph of the parent function obtained from that graph.

1) Has somehow disappeared.. I'm very confused. Oh well. Anyway, the first question's basically what we did yesterday, no transformation involved. It just has the interval from x to zero, so in other words it the widths dx are negative, therefore the sign of the areas changes. The handy diagram below shows why. SEE YELLOW BOX. That's the thing to remember.


OH I FOUND IT! So.. never mind, here's the first question.



2)


3)


If you were to look for the derivative of g(h(x)), you would use chain rule. So you would get
g'(h(x))*h'(x)
= f(2x)(2)

4)

5)


And that's about all we did. Sorry this is up so late but the closer I get to a computer it seems the more technology-impaired I become. Oh right, and the next scribe is... Anh.

Scribe Post: day 68

Peter Donnelly

Oxford statistician Peter Donnelly explores the common mistakes humans make in interpreting statistics, and the devastating impact these errors can have on the outcome of criminal trials.

You learned the math he's talking about here in your grade 12 Pre-=Cal class ... probability.

Click on the picture. (22 min. 6 sec.)

Scribe Post: Day 66

Well here I am almost the end of winter break and finally doing my scribe post. Reasons for me taking so long:

1. Basically the first week of the break I was sick and when I was getting better I'd get sick again =S

2. After that I was unable to find the sheet of notes I had jotted down during that class

3. Lastly, when I did find it I was already in that "winter break relaxed" type of state and became lazy to do it =S

If I hadn't got sick this post may have actually been up in the first few days of the break, now unfortunately I'm putting it up AFTER Christian's scribe post for....the day AFTER this class =S Well anyways I'm going to try and put together what I can remember from my notes and from the class.

We started a new unit, Chapter 6: Integrals. However before that we were debating over a question from the test on the past unit. After several minutes of debating we talked about integrals. Discussing on what the definition of an integral was we came up with these things:

- sign area under a curve

- antiderivative

- there are two types of integrals, indefinite and definite integrals

- indefinite integrals represents a family of functions with the same derivative and a vertical shift

- definite integrals represents different numbers.

- to find an integral you can use Rieman's Sum (left hand sum, right hand sum) although it won't give you an exact value, divide to sub intervals, find area under curve using rectangles.

Mr. K mentioned how we learned the first part of the Fundamental Theorem of Calculus and how we were going to learn the second part of it in this chapter, at least I think so from my notes? =S

The rest of my notes are extremely hard to read (I got to learn how to write faster and more clear when im writing fast =S) but there's one integral I wrote down because he had it on the board that he said he'll show us how get in a later unit, unfortunately it's too messy for me to put it down onto this post so if anyone has it, please comment so I can add it.

We learned the Constant Multiple Rule:

Basically to make the first integral easier to do, you bring out the 2 and work with the basic integral, this is the Constant Multiple Rule I think? =S

To top off this horrible post, at the corner of my sheet of notes in capital letters it says" NEED TO KNOW ANTIDERIVATIVE TO FIND INTEGRAL" that has to be important cause it was in capitals. Well as you all know Christian was the next scribe and yes I hope you all enjoyed your winter break see you all tomorrow!

Scribe Post: Day 67

Hello everyone! It's Christian, and I was the scribe on the last school day of '06 =(. As a summary, we didn't really start anything new. We just reviewed some things that we learned in previous classes.

Integrals As Signed Areas Under a Curve

Graphically speaking, an integral is the signed area under a curve. So, from the interval A to B on the graph above, the highlighted region is the integral. This graph is the same as the one up there, except that the axes are now labeled.

To integrate on the interval A to B, we must find the area of the red triangle. If we do that, we'll end up with an equation similar to the one in the green box. Don't mind the "1/2". What's important to note is that if we solve this equation, the s's will cancel, and we'll be left with the unit m, as seen in the purple box. As we know, "metres" is a unit that indicates "position", d. The parent graph of a velocity-time graph is a position-time graph. YAY! This shows what we learned about integrals: that they are signed areas under a curve, and that they are the total change on a parent function.

I'll explain the last concept we learned as quickly as possible. The graph above is, again, the same as the ones before, except that the integral from A to B is now divided into a blue region and green region. We learned that if we find the integral from A to Z and add that to the integral from Z to B, we'll end up with the same number as if we just took the integral from A to B.

That was the class!

Hey guys, hope you had a great winter break. Back to school now! No one skip the first day of classes! Can't wait to see all of you =)

I guess we're starting the new cycle, so the next scribe is... Crystal!