IP EMaths Notes (Upper Sec, Year 3-4): 13) Variation and Rate of Change

Variation equations compress word problems into algebra. Identify whether quantities move together or in opposition, then encode that with proportionality constants and a positive constant \(k\) that captures the context (units, efficiency, scaling).

Quick reference

• Direct variation: \(y \propto x \Rightarrow y = kx\).
• Power variation: \(y \propto x^n \Rightarrow y = kx^n\) for real \(n\).
• Inverse variation: \(y \propto \dfrac{1}{x} \Rightarrow y = \dfrac{k}{x}\).
• Joint variation: \(y \propto xz \Rightarrow y = kxz\) (extend to more variables similarly).
• Combined variation: mix the patterns, e.g. \(y \propto \dfrac{x^2}{z} \Rightarrow y = k \dfrac{x^2}{z}\).
• Rate of change from a graph: gradient \(= \dfrac{\Delta y}{\Delta x}\); include units such as \(\pu{m.s-1}\) or \( \$ \space \text{per hour} \).

Spotting variation language

• "Varies directly / proportional to" \(\Rightarrow\) quantities increase together.
• "Varies inversely / inversely proportional" \(\Rightarrow\) one increases as the other decreases.
• "Varies jointly" \(\Rightarrow\) multiple factors multiply together.
• "Varies as the square / cube / square root" \(\Rightarrow\) attach the stated power.
• "Constant of variation" \(\Rightarrow\) solve for \(k\) using the given data before predicting new values.

Workflow for variation questions

1. Translate the English statement into a proportionality sentence (e.g. \(d \propto v^2\)).
2. Replace the proportionality symbol with an equation and constant \(k\).
3. Substitute known values to solve for \(k\) (keep units consistent).
4. Substitute the new input(s) and compute the required output, rounding sensibly.
5. Check the answer against intuition (does the result get larger or smaller as expected?).

Worked example 1 - Fuel consumption model

The stopping distance \(d\) of a car is directly proportional to the square of its speed \(v\). When \(v = \pu{40 km.h-1}\), \(d = \pu{24 m}\). Find the stopping distance when \(v = \pu{70 km.h-1}\).

1. Model: \(d = kv^{2}\).
2. Substitute known values: \(24 = k(40)^{2} = 1600k \Rightarrow k = \dfrac{24}{1600} = 0.015\).
3. For \(v = 70\): \(d = 0.015(70)^{2} = 0.015 \times 4900 = 73.5\).
4. Hence \(d \approx \pu{73.5 m}\).

Worked example 2 - Inverse variation (product stays constant)

The intensity \(I\) of light at a sensor varies inversely with the square of the distance \(r\) from the source: \(I \propto \dfrac{1}{r^2}\). When the sensor is \(\pu{2.5 m}\) away, the intensity is \(\pu{180 lm}\). Find the intensity at \(\pu{4.0 m}\).

1. Model: \(I = \dfrac{k}{r^2}\).
2. Solve for \(k\): \(180 = \dfrac{k}{(\pu{2.5 m})^2} = \dfrac{k}{6.25} \Rightarrow k = 1125\).
3. At \(r = \pu{4.0 m}\): \(I = \dfrac{1125}{(4.0)^2} = \dfrac{1125}{16} = 70.3125\).
4. Quote the answer to three significant figures: \(I \approx \pu{70.3 lm}\).

Inverse variation questions often include phrases like "constant product" or "work rate". Always state the restriction \(r \neq 0\).

Worked example 3 - Joint variation with rate of change

The volume of water \(V\) flowing through a pipe varies jointly with the cross-sectional area \(A\) and the flow speed \(u\): \(V \propto Au\). When \(A = \pu{18 cm2}\) and \(u = \pu{2.4 m.s-1}\), the volume rate is \(\pu{0.0432 m3.s-1}\). Find the rate when \(A = \pu{25 cm2}\) and \(u = \pu{2.9 m.s-1}\).

1. Convert the area to \(\pu{m2}\): \(\pu{18 cm2} = 1.8 \times 10^{-3} \pu{m2}\).
2. Model: \(V = kAu\).
3. Solve for \(k\): \(0.0432 = k (1.8 \times 10^{-3})(2.4) \Rightarrow k = 10\).
4. New area: \(\pu{25 cm2} = 2.5 \times 10^{-3} \pu{m2}\).
5. Substitute: \(V = 10 \times (2.5 \times 10^{-3}) \times 2.9 = 0.0725\).
6. Therefore \(V \approx \pu{7.25 \times 10^{-2} m3.s-1}\).

Rate of change on graphs and tables

• Average rate of change between \(x = a\) and \(x = b\): \(\dfrac{f(b) - f(a)}{b - a}\).
• Gradient of a line segment on a linear graph gives a constant rate (e.g. "cost per item").
• Always pair the gradient with units: if \(y\) is measured in dollars and \(x\) in hours, the rate is \( \$ \space \text{per hour} \).
• For distance-time graphs, gradient gives speed; for velocity-time graphs, the area under the curve gives displacement.

Worked example 4 - Average rate from a data table

A technician records the volume of a substance in a tank every \(\pu{10 s}\).

Estimate the average rate of change between \(t = \pu{10 s}\) and \(t = \pu{30 s}\).

\[ \frac{\Delta V}{\Delta t} = \frac{195 - 147}{30 - 10} = \frac{48}{20} = 2.4 \pu{L.s-1}. \]

State the meaning: the tank gains about \(\pu{2.4 L}\) of fluid per second over that interval.

Try this

1. The time \(T\) required to print a batch of photos varies inversely with the number of printers \(n\) operating and directly with the number of photos \(p\). One printer processes 120 photos in \(\pu{18 min}\). Form the model and find the time needed for 3 printers to process 300 photos.
2. The brightness \(B\) of a screen varies directly with the voltage \(V\) and inversely with the square of the distance \(d\). When \(V = \pu{24 V}\) and \(d = \pu{1.2 m}\), the brightness is \(\pu{540 cd.m-2}\). Predict \(B\) when \(V = \pu{18 V}\) and \(d = \pu{0.9 m}\).
3. A cyclist's position is recorded every \(\pu{5 s}\). At \(t = \pu{15 s}\) the displacement is \(\pu{210 m}\); at \(t = \pu{30 s}\) it is \(\pu{420 m}\). Interpret the average rate of change and state its units.