Thursday, September 27, 2007

Derivative of a function at a point

Oh yeah, i forgot to mention that the next scribe will be DINO.

Wednesday, September 26, 2007

Slow driver, or maniac driver? That is the question...

Hello everybody, I'm MrSiwWy and I'll be scribing for today's (and yesterday's) class(es). I hope everyone out there is doing great and feeling well. I was going to title this post "the indrocution to the derivative", but I think that would have been a tad too bland for a class such as today's. Well, enough about the unnecessary, without further ado, here is my long awaited (Ha! ya right) scribe post.

Tuesday, September 25, 2007

During this class, we attempted to complete as much of the review as we could. And though much of the class would deem this class to be a "slack" class, I think everyone did their share of their own work which undoubtedly benefitted each of us in terms of perparation for the upcoming test. Now, during this class, in case anyone was absent (though I don't believe so) was essentially a review class consisting of a work period on questions from the textbook (pg.76-82). Though some of the time was consumed by the utterly gravitational math riddles Van put up on the whiteboard, most of us completed a considerable amount of work in class. There weren't really any problems that we encountered, except for the "bump" in question number 7, so the description of this class is actually quite compendious and sufficient as of this point. Thus, it's time to move onto the subsequent class on September 26.

Wednesday, September 26, 2007
Class began as usual, we all casually made way to our habitual abode in the calculus classroom, though some of us could vaguely hint upon an incoming lesson for the day, despite expecting another review period for this class. Many of us were proven wrong when Mr. K zealously presented to us the single, seemingly monumentous slide which our entire class would encircle upon. The slide of course consisted of only a graph, and after Mr. K quickly went over a question concerning one of the homework questions (question 7 on pg. 77)* we were fully enveloped by the day's lesson.
*Mr. K was explaining the question which I also referred to as before as the question with the "bump". Craig asked how one would come about sketching that graph without using a calculator, since the bump on the graph crossed a horiztonal asymptote and this shape would be terribly difficult to determine without a graphing instrument. Mr. K basically said that sketching this graph normally, as we have learned to do so in the past, would be sufficient to approximate specific qualities (such as asymptotes, domain, range, etc.) and thus be completely alright.

Onto this so-called monumentous slide. Though there was quite a lot of information implanted onto the graph after it's reveal, here is how the slide looked at the end of the class (I will of course outline the progression of the lecture and convey the implications and significance of the aforementioned information shortly):Mr. K began projecting to us the meaning of the distance vs. time graph displayed by assigning a story to definitively represent the data. He eventually decided upon Mark and I journeying along the road from Winnipeg to Winnipeg Beach. And he then asked the class to determine the average velocity of the graph, and as observed from the above graph, a total distance of 100 m was covered over a time interval of 2 hours, therefore the average velocity was quickly calculated to be 50 km/hr. Mr. K then elaborating upon how this average was attained, through the calculation of the slope over the interval at hand, or rather the amount of distance traveled during the given amount of time. He specifically explained (as it is a vital concept in the comprehension of a derivative) that these lines (that join the graph at exactly two points, as we shall also see later) are referred to as secant lines. It easy to think of secant lines in terms of tangent lines that meet at two points instead of one, as to determine the slope over an interval instead of at one specific point (through the direct utilization of a tangent line).
He then began asking questions pertaining to the overall volubility of the function, for one instance, he asked us what kind of driver did we conclude Mark to be with respect to the average velocity of the journey. We all essentially replied with a unified "slow driver", though Mr. K began to dissect this comment and attempted to prove otherwise. He then started to ask us for the average velocity of Mark's driving endeavor over specific intervals, such as depicted above by the green triangles. We determined the average velocities to be 40km/hr and 60 km/hr respectively.
Subsequent to the identification of the velocities over specific intervals, Mr. K then asked what happened at approximately 45 minutes into the journey, and why this occurred. The answer to this is of course due to the fact that the car stopped, since no distance was covered over the time interval from 45 minutes to about 75 minutes. We also know that Mark stopped the car because the graph representing out travel's progression began to flatten out, meaning that we didn't move for that amount of time. The class then decided that Mark and I could have stopped at a gas station, or even could have stopped to eat before we continue on our quest to Winnipeg beach.
We began to dissect the graph over specific intervals once again, but this time over smaller intervals at specific portions of the graph. As denoted by the many coloured triangles in the graph, as well as the corresponding colour-coded velocity values, these values change over different intervals on the graph. This caused the class to suggest that I took over the car after we stopped, and I of course was a maniac driver on the road. This was derived from the fact that as we calculated the average velocity for the intervals subsequent to the "pit stop"/plateau found roughly in the middle of the travel, we determined them to become increasingly grand. The velocity of the car rose very significantly as we shortened the intervals of which we analyzed. This then led to the introduction of the key concept of the day's class; the derivative.

