Thursday, January 29, 2009
Don't forget, you guys, Mr. Beaumont is interested in seeing our projects! =)
Friday, January 23, 2009
And here is your first Take Home Assignment for 100 marks.
Full solutions must be shown.
Due Date: Feb 2,,2009.
Thursday, January 22, 2009
Note: *Don't forget the + C. It was evident during the class that we constantly forgot to add the + C. It costs a mark!*
We now have x^2 /2 + lnx evaluated on the interval from 1 to e. Now we get...
After evaluating, we get the final answer of e^ + 1 / 2. This leads us to conclude to not differentiate too fast, and algebraically change it so that it's easier to differentiate it.
But on to the main dish. If we can differentiate multiplying terms, can we antidifferentiate multiplying terms as well? Well we tried that out with this question. Mr. K pretty much confused us when we asked if there was a product rule for antidifferentiating, and his answer was a "kind of, not really, definitely, maybe" answer. So, basically, yes, but mostly no. -__-"
So because there's no g'(x), we multiply by one using a full fraction using the coefficient number from g(x). From the fraction, remove the half from the integral so that we're left with F'(g(x)) * g'(x), which is sin(2x) * 1/2 + C <-- There it is again! OMG!
Anyways, that wasn't fun because I fell asleep for an hour and woke up really tired, so I'm happy that I'm finished, 'cause I have to study for Chemistry now.. Like, now. I forgot, I have to crown the next scribe. Hi, I'm Justus. No, I'm not really Justus, I'm saying that Justus is the next scribe for the last day of the semester. Have fun guy.
Wednesday, January 21, 2009
Okay so this here is the first free response questions, and is the easiest of the two we went over in class. The answer would be to sketch the graph y = (Sinx)/x. Where x cannot equal zero which means its a discontinuity at x = 0, but it also says that the graph exists at 1 where x = 0, so it pretty much fills in this discontinuity. Confusing? Yeah, I know. Part b asks for what values of x does A have a local minimum, and A is the parent function of the graph we drew, so the local minimum of the parent function would be the point where it the x values decrease then increase, and on the derivative, it would be where the values are less than zero, to when they begin to be greater than 0, this would be at 2pi, and it would repeat every cycle. Part c asks for the coordinates of the first inflection point of the parent function right of the origin, this would be at 3pi/2 where the derivative graph decreases then increases. Finally, part d asks to solve the following equation accurate to one decimal point.
This is quite simple, because we just use our calculators, and keep integrating the graph until the area we get is 1. It's pretty much trial and error, and the answer should be 1.1
I honestly don't know how to explain the answer to the 2nd free response question, so I'm not going to try, but Mr. K gave us an alternate solution, which we will learn about later on in this unit. Here it is:
Part b asks us to find where k = 2 and this is simple enough, just plug in the value 2 where k is supposed to be, the answer is 4.
Part c is here:
It's pretty much a rates problem, where we are given the rate of k in respect to time, and we have to find the rate of A(x) (parent function) in respect to time. We know when k = 2, A =4 from part b, and we have the area of A as a function of k from part a. We pretty much derived the equation of A and multiplied it by the rate given, and plugged in 2 for k, and the answer is 4.
That was it for the test questions. Mr. K also introduced us to a new site entitled Mathway which pretty much solves all our questions on calculus, of course you should use this for help, and not answers for upcoming exercises. Now onto our actual unit, "finding anti derivatives" we went over many derivative rules, and we anti derived them and don't forget to add +C, when anti deriving an equation, because the C stands for a constant, and this constant disappears when it is derived, so don't forget! We also anti derived some questions toward the end of class, and we also looked at a list of partial anti-derivatives rules. Here it is! Our next scribe will be Rence.
Tuesday, January 20, 2009
The first one that involved the the functions composed of many other functions was quite confusing. I don't know if I got the right answer, but I think my work got me some part marks I hope. The second Free Response question was a bit easier. It asked for the area enclosed by the y-axis and the function. Yes, I said y-axis. I remembered that for the inverse of a function, the x-axis of the parent function is it's y-axis and vice versa. So I used that to find the area function.
Okay that is all that I had to say about that test. So I want to hear what everyone else thought of the test. What were the easy parts? Hard parts? Also if you want to contribute with your thoughts on how you solved the questions, feel free to do so.
See you all tomorrow.
For the most part, I found this unit to be easier than the last few we've gotten. It was also a helpful review and helped clear some of the fogginess I had with the first Integrals unit.
