Calculator Tools

Two more Booklets are available in this series:

TI-Nspire CAS Probability & TI-Nspire CAS Geometry

For these and other activities please go to: http://education.ti.com/asiapacific/activities

The New Australian & New Zealand Activities Exchange

Introduction to TI-Nspire...... 6 Navigation...... 6 Applications ...... 7 Documents...... 7 Document Navigation & Page Layout...... 8 Common Operational Procedures ...... 8 Start a new document and create a calculator page...... 8 Catalogue & Help...... 9 Mathematical - CSI...... 11 The Crime Scene...... 11 Collecting new data: ...... 11 Measurements:...... 13 Setting up the calculator ...... 15 Femur ...... 16 Humerus ...... 19 Radius...... 20 Tibia...... 21 New Evidence...... 22 Teacher Notes – Mathematical CSI: ...... 23 Exchange Rates...... 24 Activity: ...... 24 Developing a rule...... 26 Teacher Notes – Exchange Rates ...... 30 Time Zones ...... 31 Activity: ...... 31 Time Zones: ...... 32 Wellington GMT + 12 ...... 32 Beijing GMT + 8 ...... 32 New York GMT - 5 ...... 33 Extension Questions: ...... 33 Teacher Notes – Time Zones...... 34 Time Zone History...... 34 Decathlon ...... 35 Activity: ...... 35 2004 Olympic Games...... 37 Investigating the 100m Event ...... 39 A Balanced Score...... 40 Investigation: Approximating Individual Event Scores ...... 42 http://education.ti.com/asiapacific Page 3 of 73 © 2008 Texas Instruments Author Peter Fox

Teacher Notes – Decathlon...... 43 Extension: Functions ...... 43 The Cube Factor ...... 44 An easier Problem:...... 44 Problem to Solve ...... 44 Teacher Notes – The Cube Factor ...... 45 What’s the difference? ...... 46 Linking algebra and Geometry: ...... 46 Instructions ...... 46 General Problem ...... 47 Teacher Notes – What’s the difference?...... 48 Vedic Mathematics...... 49 Introduction to Vedic Mathematics ...... 49 Teacher Notes – Vedic Mathematics ...... 52 Indices...... 53 Activity ...... 53 Introductory Problems:...... 53 Index Rule – One ...... 53 Index Rule - Two ...... 54 Index Rule - Three...... 54 Teacher Notes – Indices ...... 55 Paper Folding...... 57 Constructing a Locus – Stage 1 ...... 57 Step 1 ...... 57 Step 2 ...... 57 Step 3 ...... 57 Step 4 ...... 57 Constructing a Locus – Stage 2 ...... 58 Constructing a Locus – Stage 3 ...... 60 Teacher Notes – Paper Folding ...... 62 Conic Sections ...... 63 Conic Sections – The parabola: ...... 63 A backwards Look ...... 64 Extension: Finding an Equation...... 64 Teacher Notes – Conic Sections ...... 65 Transforming Parabolas...... 66 Activity ...... 66 Teacher Notes – Transforming Parabolas ...... 69 Transforming Parabolas - II ...... 70 http://education.ti.com/asiapacific Page 4 of 73 © 2008 Texas Instruments Author Peter Fox

Activity ...... 70 Complete the following table: ...... 72 Extension: ...... 72 Teacher Notes – Transforming Parabolas II ...... 73

Introduction to TI-Nspire TI-Nspire is as much a handheld computer as it is a calculator. Calculations using a traditional disappear from the screen as the subsequent calculations are performed. With TI-Nspire the entire mathematical history is stored and can be saved as a document, much the same as on a computer. The document can be sent from one handheld unit to another or saved onto a computer and emailed much the same as any other Windows™ file. Documents created on the handheld unit can be opened and modified using the computer software; similarly, documents created on the computer software can be opened and modified on the handheld unit. Settings created for the document are saved with the document; from a practical perspective this is really powerful. When documents are provided to students to work on, they do not need to spend the first five minutes of the lesson adjusting all their settings; for example if the document is set to degrees, when the file is opened again, it remains in degrees regardless of the standard mode settings on the handheld unit.

Navigation Much of the navigation within TI-Nspire can be completed through drop down menus, operating a mouse via the navigation pad or using the control key with a combination of other keys as short cuts. Many of the short cuts may be familiar such as:

Copy Ctrl + C Cut Ctrl + X

Paste Ctrl + V Undo Ctrl + Z

Save Ctrl + S Redo Ctrl + Y

New Ctrl + N Insert Ctrl + I

Other common navigational features include: d - Used to ‘back out’ of a menu, selection or dialog box e - Move through dialog boxes or around templates, press tab to move to the next selection a - Same as a single mouse click c - Returns to the home screen, much the same as a computer desktop displaying short cuts to applications b - Access applicable drop down menus, these change depending on the active application / + b Same as right mouse click, displays a contextual menu / + c File and editing tools

Navigation around TI-Nspire is more intuitive. The presence of a mouse, right and left mouse click functionality, drop down menus, common short cuts and a variety of help options reduce the necessity for individual key stroke instructions for navigational purposes. For this reason most of the activities contain limited specific key stroke instructions. In many cases instructions consist of create a new document or produce a scatter plot of your data. In the later case, it is assumed that students will: • add a graphs and geometry page • select scatter plot as the applicable graph type • select the appropriate dependent and independent variables from the drop down menus The aim is for students to eventually make their own choices about the various representations of a problem and use TI-Nspire as a tool to explore problems rather than simply a calculator used to generate answers. Experience has shown that the majority of students learn the navigational aspects of TI-Nspire very quickly; this allows both teachers and students to focus on the mathematical content. Initially students will need support, much the same as learning any new software on a computer, to this extent initial problems and explorations should be kept relatively simple.

Applications TI-Nspire contains a number of applications including:

Perform basic calculations, manipulate algebraic expressions or use a large range of higher level mathematical functionality including differential and integral

calculus, matrix and vector operations, probability and statistical calculations. The two dimensional editor for mathematical expressions combined with a range of templates makes this environment intuitive and simple to use and mathematically more relevant to by hand notation.

Draw and manipulate graphs, construct dynamic geometry objects and integrate them with the graphing environment.

Generate lists of data and use formulas within a spreadsheet. The spreadsheet also operates algebraically which allows students to explore algebraic patterns.

Students can write their own brief notes; teachers can write an introduction, instructions or questions for students to respond.

Data from the spreadsheet application can be displayed using a large array of graphical representations. One of the most significant developments of this environment is the dynamic integration of data and graph.

Documents As problems are explored and solved they can be saved as a single document. A document can contain a large number of pages consisting of a combination of applications. Pages can be collected together as a single problem. Each application within a problem is dynamically linked, automatically. If a stored value is changed on one page, it is changed in all the other applications. If a graph is changed by editing or dragging, the equation is changed on all pages within the problem. A document may consist of many problems with each problem consisting of many pages with a range of applications. Creating a new problem within a document provides a break between the dynamic linking of variables. If a = 2 on a calculator page in the first problem, creating a new problem refreshes a so that it once again becomes a variable.

Document Navigation & Page Layout As additional applications are added to a document it Problem 1 becomes necessary to move between pages. The number Page 2 of pages within problem is identifiable via the tabs on the top of each page. Displayed as: 1.2 o To move forward one page at a time press: / + ¦ (Right side of navigation pad) o To move backward one page at a time press: / + ¥ (Left side of navigation pad) If you need to move forward or backward several pages at a time, or move quickly to the next problem in the document it is sometimes quicker to go to the document view; this is similar to the slide view in Powerpoint. o To go to document view press: / + © (Top side of navigation pad) o To return to a single page, either click on the page or press: / + ¨ (Top side of navigation pad) The status of the document is also displayed on this screen; the document displayed opposite has not yet been saved “Unsaved Document”.

Common Operational Procedures This set of activities includes access to all of the applications in TI-Nspire. To get started students are generally required to start a new document and launch an application.

Start a new document and create a calculator page. o Press the home button – c o Use the navigation pad to highlight new document and press the selection arrow; alternatively press the number corresponding to “new document”.

http://education.ti.com/asiapacific Page 8 of 73 © 2008 Texas Instruments Author Peter Fox o If an active document has not been saved a dialog box will appear asking if you wish to save the document before exiting. Make your selections as appropriate

o When a new document is created, a dialog box will open listing all available applications on the handheld unit. Create a calculator page.

You are now ready to start performing calculations!

Catalogue & Help o The TI-Nspire keyboard has a key that provides quick access to many of the templates and functionality, almost like a ‘help’ button. k o The first tab contains an alphabetical listing of all the commands / functions the calculator can perform. The display line at the bottom of this dialog box shows the syntax for the command. You can also ‘check’ the Use Wizard option box as some functions include a wizard to assist in the completion of the syntax. o To get to the second tab on the catalogue dialog box, press ‘2’ on the keyboard. This listing is similar to the first, only mathematical functions or commands are stored under appropriate mathematical concepts rather than listed alphabetically. For example, if you are working with fractions and need to know what sort of fraction tools are available, rather than looking through the entire catalogue, select ‘number’ follow this list down to ‘fraction tools’.

o The forth tab in this dialog box contains commonly used symbols. Some of these symbols are on the keypad, others are short cuts and some are ‘auto corrected’ much the same as Windows automatically changes (c) to ©. On TI-Nspire, if you type >= it will automatically be changed to ≥ when enter is pressed. This symbol is also included as a short cut by pressing CTRL + > and of course is available through this section of the help menu. o The mathematical template is very useful. Many of these templates are also automatic. For example if you want to compute the derivative of an expression the calculus menu will provide the same template. The derivative template is also available as a short cut: g + - and of course through the templates menu. o To use any of the items in the catalogue, click or press enter; the dialog box closes and the corresponding command or template is pasted onto the active page. o The last tab in the catalogue help section is the variable management menu. By default, variables are defined for each problem. When a ‘new problem’ or ‘new document’ is created, previously defined variables are no longer defined for the new environment. If a variable, function or program is required for other documents or problems they can be shared using the library functionality.

Mathematical - CSI

Forensic anthropology is a combination of physical anthropology and the legal process. Forensic anthropologists assist in the identification of human remains and subsequent detection of crime. Forensic anthropologists work to suggest the age, sex, ancestry, stature, and unique features of a decedent from the skeleton.

The Crime Scene A number of murders have taken place in Waipoua forest. Recently a gruesome discovery was made by a group of hikers, a collection of what appears to be human bones scattered over a small area. As one of the leading anthropology teams in New Zealand you have been called to assist in the investigation. Upon arrival at the site you take a series of measurements. You use a GPS to pin-point the location of each bone; a set of callipers to measure bone length and a few notes about the type of bone. A summary of the notes is included here: • Specimen: WF181107B1 Location: E173°31.5538’ - S35°36.0740’ Description: Appears to be human Femur: 41.9cms • Specimen: WF181107B2 Location: E173°31.5541’ - S35°36.0736’ Description: Appears to be human Humerus: 34.1cms • Specimen: WF181107B3 Location: E173°31.5539’ - S35°36.0741’ Description: Appears to be human Radius: 23.9cms • Specimen: WF181107B4 Location: E173°31.5545’ - S35°36.0732’ Description: Appears to be a human Tibia: 31.8cms The detective in charge of the case has asked you to determine a possible height and gender for the deceased. A good forensic scientist uses the evidence to establish a possible identification before trying to match victims. Your task is to use some of the data from your files, combined with some new data to assist in the investigation.