The above image is of a zoomed-in portion of the graph (upper-right portion), which was used to then transition how we would calculate the velocity for time intervals in units of seconds (such as 1 second, or 0.5 seconds). As we may know, seconds are extremely short amounts of time, and calculating this velocity is approximately and yet increasingly similar to determing the instantaneous velocity immediate to that interval. This meaning that since we are calculating the ratio of the change in distance to the change in time, as the change in time decreases, the distance values are ever closer, narrowing into some specific value.

We attempted to absorb the true connection of the secant line and it's importance to the concept of the derivative to a certain extent. Mr. K then altered the distance and time variables of the graph help dignify this connection by explaining how we are determining the rate of change within the graph.

This graph illustrates this explanation. The alteration is visible as a function of the volume of a balloon in terms of it's radius as it is being filled. The graph portrays how we would expect to determine the rate of change of the overall filling process over intervals, on average, or as a cumulative generality. This rate of change was given by the triangles we found earlier, which were formed by specific secant lines. As shown below, and through the above explanation of determining increasingly miniscule time intervals of units in seconds, Mr. K once again elaborated on the secant line and revisited an aforementioned comparison.

From the long magenta line, which is a secant, we can see that as we found smaller and smaller intervals (leading to smaller secant lengths) we achieved larger average velocity values. But as we also outlined, examining these intervals using secant lines (determining the slope/rate of change) with an increasingly insignificant amount of change, we can see how the line then becomes closer and closer to acting as if it were a tangent line. This means that by using secant lines with an ascendingly small change in values, is like determining the instantaneous velocity approximate to that point. As this change approaches an infinitely small amount, the instantaneous rate of change becomes more apparent and more accurate. This, in essence, is the concept of the derivative.

To conclude our class, we shortly discussed what day our first calculus test of the year lands on, which is friday in case anyone missed that, and we also covered what questions will be due for homework tomorrow (chapter 1, pages 76-82) and the homework for the weekend on derivatives (exercise 2.1, all odd questions including 6 and 12). I will now conclude this scribe post by conveying the aptly subtle, yet beautiful analogy Mr. K used to contrast the significance and vastness of the concept of the derivative to another profoundly familiar concept. He stated that his son asked him what a decimal was, and Mr. K simply explained to him that a decimal is used to indicate values less than 1 instead of using fractions. He said that decimals have a more universal and prolific usage than just to indicate number less than 1, but to comprehend the subject fully, one must start somewhere and build up their knowledge. This is the same as the derivative, though we can analyze it's uses and it's essential meaning and implications, it's just the birth of our progressively intricate understanding of the derivative.

Enough of that, I think I covered the entire class now, I hope I helped anyone if they didn't understand something in class, or if anyone other than my classmates that require help, have come no further but to seek sufficiently helpful information here at our very own blog. Please, feel obligated to leave any comments, feedback, corrections or questions that you may have. But for now, I bid everyone goodday and have a great night!

The next scribe will be Mark!

Today's Slides: September 26

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Monday, September 24, 2007

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.


Test on Thursday! Be Prepared!

Today's Slides: September 24

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Saturday, September 22, 2007

Scribe's Scratches

Well there's the slideshow, didn't upload like it was on the smart software and I tried in 5 different formats...
If you want some more practice I attempted to make an applet in JAVA. I must say it's one of the ugliest computer programs I've ever made and it was a heck of a time making it work online (even though it is still sketchy even now seems to work all the time on ei6 and from time to time completely kill firefox). For any computer programmer people out there who look at my code I know it could have been done better with arrays but I remembered that JAVA arrays were kinda funky compared to VB so I didn't bother re-reading that part of my book and just avoided it. Here is the program (hopefully) the code is under it for anyone else who wants to compile and use it (but I doubt it). I'll try to make a smoother, better looking, more complete(more angles) and more stable program some other time. I didn't this time because there are just too many little things that went wrong. Spent an hour debugging and the bug was a single equals sign. ugh.

import java.applet.Applet;
import java.awt.Button;
import java.awt.Color;
import java.awt.Graphics;
import java.awt.Label;
import java.awt.MediaTracker;
import java.awt.TextField;
import java.awt.event.ActionEvent;
import java.awt.event.ActionListener;

public class TrigQGenerator extends Applet{

Boolean firstQuestionAsked = false;

Color bgColor;

URL path;

//for future versions w/ images
MediaTracker mTracker;

Button submit;
Button squareRootButton;
Button newQuestion;

TextField answerArea;

Label questionDisplay;
Label answerDisplay;

Label rightLabel;
Label wrongLabel;

int right, wrong;
int rand;

