algebraic models

7/28/2019 Algebraic Models

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Working withAlgebraic Models

Answering mathematical questions about the world

Written by Alastair Lupton and Anthony Harradine


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Working with Algebraic Models.

Version 1.01 J uly 2007.

Written by Anthony Harradine and Alastair Lupton

Copyright Harradine and Lupton 2007.

Copyright Information.

The materials within, in their present form, can be used free of charge for the purpose

of facilitating the learning of children in such a way that no monetary profit is made.

The materials within, in their present form, can be reprinted free of charge if being

used for the purpose of facilitating the learning of children in such a way that nomonetary profit is made.

The materials cannot be used or reproduced in any other publications or for use inany other way without the express permission of the authors.


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Index

Using an Algebraic Model:

Australias population in the future.

Activity Checkpoints

Developing an Algebraic Model:

Managing a gas field.


Verifying an Algebraic Model:

10th Planet Fact or Fiction?


Refining an Algebraic Model:

Obesity a global epidemic.



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- An application involving the use of algebraic models -

Copyright 2007, Harradine and Lupton Page 1 of 4

Sydney Morning Herald

February 16, 2006

Danna Vale defendsMuslim comments

A federal Liberal MP has defended hercontroversial comments on Australiabecoming a Muslim nation as being about

population figures.Sydney MP Danna Vale was criticised aftersaying this week that Australia wasaborting itself out of existence and couldbecome a Muslim nation.

Introduction

It is obvious that a nations future is affected by the size and composition of its

population. Population issues vary from country to country. In the developing

world a common issue is unsustainable population growth.

In the western world the issue is more

often an aging and declining population.

In countries like Germany population

decline is already being experienced.

Whilst Australias population is currently

growing, projections suggest a possible

decline in the near futurei. It is in this

context that vigorous debate has taken

place about the growth and composition

of Australias future population (see left).

Population projections for Australia.

The Commonwealths Department of Immigration and Multicultural Affairs spends

a lot of time and money developing projections about the future of Australias

population. These projections are based on mathematical models that combine

current trends with sets of assumptionsabout the future. Different projections are

based on different assumptions. One

projection might assume that current trends

will continue, another might incorporate a

fall in fertility levels and a third might

include an increase in net migration. These

projections present some what if

scenarios that can help governments

develop policies and plan for the future.

Aspects of population change.

The data that government departments use for population projections is supplied

by the Australian Bureau of Statistics. Their data distinguishes between two

aspects of population change. The first is Natural Increase, which is the amount

by which births exceed deaths. The second is Net Overseas Migration, which is

the amount by which immigration into Australia exceeds emigration out of

Australia. Population models deal separately with these two aspects of population

change, and then combine them to provide an overall picture of the future.


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Activity 1: Modeling population change due to Natural Increase.

Based on Australian Bureau of Statistics data, a model for Australias population

change due to Natural Increase N(in thousands per year) is the function

N =100+ 45 (0.92) where represents time, in years since 1990.

Working with this model.

A. Enter theW mode of a CASIO ClassPad 300.B. Enter the function for Ninto the y1 row.

Use theM key to enter as an exponent.C. Set an appropriate View Window.

Tap6 to see the view window settings.

Set the View Window so that the graph is drawnfrom 1990 ( =0) until to 2006.

The graph will need to incorporate populationvalues of up to N= 150.

Your choice ofscale will determine how often tick marks are made onthe axes. The dot value will be set automatically.

Tap or pressl to enter each of your settings. You can also usef andc to move within elements of this window. Tap OK to exit this window.

D. Draw the function by tapping$.

E. To obtain function values from your graph by tracing

tap u then= or tap Analysis : Trace.

Use! and$ to move from left to right. If you cannot move to the exact value you

require just enter it on your keypad.

1. Use this model for Nto determine the change in Australias population due to

Natural Increase in the yearsa. 1990

b. 2000

c. 2006

2. Describe how, according to your model, Natural Increase has contributed to

Australias population in the years between 1990 and 2006.

Checkpoint


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Activity 2: Modeling population change due to Net Overseas Migration.

Based on Australian Bureau of Statistics data, a model for Australias population

change due to Net Overseas Migration M(in thousands per year) is the function

308 += xM where represents time, in years since 1990.