Collecting new data: You need to collect some new data for bone length. Class mates, relatives and friends will make good subjects to measure. You will only be able to determine bone length approximately as flesh obscures the measurements a little. The instructions provided here will assist in identifying how to measure each of the bone lengths. Femur The Femur or thigh bone measurement should be taken from the knee joint to the hip joint. If you place your hand on your upper thigh, you will feel the top of the femur just before it bends to connect with your pelvic bone. This is where the femur should be measured. You MUST be consistent with your measurements.

Tibia The Tibia should be measured from the base of the knee cap to the base of the ankle bone. Radius The radius stretches from the wrist to the elbow. The bulge at the wrist is the end of the radius, and the base of the elbow is the start of the radius. Humerus The humerus is often called the funny bone. Anyone that has ever felt the dull ache after knocking their elbow will know that it really isn’t a very humorous bone to knock! The bone should be measured from the base of the elbow to the underside of the shoulder. You should be able to feel the edge of the clavicle. (Shown opposite)

It is important that you collect a range of measurements from classmates, teachers, friends and relatives. The more data you have, the more reliable your final conclusion. Collecting data for a range of heights will also increase the accuracy of the final conclusions. This information you collect will be combined with previously recorded data and used to produce a graph. The graph will help you determine a relationship between the length of individual bones and the height of a person, in this case, the victim.

Measurements: Complete the table below for the measurements for a range of individuals in your class: You will need to ensure you measure a relatively diverse range of individuals, varying significantly in height.

Gender Height Age Femur Tibia Length Radius Humerus Male / (cms) (Years) Length (cms) (cms) (cms) Female (cms)

The male and female skeletons are proportioned differently. It is recommended that the data be considered separately to identify any improvements in correlation. An improvement in correlation can be seen visually on a scatter graph if the points huddle closer to a single line. If the points are more dispersed the correlation is weaker.

Setting up the calculator You may be provided with the CSI – Mathematics file which includes previously collected data; or your teacher may ask you to create a new document. If

your teacher has provided you with the CSI – Mathematics file, skip to step 4. Step 1 If you need to create a new document; start by inserting four applications: • Text page – Put the activity title and your name on the text page. • Two spreadsheets: To assist in the collation and organisation of the data insert two spreadsheets, one for males and one for females. • One Data & Statistics page to display data and relationships. Note: More applications will be added later. Step 2 Type the column headings into the spreadsheet. To distinguish between Male – Femur length and Female – Femur length the column headings should be: MFemur and FFemur respectively on the two spreadsheets. The headings in spreadsheet one (1.2) should therefore be: mheight; mage; mfemur; mtibia; mradius; mhumerus The headings on spreadsheet two (1.3) should therefore be: fheight; fage; ffemer; ftibia; fradius; fhumerus

The second of the two spreadsheet images shown

opposite displays problem 1, page 3 (1.3). This spreadsheet has been set up for females and includes some data. Complete column headings can be displayed by changing the column width. (refer below) Column names are also displayed at the bottom of the page when the column name is selected. Note: To change the column width, highlight the entire column by moving to the top of the column and attempting to move up one more space. Column width can be adjusted using Resize under the Actions menu. Step 3 Enter the data provided to you by your teacher in the appropriate cells in each spreadsheet. Step 4 Enter the data you collected into the appropriate spreadsheets. http://education.ti.com/asiapacific Page 15 of 73 © 2008 Texas Instruments Author Peter Fox

Femur Place the male height data (mheight) data on the x-axis. Place the male femur (mfemur) data on the y-axis. Note: The axis scales are automatically determined. These scales are suitable to view the data, but when a relationship has been determined, it is best to view the origin also.

The data appears to be scattered around an imaginary line. It is possible to draw a line through this region by adding a moveable line.

Once the line is displayed over the data, grab the line using CTRL – Click (/ + a) and move it so that it represents the general trend of the data. The equation for the line is displayed: • mheight (Independent axis or x-axis) Variable: x • mfemur (Dependent axis or y-axis) Rule: m1(x)

The scale on the axis needs to be adjusted in order to better understand the meaning of the equation. The easiest way to change these scales is through the window settings. (refer opposite)

The origin has the coordinates (0, 0). This means if the Xmin = 0 (minimum value for x) and Ymin (minimum value for y) are both set to zero; the origin will be included in the graphing window. The Xmax (maximum value for x) and Ymax (maximum value for y) are generated automatically to include the most extreme point in the data set.

1. What is your equation for the relationship between male height and male femur length? 2. Using specimen WF181107B1, measuring 41.9cms: a. What does your equation predict for the height of a male with a femur of this length? b. Is it possible that the femur belongs to a male of height 172cms? Explain c. Is it possible that the femur belongs to a male of height 160cms? Explain d. Is it possible that the femur belongs to a male of height 150cms? Explain 3. Explain the meaning of the gradient of your line. 4. A new born baby is measured at 40cms: a. What does your equation predict for the length of the baby’s femur? b. Is your equation appropriate for predicting the length of the femur in young children or babies? It is possible to draw more lines around the data. Before fitting another line to the data it is appropriate to return to the original window settings whereby the data fills the screen and the origin is not visible. The window – zoom menu includes an option to zoom in on the data.

Add another moveable line to the data. This time adjust the line so that it sits just under the data points. (Under-fit) If there are any extreme points in the data you need to make a decision as to whether these points are to be included. The data displayed opposite does not contain any such points.

5. Explain why it is useful to zoom in on the data before drawing the second trend line.

6. What is your new equation (under-fit line) for the relationship between male height and male femur length? 7. Explain the meaning of the gradient of this line, and compare it with the gradient of the previous line. 8. Using specimen WF181107B1, measuring 41.9cms: a. What does your under-fit equation predict for the height of a male with a femur of this length? b. Is it possible that the femur belongs to a male of height 172cms? Explain c. Explain, with reference to the gradient of the line and applicable candidates, whether this new equation predicts taller or shorter people for a given femur length. 9. It is possible to include a line that just passes by the top of the data. a. Insert a line that just sits over the top of the data; record the equation to this new line. (Over-fit) b. Using specimen WF181107B1, measuring 41.9cms, what does your new equation (over-fit) predict for the height? 10. Using your three equations; best fit, under-fit and over-fit, specify a suitable range of heights for the victim. It is difficult to accurately predict the height of the victim from a single bone measurement. The previous three equations illustrate this fact. All four bones will be analysed and compared to help determine the most likely height of the victim. A chart is drawn below that will be used to compare the predictions. Using a single line, the range of applicable victim heights for a humerus bone of length 41.9cms can be displayed.

11. On the chart, draw a horizontal line aligned with the femur label that indicates the range of possible heights for the victim. In each of the subsequent investigations you will be required to determine the three equations: • Line of best fit (One that best represents the data) • Under-fit (A line that falls below most/all of the data points) • Over-fit (A line that falls above most/all of the data points.)

Humerus Create another Data and Statistics page and graph the relationship between mhumerus and mheight. 12. What is your equation for the relationship between male height and male humerus length for: a. The line of best fit? b. The under-fit line? c. The over-fit line? 13. Using specimen WF181107B2, measuring 34.1cms: a. What does your line of best fit predict for the height of a male with a humerus of this length? b. What does your under-fit line predict for the height of the victim? c. What does your over-fit line predict for the height of the victim? d. Is it possible that the humerus belongs to a male of height 172cms? Explain e. Is it possible that the humerus belongs to a male of height 160cms? Explain 14. Determine an appropriate range of heights for a male with a humerus bone of length 34.1cms.

15. On the chart, draw a horizontal line aligned with the humerus label that indicates the range of possible heights for the victim. 16. Using the combination of the two lines, humerus and femur, indicate the most likely range of heights for the victim.

Radius Create another Data and Statistics page and graph the relationship between mradius and mheight. 17. What is your equation for the relationship between male height and male radius length for: a. The line of best fit? b. The under-fit line? c. The over-fit line? 18. Using specimen WF181107B3, measuring 23.9cms: a. What does your line of best fit predict for the height of a male with a radius of this length? b. What does your under-fit line predict for the height of the victim? c. What does your over-fit line predict for the height of the victim? d. Is it possible that the radius belongs to a male of height 172cms? Explain e. Is it possible that the radius belongs to a male of height 160cms? Explain 19. Determine an appropriate range of heights for a male with a radius bone of length 34.1cms.

20. On the chart, draw three horizontal lines aligned with the radius, femur and humerus labels that indicate the range of possible heights for the victim. 21. Using the combination of the three lines, humerus, femur and radius indicate the most likely height range for the victim.

Tibia Create another Data and Statistics page and graph the relationship between mtibia and mheight. 22. What is your equation for the relationship between male height and male tibia length for: a. The line of best fit? b. The under-fit line? c. The over-fit line? 23. Using specimen WF181107B4, measuring 31.8cms: a. What does your line of best fit predict for the height of a male with a tibia of this length? b. What does your under-fit line predict for the height of the victim? c. What does your over-fit line predict for the height of the victim? d. Is it possible that the radius belongs to a male of height 172cms? Explain e. Is it possible that the radius belongs to a male of height 160cms? Explain 24. Determine an appropriate range of heights for a male with a tibia bone of length 31.8cms.

25. On the chart, draw all the horizontals lines representing the possible height of the victim using the radius, femur, tibia and humerus. 26. What conclusions can you draw from your findings?

New Evidence Forensics have just discovered some new evidence, they believe victim(s) may have been female. Conduct an investigation, using the same bone samples into the most likely height of the victim. Your investigation needs to include: • Data from your friends, relatives or classmates for bone length, plus the sample data provided. The more data you have the more confident you will be with the height prediction. • Graphs of the data clearly showing the best fit, under – fit and over – fit lines with their equations. • Calculations showing the range of heights using each bone. • A graph showing the range of height predictions using each bone. • A conclusion for the height range of the victim(s).

Teacher Notes – Mathematical CSI: The first thing you should notice about this activity is the lack of specific ‘calculator’ instructions. It is assumed that students performing this activity can: • Enter data in a list • Graph data using a scatter plot • Perform substitutions into formula (Not essential as this component can be done by hand) Encourage students to measure as many people as possible. Homework should be set to measure other family members or friends of different ages and heights in order to provide sufficient data. Increasing the amount of data collected increases the accuracy of the results. For example, in a class of 25 students, if each student measures at least four family members / friends this will provide an additional 100 data points. (Approximately 50 for each gender) Additional data can be collected from other classes, effectively doubling the amount of data. The formulas used by Forensic Anthropologists are: Male h = 69.089 + 2.238 F h = 81.688 + 2.392 T h = 73.570 + 2.970H h = 80.405 + 3.650 R Female h = 61.412 + 2.317 F h = 72.572 + 2.533 T h = 64.977 + 3.144 H h = 73.502 + 3.876 R Adjustments to these formulas are used when trying to determine an accurate estimate of the person’s height. After the age of thirty, the height of a person begins to decrease at the rate of approximately 0.06 cm per year. This shrinkage must be considered when the age of the victim is known. The television show Numb3rs includes lots of similar investigations. The Judgment day episode uses formulas to predict the height of a person based on some bones. The activity connected with the episode is located at: http://education.ti.com/educationportal/activityexchange/Activity.do?cid=US&aId=5915 Warning: Students of some religious backgrounds may find the material in this activity inappropriate.

