Student Voices Episode 2: Tim_MATH_y

In this episode Timothy came back to school on Friday afternoon to talk about his week attending the miniUniversity program at the University of Winnipeg. He talks about the differences he finds between teaching and learning at high school and university and describes learning in the university classroom using a thought provoking metaphor, listen for it. Also, we have a cameo appearance by two very special people at the very end.

Please feel free to leave Tim_MATH_y your comments here on this post.

(Download File 7.2Mb, 15 min. 3 sec.)

Photo Credit: Shadow singer by flickr user EugeniusD80

Blogging on Blogging

Oh my, it's test time tomorrow! Well, here for another bob (throws bob post onto large stack of bobs). This time it's for the test on Differential Equations! How exciting? Well anyway, this unit, like most of the other units had its ups and downs. With regards to my muddiest point, I would have to say that I don't have a specific problem. All that I'm worried about, like most of the times is the battle against time, by trying to make everything click on the test. Thats the thing though, tests aren't as difficult. Well what I'm trying to say is, the solutions aren't as bad as we usually think, but the difficult thing is, figuring the key solution that would unlock insight to finding the final answer. Asides from the likes of that, theres also remembering to attempt to solve a question using as little time as possible (such as taking advantage of your calculator). Well, thats all i really have to say about this unit! Good luck everyone! I hope people remember to bob, cause I almost forgot =(

SCRiBE: Last Of Content

Hello everyone! I'm known as Tim-math-y and I'm the scribe for Thursday's lessons. On Thursday, our topic was: Solving Differential Equations Symbolically and Newton's Law of Cooling.

Introduction:

Solving Differential Equations Symbolically and Newton's Law of Cooling:

On Thursday, we learned more about differential equations, this time using the symbolic (algebraic) approach. This topic tied into Newton's Law of Cooling. According to this law, a hot object cools at a rate proportional to the difference between its own temperature and that of its environment.

By applying this concept, we solved questions that dealt with cooling.

Content and Lessons:

First we began by deconstructing the equation of "dy/dx = ky." Mr. K mentioned in the past that normally, you are not allowed to pull apart the differential operator. However, without going through the complexities of why it is possible, we pulled it apart.

By seperating the two variables, including both 'dy' and 'dx', we found a familiar term that could be antidifferentiated.

Through algebraic play, we found the equation that defined the parent function of cooling.

*Note: the variable "c" and "C" are different in value but generally both represent a constant.

Next we looked at an example question.

• First we assigned variables, F = Temperature in Degrees of F and t = T time in Hours
• Second we applied the variables to "dy/dx=ky" to produce "dF/dt=k(F-20)" where 2o is the lowest temperature the object can attain in an environment
• By applying the same steps to seperate 'dF' and 'dt', we antidifferentiated and worked the out an equation that could be used to find 'F', the temperature, at any given time. The only unknown left is the constant that we have yet to solve.
  

By using the information given (two pieces of data that showed us the temperature at a specific point in time) we substituted the values in and solved for the final equation that represented the temperature at any given time of this situation.

At this point, we considerred both questions solved.

Next, we looked at another example question (Don't be overwhelmed by the length of the question as most of it is just a background story that Mr. K conjured up: The information required is located in paragraph 3).

By applying the same steps from the previous question, we found the solution to the problem. Again, our variable assignments were the same. The only difference is that the coldest temperature in this situation was 25 degrees. We found the equation that still included the constant value of C.

By taking this a step further, we inputted the coordinates given (temperatures at given times) and solved for the value of C to in turn, solve for the final equation.

Finally, by inputting the temperature of the cola at an unknown point in time, we solved for the uknown value of 't'.

Conlusion:

We went through two problems that dealt with Newton's Law of Cooling, solving differential equations symbolically.

******

Homework is on the end of the slides posted in the previous post! Answer is on the following slide.

Tomorrow's Scribe is... Craig!

