Scribe: Related Rates!

Introduction:

Hello, I'm known as Tim-Math-Y on our class blog and I will be the scribe for today's lessons! Today's class started with an image that introduced our content. The image was a combination of three that portrayed related rates with spheres, water ripples, and shadows. It was the precursor to the questions that we were to solve.

Key points to solving Related Rates questions:

• Find a way to relate all the variables within a given question
• Differentiate implicitly using chain rule, with respect to time

Content:

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Conclusion:

Well, I completed the scribe as soon as I could because it seemed that many people were missing today. I hope this helped!

Notice that most of these solutions include the following:

• A labelled diagram to help you visualize
• Listing given information, and finding information that will be required in the solution
  
• Differentiating implicitly using chain rule with respect to time
• A sentence answer including appropriate units

That sums it up folks! Thanks for reading =)

The Scribe for next class will be, John D. !

The Post of TODAY!!

More optimization problems

Hello everyone this is the scribe post for May 6, sorry it is late but I had to write the AP calc exam and I was not able to post it earlier. Well in Tuesday's class we only did one question, but there were two others for homework. The first question was about a man who was driving his car through the desert, and wanted to know at what distance he should drive through the desert and the paved road to get to the city B 5km south from city A. He can drive 15mph in the desert and 35mph on the paved road, he is 5 km east of city A. So the maximum distance the man could travel is 5km in the desert and 5km of pavement to get to city B (The diagram is on the slides). So that would be unrealistic, so in order to get the minimum amount of time the car must cut time through the desert and pavement alike. So the pavement equation will become 5km subtract x, x the amount of distance that is going to be cut by driving through the desert. Then the amount of distance driving through the desert is found by a^2 + b^2 = c^2. Where a and b is 5-x. So c is (25+x^2)^.5 (square root is ^.5).

As you know Rate, Distance, and Time is all related. Time is Distance over rate. So the total time is (5-x)/35 + ((25+x^2)^.5)/15 =t(x). Solving for the derivative (solved on the slides) , t'(x) =
7x-3((25+x^2)^0.5)/105((25-x^2)^0.5). Now solving for the zero's you get 40x^2-225. So the roots are -15/2((10)^0.5) and 15/2((10)^0.5). We reject the negative value because its not possible in the context of the question. So the man should go and head for the point x = 15/2((10)^0.5), as that point is the minimum value of the First Derivative Function.

Remember to do the other two questions and the homework given in class, the next scribe is Ethan.

Optimization problems

Today, we were doing optimization problems.
In order to describe how we are solving these problems, I will be using an example from the slide 5.
It is the open box question where you have to cut squares from the corners to maximize the volume of the box.

First step is to draw a picture of what you want to do. We are cutting squares from the corners and we label the square's side with x. We know that the side of the entire piece is 24 inches per side. Therefore the equation is (24-2x) for the space between the two corner squares you have to cut. Everything is a square so all sides are the same.

Next we know that the volume of the box is going to be V=L*W*H.
We know that the height(H) is x and each side (L & W） is the same and equal to (24-2x), so the equation is going to be
V(x)= （24-2x)*(24-2x)*x
which is
V(x)=((24-2x)^2) * x

Third step is to simplify the equation and you will get
V(x)=(4x^3) - (96x^2) + 576x.

We take the derivative of that to find the max.
V'(x) =(12x^2) - 192x + 576
V'(x) =12((x^2) - 16x + 48)

The derivative at the max is equal to zero.
So we let V'(x) = 0
0 = 12((x^2) - 16x + 48)
0 = ((x^2) - 16x + 48)
0 = (x - 4) (x - 12)
x = 4, 12
x can't equal to 12 because that would make the side of the box equal to zero.
Therefore x = 4.

The last step is to test it.
___+___4__-___12_____

4 by 4 squares should be cut out to maximize the volume of the box.


The next scribe is Dino.

Optimization Problems

Hey everyone this is m@rk and i'm scribing for Wednesday's class. First of we took a quiz on visual calculus about what we've learned from the first and second derivative test. Second , Mr. K introduced a new topic called Optimization Problems. Optimization problems are problems that deal with either finding the maximum or minimum of the function in the problem.

