Choosing between linear and non-linear models for HSC Maths Standard 2

For linear, check first differences: $1100 - 1000 = 100$, $1210 - 1100 = 110$, $1331 - 1210 = 121$, $1464 - 1331 = 133$. Differences are increasing, not constant, so not linear.

For exponential, check ratios: $\frac{1100}{1000} = 1.10$, $\frac{1210}{1100} = 1.10$, $\frac{1331}{1210} = 1.10$, $\frac{1464}{1331} \approx 1.10$. Ratios are constant at $1.10$, so the model is exponential.

Model: $P = 1000 (1.10)^t$. This is $10\%$ growth per year.

Markers reward both first-difference and ratio checks, the conclusion that ratios are constant, and the model with correct base $1.10$.

Check the product of speed and cost: $40 \times 18 = 720$, $60 \times 12 = 720$, $80 \times 9 = 720$, $120 \times 6 = 720$.

The product is constant at $720$, so the relationship is inverse: cost $\times$ speed $= 720$, that is $C = \frac{720}{s}$.

Markers reward checking the product, identifying it as constant, and writing the reciprocal model with the constant explicitly.

Test the first differences

The $t$ values are equally spaced, so subtract each cost from the next: $95 - 60 = 35$, $130 - 95 = 35$, $165 - 130 = 35$, $200 - 165 = 35$.

Conclude the model

The first differences are constant at $35$, so the data is linear and the gradient is $m = 35$ dollars per month.

Find the intercept and write the equation

At $t = 0$ the cost is $60, so $c = 60$ (the joining fee). The model is $C = 35t + 60$.

Predict at $t = 10$

Substitute: $C = 35 \times 10 + 60 = 410$.

State the answer

A constant first difference of $35 confirms a linear model $C = 35t + 60$, and after $10$ months the total cost is $410.00.

Test the ratios

The first differences ($-4500$, then $-3825$, then $-3251.25$) are not constant, so try ratios of consecutive values: $\frac{25\,500}{30\,000} = 0.85$, $\frac{21\,675}{25\,500} = 0.85$, $\frac{18\,423.75}{21\,675} = 0.85$.

Conclude the model

The ratio is constant at $0.85$, so the data is exponential with base $b = 0.85$. Because $1 - 0.85 = 0.15$, the car loses $15\%$ of its value each year.

Write the equation

The starting value is $30,000, so $V = 30\,000(0.85)^t$.

Predict at $t = 5$

Substitute: $V = 30\,000 \times 0.85^5 = 13\,311.16$ (to the nearest cent).

State the answer

A constant ratio of $0.85$ confirms exponential decay $V = 30\,000(0.85)^t$, and after $5$ years the car is worth $13,311.16.

Test the first differences

The minute values are equally spaced, so subtract each volume from the next: $58 - 50 = 8$, $66 - 58 = 8$, $74 - 66 = 8$.

Conclude the model

The first differences are constant at $8$, so the data is linear with gradient $m = 8$ litres per minute, and at minute $0$ the volume is $50$, so the model is $V = 8t + 50$.

Predict at $t = 6$

Substitute: $V = 8 \times 6 + 50 = 98$.

State the answer

A constant first difference of $8$ confirms the linear model $V = 8t + 50$, so after $6$ minutes the tank holds $98$ litres.

Try first differences first

$300 - 100 = 200$, $900 - 300 = 600$, $2700 - 900 = 1800$. These are not constant, so the data is not linear.

Test the ratios

Divide each count by the one before: $\frac{300}{100} = 3$, $\frac{900}{300} = 3$, $\frac{2700}{900} = 3$.

Conclude the model

The ratio is constant at $3$, so the data is exponential with base $b = 3$ (the shares triple each hour).

State the answer

A constant ratio of $3$ confirms an exponential model, and with $100$ shares at the start the equation is $S = 100(3)^t$.

Spot the context clue

A fixed total job split among a variable number of workers points to a reciprocal (inverse) model, so test the product $n \times d$.

