Using Real Life Contexts in Mathematics Teaching

QAMT June 2013

Peter Galbraith University of Queensland <p.galbraith@uq.edu.au>

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Curriculum Statements

Mathematics aims to ensure that students are confident, creative users and communicators of mathematics, able to investigate, represent and interpret situations in their personal and work lives and as active citizens. (Australian Curriculum Assessment and Reporting Authority, 2010).

Applications and modelling play a vital role in the development of mathematical understanding and competencies. It is important that students apply mathematical problem-solving skills and reasoning skills to tackle a variety of problems, including real-world problems.(Ministry of Education Singapore, 2012)

Mathematically proficient students can apply the mathematics they know to solve problems arising in everyday life, society, and the workplace. (U.S Common Core State Standards Initiative 2012)

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RATIONALE: The curriculum focuses on developing increasingly sophisticated and refined mathematical understanding, fluency, logical reasoning, analytical thought and problem-solving skills. These capabilities enable students to respond to familiar and unfamiliar situations by employing mathematical strategies to make informed decisions and solve problems efficiently. http://www.acara.edu.au/verve/_resources/Info+Sheet+-+Draft+Senior+Secondary+Mathematics+FINAL.pdf

Australian Curriculum (circa 2013)

It (the national curriculum) develops the numeracy capabilities that all students need in their personal, work and civic life, and provides the fundamentals on which mathematical specialties and professional applications of mathematics are built... These capabilities enable students to respond to familiar and unfamiliar situations by employing mathematical strategies to make informed decisions and solve problems efficiently.

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Some Specifics

Mathematics / Year 10A / Number and Algebra / Linear and non-linear relationships

Content descriptionSolve simple exponential equations

Elaborationsinvestigating exponential equations derived from authentic mathematical

models based on population growth

Mathematics B● identify contexts suitable for modelling by exponential functions and

use them to solve practical problems. (ACMMM066) ●use trigonometric functions and their derivatives to solve practical

problems. (ACMMM103) ●use Bernoulli random variables and associated probabilities to

model data and solve practical problems. (ACMMM146)

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Modelling Process: a common representation Real World Real/Math Links Math World

1. Describe the real problem situation

2. Specify the math problem

3. Formulate the math model

4. Solve the mathematics

5. Interpret the solution

6. Evaluate/validate model

7.Communicate/report. Use model to predict, decide, recommend…

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Using the modelling diagram

● Follow solid arrows clockwise from box 1 to trace out cyclic modelling process (path to a solution)

● Broken arrows indicate there is often to-ing and fro-ing by a modeller between different stages on the way to obtaining a solution. (metacognitive activity)

● The box headings are useful for structuring a report

● What is written of course varies widely and may or may not utilise boxes explicitly

● The following (artificial) example shows how the different stages occur systematically in even simple modelling tasks – another reason for providing students with a scaffolded process to follow through.

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School Fundraising

A school plans a fundraising event, for which it has a sponsor and

an equipment supplier (benches, tables etc). The sponsor agrees

to pay an amount equal to 20% of the proceeds obtained, and the

school agrees to pay the supplier 10% of the proceeds.

Does it matter whether the equipment supplier is

paid before the sponsorship money is obtained – or vice versa?

Approach to solution

Suppose that the total takings are $1000.

A. Pay supplier first:

Amount to school after costs = $1000 x (90/100) = $900

Final amount to school = $900 x (120/100) = $1080

B. Obtain sponsors money first

Amount to school after sponsor = $1000 x (120/100) = $1200

Final amount to school = $1200 x (90/100) = $1080

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Solution and Interpretation

From the school’s point of view - No ($1080 both ways)

From the viewpoint of the supplier and the sponsor – Yes.

A: Supplier gets $100 and Sponsor pays $180

B: Supplier gets $120 and Sponsor pays $200.

In B the sponsor has subsidised the supplier as well as sponsoring the school.

General insights: ● The best solution to a problem may depend on the perspective of

different persons involved: here school? supplier? sponsor? all three?

● Key terms need to be defined clearly up front – are “proceeds” to be determined before or after subtracting costs?

● Is sponsorship to be applied to gross or net profit?

