## Monday, September 28, 2009

### 1.6 Slopes of Tangents & Instantaneous Rates of Change - Period 4

Hi Everyone,

My Name is David L I will be the scribe of today's post.

Today lesson is about Instantaneous Rates of Change:

What is Instantaneous Rates of Change? How do we find it?

Instantaneous Rate of Change is an approximation of how a graph changes at a accurate instance in time.

Alternative 1: Find the slope of secant

E.g. In this scenario Point X (10,10) and Point Y (20,22) are points that the secant passes through.

Slope = delta d - delta t
= D2-D1 / T2 - T1
= (22 - 10) / (20 - 10)
= 12 / 10
= 1.2

Alternative 2: Use 2 points on the secant line (one being the tangent point and other is a point passing through secant line)

E.g. In this situation Point A (60,70) is secant point and Point B (70,80) is and regular point passing the secant line)

Slope = delta d - delta t
= D2-D1 / T2 - T1
= (80 - 70) / (70 - 60)
= 10 / 10
= 1.2 m/s

Alternative 3: Use an equation and set intervals between each point e.g. [0.1, 0.01,0.001]

e.g. The Instantaneous rate of change at 5.5 seconds:

h(t)=-4.2t^2+10t+2
h(5.5)=-4.2(5.5)^2+10(5.5)+2
h(5.5)=-70.05m

h(t)=-4.2t^2+10t+2
h(5.6)=-4.2(5.6)^2+10(5.6)+2
h(5.6)=-73.712m

h(t)=-4.2t^2+10t+2
h(t)=-4.2t^2+10t+2
h(5.1)=-4.2(5.1)^2+10(5.1)+2
h(5.1)=--70.41

h(t)=4.2t^2+10t+2
h(t)=4.2t^2+10t+2
h(5.01)=4.2(5.01)^2+10(5.01)+2
h(5.01)=-70.09

AROC(5.5,5.6)
=(-73.71--70.05) / (5.6-5.5)
=36.6m/s

AROC(5.5,5.51)
=(-70.41--70.05) / (5.51-5.5)
=-36m/s

AROC(5.5,5.501)
=(-70.09--70.05) / (5.501-5.5)
=-40m/s

Therefore the slope of tangent or instantaneous rates of change would be about -40 m/s.

### Instantaneous rate of change

Today's Lesson is Instantaneous rate of change:

Instantaneous rate of change is the rate of change that is measured at a single point on a continuous curve. Instantaneous rate of change corresponds to the slope of the tangent line at that point.

There are three ways to find the instantaneous rate of change.

Estimate an Instantaneous Rate of Change from a graph:

Method 1: Use the slope of a Secant

We use the slope of a secant passing through the point and another point on the curve that is very close to it to find the instantaneous rate of change.

Example 1:

Point S(4,22) and point Q (8,50)
The slope of the secant SQ is
m SQ= delta d/delta t
=50-22/8-4
=28/4
= 7

Method 2: Use Two Point on an Approximate Tangent Line
Another Method to find an instantaneous rate of change from a graph is sketch an approximate tangent line through that point on a second point on the tangent line.

Example 2:

Point S(4,22) and Point R(10,63)
m RS= delta d / delta t
=63-22 / 10-4
=41/6
=6.83333333

Estimating Instantaneous rate of change from a table of values:

To find the estimating Instantaneous rate of change from a table, you first need to calculate the average rate of change over a short interval.

Estimating Instantaneous rate of change from an equation:
To find the estimating Instantaneous rate of change from an equation by using a very short interval between the tangent point and a second point found using the equation.

Example1 :
Find the instantaneous rate of change at 5.5 modelled by the function h(t)=-4.2t^2+10t+2.

h(t)=-4.2t^2+10t+2
h(5.5)=-4.2(5.5)^2+10(5.5)+2
h(5.5)=-70.05

t delta t
5.5 -70.05
5.6 -73.71
5.51 -70.41
5.501 -70.09

AROC(5.5,5.6)
=(-73.71--70.05) / (5.6-5.5)
=36.6

AROC(5.5,5.51)
=(-70.41--70.05) / (5.51-5.5)
=-36

AROC(5.5,5.501)
=(-70.09--70.05) / (5.501-5.5)
=-40

Therefore I would approximate the slope of the tangent at 5.5 to be about -40.