Happy Easter? Haha, good night.

BoB!

Ah! Once again there's an upcoming test! I am confident in saying that I am prepared, but not quite.

There are still a couple of things that I'm worried about. However I can't seem to place it. It's just that feeling in the gut. Hopefully some more practice will clear out any doubts.

Confidence may be my only hope! Good luck to everyone else on the test :)

The Scribe: Slices of Bread

Hello, I am known as Tim-math-y on our blog, and I will be the Scribe for today's lessons.

The Overview:

Today, we discussed the concept of "How to Build a Solid." By rotating certain functions about the x-axis, three dimensional solids were formed, upon which we could use calculus to solve for the unknown volume. We found that the solution to calculating such values, especially with irregular shapes, involved the idea of Bread Slices. These concepts will be thoroughly discussed throughout this scribe post.

The Introduction:

To begin the class, we started off with a concept we were already familiar with: areas between curves. We were given three questions that would lead us into our main lesson of today. They all involved similar steps, and did not present any new difficulties. Calculators were not permitted to solving the first two, but for the third one, they were permitted as the algebraic calculations could become disorganized and difficult.

This was our first question. As one may see, it involves a set of three major steps.

1.) First, we set up an equation in hopes of solving for the points of intersections, shared by the two functions. We set each function on opposite sides of the equation and algebraically solve for the value of 'x'.

2.) Second, we attempt to find which function is 'on top of the other'. This may be done by plotting both functions onto a coordinate plane or using a number line.

3.) Third, we subtract the two areas underneath each curve (area1 - area2, where area1 maintains the upper position). To find the areas underneat each curve, bounded by the x-axis, we used integration, from 'a' to 'b', where a = 0, and b = 1 in this question.

By solving, we calculated the area bounded by the pair of functions.

The Next two questions given included the same ideas and concepts for finding the solution. What is more important however, is how this connected to the concept that we learned today.

The Focus: Volumes by Slicing
Included are exerpts from the textbook (Calculus; Concepts and Calculators), hopefully to provide insights

The method we use to compute the volume of certain solids revolves about using the definite integral. By dividing the solid into small pieces whose volume we can easily approximate, we can compute the volume. As the number of terms in the sum grows larger and larger, the approximation improves and the limiting values is a definite integral.

For any solid that has a constant cross-section its volume is the product of its cross-section area 'A' and its thickness '(delta) x'. Most solids do not have regular shapes and the task of computing their volumes requires calculus methods.

Imagine a loaf of bread lying along the x-axis in the xy-plane between x = a and x = b.

The loaf can be divided into several slices by making cuts perpendicular to the x-axis. The volume of the loaf is the sum of the volumes of all of the individual slices. In general, since the shape of the loaf varies, different slices have different volumes even if they all have the same thickness.

For any 'x' between 'a' and 'b', let A(x) represent the cross-sectional area created by the cut at x. Suppose we divide the loaf into n slices of thickness (delta) x = (b - a)/ n. The cross-sectional area of a typical slice is not constant. However, if we replace it with a slice of the same thickness and constant cross-sectional area given by one face of the slive, then the volume of this slice is:

Volume of one slice = A(x) * (delta) x

To estimate the volume of the whole loaf, divide it into n pieces of equal thickness: (delta) x = (b - a)/ n , by making cuts at a = x0, x1, x2 ... xn = b.Using the cross-sectional areas A(x1), A(x2), ... , A(xn) for the individual slices leads to an estimate for total volume.

Total Volume = A(x1)(delta)x + A(x2)(delta)x + ... + A(xn)(delta)x

This is a Reimann sum for the cross-sectional area function A(x) on the interval [a,b]. Thus, the limiting value as n grows larger and larger is the definite integral concluded below: Coming back to our class lesson, we looked at an example question:

In this situation, the linear function, within the interval of [0,8], was 'spun' around the x-axis to for a coned solid. We found that, by slicing the solid into smaller pieces, we could find the volume of the given piece. In this case, each piece took the shape of a cylinder, where its volume could be expressed as: area x height, where area is (pi)r^2, and height is the subinterval of (delta)x.