Slide 1

First of all you need to do is draw a diagram to help you visualize the problem. Second, construct the constraint equation. The constraint equation is the equation is something that limits the problem. In this case it is 320=2L+2W. Third, find what you are trying to optimize and construct an optimization equation. In this case you are trying to maximize area, so you need to use the area formula A=LW. Fourth, you need to solve for one of the unknown variables on the constraint equation and then plug it in on the optimization equation. We choose to solve for L in this case and we plug it in on the optimization equation. Fifth, solve for the derivative of our new optimization equation. Sixth, solve for the roots of the derivative of our optimization equation. We found out that the root is located at W=80. Seventh, use the first derivative test and find out whether it is a maximum or a minimum. Lastly, answer the question in a complete sentence.

I'm not going to go over the other question since it is just almost the same problem but instead I'm going to give you an outline how to solve an optimization problem, so after digging up the apcalc blog here it is:

Step 1: Find what you are trying to maximize or minimize. This will be stated excplicitly (in the question).

Step 2: Write an equation for it. Use V = for volume and M = for material, A for area, etc. I must insist on using descriptive variables, because in optimization, if you are sloppy, you lose track of what's going on.

Step 3: Try to get the equation into a two variable form, so you can take the derivative.
- Step 3a: To do step 1, you will often have to create a second equation from additional information given in the problem. This may require ingenuity, but it should become natural.
- Step 3b: ISolate one of the two variables in the equation drawn in step 3a.
- Step 3c: In the original equation that you are trying to minimize or maximize, replace the variable you isolated in step 3b.

Step 4: Take the derivative of your two-variable equation.

Step 5: Set the derivative to 0, and solve for the value of the remaining variable.

Step 6: Plus the value of that variable into the first two equations to find all dimensions, including the final goal, such as the amount of material, cost, or volume.

NOTE: This is just an outline so follow it loosely. There is really no chronological way to answer this problems.

Thats it , the next scribe is haiyan.

The late Thursday scribe.

Yes me again. There actually isn't a lot of information to include on this scribe.

Due to the fact, Mr. K thought it was a lot of information to absorb for the past few classes.

So we....

basically reviewed Tuesday's class.

We got into groups, did the second slide question.

And critical numbers of f, are when f' and f'' are equal to zero.

Next scribe... Kristen.

Sorry it's so short...

The Last Tuesday Scribe

Sorry it's so late. Kept forgetting and procrastinating. Yes, it was supposed to be me... but, I wasn't even participating during the class, because of APcalc in the corner of the room. But, I can at least explain the slides.

Slide 1, 2

Picture and the question.

Slide 3,4,5

Velocity is the change of distance over the change in time.

a. The average velocity, is the slope of the line, from the interval 0 to 4 of the given position function.

b. Velocity is zero when, the derivative of x(t) is equal to zero.

c. When the point is moving to the right (positive) meaning, the velocity is positive. (Derivative of x is positive)

d. Same thing as c, but moving left (negative)

e. When the derivative is equal to 3

f. Taking the derivative of the velocity function. Or the second derivative of the position function, called the acceleration function is equal to 3.

q. (Not sure why q, but okay) Draw what x(t) looks like. Red graph is the derivative of x(t)

Slide 6,7

The graph, and line test means, you are trying to find from the derivative, where there may be local minimums or maximums on the parent function.

So, when the derivative has values that crosses the x axis(zero) there is a local maximum or minimum. To determine whether it is a maximum or a minimum, on the left side of the zero, if it is a positive value then crosses the x-axis and then is negative, it is a local maximum. Vice versa, it is a local minimum.

Slide 8

The extreme value theorem, says, on a differentiated function, within a closed interval, there exists a maximum and minimum value on the parent function.

Slide 9 and 10 practice what we just learned.

That's it for that scribe. I'm onto Thursday's scribe now. Ciao.

Application of Derivatives

On Friday, our class started talking about the application of derivatives. To start off, Mr. K showed us this slide:


Mr. K then asked us when was the car moving fastest. The class decided that the car was moving fastest during t2, because the slope of the tangent line was the steepest at that point. During the interval [t1,t2], it is clear that the car is speeding up during that interval because the slope of the tangent lines are getting steeper going from t1 to t2. The car is definitely slowing down on the interval [t2,t3] as we can see that the slope of the tangent lines are decreasing.