String question = "Questions will appear here";
String answer;
String textToDisplay;

final String answer1 = "1/2";
final String answer2 = "sqr(2)/2";
final String answer3 = "sqr(3)/2";
final String answer4 = "1/sqr(3)";
final String answer5 = "1";
final String answer6 = "sqr(3)";
final String squareRoot = "sqr(";

final String question1 = "What is the sine of pi/6?";
final String question2 = "What is the sine of pi/4?";
final String question3 = "What is the sine of pi/3?";

final String question6 = "What is the cosine of pi/6?";
final String question5 = "What is the cosine of pi/4?";
final String question4 = "What is the cosine of pi/3?";

final String question7 = "What is the tangeant of pi/6?";
final String question8 = "What is the tangeant of pi/4?";
final String question9 = "What is the tangeant of pi/3?";

String[] answers;
String[] questions;

public void init(){
mTracker = new MediaTracker(this);


submit = new Button("Submit");
submit.addActionListener(new ButtonSubmitListener());

squareRootButton = new Button("Square Root");
squareRootButton.setBounds(120, 100, 100, 40);
squareRootButton.addActionListener(new SquareRootButtonListener());

newQuestion = new Button("New Question");
newQuestion.addActionListener(new ButtonNewQuestionListener());

rightLabel= new Label("Right: 0");

wrongLabel = new Label("Wrong: 0");


answerArea = new TextField("Answer Here");
answerArea.setBounds(120, 60, 100, 40);

questionDisplay = new Label("Questions will appear here");

try {
//gets the path the applet is in
path = getDocumentBase();
}catch (Exception e) {

bgColor =;

//this TC block waits for everything to be loaded
try {
}catch (InterruptedException e) {
public void stop(){

public void paint(Graphics g){

public void generateQuestion(){
String currentQuestion = questionDisplay.getText();
while (currentQuestion.equals(question)){
rand = (int)(Math.random()*9);

if (rand == 1) {
question = question1;
answer = answer1;
if (rand == 2) {
question = question2;
answer = answer2;
if (rand == 3) {
question = question3;
answer = answer3;
if (rand == 4) {
question = question4;
answer = answer1;
if (rand == 5) {
question = question5;
answer = answer2;
if (rand == 6) {
question = question6;
answer = answer3;
if (rand == 7) {
question = question7;
answer = answer4;
if (rand == 8) {
question = question8;
answer = answer5;
if (rand == 9) {
question = question9;
answer = answer6;

class SquareRootButtonListener implements ActionListener {

public void actionPerformed(ActionEvent event){
String currentText = answerArea.getText();
currentText = currentText + squareRoot;
class ButtonSubmitListener implements ActionListener {
public void actionPerformed(ActionEvent event) {
if (questionDisplay.getText() == "Questions will appear here") {
} else {
if (answer.equals(answerArea.getText())){
} else {
class ButtonNewQuestionListener implements ActionListener{
public void actionPerformed(ActionEvent event){
firstQuestionAsked = true;
public void displayQuestion(){

public void rightAnswer(){

textToDisplay = Integer.toString(right);
textToDisplay = "Right: " + textToDisplay;
rightLabel.setText( textToDisplay);
public void wrongAnswer(){
textToDisplay = Integer.toString(wrong);
textToDisplay = "Wrong: " + textToDisplay;


Friday, September 21, 2007

Today's Slides: September 21

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Thursday, September 20, 2007

"A Scribe of Excellence"

WOW!!! Very sorry for the lateness of this scribe. I have just returned from "An Evening of Excellence" (for those who don't know: our school's awards ceremony) at which, I must add, our Calculus class did quite exceptionally. =D Now here I go with my Scribelicious post:

As we all took our seats in class today, we were presented with a picture of waves, which could only mean one thing: TRIGONOMETRY. It was in fact a review of the stuff we had learned in our previous years.
The first thing that we were told to do is take a look at the page on the Weather Network's web page, choose a statistic and find a periodic (sinusoidal) function that would fit the data. Of course we used our calculators, so this wasn't too difficult. We found that the two best fits for a sinusoidal function were MAXIMUM TEMPERATURE and SUNSHINE.
Next, we entered the real review part of the lesson; we were given a graph of a sinusoidal function and were told to write two equations for it. (see questions and answers here.)
Now there are two formulae that may be used to write an equation:
f(x)=Asin[B(x-C)]+D and f(x)=Asin(Bx-BC)+D
The first is the easier one to derive from a graph and it is the way it was originally taught to most of us. The other is there because it is possible that you might see functions written either way on an exam or test.
The parameters of these formulae are as follows (just a refresher):
A = AMPLITUDE(vertical stretch)
B = PERIOD STRETCH(determines the Period via 2π/B)
C = PERIOD SHIFT(horizontal shift)
D = SINUSOIDAL AXIS(vertical shift)
***REMEMBER: when using your calculator to find an equation(STAT, SinReg), it will appear as f(x)=Asin(Bx+C)+D. The B has not yet been factored out, and C is being added, not subtracted. Therefore, to find the value of C, divide by -B!!!***
After that quick review, we received a question that gave us an equation of a function and told us to graph them...
So, when graphing these functions, take a look at the given functions and compare them to the ones above. The next step is to figure out the value of each parameter make sure B is factored outside the brackets!!! Then just follow these easy steps:
1. Using the AMPLITUDE and SINUSOIDAL AXIS, determine the max. and min. of the graph. Then scale y-axis accordingly. Draw a dotted line for the Sinusoidal Axis.
2. Using p=2π/B, determine the period of the graph and scale the x-axis from -P to P unless otherwise instructed. Scale the axis with a mark on each quarter of the period.
3. Draw dots (outline) using the PERIOD SHIFT to determine the starting x-coordinate on the Sinusoidal Axis. The starting position depends on which trigonometric function is being used and the sign of the AMPLITUDE.
4. Finally draw a smooth, fluid curve that connects each point. NO SHARP CORNERS OR EDGES!!!.
Next Mr. K. flipped quickly to a review on the Trigonometric Identities, but did not have much time to go over it. Maybe we'll go over it in class, but just in case, take a quick look at it on SLIDE 6. He also announced that we will be having a quick MENTAL MATH quiz on the values of the Unit Circle... make sure to look over that before you go to bed tonight.

