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.

• 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

Due to this cluster of due dates, I on behalf of Team FLJ, am requesting a DEV extension. I will not keep this request exclusive to us, and I'm letting other groups also request an extension via Comment.

The Scuba Steve Dillema

I knew I forgot to do something yesterday. And that was check to blog -__-"

Sorry that this is very late, some urgent things came up and I wasn't able to get home until 1, so I'll just give a quick run down of what happened since everyone was in class.

Anyways, we worked on another question, this one being about Scuba Steve. There was a lot of information, so the question had to be read carefully to fully understand it.



So as you can see, to solve it, we found that the hypotenuse of the triangle at the beginning is 30 root two. we then found the remaining area near the end of the distance to be x since we do't know how long that distance was. Because triangle is a triangle with two sides equal, the top of it would be 30, thus the 170 - x (it's 170 because 200 - 30)

So that makes the second part of the equation for part a. The next slope we found one of the sides to be 45 feet. So then we found the hypotenuse to be root x squared + 2025. And now we can build our equation.

*Red is what Paul stated. Blue = edited work.
The RDT triangle is an example of how to change from rates, to distances, to times.

The next part is done by benchmen (Yes, he's a plural. That's not weird), and he found the derivative of the function, and set it to zero to find x.

Unfortunately, this is as far as we got as we ran out of time.
Next scribe will be Ben

The Truth about Bruce Lee and Chuck Norris

Some real fighting, none of that ninja fighting bullet dodging special effects haha.

Differential BOB

So at first the unit started off nice and slow. Quite easy to understand for the most part. I liked it. Then it was weird because somewhere along the way in the middle of the unit, it was like my mind just blanked out and everything was blurry and unclear. Mind you this is just me.

The Eulers Method I will study up a bit on because I don't quite remember it... At all.

Newtons Method of Cooling I believe I had the easiest time with (being that it was taught last) and that I seemed to catch on how to do it. I felt comfortable doing the questions and they just seemed tedious, and sometimes mind wrapping.

Cooling Off

So technically, I'm not supposed to be today's scribe, but its not like anyone else took scribe notes but it doesn't really matter since today's class was pretty straight forward. And for the record, the Scribe List is always up to date 8).

Now, it's not really cooling off, since it's spring! Well, with all the left over snow, it doesn't really look like spring break, but you get what I mean. Anyways, we started the class talking about how we're in the home stretch (That's right guys, there's only less than a month left before exams) and we need to start getting some practice in. There's about 72 exam questions we can work on (found over on the side here --->) and ideally, we should be getting in about 3 questions a day. Remember to not erase any of your work, that way, when you consult with Mr. K or other students, we can analyze the work to see where you went wrong (if you were wrong at all in the first place).

Anyways, let's get to the main stuff..

We started by finishing off last week's question about the roast. We found that k = 1/2ln(5/12)

So now that we know A and k, we can now solve for any time where t = time. So we then had to find how long would it take for the temperature of the roast to fall to 21 degrees Fahrenheit.

In place of T we put 21, because that's the temperature we want to find the time for, subtract it from 20. That leaves us with one. We move the 48 from the other side to isolate e^1/2ln(5/12). We then took the ln of both sides and solved for t and got approximately 8.844 hours.

So next on slide 7, we started cooling soda's outside. So now we want to find how long would it take for the drinks to cool to 35 Degrees Fahrenheit. So now, we have some initial values that we can work with.

The temperature outside is 25 Degrees Fahrenheit(we checked the thermometer outside), and the temperature inside is 72 Degrees Fahrenheit. We put one of the cans outside for 30 minutes or 1/2(0.5) an hour and measured it's temperature to be 60 Degrees Fahrenheit.

So, T(30mins) = 60 Fahrenheit (or T(0.5hours) = 60 Fahrenheit), and T(0) = 72 Fahrenheit. Because the drink is cooling at a constant rate, we can make a function. dT/dt, where T = Temperature and t = Time.
dT/dt = k(T-25)

We antidifferentiated the function and let e^c = A. ( You will recall this procedure from the last scribe post).
So we want to know what k is. So we use T(0) = 72. The e^(k(0)) will become 1 and we're left with A. So 72 - 25 leaves us with 47. Now we have our A value. To find the k value, we now use (in this case) T(30) = 60.
So we get 60 - 25 = 47e^(k(30)). We then isolated e^30k by dividing 47 from both sides. We now have 35/47 = e^30k. Finding k, we find the ln of both sides and now have 1/30ln(35/47) = k. Using the calculator, we find that k is approximately -0.0098.