Now this was a tough graph to decipher. The AP Calculus students had no problem understanding this one, but to those of us who aren't taking AP, this was kind of difficult. I myself found it quite confusing at first, but after clarification, I understand it. Now, you have to find the derivative of f in this graph to find out where is f ' is positive. First we have to find out where the tangent lines in f are zero. Looking at the function f, we can clearly see that the tangent lines are zero in -2 and +2. This means that f ' will have roots at -2 and +2. Then, we look when the tangent lines of f are approaching zero. There are 2 instances in this graph. The first instance is from (-∞, -2]. The slope from that domain is increasing, since it's coming from a negative value to zero. But that still doesn't solve our problem since f ' is still negative at that point. Then, from (-2, 2) in f, we can see that the slope is increasing until (0,0) and it starts to decrease until it reaches (2,3), where it becomes zero. At this point, f ' is positive as the slopes of the tangent lines from (-2, -1) to (0,0) are increasing from zero and reaches its maximum at (0,0), and then it starts to decrease from (0,0) until it reaches (2,3) where the slope is zero. This then solves the first question which asks where is f ' positive, which is from (-2, o) to (2,0).

Now to find out where f '' is positive. From the graph, we can see that f is some kind of a cubic function, so we know that its derivative will be a quadratic function. Now, the derivative of a quadratic function is a straight line, so now we know that f '' is a straight line. Looking at the shape of f ' (say it is -2x²), it is definitely a negative quadratic function because it has a maximum, not a minimum. We then have to find out the derivative of f '' which is represented as -2x² (However, it IS NOT -2x², it's just a representation). Using the power rule, we can determine that the derivative of f ', which is f '', is -4x. From this we can see that the function is positive at quadrant 2.

The critical numbers (which is another term for the roots) of f are -2 and 2.


This is a little difficult...
For a), f is increasing wherever f ' is positive.
For b), f(0) is negative.

Well, that was all we talked about last class. Mr. K tried to squeeze in one more slide but we ran out of time.

The next scribe is Kristin.

Pre-test'(x)

Hey, MrSiwWy here for Monday's scribe! Now, all that we really did during class was write the pre-test, though Graeme, Mark and I all decided to rush through it and discuss it in our own group so that we could leave early to practice for the Literacy presentation in the afternoon. I know that it was my fault that we lost a few silly marks because of the fact that we rushed through everything on Monday, but since I really haphazardly rushed and finished way too early some of my solutions were flawed (and we ended up handing in my paper). Well, that's in the past, so on to the scribe. Now, usually a scribe for a pre-test class wouldn't really be that detailed, but I feel obligated to try and explain the solution process for each of the questions on the pre-test as well as I can. I'll accompany each with the corresponding slide, to show Mr. K's work (if that question was brought up in class) and to show the question itself.

1. What is dc/dx, where c is a constant?


Well, this is simply testing to see if you remember the constant rule for differentiation. Thus the answer is simply (b) which is 0. It might help if I reviewed with you the basic idea of each of the rules we have learned for differentiation, since the test is just about 12 hours away from now.
Constant Rule: The constant rule basically dictates that any function that is solely just a constant, such as f(x) = 4, or g(x) = 13, has a derivative that is 0, or f'(x) = 0 and g'(x) = 0.
Coefficient Rule: When differentiating a first-order function with a coefficient, such as f(x) = 2x, or g(x) = 215x, the derivative will simply be that coefficient. In the examples given, f(x) = 2x yields f'(x) = 2, while g(x) = 215x yields g'(x) = 215.
Sum/Difference Rule: When a function is composed of more than one term, then you can break that function up into it's constituent parts and differentiate each part separately. Examples include f(x) = 388x + 6, or g(x) = 56x² - 21x, where f'(x) = 388 + 0 = 388, or g'(x) = 102x - 21.
Power Rule: For any function where x is raised to an exponent n, or f(x) = xⁿ, the derivative of that function will be x raised to the exponent (n-1) all multiplied by n, or f'(x) = nx^n-1.
Product Rule: For any function that is composed of two functions being multiplied together, the derivative is not the derivative of their product, nor is it the product of their respective derivatives. To find the derivative of a product, the derivative will always be given by h'(x) = f(x)g'(x) + g(x)f'(x), assuming h(x) = f(x)g(x). It's basically the derivative of the first function multiplied by the other function, then add on the derivative of the second function multiplied by the first function.
Quotient Rule: To find the derivative of a quotient, all you have to do is recall the quotient rule song! Assuming that h(x) = f(x) / g(x) then h'(x) = [g'(x)f(x) - f'(x)g(x)]/g²(x). Here's the song in case you forgot: "High de low minus low dehigh, all over low low"
Chain Rule: This must be used whenever a function is actually buried deep inside another function. This can be given by h(x) = f(g(x)), such as h(x) = (2x² + 3x)³. First you must find what g(x) is, or what the inner function is, then it should be easy to determine what is the outer function. The derivative of any composition of functions (when a function has another function inside of it) is given by h'(x) = g'(x)f'(g(x)). In the given example, the derivative would be h'(x) = (4x + 3)[3(2x² + 3x)²], or (12x + 9)(2x² + 3x)² since g(x) = 2x² + 3x and f(x) = x³.