Test the product

$4 \times 15 = 60$, $6 \times 10 = 60$, $8 \times 7.5 = 60$, $12 \times 5 = 60$.

Conclude the model

The product is constant at $60$, so the relationship is reciprocal with $k = 60$, and the equation is $d = \dfrac{60}{n}$. It cannot be linear: more workers give fewer days, so $d$ must decrease as $n$ rises.

Predict for $5$ workers

Substitute $n = 5$: $d = \dfrac{60}{5} = 12$.

State the answer

A constant product of $60$ confirms the reciprocal model $d = \dfrac{60}{n}$, so $5$ workers would take $12$ days. Check: $5 \times 12 = 60$, matching the constant.

Test for linear (first differences)

$25 - 0 = 25$, $40 - 25 = 15$, $45 - 40 = 5$, $40 - 45 = -5$. The first differences are not constant, so the data is not linear.

Test for exponential (ratios)

$\frac{25}{0}$ is undefined (division by zero at the start), and $\frac{40}{25} = 1.6$ while $\frac{45}{40} = 1.125$ are not equal, so the data is not exponential.

Test for quadratic (second differences)

Take differences of the first differences: $15 - 25 = -10$, $5 - 15 = -10$, $-5 - 5 = -10$.

Conclude the model

The second differences are constant at $-10$, so the data is quadratic (a single bend with one turning point, here a peak between $t = 3$ and $t = 4$).

State the answer

Differences that change rule out linear, an undefined and non-constant ratio rules out exponential, and a constant second difference of $-10$ confirms a quadratic model, the expected shape for the height of a thrown ball.

Test the first differences

$2400 - 2000 = 400$, $2880 - 2400 = 480$, $3456 - 2880 = 576$. The differences grow, so the growth is not by a constant amount and the data is not linear; the channel's claim is wrong.

Test the ratios

$\frac{2400}{2000} = 1.2$, $\frac{2880}{2400} = 1.2$, $\frac{3456}{2880} = 1.2$.

Conclude the model

The ratio is constant at $1.2$, so the data is exponential with base $b = 1.2$, i.e. $20\%$ growth per month. Starting from $2000$, the model is $S = 2000(1.2)^t$.

State the answer

Growing first differences rule out linear and a constant ratio of $1.2$ confirms exponential, so the better model is $S = 2000(1.2)^t$ at $20\%$ growth per month. The constant percentage, not a constant headcount, is the giveaway.

Test the first differences

The side lengths are equally spaced (steps of $2$ m), so subtract each cost from the next: $640 - 160 = 480$, $1440 - 640 = 800$, $2560 - 1440 = 1120$. They are not constant, so the data is not linear and the planner's claim is wrong.

Test the second differences

Take the differences of those first differences: $800 - 480 = 320$, $1120 - 800 = 320$.

Conclude the model

The second differences are constant at $320$, so the data is quadratic. This makes sense: the area of a square is $s^2$, so doubling the side roughly quadruples the area and the cost.

Predict at $s = 10$

The pattern is cost $= 40 s^2$ (check: $40 \times 2^2 = 160$, $40 \times 8^2 = 2560$). At $s = 10$: $40 \times 10^2 = 4000$.

State the answer

Growing first differences rule out linear and a constant second difference of $320$ confirms a quadratic model, so a $10$ m side would cost $4000.

Identify the model

A fixed distance covered at a variable speed is a fixed total shared out, so test the product $s \times t$: $40 \times 6 = 240$, $60 \times 4 = 240$, $80 \times 3 = 240$, $120 \times 2 = 240$.

Conclude

The product is constant at $240$ (the distance), so the model is reciprocal, $t = \dfrac{240}{s}$.

Predict at $s = 100$

Substitute: $t = \dfrac{240}{100} = 2.4$ hours, which is $2$ hours and $0.4 \times 60 = 24$ minutes.

Show why linear fails

A straight line through the first and last points, $(40, 6)$ and $(120, 2)$, has gradient $\frac{2 - 6}{120 - 40} = -0.05$ and equation $t = -0.05s + 8$. At $s = 100$ that predicts $t = -0.05 \times 100 + 8 = 3$ hours, well above the true $2.4$ hours, because the real data curves (falls fast then flattens) while a line cannot bend.

