Exponential growth and decay | Year 12

01 / LEARN

The idea, step by step

Proportional rates

For $Q (t) = A e^{k t}$ , differentiation gives $Q^{'} (t) = k Q (t)$ . The constant $k$ is the instantaneous fractional growth rate per unit of time, not the exact percentage change over one whole time unit.

Positive $k$ means growth and negative $k$ means decay. The exact one-unit multiplier is $e^{k}$ ; the corresponding percentage change is $100 (e^{k} - 1) %$ .

Interpreting a model

A half-life or doubling time is found by comparing two quantities and taking logarithms. Use the time unit of $k$ consistently and round only after the calculation.

Exponential models can fail when resources are limited or a background level remains. State the time interval over which a conclusion is reasonable.

A common trap

Calling $k = 0.2$ exactly 20% growth over one hour.

Instead: It is a continuous proportional rate; the one-hour increase is $100 (e^{0.2} - 1) %$ .

02 / WORKED EXAMPLES

See the thinking

Example 1

An initial growth rate

A culture is modelled by $P (t) = 500 e^{0.2 t}$ , with $t$ in hours. Find its initial growth rate.

$P^{'} (t) = 100 e^{0.2 t}$ .
At $t = 0$ , this is 100.

100

organisms per hour

Example 2

A decay half-life

A substance follows $M (t) = 80 e^{- 0.1 t}$ grams. Find its half-life.

$40 = 80 e^{- 0.1 t}$ gives $e^{- 0.1 t} = 1 / 2$ .
Taking natural logs gives $t = ln 2 / 0.1$ .

10 ln 2 \approx 6.93

time units

Check your understanding

If $Q (t) = 30 e^{- 0.4 t}$ , what is $Q^{'} (0)$ ?

Show answer

$- 12$ quantity units per time unit

Differentiate to get $- 12 e^{- 0.4 t}$ and substitute zero.

03 / WORKSHEET

Your turn

6 questions · 14 marks · Work at your own pace.

Worksheet PDF Worked solutions

Read the questions online

For $P (t) = 1200 e^{0.05 t}$ people, find $P^{'} (0)$ .
2 marks
A mass is $M (t) = 50 e^{- 0.3 t}$ grams. Express $M^{'} (t)$ in terms of $M (t)$ .
2 marks
Find the doubling time for $Q (t) = 10 e^{0.4 t}$ .
2 marks
For $M (t) = 100 e^{- 0.2 t}$ mg, find the rate when $M = 25$ mg.
2 marks
Find the exact percentage increase in one year for $A (t) = 1000 e^{0.03 t}$ .
3 marks
A cooling excess above room temperature is $E (t) = 40 e^{- 0.2 t}$ degrees Celsius. Does this predict the actual temperature tends to 0 degrees Celsius?
3 marks

Read the worked solutions

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

For $P (t) = 1200 e^{0.05 t}$ people, find $P^{'} (0)$ .
2 marks
Differentiate: $P^{'} = 60 e^{0.05 t}$ .
At zero the exponential is 1.
$60$ people per time unit
A mass is $M (t) = 50 e^{- 0.3 t}$ grams. Express $M^{'} (t)$ in terms of $M (t)$ .
2 marks
The model has the form $A e^{k t}$ .
The derivative is $k$ times the original quantity.
$M^{'} (t) = - 0.3 M (t)$
Find the doubling time for $Q (t) = 10 e^{0.4 t}$ .
2 marks
Set $20 = 10 e^{0.4 t}$ .
Take logs and solve for $t$ .
$ln 2 / 0.4 \approx 1.73$ time units
For $M (t) = 100 e^{- 0.2 t}$ mg, find the rate when $M = 25$ mg.
2 marks
Use $M^{'} = - 0.2 M$ .
Substitute $M = 25$ .
$- 5$ mg per time unit
Find the exact percentage increase in one year for $A (t) = 1000 e^{0.03 t}$ .
3 marks
The one-year multiplier is $e^{0.03}$ .
Subtract 1 and multiply by 100.
$100 (e^{0.03} - 1) % \approx 3.045 %$
A cooling excess above room temperature is $E (t) = 40 e^{- 0.2 t}$ degrees Celsius. Does this predict the actual temperature tends to 0 degrees Celsius?
3 marks
Identify what $E$ measures.
A zero excess means the object has reached room temperature.
No. Only the excess tends to zero; actual temperature tends to room temperature.

Checking saved progress…

Continue learning

Derivatives of sine/cosine/tangent

Next lesson

Curriculum reference & lesson scope

SACE Stage 2 Mathematical Methods

Further differentiation and applications from the current SACE Methods outline. Logarithmic derivatives have their own chapter; integration follows later. Trigonometric calculus uses radians.

Official curriculum source ↗

Back to Differentiation and applications