2. What is dxⁿ/dx?


Again, this is just testing the knowledge of a rudimentary rule given above.

3. Given f(x) = √(1-x²) find and simplify f'(x).


Well, this is basically just applying the rules of differentiating I reviewed above in a very straight forward manner. You must be able to recognize that you must use the chain rule with the inner function g(x) = 1 - x², while the outer function f(x) = √x. Then the derivative becomes f'(x) = -2x / 2 √(1-x²), which simplifies to f'(x) = -x / √(1-x²).

4. Find the equation of the line tangent to the graph of 8xy² = (x + y)⁴ at the point (1/2, 1/2).


This was a rather tricky question, but I remember first determining how I forgot to take the derivative of x + y as 1 + dy/dx, instead of just dy/dx, and I also forgot to distribute the 8 through the left side and didn't read the question and gave him only a slope for an answer. I really need to take my time answering these questions =/. Well anyways...

The solution is shown above, but the main idea behind solving this question is just realizing that you must differentiate the equation implicitly, since y might represent any one of a variety of possible functions. If you can see that you must differentiate implicitly, it's pretty just grunt work applying the above rules and keeping in mind the fact that any time you differentiate the variable y, you must use the chain rule since it has some unknown buried inside of it, meaning that that term must be multiplied by y'. In the work you can see this being done. Say a function x + y = 4 was to be differentiated, then afterwards it would transform into 1 + 1y' = 0, or y' = -1.

Don't forget, as I did, that the question is asking for the equation to the tangent line at (1/2, 1/2), not just the slope of that tangent line. So, once you can solve for y' as shown above through implicit differentiation, you can use the point given (1/2, 1/2) to find the equation of the tangent line, thus completing the question.

5. (a) If f(x) = √(x² + 9), use the limit definition of the derivative to find f'(0). You must show all work and use the limit definition properly to receive any credit.


The work for this question was once again fully shown by Mr. K, but the key to this question is realizing that it's asking for the derivative at a point, and not the derivative function of f(x). Using this knowledge, we can apply the limit definition at f(0) instead of f(x), and f(0+h) instead of at f(x+h). What Mr. K did with his work was that he determined what f(0) was by simply plugging 0 into x inside f(x), and then determined what f(0+h) would be by plugging in (0 + h) into f(x). He did this so that he could use the values he arrived at in these initial steps, and simply apply them to the limit definition as shown near the bottom of the slide.

5 (b) If g(x) = x2 - 3x + 4, use the definition of the derivative to find an equation or formula for the derivative of x. Again no credit will be awarded unless you demonstrate competent use of the limit definition and show all your work.


Once again you can see the solution to this question in the slide above. What you had to realize to solve this question is that you have to know what g(x+h) will be when expanded, as Mr. K has shown at the top portion of the slide. Once this is known, you must plug it into the fundamental limit definition of the derivative and things should start to cancel quickly and elegantly. In the final steps of the solution, once you plug in 0 into h, then the lim part goes away and any term with an h in it leaves along with it. Thus leaving 2x - 3, which gives the derivative of g(x), or rather gives g'(x).

Well that's it for my scribe post, I think that this might be one of my shortest scribe posts since way back in grade 12 pre-cal ^^. Too bad I wasn't present for the entire class, but I guess I tried to cover everything to the best of my ability since I was only present for maybe 20 minutes. Don't forgot to study tonight folks, since the test is tomorrow. I wish everyone the best of luck!
Oh yes, almost forgot, the next scribe will be: John D.

Late Scribe x2 and Bob

Well I dropped the ball here, haven't really been paying too much attention to this particular blog of late. So by Murphy's law I was picked as scribe. Alright this isn't my best work I can tell you now, but this is how those two classes I have to scribe for go:

Day 1:

The point of this class was to come up with the equations of tangent lines and normal lines.
-Tangent lines are lines touch that touch the graph only once on a certain interval (the graph may wiggle and the tangent line can cross the graph at another point that does not matter)
-Normal lines are lines that are perpendicular to a tangent line.
-Recall that the first derivative of any funtion gives the slope of the graph at any point
-To find the equation of a tangent line at a point on a graph you need a couple things
-You need a point and you need slope
-You may use slope-intercept formula if you are given the y intercept y=mx+b
-But you'll probably be using point-slope formula (y-y1) = m(x - x1)
-So the general method to finding the formula of a tangent line to a point goes something like this:
-Find the derivative of your function
-Evaluate at your point
-Plug it into point-slope formula
and presto! There's your answer!