Exchange Rates

In the years BTI, (Before The Internet), exchange rates mainly concerned business people dealing with other countries and families going on holidays. Since the expansion of the Internet, particularly eBay, millions of people around the world purchase items from other countries.

Activity: In this activity you will determine rules for exchanging money from one currency to another. On line shopping sites may contain a currency conversion tool. Some websites monitor where the signal is being delivered and automatically display amounts in the host currency. For example, if you look in some on-line shops in the US, amounts may be displayed in New Zealand dollars. This is because the website knows where the signal is being sent and applies a currency conversion to the local dollar. Another way some international companies deal with the different currencies is to have a ‘mirror’ site. An example of such a website is the Air New Zealand website: http://www.AirNewZealand.com and http://www.AirNewZealand.com.au Both of these websites are ‘official’ Air New Zealand websites, the later is the Australian website where Australian residents can book flight to New Zealand and see the prices in Australian dollars. For the websites that don’t display this sort of information, you need to know how to convert from one currency to another. If you are planning an overseas holiday you will definitely need to understand the buying power of your dollar. If you are purchasing items directly from shops you will most likely need to convert the amounts without the aid of a computer. Currency exchange rates vary from day to day. Most news broadcasts mention the current exchange rates. News papers, travel agents and the internet are all sources of exchange rates. You can read more about how exchange rates work at: http://money.howstuffworks.com/exchange-rate.htm 1. Your first task is to convert New Zealand dollars (NZD) into US currency (USD). Suppose the exchange rate is: 0.7547. This means for each New Zealand dollar converted, you will receive $0.7547 US dollars. Complete the conversion table below:

New Zealand Dollars 0 100 200 300 400 500

US Dollars

Another way to represent this information is on a graph. Follow the instructions included here to draw a graph of your data. Start a new document; create a text page with your name and activity title on the page.

Insert a spreadsheet and put column headings: • NZD • USD Enter the values: 0, 100, 200, 300, 400, 500 In the NZD column. Enter the equivalent dollar amounts in USD in the USD column.

Insert a data and statistics page. Draw a scatter plot with NZD on the x axis (independent axis) and USD on the y axis (dependent axis).

The graph is useful for showing the nature of the relationship. The scatter plot only shows ‘discrete’ amounts. What happens if you want to know how many USD you would get for $345.00 NZD? Return to the spreadsheet page and add another data point to the list: NZD 345 USD 200 The amount of USD is not the correct amount; it can be adjusted from the data and statistics page.

Return to the data and statistics page. The new point should be obvious as it does not fall exactly into the pattern. Click and grab the new point and move it until it falls into the same pattern as the other points.

2. What is the approximate amount of USD you can purchase for $345.00 NZD? (Using the data and statistics application) 3. Using the currency conversion rate of 0.7547 (NZD – USD), determine the actual amount of USD. It is possible to put a formula in the cells in a spreadsheet. In cell B1 on the spreadsheet page, put the formula: = A1 * 0.7547 Now put appropriate formulas in each of the cells B2 to B6

The advantage of using formulas in a spreadsheet is that new amounts can be computed quickly by changing the amount in the first column. Change the amount in cell A1 to 345 and observe what happens in the amount in cell B1.

Developing a rule It is quicker and easier to determine a rule for doing the conversions, particularly once some of the practical issues associated with currency exchange are included. These complications will be ignored initially. Return to the Data and Statistics page and insert a ‘moveable’ line. Use this moveable line to help determine a rule relating NZD to USD.

NOTE: It is difficult to place the line exactly, observe the rule closely and you may be able to deduce the actual rule.

Another option exists for determining a ‘general’ rule for converting NZD into USD. On the spreadsheet, change the 0 in cell A1 to an x. (Where x represents the amount of NZ dollars to be converted.)

Note: One you have seen the result, return cell A1 back to a zero.

One more option will be explored here for establishing the rule. Even though you may have already established the rule, it is worthwhile completing this task as it increases your options for further explorations. Insert a Graphs and Geometry page. Change the graph type to ‘scatter plot’ and select NZD for the x axis and USD for the y axis.

Change the window settings to match those shown opposite.

In this environment, lines can be attached to points. Use the line tool from the points and lines menu to draw a line through two of the points. The points will flash when they are selected; this ensures the points go exactly through the point. Selecting the point on the y axis (in this case, at the origin) ensures a more accurate equation.

From the actions menu select the Coordinates and equations command. Click on the line and the equation will be displayed. The equation will change only if the points are moved via the spreadsheet. Note: The points drawn via the scatter-plot can’t be moved directly, this option is only available to the data and statistics page.

4. Determine a rule relating NZD (x) and USD (y). Graph your rule to make sure it models the data precisely. Write down the rule you established.

Once the rule has been established, the calculator can be used to convert any amount. Place a point on the line and measure the coordinates of the point. The point can be dragged along the line or values can be typed in directly for x (NZD) or y (USD). Try entering $345.00 NZD to confirm your answer to one of the previous questions.

5. Use your rule to determine how many NZ dollars you would receive for US$250.00 6. An item in an on-line store costs US$126.00. If the same item can be purchased from a New Zealand on line store for NZ$165.00, which item is cheaper? (Both items are delivered free and neither requires payment of any sales tax.) 7. An item in an online store costs US$120.00. The same item costs NZ$215.00. The New Zealand item includes free postage to anywhere in NZ. How much could you afford to pay for postage before the US item would cost you more? 8. The exchange rate between NZ dollars and Australian dollars is 0.9254. This means for each New Zealand dollar you convert, you will receive $0.9254 Australian dollars. Complete the table below:

New Zealand Dollars 0 100 200 300 400 500

Australian Dollars

9. Draw a graph of the currency exchange on the graphs and geometry page. Leave the NZ – US conversion data and graph on the same screen. Write down the equation used to convert NZ dollars into AUS dollars. Include a graph of this equation on your screen. 10. Which graph is steeper? Explain why. 11. What does it mean if the slope of the graph exceeds one? 12. What would you expect the exchange rate from USD to AUD to be? http://education.ti.com/asiapacific Page 28 of 73 © 2008 Texas Instruments Author Peter Fox

13. An item purchased for $200.00 in an overseas currency cost the customer $280.00 in NZ dollars. Use either the internet; travel agents or a newspaper to find at least one country where the item may have been purchased. (Remember exchange rates fluctuate daily, an approximate conversion is sufficient.)

Exchanging currency is not generally done for ‘free’. While some currency exchange operators do not charge an “advertised fee”, their costs are built into the exchange rate. This means they may not give you the same exchange rate as a company that has a set fee for currency exchange.

14. Suppose a Bank carries a currency exchange fee of $28.00. This means, if wanted to convert $100.00 NZD into USD, the fee of $28.00 (NZD) would be deducted first. Only $72.00 NZD would be converted. The bank however offers a better conversion rate than that previously quoted. Use a currency exchange rate of 0.850 Determine how many US dollars you would receive.

15. Complete the following table for the conversion from NZD to USD taking into account the conversion fee of $28.00 and the more favourable exchange rate of 0.850

New Zealand Dollars 0 100 200 300 400 500

US Dollars

16. Draw a graph of the conversion amounts. (Remove the AUD graph) 17. Determine an equation for the currency conversion for the Bank. 18. Compare the amount of USD you would receive for converting each of the following amounts: a. $200.00 using both the currency exchange centre and the bank. b. $300.00 using both the currency exchange centre and the bank. c. Which provider gives you more USD for your NZD? 19. At what point is it cheaper to pay the bank fee of $28.00 to perform the currency exchange?

Teacher Notes – Exchange Rates There are a variety of methods presented in this activity for determining the appropriate equation for the data. Teachers may elect to have students complete only one method. The intention is to provide students with a number of solution processes and present the information in a variety of ways. This task could easily be turned into an extended investigation requiring students to plan a brief work visit. The internet provides a range of opportunities: • Airfares • Hotel prices • Tours • Gifts • Transport It is easy to obtain prices for all of the above without leaving your computer screen. The advantage of doing the research on-line is that for items such as international hotels, their prices are often quoted in ‘local’ dollars; depending on the website. Students can also be required to obtain a list of the fees associated with purchasing goods from overseas. Credit card transactions are not exempt from international fees. Whilst credit card exchange rates are quite competitive, a premium is paid for every transaction. Any business person will be familiar with the international transaction feeds attached to every credit card purchase made overseas. Another opportunity for extension is to ‘visit’ more than one country and have the money exchanged in each venue. For example: NZD – USD – AUD Consider the following: • Transaction fee: $28.00 every time a currency is exchanged. (For amounts up to $10,000) • NZD – USD currency exchange rate: 0.7547 • USD – AUD currency exchange rate: 1.032 Suppose a tourist starts off with $1000.00 NZD and travels to the USA where the money is exchanged to local currency. The tourist does not spend any money in the USA but then needs to exchange the USD into AUD. How much will they have in AUD? The tourist buys a camera in the US for $500.00 (USD) and then sees the same camera in Australia for $530.00 (AUD), which place is the camera cheaper?

Further extension opportunities exist to explore what happens when exchange rates vary. For example, you purchased a MP3 player on-line for $150.00 USD last week. The exchange rate just changed from 0.7547 to 0.8421, how much is would your MP3 player cost now. (Given it is still the same price in the USA). These types of questions can then lead to a discussion of ‘global economies’.

Time Zones

The world is becoming smaller! The physical size of the globe remains constant, but the abilitiy to communicate on an international basis has meant global relationships for both business and pleasure are becoming more common.

Activity: Regular and frequent global communications include air travel, phone calls, satelite transmissions, email and an enormous array of business transactions. It is therefore necessary that most people develop an ability to perform appropriate time conversions. Most international businesses executives include a signature on their email. The signature generally includes something like: GMT + 10. What does GMT+10 mean and why is it included in the email? New Zealand, technically speaking, is the most advanced country in the world! New Zealand is the first major country to see in the new day, every day of every year. Monday morning in New Zealand is Sunday mid – morning on the west coast of the United States. What time is it London? What time is it in Tokyo? Time zones are well documented and clearly indicated on world maps. Even experienced business executives can incorrectly calculate the time in another country. For a simple telephone conversation between friends, this is not such a problem. If the telephone conversation is part of a business meeting it poses a more significant problem. For simplicity, this investigation ignores problems such as day light saving. For example, daylight saving in New Zealand ends on 19th March. Three time zone hours away in Melbourne (Australia), daylight savings ends on 2nd April. The equations generated in this investigation will NOT take into consideration daylight savings times. 1. Your first task is to construct a table of values for converting Wellington – New Zealand (local) time into Melbourne – Australia (local) time. Melbourne is three hours behind Wellington. Complete the table of values for the time conversions. (24hr time) Note: For times ‘before’ midnight a negative value is used to indicate ‘previous day’.