Before you start work with this new model, deselectthe

previous model by tapping on the ticked box.

The tick will disappear, showing that the function is no

longer selected. Repeating this process will reselecty1.

Working with Net Overseas Migration

1. Draw a graph of this model for M, Australias population change due to NetOverseas Migration for the period 1990 until 2020. You will need to adjust

your View Window.

2. Use this model to determine Australias population change due to Net

Overseas Migration in

a. 1990b. 2000

c. 2006d. 2020

3. Describe how, according to your model, Net Overseas Migration has

contributed to Australias population over this period.

Checkpoint

Activity 3: Representing total population change.

1. By reselecting your model for N(and leaving your model for Mselected) draw

a graph ofboth models for the years 1990 until 2020 on the same axes.

2. Find the year in which these two models intersect.

3. Interpret what this result means about Australias population change, at and

after that time.

Checkpoint


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Activity 4: Modeling total population change

1. Write down an algebraic model for T, Australias total population change in the

years since 1990.

2. Represent this algebraic model graphically.

3. Determine the year in which Australias total population will first increase by

more than 300 000 individuals, according to your model.

4. Of these 300 000 person increase in population, what percentage will come

about by Net Overseas Migration, according to your models?

Checkpoint

i The information about, and graph of, long-term population projections comes

from theAustralian Immigration Fact Sheet Population Projections. This is

published by the Australian Governments Department of Immigration and

Multicultural Affairs and can be found at

http://www.immi.gov.au/facts/15population.htm


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Australias population in the future.Checkpoints


Activity 1: Modeling population change due to Natural Increase.

Part B Entering a function.

Part C Setting a View Window.

Starting from the Default view window (shown left),

Set the xmin as 0l, xmax as 16l and a scale of 1l.

Arrow upE or downR to move between rows. Set the ymin as 0, the ymax as 150 and a scale of 10.

Part D Drawing a graph

Part E Tracing to obtain function values

Tapping u then= or tapping Analysis : Trace starts the tracing process,

which always starts in the middle of the graph (horizontally speaking).


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Copyright 2007 Harradine and Lupton Page 2 of 5

Arrowing left! and right$ will get you to x=0 and x=16

To get to x=10 you just need to type10 while in Trace

Answers.

1. a. In 1990 Natural Increase caused Australias population to grow by 145 000

according to the model for N.

b. In 2000 Natural Increase caused Australias population to grow by 119 550

according to the model for N(to 5 significant figures).

c. In 2006 Natural Increase caused Australias population to grow by 111 850

according to the model for N(to 5 significant figures).

2. Natural Increase has caused Australias population to grow in the years from

1990 to 2006, but in each successive year it has caused the population to

grow by less than the growth of the previous year, according to our model.

Activity 2: Modeling population change due to Net Overseas Migration.

Deselecting a function.

By tapping the ticked/unticked box a function can be deselected and reselected.

Only selected functions are drawn.


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Answers.

1. The View Window will need to be changed to include the years up to 2020

(x=30) as well as Nvalues of up to at least N=270. A set of View Window

settings are shown below. Tap OK and then$ to draw the graph.

2. Using Trace, and entering the x values required, the following information can

be obtained

From this we can see that,

a. In 1990 Australias population change due to Net Overseas Migration was

30 000 individuals, according to the model.

b. In 2000 Australias population change due to Net Overseas Migration was


c. In 2006 Australias population change due to Net Overseas Migration was


d. In 2020 Australias population change due to Net Overseas Migration will be


3. Over this period Net Overseas Migration has made an ever-increasing

contribution to Australias population growth, according to the model.


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Activity 3: Representing total population change.

Answers.

1. By reselecting and drawing we get

2. The year when Nand Mintersect can be found using

Trace. When using Trace with two or more graphs

drawn theE andR keys (or screen icons) allow

you to move from one function to another.

Note: If you wish to take advantage of the full size graph (as above) you

need to reset the view window after resizing usingr.

The time of intersection can also be found by using the

Intersect command, part of the G-Solve menu. This,

and other useful commands, can be obtained by

tapping Analysis : G-Solve and choosing

Intersect.

This confirms that the models predict that Nand Mwill

intersect near the time =11, corresponding to the

year 2001.