What I found to be the most challenging parts were the Accumulation parts at first but now I think I kind of got that down now. I also remember looking at some questions in the exercises where it talked about Non-Odd or whatever it was and being completely confused so I'm still kind of iffy on those parts. I also remember being confused about that rational rule thingie (I also saw it in the exercises..it was asking to explain how this is greater than that in words or something like that >__>). Yeshh..
The easiest parts were definitely finding the derivative and such from the integral and also finding the area between two curves, which I found kind of fun :P.
Well, then. There are still some things that are still a bit foggy like that part with underlying functions..but if I think hard enough I can get it to wrap around me head. Also, after looking at the pre-test and such, I can say that I'm in good shape...for now. The last question was a bit hard but the rest was pretty good. Anyways, I should get going now! Good luck on the test everyone :D
Monday, January 19, 2009
Anyways, this unit. Not to bad I must say. I kinda like this stuff, although it is a bit confusing at times. Overall, not my strongest unit (the first integrals unit gets that award), although definitely not my weakest (since it IS a cousin of the other one)
Easiest bit: The mechanical process of taking the inner function and replacing things and mixing it around. It's mah fav :3
Hardest bit: Remembering that the inner function is the derivative of the accumulation function, and the parent function is found by anti-differentiating the inner function D: Mind twisting stuff there.
Overall: I'm ready for round 2! :p
Anyways, I need to go to bed, so I can have a decent sleep in time for tomorrows test D:
night guys, and good luck!
The Second Fundamental Theorem of Calculus, as pompous as it may sound, took a bit getting used to. I found the part where we had to prove (or was it argue?) that the derivative of the accumulation function is the parent function was easier to grasp than I anticipated it. Also, this is where I "lix up the metters," as Mr.K would say, talking about the variables of the accumulation function and the original function.
Visualizing the areas that's bounded by graphs helped a LOT. (Hence, I am a visual learner and I like to see things visually.) From the graphs, all I really need to do, basically, is to subtract or add the areas, however the case may be. Although the graphs dealing with the trigonometric function and integrating them or finding the areas, I don't consider them as my friends as of the moment. Sorry! (Yes, I apologized to the trig functions.) =)
Mr.K if i'm a little late tomorrow it's cuz my parent have some home inspection appointment tomorrow at 8:30 far far away.I have no way of getting to school other than my mama and papa. Yippee less time for me.
well, this is a short unit, easier then other unites but also hard in some way.
I think i understand everything in 6.1 (might forget some part,but I'm try to review)
6.2 was all about accumulation function which a function in side a function. I get the idea o it, but may be not the calculation part yet. have to review that over again.
6.3 the fundamental theorem.i think i get 1/2 of it. need some more time for it too.
6.4 the area, i think I'm fine with it.
hope i can do better on this test ^_^
But otherwise...back to blogging on blogging. I sort of eavesdropped on something like.. BLOBBING but I didn't catch that. eh. I quite enjoyed this unit. I enjoyed learning about accumulation functions and looking back, I can't believe how juvenile I was being afraid of the integral notation.. because I couldn't understand what it meant. Now it's just like.. really? That's all there is to it? Actually a lot of this unit was like that. I think what made it easier was the fact that we now know more about material like derivatives and Riemann sums which were incorporated into certain aspects in this unit.
One difference I'd like to note though. I don't seem to be having much difficulty with the patterns in the math and notation, but rather.. more the theorems, and what they represent. Like...the extreme value theorem. I mean I know what it is and what it does, but thinking about that again tonight, took me some time to REMEMBER what it was and how to explain it. You don't fully understand something until you can explain it. I also can't fully explain the 1st and 2nd fundamental theorem of calculus, in words, but I'm sure with notation I understand it. I guess I'm just seeing things in another perspective again, but totally missing the other understanding. I mean. If I know how to explain it, then it's engraved in my head then I would know where to use it. But since I can't right now, fully, then I might end up using the theorem and not realize that I just applied it. Does that...confuse you?
Well that's my BOB for tonight. I hope everyone does well on this test. Last one for the semester. I better not muck this up.
Anyways, I was lost for a minute because everything I was typing was turning into Hindi and I was literally tripping out, taking off my glasses and all. However I found that blogger added a nifty new function that translates what you're writing into the chosen language.
But besides that, I found that Integrals were not too difficult and that I won't struggle too much on the test, but either way, the way I did the pre-test today, I will be studying anyway. The accumulation and the concepts in this chapter were quite simple when thought through thoroughly. (Tongue-Twisting much?) Like Mr. K said, Calculus is more about understanding rather than Calculating. Like the Matrix. -__-"
And I demand assignments for marks ":O. It's not that I'm all about the marks, I'm just more driven that way.