Wellington 0 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 Time

Melbourne -300 -200 -100 Time

Wellington 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 Time

Melbourne

Time

2. Produce a graph of the data using an appropriate application. 3. Use the graph to locate midday in Wellington and find out the corresponding time in Melbourne. http://education.ti.com/asiapacific Page 31 of 73 © 2008 Texas Instruments Author Peter Fox

4. Use the trace feature to determine the time in Wellington if it is 6:00pm in Melbourne.

In 1884 a meeting of representatives from 25 nations decreed that world times would be measured against a single ‘standard’. The meridian, passing through the observatory in Greenwich was set to be zero. All other meridians would be measured from this point. Time zones at 15o intervals that approximately follow the meridians, allowing for country borders, were established. Each time zone is specified as being a certain number of hours ahead, or behind, Greenwich. For example, Melbourne is located at GMT + 10, which signifies that it is 10 hours ahead of Greenwich.

Time Zones: 5. Due to the declaration signed in 1884, it makes sense to produce graphs of time zones relative to Greenwich. Time zones for a range of locations have been provided. Complete a table of values and subsequent graph for each location. Use Greenwich times on the x – axis and local times at each of the destinations on the y – axis.

Wellington GMT + 12

Greenwich 0 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 Time

Wellington

Time

Greenwich 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 Time

Wellington

Time

Beijing GMT + 8

Greenwich 0 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 Time

Beijing

Time

Greenwich 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 Time

Beijing

Time

New York GMT - 5

Greenwich 0 0100 0200 0300 0400 0500 0600 0700 0800 0900 1000 1100 1200 Time

New York

Time

Greenwich 1300 1400 1500 1600 1700 1800 1900 2000 2100 2200 2300 2400 Time

New York

Time

6. Determine the equation to for each of the graphs plotted in the previous question. 7. Without producing a table of values, determine the equation for each of the following locations: a. Melbourne GMT + 10 b. Moscow GMT + 3 c. London GMT – No offset d. Los Angeles GMT – 8 8. Explain why each graph has the same slope. 9. Suppose, at a meeting 25 nations it is decided that all times should be measured against Wellington local time. The new system is called: WMT. Sketch a time conversion graph for each of the previous locations. 10. Under the GMT system, what happens to the graph for Wellington during day – light savings? Note: Clocks are wound forward one hour at the start of daylight savings. Extension Questions: 11. An aeroplane is flying from Melbourne to Wellington. The flight leaves Melbourne at 6:30pm and takes 3 hours to reach Wellington: a. What time is it when the plane touches down? b. Sketch a graph of the time on the aeroplane against Wellington Local Time. 12. An aeroplane is flying from Wellington to Melbourne. The flight leaves Wellington at 6:30am and takes 3 hours to reach Melbourne: a. What time is it when the plane touches down? b. Sketch a graph of the time on the aeroplane against Melbourne local time. 13. An aeroplane flies from Melbourne to Beijing. If the flight leaves Melbourne at 8:00am and takes 6 hours, what time will the plane touch down in Beijing? 14. An aeroplane flies from Wellington to Los Angeles. If the flight leaves Wellington at 8:00am (Wednesday Morning) and takes 12 hours, what time will the plane touch down in Los Angeles? Comment on your answer. 15. Where is Greenwich?

Teacher Notes – Time Zones This activity contains NO screen shots to show students (or teachers) how to operate Nspire; this has been done to see if students are able to use Nspire as a learning / exploratory tool independently. It is important that this be done on occasions as it helps identify how comfortable the students are with using the platform.

Time Zone History GMT dates back to October 1884. At the request of the President of the United States of America, 41 delegates from 25 nations met in Washington DC for the International Meridian Conference. At the Conference the following important principles were established: • It was desirable to adopt a single world meridian to replace the numerous one's already in existence, • The Meridian passing through the principal Transit Instrument at the Observatory at Greenwich was to be the 'initial meridian', • That all longitude would be calculated both east and west from this meridian up to 180°. • All countries would adopt a universal day, • The universal day would be a Mean Solar Day, beginning at the Mean Midnight at Greenwich and counted on a 24 hour clock, • Nautical and astronomical days everywhere would begin at mean midnight, • All technical studies to regulate and extend the application of the decimal system to the division of time and space would be supported, • Resolution 2, fixing the Meridian at Greenwich was passed 22-1 (San Domingo voted against), France & Brazil abstained.

Decathlon

Decathlon sports are an ultimate test of an athlete’s speed, strength, agility and endurance. Decathlon is a two-day track and field meet that encompasses 10 individual athletic sport events -- 100-meters, 400- meters, 1500-meters, 110-meter high hurdles, javelin, discus, shot put, pole vault, high jump, and long jump.

Activity: The Decathlon is a track and field event for men with five events on each of two days. An equivalent for women is the heptathlon consisting of four events on the first day and three on the second day. The decathlon has been included in the Olympic games since 1912. Points are awarded for each event. The events include:

Track: 100m, 400m, 1500m 110m Hurdles

Long High Pole Field: Shotput Discuss Javelin Jump Jump Vault

Points are scored for each event. The person with the most points at the end of the two days is the winner. The challenge for scoring the event is trying to compare winning heights or distances with winning times. Suppose you are a champion competitor in the decathlon. On the first day of competition you win the high jump with a height of 2.25m. Your next closest competitor cleared 2.05m. You are the clear winner. In the 100m you run a time of 10.9s but the winner obtained a time of 10.1s. At the conclusion of these two events who should be in the lead? In the high jump you jumped almost 10% higher than your competitors. In the 100m your time was almost 10% slower. Is this a fair comparison? A series of formulas exist for the decathlon. Each event has a specific formula that converts the time, distance or height into a quantity of points. At the conclusion of the event, the person with the highest score wins. The formula for each event is included below: Track 100m: 25.4347×− (18x )1.81 where x = time

400m: 1.53775×− (82x )1.81 where x = time

1500m: Round-down(0.03768×− (480 x )1.85 ) where x = time

110m Hurdles: Round-down(5.74352×− (28.5 x )1.92 ) where x = time

Field Long Jump: 0.14354×− (x 220)1.4 where x = distance

High Jump: 0.85465×− (x 75)1.42 where x = height

Shot Putt: 51.39×− (x 1.5)1.05 where x = distance

Javelin: 10.14×− (x 7)1.08 where x = distance

Discus: 12.91×− (x 4)1.1 where x = distance

Each of the point scoring equations for the decathlon can be entered as a ‘function’ on the calculator. The points for each event can be calculated using the corresponding formula.

Function definitions are very useful when an equation will be used repeatedly. When a function is defined it remains in the calculator after a ‘clear home’ command and also after ‘clear a – z’. Using a function makes substitution and solving quick and easy.

Start a new document and create a title page (as shown opposite). Once the title page is complete insert a calculator page. This page will be used to store the rules for the decathlon.

To make the functions easy to recognise, each one should have a meaningful name, the One Hundred Meter function can therefore be defined as: ohm( x )=×− 25.4347 (18 x )1.81 The define command can be typed in directly using the keyboard or accessed through the Actions menu. Define ohm( x )=×− 25.4347 (18 x )1.81

To check how many points is scored for running the One Hundred Meters sprint in 11 seconds type: ohm(11) Compare this result with running the One Hundred Meters in 12 seconds. ohm(12)

Suppose an athlete wants to work out how fast they need to run in order to earn 900 points. From the result of ohm(11) it can be seen that they need to run faster than 11 seconds, but how much faster? Use the algebra menu and type: solve(ohm(x) = 1000,x) As x represents time, this statement reads as “solve, one hundred meters result equals 1000pts, solve for time (x)” Define functions for each of the events using the rules provided:

One Hundred Meters Four Hundred Meters define ohm( x )=×− 1.53775 (82 x )1.81 define fhm(x)=1.53775×− (82 x )1.81

One Thousand Five Hundred Meters Hurdles define ofhm( x )=×− int(0.03768 (480 x )1.85 ) Define hur( x )=×− int(5.74352 (28.5 x )1.92 )

Long Jump Shot Putt define lj( x )=×− 0.14354 ( x 220)1.4 define sp() x=×− 51.39( x 1.5)1.05

Javelin Pole Vault define jav() x=×− 10.14( x 7)1.08 define pv( x )=×− 0.2797 ( x 100)1.35

Discuss High Jump define dis(x)=12.91×− ( x 4)1.1 define hj(x)=0.85465×− ( x 75)1.42

Round – Down means to drop the decimal place. If a calculation returns a result of 894.89, the result is rounded down to: 894. To achieve this on the calculator

use the integer command: int( )

2004 Olympic Games 1. In the 2004 Olympics, Dmitriy Karpov came third. His times, distances and heights for each event are listed below in the table. Determine how many points he scored for each event and the subsequent total points for the competition.

High Long Shot Pole Event 100m 400m 1500m Hurdles Javelin Discuss Jump Jump Putt Vault

Result 10.5s 46.81s 4:38.11s 13.97s 2.09m 7.81m 15.93m 55.54m 51.65m 4.60m

Score

The ‘solve’ command on the calculator can be used to determine the solution to a problem. For example, if you need to work backwards from a score in order to determine a time you could type: solve(()800,) ohm x= x This instruction would determine how fast you would need to run the 100m (OHM) event in order to score 800 points. 2. At the 2004 Olympic Games, Bryan Clay from the USA scored a total of 8820 points to finish second in the Decathlon. His score for each event is included in the table below. Determine the time / distance / height recorded for each event.

High Long Shot Pole Event 100m 400m 1500m Hurdles Javelin Discuss Jump Jump Putt Vault

Result

Score 989 852 670 958 859 1050 804 885 873 880

3. Roman Sebrle won the Decathlon at the 2004 Olympic Games with a score of 8893pts.

High Long Shot Pole Event 100m 400m 1500m Hurdles Javelin Discuss Jump Jump Putt Vault

Result 48.36s 2.12m 16.36m 5.00

Score 894 968 1020 897 844

a. Determine the missing scores, times and distances that earned him a gold medal. b. Prior to the last event in the decathlon (1500m), Dmitriy was on 8033pts and Bryan Clay was on 8150pts. Roman Sebrle was confident of achieving a time of 4min 45sec, or better. i. What score would Roman have achieved overall if he completed the 1500m in a time of 4min 45sec? ii. If Roman had finished the 1500m in 4min 45sec, what time would Dmitriy have needed to complete the 1500m event in order to win the decathlon? iii. If Roman had finished the 1500m in 4min 45sec, what time would Bryan have needed to complete the 1500m event in order to win the decathlon? iv. How safe was Roman’s gold medal heading into the last event? Note: The winner of the 1500m in the decathlon finished in 4min 23sec. v. If Roman had finished the 1500m in 5 minutes (exactly), what would have been the times required by Dmitriy or Bryan to snatch the gold medal. 4. Suppose, prior to the Olympics, a competitor figured a score of 9000pts would win the decathlon. If the athlete was ‘perfectly’ consistent across all events, determine the time / distance / height for each event required to win the event.