The HOMEWORK for tonight is: Exercise 1.10 Questions #1,2,3,5,6,13.


Now playing: Linkin Park - Shadow Of The Day
via FoxyTunes

Today's Slides: September 20

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Wednesday, September 19, 2007

Scribe Post

that by blogging you Hello once again, this is Dino and I was today's scribe. well i have to say that today's class was not as extravagant as the other class that we have been having throughout the past weeks. Mr. K started of the class with a podcast recording that referenced Mr.k and all the class blogger's receive a world wide audience, which is so very true especially after hearing that podcast. Mr. K then began explaining to the class that he was able to get each member of the class smart board programs which he was going to give us through the means of the blog, which to me sounded pretty cool. Then it was time to get into the inverses workshop he had planned for us today. He began with this problem:

Problem: Given the function h(x) = (3x)/(x+5)

Explain how you know that h has an inverse? The function passes the horizontal line test.

Find a formula for h^-1(x). h(
h^-1(x)) = x To get the inverse put the inverse of h into the (x) value of the function h(x). So intern the function h, (inverse of h) = x.

h(h^-1(x)) = 3(h^-1(x)) / (h^-1(x))+5

x =
3(h^-1(x)) / (h^-1(x))+5

x (
(h^-1(x))+5) = 3(h^-1(x))

(h^-1(x))x + 5x = 3(h^-1(x))

(h^-1(x))x - 3(h^-1(x)) = -5x

(h^-1(x))[ x - 3] = -5x

(h^-1(x)) = -5x / (x - 3) This is the inverse of the function h.

Suppose f & g are inverses of each other, what is true about their composition?

f(g(x)) = x
g(f(x)) = x Both functions are equal to x.

That was it for class today, although remember to do questions 1,3,5,7,11,13 in section 1.9 for homework. Tomorrow's scribe is going to be......Craig.

Today's Slides: September 19

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Tuesday, September 18, 2007

Homework Question

As time ran short, many of us were unable to copy down the final question given to us at the end of the class, let alone actually complete it. The question is as follows (though only paraphrased):

Show in as many ways as you can, whether the following function f is even, odd, or neither.
f(x) = x3 - 2x

Good luck and have a great night everyone! See you all in class tomorrow!


Hello everybody! I'm MrSiwWy, and today, as the scribe, I will convey the events and educational endeavours that took place during today's class. Nothing more to say but that I shall continue on with this scribe!

The class began with Mr. K commenting on why we all should have posted our answers for the previous class' assignment so that the rest of the class can comment on each groups answers by tomorrow. Then we proceeded to our lesson plan for the day.

Operations on Functions
We began by conversing upon the utilization of numbers in terms of our progression of mathematical knowledge. Not only can be implement operations on these numbers, but on the functions used to represent and manipulate numbers as well. Functions and numbers can cumulatively be manipulated for any specific use, as we have learned through the wonders of mathematics.

Mr. K continued on with the slides on the smart board by displaying three pairs of functions to us. We were then given time to implement the specific operations within each pair, and were also told to specify the domain for the output function. These function pairs were given as:

f(x) = x2 , g(x) = x + 5.
f(x) = √(1+x) , g(x) = √(1-x).
f(x) = √x , g(x) = ∛x.

The operations we were asked to perform on the above pairs of functions are as follows:

f + g (which can also be written as (f + g)(x)) .
f - g (which can also be written as (f - g)(x)) .
f g (which can also be written as (f*g)(x)) .
f/g (which can also be written as (f/g)(x)) .