3. This result means that, in 2001, Natural Increase and Net Overseas Migration

made equal contributions to Australias population increase. After that time

Net Overseas Migration makes a greater contribution to Australias population

growth than Natural Increase, according to our model.


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Activity 4: Modeling total population change

Answers.

1. Using the idea that

Total Population Change = Natural Increase + Net Overseas Migration

we can define a model for Australias Total Population Change T(in thousands

per year) as the function

T= N+ M

which, in terms of , can be written as

308)92.0(45100 +++= xTx

where represents time, in years since 1990.

2. Either version of the model for Total Population

Change can be graphed as shown.

To enter the first version, y needs to be obtained from

the abc keyboard (not they key), as it is the name

of a function.

The second version can be easily entered by

highlighting the first expression required, then tapping

on it and dragging it into y4, then entering the+

and repeating the procedure for the second expression.

3. The year in which Twill exceed 300, according to our model can be found,

using Trace, to be 2010.

4. In 2010 Net Overseas Migration is predicted by our model to be 191 900,

roughly 64% of Australias Total Population Change.


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- An application involving the development of algebraic models -


Introduction

There is a great deal of planning

involved in running a gas

production site such as the one

pictured.

Once a site has been chosen

and gas wells have been drilled,

productivity is monitored by

measuring the rate of flowof

the gas out of each well.

The scenario (a real one)

A gas production site in central Australia contains, potentially, up to six wells. At

five of these, wells are already installed and producing gas.

After considering demand levels and production costs the

Reservoir Engineer decides that, for the site to be considered

viable, the average daily rate of flow from the entire site in any

given month must be at least 5 MMscf/day(millions of cubic feet

per day).

If the average daily rate for a given month falls below this, the

sixth well will be installed to increase gas production.

The table below gives the actual average daily flow rate from the site for the

months shown. During this period only five wells are installed and producing gas.

Month(end date)

Relative timet (months)

Rate of Gas Flowf (MMscf/d)

5/31/1998 51.717

6/30/1998 47.724

7/31/1998 36.717

8/31/1998 31.755

9/30/1998 28.066

10/31/1998 22.248

11/30/1998 22.199

12/31/1998 19.154

1/31/1999 16.377

2/28/1999 14.611

3/31/1999 13.403

4/30/1999 12.72

5/31/1999 11.285


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Reservoir Engineers dont wait until the rate falls below the value they have set.

They will use the data to predict when the sixth well should be installed.

Activity 1: Viewing the gas flow data.

A. Fill in the Relative time (months) column of the table above so that the

variable t represents the number of months that have passed since the

commencement of data collection.

B. Enter the data intoI mode in a CASIO ClassPad 300 in the followingmanner

Tap on the heading row oflist1 Use the abck to name this list as time

pressw.

Enter the relative time values in list1, pressingw after each value.

C. In a similar fashion enter the values for f, the

monthly rate of gas flow values into an appropriately

named list2.

D. To represent this data graphically

Tap onG or tap SetGraph then Setting Once satisfied that StatGraph 1 is set appropriately

tap Set.i

Now tapy to drawn the graph as set.E. To have a look at the data represented by the marks

on the screen

Tap on Analysis : Trace or tap=. Press! and$ - on arrow pad or screen - to

move through your scatter plot. Time (x) and flow

(y) values are shown at the bottom of the graph.

1. Describe, in words, what happens to the flow values as time passes.

2. Do you think that a sixth well will be required?

3. At what time would you estimate that it will be required?

Checkpoint


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Activity 2: Developing a model for the gas flow data.

These questions might be more easily answered if we had an algebraic modelfor

the relationship between gas flow and time, for the gas field in question.

To develop such an algebraic model,

Tap Calc to see the types of algebraic models that the ClassPad 300 canfit to data.

By tapping on your choice of model type, the settings for that regressionare confirmed, the co-efficients of the best fitting model of that type are

calculatedii and then this model is drawn.

Shown here is the process for a Linear Reg(ression) moldel for this data

In the Set Calculation screen you have the option of copying the modelinto a selected row ofW mode, as well as copying the residuals ofyour model into your choice of list.

Clearly a linear model does not accurately represent the relationship between flow

and time.

1. Experiment with other types of algebraic models.

Use Copy Formula to store the equation of your choice of model inW.2. Use your choice of model to predict in what month the rate of gas flow will

drop to below 5 MMscf/day and hence, when the sixth well should be installed

to boost gas production.

