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I thought that this was just absolutely HILARIOUS.
Wednesday, May 27, 2009
Monday, May 11, 2009
DEV Status Report
Hey Mr. K,
As you know, everyone is working on their DEV's however, due to May-Madness, which includes the arrival of AP exams including Bio, Calculus, and Chem, our DEV's progress has hit a snag. Although I only speak on behalf on Team FLJ, I can undoubtedly say that all other groups have been hit with scheduling crisis's.
I myself have been kept busy with Career Internship, work, and projects from other classes. Unfortunately, all the due dates and AP exams are clustered up in the same region, leaving little or no room for flexibility. Other members of my team have also been busy with their own studying, individual class projects and others with myself included took part in this year's Fashion Show.
As you know, everyone is working on their DEV's however, due to May-Madness, which includes the arrival of AP exams including Bio, Calculus, and Chem, our DEV's progress has hit a snag. Although I only speak on behalf on Team FLJ, I can undoubtedly say that all other groups have been hit with scheduling crisis's.
I myself have been kept busy with Career Internship, work, and projects from other classes. Unfortunately, all the due dates and AP exams are clustered up in the same region, leaving little or no room for flexibility. Other members of my team have also been busy with their own studying, individual class projects and others with myself included took part in this year's Fashion Show.
- Calc exam = May 6
- Fashion Show = May 7
- Bio Exam = May 11
- Chem exam = May 12
- English project = May 15
- Career Internship Portfolio = May 20
Tuesday, May 5, 2009
Twizzler MANIA!
As we all know, tomorrow MORNING, the bomb goes off. The AP is here! HOLY GOODNESS. Of course, anxiety is kicking me in the head right now. I'm slowly snapping back to reality.
Today was, THE LAST DAY before the exam. I still can't get that thought out of my head. It's so close. It seemed like only yesterday that we first started the class.
BEFORE we officially started the class, we were reminded of the really important rules we should follow to avoid automatic failure. As well as that, we were given the following tips since our exam would be running from 8 AM [be there by 7:45] until noon.
Make sure to:
As for the actual exam itself, it will run approx. 4 hours in four sections.
NEVER recopy something that is already defined in the question. Also, don't feel the need to show off your algebra skills. This is already expected of you. Plug things into your calculator for efficiency. Don't do more than you're asked to do.
Finally, we started on the actual review section of the class. It was sort of a race, and whoever had the answer buzzed in with the phrase "Got it."
There was an exquisite blend of both calculator and non-calculator questions. We had fun with it, and I don't know about everyone else, but I somewhat forgot why we were doing it. We were all about business, but simultaneously having a good time. Especially when playing Jenga with the pile of Twizzler packs on my desk.
The questions were pretty straight forward multiple choice, and by now, we should know what we're doing given the circumstances of each question. We might not be as speedy as Benchmen, but I think we've proven that we could get there. I believe in you guys and I just don't feel like explaining slides right now. haha. I will though. I think.
I'm going to follow the advice given to us, and rest up, get nourished and do some light reading and test myself and see if I know what I'm doing.
But for now, while I'm on "standby" mode, enjoy this vid. It's not precisely funny, but it was part of our school's production of The Wiz. AWESOME JOB and the rest of the videos are on the channel.
The sound isn't really that good, and I think the footage is from the first matinee show. But still I miss the days where I eased on down the road. xD
...and here is one last kicker for you guys. This song is called "Believe In Yourself". I guess it's a really nice inspirational message before something like an exam comes up, don't you think?
Today was, THE LAST DAY before the exam. I still can't get that thought out of my head. It's so close. It seemed like only yesterday that we first started the class.
BEFORE we officially started the class, we were reminded of the really important rules we should follow to avoid automatic failure. As well as that, we were given the following tips since our exam would be running from 8 AM [be there by 7:45] until noon.
Make sure to:
- Sleep reasonably early, preferably before midnight-- don't DARE to do last minute cramming, it's way too late to do that, but sleep EARLY to wake up EARLY.