● Team approaches can ensure that the interests of all players are considered

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1. Describe the real problem situation

Find the outcome for school from a fundraising event.

2. Specify the math problem

Find the funds raised (in $) for the school after both costs and sponsorship have been allowed for.

3. Formulate the math model Suppose $1000 is raisedA: Assume: Pay costs firstProfit = (funds raised – costs) + sponsorship.B: Assume: Obtain sponsor first Profit = (funds raised + sponsorship) - costs

4. Solve the mathematics

A. Profit = $(1000x90/100) x 120/100 =$1080

B. Profit = $(1000x120/100) x (90/100) = $1080

5. Interpret the solution

From the school’s point of view the outcome is the same.

6. Evaluate/validate model

Deciding whether the order matters involves also finding the impacts on the supplier and the sponsor

7. Communicate/report. Use model to predict, decide, recommend…

Conduct further modelling to find impacts on supplier and sponsor

Modelling Diagram (1)

Real World Real/Math Links Math World

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Modelling Diagram (2)

Real World Real/Math Links Math World

1. Describe the real problem

Find the impacts on the supplier and the sponsor of changing the order of calculating costs and sponsorship.

2. Specify the math problem

Compare the amounts paid to the supplier - and by the sponsor -under the two arrangements.

3. Formulate the math model Assumptions as before: A: Pay costs first. Supplier gets 10% of funds raisedSponsor pays out 20% on 90% of funds raised B: Obtain sponsorship first. Sponsor pays out 20% of funds raised Supplier gets 10% of 120% of funds raised

4. Solve the mathematics

A. Supplier gets $1000 x 10/100 = $100 Sponsor pays $900 x 20/100 = $180

B. Sponsor pays $1000x20/100 = $200 Supplier gets $1200x10/100 = $120

5. Interpret the solution While giving the same outcome for the school, A and B have different outcomes for supplier and sponsor.Method A is fairer.In B supplier is subsidised by $20 from the school sponsor!

6. Evaluate/validate model

While the outcome for the school is the same for both methods, Method A provides a fairer outcome for all parties.

7. Communicate/report. Use model to predict, decide, recommend…

Method A is the preferred approach - it satisfies both mathematical accuracy, and notions of fairness for all parties.

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Authentic Modelling and Applications: The problem drives the process

1. Content authenticity Does the problem satisfy realistic criteria (involve genuine real worldconnections)? Do the students possess mathematical knowledge sufficient to support aviable solution attempt.

2. Process authenticityDoes a valid modelling process underpin the approach?

3. Situation authenticity Do the requirements of the modelling task drive the problem solving process, including pedagogy, location, use of technology…

4. Product authenticityGiven the time available: Is the solution mathematically defensible?Does it address the question asked?

Note: refinements following a first attempt are an expected modelling outcome, and additional questions are often suggested by earlier attempts

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A Question of Sag

The Zhoushan Island Overhead Powerline connects the power grid of Zhoushan Island with that of the Chinese mainland. It runs over several islands and consists of several long distance spans, the longest with a length of 2.7 kilometres south of Damao Island. This span uses two 370 metres tall pylons, which are the tallest electricity pylons in the world.

Problem: 1. Estimate the amount of sag if a cable of length 2701 m is used in the longest span for Zhoushan Island. 2. Estimate the sag in general when a line is strung between two pylons a known distance apart.

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Considering the Zhoushan Island example, suppose we consider a cable length 1 metre longer than the distance (2.7 km) between the pylons? Estimate the amount of sag?

Pythagoras approximation:

AD = 2701/2 = 1350.5 m

AD2 = AC2 + CD2 gives s2 = (1350.52 – 13502) = 1350.25

s ≈ 36.75 (m)

(approx 1.4% of distance between pylons)!

Suppose we just have 100 m between pylons a= 50, l = 50.5 gives s ≈

7.1 (m) (approx 7% of distance between pylons)!

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Follow up question;

Can we find a formula that gives a quick estimate of the sag?

Let ‘2d’ be the extra length of the cable to link two towers distance ‘2a’ apart

s2 = (a +d)2 –a2 gives s2 = 2ad + d2

s = (2ad + d2)1/2

Here d2/2ad = d/2a = 0.5/100 = 0.5%

Answers are 7.09 (full) and 7.07 (approx) – within 2 cm.

s ≈ (2ad)1/2

Use whenever d/a is small (less than 10%)

For the record: Calculations based on a catenary shape give a

value for ‘s’ of approximately 6.15 (m) when a =100; 31.8(m) when

a= 1350.