To find the equation of a normal line (a line that is perpendicular to a tangent line) there is only one more step. When you find the slope you just need to take the negative reciprocal of that and then plug it into our formula.

The last question we had that class asks us to find when the tangent line is horizontal. Well that is when the derivative equals zero. So you come up with the formula of the derivative then solve for the zeroes. Then you plug those x-co's back into the original function to find the points where the tangent line equals zero.

Day 2:

This class is a little more complicated.
-Mr.K started off with a talk about questions that the answers to were that just because we don't know for sure doesn't mean that something is or isn't there or happening.
-The shadow of the balcony, we would infer that there is a balcony casting that shadow, doesn't have to be, could be something that looks like a balcony.
-Then is that roof insulated? Well there is melted snow in places so we could assume no.
-Then do those people agree with one another? Through the story of Mr.K since their body language is all somewhat the same they probably are, but we don't know for sure.

-We then found the derivative of a semicircle.
-Mr.K showed us that we could define this function in other ways to get an infinite amount of other circle bits-and-pieces functions (see the slides)
-This was to show us that there could be many functions buried within another and that even though we don't know which one there may be we have to treat it as if there is another function within it and therefor when we differentiate we have to use the chain rule.
-One thing to notice about this is that the derivative might be in terms of y and x so you may need a set of coordinates to solve for the slope at a certain point instead of just an x-co.


Bob

I feel fairly confident for the upcoming test. No worries at all really. Schedule should die down a little so I wont' miss any more scribes or bobs!

G'night! (might edit this sometime to add some examples but I am *really* tired right now so sleep is a must... after bio...)

Scribe posts for: March 19 & 28

Hey everyone! I apologize for the super super late scribe post. I know you are all waiting for it so here it is.

March 19.
The class started off with a discussion about Article 13 and the students' comments on their class blog. Moving on from that, we reviewed the sum/difference rule and then moved on to the new topic. The class topic was mostly about the product rule. All the information about the class work and the rule and how to apply it is in the slides below. Also we learned about the quotient rule but we ran out of time so it was continued in the other class.

March 28.
This was our last class before the spring break. During this class we reviewed the quotient rule and Mr K. taught the class the quotient rule song. The song goes like this: "high de low minus low de high all over low low" which translate to: [f(x)^1 * g(x) - g(x)^1 * f(x)]/g(x)^2.
But this class' real topic was... an introduction to the chain rule! The chain rule basically states that the whole function is called F(x). This F(x) is composed but different functions which are f(x) and g(x) which g(x) is the inner function and f(x) is the outer function. If examples are need it is in the slides below.

Thank you for your time and I do apologize for the super super late blog post. Now that I finished this scribe post the next scribe shall be... GREY-M.

March 17, 2008

Today's class went similar to the class last Friday. Mr. K assigned people into different groups of three or four students, each group having at least one AP Calculus student. He wants the non-AP students to learn from their peers and the AP students to apply what they've learned before. The first proof that was done was the derivative of the power function, but before that Mr. K refreshed our memory about the binomial theorem. Next we worked on the proof of the derivative of a constant times a function. Then, we applied what we know about limits to prove the derivative of sum and difference of functions. Thats about everything that happened in class today. The next scribe will be rusz.l.

Scribe #1

Sorry for the late scribe everyone, but this week has been very hectic for me. Well Tuesday's class began very differently than usual. The class began with Mr. K splitting everyone into groups of two or three students, with each group having at least one AP calculus student. Mr. K wanted the AP calc students to be the teachers and the intro calc students to be attentive and learn what the AP calc students were teaching them. The first proof that was to be solved was (F(x+h) - F(x)) /h where (mx+b) was plugged in for (x). The derivative of mx+b is m, as shown in the slides from Class below this blog post. The second proof was plugging in a constant (K) in for x. Where the derivative of the constant is always zero (0). The derivative of K is zero (0). That was the class for the shown proofs scroll down below this post to see the proof slides. The next scribe is M@rk.

Limits: The Scribe.. Continue or Discontinue?

Hello! On our blog, I am known as Tim-math-y, and I will be your scribe for today's lessons.

Introduction:

We started off the class with a brief discussion on our del.icio.us accounts and homework. We are to find, with effort, atleast one site that we can learn from and that can be leveled as a quality find. Then we 'tag' it with: cal45sw08, so that it will be added to our blog's bucket. Remember that it may not only aid in developing our learning outside the classroom but also, may prove to be great resources for others reading our blog.