CheckpointActivity 3: Further modelling.

This data covers the next 18 months of gas flows.

Month(end date)

Relative timet (months)

Rate of Gas Flowf (MMscf/d)

6/30/1999 12.992

7/31/1999 9.21

8/31/1999 8.836

9/30/1999 5.87410/31/1999 4.938


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11/30/1999 11.775

12/31/1999 16.709

1/31/2000 15.579

2/29/2000 14.861

3/31/2000 14.067

4/30/2000 26.285

5/31/2000 28.882

6/30/2000 24.963

7/31/2000 23.124

8/31/2000 20.43

9/30/2000 18.963

10/31/2000 17.335

11/30/2000 15.61

12/31/2000 14.516

1. Use this data to suggest the month in which the sixth well was actually

installed. Compare this with the prediction you made in the previous activity.

The site that we have been studying only had the potential for six wells. Hence,

when the average daily rate of flow falls below 5 MMscf/day after the installation

of the sixth well, the site will be closed down. It is very important for companies

to be able to forecast when such an event will occur.

2. Use the extra data supplied above to develop a model of the flow of gas from

the site in the time after the installation of the sixth well.

3. Use this model to predict when this site will be shut down.

Checkpoint

i Note: Upon entering Set StatGraphs it is possible to set up and turn on or off

up to 9 graphs, change Graph Type, change the list data on which these graphs

are based and change Mark Type. The default settings will need to be changed to

generate graphs of different types in different circumstances.

ii Also displayed are the r2 value and MSe.

The r2 value known as the Co-efficient of Determination represents thepercentage of variation in the dependant variable that can explained byvariation in the independent variable.

The MSe value is the sum of the squares of the residuals of the modelunder consideration, corresponding to the quantity that is minimised whendetermining the co-efficients of the model of best fit. Hence MSe is the

minimum sum of squared errors.


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Managing a gas field.Checkpoints

Activity 1: Viewing the gas flow data.

Parts B and C Data entry

The input of headings and data should look like for time

and similarly for flow

Part D Drawing a graph.

TappingG (or tapping SetGraph and then Setting) allows us to set thegraph up appropriately. The graph is then drawn by tappingy.


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Part E Tracing through the graph.

Tap on Analysis : Trace or tap= then press!

and$.

Answers

1. It falls/decreases in a fairly consistent manner, quickly at first, then more

slowly as time passes

2. Yes, it looks like it will keep falling until it falls below 5 MMscf/day.

It might level off below it reaches 5 MMscf/day, but in the context of gas

extraction this is unlikely.

3. In around 6 months time (i.e. December 1999) is a very rough estimate.

Activity 2: Developing a model for the gas flow data.

Tap Calc to see some options

Answers

1. Investigating a quadratic model, for example,

by tapping Quadratic Reg provides the following model

We can see that the model, in general, fits well but clearly is about to turn and

start increasing towards the end of the time period. This makes extrapolation to

find when the flow falls below 5 MMscf/day impossible, as a quadratic model says

that it never will, but the scatter plot and common sense suggest otherwise.

Looking for a model type that decays to a flow of zero, two algebraic model types

could be considered, a simple exponentialmodel and apowermodel.


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Investigating a simple exponential model by tapping on Exponential Reg,

We can see that an exponential model fits the data well (after the first few

points). It fits particularly well as time passes, making it suitable for the

extrapolation that we require.

Investigating the power model by tapping on Power Reg gives,

This shows that this type of model fits the shape of the data less well.

If a visual comparison of two or more models is required

it is best to copy their formulae into rows ofW,tap on! and check that they are selected, and then tap

ony to add these graphs to your scatter plot.

2.

With our chosen model stored, we can work with it in

W, using the full range of graphical tools that areavailable in that mode.The view window settings from the last graph drawn,

the scatter plot drawn inI mode, are unchanged bychanging modes. All we need to do to graph the model is

make sure that it is selected (tap the box to tick it) thentap$, getting this result


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The extrapolation that we wish to do, to find when the model predicts that the

gas flow falls below 5 MMscf/day, is equivalent to extending the graph to the

right.

By tapping6 we can change the X max up to 25 months.