Investigating the 100m Event The 100m sprint in the Olympics is generally won in a time close to 10 seconds. Competitors in the decathlon are not generally able to perform at this level due to the large number of events for which they must train. It is worthwhile investigating how much difference is made to the score for small changes in time. Formula for scoring the 100m event: 25.4347×− (18x )1.81 where x = time 5. Complete the score table below:

Time 10.0 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 11.0

Score

Note: Most decathlon competitors complete the 100m sprint in a time of 10 to 11 seconds. a. Does a 0.1 second improvement in time equate to 25 point increase in the score? (Explain) b. Draw a graph of time versus score for the 100m event for range of times produced in the table. (above) c. Draw a graph of time versus score for the 100m event for times ranging between 9 seconds and 20 seconds. 6. A straight line graph is to be used to ‘approximate’ competitor’s scores. Most competitors complete the 100m event in a time ranging between 10 seconds and 11 seconds. a. Select two points from the table for your straight line equation to pass through. b. Determine the equation to the straight line. c. Complete the score table using your straight line equation.

Time 10.0 10.1 10.2 10.3 10.4 10.5 10.6 10.7 10.8 10.9 11.0

Score

d. Compare the scores produced using your ‘approximate’ rule and the rule used by the officials.

A Balanced Score Suppose an athlete considers themselves to be a ‘perfect’ all rounder. The plan is to achieve the same score in each event. With ten events and a target score of 8,000 points, the aim is to get a score of 800 points in each event.

Insert a Graphs and Geometry page. Enter the One Hundred Meter function in the equation editing line: f1(x) = ohm(x) Enter the Hurdle function in for f2: f2(x) = hur(x) Set the window settings as: xmin = -4 xmax = 25

ymin = -200 ymax = 1500

Use the Points and Lines menu to place a point on the graph of the One Hundred Meters race. (ohm(x)) The coordinates of the point are displayed automatically. Press [esc] to release the point tool. Click once on the y coordinate of the point, this selects the value, click again to edit the value. Change the y value to 800.

Use the Construction menu to create a Perpendicular line. Click on the point followed by the y axis. Any point along this line will have the same y value. From the Points and Lines menu, select Points of Intersection. Mark a point of intersection between f2(x) = hur(x); the hurdles function.

Measure the coordinates of this point using the Actions menu – Coordinates and Equations

7. According to your measurements, how fast will the all rounder need to complete the hurdles race in order to obtain a score of 800 points? 8. Enter the four hundred meter equation for f3. Change the window settings so that xmin = -10 and xmax = 70 a. Extend the horizontal line across to meet the four hundred meter line and determine the time required for this event in order to achieve a score of 800 points. b. Suppose an athlete needs to obtain a score of 900 points for each individual event. Adjust the original coordinate and hence determine the time required for each event.

The equations turn back up, just like a parabola. (They are NOT parabolas). This ‘problem’ is caused by an inappropriate selection for the scoring system. The error can be corrected by limiting the function. For example, the hurdles rule is not supposed to score if the time is longer than 28.5 seconds. Recall the function definition, at the end of the definition put: |x<28.5 This means the function will only be defined when x < 28.5

An example of the above tip is shown opposite. The result on the graph page can be seen clearly. 9. Suppose the perfect athlete completes the one hundred meter event in 11 seconds. If the other events received the same score: a. What time did the athlete run the hurdles in? b. What time did the athlete run the four hundred meters event in? c. Explain why each of these three graphs slopes downwards. d. By considering the previous answer, which way will all of the field events slope? e. Look carefully at the field event equations and compare them to the track event equations. Describe the difference between these sets of equations, and corresponding graphs.

Investigation: Approximating Individual Event Scores Draw a graph of each event’s scoring rule. Comment on the suitability of approximating the scoring system using a ‘linear’ relationship. (Straight line graph) Your comments should include a comparison between approximated score and the actual score using the official formula. Your comments should reflect the typical times, distances and heights scored in the Decathlon at the Olympic level.

Teacher Notes – Decathlon The bbc news website contains very detailed information about the 2004 Olympic decathlon, including individual scoring for each event combined with progressive totals.

Visit: http://news.bbc.co.uk/sport1/hi/olympics_2004/athletics/results/3532992.stm

2004 Olympic Games Results – Men’s decathlon

First Second Third Roman Sebrle (Cze) 8893pts Bryan Clay (USA) 8820pts Dmitriy Karpov (Kaz) 8725pts Olympic Record Personal Best Area Record

100m: (10.85s, 894pts) 100m: (10.44s, 989pts) 100m: (10.50s, 975pts) Long Jump: (7.84m, 1020pts) Long Jump: (7.96m, 1050pts) Long Jump: (7.81m, 1012pts) Shot: (16.36m, 873pts) Shot: (15.23m, 804pts) Shot: (15.93m, 847pts) High Jump: (2.12m, 915pts) High Jump: (2.06m, 859pts) High Jump: (2.09m, 887pts) 400m: (48.36s, 892pts) 400m: (49.19s, 852pts) 400m: (46.81s, 968pts) 110m hurdles:(14.05s, 968pts) 110m hurdles:(14.13s, 958pts) 110m hurdles:(13.97s, 978pts) Discus: (48.72m, 844pts) Discus: (50.11m, 873pts) Discus: (51.65m, 905pts) Pole Vault: (5.00m, 910pts) Pole Vault: (4.90m, 880pts) Pole Vault: (4.60m, 790pts) Javelin: (70.52m, 897pts) Javelin: (69.71m, 885pts) Javelin: (55.54m, 671pts) 1500m: (4:40.01s, 680pts) 1500m: (4:41.65s, 670pts) 1500m: (4:38.11s, 692pts)

Extension: Functions A single rule can be defined to determine the score for the entire decathlon. A different variable must be used for each event. For example d(x,y,t,h,l,…) = 25.4347×− (18x )1.81 + 1.53775×− (82y )1.81 …x would represent the time for the 100m, y would represent the time for the 400m and so on. This single rule would need all ten measurements to determine the final score. Alternatively, if the final score is ‘known’ and one event remains, it would be possible to determine the required time / distance / height for the single event.

The Cube Factor

What is the value of mathematical formulae? The value of formulae is not as an incantation, but that in all subjects we can use abstraction and logic to solve a whole range of problems rather than solving them one at a time. In this challenging problem, you will be required to identify a particular solution, it is the general solution however that is more powerful and the means by which you arrive at the formula that is most powerful; a journey that should involve mathematical thinking.

An easier Problem: A good problem solving technique is to start with an easier problem. The easier problem can lead to an insight into a more complex situation. In this problem solving task you will start with a familiar dice. An understanding of this problem will lead to an understanding of the more complicated problem. 1. What number is opposite the 2 on a die? 2. What number is opposite the 4 on a die? 2 3. What number is opposite the 1 on a die?

Imagine the corners of the die also contained a number; this number is equal to the product of This corner is the three sides forming the corner. Two of the equal to: corners have already been computed below. 1 × 2 × 4 = 8 4. What numbers will be on the front corners of the die; as displayed? 5. What numbers will be on the back corners of the die; as displayed? 2 This corner is 6. What is the sum of the eight corners? equal to: 7. Is your answer a ‘perfect cube’? 2 × 4 × 6 = 48

Problem to Solve The corners of the die shown are equal to the product of the numbers on the adjacent sides. The sum of the corners is equal to 1001. 8. Identify 6 numbers that could appear on the sides of the die. 9. Identify another set of numbers that will satisfy this condition. 10. How many solutions are possible? a

Teacher Notes – The Cube Factor The cube factor is one of a handful of examples included in this booklet that focuses on students using algebra to solve a problem. It does not necessarily require students to be able to perform the algorithms on which the solution is based; rather focuses on establishing the associated algebraic equations and expressions that help in the determination of a solution. In this problem, students see the use of factorising. In the case of the dice, the sum of the numbers on each vertex is equal to 73. The significance of this result is that the sum of the opposite faces on a regular dice is 7. This extends to the next problem where the sum of the numbers on each vertex is 1001. While this number is not a perfect cube, it has exactly three prime factors. The three prime factors represent the sum of the numbers on opposite faces of the dice. One very important characteristic of this problem solving task lies not just in the acknowledgement of the importance of factorising, but also in student identification of the generalisation that algebra represents in the solution. It is an important component of this task that teachers seek student ideas on the number of solutions. It has been my experience that students focus on integer solutions. For example; the sum of the opposite sides on the dice needs to be 7, 11 and 13. It is therefore possible that opposite sides of the dice could be paired up as: (1, 6); (5, 7); and (9, 4), where the numbers in the brackets represent numbers on opposite sides of the dice. In this case none of the sides of the dice share the same number. This may be a restriction that the students impose, but is not stated explicitly in the question. Student may pick this restriction up implicitly from the original problem involving a regular dice. For students that do not necessarily make the connection with the dice, they may suggest that other solutions such as: (3, 4); (5, 7) and (4, 9) are also acceptable. In this case the number 4 appears on two different sides. One thing that is consistent through these solutions is that they are also integer solutions. It is important for teachers to allow students to explore possible answers and also to discuss any ‘assumptions’ that student may have made about acceptable solutions. As students start to check solutions a question that is often asked is “can we include zero?” The answer is of course “Yes”. The question relating to the inclusion of zero may lead to students checking if negative numbers are allowable, to which the answer is also “Yes”. In my experience, I have not yet had a student ask about decimal numbers: (2.9, 4.1); (3.2, 7.8) and (9, 4). When this solution is suggested to students they believe that it will ‘not’ work since multiplication of these decimal numbers will only lead to more decimal numbers and not a whole number. This restriction is not indicated in the algebraic solution. Students should be encouraged to check the solution; indeed some adults feel the necessity to check the solution. This helps students see that using an ‘a’ or ‘x’ provides a truly general answer.

What’s the difference?

Perhaps the best indication of what Archimedes truly loved most is his request that his tombstone include a cylinder circumscribing a sphere, accompanied by the inscription of his amazing theorem that the sphere is exactly two-thirds of the circumscribing cylinder in both surface area and volume!" As long as algebra and geometry have been separated, their progress have been slow and their uses limited, but when these two sciences have been united, they have lent each mutual forces, and have marched together towards perfection. Joseph Louis Lagrange, 1736 – 1813 French Mathematician

Linking algebra and Geometry: There are clear and common links between algebra and geometry. In this activity you will identify a common algebraic form that is used extensively and at the same time demonstrated elegantly using geometry.

Instructions Step 1. To begin, cut out a square piece of paper 20cms × 20cms. Step 2. Fold the piece of paper along the diagonal of the square. Step 3. With the paper still folded, measure 5cms along the base of the triangle. Cut this similar triangle off. (As shown in step 3 below) Step 4. Unfold the paper and cut along the fold.

You will now have three pieces of paper, one small square and two trapeziums. Rearrange the two trapeziums to form a rectangle.