So now that we have k and A, we can now find the time it takes for the drinks to cool to 35 Degrees Fahrenheit. So we set up the equation above. The first two lines are pretty straight forward, so I'll just tell you what's happening in the last three.
After plugging and chugging, we get 10/47 = e^-o.0098t and take the ln of both sides to get ln(10/47) = -0.0098t. Divide 0.0098 from both sides and you get t is approximately 157.486. Now that's not a pretty number to say unless you're talking movies, so, we convert it to be 2 hours, 32 minutes and 30 seconds. Voila!

Homework is Exercise 9.4.

Next scribe is Paul.

Now I gotta jet. Bye.

~Rence - Out

A New Unit: Differential Equations

So today was a new unit.

Gundam Unit Exia

But apparently, it's something new, but something old. Like a nice pair of chucks. :)

We started with an equation to see if we could grasp the concept before he formally told us.

So Joyce did this question for us. Basically what she did is she anti differentiated a(t) to get v(t). She determined C to be ten by setting v(0) = 10 so then the 0 made the first and second term in v(t) to be 0 so then C = 10.


She then found the roots at 5 and 1 by factoring the function. v(t) = 2(t-5)(t-1). She stated the particle (Why are particles moving along x-axis' again??), changes direction at one and five, but Mr. K asked her to verify it. So she drew a graph, but he then again asked her how she could back that up, and she used the first derivative test.

Why?

Because velocity is the derivative of position. So from negative infinity to 1, the function is positive, from one to five the function is negative and finally from five to infinity, the graph is positive. Therefore, the particle changes direction at one and five.

This is when thing's started clicking. We realized that..

Velocity is the derivative of Position
Acceleration is the derivative of Velocity.

We realized this because due to the given values, they were all related to each other and they are rates of change of each other.

So we tried the theory and antidifferentiated velocity.

This is when he introduced that these were differential equations, and introduced five words to us.

Differential Equation
Order
General Solution
Initial Conditions
Initial Value Problem

So if there is a derivative of the equation and is represented by a variable, we call it differential equations. But wait! There's more. There's flavours but we'll get into that later.

Next are the order.

Obsessive Pencil Ordering by mimi.amoure

Order tells you which derivative is in the equation. For example. v(t) would be the first order, and a(t) would be the second order. We could talk about the degree (here comes the ripples...), in which it would be an exponent on the function for example: v(t)^2.

The general solution is the family of functions, which is the function plus the vertical shift [ + C]

Initial conditions are the one value that picks a certain function from the family. Using conditions like v(0) = 10, which would pick out that one function from the whole family, like Tito Boy (Filipino Joke :P).

The initial value is determined by the initial conditions of the differential equations.

So learning all tha simple stuff, we were given a differential equation to solve given the intial conditions.

So Justus did the courtesy of doing this question for us (which I assume everyone did correctly).
First the anti derivative of f'(x) was found and then we subbed in f(1) to find C to find the parent function. Pretty simple right? Yeah, I liked it too.

Anyways, zeph's the new Scribe. :)

Now to end it off with the new trend... YouTube Videos!!! I don't know how old school some of you people are, but I'm upset that they don't do re-runs of "Whose Line Is it Anyway?". Best Improv show I've watched.



One of my favourite episodes. XD

Application of the Integral BOB

So, it's been pretty hectic recently so I haven't exactly been able to put 100% into everything, which is why I believe I've had quite an amount of trouble. I had no problems visualizing washers, shells etc, but I have to brush up on the mechanics of the applications. Some were simple, but i basically forgot it because I will admit, I haven't studied for the last few days. I tried, but too many things have been coming up which I can't say as it's quite personal. I actually kinda had some problems with the density and had trouble understanding it but I think I got the gist of it.

I hope I do alright and MacGyver it tomorrow. I'm actually cramming right now while trying to finish a bunch of other homework, so I'll see how it all goes.

Cylindrical Shells

Something important came up so this'll be brief and not too fancy, but I'll try my best to make it good (Don't see a reason why to be so detailed when everyone was in class.)

Anyways, we started by reviewing the disk...

and the washer method...
We talked it can be applied to real life situations and the number of questions it can be used. He used Boston as an example as the city grows in semi-circles. He also talked how these are the inverse of related rates problems and how it differs how we're trying to find out "how much" of a thing we have rather than how it's changing.