To find when the graph of the flow falls below 5 MMscf/day we can do an

xcalculation. Tap Analysis : G-Solve : x-Cal then enter the value y=5.

The screen sequence looks like this

This provides the crucial x value of 18.7, corresponding to a time of November

1999 when the sixth well will be required.

If, when performing an x-cal the ClassPad 300 responds with Not Found thenthe y value entered does not occur in the current view window and you may need

to widen your view!


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Activity 3: Further modelling

The output fell below 5 MMscf/day in October 1999 and the sixth well seems to

have started producing in November 1999. This corresponds quite well with the

prediction that was made.

Using the data from April 2000, when the sixth well seems to be in full

production, the data we need to model and its scatter plot appears on the left,

..and the exponential regression model for this data is drawn on the right.

This model allows us, when stored inW mode, to predict the time when thegas production site is no longer viable.

This equates to a prediction of November, 2001.


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10th planet fact or fiction?

- An application involving the verification of an algebraic model -


2003UB313 (artists impression)

with the Sun in the background

Introduction

The planets of our solar system are more than just

well-known aspects of astronomy. They are a part

of popular culture. However, memory aids such as

My Very Easy Mnemonic Just Summed Up Nine

Planets

may need to be revised, with the recent discovery

of what has been claimed to be the 10th planet of

our solar system.i

This new discovery is currently called 2003UB313 and has been nick-namedXena

while it waits for its official name. At present it is roughly 15 billion kilometres

from the Sun, 100 times more distant than the Earth. It has a very elliptical 560

year orbit, which is inclined at

nearly 45o

to the orbit of the

other planets, as shown here.

This unusual orbit suggests the

possibility that this new planet

may not conform to the laws of

planetary motion, as known to

humankind since the 16th century.

Keplers Laws of Planetary Motion

The three fundamental laws that describe planetary motion

were determined by Johannes Kepler, born in Germany in

1571. His Third Law, when simplifiedii says that, if R is the

average radius of a planets orbit, measured in Astronomical

Units (AU) and P is the period (length of time) of a planets

orbit in earth years, then

R3= P2

So, how well does 21st century information about the nine planets fit this 400

year old law, given that three of these planets were unknown at the time it was

formulated? How well does Keplers law describe 2003UB313 with its unusual orbit?


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Activity 1: Viewing the data - theory vs practice

The table below contains data relating to the orbit of the nine established planets

of the solar system

Planet Orbital PeriodP(earth years)

Orbital Radius averageR (AU)

Mercury 0.241 0.387

Venus 0.615 0.723

Earth 1 1

Mars 1.88 1.524

Jupiter 11.86 5.203

Saturn 29.46 9.539

Uranus 84 19.18Neptune 164.8 30.06

Pluto 247.7 39.53

Before we view the data in scatter plot form we need to decide which variable we

will represent horizontally and which we will represent vertically. As there is no

obvious independent or dependant variable, the choice is somewhat arbitrary.

Whilst either representation will suffice, we might choose to allocate the more

likely input variable to the horizontal axis. The period is more easily measuredthan the radius, as it can be calculated based on movement through the sky. This

makes it more likely to be input, so we will allocate it to the horizontal axis.

A. Draw a scatter plot of period verses radius, using a CASIO ClassPad 300.

To see how well Keplers third law fits this data we would like to graph an

algebraic model upon the scatter plot.

To be graphed on the CASIO ClassPad 300, Keplers third law needs to written in

the form ...=P . By taking the cube root of both sides of23

PR = we obtain

3 2PR = which simplifies to 3

2

PR = .

B. Graph this theoretical model on your scatter plot

Tap on!. In thisW window, enter Keplers third law. Tap ony.

Checkpoint


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Activity 2: Looking closer how well does the theory fit the practice?

By inspecting the graph on the previous page it seems that Keplers law fits the

data on planetary orbits very well, but what if we looked

more closely at the differences between what Keplar

predicted and what has been observed?

Tap in the Cal row at the bottom of theStat window underneath list3.

Enter period^(2/3).1. For which planets orbit does Keplers third law fit leastwell?

2. How can this discrepancy between the observed data and the models

prediction be best represented?

Activity 3: What about 2003UB313 ?