Sweeping that discussion aside, we started off our pre-test on the unit of limits! The pre-test consisted of 5 questions in total. For those who do not know the procedures of a pre-test, it is an effective practice worth marks where a short test is written. After a set test-writing duration, we are placed into even groups where we share our answers to compile the best solutions onto one paper, as a team. This individual test paper is handed in before Mr. K reveals and explains the correct solutions.

The Pre-test:

As mentioned earlier, this Pre-test consisted of 5 questions: 2 multiple choice questions, 2 short answer questions (where work was required to show), and 1 long answer question.


The first question included an error that stumped everyone. The x^4 in the numerator was supposed to be x^2. Because of this unintentional error, this question was ommitted, as far as marks go.

However, this question could still be solved by exploring the function. This is shown on the slide. First, we notice that there is a vertical asymptote at x = 4 (Remember that when a question is asking for a limit, it is essentially looking for a horizontal asymptote).

By creating a number line, one will find that as 'x' approaches 4 from the negative side, the function goes to positive infinity. One would also find that as 'x' approaches 4 from the positive side, the function goes to negative infinity. Because of this occurrence, the limit, as 'x' approaches 4 from the positive and negative side DOES NOT EXIST.

This question is simply a give-away, as many may describe it. As 'x' approaches the value of 1 from the positive and negative side, the value is 1.

A number of groups faltered on this question simply because of the nature of previous 'short answer' questions. In the past, short answer questions were marked based on the final answer only, with a chance to earn partial marks for work shown. However, this question stated: Evaluate using the Limit Theorems, upon which many did not. This question is extremely simple yet painful. As long as you know your limit theorems, you should be fine. Listed are the limit theorems from [visual calculus]:


These are the main limit theorems we are required to know.

To start off this question, we chose to solve for the horizontal asymptote first. By dividing each term by the variable with the highest degree, we found that the horizontal asymptote y=0, when the value of 'x' approaches infinity (any number divided by infinity is extremely close to zero, therefore in this method, terms are reduced substantially).

Next we solved for the vertical asymptotes. This is done by factoring the denominator and solving for restrictions (the denominator can not equal to zero). We found the vertical asymptotes to be @ x=-9, 0.

Finally, to help visualize the graph and sketch it, a simple method of finding out the positions around the asymptotes is by creating a number line:

• As 'x' approaches -9 from the negative side, the limit is +infinity
  
• As 'x' approaches -9 from the positive side, the limit is -infinity
• As 'x' approaches 0 from the negative side, the limit is -infinity
• As 'x' approaches 0 from the positive side, the limit is +infinity

Thus, the graph can be sketched.

In the final question, we started off by running the piece-wise function through the three steps of continuity testing.

• Does f(a) exist?
• Does the limit as 'x' approaches 'a' exist?
• Does f(a) = L?

If not, the function is discontinuous.

• f(a) = f(2)
  f(2) = 2
• the limit as 'x' approaches 2 is 5
• f(2) does not equal L: 2 does not equal 5

Therefore, this piece-wise function is not continuous. By discovering this, we found that this function is a removable discontuity.

Finally, we had to sketch this piece-wise function. The graph maintains the shape of (x+3). However, it has a hole at x = 2 because there was a reduction in the factors of (x-2). Remember: when there is a reduction in factors, there is a hole at that point rather than an asymptote. Because the function of f(x) has a value of 2 @ x = 2, there is a black dot at that location.

The Conclusion:

Well that was our pre-test! To sum things up, there were multiple things that should be remembered.

• A limit as 'x' approaches a value from both sides must meet at the same point, otherwise, the limit does not exist
• Remember how to solve using the painful work of writing out all of the evaluation steps using the limit theorems
• A number line really helps in determining the shape of the function
• Remember the three steps to testing continuity
• When factors reduce a restriction in the denominator, there is a hole at that value of 'x' rather than a vertical asymptote

I hope this scribe helped any of the readers! There will be a test on wednesday so DON'T DON'T DON'T DON'T DON'T FORGET TO "BOB" !

Good luck everyone on the test! Do not forget to study either! =) Have a great night everyone.

OoOoooOOo! And the scribe for the next class will be: (Give me a sec while I find the scribe list)

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John D. !!!!!