In Conclusion, You can find the Instantaneous rate of change through three methods:
1) from a graph,determine the slope of a secant passing through a point and another point on the curve that is very close to it
2) table of values, calculate the average rate of change over a short interval by using points in the table that are closest to the tangent point
3) from an equation, determine the average rate over shorter and shorter intervals (0.1,0.01, 0.001, etc)

Hope this helps! =)

### Period 3 - IROC

Please contribute your calculations for the instananteous rate of change of the instance of Usain Bolt's 2008 Olympic run that you were responsible for. Please remember to demonstrate good mathematical form including proper units.

(Complete the homework from today's entry in your homework log.)

### Period1: IROC

Please contribute your calculations for the instananteous rate of change of the instance of Usain Bolt's 2008 Olympic run that you were responsible for. Please remember to demonstrate good mathematical form including proper units.

(Complete today's homework assignment from the homework log.)

## Thursday, September 24, 2009

### Secants, and Average Rates of Change

Hello everyone,

Today, we did further investigation into secants, and how they determine average rates of change in a given interval of a function.

For linear polynomial functions, it is easy to determine the average rate of change; you simply look at the leading coefficient of the function.

For any other polynomial function, (and ill use the example of f(x) = x^2, x e [0,10]) however, the rate of change is different when x=5, than it is when x=6. It changes constantly throughout.

What can be determined however, is the average rate of change from two points along the graph. This can be done using the formula, where m= slope of secant:
m=(y2-y1)/(x2-x1)

(my appologies; the formula maker is not agreeing with me, so just bear with my format)

for example, in the given polynomial function above, if i wanted to find the average rate of change, for the given domain, I would calculate the slope of the secant that passes through the points, (0,0) and (10,100).

By plugging the points in to the formula:

m=(y2-y1)/(x2-x1)
m=(100-0)/(10-0)
m=10
So there you have it; the average slope of the stated domain of the function x^2, is 10. Sure the slope of the secant of the whole domain tells me the average rate of change, but it does not tell me the slope at more specific areas on the graph.
I am going to split the domain of this function into two seperate domains: x e [0,5] and x e [5,10]. This will result in seperate secants for each new domain.
m, x e [0,5]
m=(y2-y1)/(x2-x1)
m=(25-0)/(5-0)
m=5
So the slope for this secant is 5
m, x e [5,10]
m=(y2-y1)/(x2-x1)
m=(100-25)/(10-5)
m=15
The slope of this secant is 15
What if, however, i want to find the slope, not for the average of an interval, but for a specific point on the graph? For the example function that i provided above, what if i wanted to find the slope when x=7? What i will do first, is i will calculate the slope of the secant between the x values of 7 and 8, (7,49) and (8,64)
m, x e [7,8]
m=(y2-y1)/(x2-x1)
m=(64-49)/(8-7)
m=15
Then between the values of 7 and 7.1
m, x e [7,7.1]
m=(y2-y1)/(x2-x1)
m=(50.41-49)/(7.1-7)
m=14.1
You can now observe that the secant is becoming a tangent, and y2 approaches 49, and as x2 approaches 7.
Now lets go between the x values of 7, and 7.00001
m, x e [7,7.00001]
m=(y2-y1)/(x2-x1)
m=(49.00014-49)/(7.00001-7)
m=14.00001
As y2 approaches y1, and as x2 approaches x1, secant approaches tangent, the instantanious slope, which we will probably cover in the next lesson. The homework touched base on this, kind fo forshadowing in a way, so i figured id make the connection: secants can become more specific, untill the tell the slope at an exact point.
Enjoy the University fair tomorrow,
David

### Period 1: AROC and Secants

Now, let us begin.

A line passing through at least two different points on a curve is called a secant. The average rate of change, abbreviated as AROC, is found by determining the slope of a secant on the graph. Therefore, Average rate of change=Slope of Secant.

In the investigation, Usain Bolt's 100m sprint in the 2008 Olympics was graphed. To determine the slope of a graph, we would normally use y2-y1/x2-x1. However, because we are looking at the average change in speed, it will be d2-d1/t1-t2. (final distance minus the original distance, divided by the final time minus the original time)

change in distance
change in time

Δ d
Δ t

By calculating the AROC per 10m interval, we are able to find the rates of change for each segment of his race (m/s) - 5.4, 9.8, 10.99, 11.49, 11.76, 12.2, 12.2, 12.2...