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Related problem:Boundaries are sometimes marked by decorative fencing in the form of hanging chains.

It is intended to border a path with decorative ‘fencing’ in the form of chains to be hung between adjacent posts 4 metres apart. Decide on the height of the posts and the lengths of chain needed.

A

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L

M Ba

sl

BL2 ≈ ML2 + MB2 gives l2 ≈ s2 + a2

If ‘h’ is the height of attachment to the poles, and ‘s’ the depth of sag’

Example: Set h = 1m, and clearance above ground (h – s) = 0.5m, so s = 0.5

l2 ≈ 0.52 + 22 = 4.25

l ≈ 2.06

So ALB ≈ 4.12 (12 extra cm).

We might want to work by ‘eye’ to choose the appropriate sag. While the extremes (s =0 and s = h) might in general be of little interest, we note that choosing s = h enables the maximum estimate for arc ALB (4.48 m per segment) – allowing for all choices of sag.

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Generalise

● Where there are arithmetic calculations, there are formulae -

and where there are formulae there are functions lying behind them.

So let s = kh where 0<k<1

l2 ≈ a2 + k2h2

l2/a2 - k2/(a2/h2) =1

Hyperbola with semi-major (l) axis = ‘a’, and semi-minor (h) axis = ‘a/k’.

PQ is the segment of the hyperbola that gives the ‘l’ values for different proportional sags (k)`.

Q

P

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Forest Mathematics

Today, 90% of paper pulp is created from wood. Paper production accounts for about 35% of felled trees, and represents 1.2% of the world's total economic output. (Wikipedia)

How many trees make a ton of paper? Claudia Thompson, in her book Recycled Papers: The Essential Guide (Cambridge, MA: MIT Press, 1992), reports on an estimate calculated by Tom Soder, a student in the Pulp and Paper Technology Program at the University of Maine. …. that, based on a mixture of softwoods and hardwoods 40 feet tall and 6-8 inches in diameter, it would take a rough average of 24 trees to produce a ton of paper. If we assume that the process is now about twice as efficient in using trees, then we can estimate that it takes about 12 trees to make a ton of newsprint. http://conservatree.org/learn/EnviroIssues/TreeStats.shtml(1 tonne =0.98 ton)

Problem: How much timber is needed to provide a years circulation of a major newspaper (excluding inserts) such as the Courier Mail?

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Measurement data

Width(cm)

Height(cm)

Top margin (cm)

Bottom margin (cm)

Left margin (cm)

Right margin (cm)

Courier Mail 29 40 1.8 1.6 1.8 1.8

Australian 41.7 58 1.3 1.3 1.3 2.3

Number of pages (Courier Mail ) - excluding inserts and magazinesSun (96); Mon (64); Tue (56); Wed (80); Thu (80); Fri (112); Sat (96)

Circulation: http://www.newsspace.com.au/the_sunday_mail_qld Courier Mail: M – F: 185,770 Sat: 237,798; Sunday Mail: 438, 994

(not allowing for additional copies printed)

Grammage (or weigh some!)The mass per unit area of all types of paper and paperboard is expressed in terms of grams per square meter (g/m²). Since the 70’s, the grammage of newsprint has decreased from the global standard of 52 g/m2 to 48.8, 45 and 40 g/m2 – cheaper and also felt to be a necessary solution to the problem of optimizing the use of forest resources.

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Calculating usage

Courier Mail single copy (pages/week): 584

Area of page 1160 sq cm = 0.116 sq m

Circulation (pages/week) = 137 492 032

Total (pages/year) 137,492,032 x 52.28 = 7,188,083,433

Total area (per year) =0.116 x 7,188,083,433 = 833,817,678.2 sq m

Assume grammage = 45 g/m2 = 0.045 kg/m2 (website data)

CM for year weighs 833,817,678.2x0.045 = 37,521,795.52 kg = 37,521.8 tonne =

36, 851.8 ton

1 ton of newsprint uses 12 trees (from website data)

For a paper with size and circulation similar to the CM usage rate would be

36,851.8x12 = 442, 222 trees if no recycling.