- EAT a really healthy meal before [not immediately before] your bedtime and have a really hearty breakfast. Eggs are good brain food, perhaps pack a granola bar, dipped in chocolate. Get Twizzlers, if you wish. HAHA
As for the actual exam itself, it will run approx. 4 hours in four sections.
- Multiple choice, calculator [50 mins]
- Multiple choice, non-calculator
- Free response, calculator [45 mins, 3 questions]
- Free response, non-calculator [45 mins, 3 questions]
NEVER recopy something that is already defined in the question. Also, don't feel the need to show off your algebra skills. This is already expected of you. Plug things into your calculator for efficiency. Don't do more than you're asked to do.
Finally, we started on the actual review section of the class. It was sort of a race, and whoever had the answer buzzed in with the phrase "Got it."
There was an exquisite blend of both calculator and non-calculator questions. We had fun with it, and I don't know about everyone else, but I somewhat forgot why we were doing it. We were all about business, but simultaneously having a good time. Especially when playing Jenga with the pile of Twizzler packs on my desk.
The questions were pretty straight forward multiple choice, and by now, we should know what we're doing given the circumstances of each question. We might not be as speedy as Benchmen, but I think we've proven that we could get there. I believe in you guys and I just don't feel like explaining slides right now. haha. I will though. I think.
I'm going to follow the advice given to us, and rest up, get nourished and do some light reading and test myself and see if I know what I'm doing.
But for now, while I'm on "standby" mode, enjoy this vid. It's not precisely funny, but it was part of our school's production of The Wiz. AWESOME JOB and the rest of the videos are on the channel.
The sound isn't really that good, and I think the footage is from the first matinee show. But still I miss the days where I eased on down the road. xD
...and here is one last kicker for you guys. This song is called "Believe In Yourself". I guess it's a really nice inspirational message before something like an exam comes up, don't you think?
Monday, May 4, 2009
Exam Review: Applications of Integrals
We started off class by talking a little bit on how sushi is made. It's a roll, but when sliced (hondamitsubushi!) its made into cross-sections. By knowing the area of these cross sections, we can find the total volume of the initial roll if we have the length. This class was pretty much dedicated to applications of integrals for our exam review.
We started with:
This was easily found with the use of a calculator. But if not given use to a calculator, it's good to know how to solve this question.
We found the antiderivative of the given function over the interval of 0 to pi and replaced all the values of x with the values of the interval to find the difference in area. Now if no calculator is given, it's simple, just simply simplify it to the simplest form. Then repeat for the other question.
Our next question had to deal with the same functions as the last, but we have to find the integral equation of the volume of the shape over the same interval of 0 to pi, but when it's generated over the x-axis, and then the we found the volume when generated over the y-axis.
Now we cut this shape into tiny cylinders and find the volume of each cylinder, given by V = (pi)r2h but because we are dealing with 2 different functions, and finding the difference in the volumes, each graph has its own radius. The volume is then simply stated as V = pi(R2 - r2)h, we use an infinite amount of cross sections, and because of this the height is quiet small, its so small that we call it dx. Now we throw the functions' radius' into the equation and voila.
When revolving around the y-axis, it's the same deal, but with the use of a different equation. Note: I'm not sure what it is but I'll edit it in when I find out.
Our next question involved trapezoid sums.
The question shows a table with 9 sets of information when x is various values. We need to find the difference of area between the 2 given functions. Using a trapezoidal approximation, we find the upper limits and lower limits of f(x) and find the differences between these limits of g(x) and find the average. Quite simple.
Next question deals with finding the area of the region of a graph.
Now because the graph has a root between the given interval, we split the region into 2 seperate intervals. One finding the area of the region between the root at -1.100457 and -1.5 (I'm not sure why we're finding the area seemingly backwards on the interval) and we add this to the area of the function in the interval -1.100457 to 1.5
This is all we did in today's class. Review questions are in today's slides. Our exam is in 2 days and good luck to everyone. Next scribe will be .:. J + ME .:. or better known as Jamie.