Basically, today at my basketball game, we beat Oak Park by about 4 points. Now this might not mean anything to you, but it does to us (the basketball team.) On paper, Oak Park should destroy just about any team in manitoba. Hence the DvG story.
Anyways onward with the math. Today we had a revised version of the pre-test, which I liked much better. Coming out of that pre-test I felt much more prepared for the test tomorrow then I usually do, which is a good thing! :p Basically, what we changed was, instead of working by ourselves for a large portion of the pre-test, we worked on it independently for about 20 mins. We then went (read: Were placed) into small groups, and treated it somewhat like a workshop. After about 10 minutes of that, we spent the rest of the class going over the answers. It's those answers/ After pre-test discussions I'll be going over now ;p
What the writing means:
Essentially, how the problem is set-up means that F(x) = the stuff in the slide written in black.
Now we know that the derivative is the underlying function. It is from this notion that we end up with the next bit, (starting with F'(x))
What see you there in blue is the anti-derivative of F'(x) and therefore is F(x).
From there it's simply a matter of finding the values of those functions at (2) (the bit you see in the green there.) One thing to remember for this last step here, is that the integral of any function from x to x, is 0.
Alright, as you can probably guess by all the blue circles the main idea here is that the accumulation function is actually a COMPOSITE OF FUNCTIONS. This means, that you need to use chain rule when differentiating.
SO the procedure here is fairly standard, take the derivative of the outer at the inner, multiplied by the derivative of the inner. The result is the green stuff below. Then evaluate this function at -1 and you have your answer!
Okay, lets see if I remember this. Basically, the first thing to do, was sketch the graph (which is usually a good move.) Next you need to do some logistics work, by looking at the actual integral itself. You should notice that the inner function f(x) is all that there really is, and this is given in the question, so you don't need to do any guess work there. Next, start going through the options.
Option I : F(1) = 6.5
This is confirmed by counting up the squares. If you did it right you should end up with 6.5.
Option II : F'(1) = 3
The thing to remember here, is that the derivative of F(x) is f(x) which is shown in the graph. Therefore, all you need to do to confirm this, is go over to 1, and read the according value on the y axis.
Option III : F''(1) = 1
To check this one, you need to remember a continuation on the first idea from option 2. As said before, the derivative of F is f(x), so the next step would be to look at the second derivative of F. It just so happens that the second derivative of F is the first derivative of f(x). Looking at the function for f(x) and taking the derivative of it's two pieces you get f'(x) = 0 and f'(x) = 1. It just so happens, that f'(x) = 1 is just what we were looking for. Thus, myth confirmed!
This one was easy enough, a fact confirmed by the lack of needing an explanation.
Basically, use your calculator for this one. Plot the graphs, find the interval in which they overlap. 2nd calc > 7 select your lower and upper bounds. Repeat for the other graph, and subtract the values. If that all went according to plan you should have something like, 1.66666666666666666666666666666666666666666666666666667, or 4/3.
Answer : C
So, this problem is a simple definite integral.
Start by anti-differentiating, to end up with the first bit of blue seen in the slide.
After you anti-differentiate, substitute in the values seen on the definite integral (-2 and 2), and subtract. Its all simple algebra from there.
If it all went well, you should get k = 4, which happens to be answer d.
Answer : D
Question Free Response
So, you need to find the value of the accumulation function, G, on the interval from - 4 to 4 at -4. This should be simple enough, since we know that if the interval of an integral is from x to x, then the integral is 0.
This is a bit trickier, although not much moreso. G'(-1) is the same thing as asking for the value of the inner function (f in this case) at that point. Since we have the graph of f right here, lets do that shall we? Simple look on the graph to discover that f(-1) = 2 so therefore, G'(-1) also is 2.
on the graph you may notice that it on the intervals (-4 , -3) U (-1, 2) the graph of f is decreasing, and therefore the first derivative of G is decreasing on those intervals as well. So, by taking the derivative of f(x), you get the second derivative of G, and can go ahead and apply the second derivative test. (I think D:)
You also have to justify why this is true (as in write a short little ditty about why its decreasing on those intervals and things.)
Sorry if this doesnt make sense, I kinda need to clarify some things here D:
This bit is also somewhat hazy for me, but what I have in my notes here, is that you need to use extreme value theorem, and examine both the endpoints, and where the derivative is zero.