Given that 2003UB313 has an orbital period of 560 years and an average orbit

radius of 67.5 AU

1. Add 2003UB313 to the data on the other 9 planets.

2. Redraw the scatter plot to include 2003UB313.

3. Draw Keplers third law on your new scatter plot.

4. Look at how well Keplers law describes the data on the orbit of 2003UB313.

5. Does Keplers law work as well for 2003UB313 as it does for the other nine

planets? Explain your answer in detail.

Checkpoint

iWhat constitutes a planet is a surprisingly complex question. Whilst the

International Astronomical Union is yet to make a ruling, the finders of 2003UB313

consider that anything larger than Pluto should be considered a planet. On that

basis 2003UB313, which approximately 5% bigger than Pluto, is the Suns 10th

planet.

ii Keplers third law states that the ratio of the square of a planets orbital period P

to the cube of its orbital radius R is constant so, for two planets a and b, 23

2

3

b

b

a

a

P

R

P

R= .

If we take earth as one of the two planets, with an orbit of 1 earth year and an

orbital radius of 1 AU (where AU the Astronomical Unit - is defined as the

orbital radius of the Earth) then the laws simplifies to1

12

3

=a

a

P

Rimplying

23

aP

aR = .


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10th planet fact or fiction?Checkpoints

Activity 1: Viewing the data - theory vs practice

Part A Drawing the scatter plot.

Enter the data intoI then tapG (or tap SetGraph then Setting ) to set up

your graph. Tap Set.

Tapy to obtain the scatter plot of radius R against period P.

Part B Graphing the theoretical model.

Tap! to open the Graph Editor.

Enter Keplers third law.

Tapy to add its graph to your Scatter plot.


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Activity 2: Looking closer how well does the theory fit the practice?

We can obtain the radii that Keplers law predicts for the

nine planets by entering his formula as a calculation in

list3. This is done by tapping in the Cal row at the

bottom oflist3 and then entering period^(2/3).

Once this is done the variation between the observed

radii and the predicted values in list3 can be studied.

Answers

1. Keplers third law fits Pluto worst.

2. The discrepancy between observed and model value is 0.09 AU. This seems

very small in absolute terms. In relative terms, expressed as a percentage of

the observed value of 39.53 it equates to an error of 0.22% (less than a

quarter of a percent error). This discrepancy is not trivial as, because an AU is

roughly 150 million kilometres, it equates to around 13.5 million kilometers.

Activity 3: What about 2003UB313 ?

Answers

1. 2.

3. 4.


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5. A more detailed analysis of the errors in the way a model fits given data can

be completed inI mode, using the ideas that

Error = measured value model value

Percentage Relative Error = %100valuemeasured

error

These values can be calculated in following way:

Name the list you plan to use (if desired).

Tap on the Cal row of the list you are using.

Enter the formula required.

The errors in the values given by Keplers model can found by,

This allows us to determine the percentage relative error for Keplars model

The abs( command is found in the cat menu on the keyboard, seen above left.


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This information can be seen more clearly in a table like the one below

Based on this table, Keplers law does not fit 2003UB313 as well as it does for the

other planets. The laws relative error for 2003UB313 is three times bigger that its

relative error for Pluto, which in turn is three times bigger than the relative error

for any of the other 8 planets.

It does, however, still have a greater than 99% accuracy for 2003UB313. Further,

given that the orbital data for 2003UB313 is first of many measurements that will

be taken of this new planet, it may turn out that error is with the initial

measurements and not with Keplers law!

Planet PeriodRadius

(AU)

Radius

(Kepler)Error

Relative Error

%

Mercury 0.241 0.387 0.38726778 0.00026778 0.06919282Venus 0.615 0.723 0.72318611 0.00018611 0.02574158

Earth 1 1 1 0 0

Mars 1.88 1.524 1.5232525 0.0007475 0.04904857

Jupiter 11.86 5.203 5.20063602 0.00236398 0.045435

Saturn 29.46 9.539 9.53868473 0.00031527 0.00330503

Uranus 84 19.18 19.1801879 0.0001879 0.00097965

Neptune 164.8 30.06 30.0587884 0.00121157 0.00403052

Pluto 247.7 39.53 39.4412501 0.08874995 0.22451289

2003UB313 560 67.5 67.9399701 0.43997009 0.65180754


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Obesity a global epidemic

- An application involving the refinement of an algebraic model -


Prevalence of Overweight and Obese

Children aged 7 to 15 years.