State the answer

A constant product of $240$ gives the reciprocal model $t = \dfrac{240}{s}$; the time at $100$ km/h is $2.4$ hours, and a linear fit overstates it as $3$ hours by ignoring the curve.

Rule out linear

First differences are $60 - 80 = -20$, $45 - 60 = -15$, $33.75 - 45 = -11.25$. They are not constant, so the data is not linear.

Confirm exponential

Test ratios: $\frac{60}{80} = 0.75$, $\frac{45}{60} = 0.75$, $\frac{33.75}{45} = 0.75$. The ratio is constant at $0.75$, so the data is exponential with base $b = 0.75$.

Interpret the base

Since $1 - 0.75 = 0.25$, the body eliminates $25\%$ of the remaining drug each hour (so $75\%$ stays).

Write the model and predict

The starting mass is $80$ mg, so $A = 80(0.75)^t$. At $t = 5$: $A = 80 \times 0.75^5 = 18.98$ mg (to $2$ decimal places).

State the answer

A constant ratio of $0.75$ confirms exponential decay $A = 80(0.75)^t$, with $25\%$ eliminated hourly, leaving about $18.98$ mg after $5$ hours. Falling by a constant percentage, not a constant amount, is why a line does not fit.

Find Model A's predictions and residuals

$y = 6x + 6$ gives, at $x = 1, 2, 3, 4$: $6 + 6 = 12$, then $18$, then $24$, then $30$. The residuals (data minus prediction) are $12 - 12 = 0$, $18 - 18 = 0$, $24 - 24 = 0$, $30 - 30 = 0$.

Find Model B's predictions and residuals

$y = 8(1.5)^x$ gives $8 \times 1.5 = 12$, $8 \times 1.5^2 = 18$, $8 \times 1.5^3 = 27$, $8 \times 1.5^4 = 40.5$. The residuals are $12 - 12 = 0$, $18 - 18 = 0$, $24 - 27 = -3$, $30 - 40.5 = -10.5$.

Compare

Model A has residuals of $0$ at every point, while Model B drifts to $-3$ and $-10.5$. The smaller residuals across the whole dataset belong to Model A, the linear model. (The first differences $6, 6, 6$ confirm linear directly.)

Predict with the better model at $x = 8$

$y = 6 \times 8 + 6 = 54$.

State the answer

Model A fits better (residuals all $0$ versus Model B's growing gaps), so the prediction at $x = 8$ is $y = 54$.

(a) Test Town A

The years are equally spaced, so check first differences: $12\,500 - 12\,000 = 500$, $13\,000 - 12\,500 = 500$, $13\,500 - 13\,000 = 500$. They are constant, so Town A is linear, $A = 12\,000 + 500t$ (a fixed $500$ people added per year).

Test Town B

The first differences ($480$, $509$, $539$) grow, so it is not linear; test ratios instead: $\frac{8480}{8000} = 1.06$, $\frac{8989}{8480} \approx 1.06$, $\frac{9528}{8989} \approx 1.06$. The ratio is constant at $1.06$, so Town B is exponential, $B = 8000(1.06)^t$ (growth of $6\%$ per year).

(b) Compare the models year by year

Town B catches up because a constant percentage beats a constant amount. Evaluate near the crossover: at $t = 15$, $A = 12\,000 + 500 \times 15 = 19\,500$ while $B = 8000 \times 1.06^{15} \approx 19\,172$, so B is still behind. At $t = 16$, $A = 12\,000 + 500 \times 16 = 20\,000$ while $B = 8000 \times 1.06^{16} \approx 20\,323$, so B has overtaken A.

State the answer

Town A is linear, $A = 12\,000 + 500t$; Town B is exponential, $B = 8000(1.06)^t$. Town B first exceeds Town A in year $16$ (about $20\,323$ versus $20\,000$), the year the accelerating curve passes the straight line.