Once the class was given sufficient time (of course deemed sufficient by Mr. K) to complete the above, we all went over the answers to each with Mr. K as a class. The answers as well as the actual slides can be seen here (have to wait for Mr. K to post slides). As we analyzed the answers to the given problems, Mr. K had a unique and very intriguing utility to show us on the net. This site of course being

The site offers a utility that is not only accessible (via an example of how it works and how to use it) on the website, but the tool is available for download to your very own cell phone. The tool transforms your cell phone from the once inept, feeble calculating device into a mathematical modeling and graphing calculator extraordinaire. Mr. K then briefly demonstrated how the graph would portray an example function. He also vaguely noted that in the example function we used (x2 +x + 5), altering the magnitude of the coefficient for the term containing x1 would cause an "oblique translation", or rather a diagonal shift. Though, we did not establish the significance of this concept, and we did not go further into detail concerning the matter so it might not be important as of yet.

After the short math4mobile presentation, Mr. K then transitioned the class to continue with our lesson by presenting us a question. The question proposed that we determine the turning point of the function f within quadrant II for:

f(x) = x3 - x2 - 3x + 7.

To determine this, it's simply achievable by graphing the given function on a calculator and then finding the local max for the function, seeing as if you examine the graph, the point at which the function will turn (change from increasing to decreasing, or vice versa) found in quadrant two is a maximum point. The point was determined to be (-0.72, 8.27).

As Mr. K was dissecting the given graph up on the smart board, he then digressed and began explaining that specific functions can represent other functions on specific intervals. If you imagine viewing the given function, or any cubic of a similar shape, from one turning point to another, you might be able to visualize many functions that behave similarly on the same domain, such as sin(x). Polynomials are often used to approximate many functions on specific domains, seeing as polynomial functions are often more simplistic and are relatively less complicated to conduct calculations on. After we absorbed the useful tidbit conveyed to us, we continued our lesson with yet another question.

This question, required us to analyze the function and determine not only the domain, but the range and the left and right-hand asymptotes of the following function:

f(x) = (x2 - x)/(x2 - 1).

We were given time to perform such duties, and we determined the following:

domain: {x l xeR;x≠±1} or (-∞,-1)U(1,∞).
range: {y l yeR;y≠1} or (-∞,1)U(1,∞).
left-hand asymptote: 1.
right-hand asymptote: 1.

As Mr. K explained why each answer was true, we somewhat diverged to a quite elaborate exploration of horizontal asymptotes. We in turn learned once again how to determine the horizontal asymptotes for a function. To determine such, we must analyze the degree's of both the numerator and the denominator. Here are some rules which Mr. K projected to us to signify the relevance of finding these degree's:

- If the degree of the numerator and the denominator are equal, then to determine the horizontal asymptote, you must simply divide the leading coefficients within the function. This meaning that if the degree's are equal, divide the leading coefficient of the numerator by the leading coefficient of the denominator to determine the horizontal asymptote of the function.
- If the degree of the numerator is greater than that of the denominator, then there is no horizontal asymptote for the function.
- If the degree of the numerator is less than that of the denominator, then the asymptote can be found at y=0. The due to the fact that is that when a horizontal asymptote occurs, we can then use the fact that the x value for that asymptote will be approaching infinity. If we then substitute infinity into the given equation, the resultant will yield a final value of 0.

And as dismissal became imminent, we briefly went over what classifies functions as either even or odd. Here are some quick reminders concerning both.
For an even function, the function must be symmetrical about the y-axis. This meaning that the function lying left of the y-axis, can be flipped and will give exactly what is present to the right of the y-axis.
Algebraically, a function can be determined to be even if:
f(x) = f(-x)

ex. f(x) = x2 + 2
*To check if the function is even, then substitute -x into each x in the equation. If the new equation is equivalent to the original, the function is said to be even.
f(-x) = (-x)2 + 2
f(-x) = x2 + 2

For an odd function, the function must be symmetrical about the origin. This meaning that the function can be flipped over the x-axis then the y-axis, or rotated 180o and the graph will stay the exact same.
Algebraically, a function can be determined to be odd if:
-f(x) = f(-x)

ex. f(x) = x3
*To check if the function is odd, then substitute the -x into each x in the equation to get f(-x). Then, multiply the original function by -1 to determine -f(x). If both functions are equal to each other, then the function is odd.
f(-x) = (-x)3
f(-x) = -x3

-f(x) = -(x3)
-f(x) = -x3
Since -f(x) = f(-x), the function can be deemed to be odd.

Now, that was all that we did today in class! Don't forget to complete your homework (a question that I will be posting subsequent to this very post) as well as the exercise 1.8 questions 1, 3, 5, 6, 11, 19 and 21. Have a great night everyone! Hope this post helped anyone who missed class (though I don't think anyone did). Oh yes, and the next scribe will be... surprise surprise...