He returned to the "paper towel" idea so that we'd understand it better. At first we took a solid and rotated it to get a solid, but now we're taking arbitrary solids, and instead we'll get one piece of "paper towel". If we can get the volume of one of them, we can extend the idea of how the it's changing. This is called Cylindrical shells. So we'll just be adding up the shells.

So first we get the volume of one shell. If we open it up, one side of it will be 2*pi*r because it's about half of the circumference. The other will be the function, which is x^2 right now. the depth of it will be dx. To get the volume it will be l*w*h, so...

We have a problem though because one piece is a little like a trapezoid. If he took the end piece and put them together, you'd get a square. So the little piece will be delta little r square. It will get small because the change isn't 1. Even 0.1 is large. That squared is going to be .001, so it's still large, so if you square it again, you'd get .0001. It approaches zero, so the difference between them is negligible. As the change in 'r' gets smaller and smaller, they approach the same length. Because the area is delta r squared, we can regard it as insignificant, because the change is so small. So that difference, "r" can just be called x.

Mr. K tol us that the length will always be circumference (2*pi*r), and x will always be the radius of the shell, and height will be determined by any function that is present.

Anytime you rotate horizontally, circumference will be the length. The radius can be different things, for example, if rotated at -1. radius will be larger. if it's rotated around x = 5, the outside would have curves all around.

Mr. K informed us that the First question on the free response portion of the exam is commonly like this: Find the area between two functions, and that area will be revolved around the x axis, and then find the volume.


All of a sudden we started talking about the 'z' axis, which is an axis coming toward you and out. So the thickness on the z axis is zero... Always. It's the easiest haha.

We find 'S' is the area inbetween the two functions. We'll take S and rotate it. Intersects at zero and four. We then rotated it around the x axis. Then we took a slice and we got a washer and we found the radius was was f(X) and the other part of the washer was g(x) so to find the area we piBIGrsquared minus piSMALLrsquared, and then we found the area. *

We stop here because it's tedious, so then instead we put it into our calculator. Then we got 768pi/5

Next scribe is... Kristina. And remember, you can't talk to me, I'll be silent the whole day. :)

Chapter 7 BOB

I liked this chapter, kinda. There's never gonna be a chapter that I absolutely like without problems. However I was relatively comfortable with the unit with the techniques of antidifferentiation like substitution and integration by parts. The only real problem is having to recognize which one to use.

Next was the Midpoint and Trapezoid sums were pretty straight forward. I think this test will be okay, but a hard okay. Lol.

Bad News Bears: Mr.K as - The Rock 'n Rolla

Today, we have received some news. Some serious news. As those who were in class, already know, but Mr. Kuropatwa will be gone next week for about 9 days. He will be gone for reasons I will not say, for they are very personal. We are all praying and wishing him the best.

Due to these events, Dr. Eviatar will return to substitute for 3 lessons of Chapter 8. Mr. K also gave us tinyurl's to assist us during that week should we have any concerns or confusions. I will post the tinyurl's now.

http://tinyurl.com/b94cgo - Chapter 8.1
http://tinyurl.com/c5c2nm - Chapter 8.2 & 8.3

All of them are interactive lessons. When he returns, we should be finishing up Chapter 8.3 or starting Chapter 8.4

Oh yeah, we have a test on Friday, be there or be cubed!

Starting off our 'Spare Lesson', we reviewed the Left and Right hand sums. After that we reviewed the midpoint and trapezoid sums. (I find it that I shouldn't have to really explain it here as it is explained in detail in the scribe post below)

Now after doing the homework that Mr. K assigned us, we found that the Midpoint errors were approximately half of that of a Trapezoid sum, therefore Midpoint sums are more accurate by approximately double that of a Trapezoid sum.

Error Estimates for Trapezoid & Midpoint Sums

The Equation for Trapezoidal Sums

The Equation for Midpoint Sums

Where...

• a & b are endpoints of an Interval
• n is the number of sub divisions
• M2 is the maximum value of the second derivative on the given interval.
  

For the Midpoint sum equation, it's 24n^2 because it is more accurate.

So to use this, we 'choose' the accuracy then find n, so we know how many rectangles there are.