I also recall hearing that if you dont examine the endpoints (read: forget them)
Okay guys, I think that about wraps it up for todays scribe post. Since it's a new cycle I think I will pick francis to be the next scribe mmkay? Hopefully you kinda understood what I was talking about here. I apoligize if you didn't. Any comments/critique would be appreciated!
That was my BOB and good luck tomorrow.
Friday, January 16, 2009
On Friday's class, Mr. K told us that this was the last class for this unit and that the pretest would be on Monday with the test coming Tuesday. I'm sure he said something encouraging us about scholarships or maybe that was another class. But we did also talk about how it was nearing the conclusion of the first semester. My my, how time flies by.
Alright where was I? Right. Second semester. The second half. The beginning of the end. It scares me that high school is ending...it's also somewhat of a thrill. But the main thing we concentrated on for the moment was our scheduling. Even if it wasn't necessary to come to calculus class because of the extra spares, it was, well encouraged. It was a choice. No one is forced to do anything. Lovely words.
After sorting that out, with a fair amount of people agreeing to come into calculus class, or at least that's the impression I got, we split up into groups and began a workshop type class.
In our groups, we concluded and REVIEWED the "total area under a curve question" from the end of Thursday's class.
Just to jog some memories, we used the method of SYNTHETIC DIVISION to find the roots of this cubic function. After the roots were found, we continued on to find the integral of this function; the SIGNED area under the curve. But seeing as part of one curve was under the x-axis, things got complicated. Just a bit. If we were to add the area of the curve in the interval of [1, 2] it would be adding a negative, which is just subtracting. But that is WRONG since it would be subtracting that small area from the larger area. We are trying to find the TOTAL area. This is why the idea of "SIGNED" is fundamental.
The solution? Take the absolute value of the function, so that all of the values would be positive and over the x-axis, so in the case of adding these areas, these areas would both be positive. In finding it out, the class did it both ways and found the values using a calculator.
Below is the graph we put into our calculators, using the same window so that the function was zoomed.
To find the integral of a function, type the function in the Y1 window and press 2nd + calc + 7 [integral function]. It will then ask for a number to act as a lower limit. In this case put -2 then press enter. Upper limit, is 2. Then for the interval you've want the integral to be calculated, the area under the curve will be shaded in black and it will give you a value. We were given 10.66 repeating. This value is incorrect but this teaches us how to use the integral function on our graphing calculators.To get the absolute value of the function, all we have to do is go to the y-variable window and insert the absolute value function on our calculators represented by "abs (". Close the parentheses at the end of the function and integrate using the process above. The value we end up getting is a little more, since the values are positive. This gives us a value approx. 11.83 repeating.
Then we caught up to the material we were supposed to learn that day. Finding the area between 2 curves.
We were given 2 functions. We find the area in a similar fashion as finding the area under one curve except for a few differences.
I apologize if this is really LOOSELY EXPLAINED. I'm not trying to be lazy I swear. I'm doing my best. With the time I can spend on calculus. I can always edit this post afterwards.
But the main thing to know is the main pattern for integrating "in-between functions". That is: TOP FUNCTION - BOTTOM FUNCTION. This is said because the "top" function is usually the function with the larger area.
These functions also have values that make a curve under the x-axis making those values negative. In order to make those positive, again, find the absolute value of those functions and integrate.
This unit is kaput!
THE THINGS WE LEARNED [I know I'm missing something on this list..somehow]:
- definition of an ACCUMULATION FUNCTION and how to use them and how they are involved with integrals.
- 2nd Fundamental Theorem of Calculus [or first] which states that the derivative of an accumulation function [integral] is equal to the original function, f(x)
- integral is the SIGNED area under a curve
- how to integrate/find area between two curves; and a helpful tip in doing so is top function - bottom function.
Finally, the next scribe is Hi, I'm Justus. You're the last one yet again. And I'm picking you. Yet again. Déjà vu, I'm telling you. I know you're really busy lately and if you can't handle it and you need help, I'm willing to do a collaborative slide [somehow] to make things easier for you. But then again, I think you have pre-test day, so that's pretty good. It's entirely up to you. Let me know Monday.
As for this disappointing scribe post.. I'll be back to fix you. I'm not happy with the way I made you. I'll be back. Arnold Schwarzenegger said so.
Thursday, January 15, 2009
Wednesday, January 14, 2009
In the graph the blue area is the difference between green and orange area.
h is a tiny number that is close to 0, but not 0. Therefore x+h.