Exponential Functions Lab Solutions

Question 1

  • Range is infinite with the exception of horizontal lines
  • Domain is infinite
  • both have y-intercepts
  • can both be transformed
  • both can be used to model real life situations
  • linear function has common difference ,exponential has common ratio
  • linear function has constant slope
  • we need the use of a tangent line when finding the slope of a exponential graph
Question 2

The y-intercept is at 3 because of you let x=0 you will get a answer of 3. The growth factor is 4.

Question 3

The value of a is 6,because anything with the base of 1 is just itself.

Question 4

It means that a is less than 1 because the graph is decreasing.

Question 5


The 2 signifies the y-intercept and the 5 is the growth factor.

Question 6

It is equivalent to 16 fold because in one growth period it doubles, so growth periods it would double 4 times or in other words 2^4 which is equal to 16.

Group: Graeme, Mark, Ethan (I'm apologizing if i misspelled someone's name)

Exponential Functions Lab

I'll do this by writing out the lab questions and putting the answers beside the questions.

Introduction: In this assignment you will explore what you can recall about exponential functions.

Background: Definition Let a be a positive real number. The exponential function with base a is the function: f(x) = a^x

- The domain of f(x) = a^x consists of all real numbers
- The range of f(x) = a^x is the collection of all positve real numbers
- If a<1 color="#ff0000">a>1 then f is an increasing function.
- Exponential functions satisfy the following properties:

1. a^r x a^s = a^(r+s)
2. a r/as = a^(r-s)
3. (a^r)^s = a^rs
4. a^r x b^r = (ab)^r


1. How is a linear function similar to an expoential function? How is a linear function different than an exponential function?


input and an output
rate of growth


no slope (changing slope)
no common differences, but a common ratio
arithmetic and geometric

2. Let f be the exponential function f(x) = 3 x 4^x. What is the y-intercept? What is the growth factor?

y-int: y = 3
3-> 12 -> 48

Common ratio = 4

3. Let f(x) = a^x such that f(1) = 6 What is the value of a? How do you know?

f(x) = a^x f(1) = 6
6 = a^1 therefore a = 6

4. Let f(x) = a^x. In the following graph of f, is a greater than one or less than one? How do you know?

f(x) = a^x
Since it's always decreasing, a must be multiplying each interval by a value less than 1.

5. Give the equation for the exponential function whose y-intercept is 2 and whose growth factor is 5.

f(x) = 2 x 5^x

6. Some chicken left on the kitchen counter has salmonella bacteria growing on it. The bacteria doubles every 6 hours. Explain why this is equivalent to a sixteen fold increase each day?

Four times in one day it would double, so 2^4 would be how much the bacteria increased, which is the number of bacteria per day, x 16.

"That's it, that's all, it's been a pleasure having you, go out there and commit random acts of kindness and have yourself a holly jolly great day." <-- Mr. K

I think that's how it goes... I'm not entirely sure >_>'

Group: Dino, Chris, Timmy, Van

Sorry if a I spelled your name wrong Dino. (Deno?)

The Super Late Friday Scribe Post

Sorry for the great delay in posting the scribe. It was a hassle just getting onto the blog. Anyways... on to the scribe.

For September 14, 2007, we had a substitute. Ms. McQueen gave us a quiz, and then assigned homework for the weekend. They were...

Read pages 39-41

and Questions 1,5,9,10,18,28,30,31 from exercise 1.6

The next Scribe will be Sandy, and her's is already posted!

Today's Slides: September 18

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Monday, September 17, 2007

Sorry for the late scribe.

Exponential Functions Lab: Team Sandy/Robert/Aichelle

Our group consisted of Sandy, Robert and me, Aichelle.

For question one, we came up with:
similar---> both have inverses.
different---> linear functions have a common difference; exponential functions have a common ratio.

For question two, we said the y-int. was 3 and the growth factor was 4.

We found the value of a to be six in question three.
We know this because:
we were given: f(x) = a ^ x and f(1)=6
so that means a ^ x = 6
and if x = 1
then, a ^ x = a ^ 1
a ^ 1 = 6
a ^ 1= 6 ^ 1
since they have the same exponent [1] we can say a = 6.

We said the graph is less than one because the graph is decreasing. As the x-value increases, the y-value decreases. Therefore, a has to be less than 1 in order for that to occur.

For question five we came up with an equation of f(x) = 2 * 5 ^ x.

And lastly, for question 6, we said that the salmonella bacteria growth that doubles every six hours is equivalent to a sixteen fold increase each day because:
there are 24 hours in a day and it takes 6 hours for the bacteria to double.
24hrs/6 hrs = 4 times the bacteria doubles in a day.
since it is doubling we must square how many times in doubles in a day.
4² = 16 times or 16 folds in a day.


Today's Slides: September 17

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

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 accesed 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.