5cms

Step 1 Step 2 Step 3 Step 4

1. What is the area of the original square? 2. What is the area of the square that was removed? 3. What is the area of the shape (before cutting) in step 4? 4. When the two trapeziums are arranged to form a rectangle; a. What are the dimensions of the rectangle? b. What is the area of the rectangle? http://education.ti.com/asiapacific Page 46 of 73 © 2008 Texas Instruments Author Peter Fox

General Problem Investigating the general solution to the previous problem provides a more powerful insight to a range of mathematical processes. 5. Suppose the original dimensions of the piece of paper were a cms × a cms. What would be the area of this piece of paper? 6. A square is cut out from the corner, as per step 3. This square has length b cms. a. What is the area of the square removed? b. What is the area of the shape remaining? 7. The trapeziums in step 4 are used to form a rectangle; a. What is the length of the rectangle? b. What is the width of the rectangle? c. Write an expression for the area of the rectangle. d. Compare this area with that obtained for question 6b. e. Write an equation relating the two area expressions from 7c and 7d. 8. Andrew is very good at mathematics and notices a pattern when he is squaring two digit numbers ending in 5, for example: 452. Andrew has discovered a quick way to do the calculation. “Multiply 40 and 50 then add 25 to the result.” ie: 40 × 50 + 25 = 2000 + 25 452 = 2025 Andrew says you can do this will all two digit numbers ending in 5. “If you want to calculate 652; 60 × 70 + 25 = 4200 + 25 = 4225.” Andrew’s friend Jan wants to know where the 60 and 70 come from.” Andrew replies, “you go down from 5 from 65 and up 5 from 65. That gives you 60 and 70; and 60 × 70 is no harder than 6 × 7.” a. Use Andrew’s method to calculate the answer to 852. (Check your answer with a calculator.) b. Use Andrew’s method to calculate the answer to 752. (Check your answer with a calculator.) c. Use algebra to show that Andrew’s system does work. 9. Dale is a friend of Andrew’s and has watched Andrew do this mental arithmetic a number of times. Dale tries to apply this method to other problems. To test his idea he tries 222. Instead of going up / down by 5, he tries going up and down by 2. This leads Dale to the following: 222 = 20 × 24 + 22. Dale figures that 20 × 24 is no harder than 2 × 24, therefore 222 = 484. Dale decides to test another: 242 = 20 × 28 + 42 Once again Dale does the mental arithmetic: 560 + 16 = 576. a. Using Dale’s method, show how he would calculate 232 (Check your answer with a calculator) b. Use Dale’s method to calculate 312. (Check your answer with a calculator) c. Use algebra to show that Dale’s method also works. 10. Explain how the paper folding problem is similar to Dale’s technique.

Teacher Notes – What’s the difference? The purpose of this activity is to introduce the difference of perfect squares in a visual way. Students can see the geometrical relationship, one square minus another square. Students are generally surprised that the two trapeziums can be placed together to form a rectangle. Using numbers in the rule provides a simple application of the difference of perfect squares. If this activity is combined with some of the Vedic mathematics techniques, students will begin to develop a range of strategies to multiply large numbers without the aid of a calculator. Students should also be given the opportunity to explore some of these options for themselves. For example, the technique used to determine 652 = 60 × 70 + 25 = 4225; can this be used to find for example: 1252? 1252 = 120 × 130 + 25 This problem looks easier as: 1252 = 12 × 13 × 100 + 25 1252 = 15600 + 25 1252 = 15625 Another way to explore this problem: 1252 = 100 × 150 + 252 1252 = 15000 + 625 1252 = 15625 If the problem is a little harder:

1852 = 100 × 270 + 852 (185 – 85 = 100 and 185 + 85 = 270, but the 852 may be difficult)

1852 = 100 × 270 + 80 × 90 + 25 (Students can break down 852 using previous techniques) 1852 = 27000 + 7200 + 25 1852 = 34225 This problem may be tackled an easier way:

1852 = 170 × 200 + 152 (185 – 15 = 170 and 185 + 15 = 200, 200 is easy to multiply) 1852 = 34000 + 225 (152 should, by now be committed to student memory) 1852 = 34225 This selective technique makes other problems easier: 3752 = 350 × 400 + 252 3752 = 350 × 2 × 2 x 100 + 625 3752 = 700 × 2 × 100 + 625 3752 = 140000 + 625 3752 = 140625 As ‘trivial’ as these techniques may appear, it has numerous benefits: o Increases student numeracy o Develops ownership as students choose the best way to calculate the answer o Includes another form of ‘factorising’.

Vedic Mathematics

People who can perform astonishingly long or complex calculations in their head are often revered. History is filled with people that were identified as ‘human calculators’. Karl Friedrich Gauss was a mathematician ranked with Newton and Archimedes. History states that at the age of three Gauss corrected is father’s mistake in a payroll reckoning. A variety of mental arithmetic techniques have been used by these mathematicians. Vedic mathematics is one such technique. It consists essentially of sixteen sutras. These principles (sutras) are general in nature and can be applied in many ways.

Introduction to Vedic Mathematics The examples provided here show a technique used to multiply some two and three digit numbers. If you practice this technique for a while you will be able to determine the answer to questions such as: 98 × 97 in your head within seconds.

100 98 × 97 = ? 100 is selected as the reference number, it is close to 98 & 97 and is easy to multiply.

2 × 3 = 6 100 – 98 = 2 and 100 – 97 = 3 Multiply these two numbers: 2 × 3 = 6

95 × 100 = 9500 97 – 2 = 95. Multiply this result by the reference number: 100, this gives 95 × 100 = 9500

98 × 97 = 9506 Add the two results together: 9500 + 6 = 9506

Here is another example, using 94 × 98:

100 94 × 98 = ? 100 is selected as the reference number, it is close to 94 & 98 and is easy to multiply.

6 × 2 = 12 100 – 94 = 6 and 100 – 98 = 2 Multiply these two numbers: 6 × 2 = 12

92 × 100 = 9200 98 – 6 = 92. Multiply this result by the reference number: 100, this gives 92 × 100 = 9200

94 × 98 = 9212 Add the two results together: 9200 + 12 = 9212

A final example, using 92 × 97:

100 92 × 97 = ? 100 is selected as the reference number, it is close to 94 & 98 and is easy to multiply.

8 × 3 = 24 100 – 92 = 8 and 100 – 97 = 3 Multiply these two numbers: 8 × 3 = 24

89 × 100 = 8900 97 – 8 = 89. Multiply this result by the reference number: 100, this gives 89 × 100 = 8900

92 × 97 = 8924 Add the two results together: 8900 + 24 = 8924

1. Use the Vedic mathematics technique to do the following multiplication problems: a. 97 × 91 = b. 95 × 99 = c. 92 × 96 = d. 91 × 97 = e. 89 × 99 = f. 84 × 98 = g. 86 × 96 = Comment on why this last question is more difficult than the previous questions.

This number technique can also be used on other numbers, including multiplying numbers larger than 100. The example below shows the calculations for 102 × 104

Reference Working out in this problem is done ‘upwards’. The numbers are larger than the reference number so they are written above. Number

106 × 100 = 10600 106 = 102 + 4 which is the same as 106 = 102 + 4; and 106 × 100 = 10600

2 × 4 = 8 102 – 100 = 2 and 104 – 100 = 4. Multiply these two numbers; 2 × 4 = 8

100 102 × 104 = 10608 102 × 104 = 10600 + 8 = 10608

Another example: 103 × 112

Reference Working out in this problem is done ‘upwards’. The numbers are larger than the reference number so they are written above. Number

115 × 100 = 11500 115 = 103 + 12 which is the same as 115 = 103 + 12; and 115 × 100 = 11500

3 × 12 = 36 103 – 100 = 3 and 112 – 100 = 12. Multiply these two numbers; 3 × 12 = 36

100 103 × 112 = 11536 103 × 112 = 11500 + 36 = 11536

2. Now its time for some practice; use the Vedic mathematics technique to multiply the numbers below: a. 105 × 109 b. 113 × 102 c. 109 × 104 d. 107 × 111 e. 128 × 102

3. Suppose the two numbers to be multiplied are a and b. These two numbers are less than 100. (ie: The first technique is used) a. Write an expression for the difference between a and 100. b. Write an expression for the difference between b and 100. c. Write an expression for the product of your answers to questions 3a and 3b? d. Show algebraically that the Vedic mathematics technique works.

The Vedic mathematics technique works for numbers close to 10.

This number technique can also be used on other numbers, including multiplying numbers larger than 100. The example below shows the calculations for 102 × 104

Reference Working out in this problem is done ‘upwards’. The numbers are larger than the reference number so they are written above. Number

26 × 10 = 260 26 = 19 + 7 which is the same as 26 = 17 + 9; and 26 × 10 = 260

9 × 7 = 63 19 – 10 = 9 and 17 – 10 = 7. Multiply these two numbers; 9 × 7 = 63

10 19 × 17 = 323 19 × 17 = 260 + 63 = 323

4. Show using algebra that the Vedic mathematics techniques will also work for two digit numbers close to 10.

Note: Whilst these techniques work for all numbers, the multiplication becomes more complicated when numbers further from the reference number(s) are used. For example, 82 × 77 would involve problems almost as complex such as: 18 × 23. To solve this problem extra steps are introduced.

Teacher Notes – Vedic Mathematics The techniques of Vedic mathematics have been around for a long time. There are many books that use these techniques, such as Speed Mathematics (Bill Handley); many of them use variations of the techniques of Vedic Mathematics. While the students may be impressed by the simplicity of the techniques, it is an opportunity to demonstrate algebraically that this system will always work. The first technique explored is multiplication of numbers ‘just under’ 100; for example 98 × 97. The method can be mirrored algebraically:

100 94 × 98 = ? a × b = ?

6 × 2 = 12 100 – a × 100 – b = (100 – a)(100 – b)

92 × 100 = 9200 b – (100 – a) × 100 = 100(b – (100 – a))

94 × 98 = 9212 a × b = ab

There are several interesting aspects to the algebraic ‘proof’ of the Vedic Mathematics technique: o It is not obvious how the two expressions: (100 – a)(100 – b) and 100(b – (100 – a)) add together to get ab. It is worthwhile have students expand the expressions first, either manually or using the CAS. It is only then students see how the terms cancel and leave ab. o In the algebraic proof, values are not restricted for a and b. It therefore follows that if a is more than 100; and similarly with b, calculations such as 100 – a and 100 – b will result in negative numbers. When a negative is multiplied by another negative number the result is positive and hence the same technique is applicable to numbers larger than 100. This also leads to the observation that if one of the numbers is larger than 100 and the other is smaller, the product must be subtracted.

100 × 102 = 10200 104 – 2 = 102. Multiply 102 by the reference number: 102 × 100 = 10200 Note: 98 + 4 = 102

4 104 is 4 above 100

100 104 × 98 =

–2 = - 8 98 is 2 below 100. The +4 and – 2 are multiplied: 4 × –2 = – 8

10192 Subtract 8 from 10200 to get the answer: 10200 – 8 = 10192

o Further to the realisation that a and b have no restrictions leads to the conclusion that this technique works for 67 × 78. Students can explore this as an option and see why it is not ideal to use this technique to compute the answer; further techniques are applicable for such problems. o The algebraic solution can be modified easily to accommodate for numbers close to 10 or indeed any other number such as 20, 1000 and so forth. o Students could explore options for 14 × 97. Is it possible to have two reference numbers?