MORE LIMITS

THURSDAY'S CLASS

We started off class with a quiz. It had two graphs and we had to find the limit of this and that. We marked them in class and went over some of them. We mainly focused on the different type of graphs that are discontinuity. The three different types of discontinuity are:


1. Removable Discontinuity:
A hole in a graph. That is, a discontinuity that can be "repaired" by filling in a single point. In other words, a removable discontinuity is a point at which a graph is not connected but can be made connected by filling in a single point.



2. Jump Discontinuity:
Jump discontinuities occur where the graph has a break in it is as this graph does. It can't be fixed or repaired so that the graph is continuous.





3. Infinite Discontinuity:
A discontinuity of a function for which the absolute value of the function can have arbitrarily large values arbitrarily close to the discontinuity. Can't be fixed either.

The formal way of telling what kind of discontinuity it is:


Then we worked on some questions in class.


We also did some work on Mr. K's favorite website for limits ( i think ). We worked together as a class to solve them and they're all on the slides that Mr. K posted up on the 28Th. It also has explanations and the homework assignments. And that was Thursday's class.
The next scribe is KIM POSSIBLE. Ha ha i totally ripped that off from your brother but he's not around.

L-I-M-I-T-S

TUESDAY'S MATH CLASS

Someone forgot to scribe for Tuesday's math class and she's making up for it by scribing for Tuesday and Thursday. Sorry. Well, on Tuesday we mainly focused on limits in a symbolic approach. The first question was....

1. Factor both the numerator and denominator. If you don't factor the numerator and denominator and go straight to substituting in the value 2 for x, you'll end up with zero in the denominator making the whole thing undefined.
   
2. Reduce
   
3. After reducing the same terms in the numerator and denominator you substitute the value two in everywhere there is an x.
   
4. Voila, you end up with the answer.

1. Rationalize the numerator because if you do then you can reduce 25 - x in the numerator and denominator.
2. After reducing, you're left with 1 over 5 + √x. You can now substitute the value 25 for x because it's in its most reduced form.
   
3. Now simplify, the √25 is 5. What is left is 1 over 10.
   




The whole thing is undefined because in the end your gonna have to substitute zero in for x and 3 over 0 is undefined. Any number over zero is undefined.




1. The numerator can be factored so that something can be reduced from the bottom.
2. After reducing to the simplest form you now can substitute the value nine for where there's an x.
   
3. The end result is negative 6.





That would have been marked wrong if it was on a test or exam. Why?
Well, in the end since it's just the notation and you've solved for it, you don't need the lim thing.


For the rest of the class, we briefly talked about horizontal and vertical asymptotes. Homework was posted in the slides.

Whoa! I can’t remember the last time I blogged for Math! But, yes, again, blogging is a very important instrument to enhance knowledge and develop self responsibility (remember the escalator video..). It is our utmost responsibility to do things ourselves even though we tend to lean on others – which means:

• do our homework
• try to blog as extensive as possible for it determines how well you know the material
• read scribe post everyday
• ask questions if necessary
• always check for misinterpretation by peers so that no one else will be confused and eventually mess up a test or exam in the future.

So, have you read the blog-online rules? Have you watched the video? Have you finished your homework? If you haven’t done so, please do.

Talking about unlimited amount of math, Mr. K introduced a very interesting approach to

UNIT 1: LIMITS

What is a limit? The first time I heard it was when I watched “Mean Girls” back then. I didn’t know what it was. Yes, 5 years after… here it comes, slowly making sense. Yes, AP CALC people, you can laugh since we’re only at the threshold of the house you have already explored three times! Here it goes.

This notation is used to express LIMITS, which means if you do not have this in every line, the entire thing is incorrect!

PROBLEM #1:

What do we first notice about this equation?

• a second degree function (x^2 : parabola) over a first degree (linear)
• a difference of square in the numerator
• if we graph this on our graphing calculator, it’s a straight line, opposing the fact that it should be some kind of a parabola. Hmmm, weird.

What happens if we factor the numerator?

• the (x-1) reduce
• we’re left with f(x) = x + 1
• graphically, it is identical to the graph we had earlier when we graphed the original equation

What if x = 1?

• Most of the students would say, “IT’S 2!” because of the equation: f(1) = 1 + 1 = 2
• However, some might disagree and say, “It’s undefined, buddy!”

The question now is WHY? Well, f(x) = x + 1 isn’t the original equation.Therefore, substituting 1 for all x gives us:

This brought the discussion about the very round number called ZERO. Usually when we divide any number by zero, we say “YOU CAN’T!!!”. It’s very hard to explain. Actually it’s pretty simple. You can’t divide by NOTHING! This follows the same curvature of the ball of wax. Anything over ZERO is undefined… it’s not two… again, it’s undefined!