We notice that his rate of change is increasing (he is getting faster). This makes sense because, the slope of the graph gets steeper and steeper.

Have a great week :)
-Ana

## Wednesday, September 23, 2009

### Period 4 Scribe: AROC & Secants

Hi class!

Today we learned about AROC, which is an acronym for Average Rates of Change. This comes from the physics equation of Average Velocity which calculates "change in distance/change in time" which means delta "d" over delta "t". Delta means "change in" so an example would be two different distances of 2 m and 10 m. 2 m would represent d1, and 10 m would represent d2. Now that covers the distance. For time, lets say t1 is 2 seconds, and t2 is 4 seconds. Now since delta means "change in", the equation with the variables plugged in will look like this (d2-d1)/(t2-t1) -> (10m-2m)/(4s-2s). The final answer will be 4 m/s.

AROC is essentially the same. It uses the same delta d over delta t equation to find the average rate of change. On a distance-time graph, we can get our average rate of change by using the slope of a secant. This means that on a graph, we draw a straight line that passes at least 2 different points on a curve. We start at (0,0) and we end at the endpoint. This represents our secant and with this, we can use it for average velocity calculations.

J.

### Per4: Usain Bolt's 2008 Race Analysis

Today in class, we examined the concept of an average rate of change by analyzing different intervals of Usain Bolt's 100m World Record run at the 2008 Beijing Olympics. We used data from the article, "Bolt's 9.69s. Analysis of speed during the world record. How fast did Bolt run?"

Please add a comment regarding the race interval that your group was responsible for. Include the calculation that you used to determine the average velocity. This may be a good opportunity to try out the sitmo link on the right-hand side of the blog.)

(You can view the video in the post below.)

### PER.1: Usain Bolt's 100m World Record Analysis

Today in class, we examined the concept of an average rate of change by analyzing different intervals of Usain Bolt's 100m World Record run at the 2008 Beijing Olympics. We used data from the article, "Bolt's 9.69s. Analysis of speed during the world record. How fast did Bolt run?" Please add a comment regarding the race interval that your group was responsible for. Include the calculation that you used to determine the average velocity. This may be a good opportunity to try out the sitmo link on the right-hand side of the blog.) See video.

## Monday, September 21, 2009

### Period 4: Odd and Even Functions

Odd Functions

Graphically:

- We reflect the function f(x) first in the y-axis, then in the x-axis, so we reflect it twice on two axis, and we can say that the function is odd.

-After that we can check by rotating the function 180 degrees about the origin!

f(x) = x^9 is a typical odd function because the degree is odd

Algebraically:

f(x) = 3x^9
f(-x) = 3(-x)^9
f(-x) = -3
^9

it does not equal to f(x) so it is an odd function!
-Remember: a function can not be odd and even at the same time!

Even Functions

Graphically:

- We reflect the function f(x) in the y-axis, and it is identical to the original graph, and the graph is even.

Algebraically:

f(x) = x^4
f(-x) = (-x)^4

f(-x) = x^4

f(-x) = f(x)

Because any number no matter positive or negative, to the power of an even number, it will turn positive, therefore it will be identical to the original function.

### Period 1: Odd and Even Functions

Even Functions

- a function that can be reflected in the y axis and remain unchanged (symmetric about the y axis)

Algebraically to find if a function is odd, one must sub f(-x) into the function

For ex: f(x) = x^2 - 3
f(-x) = (-x)^2 - 3
(watch for negatives since (-x)^2 = (-x)(-x))
f(-x) = x^2 - 3 = f(x)

Because f(x) is the same as f(-x) then it's an even function.

Odd Functions

- a function that can be reflected first in the y axis then in the x axis and remain unchanged (basically rotating at the origin for 180 degrees)

f(x) = x^3 is a typical odd function

To find out if a function is odd algebraically, one subs -f(-x) into an equation:

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

f(-x) = -2x^3 which does not equal f(x) so it's not even
-f(-x) = -1[-2x^3]
-f(-x) = 2x^3 = f(x) therefore since f(x) equals -f(-x) then the function is odd

It is also possible for a function to be neither even or odd.

