Year 10AC9M10M05

Modelling with Proportion and Scaling

Use mathematical modelling to solve practical problems involving proportion and scaling of objects; formulate problems and interpret solutions in terms of the situation; evaluate and modify models as necessary, and report assumptions, methods and findings.

Scaling is not one rule but three, and using the wrong one is the standard mistake. Double the length of a box and the surface area quadruples while the volume goes up eight times.

Builds onYr 10 · Surface Area and Volume of Composite Solids

Method

The three scale factors

k \qquad k^2 \qquad k^3

Lengths scale by $k$, areas by $k^2$, volumes by $k^3$.

Ratio must be like for like

Compare lengths with lengths. Comparing a length with an area is the commonest slip.

State your assumptions

A model is only usable if the reader knows what you assumed. This is directly assessed.

Check the answer against reality

If a model says a scaled-up animal weighs more than a building, the model, not the arithmetic, is wrong.

Worked example

e.g.

\text{A model car is built to a scale of } 1:18. \text{ The real car is } 4{,}5 \text{ m long and its bonnet has area } 1{,}8 \text{ m}^2.

Write the scale factor as a fraction. $k = \tfrac{1}{18}$
For a length, multiply by $k$ . $4{,}5 \times \tfrac{1}{18} = 0{,}25 \text{ m}$
For an area, multiply by $k^2$ — never by $k$ . $1{,}8 \times \left(\tfrac{1}{18}\right)^2 = 0{,}00556 \text{ m}^2$
For a volume or a mass, multiply by $k^3$ . $\left(\tfrac{1}{18}\right)^3 = \tfrac{1}{5832}$
State the assumption your model rests on. $\text{assumes the model is made of the same material throughout}$
Evaluate. Does the answer make sense in the situation? $0{,}25 \text{ m is a sensible model car length} \quad\checkmark$

Practice

Decide first whether the quantity is a length, an area or a volume.

\text{A photo is enlarged by } k=3. \text{ What happens to its area?}

Answer

\times 9

\text{Two similar cans have heights } 6 \text{ and } 9. \text{ Find the ratio of their volumes.}

Answer

\left(\tfrac{9}{6}\right)^3 = \tfrac{27}{8}

\text{A map scale is } 1:50\,000. \text{ How many km is } 4 \text{ cm?}

Answer

4 \times 50\,000 = 200\,000 \text{ cm} = 2 \text{ km}

\text{A tin's label area doubles. By what factor did its height grow?}

Answer

k^2 = 2 \ \Rightarrow\ k = \sqrt{2} \approx 1{,}41

\text{Why does a model elephant scaled up ten times not work as a real animal?}

Answer

\text{Mass grows by } 10^3 \text{ but bone cross-section only by } 10^2 \text{, so the legs cannot carry it.}