We started with:
This was easily found with the use of a calculator. But if not given use to a calculator, it's good to know how to solve this question.
We found the antiderivative of the given function over the interval of 0 to pi and replaced all the values of x with the values of the interval to find the difference in area. Now if no calculator is given, it's simple, just simply simplify it to the simplest form. Then repeat for the other question.
Our next question had to deal with the same functions as the last, but we have to find the integral equation of the volume of the shape over the same interval of 0 to pi, but when it's generated over the x-axis, and then the we found the volume when generated over the y-axis.
Now we cut this shape into tiny cylinders and find the volume of each cylinder, given by V = (pi)r2h but because we are dealing with 2 different functions, and finding the difference in the volumes, each graph has its own radius. The volume is then simply stated as V = pi(R2 - r2)h, we use an infinite amount of cross sections, and because of this the height is quiet small, its so small that we call it dx. Now we throw the functions' radius' into the equation and voila.
When revolving around the y-axis, it's the same deal, but with the use of a different equation. Note: I'm not sure what it is but I'll edit it in when I find out.
Our next question involved trapezoid sums.
The question shows a table with 9 sets of information when x is various values. We need to find the difference of area between the 2 given functions. Using a trapezoidal approximation, we find the upper limits and lower limits of f(x) and find the differences between these limits of g(x) and find the average. Quite simple.
Next question deals with finding the area of the region of a graph.
Now because the graph has a root between the given interval, we split the region into 2 seperate intervals. One finding the area of the region between the root at -1.100457 and -1.5 (I'm not sure why we're finding the area seemingly backwards on the interval) and we add this to the area of the function in the interval -1.100457 to 1.5
This is all we did in today's class. Review questions are in today's slides. Our exam is in 2 days and good luck to everyone. Next scribe will be .:. J + ME .:. or better known as Jamie.
Friday, May 1, 2009
3rd Last Class Before the Exam
WOW! Just 2 more classes until the big exam. That was very fast.
Today we reviewed Techniques of integration.
The first one we looked at was the U-Substitution technique. This can be used when you are antidifferentiating a function that is composed of a factor that is a composite function and the other factor is the derivative of the inner function of the composite function.
I did this question on the smartboard. First I chose a part of the function to be substituted for U. I saw that the denominator and the exponent of e was root x, so I chose that one.
So I let u equal root x and differentiated u as shown in the image above. I also solved for solved for x using the substitution, just in case I needed it. So I substituted all values necessary and I got an antiderivative. Great we found an antiderivative, how do we know if it correct? Just differentiated the antiderivative that you found and if you get the function that was to be antidifferentiated, then you got the right antiderivative.
This question is similar to the previous one but this one is an integral. What is the difference you ask? An antiderivative will give you a function, an integral will give you a number. Even though they are different, they are also related.
To answer this question, just antidifferentiate the function and substitute the limits into the antiderivative to find the answer. This is the definition of the second part of the Fundamental Theorem of Calculus.
Oh no! There is an antiderivative that we've never seen before. What is the antiderivative of cosecant, let alone the antiderivative of cosecant squared. I can't remember how we thought of this but Mr K got us to differentiate Cotangent. Using the Quotient Rule because cotangent can be written as cosine divided by sine as shown in the image above using abbreviations. Don't try to memorize the antiderivative/derivative relationship. Just understand that you can easily build it when needed.
Mr. K began talking about the antiderivative of x to the power of some number times e to the x. He said that it was a telescopic function which means as you evaluate it, the function will get longer, but things can be reduced or factored in such a way that it will shrink like a telescope.