Cycle 8

etimz (-1)


Thursday, September 13, 2007

First quiz of the Year

Well, we started off today's class with a little quiz about functions. Then after that we spent some time on answering some of the questions from lesson 1.6 , which is all about inverse functions. We were supposed to do the odd numbered question from lesson 1.5 but we already finished that yesterday. The main concept in lesson 1.6 is that in order to get the inverse of a function you only need to switch the inputs and outputs. Only one-to-one functions have inverses. One to one function means there are different outputs for every different inputs. Thats all we need to remember in lesson 1.6. Do not forget that we will be having another quiz tomorrow about functions.

The next scribe will be .......... Van or is it Phan? I can't spell so whatever you asked for it. Good night everyone and study .

EDITED: On exercise 1.6 we have to do questions number 1,5,9,10,18,28,30 and 31. That is for the people who missed today's class.

Wednesday, September 12, 2007

1.5 Compound Interest and the Number e

Well today started off with everyone walking into class and wondering where Mr. Kuropatwa went. He was sick today, so we had a sub. The sub told us that we were to work on exercises (1.5) Compound Interest and The Number e. We then all decided to get into a circle and do the exercises together to help each other out. As a class we got through them all besides number 17. Maybe Mr. Kuropatwa can go through # 17 in class.

The main thing to remember about this section in chapter one is the two formulas. The two formulas are a key part in getting the right answers. Here they are:

A(t)=P(1+r/n )^n t


A(t)=Pe^r t

Well that is it folks. See you guys tomorrow in class. I almost forgot that the next scribe is........mark!

Tuesday, September 11, 2007

Late Scribe... But I did say it would be.

Well, well, well. Where to start? i guess the beginning of class would be a good place, if I can remember it... (please bear with me, I'm a little worn out)

So however it started, it did. Mr. K. introduced the day's topic as Exponential Modeling. We began this familiar unit by adding the new stuff that we had learned yesterday to it.SLIDE2 We were told to use our new TI-83 skills to plot the data, and then graph an exponential curve that relates to the data. We were kind of confused with this considering the scatter plot curve didn't resemble the exponential one very well. It was then that Mr. K. explained that when using "REAL DATA" (as we were), the graph may seem less accurate and straight forward as simulated data. this is because REAL DATA must deal with unexpected factors that interrupts the pattern (war, disease, etc.). Shortly after that, he began to see if we had retained any idea of the concept in our brains over the summer. For some very tired students (me), it took a little while, but we (I) got it nonetheless. When given the next example and told to write the equation for the function, we used the formula:


A= final amount (amount present after t has passed)
A(naught)=the original or present amount
model=the percent growth expressed as a decimal +1
t = time (time needed to reach A)
p = period (amount of time between each application of the model)

Using this formula we constructed a model for the price of a BigMac in Canada in the future (Slide3). This was a really cool question, the study was really original. If you want to have a look at it, click here.

Next, Mr. K. took a little detour and talked a little bit about how the value of currency can change. He used the fact that the German mark, from the time it was offered as a reward for Fermat's last proof to the present day, has decreased in value so massively that 100 000 then is probably worth about 5 cents now.

Finally, we continued our "memory re-jogging" with one final question about the half-lives of chemicals (Slide5).

T'was around time that Mr. K. was putting the finishing touches on the solutions that the bell rang. Before we left, he added that Exercise 1.4: Odd Questions + 6, 12, and 14. were for HOMEWORK!!!

So before you leave this post, I will add that Vincent, no... John D, no... Jojo? no...

...hmmm who is left in this class???

Oh I know! Robert P is the scribe for tomorrow!!! =P

Good night! =D

Now playing: Framing Hanley - 23 Days
via FoxyTunes

Today's Slides: September 11

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Monday, September 10, 2007

Today's Slides: September 10

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

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!

Students Made This

Blogging is a very public activity. Anything that gets posted on the internet stays there. Forever. Deleting a post simply removes it from the blog it was posted to. Copies of the post may exist scattered all over the internet. I have come across posts from my students on blogs as far away as Sweden! That is why we are being so careful to respect your privacy and using first names only. We do not use pictures of ourselves. If you really want a graphic image associated with your posting use an avatar -- a picture of something that represents you but IS NOT of you.

Here are a few videos that illustrate some of what I want you to think about:

Two teachers in the U.S.A. worked with their classes to come up with a list of guidelines for student bloggers.

One of them, Bud Hunt, has these suggestions, among others:

  1. Students using blogs are expected to treat blogspaces as classroom spaces. Speech that is inappropriate for class is not appropriate for our blog. While we encourage you to engage in debate and conversation with other bloggers, we also expect that you will conduct yourself in a manner reflective of a representative of this school.

  2. Never EVER EVER give out or record personal information on our blog. Our blog exists as a public space on the Internet. Don’t share anything that you don’t want the world to know. For your safety, be careful what you say, too. Don’t give out your phone number or home address. This is particularly important to remember if you have a personal online journal or blog elsewhere.

