Using Real Life Contexts in Mathematics Teaching

Toowoomba March 2013

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

2

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, 2006)

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)

3

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.

4

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)

5

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…

6

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

7

Fermi Problems

Typical (Fermi) problem involves estimating the number of service

providers appropriate for some population – such as a city.

Order of magnitude estimates are often sufficient.

Prototype: How many piano tuners are in Chicago?

Problem Stimulus : Waiting lists continue to grow

http://www.thechronicle.com.au/news/dental-waiting-list-continues-to-grow/1623901/

MORE than 1200 Toowoomba patients are waiting longer than five years for general dental work through the public health system.

Figures show queues in Toowoomba has reached a 5088-patient peak, some waiting more than six years to see a dentist.

Problem: Estimate how many dentists Toowoomba needs. Now? In approximately 20 years (2031)?

8

● What key quantities would we need to include in a mathematical relationship to enable us to estimate the number of dentists

needed to service a population like Toowoomba?

● What key assumptions do we need to make?

● What are some defensible numerical values associated with these assumptions?

● How sensitive are our estimates to these assumptions?

● How would changes in circumstances (decide which?) affect the estimates?

Approach

Choose time period for focus (e.g. week, month, year…) YEAR

9

Needed information:

dentally relevant population (Pd ) persons

average visits per year per person (n)

average duration of a visit (d) hr

typical hours worked by dentists (h) hr

*********************************************************************With year as the timescale:

N = (Pd x n x d) /h Quantities need to be researched (Internet etc) or else sensibly estimated

10

Populationhttp://www.thechronicle.com.au/news/city-prepares-manage-explosion-population/283222/

AN extra 75,000 people are expected to call Toowoomba home by 2031. This is a prediction for the future growth in the Toowoomba region released yesterday in the South East Queensland Regional Plan 2009-2031. The population of Toowoomba is expected to grow to about 197,000 by 2031 requiring an additional 31,000 dwellings.

Hours workedhttp://www.arcpoh.adelaide.edu.au/publications/report/research/pdf_files/rr23_productivity_trends.pdf

An average of 1700 to 1800 hours per year is worked by individual dentists according to the findings of a research project (2008)

Key survey findings http://www.thechronicle.com.au/news/toowoomba-residents-avoid-dentist-due-hip-pocket/1609590/

65% of Toowoomba residents avoid the dentist due to cost60% of Queensland residents and 65per cent of Toowoomba residents have no plans to visit a dentist for at least 12 months,and 12 per cent for at least two years

11

Consider period of 1 year

Assume Pd = 122 000 x 0.95 (persons - estimated from population)

n = 0.75 (dimensionless - estimated from survey data)

d = 0.67 (hours - estimated from consultation length chit- chat on Web)

h =1750 (hours - estimated from working hours research data)

N = Pd x n x d/h gives N ≈ 33

If n = 1, N ≈ 44 (more visits/person)

If n = 1, d = 0.5, N ≈ 33 (more visits and shorter consultations)

If n = 1.5, d=0.75, N ≈ 74 (Medical fund supports costs – system wouldn’t cope)

And so on

Other examples:

Estimate the number of petrol service station outlets

Estimate the number of fast food outlets etc

12

Goldilocks Principle (not too big, not too small…just right)

In Fermi problems orders of magnitude are often sufficient

When statistics or information are not available we need to estimate

quantities.

For example (Service stations) what is average kilometres travelled per

vehicle?

Minimum – perhaps only used to travel to shops. (M1)

Maximum – useful to think as multiple of minimum rather than guess a

number (M2)

Then geometric mean G = √(M1M2) is a more measured estimate that the

arithmetic mean A = (M1 + M2)/2.

Note G ≤ A.

Field of Dreams Principle (build it and they will come)

Will they? McDonalds? dentists? taverns? football teams?

13

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: Estimate the amount of sag in a cable of given length (2l) when strung between two pylons a known distance apart.

14

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 1.36% 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)!

15

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.

16

A

Related problem:Boundaries are sometimes marked by decorative fencing in the form of hanging chains – car sales yards and parks feature this design.

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.

17

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.

18

Generalise

Where there are arithmetic calculations, there are functions (formulae)

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

19

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)

So how many trees would 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, then a graduate student in the Pulp and Paper Technology Program at the University of Maine. …. we can estimate that it takes about 12 trees to make a ton of newsprint. (1 tonne =0.98 ton)

Problem: How much timber is needed to provide a years circulation of a major newspaper (e.g. Courier mail)?

20

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

ICEM text 24.3 1.7 1.7 2 2

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.

21

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

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 data)

CM usage rate would be 36,851.8 x 12 = 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 = 7249 x 12 = 86,988 trees

22

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

23

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.

24

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

25

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)

26

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.

27

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.

28

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

● All floors have equal numbers of residents

● 1,2,3 bedroom apartments are equally distributed

● Number of stairs between levels is the same throughout

29

30

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 (n = 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)

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.

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

descent, delay times etc?

31

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.

32

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)

33

N (vehicles/hr) = (vehicles/km) x (km/hr)

km/vehicle = (d + s)/1000 = (d + 5V/9)/1000

= (9d + 5V)/9000

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)

34

Graph of N versus V

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

35

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)

36

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)

Stopping distance: s = 0.42v + 0.0086v2

Alternatively: Calculate reaction distance formula for individual cases:

e.g. when v = 60, r = 5 *60/12 = 25 etc (note 5/12 = 0.4167)

r = 1.5 x 5v/18 = 5v/12

For braking distance (b): assume b = kv2

v = 60, b = 31 gives k = 31/3600 = 0.0086 and similarly for the others.

37

Traffic flow using stopping distance as separation between vehicles

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

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

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

Maximum flow is about 1280 vehicles/hr when v = 25km/hr

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

38

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

39

The End