To approximate the blue area by multiply (f(x))(h).
This is not a proof, but a really good augment.
We went back to yesterday's slid, the one that Mr.k made a mistake.
The statement is wrong. It suppose be
What ever you put in f(x)will end up be in f(t).
Then we worked on this problem.
1st find x^2, then us chain rule.
The inner function is x^2, outer function is f(x).
x^2 determines the limits of the integrals.
Use integrals to find the area. Accumulation function is always involve with integrals, it always requires 2 function.
t is always depend one x, always find x first.
At the end we did some questions depends on what we had learned.
OK. Next scribe is Not_Paul again. Hope you are not late again. Good luck ^_^.
Tuesday, January 13, 2009
We were also shown a new program in our calculators called fnInt (math 9 on calculator). This is able to graph an accumulation function. The slide above shows the parameters of the program. The default h value for this is .001 if I heard correctly. One thing we were told was to never write fnInt on a test EVEER. I'm surprised I've never done something like that on a test.
Then we found out there was more than 1 fundamental theorem of calculus. The 2nd fundamental theorem of calculus is displayed above. All it says is that the derivative of an accumulation function is equal to the original function f(x). An accumulation function is a function made from using the area under the curve. Like the previous example before using fnInt, we saw that the graph it created was the antiderivative of the function x. So the derivative of the accumulation function would be the original function...(in terms of the example the original function would be x) That shows a connection between derivatives and integrals. They are like multiplying and division or adding and subtracting. Differentiation and integration are inverses of each other.
In between the learning there was chit chat about scholarships, Mr. K's new house, and other things I unfortunately didn't write down =(. Anywho next scribe will be Paul or Not_Paul...whatever name you go by XD.
Monday, January 12, 2009
I'll be your scribe for today.
Today we began with talking about Googling ourselves. Remember that every time we post something anywhere at anytime on the internet, we are leaving an online "footprint". Mr.K used an example when someone would search your name on Google or on any other online search engine. If an employer or future employer were to search your name online, would they find anything that you do not want to be known about yourself? So becareful with what you leave online, just incase someone searches you up online.
Okay back to math.
We began with a little reminder about what the Inequality Rule of Integrals was about. The Rule says that if a function, for example, f(g) ≤ g(x), than between the same intervals, the integral of f(x) is also less than or equal to the integral of g(x). Refer to the image above for a visual.
Introducing the mighty Obi-Wan Kenobi. Just so everyone knows Darth Vader is better, and if he had Darth Maul's Double Lightsaber, he would own everyone.
On this slide we applied the Fundamental Theorem of Calculus (FTC). If you want more info on the FTC then click here. But I'll explain it anyways. The fundamental theorem says that the integral of any function is equal to the total change in output (y-values) of the antiderivative function. For example if you take the integral of a velocity function/curve, it is also equal to the change in position of the object moving. The downside to this is that we won't know where the object is or where it began, but only how far it moved.
So to solve the problem we have to realize that we are looking for a change in values on the antiderivative of th given function. That is the Fundamental Theorem of Calculus. So what we can do is apply the constant multiple rule to make some antidifferentiating a little light for us. Then we apply the Fundamental Theorem of Calculus. So we have to find the antiderivative of the function. Once that is done we can find the total sales between year 2 and year 4. Remember that we cannot have part sales so we have to round down to the nearest integer for the final answer.
This next part is the introduction of a new topic in the unit of integrals. But it isn't really a new topic because it is putting some visuals to the antiderivatives. The function on the above slide may be a bit ugly but I'll pick out the little things to make bit nicer. Well the function involves 2 variables, x and t. t is the independent variable in the inner function, f, and x is the independant variable for the outside function, A. Since f is a function that is a nice and smooth function so we don't have to worry about any conflicts with arguements. So is the interval of the integral of the integrand, f(t). The variable, x, increases or decreases the size of the interval of the integral. So we were told to fill in the tables on the slide. For each value of x the size of the interval increases or decreases changing the integral of f(t). Since f(t) is a straight line, we can just count squares and triangles.
Oh no you have just ran into an area below the x-axis. What do you do? I'll tell you. Notice that there is a negative change in x and also a negative change in y. So the area can be multiplied by multiplying the y value and x value depending on either the area is a triangle or rectangle. So when you multiply a negative by a negative, you get a positive.
Now let`s graph the numbers found. Notice that it forms a parabola. This should make sense because the antiderivative of a line is a parabola.
