Comparing linear with exponential change | Year 11

01 / LEARN

Understand the idea.

Different mechanisms

For equal input steps, a linear model has constant differences; a positive exponential model has constant ratios. Calculate both when identifying a pattern. A short curve can look almost straight, so appearance alone is weak evidence.

Adding twenty units and adding twenty percent are different instructions. The first adds a fixed amount; the second multiplies the current amount by $1.2$ , so its added amount changes as the base changes.

Compare fairly

Compare models at the same input values with the same units and allowed domains. A table can reveal when their order changes. For continuous models, opposite order at two inputs establishes a crossing between them because their difference is continuous.

Choose a model using its mechanism and agreement with measurements. A larger long-term forecast does not make a model more realistic. Both linear and exponential rules can cease to apply when conditions change.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Same start, different change

Compare $L = 100 + 20 n$ and $E = 100 {(1.2)}^{n}$ at years 0 to 4.

Predictions at integer years zero through four are joined as in the original. The sage linear values are 100,120,140,160,180; the gold exponential values are 100,120,144,172.8,207.36. The original x window zero to five and amount scale zero to 250 remain.

n	0	1	2	3	4
L	100	120	140	160	180
E	100	120	144	172.8	207.36

Worked solution

1
Calculate the linear values.
$\begin{aligned} L (0), L (1), L (2) & = 100, 120, 140 \\ L (3), L (4) & = 160, 180 \end{aligned}$
The linear model adds twenty at each yearly step.
2
Calculate the exponential values.
$\begin{aligned} E (0), E (1), E (2) & = 100, 120, 144 \\ E (3), E (4) & = 172.8, 207.36 \end{aligned}$
The exponential model multiplies by one point two each year.
3
Compare at matching years.
$\begin{aligned} E (0) & = L (0), E (1) = L (1) \\ E (2) & > L (2), E (3) > L (3) \\ E (4) & = 207.36 > 180 = L (4) \end{aligned}$
Agreement at the first two inputs does not continue at the remaining listed years.

Answer

At year 4: L = 180 and E = 207.36.

EXAMPLE 02

A later change of order

Compare $L = 200 + 30 t$ and $E = 200 {(1.1)}^{t}$ at times 8 and 9.

Worked solution

1
Evaluate both models at eight.
$\begin{aligned} L (8) & = 200 + 30 (8) = 440 \\ E (8) & = 200 {(1.1)}^{8} \approx 428.72 \end{aligned}$
The linear prediction is larger at this common input.
2
Evaluate both models at nine.
$\begin{aligned} L (9) & = 200 + 30 (9) = 470 \\ E (9) & = 200 {(1.1)}^{9} \approx 471.59 \end{aligned}$
The exponential prediction is larger at the next whole reporting time.
3
Interpret the reversal.
$E (8) < L (8), E (9) > L (9)$
If both rules apply continuously, a crossing occurs between eight and nine; the listed whole-time order has reversed by nine.

Answer

L is larger at 8; E is larger at 9.

03 / YOUR TURN

Try it yourself.

Check your understanding

At consecutive inputs, outputs are 10, 15, 20, 25. Which simple model fits exactly?

Show answer & reasoning

A linear model.

The consecutive differences are $15 - 10 = 5$ , $20 - 15 = 5$ and $25 - 20 = 5$ . The fixed increase identifies a linear model; the ratios are unequal.

Keep practising

6 worksheet questions · 13 marks

Open worksheet Worked solutions

Read the questions online

Classify 40, 50, 60, 70 and give a rule starting at $n = 0$ .
2 marks
Classify 40, 60, 90, 135 and give a rule.
2 marks
Compare $A = 300 + 30 t$ and $B = 300 {(1.08)}^{t}$ at $t = 2$ .
2 marks
Find the differences for 100, 110, 121, 133.1 and identify its change.
2 marks
One tank loses 4 L/min; another loses 4% of its current volume per minute. Identify the model types.
2 marks
Why might neither an unrestricted linear nor an exponential model describe a population indefinitely?
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Classify 40, 50, 60, 70 and give a rule starting at $n = 0$ .
2 marks
Constant difference is 10.
Linear; $L = 40 + 10 n$ .
Classify 40, 60, 90, 135 and give a rule.
2 marks
Each ratio is 1.5.
Exponential; $E = 40 {(1.5)}^{n}$ .
Compare $A = 300 + 30 t$ and $B = 300 {(1.08)}^{t}$ at $t = 2$ .
2 marks
Substitute the same time in both models.
$A = 360; B = 349.92; A$ is larger.
Find the differences for 100, 110, 121, 133.1 and identify its change.
2 marks
Differences vary but consecutive ratios are all 1.1.
10, 11, 12.1; exponential factor 1.1.
One tank loses 4 L/min; another loses 4% of its current volume per minute. Identify the model types.
2 marks
One removes a fixed amount; the other applies factor 0.96.
The first is linear; the second is exponential.
Why might neither an unrestricted linear nor an exponential model describe a population indefinitely?
3 marks
A simple rule may ignore crowding, migration or limited food.
State a useful domain and check new observations.
Resources, capacity and conditions may change.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: algebraic and graphical modelling of linear/exponential change, with interpretation and evaluation.

Official curriculum source ↗

In this lesson

Understand the idea.

Different mechanisms

Compare fairly

See the method in action.

Same start, different change

Calculate the linear values.

Calculate the exponential values.

Compare at matching years.

A later change of order

Evaluate both models at eight.

Evaluate both models at nine.

Interpret the reversal.

Try it yourself.

Check your understanding

Keep practising