0.0

10.0

20.0

30.0

40.0

50.0

60.0

70.0

80.0

90.0

100.0

1985 1995

Percentage

Healthy

Overweight

(not obese)

Obese

Introduction

Obesity and the health problems that it causesi,

are widely acknowledged as an epidemic of

global proportionsii. As part of this epidemic,

Australia is currently experiencing significant

increases in overweight and obesity levels.

The percentage of Australians over 25 who are

overweight has more than doubled in 15 years,

from 12.1% in 1988 to 26.7% in 2003. In the

same time period the percentage of Australians

over 25 who are obese has increased by more

than 360%, from 1.7% to 7.9%iii.

Obesity in the young

- tomorrows health crisis today -

Perhaps of greatest concern, some of the most

dramatic increases in obesity are seen amongst the

young. In Australians aged 7 to 15 years, obesity

tripled between 1985 and 1995, affecting 5.1% of

children; with a further 15.7% overweight.

This trend has continued and accelerated, with

recent studies suggesting that obesity levels amongst children have now reached

nearly 8%. Even more worryingly, obesity is starting to occur earlier in life, with

around 5% of preschoolers now considered obese.

Measuring Obesity.

Gathering data like the information presented above requires an accepted

definition of who is overweight and obese. The accepted definition involves an

individuals Body Mass Index (BMI).Persons over 18 having a BMI of, greater than 25 are considered overweight. greater than 30 are considered obese.

There are similar cut-offs for children.

What is the BMI?

The Body Mass Index is an approximation for an individualspercentage body fat.

An individuals BMI can be found by dividing their weight (in kilograms) by the

square of their height (in metres) i.e.

2)(

BMIheightweight=


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Activity 1: Working with the BMI.

1. I am 180 cm tall and weigh 80 kg. What is my BMI? Am I overweight?

2. A healthy BMI range is considered to be between 20 and 25.

If my partner is 170 cm tall, what is her healthy weight range?

3. OptionalCalculate your own BMI.

Checkpoint

As mentioned previously, BMI is used to classify obesity because it approximates

an individuals percentage body fat. The question is, how well can such a simple

formula approximate a complex quantity like an individuals percentage body fat?

Activity 2: Evaluating the BMI model for percentage body fat.

To evaluate how well BMI approximates percentage body fat requires the

calculation of BMIs for individuals with known percentage body fat, measured

using a different, more accurate methodiv. Fortunately such data was the subject

of a study by Roger W. Johnson, published in the Journal of Statistics Education

and at http://www.amstat.org/publications/jse/v4n1/datasets.johnson.html

In this study the percentage body fat of 252 men was measured, along with their

height, weight and 10 other body measurements. To see how well the BMI

approximates percentage body fat we are going to study a random selection of 30

of these men, whose data appears below.

Age Weight (kg) Height (m) BMI % Body Fat

22 69.9 1.68 25.0

23 89.9 1.87 11.9

32 81.9 1.77 20.7

28 68.6 1.72 14.1

41 112.2 1.87 31.7

46 68.3 1.73 28.0

48 98.4 1.78 31.0

62 98.0 1.86 25.8

72 71.6 1.71 15.0

46 80.3 1.78 20.4

48 80.4 1.85 20.042 80.5 1.75 26.8

47 89.4 1.86 23.4

49 96.5 1.91 20.3

40 80.2 1.80 24.6

23 85.3 1.97 10.6

26 69.1 1.75 9.7

27 90.8 1.87 20.8

33 88.9 1.85 14.7

35 98.4 1.87 19.1

35 103.5 1.77 34.5

35 80.4 1.80 20.4

37 109.4 1.82 29.4

41 105.6 1.89 23.3

42 110.8 1.93 37.3

42 101.9 1.90 24.4

50 78.4 1.85 19.4

51 67.7 1.77 13.7

54 69.5 1.79 12.6

68 70.5 1.76 15.3


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Activity 2 (continued): Evaluating the BMI model for percentage body fat.

A. Enter these weights and heights into a CASIOClassPad 300.

B. Calculate the BMI of these men inI mode inthe following way,

Tap on the heading row ofList 3 and name it asBMI.

Tap on the Cal row of this column and enter theBMI formula as weightheight^2

You may wish to copy these BMI values into theprevious table.