Margins

Area of print in page = 25.6 x 36.4 = 931.84 sq cm

Fraction ‘wasted’ = (1160-931.84)/1160 = 19.67%

Annual weight not used for print = 0.1967 x 36,851 = 7249 ton

Trees used for blank space = 7249x12 = 86, 988

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How much treed area?

PlantationsGenerally, the ideal initial stocking rate for most eucalypts in South Australia is around 1000 stems per hectare. Typically seedlings are planted 2.5m apart along rows that are 4m apart.

http://www.pir.sa.gov.au/__data/assets/pdf_file/0015/75021/FS16_Eucalypt_Seedling_Information_and_Planting_Tips.pdf

N = 442 222 trees

√ 442 222 = 664.0

‘Square area’ : (663 x 2.5) x (663 x 4) = 4 395 690 (m2)

= 439.6 hectares

= 4.4 sq km.

Additional: Suppose planted in 3m x 3m format (calculate area)

Can also use Pick’s rule!

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What about recycling?

Australia leads the world in newsprint recycling http://www.proprint.com.au/News/266132,australia-leads-the-world-in-newsprint-recycling.aspx Australia's newsprint recycling rate is almost 10% higher than in Europe, according to a new report. The survey, commissioned by the Publishers National Environment Bureau (PNEB), showed that 78.7% of all Australian newsprint was recovered. How many trees are saved annually by using 78.7% recycling of newsprint?

Trees saved for CM usage rate = 0.787 x 442, 222 = 348, 029 (3.46 sq km)

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How much wood is in a tree?Estimating the Volume of a Standing Tree Using a Scale (Biltmore) Stick See : http://www.ces.ncsu.edu/forestry/pdf/WON/won05.pdf

Measuring DiameterTree diameter is the most important measurement of standing trees. Trees are measured 4½ feet above ground-level, a point referred to as diameter breast height or DBH. Diameter breast height is usually measured to the nearest inch; but where large numbers of trees are to be measured, 2-inch diameter classes are used. To measure DBH, stand squarely in front of the tree and hold the scale stick 25 inches from your eye in a horizontal position against the tree at 4½ feet above the ground. Shift the stick right or left until the zero end of the stick coincides with the left edge of the tree trunk. Without moving your head, read the measurement that coincides with the right edge of the tree trunk. This measurement is the tree’s DBH, including the tree’s bark.

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DBR = 2[x2 + x√(a2 + x2)]/a

Find correct formula for DBH (r =radius at measured height)

Generalise the (25”) distance to ‘a’.

(l + x)2 + r2 = (a + r)2 (1)

l2 = a2 + x2 (2)

(1) and (2) give r = [x2 + x√(a2 + x2)]/a, so

Can individualise by finding the personal value of ‘a’.

Check e.g. using a down pipe of known dimensions

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Robust measures

Since cross section is not a perfect circle it makes sense to obtain two radius

estimates at right angles. Their geometric mean r = √(r1r2)gives an appropriate

average. (Noting for an ellipse A = πab where ‘a’ and ‘b’ are the major and

minor axes). Graph of r = f(x) (a = 60)

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Create Conversion table (a = 60)

x (cm) r (cm) x (cm) r (cm)

0 0 16 20.8

2 2.1 18 24.2

4 4.3 20 27.7

6 6.6 22 31.5

8 9.1 24 35.4

10 11.8 26 39.6

12 14.6 28 44.0

14 17.6 30 48.5

Measuring Merchantable Height

Merchantable height refers to the length

of usable tree and is measured from

stump height (1 foot above ground) to a

cut off point in the top of the tree. (use

clinometer). How to Make a Clinometer With a Straw Using Trigonometry | eHow.com http://www.ehow.com/how_10049738_make-clinometer-straw-using-trigonometry.html#ixzz2N2mZfAp5

Estimating Tree volumehttp://www.fao.org/forestry/17109/en/Trees are neither cones nor cylinders, but empirical analyses often indicate that the volume of a single-stemmed tree is between that of a cone and a cylinder, with tree volume often lying between 0.40 and 0.45 times that of an equivalent cylinder.

