AO2 - Building a mathematical model (Extended) | Cambridge IGCSE International Mathematics

A curve can match the first few measurements and still make poor predictions later. How do you choose a useful relationship and decide where to trust it? AO2 develops variables, assumptions and several model families, then tests predictions against data and practical limits.

Variables and assumptions

A mathematical model represents selected features of a situation. Start by defining each quantity, its unit and which quantities can change. Include information that affects the required result and state assumptions that make a calculation possible.

Mathematical model

A mathematical model is a representation of a situation using quantities and mathematical relationships. An assumption is a simplifying condition used to build it, such as a constant flow rate or a fixed cross-section. Its predictions depend on those conditions remaining appropriate.

For filling a tank, volume, height and time are different quantities. If the horizontal cross-sectional area is constant, $V=Ah$ . A volume flow rate multiplied by time gives volume, not height; divide by the cross-sectional area to obtain a height change.

Worked example 1: from flow to height

Question. An empty tank has constant horizontal cross-sectional area $2\,\mathrm{m}^2$ and height 3 m. Water enters at $0.3\,\mathrm{m}^3$ per minute. Build a model for water height $h$ metres after $t$ minutes, stating an assumption and the time range before it fills. Use $V=Ah$ .

Plan. Find volume from rate and time, then divide by the fixed area.

\begin{aligned} V&=0.3t &&\text{volume in cubic metres}\\ h&=\frac{0.3t}{2}=0.15t &&\text{height in metres}\\ 3&=0.15t\quad\Rightarrow\quad t=20 &&\text{full-tank boundary} \end{aligned}

Answer. $h=0.15t$ for $0\le t\le20$ , assuming the inflow stays constant and no water escapes before the tank is full.

Check. The tank capacity is $2(3)=6\,\mathrm{m}^3$ . In 20 minutes the inflow supplies $0.3(20)=6\,\mathrm{m}^3$ , independently checking the boundary time.

Keep the same tank and inflow but add a steady outlet. Net inflow, rather than the inlet alone, determines the height change.

Worked example 2: revising the net rate

Question. The same empty tank also releases water at $0.06\,\mathrm{m}^3$ per minute throughout filling. Build the revised height model and give its filling time. Assume both rates are constant.

Plan. Subtract the outlet rate before converting volume to height.

\begin{aligned} \text{net rate}&=0.3-0.06=0.24\,\mathrm{m}^3\text{ per minute}\\ h&=\frac{0.24t}{2}=0.12t\\ 3&=0.12t\quad\Rightarrow\quad t=25 \end{aligned}

Answer. $h=0.12t$ for $0\le t\le25$ ; the tank fills in 25 minutes under the stated steady-rate assumptions.

Check. Over 25 minutes, 7.5 cubic metres enter and 1.5 leave, giving $7.5-1.5=6\,\mathrm{m}^3$ , the tank capacity. An outlet should lengthen, not shorten, the original 20-minute filling time.

Real outlets can depend on water depth; the constant outlet is an assumption in this question. A more accurate measurement of the inlet does not remove an unsuitable assumption about the outlet.

Watch out

Do not mix a rate of volume with a rate of height. Units expose the error: cubic metres per minute must be divided by square metres to give metres per minute.

Define the new variables and use the non-zero initial level.

Choosing a model

The appearance of “increasing” or “decreasing” does not identify a model. Compare mathematical signatures across the data, using equal input steps for difference and consecutive-ratio tests. State whether the values fit exactly or approximately.

Candidate relationship	Useful evidence
Linear, $y=mx+c$	Constant changes in $y$ for equal changes in $x$
Quadratic, $y=ax^2+bx+c$	Constant non-zero second differences for equally spaced inputs
Variation, such as $y=k/x$	A constant appropriate product or ratio, here $xy$
Exponential, $y=Aq^t$ , $A>0,q>0$	A constant multiplier for equal time steps

These tests help select candidates. A finite table alone does not prove a relationship outside its observations. Prefer a simple model supported by the situation as well as by the pattern. Four short examples distinguish the four families.

Worked example 3: a constant addition

Question. Journey distances $d=1,2,3,4$ km have charges $C=8,11,14,17$ dollars. Choose a linear model and interpret its constants.

Plan. Use the common increase as the gradient and find the starting charge by substitution.

11-8=14-11=17-14=3

Write $C=3d+c$ . Using $d=1,C=8$ gives $c=5$ .

Answer. $C=3d+5$ : a 5-dollar fixed charge plus 3 dollars per kilometre. It fits every supplied pair.

Check. Substituting $d=4$ gives $12+5=17$ . Extending to $d=0$ predicts the fixed charge 5, but whether that is actually charged for no journey depends on the pricing rules.

Now the additions themselves increase. A straight model cannot preserve that pattern, so compare with squares rather than forcing a constant gradient.

Worked example 4: a square contribution

Question. For inputs $x=1,2,3,4,5$ , measured areas $A$ are $6,18,38,66,102$ square metres. Select a simple quadratic model and predict the area at $x=6$ .

Plan. Examine second differences, then compare the areas with a multiple of $x^2$ .

\begin{array}{c|rrrrr} A&6&18&38&66&102\\ \text{first differences}&12&20&28&36&\\ \text{second differences}&8&8&8&& \end{array}

Four times the squares gives $4,16,36,64,100$ , leaving 2 each time. Thus a candidate is $A=4x^2+2$ .

Answer. The quadratic model fits all five values and predicts $A(6)=4(36)+2=146$ square metres.

Check. Continuing the first differences gives next increase 44, so $102+44=146$ . The prediction agrees with that separate calculation, although no area at input 6 has yet been measured.

A falling output may follow a constant product. The next relationship comes from sharing a fixed amount, rather than subtracting a fixed amount each step.

Worked example 5: inverse variation

Question. A fixed quantity of liquid is pumped at steady rates $r=3,6,12$ litres per minute, taking $T=24,12,6$ minutes. Select a variation model and predict the time at 8 litres per minute.

Plan. Test the rate–time product across all observations.

3(24)=6(12)=12(6)=72

The constant product gives $T=72/r$ , for positive rates.

Answer. An inverse model fits, predicting $72/8=9$ minutes. The constant represents 72 litres of liquid.

Check. Pumping at 8 litres per minute for 9 minutes supplies 72 litres. A zero rate is excluded: it would not finish the task and the expression would divide by zero.

Finally, the amount falls by the same fraction in each equal time interval. That calls for multiplication, rather than the fixed-total inverse relationship.

Worked example 6: a constant percentage decrease

Question. At times $t=0,1,2,3$ hours, masses are $160,120,90,67.5$ grams. Choose an exponential model and predict the mass at 4 hours.

Plan. Find the initial amount and check every consecutive multiplier.

120/160=90/120=67.5/90=0.75

The initial value is 160, so $M=160(0.75)^t$ .

Answer. The model predicts $M(4)=50.625$ grams, or 50.6 grams to 3 significant figures. It represents a 25% decrease each hour.

Check. Applying one more multiplier to the last supplied mass gives $67.5(0.75)=50.625$ . The decreases $40,30,22.5$ are not constant, so the same table does not fit a straight-line decrease.

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AO2 - Building a mathematical model (Extended)

Variables and assumptions

Choosing a model

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