As the amount of slices grew to infinity, or in other words, let (delta)x become infinitely small, the integral of the area would compute the volume of the solid.

We then continued our calculations to solve for the volume:
Here, we found the volume of the coned solid to be 128(pi)/3. Referring back to the original equation for solving a volume of a cone, we checked whether or not our solution was correct.

By inputting the given values of 'r' (4) and 'h' (8), we found that our calculations were indeed correct.


Finally, there was one final concept that Mr. K parted to us. It was a brief explanation as to how to solve these type of questions:


The solution is as simple as the slide. To solve for the volume of a given solid, such as this, we utilize the same techniques. To solve this question, we find the area of the shaded portion. Simply put, we find the area by using subtraction: A2 - A1.

That concludes our lesson today. I hope that any of the readers found this somewhat helpful, if required. I just thought I'd add a slight more effort in today's scribe post.

I found a link to a site that provided me with some animations, to help anyone who wishes to visualize such shapes to a greater extent:

http://curvebank.calstatela.edu/volrev/volrev.htm

I'm not sure if we were assigned homework. However, I'm guessing exercise 8.2. We will continue this topic the following day, said Mr. K.

That's all everybody! Tomorrow's scribe will be: Dino, the first name that popped into my head. Good night everyone!

BoB ing.

Personally, I found this unit to be one of the most difficult units. A large portion of the questions were undeniably difficult to solve, such as those where "algebraic massaging" or "adding/subtracting zero or multiplying/dividing by one" in various forms were required. I still have many questions left unanswered that I hope to be assauged tomorrow during my spare and at lunch. I know much of the mechanics, but much still remains clueless for myself. Thats my reflection on this unit. The pretest wasn't too bad either - lets hope that the test will be quite similar.

Good luck to everyone on the Test tomorrow! I'm glad that I rememberred to bob.

SCriBE O_O

Hello everyone, Im Tim-Math-Y your scribe for todays session. I do apologize for the late scribe.

| View | Upload your own

Today, we applied the lessons that we've learned during the previous day. We found the derivatives of arcsinx, arccosx, and arctanx. With this new knowledge, we have discovered that underlying, are the rules for antidifferentiating these functions. Truly, we only need to remember the derivative of arcsinx, as the derivatives of arccosx and arctanx are only slightly changed forms of the former.

We started out by taking a look at multiple questions where we were instructed to solve them freely; use any method you would think is the least difficult. These include: substitution, antidifferentiating by parts and the new method that we have discovered.

As we made progress, we found that the questions became more complicated. These questions involved algebraic massage, where we have to work the questions to find the solution. By looking at the expressions, we found that most of the times, the questions were near perfect from a simple antidifferentiation. By algebraicly massaging, we can add the value of 0, multiply by 1 or divide by 1 in various ways to solve each question.

Truly it comes down to that.

To end off the class, we were given a handout that included 16 questions. It should be completed for tomorrow, as it would greatly aid in your practice experience.

Tomorrow's scribe is: MrSiwwy?? I don't know who else to pick =)

Again, I apologize for the late scribe and, the lack of detail incorporated in the explanations.

Have a great night everyone, see you guys tomorrow.

BOB

Almost forgot to reflect.. again! THANK YOU GREYM I LOVE YOU MAN for reminding me to bob =). Okay well this unit, applications of integrals, was very short and shouldn't be a real problem to me. I understand everything well. I think the only problem that I have is seeing the big picture of this concept. Including how to write all the steps properly, if you know what I mean. Like all the steps in solving for a derivative are distinguishable and I'm scared I might make an improper mathematical expression of some sort. Ah well, I'll just piece it together tonight. Everything else seems fine. Good Luck to everyone else on the pre-test tomorrow and test of Thursday! I CAN'T WAIT UNTIL WINTER HOLIDAYS! YAY!