We also had a discussion about 0/0 is not 1, why is it so different from 2/2 = 1?, when both 0 and 2 are numbers? Isn’t a number divided by itself equal to 1? Why is zero such an exception? Well it could mean NOTHING, INFINITY, or ZERO. It all depends on the hwo you look at it. Interesting… Mr. K, took out his “block of wood” to further discuss how you can look at something at different ways but it still refers to the same thing.

Consider SLOPE. It can be represented in three ways:

• m
• y = rise / run
• y = delta y / delta x

So, why do we have three ways to describe slope? ANSWER: because we have three ways to illustrate a function:

• equation – where ‘m’ is present in the standard form of a line (y = mx +b)
• numerically (table of values) – where we can take two ordered pairs and put them into the equation
• graphically – where we can locate two a point and use y = to find the slope.

In this case, f(1) is in an indeterminate form, which means, when x = 1, it is undefined. To prove this, we can graph the equation one more time. But this time, hit ZOOM 4, which will provide you with a much closer scrutiny at the graph.

Look at the gap on the linear equation. Isn’t it weird? Well that’s exactly what we had earlier. f(1) ix not 2. Because the point missing on the graph. Try tracing any integer greater or less than 1. It will give you the y-value but will not do it so if you enter x = 1. This squeezes out the value of 2 from both sides - >2 and <2.>

Let's take a look at a very similar problem:

As the end of class approaches its LIMIT, Mr. K, very quickly went through the laws of limits with all the mathematical operations (addition, subtraction, multiplication, and division). They are on the slides posted of Friday! It's pretty simple. It's somehow like logarithms but not really! AND AGAIN, MAKE SURE TO INCLUDE THE PROPER NOTATION FOR LIMITS OR YOUR WORK WILL LOSE A VERY FRUSTRATING AMOUNT OF MARKS!

We had a glimpse of the graphs of limits at the end of the period but since I was not sure about how it goes, I chose not to include it here. I hope Mr. K further go into details about that one in class next next class (he won't be here on Wednesday, which is the only class we have this week). This concludes my scribe post and I hope everyone had a great 3-day weekend! If you didn't, don't worry, there IS another one! YES! Anyway the next scribe will be...

K r i s t i n !

The Scribe List

This is The Scribe List. Every possible scribe in our class is listed here. This list will be updated every day. If you see someone's name crossed off on this list then you CANNOT choose them as the scribe for the next class.

This post can be quickly accessed from the [Links] list over there on the right hand sidebar. Check here before you choose a scribe for tomorrow's class when it is your turn to do so.

IMPORTANT: Make sure you label all your Scribe Posts properly or they will not be counted.

Cycle 1

m@rk haiyan 1 SAMUS MrSiwWy

Grey-M John D. «Craig» vincentr Dino

Tim_MATH_y rusz.l Kristin_R (update your posts)

February 13 / 08 - Scribe

I will be taken Vincent's scribe duties as he requested.

Mr. K was not present in class today so we had a sub. This class was a working period to do the review booklets. (Blue, White/Orange)

The new assignments are: Read pages 30-45 (White/Orange book), Do page 5 (blue book).

That concluded the class of February 13!

The next scribe will be: Vincentr!

First Scribe!!!

Oh Boy! Here we go, I'm scribing again.

This is my fourth class with Mr. K., all of which I had to scribe for at least thrice, so I'm used to it.

Anyway, today's class was basically our first official class of learning. However, it ended up being a review of grades 10 through 12. Mr. K. went over the Review Test we had the previous class, touching on the questions we seemed to have the most trouble with.

These questions included the concepts of:
•Inequalities
•Fraction Operations
•Factoring
•Rational Functions
•Function Notation
•Trigonometry

As well, there is a list that may be seen here(slides 14 and 15) that goes over the things we should already know.

I believe Mr. K. did a sufficient job in discussing the problems we may have had in class, as no one seemed to ask questions.
***But remember, if you need any help in dealing with any of the content in the course, arrange a time with Mr.K. or even ask some other students to see if you can get the advice you need***

As well, I suggest if you got some questions wrong on this quiz that you should go back to the questions and try them again. It should help. Or, just open your booklets and do the homework; it provides tons of practice questions dealing with the above concepts and more.

Homework: •All exercises up to page #30 in the ORANGE BOOK
•Pages 2, 3, 4, and 5 in the BLUE BOOK


Last but not least, Wednesday's Scribe will be:......


VINCENT!!! (he does awesome Scribes =D)