January 23, 2007
Today, we practiced more antiderivative questions using "Integration by Parts".
Here we go...
1. Find the antiderivatives of the following:
Q: How can we decide which function should be "u" (aka "f") and which one should be "dv" (aka "g' ")??
A: Good question! Today, we learned a new "rule". It's the...
This rule is used to decide which of the two functions will be "f" and "g' ". This rule is followed from top to bottom (or left to right if you wrote it in that type of manner). For example, if an "algebraic" and a "trigonometric" is present in the problem, then "f"= "algebraic" and "g' "= "trigonometric". According to the rule, "A" comes before "T".
Q: How come we did the "Integration by Parts" twice?
A: We did the technique twice because we dont know the antiderivative of "xcosx". We have to use the technique again to find the antiderivative of this expression.
Then, we answered some trig problems...
Note: "Arc(cos or sin)" is the same as "sin or cos"^-1.
The answer to letter "c" is undefined because the value is greater than 1. The answer to letter "d" is -3/10 because cos and arccos cancel each other out.
Finally we looked at this graph...
This is the graph of y= Sine. It is fixed in the domain, (-pi/2, pi/2). By doing this, the graph of sine is now a "One-to-one" function. It means that each input has 1 and only 1 output.
1. Vertical Line Test- this test is used to determine if the graph is a function or a relation. If the line test touches the graph more than twice (1 input has 2 or more outputs), then it is a relation.
2. Horizontal Line Test- this test is used to determine if the inverse of a function is a function or a relation. If the line touches the graph of the parent function more than twice, then the inverse of the graph is a relation.
I hope I covered most of the things we did for this class. XD
Next scribe will be Danny.
January 22, 2007
January 21, 2007
There are 3 main steps to this method:
-you let the inner function g(x) = u
-differentiate that inner function and let it = du/dx
-we solve for du by multiplying dx to the other side (cross multiply):
du = 3
du = 3dx
-substitute u and du in original given integral.
-antidifferentiate the new integral that consist of u and du.
-substitute g(x) the inner function in for u to get your final answer, an antiderivative.
*REMEMBER: It doesn't hurt to check your final answer by differentiating!!!
If it matches the given integral then your answer is right!!!
The following examples we did in class will show and explain this method:
1) the integral of sin (3x) dx
Click--> on pic so it's easier to read
ANSWER: -1/3 cos(3x) + C
*DON'T FORGET + C
2) the integral of 5e^5x dx
Click--> on pic so it's easier to read
ANSWER: e^5x + C
*DON'T FORGET + C
3) the integral of (x^3+2)^4 ( 3x^2) dx
Click--> on pic so it's easier to read
Answer: (x^3+2)^5 / 5 + C
*DON'T FORGET + C.
3) the integral of (x^3+2)^4 ( x^2) dx
- similar to question #3 but WITHOUT the 3 in front of x^
Click--> on pic so it's easier to read
Answer: (x^3+2)^5 / 15 + C
*DON'T FORGET + C.
4) the integral of (x / √x+7 ) dxClick--> on pic so it's easier to read
Answer: 2/3(x+7)^3/2 - 14(x+7)^1/2 + C
*DON'T FORGET + C.
HOMEWORK: 7.3 Exercises --> odd #'s
The next scriber is MARK =]
January 18, 2007
Remember in general,
If g is any particular antiderivitive of f, then we write
S f(x) dx = G(x) + C
Today we basically did seven questions on the board and discussed about how finding the antiderivitve of a function was not that easy than finding it's derivitive.
Sorry everybody but I'll finish this later tonight. My internet just connected. :(
Homework: EXERCISE 7.1 AND 7.2 ODD NUMBERS
The scribe for tomorrow is... Katrin
January 17, 2007
The first thing we did in class was discuss our picture assignment. We were assigned to capture a scene that depicts "integrals", and were given since the beginning of the winter break to do it. We all found it tough, so much so that we asked Mr. K to change the theme. So he did. The new theme for our photo assignment is related rates.
We are to take a picture of something that shows related rates. The photo must be uploaded on Flickr, tagged "apcalc06", "related" and "rates". We are given over a week to do this!
After that, we continued writing our test. Mr. K originally gave us 10 minutes to do it, but we ended up taking all but 5 minutes of the class, so we didn't really start anything. We were just reminded that when antidifferentiating, we use the following formula for polynomials:
We add "+ C" at the end of whatever terms we get, because we know that unless we're given specific points, we don't know which of a family of functions the antiderivatives (or derivatives) are talking about. Also, there are functions that just aren't antidifferentiable (I don't know if this word exists). Sometimes we just have to guess and check. An example is e2x. That's pretty much it.