SCRiBE

Hello! I'm Tim-Math-Y and I will be your scribe for today's lessons. Today we had a workshop to prepare us for the pre-test tomorrow and test on Thursday. It was centralized around Integrals.

| View | Upload your own

Slide 2:

1.a) - we filled in the chart by calculating the integrals (signed areas) under the graph, start from in this case, -2 because the integral is from -2 to x.

1.b) - we sketched the graph simply by plotting the points we gathered from our table in question a.

1.c) - we analyzed the graph for local extrema
- the only critical number is x=2 because at that value of x, a local minimum is present
- end points are not critical numbers because there is no way of distinguishing if it is a local maximum or minimum without checking the slope on both sides of the point

1.d) -we simply found where the graph was increasing by looking for a positive slope

Slide 3:

a, b.) - to obtain these solutions, we can simply replace the variable t with x, included in the domain of the integral, due to the fundamental theorem of calculus

brief description of process:
- F(X) = f(b) - f(a)
- from this we get: (-cos(x^3)) - (-cos(pi))
- now we want to find the derivative of this integral (stated in the quesiton)
- we get: sin(x^3)
- the derivative of a constant is 0

c.) -the solution to this question was similar, however, since the integral is from x to 1, we can make the integral NEGATIVE to make the integral from 1 to x instead

Slide 4:

d.) - the solution is similar to that of question c on slide 2

e, f.) - these questions include a slight different solution

- since they are an accumulation of functions, to find the derivative, we must apply the chain rule: (f'(g(x)) (g'(x))

Slide 5:

- to find the amount of gallons using the rate (derivative), we integrate it from 0 to 4 hours
- yes, it's that simple =)

Slide 6:

- the only trick to this question is the +3 included in the integral
- we solve this part by find the integral of 3, from 4 to 7

Slide 7:

- points of intersections are found by making the two functions equal each other
- then, but inputting values between the pair of intersections (-2, 0) & (0, 3) we find which graph is on top of the other, in each case (in this case, twice)

Slide 8:

- we applied: (integral from (-2) - 0) ((top function) - (bottom function)) + (integral from 0 - 3) ((top function) - (bottom function))
- solve algebraically, applying rules of finding the derivative of an integral
- grunttt workkk

THE END

Okay I hope you guys found this slightly helpful. It wasn't as specific as it should be but yeahh. Tomorrow's scribe is.. Etimz since he was the only one who hasn't done his fifth scribe yet.. or something like that.

Good luck on the pre-test tomorrow! bye =)

Blogging On Blogging = BOB = 2152 - 2146 = 6!

WOW! BOB! I almost forgot again! =) =) good thing i didn't! I'm surprised that I forget to check the blog only during days previous of the test. Probably because i'm usually stuck worried and am studying sooooooo hard =)???? Anyways Applications of Derivatives! WHOO, was it ever difficult! Well.. atleast most of the stuff. I find that I understand the lot of it, but when it comes to solving it myself, i'm stuck! Gosh.. that mostly happens with optimization problems though. I'm truthfully going to admit that.. this test.. seems.. overwhelming.. and that im not going to ace this test. BUT! im not going to give in and still, try my best.. as always!

Strengths: Asymptotes, First + Second Derivative Tests
In Betweens: Anti-derivatives, Mean Value Theorem
Weakness: Optimization Problems

Well.. goodluck everyone.. let's hope we get out of this alive and in one piece! Bye

SCRiBE =)

Hello! I'm known as Tim-Math-Y and I will be the scribe for today's lessons! Today, Mr.Kuropatwa was victim to a fourth computer crash and thus lost most of his data, including that of today's lesson! However, Mr.Kuropatwa is quite the resourceful teacher and came up with problems for us to place our minds into. We were separated into work groups in which we aided each other in solving optimization problems, a continuation.