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Evacuating Q1

Fire safety problems in Australia’s tallest apartment block

http://www.wsws.org/en/articles/2012/11/fire-n01.html?view=print

1 November 2012

ABC Radio National’s “Background Briefing” revealed last Sunday that

Q1, an 80-storey $260 million apartment block on Queensland’s Gold

Coast, has fire safety problems. Completed in 2005, Q1 has 527

apartments and over 1,000 residents. It is Australia’s tallest building

and one of the highest residential blocks in the world. The weekly

current affairs show reported that Q1’s northern fire escape stairwell is

unsafe and could quickly fill with smoke, endangering hundreds of

people if the building were hit by a major fire.

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Q1 data

Over 1000 residents (plus staff)

76 residential floors to be evacuated

527 residential – 1, 2, 3- bedroom apartments

1430 stairs and 2 stair wells (one non-operational) - 11 lifts

Problem

A bomb threat is received at 2 am and the building must be evacuated –

how long would this take?

Situational Assumptions

● Building has two exit stair wells (one working)

● Lifts are closed to avoid failure through panic – overcrowding

● All residents are mobile and hear the evacuation call

● Building is full and all floors have equal numbers of residents

● 1,2,3 bedroom apartments are equally distributed

● Number of stairs between levels is the same throughout

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Mathematical Assumptions

● Number of apartments /floor = 527/77 = 6.8 (7 approx)

● Average number of evacuees per apartment =3.5

● Number of evacuees per floor using stair well (N = 7 x 3.5 ≈ 25)

● Number of steps per floor (s = 715/76 = 9.4 ≈ 10)

● Number of floors (f = 76)

● Average speed on steps (v = 0.5 steps/sec) - careful

● Time delay between successive evacuees when moving (d =1 sec)

● Time for first occupant on floor 1 to reach stairs (t = 10 sec)

Notes

● values of v, d chosen to reflect a balance between speed and safety

● evacuation rate is governed by access to stairs – what happens in

corridors is effectively irrelevant because of wait times.

Estimation of evacuation time (T) ??

Time for first occupant (floor 1) to leave building + delay until last occupant can move

+ time for last occupant to leave building

● T = (t+s/v) + (fN – 1)d + fs/v (67 sec approx??)

● how are outcomes affected by changes in: numbers of people,

speed of descent, delay times etc?

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Airport Tunnel

The reality was much different, with traffic volumes peaking at 81,500 in September before plunging to 47,102 in December.Although experts repeatedly cast doubt on the projections compiled by consultants Arup, BrisConnections' boss Ray Wilson insisted they had "done their homework" and the forecasts were achievable.Griffith University Urban Planning researcher Matthew Burke warned that each freeway lane could only take a maximum of 1700 vehicles an hour, and a six-lane road would need to be at peak conditions for 24 hours to achieve 250,000 cars. (Courier Mail February 20, 2013)http://www.couriermail.com.au/news/queensland/experts-say-traffic-projections-on-the-number-of-cars-using-the-airport-link-tunnel-were-simply-unrealistic/story-e6freoof-1226581508721

TRAFFIC projections have been blamed for Bris Connections' rapid demise, less than seven months after opening the $4.8 billion Brisbane Airport Link. Shortly after opening, the company was expecting 135,000 vehicles a day through the tunnel, rising to 190,000 within six months.

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Problem: How realistic were the expectations?

Mathematical Question: What is the maximum number of vehicles per

hour to be expected in a tunnel lane?

The two second separation rule

When travelling in a tunnel in Queensland, you should: keep a safe

distance from the car in front (at least a two-second gap)

http://www.tmr.qld.gov.au/safety/driver-guide/tunnel-safety.aspx

Using two second rule

vehicle 2

vehicle 1separation (s)

V km/hr

d d

d = (average vehicle length) (m)

s = 2 x V x (1000/3600) = 5V/9 (m)

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N (no of vehicles/hr) = n (no of vehicles/km) x V (km/hr)

n = 1000/(d+s) = 1000/(d + 5V/9)

n = 9000/(9d + 5V)

N = 9000V/(9d + 5V)

Hyundai Elantra (4.5m); Terios (3.5 m)

Suppose d = 4 m (average small car)

Tunnel speed limit: V = 80 (km/hr)

N = 9000 x 80/436 = 1651 (vehicles/hour)

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Graph of N versus V

Maximum flow at 80 km/hr = 1650 vehicles/hr (approx).