I was playing around with some examples and I discovered the pattern. It is such a cool antiderivative. I'll give an example:
I got to this part by using integration by parts. So f = x³, my f' = 3x², g' = e^x, and g = e^x. Notice that the second term is still an antiderivative that we cannot find easily, but it is a factor of 2 functions, so let's use integration by parts again. So we'll get:
Once again we cant antidifferentiate one of the functions so let's integrate by parts once more:
]%2Bc)
Now we can finally evaluate that antiderivative. Once you do that you'll notice that there is a common factor of e^x. So let's factor that out and you might see something completely amazing. I'll show you:
%2Bc)
Okay look at the polynomial in the brackets. When have you seen those specific terms before? If you think and look carefully the consecutive terms are derivatives of the previous term. I tried this again with another power of x and I got the same pattern. So when you are asked to find the antiderivative of x to the power of some number times e^x, the antiderivative will be e^x times the power of x in the function being antidifferentiated minus the derivative of the previous function until you reach a constant. Don't forget the plus c.
%2Bc)
tn-1 means the previous term.
Sorry if my algebra was bad on the image above. But if you try it yourself, it really is a beautiful thing :P
The next few slides on the slide show posted by Mr K is for homework. Study for the exam on Wednesday. Next Scribe will be Francis.
Here is the video of the Scribe:
These guys have very funny podcasts.
That's all folks.
Today we reviewed Techniques of integration.
The first one we looked at was the U-Substitution technique. This can be used when you are antidifferentiating a function that is composed of a factor that is a composite function and the other factor is the derivative of the inner function of the composite function.
I did this question on the smartboard. First I chose a part of the function to be substituted for U. I saw that the denominator and the exponent of e was root x, so I chose that one.
So I let u equal root x and differentiated u as shown in the image above. I also solved for solved for x using the substitution, just in case I needed it. So I substituted all values necessary and I got an antiderivative. Great we found an antiderivative, how do we know if it correct? Just differentiated the antiderivative that you found and if you get the function that was to be antidifferentiated, then you got the right antiderivative.
This question is similar to the previous one but this one is an integral. What is the difference you ask? An antiderivative will give you a function, an integral will give you a number. Even though they are different, they are also related.
To answer this question, just antidifferentiate the function and substitute the limits into the antiderivative to find the answer. This is the definition of the second part of the Fundamental Theorem of Calculus.
Oh no! There is an antiderivative that we've never seen before. What is the antiderivative of cosecant, let alone the antiderivative of cosecant squared. I can't remember how we thought of this but Mr K got us to differentiate Cotangent. Using the Quotient Rule because cotangent can be written as cosine divided by sine as shown in the image above using abbreviations. Don't try to memorize the antiderivative/derivative relationship. Just understand that you can easily build it when needed.
Mr. K began talking about the antiderivative of x to the power of some number times e to the x. He said that it was a telescopic function which means as you evaluate it, the function will get longer, but things can be reduced or factored in such a way that it will shrink like a telescope.
I was playing around with some examples and I discovered the pattern. It is such a cool antiderivative. I'll give an example:
I got to this part by using integration by parts. So f = x³, my f' = 3x², g' = e^x, and g = e^x. Notice that the second term is still an antiderivative that we cannot find easily, but it is a factor of 2 functions, so let's use integration by parts again. So we'll get:
Once again we cant antidifferentiate one of the functions so let's integrate by parts once more:
Now we can finally evaluate that antiderivative. Once you do that you'll notice that there is a common factor of e^x. So let's factor that out and you might see something completely amazing. I'll show you:
Okay look at the polynomial in the brackets. When have you seen those specific terms before? If you think and look carefully the consecutive terms are derivatives of the previous term. I tried this again with another power of x and I got the same pattern. So when you are asked to find the antiderivative of x to the power of some number times e^x, the antiderivative will be e^x times the power of x in the function being antidifferentiated minus the derivative of the previous function until you reach a constant. Don't forget the plus c.
tn-1 means the previous term.
Sorry if my algebra was bad on the image above. But if you try it yourself, it really is a beautiful thing :P
The next few slides on the slide show posted by Mr K is for homework. Study for the exam on Wednesday. Next Scribe will be Francis.
Here is the video of the Scribe:
These guys have very funny podcasts.
That's all folks.
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