Using Real Life Contexts in Mathematics Teaching

Using Real Life Contexts in Mathematics
Teaching
QAMT June 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, 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)

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+M
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 description
Solve simple exponential equations
Elaborations
investigating 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 mathematics5. Interpret the
solution
6. Evaluate/validate
model
7.Communicate/report. Use
model to predict, decide,
recommend…

6
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.

7
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

8
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

9
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 raised
A: Assume: Pay costs first
Profit = (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)

10
Modelling Diagram (2)
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 raised
Sponsor 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.

11
Authentic Modelling and Applications: The problem drives the process
1. Content authenticity
Does the problem satisfy realistic criteria (involve genuine real world
connections)?
Do the students possess mathematical knowledge sufficient to support a
viable solution attempt.
2. Process authenticity
Does 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 authenticity
Given 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

12
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.

13
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)!

14
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.

15
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

16
L
M Ba
s
l
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.

17
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
+ k2
h2
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

18
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?

19
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 magazines
Sun (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.

20
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

21
How much treed area?
Plantations
Generally, 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_Euca
lypt_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!

22
What about recycling?
Australia leads the world in newsprint recycling
http://www.proprint.com.au/News/266132,australia-leads-the-world-in-newspri
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)

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 Diameter
Tree 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 h
ttp://www.ehow.com/how_10049738_make-clinometer-straw-u
Estimating Tree volume
http://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
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
● 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

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 (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?

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 1
separation (s)
V km/hr
d d
d = (average vehicle length) (m)
s = 2 x V x (1000/3600) = 5V/9 (m)

33
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)

34
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

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
)
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

37
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

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
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.

40
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
●●

41
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.

42
Example: Walkers stay together for an hour: t =1 so T = (k–1)/2k
Suppose k =2: T = ¼ (‘f’ should continue on for 15 minutes) etc
Relevant section of k – T graph (k ≥1)

43
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.

44
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.

45
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

46
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)

47
(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

48
(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

49
(b) By Geometric Series
Proceeding as above (annual increments)
P1 = P0R + I
P2
= P1
R + I = P0
R2
+ I (R + 1)
.
.
Pn
= P0
Rn
+ 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)

50
∫ + )(Pr I
dp
∫dt
(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

51
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?

52
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

53
Mathematics used (apart from problem solving /modelling)
(non – exhaustive
Sag
Pythagoras
Estimation
Percentage
Approximation
Function (domain, range, rule)
Hyperbola
Proportion
Q1
Basic arithmetic
Choosing units
Speed
Averages
Estimation
Formulae
Forest mathematics
Length measurement
Area
Percentage
Basic arithmetic operations
Metric and British units
Conversion of units
Area density
Volume
Angle in semi-circle
Equal tangents
Pythagoras
Function
Graph
Table of values
Area of ellipse
Geometric Mean
Pick’s rule

54
Bush Walking
d = v x t
proportion
ratio
equations
rational functions - hyperbolas
estimation
inequalities
limit
Population
Basic arithmetic
Population & rates of growth (principal/interest)
Rates of change timescales of change
Linear functions
Spreadsheet
Geometric series
Integration
Exponential functions
Graphing
Solution of transcendental equations
(technology)
Tunnel
Formulae
Rates
Metric units
Conversion of units
Functionality
Rational functions
Graphs
Curve fitting
Regression
Quadratic function
Estimation
Derivative (quotient rule)
Maximum value

Using Real Life Contexts in Mathematics Teaching

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