Projections:

135 000/day = 5625/hr (6 lanes) = 938/hr per lane.

190 000/day = 7916/hr (6 lanes) = 1320/hr per lane

Suppose runs at capacity from 8 am to 6 pm: 10 x 1650 = 16500

(190 000 – 6 x 16500) = 91 000 in off peak 14 hours = 91 000/(14x6) = 1083 per

hour = 65.6% 0f capacity

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Using Stopping Distance data (Department of Transport, QLD)

http://www.tmr.qld.gov.au/Safety/Driver-guide/Speeding/Stopping-distances.aspx

Speeding is dangerous because the faster you go, the longer your stopping

distance and the harder you hit. In an emergency, the average driver takes

about 1.5 seconds to react.

[Stopping distances increase exponentially the faster you go (see graph

below)]!!! (Retrieved March 2013)

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Stopping Distance formula (distances in metres; speeds in km/hr)

1 km/hr = 1000/(60 x 60) m/sec = (5/18) m/s

v km/hr = 5v/18(m/s)

Use regression (spreadsheet or calculator) to find

Reaction distance = 0.42 v (proportional to v)

Braking distance = 0.0086v2 (proportional to v2)

Alternatively: Assume braking distance (b) = kv2

V (km/hr) b (metres) b/V2

60 31 0.00861

70 42 0.00857

80 55 0.00859

90 70 0.00864

100 85 0.00850

110 104 0.00860

b = 0.00858V2 = 0.0086V2

Stopping distance: s = 0.42v + 0.0086v2

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Traffic flow using ‘stopping distance’ as separation between vehicles

Again vehicle/km = 1000/(d + s) = 1000/(5 + 0.42v + 0.0086v2 )

Flow (N): vehicles/hr = (vehicles/km) x (km/hr) gives

N = 1000v/(5 + 0.42v + 0.0086v2)

Maximum flow is about 1200 vehicles/hr when v ≈ 24km/hr

dN/dv = 0 when v = 24.1(km/hr)

Gives N = 1198 (vehicles/hr)

*Flow at 80 km/hr = 864 vehicles/hr - not even close

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Notes:

● Travelling at the recommended stopping distance apart, gives a much smaller

flow in vehicles/hr than using the 2-second rule.

● A six lane road with an 80 km/hr speed limit, operating at maximum safe

capacity, would sustain flows of 9906 average cars/hr using 2-second rule, and

5124 average cars/hr using the stopping distance provision.

● 135,000 vehicles per day, requires an average hourly flow of 5625 vehicles/hr

over 24 hours!

● 190,000 vehicles per day, requires an average hourly flow of 7917 vehicles/hr

over 24 hours!

● Change ‘d’ to explore effect of larger vehicles.

Short Safety Film (influence of speed on stopping distance)http://www.youtube.com/watch?v=AU3TY3JmM64

Shows impact at higher speeds on an object placed at the stopping distance of a vehicle travelling at 50 km/hr

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Bush Walking

The Source (Aoraki - Mount Cook)

It is not uncommon for companions who enjoy bush walking to differ in fitness and energy. On tracks that lead out and back they will often walk together for a time at the pace of the slower walker – until s/he indicates an intention to turn around and return to base.

The faster walker has the choice of following the same action, but alternatively may decide to push on for a time at her/his faster pace before also returning. Especially if the opportunity to travel further and faster is appreciated, the faster walker will want to go as far as possible.

However they will not want their companion to have to wait around too long at the end of the walk for them to get back.

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Problem: When the slower walker starts the return trip, for how much extra time should the faster walker travel on the outward path before turning for home – so that both will arrive at the starting point at the same time.

BF and FB are outward and return paths along the same track : Assume that after parting at C both walkers maintain their respective average walking speeds?

B = starting point (Base)

C = point that walkers reach together after walking for time (t) at the speed (V) of the slower walker.

Slower walker (s) now starts to return while the faster walker (f) continues on at speed (kV) where k > 1.