The Simpson Sum


No, Not Homer Simpson. The next sum was made by a mathematician who found a way to combine the two sums. He tried it in a different way, he tried to calculate using little parabolas.
So for example, we'll use M4 and T4, because if you have both, you can find a Simpson Sum.
M4 = 0.2422
T4 = 0.2656
So just plug it in and...

((2(0.2422)) + (0.2656)) / 3 = 0.25

0.25 is the exact point of the integral we were using, but even though it was exact in this case, we'll call it an exact approximation. This tells us that the Simpson Sum is even more accurate than both sums.

Approximate

to within an error of 0.001 using a:
a)trapezoidal sum - How many intervals, n, are required?
b)How many intervals are required using a midpoint sum?

A)

f(x) = x^-2 f'(x) = -2x^-3

f"(x) = 6x^-4, so max of f" on [1,2] occurs at 1.
f"(1) = 6, therefore M2 = 6

0.001 = 1/(2n^2)
500 = n^2
22.3607 = n

From here, we round up because it can't be less than the error that we wanted. Less Intervals = Less Accuracy. So, n = 23.

B)

0.001 = 6(2-2)^3/24n^2
0.001 = 1/4n^2
250 = n^2
15/822 = n --> 16 = n

Wow, did that honestly take me two hours? Geeez ¬__¬" Anyways, I still have loads of homework to do, so I'm gonna go do that and stay up for a few hours. Expect me to be either grumpy or tired so yeah :D. Oh yeah, next scribe will be Francis.

Also I thought I'd bring this to everyone's attention.

Take the Vow

Over one billion children live in poverty. Hunger is only one of their challenges; exploitation, abuse and discrimination haunt them on a daily basis. Many live in remote areas of the world and have little, or no, education.

They have no rights. They have no voice.

On March 5, 2009, people around the world will remain silent for 24 hours in support of those who are unable to speak up themselves.

Join the quiet revolution.

Take the Vow of Silence.

Go to I am Silent for more information.

The Chain Rule Reversal

We started the class, reviewing the format we should post our blogs in. Key points would be white space (Paragraphs), because the brain digests it easier instead of looking at it like a huge beast. Another would be to post pictures in the centre format, so that it's separated from the text and easier to see.

Moving along, we started the lesson with different ways to antidifferentiate, for example, breaking up the terms so that it's easier to antidifferentiate, as well as algebraically massaging the expression into a form that it's easier to antidifferentiate. Speaking of massages, I could use one right now thanks to all that working out, making my muscles sore. Here, we separate the terms, and move the constants (2 and 3) outside of the integral. And then we can antidifferentiate sinx and e^x easily, and after antidifferentiating it, it becomes...
Note: *Don't forget the + C. It was evident during the class that we constantly forgot to add the + C. It costs a mark!*

Now, looking at the slides, it didn't turn out so well because the transparency effect didn't work so you can't really see whats there, but I'll spell it out for you.

So we had the integral of the function (x^2 + 1)/x on the interval from 1 to e with respect to x. Because we don't know how to antidifferentiate quotients, we can massage that to separate terms so that they're both divided by x. The x's reduce on the first term and that leaves us with the integral of the function x + 1/x on the interval from 1 to e with respect to x. From here, we can antidifferentiate this, a LOT easier than before.
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. -__-"

As you can clearly see, it didn't work out so well when we tried multiplying and antidifferentiating. We even separated the terms! However, there is a product rule for antidifferentiating.

We were introduced to the Product Rule. Now, we weren't exactly sure if this would work, but it works like this. The integral sign cancels the derivative sign, so we're left with the Integral of the derivative of the outer function at the inner function multiplied by the derivative of the inner function, which should equal the parent function. So then we put it to the test.

Yay, it works. The F'(x) = cosx. g(x) = 3x and g'(x) = 3. So working that out, we get sin(3x) + C. That's right, + C and don't forget it. Always makes sure to analyze to see if they're composite functions. But that led us to the thought, what if that 3 wasn't there? How would we know how to do it? Well we tried just that too.

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.

BOB

So this is my blog on blogging on blogging on blogs... blog. :D

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.

Team FLJ Timeline

This is FLJ's Timeline as of today. Timeline is subject to change.

TIMELINE:

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.

APRIL 13 - Begin DEV Blog Construction

• A separate blog will be created for the DEV to be published to. Will consist of posts for Solved questions, Reflections, Sources and Credits.

MAY 1 - DEV Revision

• 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.
  