Quick Check

Even function: if every term in the function has a degree with an even number then the function is even

Ex: f(x) = x^6 + x^4 + 3 (3 = x^0)

Odd function: if every term in the function has a degree with an odd number then the function is odd

Ex: f(x) = x^5 + x

## Saturday, September 19, 2009

### 1.3.2: Factoring in our Graphs (Overview)

Hey, everyone it’s Shanise from Ms. Bruchat’s period 3/4 class. I hope everyone is enjoying and understanding polynomial functions so far! Here is a review of Friday’s lesson.

An equation in factored form can reveal a lot about the equation:
1. each factor gives us the value of the graph’s zeros(x-intercepts)

Ex. y= (x-3)(x+1)(x-4) Make y=0 and solve for x to find the zeros

Values of zeros are x= 3,-1, 4

2. determines the number of zeros

For example the same equation from above has three brackets or three zeros. This tells us the number of zeros in this equation is three.

Ex. y= (x-3)(x+1)(x-4)
First zero- (x-3) =0
x= 3
Second zero(x+1) =0
x= -1
Third zero (x-4) =0
x= 4

3. helps us to determine the degree of the polynomial

In this example to determine the degree of a polynomial you must count all the x’s.
Ex. y= -4(x+3) (x-2)^2(x+6) (x+2)

We must take the exponent “2” into account on the second bracket, when we are counting. In all there are four x’s, which means that this is a degree 4 polynomial.

4. the multiplicity(order) of each factor tells us how the graph interacts at the x-intercepts

To find the order of each factor, it is the exponent after the brackets and if there is no exponent on bracket/brackets then the order is one. For example this equation y=(x-1)(x+2) has an order of one.

Multiplicity of 1- a straight line travels through the zeros

Multiplicity of 2(4, 6, 8…) - bounces at the zeros
Take this equation for example y=(x-1) ^2 (x+2) at the x-intercept x=1, it will bounce and a straight line will go through the other zero x= -2

Multiplicity of 3(5, 7, 9…) - inflects at the zeros
For this equation y=(x+1) ^3 (x-3), at the x-intercept x=-1 the line going through it will not go straight through but it will be inflected. However for the second zero x=3 the line will go straight through.

## Friday, September 18, 2009

### 1.3.2 Factoring in our Graphs (Summary)

The factored form of an equation can tell you
1. number of x-intercepts
2. where they are

Ex. y = (x-5)(x+3)(x-7)

There are three x-intercepts in this equation (the three x's in the brackets). By making y=0, we can find out where the x-intercepts are. In this case, there are three ways to do so: x=5, x=-3, and x=7.

The total number of x's in the equation can tell us how the end behaviours act and the sign of the leading coefficient tells us if it is normal or reflected. Using the same example, we can see there are three x's and therefore we know the line will act like a degree 3 function (with opposite end behaviours) and will not be reflected (positive leading coefficient).
<-- y= (x-5)(x+3)(x-7)

The order (or multiplicity) of the factor tells us how the graph interacts at that particular intercept.
Ex. Order 2 -> "bounce" off the x-intercept
Order 3 -> "inflects" through the x-intercept

Ex. y= (x-2)^3(x+5)^1

In this example, we have x^3 and x^1 which multiply together to become x^4. The overall graph will act like a quartic and will have same end behaviours.

click image to see explanation.

## Thursday, September 17, 2009

### Period4 - Factoring In

As a class, I would like you to briefly comment on the findings of your investigation today.

In your response, give me an example of a function (minimum degree 4) with 4 x-intercepts. Justify how you know it has 4 x-intercepts.

There is an applet on explorelearning.com that can help you play with the functions. Enter "Polynomials & Linear Factors" into the search box and you will have a 5 minute pass to explore.

### Period1 - Factoring In!

As a class, I would like you to briefly comment on the findings of your investigation today.

In your response, give me an example of a function (minimum degree 3) with 3 x-intercepts and justify how you know it has 3 x-intercepts.

TThere is an applet on explorelearning.com that can help you play with the functions. Enter "Polynomials & Linear Factors" into the search box and you will have a 5 minute pass to explore.