F = point at which ‘f’ turns back after travelling for an additional time (T) at the faster speed.

S is the point reached by s on the homeward path when f turns for home at F.

F

S

B

C

●●

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For the walkers to arrive at B together:

BC = Vt; CF = (kV)T; CS = VT

So FB/kV = SB/V

FB = BC + CF = Vt +kVT

SB = BC – CS = Vt –VT

Hence V(t + kT)/kV = V(t–T)/V

So (t + kT)/k = (t–T) 2kT = (k–1)t T = [(k–1)/2k]t

So it is sufficient to know the time from the start of the walk to the separation point (t), and the relative walking speeds (k) of the two individuals.

Check: k=1 gives T = 0 – both walkers turn together if they walk at the same pace – as should happen.

F

S

B

C

●●

time taken for ‘f’ to cover the distance FB at speed kV = time taken for ‘s’ to cover the distance SB at speed V.

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Example: Walkers stay together for an hour: t =1 so T = (k–1)/2kSuppose k =2: T = ¼ (‘f’ should continue on for 15 minutes) etc

Relevant section of k – T graph (k ≥1)

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Further refinement

It is more realistic in practical terms for ‘f’ to aim to arrive at base so that ‘s’ doesn’t have to wait ‘too long’ – that is allowing a time window ‘w’.

Then we want 0 < (time for ‘f’ – time for ‘s’) < w

That is 0 < (t + kT)/k – (t–T) < w

0 < [2kT – t(k – 1)]/k < w t(k– 1)/2k < T < [k(t+w) – t]/2k

For example if k =2, t =1, w = 0.2 (12 minutes) we need 0.25 < T < 0.35

So ‘f’ should continue on for a time between 15 and 21 minutes.

* k > 0 from its real world property – important in manipulating the inequalities.

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Wings on the heels: Investigate limiting behaviour

Here we assume that ‘f’ can travel at any multiple of the speed of ‘s’ : k has no upper bound. Taking the case where the walkers stay together for an hour (t = 1) we have T = (k-1)/2k = ½ - 1/2k

As k ∞, T ½ (see graph).

If T = ½ there is nothing ‘f’ can do to catch up as ‘s’ is already back at the base when ‘f’

reaches C on the return trip.

Now suppose ‘f’ turns after 27 min (T = 0.45) (‘s’ is 33 minutes from base)

Need ½ - 1/2k = 0.45 1/2k = 0.05 2k = 20 k = 10

If T = 0.4833 (29 min) k = 29.94 If T = 0.499722 (29 min 59 sec) k= 1799.998

Nominate any value of time (T) less than 0.5, and we can always find a speed multiple (k) that will enable ‘s’ and ‘f’ to finish together.

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Australian population to top 23 million tonight (23 April 2013, 10:55 AEST)● Australia's population will reach an estimated 23 million people some

time tonight, and demographers say it's on track to hit 40 million by the middle of the century.

● The Australian Bureau of Statistics says the projection is based on last year's population estimate and takes into account factors such as the country's birth rate, death rate and international migration. ABS figures show that around 180,000 people move to Australia each year.

● With births outnumbering deaths two to one and a 14 per cent increase in migration, Australia's population is now growing by more than 1,000 people per day.

● It estimates that with a birth every one minute and 44 seconds, a new migrant arriving every two minutes and 19 seconds, and a death every three minutes and 32 seconds, the 23 million mark will be reached just after 10:00pm (AEST).

● That means our population increases by one person every minute and 23 seconds. http://www.radioaustralia.net.au/international/2013-04- 23/australian-population-to-top-23-million-tonight/1120164

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Check the statements:

(1) “Australia's population is now growing by more than 1,000 people per day”.

(2) “… our population increases by one person every minute and 23 seconds.”

(3) “demographers say it (population) is on track to hit 40 million by the middle of

the century.”