  3. Again, your blog is a public space. And if you put it on the Internet, odds are really good that it will stay on the Internet. Always. That means ten years from now when you are looking for a job, it might be possible for an employer to discover some really hateful and immature things you said when you were younger and more prone to foolish things. Be sure that anything you write you are proud of. It can come back to haunt you if you don’t.

  4. Never link to something you haven’t read. While it isn’t your job to police the Internet, when you link to something, you should make sure it is something that you really want to be associated with. If a link contains material that might be creepy or make some people uncomfortable, you should probably try a different source.

Another teacher, Steve Lazar, developed a set of guidelines in consultation with his students. You can read them here.

Look over the guidelines and add the ones you like in the comments section below this post; either from one of Steve's students or one of your own. I think Bud's suggestions are excellent. We'll be using the one's I highlighted above as a basis for how we will use our blog.

Mr. K.

Friday, September 7, 2007

Hello everyone!

Today started off with an in depth study of our homework. The first question we looked at was well... number 1...

The point of doing number one was to show how to find a good domain in a problem that does not give any concrete numbers to work with. The example was talking about the number of daylight hours per day. Mr.K then dived right into the in depth analysis of the problem. We discussed the day with the most daylight (Summer solstice), the day with the least light (Winter solstice) and the days which the amount of daylight is equal to the amount of night (cannot remember what they were called). With this information in hand and the assumption that we are in Saskatchewan and at the equator to simplify this we delved into finding a suitable domain and range. Since the cycle comes full circle from summer solstice to summer solstice it was decided that a year would be a fair domain. For range we estimated that the daylight would shift by three hours off dead even at the solstices (at this point we also realized that this is a sine or cosine graph pretty much straight out of last years questions). So we found the domain and range and the shape of the graph which satisfied the question. But Mr.K is not satisfied with teaching us merely the rudiments of what we need to know so to the weather network statistics page we went and we observed all the different patterns Winnipeg goes through.

The second question that we moved onto goes as follows:

An open box is made by cutting squares of side x from teh four corners of a sheet of cardboard that is 8.5 by 11 inches and then folding up the sides.

a) Express the volume of the box as a function of x.

No without crunching numbers terribly much we came up with the domain. We knew that it could not be 4.25 inches or more because we'd lose the third dimension therefor losing the volume and you cannot cut 0 or less because you cannot cut negative distances and if you don't cut at all you don't have a third dimension and therefor no volume.

From there we continued to make x useful and make it the size of our cut. From there we deduced that the 8.5 inch side must be 8-2x because two sides will be cut from it and same with the 11 inch side as shown in the diagram below.

Then using what we know about volume, we found the function that related x to the volume. All we had to do is multiply the L x W x H. So our function ended up as f(x) = x(8.5-2x)(11-2x). (Though the slides show it as 8 - 2x).

Then Mr.K got rid of a wasp.

We then moved onto the golf ball question which ended up looking very similar to many questions we have answered in the past years and we quickly caught the drift of it. The only thing that was a little bit tricky was the fact you had to look back in your work for the answer (view slide 2).

The last question that we handled from our homework was as follows:

A wire 6 meters long is cut into twelve pieces. The pieces are welded together to form the frame of a rectangular box with a square base.
a) Define a function that relates the height of the box to the length of one edge of the base.
Well first off we knew that all the sides of the base and top must be the same because it is a square and the four sides connecting the squares must be the same length as well, so our equation for the perimeter became 4H + 8L = 6.
b) Define a function that relates the total surface area of the box to the length of one edge of the base.
Well we quickly came up with a formula for the surface area 2L^2 + 4LH = SA.
But we had to express it in terms of one length of the base. So finagling with our first equation we isolated H and substituted it in and took it as solved (slide 3).

From this point on we were just drawing graphs...
The only one that was tricky was the graph of f(x) = sqr(-x-2). Within it you have to factor out the -1 in both terms or else when you do your flip then translation it will be wrong.

Remember to do your homework! If I can this weekend I'll update this to try out a JAVA applet I'll create (found a couple stumbling blocks). Have a good weekend!

The next scribe is......

John D!

Today's Slides: September 7

Here they are ...

To see a larger image of the slides go here. When you get there you'll see a button in the bottom right-hand corner that says [full]. Click it and the slides will display in full screen mode.

Muddiest Point Anyone???

Well, I've been slaving over this homework since I got home from work (11:15) and even though I find myself able to do most of the questions, I would still like Mr. K. to go over a couple things in class tomorrow.
The first question was okay except for the sketching and estimating domains and ranges of functions which do not have any data (numbers) to derive it from.
Also, I would like to see him go over some of the word problems (17, 19, 21, etc.) just to make sure I'm doing it right.
I doubt anyone else is up this late, but if you have something to add, you can put YOUR Muddiest Point in the comments. =)

Let's see if we can do this everynight, it kinda sputtered out last year, and I think it will be very helpful. (to both Mr. K. and the rest of the class)