Indices

Indices represent a shorthand technique of communicating otherwise very large expressions; for example b5 = b × b × b × b × b. Imagine having to write b23. While the shorthand way of representing such expressions is convenient, it leads to a whole new set of simplification processes, for example, it is obvious that 8 ÷ 4 = 2, but less obvious that b3 ÷ b2 = b

Activity The purpose of this activity is to identify a set of rules that are applicable to working with index notation so as to avoid having to represent expressions such as: b5 in longhand form. An understanding of the order of operations is essential before commencing the activity. Complete the introductory numerical problems before starting the investigation.

Introductory Problems: i) 5 × 12 ÷ 3 × 2 = ii) (5 × 12) ÷ (3 × 2) = iii) 5 × (12 ÷ 3) × 2 = iv) Which, if any, of the questions above result in the same answer? Explain v) 7× 24 ÷ 3 × 2 × 2 = vi) (7× 24) ÷ (3 × 2 × 2) = vii) 7× (24 ÷ 3) × 2 × 2 = 724× viii) = 322×× ix) Which, if any, of the questions above result in the same answer? Explain

Index Rule – One 1. Use the following questions as a guide to determining the first index rule.

a) x22×=x b) x34× x = c) x25× x =

d) bb56×= e) aa24× = f) ab32× =

2. Use your results to predict a answers for:

a) aamn×= b) abmn× =

3. Explain your rule for aamn×=.

Index Rule - Two 4. Use the following questions as a guide to determining the next index rule.

a) 2 2 b) 3 4 c) 2 5 ()x = (x ) = (x ) =

5. Use your results to predict an answer for:

a) m n ()a =

n 6. Explain your rule for: ()am = .

Index Rule - Three 7. Use the following questions as a guide to determining the next index rule.

a) x4 b) x7 c) x5 = = = x2 x4 x2

d) b5 e) a2 f) a3 = = = b6 a4 b2

8. Use your results to predict answers for:

a) am b) am = = an bn

am Note: Include a comment in your rule stating what happens if = and n > m. an am 9. Explain your rule for = . an 10. Some rules have exceptions. Can you find a value for a for which the rule does not work?

Teacher Notes – Indices This activity covers only the basic index laws. Further exploration needs to be covered with students. The intent of the first group of questions is to ensure students have a knowledge of the order of operations. It is important that students understand the difference between: 6x7 62x73÷ x and 2x3 The simplification of these expressions is different because the order of operations demands that in the first case the division take place before multiplication, whether working specifically with division first or working from left to right for division followed by multiplication, the answer is: 623x7310÷=xx The calculator’s interpretation of the input is displayed clearly, automatically and using two dimensional notation. This result is certainly not trivial and leads to errors in many aspects of mathematics, particularly senior mathematics classes. Common errors such as: Scientific Notation Errors (Order of operations) 6 × 104 ÷ 3 × 102 = 2 × 106 In this example the student is using scientific notation. The student enters the expression on their calculator and diligently copies the result down with little or expectation of the correct answer; subsequently the student gets the wrong answer. The automatic two dimensional display of computations on Nspire should at least ring alarm bells for students that the expression does not appear at they had intended. Graphing Errors (Order of operations) x In the example shown here students are required to draw a graph of: fx()= x + 4 If the student has little understanding or expectation of what the graph should look like, they will not be totally surprised by the graphical result. However, the calculator’s interpretation of the result is displayed on the next screen. The interpretation is clearly different from the original equation; students should begin to challenge the result. In this case the student has simply got lazy, forgetting the brackets or failing to use an appropriate template.

In the situation shown here the student has not given any thought to the order of operations. The linear graph is the correct interpretation from the input. What the calculator doesn’t make obvious is the point of discontinuity which lies on the y axis. Tracing the function will provide the missing piece of information.

Another aspect to indices that is often lost in teaching middle school mathematics classes is the domain restrictions that coincide with the index laws: 6x7 ≠ 30xifx4 = 2x3 Division by zero is an area that often confuses student in senior mathematics classes. Errors often occur or solutions go missing when equations are divided by a variable whilst students are trying to transpose. When students perform these types of calculations (potential division by zero); an error message is displayed on the bottom of the screen. In the design process of a CAS, the programmers need to make decisions. In this case the decision has been made that the most likely expectation is that of simplification. However, to ensure mathematical accuracy the warning message is displayed automatically. More of the error message can be displayed by pressing CTRL + ? ie: / + ? The complete message is shown on the next screen, it warns about the division by zero; the domain of the result may be larger than the domain of the original function. It is possible to perform the simplification without the error message being displayed: 6x7 |0x ≠ 2x3

In this case the error message is not displayed as the information: x ≠ 0 is given with the original expression. Similar results are found when expressions containing surds are not simplified:

ab×≠ ab An example to consider here is if both a and b are negative, the left hand side involves complex numbers whereas the right hand side is real. For example if a = -2 and b = -8 then the left hand side is equal -4 but the right hand side is equal to +4.

Paper Folding

A locus is the set or configuration of all points whose coordinates satisfy a single equation or one or more algebraic/geometric conditions. In this activity a locus will be created using paper folding followed by the geometry environment in TI-Nspire.

Constructing a Locus – Stage 1

Step 1 Use an A4 piece of paper and place it in ’landscape’ orientation. Place a clearly visible dot centred horizontally on the page and between 2 to 8cms from the base of the page. Label the dot: Focus Focus Label the bottom edge of the page: Directrix Directrix As shown in the diagram opposite.

Step 2 Make a fold in the page such that the bottom edge of the paper (the directrix) just touches the dot (focus). Press firmly on the fold so that it leaves a permanent crease in the paper. As shown in the diagram opposite. Place a small mark on the bottom of the page where the Focus base of the page touched the focus. Directrix

Step 3 Unfold the paper to reveal the crease. Use a pen or pencil and a ruler to trace over the crease. As shown in the diagram opposite.

Focus Step 4 Repeat steps 2 and 3 until at least 30 different creases are formed on the page; approximately 15 on each side of the focus. 1. Describe the general shape formed by the creases in the paper. o Using a fresh piece of paper, create a single fold. o When the paper is folded put a visible dot on the bottom of the page where it touches the focus. o Draw a line from the focus to this new point.

2. What is the angle between this line and the fold? 3. Measure the distance from the focus to the fold. Measure the distance from the point to the fold. Compare the two answers.

Constructing a Locus – Stage 2 Start a new document (or open the “Paper Folding” file that your teacher has provided). The first page should be a text page with the appropriate information displayed. Make sure you put your name on the first page. (Refer opposite)

Insert a Graphs and Geometry page for the second page and change the View to Plane Geometry. Use the Points and Lines tool to create a point near the centre of the screen. Use the Actions option to label the point f.

Use the Points and Lines tool to draw a horizontal segment across the screen below point f, label the line: ”directrix”.

Use the construction menu to locate a perpendicular bisector between the point f and the directrix line. Drag the point on the segment to see how the perpendicular bisector moves. While the perpendicular bisector moves, contrast its movement with the paper folding activity.

The movement of the perpendicular bisector can be tracked. Leaving a trail behind provides a more permanent reminder of its general location. To achieve this result a locus of its path can be created. From the construction menu, choose locus, click on the perpendicular bisector followed by the point on the line segment.

4. What shape is formed by the envelope? 5. Drag point f around the page. What happens to the shape of the locus? 6. Select the horizontal line segment and place it above the focus. What happens to the shape and position of the locus? 7. Explain how this construction is related to the paper folding activity. From the view menu, select Graphing View.

This will return the Cartesian Plane (x and y axis) to the active Graphs and Geometry screen and therefore allow equations to be entered.

In the equation entry line type the equation: f 1(xx ) = 2 Use the selection arrow to adjust the width and position of the parabola. Note: You can change the thickness of the parabola so that it is easier to see.

8. Can you move the parabola so that it represents the envelope of the locus? 9. Change the parabola equation back to f 1(xx ) = 2 . Change the window settings such that xmin = -5, xmax = 5, ymin = -3 and ymax = 5. Shift point f and the directrix to try and get the envelope to match the parabola with equation: f 1(xx ) = 2 a. Determine the coordinates of point f. b. Determine the equation to the directrix.

Constructing a Locus – Stage 3 Insert another graphs and geometry page. Change to the Plane Geometry View. Put a point in the middle of the screen, label the point f. Draw a line segment near the bottom of the screen. Label the point f and the line segment directrix as before. Insert a perpendicular bisector.

From the construction menu, insert a perpendicular line to the directrix, passing through point q. (q is labelled in the diagram opposite) Note that when the perpendicular line is to be attached to point q, the pencil changes to a hand pointer. It is important to ensure objects are attached when required to do so, otherwise drawings/diagrams will not function as required.

From the points and lines menu, select Intersection Point(s) and create a point of intersection between the perpendicular line and the perpendicular bisector. Label this point P.

Connect points f and p using a line segment. Note that the perpendicular bisector and perpendicular lines have been changed to dotted lines using the attributes menu. (Right click) As construction lines the aim is to make them less prominent, changing them to dotted lines makes them less obvious.

Measure the distances: fp and pq. Drag these measurements aside so that they will not move with the diagram. Drag point q along the line segment and observe how the distances fp and pq change.

10. Compare distances fp and pq. Explain your observations. Hint: Construct a line joining f and q. 11. Construct a locus of point p as point q moves along the line segment. a. What shape is the locus? b. Compare the locus formed in this example with the previous envelope formed by the locus of the perpendicular bisector. 12. Point p is on the locus, given your answers to the previous question, complete the following sentence: A parabola is a set of points equidistant (equal distance) from…

Teacher Notes – Paper Folding Paper folding is a great way of introducing concepts such as perpendicular bisector and angle bisector. When two points are placed on a piece of paper, a single fold that joins the two points creates a perpendicular bisector. When the paper is folded such that the two points are together, the crease represents a set of points that are equidistant from the two points. This can be seen below: This concept is useful for a range of explorations including finding the Paper fold or Crease circumcentre of a triangle. Similarly, bisecting angles using paper folding is useful for finding the incentre of a triangle. In this example the paper fold finds a set of points that are equidistant from a point on the line segment (labelled “Directrix”) and a single point, the focus labelled f). A set of points equidistant from a single point and a line is called a parabola. Other conic sections can be treated the same way. As CAS reduces emphasis on factorising and expanding by hand (doesn’t eliminate), time should be redirected towards an increased understanding of what these curves are all about. Too often factorising and expanding related to quadratic expressions are treated in isolation with little or no time devoted to understanding the connectedness of the various algebraic representations of a parabola and the corresponding significance of the graphical representations. The paper folding activity included here is accessible to a very large range of students. It is important that students realise that a parabola isn’t just ‘any curve’. This lack of understanding often occurs when students identify a whole range of curves as ‘parabolic’ with little other justification other than, its not straight and it sort of looks like a parabola. It is important that students understand that a parabola has a definition both geometrically and algebraically and that visual representations may simply be a similar shape.