| View | Upload your own

SLIDE1:
-the first step, which is nearly undeniable as the most important step in solving these types of problems, is picturing the problem, most commonly through a drawn diagram of the situation
-second, we set up variables and expressions for the unknown distances of the diagram
-third, referring to the rate-distance-time triangle on slide 3, we set up a a function where X is a function of Time (the time taken to travel in water + the time taken to travel via walking)
-fourth, we find the first derivative of the function using chain rule and power rule, then simplify using LCD (Lowest Common Denominator)
-fifth, we find the roots of the first derivative in order to locate where possible mins or maxes may occur. In this situation, we are looking for a possible minimum. Roots are located at +/- 3/2.

SLIDE2:
-first, we depict a number line of T'(X), displaying the two possible mins or maxes
-second, we use the first derivative test to identify whether the extrema are mins or maxes and discover what we hope to attain: a minimum

SLIDE3:
-Simply put, we imputted the minimum value of 3/2 into the parent function T(X) to find the minimal amount of time in hours to complete the trip

Afterwards, we were handed a worksheet to work in our separate groups. It was not required to be handed in today.

Question:

Scuba Steve's Shark Cages

Scuba Steve is enclosing an area of his harbor for two sharks that we wants to keep as pets (Don't try this at home). He has 450 ft. of fencing for the sides of two cages. He needs to separate the two so they won't kill each other, but they need to have the same sized cages. What is the maximum area of each section of the cage that he can build?

If you haven't already solved this problem, here are some guidelines to follow:
-draw a labelled diagram
-apply variables (there will be 3 widths, and 2 lengths)
-find a function relating to area (length x width)
-find the first derivative
-find the roots of the derivative
-run the first derivative test to identify whether it is a min/max
-substitute your minimum x value into the parent function to find the max area
-top it off with a sentence solution

Answer: 4218.75 ft^2

And that, ladies and gents is today's scribe.

Thursday's scribe will be... Phuong? It has a (-) beside your name on the scribe list ... sorry! =)

Adieu, adieu.. adieu!

... Quod Erat Demonstrandum - And Thus it is Proven

SCRIBE ( ^_^)

Hello! I hope everyone is having a great evening! I'm known as Tim-Math-Y and I shall be your scribe for today's classroom activities!

Today Mr. Kuropatwa was away and so, we had a substitute teacher. The instructions that were left for our class were to read through pages 231-234 and work/complete Exercise 4.6 numbers 1-10.

This chapter is known as Related Rates of Change. Here is the intro to this chapter: "Frequently, in trying to solve a physical problem we must establish a connection between rates of change (Derivatives). This connection is usually made by using implicit defferentiation and the Chain Rule.

The questions involve a series of rates where we must find the relationship that connect them. Your solutions may include the use of basic differentiation, product rule, quotient rule, chain rule, and more.

Good luck and Have fun!

Tomorrow's scribe: MrSiwwy? I really don't know if that scribe list is updated or not =), if not, we shall find a solution tomorrow =)

Have a Great night people!

BOB

Well, its just that time again. Another test has arrived; definite integrals. There isn't much for me to say other than that this unit was a breeze for me and so i shouldn't run into much problems on the test. Can't believe i almost forgot to do my reflection.

Goodluck everyone.

Scribe (O.o)

Hello, I'm Tim-Math-y and I will be your scribe for today's class lessons. Today, we started off class by going over the requirements for scribe posts.

Remember that you are required to label your posts accordingly, especially your scribe posts. You are required to label your scribes with your name, unit, and most importantly, 'scribe'. Without labelling your scribe with 'scribe' is equivalent to not placing your name on your work; you will not receive credit due to completion on the scribelist (If you have not already fixed your labels, you should as soon as possible).

Next, we talked about del.icio.us accounts. If you have not already signed up for one such account, it is required as soon as possible. Remember to check out other tags in math, calculus, and the such for sites that other people found extremely useful. These sites will be helpful in contributing aid in our learning outside of the class.