C. Enter the percentage body fat data into List 4.

1. Examine the BMI and % body fat data. Describe what you notice.

2. If there were a perfect correspondence between BMI and percentage bodyfat, what would a scatter plot of BMI against percentage body fat look like?

3. What do you think the scatter plot of our data will look like?

4. Draw a scatter plot of our data.

5. Hence comment on the relationship between BMI and percentage body fat

for the data in our sample.

6 Discuss how well BMI predicts percentage body fat across the range of

BMIs observed in the data set.

Checkpoint

Activity 3: Doing better than percentage body fat = BMI

Given the limited validity of the algebraic model % body fat = BMI,

1. Develop a better algebraic model for percentage body fat in terms of BMI.

2. Comment on the degree to which this new model can be used to predict

an individuals body fat based on their BMI.

Checkpoint

i These consequences include heart disease, type 2 diabetes and cancer.iiWorld Health Organisation; Global Strategy on Diet, Physical Activity and Health.

iii All Australian data was obtained from the Australian Governments AustralianInstitute of Health and Welfareiv Percentage body fat is most accurately determined using an underwater

weighing method. It can also be calculated based on skin fold measurements.


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Obesity a global epidemicCheckpoints


Activity 1: Working with the BMI.

Answers.

1. My BMI can be calculated to be 24.7, meaning that I

am notoverweight just.

2. Someone 1.7 metres tall with a BMI of 20 to 25 would

weigh in between 57.8 kg and 72.25 kg.

Activity 2: Evaluating the BMI model for percentage body fat.

Part A Data Entry

Firstly naming list1 and list2 and then entering the

weight and height data gives,

Part B Calculating the BMI

Tap in the heading row oflist3 and name it as BMI.

Tap in the Cal row oflist3 and enter the BMI

formula as weight/height^2.

Answers.

1. Comparing the values ofBMI and p_b_f (% body fat)

we can see that BMI is similar to percentage body fat

for some individuals but not for others.

2. If there was a perfect correspondence between BMI

and percentage body fat then all the points in the

scatter plot would fall on the line y=xas this

representspercentage body fat = BMI.

3. For our data some points would fall on this line but many would not. It is hard

to tell exactly how the points would vary for this line.


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4.

5. Due to the generally linear shape of the scatter plot there seems to be some

degree of linear correlation between percentage body fat and BMI. Whether or

not this linear correlation represents equality, i.e. the linear relationship

percentage body fat = BMI, is unclear from the graph above.

One very useful way to investigate this question

further is to draw the line y=xon the scatter plot.

This can be done by tapping! and entering the

function y=xand then tappingy

The fact that the line y=xdoes not represent theshape apparent in the scatter plot suggests that the

relationship of equality does not exist between BMI

and percentage body fat.

6. The number of points beneath the line y=xin the lower BMI region (left hand

side of the graph) shows that, for individuals with a low BMI, BMI frequently

over predicts percentage body fat. The closer proximity of points to the line

y=xfor larger BMI values suggests that, for individuals with larger BMIs, their

BMIs are a better estimate of percentage body fat.

Activity 3: Doing better than percentage body fat = BMI

Answers

1. Choosing a linear algebraic model, because of the generally linear shape of

our scatter plot, the co-efficients of this model can be found by tapping Calc,

Linear Reg , and setting up this calculation gives us,


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This linear relationship,

percentage body fat = 1.65BMI 21.56,

fits the data that we have used much better than the idea

thatpercentage body fat = BMI. This can be confirmed by

generating a series of values and and comparing these to

percentage body fat.

The values given by the rule 1.65BMI 21.56

seems to correlate to percentage body better that BMI.

However, there is still a significant degree of variation

in percentage body fat that cannot be attributed to the

new rule. This variation could be caused by factors like

fitness level and body type, factors that are not

incorporated into the BMI calculation but obviously

have a bearing on percentage body fat.

It should be noted that this variation is less for

individuals with higher percentage body fat, making

judgments using BMI-based rules more appropriate in cases of excess weight and

obesity. At best, however, a BMI-based calculation can only provide a warning

sign of weight problems. A more detailed analysis of an individuals percentage

body fat, and of the potential health implications, should then be undertaken by a

health care professional.


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algebraic models

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