## Wednesday, September 16, 2009

### 1.2 Characteristics of Polynomial Functions

Hi everyone!
today we had the privelege of using the Senteo class things. They were so cool 8)
Anyways, here is the summary of today's lesson:
• We looked at TOV's (table of values) and the differences columns. We observed that the number of degree of the function will be the same as the number of differences columns. For example: if the function was f(x)=x^3, there would be a third differences column.
• We also learned about increasing and decreasing functions. A function is INCREASING if the graph RISES from left-right, and it is DECREASING if it FALLS from left-right.
• A turning point is the a point on the graph when the function changes directions (ie: from increasing to decreasing)
• LOCAL MAXIMUM: when the function changes from increasing to decreasing.
• LOCAL MINIMUM: when the functin changes from decreasing to increasing.
• the ABSOLUTE MAXIMUM point is the highest max. on the function. the ABSOLUTE MINIMUM is the lowest point on the function.

I hope everyone understands the lesson. Have a great day and don't forget to do the homework assigned (pg. 26-29 # 1, 2, 3(odd), 5, 6, 7b (part i, ii), 8, 11, 15. As well as the investigation on the back of the handout from today. Have fun!

### 1.2 Characteristics of Polynomial Functions

Today, we learned that there are 3 key features of a polynomial function.

#1: The degree of a polynomial function can be used to determine the finite differences. How?

General Rule: A polynomial function of degree “n” (“n” is a positive integer, the “nth” differences are constant and have the same sign as the leading coefficient.

Example #1: f(x)=2 x^3 +5

• The degree is 3, which means the 3rd finite differences will be constant.
• The leading coefficient (2) is positive, which means the 3rd finite differences are positive.

Example #2: f(x)= -5 x^24 +2

• The degree is 24, which means the 24th differences will be constant.
• The leading coefficient (-5) is negative, which means the 24th differences are negative.

* That is, for example, if a function has a degree of 123, you do not need to calculate 123 finite differences using the table of values! All you need to do is look at the degree and the leading coefficient to determine when the differences are constant (in this case in the 123rd differences) and if it is positive or negative.

#2: You can classify a function as increasing or decreasing.

a. The graph rises from left to right - increasing
b. The graph falls from left to right - decreasing

For example:

*When writing in interval notation, make sure you include "xE" in front of the brackets!

#3: A turning point is used to describe the point where the function changes from increasing to decreasing (or vice-versa).

Some new terminologies:

## Tuesday, September 15, 2009

### 1.1 Power Functions (Summary)

Polynomial Functions: has independent variable "x" raised to non-negative integer exponent
Ending Behavior: how the graph behaves as we look at very small and large values of x
Click on image to see the summary of Odd Degree Functions vs. Even Degree Functions

To Conclude:
By looking at the "leading coefficent" of a polynomial and the degree, it can determine the behavior of the graph

Even Degree Function:
• same end behavior
• act similarly to y=x^2 (and y= -x^2)
• can reflect on y-axis and reflect onto itself
• ex. y = 7x^8 + 6x^2 + 16

Odd Degree Function
• opposite end behavior
• act similarly to y= x (and y= -x)
• can be reflected in the y-axis followed by a reflection in the x-axis and be back to its original location
• ex. y = 12x^5 + 4x^2 + 19

### Characteristics of Functions Lesson

Hey everyone! How's everyone doing? Hope you enjoyed your first week + a day at school so far =) I'm Grace Leung from Ms Burchat's third period Advanced Functions class and this is my and the blog's first scribe post!

In today's lesson, we delved a little further into the characteristics of functions. The main idea behind the lesson were the end behaviours. By looking at the end behaviours, we can describe how the graph behaves when the x-value is a large positive or negative number. By using graphing calculators, we were able to visually see and compare functions of different degrees. With the people we were sitting with, a set of functions were assigned to us and we used the calculators to graph them and try to come up with a hypothesis on how the functions related to each other. Though time ran short, groups shared their thoughts and hypotheses. Some groups (mine in particular =P) had difficulty in coming up with a hypothesis, but later on we learned that the degree of the polynomials as well as the sign of the leading coefficients can help us determine what the end behaviours will be without actually seeing the function on a graph. Remember, all even degree functions will have the same end haviours and act in likeness to y = x^2 and y = -x^2. On the other hand, all odd degree functions will have the same end behaviours and be similar to y = x and y = -x.

Does it sound like I understand the lesson? To be honest, when I look at the homework I want to take a Tylenol and go to sleep instead. Share your thoughts with me, guys =) That's all for now... see you guys tomorrow! Have a great day... enjoy the sunshine while it lasts~