That little experiment brought us to the Sign Convention. No there is no event at the MTS Centre about signs, but a sign rule. The convention says that the integral of a function is equal to the negative of the integral of the same function if the interval is reversed.
The next two questions involve the same process as the experiment from earlier. Notice that the fixed interval endpoint is changed in each question.
That was my scribe. As you can see on the last image written by Lawrence with his left hand, homework is Exercise 6.2 all odd questions and omit questions 7 and 11.
The next scribe will be Joyce.
Good night, I got to get studying. I have a lot of APs to study for.
Saturday, January 10, 2009
Chapter 1- IX (9): The Selection
Decide on the 6 sample questions from the textbook (questions or theme may change)
Chapter 1- XVIV (19): The Contrivance
Brainstorm and discuss about the kinds of questions should we make, using reference of the six sample question and try to make a few of them.
Chapter 2- X (10): Genesis
Create the six questions that will be used for the DEV
Chapter 2- XX (20): Decryption
Solve the 6 created question and edit the ways to solve them, making it easier to understand and cleaner to look at
Chapter 3- V (5): Initiation
We begin the making of the DEV
Chapter 3- XX (20): Sound Chronicle
We begin voice over of the DEV
Chapter 3- XXX (30): Fine Tuning
Edit what we have of the DEV
Chapter 4- VI (6): Scrutiny
Look over the DEV and discuss if it needs more work or not
Chapter 5- III (3): The Final Battle
Complete and submit the final copy of the DEV project
Chapter 5- IV (4): Epilogue
Its over!! However we have to study for the upcoming exam D:
J2K Timeline [SORRY FLJ for the letter stealing. At least we acknowledge our lack of creativity for the name and at least we used a number.]
Here's our timeline, also subject to change.
JANUARY 8: Making of timeline COMPLETE
Record copied examples of questions for project reference. COMPLETE
- Copying a rough draft of 6 questions from textbook/resources to use as references for the types of problems to use in DEV project; chosen based on difficulty we had with these questions so that we’d spend time learning on how to solve these problems.
FEBRUARY 5: Continue brainstorming possible theme ideas and have made the decision on what to use.
FEBRUARY 6: Create 6 questions from reference problems.
- Start the process of creating six of our own word problems that relate to a theme; using the type of questions we chose, with first, learning how to solve these problems.
- Answer as many questions as possible and make sure that we're getting the right idea by using resources, such as asking other people, reading textbook, internet and asking Mr. K for greater understanding.
- This is for the "look" of the project, the part that will make it visually easier for people to learn the material. Basically getting aspects of the theme together using videos, slides, pictures etc, that abide by copyright rules.
- By this time, hopefully we'll have all of the questions answered and explained thoroughly enough to be taught to readers of our project.
APRIL 17: Merging of all aspects of project for preparation of final structure.
MAY 1: Finalization, adding finishing touches getting ready for immediate publication on the blog.
MAY 2-3: Submit to Developing Expert Voices blog as a big finale.
There's our timeline and we hope to follow it, and finish this ASAP.
Friday, January 9, 2009
JANUARY 9 - Record Copied Examples of Questions for Project Use -Complete-
- Six examples are to be copied from Calculus textbook to be used as models for questions to be created.
JANUARY 23 - Create Six Questions from Copied Questions for Project Use
- From the six examples taken from the book, they will then be used to create questions that are going to be displayed in the project.
FEBRUARY 13 - Solve and Annotate the Six Created Questions
- The six created questions are to be solved, corrected, edited and annotated as to be properly displayed in the project
MARCH 6 - Begin Filming Gameplay
- Gameplay filming will begin via Xbox 360 and Halo 3. Filming will begin as soon as a HDPVR is acquired. Many long nights of filming will occur.
MARCH 31 - Complete all Gameplay Film
- It is expected that by this date, filming will be completed
APRIL 1 - Begin Editing Gameplay Film
- Post-production of the project will begin. Special effects, voice acting and video editing will take place. Many long nights will again, also occur.
- A separate blog will be created for the DEV to be published to. Will consist of posts for Solved questions, Reflections, Sources and Credits.
- Prior to "handing the project in", the blog will be revised that it is in correct layout, questions will be revised for errors, and links and videos will be tested for optimal viewing.
- Film will be completed and fully uploaded. Link to DEV blog will be revealed on the Expert Voices blog. Digital copies will be created and will be published via DEV blog. Project will be "handed-in" for evaluation.
This project is a sequel and will most likely require you to view the previous project :