MAY 3 - 100% DEV Project Completion

• 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 : MITSOP.

This Project is Rated EG for Educational Guidance

This film contains scenes of Intense Math, Frightening Equations, and the Use of Patterns

The New Years BOB

Unfortunately, like everyone else, I have forgotten a majority of what we learned and it remains faint in my mind. It's there, but I can't access it. All I can think of is food, eggnog, present's and fireworks. Really.

The first and second derivative test however seems to stay with me, and I still kinda understand it. I always had problems with the optimization questions and always hit the wall hard. I have scraps of pieces of the Mean Value theorum, but after looking at a few scribe posts, it still doesn't ring bells. I must've lost too many brain cells over the break.

I honestly feel very uncomfortable going into this test (I didn't even know it was tomorrow until I looked up the blog just now), due to the fact that I did so horribly on my pre-test and guessed most of it. Woe is me.

~Rence, Out.

Ryon Lovette - Equation

So I was bored while studying, and I found Math and R&B. =)

Equation - Ryon Lovette

Couldn't help to notice you standing in the hall way. Tears rolling down your face, girl. I heard him tell you that he was sorry for breaking up with you But you're way to beautiful to cry. Soon as he walked away, I came up from behind. I told you a joke to make you laugh. I'm nothing like him girl, you do the math. (Chorus) Me plus you, I'll take that number. Multiply your smile, minus the drama. Give me a fraction of your heart. I'll solve your problems. Now put that together. We make up a perfect equation, equation, equation. Me and you make up a perfect equation, equation, equation. Me and you make up a perfect equation.

And you can count on me. Forget about your past. I'll always put you first, never put you last. Girl, you're my absolute value. Have no fears, I'm here to save you. Promise I won't change, I won't play those games. Even when it rains, my feelings they stay the same. Put me to the test I haven't failed yet. Trust me I'll pass girl, just do the math cause. (Chorus) (Me and you make up a perfect equation) You deserve the best, nothing less from me. Hold your hand, open doors, treat you like a lady. All your dreams will come true, baby when you're with me. Don't fight yourself from him, I'm the half you need. (Chorus)

The Absence of Evidence

In Absense of Spring by flickr user abstract splotcHes'
Anyone else feelin' a little chilly?

We started off the class with a pretty funny clip which featured Calculus and the Numa, Numa song. Basically the person in the video showed two pieces of paper, one a function and one an anti-derivative.

Following that, we got into the stuffing of the class. We reviewed derivatives and what they are in order to grasp the concept of an anti-derivative. We were told to remember something when we do anti-derivatives, which is "If you can do something, you can undo it." For anti-derivatives, this means that any derivative can be anti-differentiated.

A key point is to know that an anti-derivative is a family of functions, however, if you have a given point, you can find one particular function within the family of functions.

An Indefinite Integral is not like a definite integral. It picks our a family and if a point is known, it picks a particular one. This sounds a lot like an Anti-Derivative.

Moving along, we are told that for every derivative rule, there is an anti-derivative rule, as this is automatic when a derivative rule is made.

So, knowing that, can we find the parent function with first and second derivatives?

Let's use an example.

If you know the derivative of sine, you know the anti-derivative of cosine.
Therefore if you know the derivative of sine, you know the anti-derivative of sine.

In words: The derivative of sinx is cos x, so the anti-derivative of cosx is sin x + C where C is an arbitrary constant. This basically means that C can be any value, because the function can exist in a number of places in the "family of functions."

The derivative of cos x is negative sin x, so the anti-derivative of sinx is negative cosx + C.

After this, Mr. K gave us a question so that we can apply this.

Consider the function (not the way it's supposed to be written but sitmo won't let me write it properly.) * "k" is a parameter and can be any value.*Using the first and second derivatives, describe the features of the family of functions generated for different values of "k".

Bench did the honours of doing this question.

Using the quotient rule and the chain rule...and factoring out...and reducing...and then we let it equal 0, so that we can find x.

o = 1 - kx
kx = 1

*Remember that critical numbers are at zero and undefined points, however, for this function there are no undefined points.
Therefore, this function will have no discontinuities (cusps, corners, magic tricks, hat tricks, pen tricks, kick flips etc.)

So by the first derivative test, when x = 1/k, the function is at a max.

Okay, it's 12 AM. I need to study for Physics... Ugh.

Next scribe will be Zeph!

Rence ~ Out