Exponential growth and decay | Year 12

01 / LEARN

Understand the idea.

Proportional rates

For a positive quantity $Q (t) = A e^{k t}$ , differentiation gives $Q^{'} (t) = k Q (t)$ . Thus $k$ measures the instantaneous fractional rate: the rate at a given moment divided by the quantity present then.

The exact multiplier over one time unit is $e^{k}$ . Positive $k$ gives growth and negative $k$ gives decay; the one-unit percentage change is $100 (e^{k} - 1) %$ , rather than exactly $100 k %$ .

Interpreting a model

To find a doubling time or half-life, set the final quantity to the required multiple of its initial value. Dividing removes $A$ , and natural logarithms bring the unknown time out of the exponent.

A rate has quantity units per time unit; a doubling time has time units. Use the time unit attached to $k$ and round at the end. Resource limits, background levels and the fitted time interval affect whether a prediction remains credible.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

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.

Worked solution

1
Differentiate the population model.
$P^{'} (t) = 500 (0.2) e^{0.2 t} = 100 e^{0.2 t}$
The derivative of the exponent supplies the factor 0.2.
2
Evaluate at the initial time.
$P^{'} (0) = 100 e^{0} = 100$
Initial means t=0, when the exponential factor is 1.
3
Interpret the rate.
$P^{'} (0) = 100 organisms per hour$
The positive derivative means the culture is initially increasing at this instantaneous rate.

Answer

$100$ organisms per hour

EXAMPLE 02

A decay half-life

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

Worked solution

1
Set the mass to half its initial value.
$M (0) = 80, 40 = 80 e^{- 0.1 t}$
Half-life is the time for 80 grams to fall to 40 grams.
2
Isolate the exponential and take logarithms.
$e^{- 0.1 t} = \frac{1}{2}, - 0.1 t = ln \frac{1}{2} = - ln 2$
The natural logarithm reverses the exponential; the negative log matches decay.
3
Solve and report the time.
$t = \frac{ln 2}{0.1} = 10 ln 2 \approx 6.93 time units$
The positive time is measured in the units used in the model exponent.

Answer

$10 ln 2 \approx 6.93$ time units

03 / YOUR TURN

Try it yourself.

Check your understanding

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

Show answer & reasoning

$- 12$ quantity units per time unit

Differentiating $30 e^{- 0.4 t}$ gives $Q^{'} (t) = - 12 e^{- 0.4 t}$ . At zero this is $- 12$ quantity units per time unit, so the quantity is initially decreasing.

Keep practising

6 worksheet questions · 14 marks

Open worksheet 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 $t$ .
$\frac{ln 2}{0.4} \approx 1.73$ 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.

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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 ↗

In this lesson

Understand the idea.

Proportional rates

Interpreting a model

See the method in action.

An initial growth rate

Differentiate the population model.

Evaluate at the initial time.

Interpret the rate.

A decay half-life

Set the mass to half its initial value.

Isolate the exponential and take logarithms.

Solve and report the time.

Try it yourself.

Check your understanding

Keep practising