Justifying the Obvious. Some Teaching Notes

Justifying the Obvious. Some Teaching NotesAuthor(s): George BallSource: Mathematics in School, Vol. 14, No. 5 (Nov., 1985), pp. 19-21Published by: The Mathematical AssociationStable URL: http://www.jstor.org/stable/30214025 .

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JUSTIF 11iT HE1OBUIOUS

SOME TEACHING

NOTES by

George Ball, Department of Mathematics and

Computer Science, Polytechnic of Wales

Context It was my first session with the Diploma in Mathematical Education' class. The group, 12 in total, consisted of infant and junior school teachers, all with differing backgrounds, and all a little unsure of themselves mathematically.

Assumptions Real Mathematical problem I model

Interpretation

Fig. 1 Mathematical modelling as a two-way process3.

We had been discussing the role of mathematical modelling in the solution of real problems'. I had pointed out the two-way nature of the modelling process (see Figure 1) and had illustrated the stages involved with a diagram similar to Figure 2.

REAL WORLD ABSTRACT WORLD Formulate

Real Problem Mathematical Model

Validate Analyse

Solution to Solution to Real Life Mathematical Problem Problem

Interpret

Fig. 2 Stages in the development of a mathematical model'.

We drew up our own flow chart before moving on to look at some "case studies".

As a first illustration, I had chosen to adapt an example due to Daviess, which I felt demonstrated well the essential features of modelling. The example also had the advantage of allowing the class to do some mathematics "at their own level". The intention was that the group should involve themselves with some number work, some elementary algebra, and some simple work on arrangements by way of tree-diagrams. However, the mathematics which actually materialised from this session was much more far-reaching and important than I had envisaged.

The Original Problem Showering in a cold changing room can be unpleasant. When drying, body-heat is lost, and this leads to discom- fort. Assume

(i) body-heat lost per unit time is proportional to the area exposed;

(ii) body-heat is only lost from a wet surface, and heat loss ceases immediately the surface becomes dry.

Thus, for a body with a surface area of A square units which is left wet for t seconds, the heat loss H, is proportional to A x t, or, in appropriate units,

H= At.

Now consider the manner in which a body is dried. Suppose drying is accomplished at a rate of d square units per second, then the total time to dry A square units is given by

t = (Ald) secs.

Since drying is a continuous process in which some parts of the surface are dried before others (see Figure 3), it is reasonable to suppose that:

2-1At

where the 2 can be considered as an averaging factor.

1 This region dried

Directionalmost immediately

of drying Area A sq. units

--- This region remains wet until near the end of the drying process

Fig. 3 Drying is a continuous process.

It was at this point that the class voiced their first real objections. It was not clear to them that the factor of -was legitimate. They appreciated that since drying was a continuous process, H would not be equal to At. But why 12?

The session was coming to an end and so I promised to prepare some material for the following week. I give below an outline of the work which originated from a statement which many of us would consider to be intuitively obvious.

Mathematics in School, November 1985 19



Investigation 1

A Numerical Approach

(i) Consider a surface area of 120 sq. units which is dried at a rate of 10 sq. units per sec. The surface is "all-dry" in

120 t - 12 secs. Now if the whole area had remained 10 wet for the full 12 secs, then H would have equalled 120 x 12 = 1440 units. As the surface is dried continuously the value of H will be less than this value. Nevertheless, let us take H1 = 1440 as a first approximation for the heat lost H.

(ii) Consider the surface area of 120 sq. units split into

two equal parts. 60 60

direction of drying

60 The first half takes -60 = 6 secs to dry and consequently 10

we can approximate the heat lost from this part by 60 x 6 units. The second half remains wet for a full 6 secs before

60 drying begins and it is then dried in a further - = 6 secs. 10 Thus heat lost from second half is approximately 60 x (6 + 6) units and total heat lost is given approximately by

H2= [60 x 6] + [60 x (6 + 6)] = 1080.

H2 should be a better approximation to H as we have started to take into account the continuous nature of the drying process.

(iii) Consider the area of 120 sq. units split into three equal parts.

40 40 40

direction of drying

This time H can be approximated by H3 where

H3 = [40 x 4] + [40 x (4 + 4)] + [40 x (4 + 4 + 4)]

= 960

and once again H3 will be a better approximation than either H1 or H2. Use a similar method to show that if the area is split into four equal parts, then H4 = [30 x 3] + [30 x (3 + 3)] + [30 x (3 + 3 + 3)] + [30 x (3 + 3 + 3 + 3)].

(iv) Obtain similar expressions for H5, H6, Hs and Hio0

By dividing the total area under consideration into smaller and smaller parts, we are modelling more closely the drying process. The series of values obtained, namely H1, H2, H3, ..., Hio ..., should approximate more and more closely to the actual heat lost H. Do the numerical values you have obtained enable you to suggest a value for H.?

The results shown in Table 1 were obtained without too much trouble.

Although a general downward trend was appreciated, it is not clear that the limit of this sequence is 720. A graphical representation was suggested, but did not find favour with the majority of the class.