Births/day = 24x60x60/104 = 830.77

Deaths/day = 24x60x60/212 = 407.55

Net migrants/day = 24x60x60/139 = 621.58

(1) Increase/day = births – deaths + immigration = 1044.80 people/day

(2) ‘Arrival interval’ = 24x60x60/1044.80 = 82.70 sec (1 min 22.7 sec)

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(3) Forecast the population in 2053: (n = 40) years

Present population (2013): 23000000

Births per year = 830.77 x 365.25 = 303 439

Deaths per year = 407.55 x 365.25 = 148 858

Immigrants per year = 621.58 x 365.25 = 227 032

Birth rate: b = 303 439/23000000 = 0.0132 per year (yr-1)

Death rate: d = 148 858/23000000 = 0.00647 per year (yr-1)

Immigrants (net): I = 621.58 x 365.25 = 227 032 per year (yr-1)

Let P0 = initial population (in 2013)

Let Pn = population in year n

Let r = natural population change rate = (b – d)

Let I = average net annual immigration intake

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(a) By Spread Sheet

P0 = 23000000; b = 0.0132; d = 0.00647; r = b - d = 0.00673; I = 227 032

P1 = P0 +rP0 + I = P0 (1 + r) + I = P0 R + I (where R = 1 + r)

P2 = P1 R + I etc

.

Copy down

.

.

P40 = 40 459 252

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(b) By Geometric Series

Proceeding as above (annual increments)

P1 = P0R + I

P2 = P1R + I = P0R2 + I (R + 1)

.

.

Pn = P0Rn + I (Rn-1 + …R2+ R +1)

Pn = P0 Rn + I(Rn – 1)/(R – 1)

P40 = 40 459 252 (annual increments)

P40 = 40 523 429 (monthly increments)

P40 = 40 523 960 (daily increments)

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)(Pr I

dpdt

(c) By Calculus

δP ≈ Pr δt + I δt

= (Pr + I) δt where r = b – d.

In the limit

dP/dt = Pr + I

=

ln(Pr + I) = rt + constant, leading to

P(t) =P0ert +I(ert -1)/r where P(0) = P0

P40 = 40 526 191

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BUT!!!

Death rate : A life expectancy of ‘L’ means that on average a fraction ‘1/L’ of the

population die each year. Figures imply average lifetime = 1/0.00647 = 154.6 (yrs)!

(The given figure might apply over a 24 hour period as here, or even over short

periods of time, but is no basis for robust population predictions. )

Current life expectancy is 81.5 years:

http://www.australiandoctor.com.au/news/latest-news/australia-on-top-in-life-

expectancy

Then long term average death rate = 1/81.5 = 0.0122699 yr-1

Also the net immigration rate is quoted at about 180 000 per year

Using these figures we get: P40 = 31 200 000 (approximately).

There are implications for: jobs, housing, health, education…

Are the demographers wrong?

Are the data robust?

Has the media been creative with facts?

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Suggests 4.

What combinations of average natural growth rates and immigration rates

would be needed for a population of 40 million in 2053 (40 years time)?

Using P =P0ert +I(ert -1)/r we need

40 000 000 = 23 000 000 e40r + (e40r -1)I/r

Using CAS (Maple) gives (noting MIG =I) (blue)

Linear approx: I = 430 652 – 31 500 000r (red)

Consider implications as they relate to family planning decisions (r) (personal) and migration rates (I) (national).

e.g. replacement rate (r =0) of population

requires I = 430 652 - gives population

(2053) = 40 226 080

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Mathematics used (apart from problem solving /modelling)(non – exhaustive

Sag

PythagorasEstimationPercentageApproximationFunction (domain, range, rule) HyperbolaProportion

Q1

Basic arithmeticChoosing unitsSpeedAveragesEstimationFormulae

Forest mathematics

Length measurementAreaPercentageBasic arithmetic operationsMetric and British unitsConversion of unitsArea densityVolumeAngle in semi-circleEqual tangentsPythagorasFunctionGraph Table of valuesArea of ellipseGeometric MeanPick’s rule

54

Bush Walking

d = v x tproportionratioequationsrational functions - hyperbolasestimationinequalitieslimit

Population

Basic arithmeticPopulation & rates of growth (principal/interest)Rates of change timescales of changeLinear functions SpreadsheetGeometric seriesIntegrationExponential functionsGraphing Solution of transcendental equations(technology)

Tunnel

FormulaeRatesMetric unitsConversion of unitsFunctionality Rational functionsGraphsCurve fittingRegressionQuadratic functionEstimationDerivative (quotient rule)Maximum value

55

That’s all folks