Homework: 7.1 do odds from 1-17
Next Scribe: Crystal
January 16, 2007
January 15, 2007
Some thoughts =D
Some people are just gifted and interested in mathematics. I am not one of those people. It was actually my mom who made me take this course. I thought that I should go and take it b/c I get good grades and all. But grades are beside the point. Grades don't measure you intelligence! It's just a way of measuring the amount of work your willing to put into it. I think anyone could be intelligent and gifted if they truly worked at it. If you really want to be there and you are truly interested in what you are doing then you are willing to put a lot of effort in what YOU want to do. Not just because you think its what the "SMART" people do. I'm not labeling people but it is what people really think. It's just frustrating to have people think that you have to have high grades to be considered smart. I just don't think thats true. It's for these reasons why it hurts my marks and my motivation to do my work. Now don't get me wrong, I think mathematics is brilliant in many ways! However, it's just not for me. I'm the one who loves English, art and things like that. But I've done all this work and it's way too late to turn back. I don't want to turn back, I want to see what I can do. I think what I'm getting at is finding out what your interested in and this class actually helped me figure that out. Make your own choices and do what you want to do (as long as you know it will truly help you). It's all about "finding out who you are," so corny but true.
What's different about this unit for me, I guess, is that I'm having a little bit more fun. I was on the bus to school one day, when I asked myself, "Why was calculus invented??!! WHYY?!!," when a thought came to mind. Maybe my attitude towards Calculus affects my learning, and therefore my marks in the class. If I don't treat Calculus as an enemy, maybe I'll be better at it. When I thought of this, I remembered my marks in Chemistry and Physics. They were a lot different ;). The difference between the two was that I thought of Chemistry as a fun subject while in Physics, even before I learned anything, I already thought it was going to be hard and boring. My mark in Chemistry was much higher, I think partly because I was motivated to do all the homework, not actually treating it like "work".
The day this happened, I told myself Calculus should be my friend, and from then on I've been having more fun in Calculus. That was an "ah hah moment" for me. Hopefully the mark I get on the test will reflect the fun I've been having =).
P.S. guys don't forget to label your posts =)
I was thinking earlier about how this course is progressing, and I realized that we really learned a few main things- about derivatives and integrals. The whole course is based on these things, and each chapter we're just looking at them in a different light, and in more depth, and connecting them in a different way. It's unlike previous courses where we cover such a broad range of topics. So I guess that means we get a good overall understanding of these concepts. I think it's great to be able to look at these things in many different ways - I think we're learning to connect what we learn to other concepts better. Which is definitely a good skill to learn. Well, good luck on the test tomorrow everyone!
January 14, 2007
We're finally going to solve the amusement park problem... it makes me feel good because we can finally apply our learning to one huge question. It was only a couple of months ago, when Mr. K told us that we'd get here... and we're here now. Things are happening so fast.. new year, new unit. And, we're going on a field trip.. woohoo!!!! Math has been a great (although challenging) adventure so far, I can't wait for the rest of it..
Speaking of which, I shall go study. I'm still frustrated about my scribe post, because I have to make a lot of changes and I wasn't quite done explaining things. Nonetheless, I can't do anything about it for now, so I shall use my time to study before it's too late. GOOD LUCK TO ALL, and sorry again about the post.. I promise I'll fix it when I can.
First of all, Mr. K gave us a five minute lesson on what we were learning:
We then had the following questions on the board:
- always figure out which function is on top
- find the roots by letting the functions equal each other
- define the integral that you are looking for
- solve the integral
We started the fourth question, which was about the amusement park. Its super long, and looks like its a lot of work, but we're finally going to solve it, woohoo =) Since we didn't finish, I will add it after tomorrow's class.
Speaking of which, TOMORROW is the PRE-TEST and TUESDAY is the TEST.
With that being said, HMWK for the weekend is to finish 6.4 odds and Supplementary Problems.
Also, don't forget about about the integral picture due next week, and and your BOB's.
Tomorrow's scribe is CHARLENE =)
January 11, 2007
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
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.
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?
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
January 10, 2007
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
January 09, 2007
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.
If you were to look for the derivative of g(h(x)), you would use chain rule. So you would get
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.
January 08, 2007
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.)
January 07, 2007
1. Basically the first week of the break I was sick and when I was getting better I'd get sick again =S
- sign area under a curve
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.
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!
January 06, 2007
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!