After the brief run overs of these topics, we continued onto learning MATH! Today, we had a workshop to prepare us for a Pretest tomorrow and a Test on Thursday.

Workshop

1.a) For this question, we simply solved the lower and upper estimates using our calculators. We placed the equation f(x) = 4x - x^2 into y1 and used our RIESUM program to solve from the interval of [0,1]. Our solutions were L(4) = 1.28125 and U(4) = 2.03125.

1.b) We simply subtract the lower estimate from the upper estimate.

1.c) Here, we sketched 4 sub intervals on the interval of [0,1] of f(x). Here we geometrically sketched the upper and lower estimates to represent U(4) - L(4). The left sketch is incorrectly scaled whereas the right sketch is a correct drawing.

1.d) We used the equation: Error = |f(b) - f(a)| x ((b-a)/n) to solve. Here, 'b' is equal to 1 where 'a' is equal to 0.

2.) Here, we analysed the table to find out how we could estimate the integral on the interval of [-2,4]. Considering the fact that we need 3 sub intervals, we found that from (-2) -> (4) was 6 and so we could easily divide the information into 3 sub intervals. Here, we found the midpoint of 1.98 and 2.04 which was 1.03, midpoint of 2.04 and 9.64 which was 5.74, and the midpoint of 9.64 and 26.29 which was 16.82. We then added the three numbers and multiplied the sum by 2, because the subintervals maintained the length of 2. We found the result of 47.18 (Credit goes to Craig's group)

Another group attempted something different. They plotted the points onto the statplot using their calculators and simulated a similar quadratic graph. The solution was fairly close (46.5248) however despite the 99% accuracy, the answer was incorrect as it was not as accurate using the information given.

3.) Here, Chris' group attempted to solve the problem. He came up with the idea to use four subintervals, seeing that the information given could not be broken up properly into 5s nor 10s and so he broke them into groups of 15s. This brought forth much sense. However, the answer was not the best approximation as he did not use all the information given.


Here, Craig's group again brought forth a solution. They used the RIESUM formula to find the best approximations on seperate intervals (the riesum finds the sum of the lower and upper estimates and divides it by 2). The intervals were: [0,20] which had intervals of 5 minutes, [20,30] which had an interval of 10 minutes, [30,45] which had intervals of 5 minutes, [45,55] which had an interval of 10 minutes, and [55,60] which had an interval of 5 minutes.

Adding together all the solutions, he found a more accurate approximation because he used all of the information given, despite the extreme excess of work compared to Chris' method.

End of Workshop

Word of the Day: Ginkgo - a herbal remedy derived from Jap/Chi tree to improve mental function and circulation.

Tomorrow's Scribe: aichelle s. Show them a nice scribe =)

Reminders: Tomorrow Pretest, Workshop Wednesday, Test Thursday

Have a great night everyone! Good luck on the pretest and test this week! Don't forget to BOB also! Good night!

Blogging on Blogging..

Hey! It's Tim-Math-Y and its time to b.o.b. again. Well this new unit was definitive.. different, as it included a new idea, Derivatives. At first this unit really had my mind twisting up on itself trying to visualize a derivative of a function, then holding that image, attempting to visualize the second derivative. I will admit two things. That wasn't a good idea to imagine similtaneously and second, that I'm still a little confused about the matter.

However, thinking back of the leaked characteristics of the power law, which I also admit that I don't know how it works.. yet, it helps me remember that each derivative is a degree down from its parent (cubic>quadratic>linear etc.), and thus aids me in picturing the graphs.

Now with the knowledge of derivatives, it brings fourth questions that we deal with in physics, such as comparing two variables: time and distance, time and velocity, time and acceleration etc. I find that I don't have many problems with these types of Qs because I understand the concepts to a more complete level.