Conic Sections

A family of specific cone dissections, referred to as conic sections, are studied in high school mathematics courses. The circle is the generally the first one studied, but only from a geometrical perspective. The second one studied is the parabola, but this is often studied from a purely algebraic perspective. Awareness of the relationship between the conic sections is the first step towards appreciation, this awareness begins with geometry. This activity includes an extension component which unites geometry and algebra to algebraically define a parabolic curve. This component leads to an understanding of a parabola; appreciation generally comes much later as the other sections are studied. “Mathematics without understanding is a set of disconnected facts”. Peter Fox - 2007

Conic Sections – The parabola: 1. The diagram below is of a parabolic curve. Measure the distance from the focus to point A and from the directrix to point A. (Along the dotted lines). Compare the two answers. 2. Repeat the previous question for the points B to F.

10 C

F 8

D 6 A Focus B

4 E

Directrix

-10 -5 5 10

A backwards Look The focus and directrix for a parabola are shown below. The parabolic curve is not shown. Use your knowledge and understanding to determine which points lay on the curve. Explain, in detail, how you determined which points are on the parabolic curve.

L 5 N

J 4 Focus H D A M

3 C I B K

2 F E

1 G

Directrix

-6 -4 -2 P Q 2 4 6

3. Write down your definition of a parabola.

Extension: Finding an Equation 1. The focus of a parabola is located at the point (0, 4). The directrix is the x – axis. A point on the parabola is labelled Pxy(, ) a. What is the distance from the x – axis to the point y? b. Determine an expression for the distance from the ‘focus’ to the point P on the curve. c. Use the CAS calculator to determine an expression for y in terms of x. 2. The focus of a parabola is located at the point (0, 4). The directrix is the x – axis. a. What are the coordinates of the point the curve passes through the y – axis? b. Write down the coordinates of two other points the parabola would pass through. Hint: A parabola is symmetrical. c. Now that you have three points on the parabola, determine the equation to the parabola. Hint: Use the solve command to solve three sets of simultaneous equations. 3. The focus of a parabola is located at the point (0, a). The directrix is the line y = d. A point on the parabola is labelled Pxy(, ) a. What is the distance from the line y = d to the point y? b. Determine an expression for the distance from the ‘focus’ to the point P on the curve. c. Use the CAS calculator to determine an expression for y in terms of x.

Teacher Notes – Conic Sections It is not expected that many students will attempt the extension section of this activity. It is however important that these types of challenges be provided to students occasionally, particularly when they have a particularly powerful too in their hand. (CAS calculator) The remainder of the activity however is important if students are to test their understanding of a parabolic curve. Students need to understand that a parabolic curve has specific properties. Including further examples such as how the focal point is used in everyday situations helps students appreciate the relevance of this very important curve. The diagrams included in this activity force students to use ‘reverse’ logic to answer the questions. These questions also help identify those students that have a firmly planted definition of a parabola and may therefore be replicated in some form on an assessment item with or without CAS. Simple factorising / expanding problems are not as relevant when students have access to a CAS calculator. It is important in such an environment that the basic algebraic manipulation skills are not the ‘focal’ point of an assessment item.

Transforming Parabolas

The dynamic graphing functionality of TI-Nspire makes learning about transformations easier than ever before. In this activity you will be responsible for making your own notes about what you have learned. Use the questions as a guide to inform you.

Activity Start a new document and create a graphs and geometry page. In the equation editing line, enter the equation f 1(xx ) = 2 . The graph will be dragged around the screen so it is best to shift the label for the equation. Drag the equation label to the corner of the screen. Drag the equation around the screen, observe the equation carefully. 1. As the graph is made wider (dilated) how does the equation change? 2. When the graph is shifted vertically up and down only, (translation parallel to the y axis); how does the equation change? 3. When the graph is shifted horizontally left and right only, (translation parallel to the x axis); how does the equation change? The general equation for a parabola that is translated / dilated is given by: yaxh= ⋅−()2 + k Dragging the parabola around the screen reveals that each of these parameters; a, h and k have an impact on the graphs position / shape. To investigate each of these parameters independently, sliders can be used. o Change f 1(x ) to: f 1(xax ) =⋅2 o Insert a slider (Actions – Slider) o Use “a” as the control for the slider. o Use the contextual menu on the slider (‘right’ click on the slider using CTRL + Menu) to change the settings as shown below. Change the minimum to -3 and the maximum to +3.

4. If the equation is of the form: y = ax. 2 ; use diagrams to show what happens when:

a. The parameter is larger than zero. (a > 0) b. The parameter is less than zero. (a <0) c. The parameter is equal to zero. (a = 0)

The slider control can be changed from ‘a’ to ‘h’. Select the slider control and change it to: ‘h’. Change the equation in f1(x) to f1(x) = (x – h)2. Change the slider settings to match those shown opposite. Drag the slider and observe the affect it has on the graph. Use your observations to respond to the following questions: 5. If the equation is of the form: y =−()xh2 ; use diagrams to show what happens when:

a. The parameter is larger than zero. (h> 0) b. The parameter is less than zero. (h<0)

Change the slider control from ‘h’ to ‘k’. Select the slider control and change it to: ‘k’. Change the equation in f1(x) to f1(x) = x 2 + k. Change the slider settings to match those shown opposite. Drag the slider and observe the affect it has on the graph. Use your observations to respond to the following questions: 6. If the equation is of the form: y = xk2 + ; use diagrams to show what happens when: a. The parameter is larger than zero. (k> 0) b. The parameter is less than zero. (k<0)

7. The thin dotted line in the graph opposite has the equation: yx= 2 . Determine a possible equation for the thick line?

Teacher Notes – Transforming Parabolas The dynamic nature of TI-Nspire makes learning about transformations of graphs rather unique. Students do not need to be given the syntax for the transformational form of a function, instead students can click, drag and observe. The new focuses in such an environment are now: o Understanding why these parameters affect the graph the way they do. Much of this is built up prior to this activity through lessons such as ‘handy reflections’. Further understanding may be achieved by having students become a ‘human’ graph. Each student is a single point responding to instructions provided by the teacher then moving to the corresponding coordinates. Similar experiences to these are now available through TI-Navigator™ o Another opportunity provided here is for students to become independent learners, taking responsibility for the information they record. From an overall learning experience this is a very powerful lesson for students, a life long lesson. Many of the students we teach will never use skills relating to the transformation of a parabolic function. However, most of them will be required to study something they have not seen or used before and teach themselves ‘how to’. This is becoming increasingly more common. Watch a student explore a new video game or piece of computer software. o As students learn faster through this dynamic environment more time can be dedicated to understanding and appreciating the significance of this representation. Furthermore students should spend more time understanding how each of the algebraic representation of a parabola are connected. To help assess students in this area, the teacher can use the overhead screen and give students graphs to guess. Without revealing the ‘teacher equation’ students are required to guess the equation. With the class split into two teams and points obtained for guessing the equation correctly the competition helps drive student desire for understanding. In this activity it is important the teachers hide the axis so that students focus on the translation / dilation of the equation. The screen shot shown here displays the teacher’s equation as a thick dotted line. The student’s equation is shown before they begin guessing. Notice how the absence of the axis forces students to focus on the translations and dilations rather than the specific coordinates of the turning point.

Transforming Parabolas - II

The variables connectivity is another way to explore the dynamic nature of graphs on TI-Nspire. This activity explores a different expression for a parabola.

Activity There are three ways of representing an equation for a parabola (quadratic equation):

o y =++ax2 bx c

o y =−+ax() h2 k o y =−ax()() m x − n This activity initially explores the third representation of a quadratic equation. By the completion of this activity you should be able to identify the benefits of each representation, how they are useful algebraically and how they relate to the graphical representation. Start a new document with a graphs and geometry page. Place two points on the x axis. For the first part of this investigation, ensure the points are placed on the hash marks on the x axis. Placing points on the hash marks means they move according to the scale on the axis.

Change the scale on the x axis by dragging a hash mark on the x axis. Drag the scale until it reads ‘1’.

Measure the coordinates of the points using the Actions menu – Coordinates and Equations.

Note: When you measure the coordinates or equation of an object, the first click selects the object; the mouse can then be moved to locate the region on the screen where the coordinate or equation is to be displayed then click a second time.

Click once on the x coordinate of one of the points. The x coordinate appears shaded. Press the VAR button to assign a variable to the x coordinate. Link the variable to m.

Link the other x coordinate to the variable n.

In the equation editing line, type the equation: f1(x)=(x – m) × (x – n)

Drag the points on the x axis and observe the affects on the graph.

Drag the points so that are located at: (-1, 0) & (-5, 0) Insert a calculator page and type: f1(x) then press [enter] Expand ( f1(x) ) then press [enter] The questions that follow require this process to be repeated several times. Swap between the Graphs and Geometry page and the Calculator page each time. The aim is to identify a pattern or rule between the ‘expanded’ form and the form provided on the graphs and geometry page.

Complete the following table:

Form: y = (x – m)(x – n) Calculator page: Calculator page: Values for m & n y = (x – m)(x – n) Expanded Form ax2 + bx + c

m = -5, n = -1 (5)(1)xx+ +

m = -4, n = -1

m = -3, n = -1

m = -2, n = -1

m = -1, n = -1

m = -1, n = -2

m = -1, n = -3

m = 1, n = -1

m = 1, n = 2

m = 1, n = 3

m = 1, n = 4

m = -2, n = 1

m = -2, n = 2

m = -2, n = 3

m = -3, n = 1

Determine a relationship between the values for m, n and b and c.

Extension: Explore what happens when ‘a’ has a value other than 1.

Teacher Notes – Transforming Parabolas II This activity has been provided in a much contracted form. It is challenging to write a document that requires students to explore, provide appropriate instructions without giving away the answers. Furthermore, student progress needs regular evaluation before moving on. To this extent the remainder of this investigation is left to the teacher (and students) The overall aim is to get students to identify the relationships between the various representations of a quadratic equation and their relationship with the graphical representation. Students should be shown the ‘factor’ command and the ‘expand’ command. Under what circumstances can ax2 ++ bx c be represented in the form: ax()()−− m x n?

Under what circumstances can ax()()−− m x n be represented in the form: ax2 ++ bx c ?

Under what circumstances can ax2 ++ bx c be represented in the form: ax()−+ h2 k?

Under what circumstances can ax()−+ h2 k be represented in the form: ax2 ++ bx c ? A concept map is a very useful tool for students to represent visually their understanding of the various forms. Dotted and solid lines should be used to show the connectivity between the various representations of a quadratic equation. If the line is dotted, the pathway can not always be travelled. If the line is solid, it can be travelled always. On the concept map, students can also include graphical representation of the various forms. What information does each form provide about the graph? Does the value of a mean the same thing in each representation? What do m and n tell you about the graph? At this point introducing the discriminant is a useful exercise. It is possible to display the discriminant live on the graphing screen so students can see its value as the graph is dragged around the screen. A good evaluation tool in this exercise is to provide students with a piece of paper that has one of the following: o A graph o An equation (in one of the three forms) o Information about the graph ‘turning point at (2, 3)’ or x intercepts at (-3, 0) and (1,0) o A discriminant of 9 Once each student has such a piece of paper, their task is to find at least one other person that could potentially have the same equation. It is important here to have some students that could fall into a number of different groups.