Table 1

n Hn

1 1440 2 1080 3 960 4 900 5 864 6 870 8 810

10 792

Investigation 2

Pattern Power

Consider the expressions obtainedfor Hk (k = 1, 2, 3 and 4):

H, = 120 x 12 H2 = [60 x 6] + [60 x (6 + 6)] H3 = [40 x 4] + [40 x (4 + 4)] + [40 x (4 + 4 + 4)] H4 = [30 x 3] + [30 x (3 +3)] + [30 x (3 +3 +3)]

+[30 x (3 +3 +3 +3)] These expressions can be re-written:

H, = 120 x 12 = 120 x 12 H2 = [60 x 6] + [60 x 6 x 2] = 60 x 6 x (1 + 2) H3= [40 x 4] + [40 x 4 x 2]

+[40 x 4 x 3] =40x 4x (I +2+3) H4= [30 x 3] + [30 x 3 x 2]

+ [30 x 3 x 3] +[30 x 3 x 4] =30 x 3 x (I + 2 +3 + 4)

(i) Write down similar expressions for H5, H6, H8 and

H1o, and evaluate. (ii) Find the values of H20, H60 and H120.

(iii) Can you now suggest a numerical value for H?

This exercise proved more stimulating. The class were interested by the use of the distributive law for factoris- ation, and the fact that I had not summed (1 + 2), (1 + 2 + 3) etc. They found it easy, and acceptable, to write down similar results. While evaluating H60 on a calculator, a class member volunteered the opinion that "there must be an easier way". The focus of attention then turned to the number pattern

1 + 2 = 3 (2 terms) 1 + 2 + 3 = 6 (3 terms) 1 + 2 + 3 + 4 = 10 (4 terms)

1+2+3+4+ ...= ? (nterms)

It was observed that 3, 6, 10, ... were all triangular numbers (we made patterns with small cubes). "Triangles have half the area of rectangles", it was suggested (we checked with some pin-board examples). I wrote:

2(1 + 2) = 6 (2 terms) 2(1 + 2 + 3) = 12 (3 terms) 2(1 + 2 + 3 + 4) = 20 (4 terms)

I emphasised: 2 terms- 6 3 terms 12 4 terms 20

n terms - ?

20 Mathematics in School, November 1985



Table 2

n H.

20 756 60 732

120 726 240 723

The pattern was discovered! H20, H60, H120 and H240 were quickly added to the table of results (see Table 2):

The class were still not prepared to accept that H would equal 720. I focussed attention on values of (Hi - H2i)

H60 - H120 = 6 H120 - H240 = 3

What did they think (H240 - H480) would be? Everyone agreed that it should be 11. We verified this result much to their obvious pleasure. What about (H48o - H960) etc? A discussion of infinite series followed. Everyone agreed that H would have a value of 720!

Investigation 3 The General Result

(i) Suppose the area of A square units is dried at a rate of d square units per second. Then, total time of drying

=t secs. Proceeding as before, take H,, the first (d

approximation to H, to be At. (ii) Now consider the surface split into two equal parts.

t The first half is dried in - secs, and consequently we can

2 A t

approximate the heat loss from this part by - 0-

units. 22

t The second half remains wet for - secs before drying 2

t begins, and is then dried in a further secs. Thus

2

H 2 2 - 2 -2 - At A t A t

....-+ ---x2 = --(1+2) 22 22 2-2

(iii) By dividing A into three equal parts and using a similar argument, show that

A t+ A t + +

A t +t+ H3

3 3 3 3 3 3 (3 3 3

A t A t 2

A t 3 + -i-- x2 + -.--x 3

3 3 3 3 3 3

A t

(1 + 2 + 3) 3 3

(iv) Write down the corresponding result for H4, H5 and H6.

(v) What will be the corresponding result when A is divided up into n equal parts?

(vi) By dividing the area into smaller and smaller parts (i.e. by making n larger and larger) we are modelling more closely the continuous process of drying. Show that as n becomes very large the value of H, obtained in (v)

1 approaches -At. 2

The class were happy to work with letters instead of numbers. They moved quickly to the result

At n(n + 1) n + 1 1 H = J At

-=At 1 + -

n 2 n n

1 All agreedthat 1+ approached 1 as .n became very

n

large; all agreed that H= At. I was suitably pleased and felt that the time and effort put

into the investigations had been well rewarded. Further, I resolved that in future I would try not to let my intuition dictate to a class which results are obvious and which results need some justification!

The students were asked to consider situations in which they might have used "pattern work" in their own teaching of mathematics. Subsequently each member of the class produced a "pattern poster", and these visual aids are now on display in the Diploma Classroom at the Polytechnic.

References 1. Diploma in Mathematical Education - a Mathematical Association

Course for experienced teachers of children in the 5-13 age range. This particular course is taught by members of the Department of Mathema- tics and Computer Science at the Polytechnic of Wales, Treforest.

2. See, for example, H. Burkhardt (1982) The Real World and Mathema- tics, Blackie.

3. Burghes, D. N. and Read, G. A. (1978) Journal of Mathematical Modellingfor Teachers 1, 1.

4. 1980 "Mathematics Across the Curriculum", An Open University Second Level Course, PME 233.

5. Davies, M. J. (1978) Journal of Mathematical Modelling for Teachers, 1, 1.

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Mathematics in School, November 1985 21



Justifying the Obvious. Some Teaching Notes

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