So I can say that I'm semi-prepared for derivative functions and their like. Other than that, I still don't really understand the exact purpose and use of finding limits. Limits still greatly puzzles me as, logically, it just isn't registering for me. Besides that, Continuity, I understand and I'm glad I can say that atleast...

Wow, my bob is long? =\ Good luck on the test on friday guys!

Derivative Assignment

| View | Upload your own

Wow, so many technical difficulties gave me headaches.. ahhh

Scribe (O.o) Pre-test

Hello my name is Tim-Math-Y and I will be today's scribe, for my second time. Today was quite the surprise. We started and ended our class with a PRETEST of Functions.



Most of the questions were not a large surprise nor were they difficult except a few.

One of them was Question 2 on Slide 2/7. The graph of this function displays a beautiful parabola with two roots. However, if you zoom out on your calculator multiple times or analyze the equation: f(X) = X^2 - e^(0.1X) you will notice that e^(0.1X) will eventually become so large that it will guide the graph down to form the third root.

A second and last question that involved issues was Question 1 on the second page (Slide 7/7). For Question 3a, it involved simple work of finding the area of three triangles without the shaded area, within the rectangle. After finding those three areas, we algebraically subtract the sum of the trio from the total area of the rectangle (LW = 8x6 = 48).

For Question 3b, we find that the domain is [o,6]. Sadly however, I cannot remember how.

For Question 3c, the least and greatest values of the triangle is found via the graphing calculator for quickest solutions. Within the domain of [0,6], we find these two solutions by solving for the minimum and the maximum value.

TOMORROW's SCRIBE IS: MrSiwwy!

Test on Thursday! Be Prepared!

Scribe: Mathematical Modeling

Hello! My name is Tim-Math-Y and I will be your scribe for today's AP Calculus class activities.

To start off the class, we watched a video (located on the post below) about internet safety and etiquette. Remember to be careful as to what you post, because once published, it cannot be removed and any one may view it.

After those couple of minutes, we then laid our eyes on the focus of this class: sets of numbers.

1.) X : 1.1, 1.2, 1.3, 1.4, 1.5

2.) f(X) : 6.341, 5.328, 4.335, 3.362, 2.409

3.) g(X) : 7.050, 5.875, 4.896, 4.080, 3.400

4.) h(X) : 6.340, 5.423, 4.506, 3.589, 2.672

Mr. K seperated us into groups to analyze these sets. Then we began take into account the differences between each term in each set to attempt to discover an underlying function.

1.) X : A simple linear function with a common difference of 0.1.

2.) f(X) : A slightly curved function with a declining graph. It included the differences:

---> 1.013, 0.993, 0.973, 0.953 (We then noticed a 'common differences' of the differences of 0.02)

---> 0.02, 0.02, 0.02, 0.02 (Mr. K stated that the number of 'difference sets' represents the 'degree' of the function)

f(x) in this case, is a function with degree 2.

3.) g(X) : A function that displays a declining curve similiar to an exponential function. It included the differences:

---> 1.175, 0.979, 0.816, 0.68 (which by Chris, says are 'arbitrary' which means: 'to be subject to individual will/judgment with no restriction')

4.) h(X) : A declining linear function with a common difference of 0.917.

After finding these characteristics, Mr. K began to guide the class through the calculator processes to set up these graphs and attempt to find out exactly the type of functions these were. The outcomes were produced with % accuracies that were indeed very close to 100%. However, with mathematical modeling, Mr. K states that there should not be a 100% accuracy.

Quickly referencing his most prized teaching object, the 'block of wood' ("the rectangular prismatic shape has multiple indistinct sides upon represent different representations but are all part of the same block of wood") to the different styles of representation and assigning us Exercise 1.3, questions number 2 and all of the odd numbers, the class ended.

And tomorrow's scribe isssss......... hmmmmmm..... CRAIG! yes yes, him! Show us your magic bro.

Have a great night everyone!