Exponential and logistic models | Year 12

01 / LEARN

Understand the idea.

Proportional growth and decay

In $P^{'} = k P$ , the instantaneous rate is proportional to the amount present. Separation gives $P = P_{0} e^{k t}$ , where $k$ has reciprocal-time units. This continuous growth parameter differs from a fixed percentage added once each whole period.

For growth, $k > 0$ and the doubling time is $\frac{ln 2}{k}$ . For decay, $k < 0$ and the half-life is $\frac{ln 2}{- k}$ . Fit parameters using exact logarithms and round only the requested final quantities.

Growth with a carrying capacity

For $r > 0$ and $K > 0$ , the logistic equation $P^{'} = r P (1 - \frac{P}{K})$ has equilibria 0 and $K$ . Positive populations below $K$ grow, while those above $K$ decrease. The parameter $K$ is the carrying capacity.

With positive initial population $P_{0}$ , the model is $P = \frac{K}{1 + A e^{- r t}}$ , where $A = \frac{K - P_{0}}{P_{0}}$ . The rate, viewed as a quadratic in $P$ , peaks at $K / 2$ ; a particular future trajectory reaches that population only if its initial condition permits it.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example

A population starts at 200 and reaches 300 after 4 years under $P^{'} = k P$ . Find k and the doubling time.

Worked solution

1
Fit the exponential model.
$P (t) = 200 e^{k t}, 300 = 200 e^{4 k}$
The initial population fixes the multiplier, and the later measurement determines the continuous rate.
2
Solve for the rate constant.
$e^{4 k} = \frac{3}{2}, k = \frac{ln (1.5)}{4} \approx 0.1014 {yr}^{- 1}$
Taking natural logarithms removes the exponential, with time measured in years.
3
Solve the doubling condition.
$2 = e^{k T}, T = \frac{ln 2}{k} = \frac{4 ln 2}{ln (1.5)} \approx 6.84 yr$
A doubling time is measured relative to any starting amount in this constant proportional-rate model.

Answer

$k = ln (1.5) / 4 \approx 0.1014$ yr $- 1$ ; doubling time $\approx 6.84$ yr.

EXAMPLE 02

Worked example

For $P^{'} = 0.4 P (1 - P / 1000)$ and $P (0) = 200$ , find P (t) and the largest possible growth rate.

Worked solution

1
Identify the logistic parameters.
$r = 0.4, K = 1000, A = \frac{1000 - 200}{200} = 4$
Comparing the equation with the standard logistic form identifies its carrying capacity and growth parameter.
2
Write and check the population model.
$P (t) = \frac{1000}{1 + 4 e^{- 0.4 t}}, P (0) = \frac{1000}{5} = 200$
The initial condition fixes the denominator constant.
3
Maximise the rate as a quadratic.
$P^{'} = 0.4 P (1 - \frac{P}{1000}) = 100 - 0.0004 {(P - 500)}^{2}$
The squared term is nonnegative, so the rate is at most 100, achieved at P equals 500.
4
Evaluate and confirm that the trajectory reaches it.
$P_{max}^{'} = 100 units/time, P = 500 \Rightarrow t = \frac{ln 4}{0.4} > 0$
The initial population lies below 500, and the exact model reaches the maximising population at a positive time.

Answer

$P = 1000 / (1 + 4 e^{- 0.4 t})$ ; maximum rate $100$ units/time.

03 / YOUR TURN

Try it yourself.

Check your understanding

For $P^{'} = 0.2 P (1 - P / 600)$ , where is the maximum growth rate?

Show answer & reasoning

At $P = 300$ , with rate $30$ units/time.

Here $K = 600$ and $r = 0.2$ . The concave quadratic rate peaks at $P = K / 2 = 300$ , giving $P^{'} = 0.2 (300) (1 - \frac{300}{600}) = 30$ units per time.

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6 worksheet questions · 12 marks

Open worksheet Worked solutions

Read the questions online

An amount decays with half-life 5 hours and starts at 80 g. Find the exponential model.
2 marks
$N = 120 e^{0.08 t}$ , $t$ in days. Find the time to reach 300.
2 marks
For $P^{'} = 0.3 P (1 - \frac{P}{800})$ and $P (0) = 100$ , find $P (t)$ .
2 marks
For the preceding logistic model, find the maximum growth rate.
2 marks
$P = \frac{500}{1 + 9 e^{- 0.2 t}}$ . Find the initial amount and long-term limit.
2 marks
Under $P^{'} = 0.4 P (1 - \frac{P}{1000})$ , state the predicted direction of change at $P = 1200$ .
2 marks

Read the worked solutions

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

An amount decays with half-life 5 hours and starts at 80 g. Find the exponential model.
2 marks
Half-life means $e^{5 k} = \frac{1}{2}$ .
$k = - \frac{ln 2}{5}$ per hour.
$M = 80 e^{- \frac{(ln 2) t}{5}}$
$N = 120 e^{0.08 t}$ , $t$ in days. Find the time to reach 300.
2 marks
Divide by 120 and take logarithms.
$0.08 t = ln 2.5$ .
$t = \frac{ln (2.5)}{0.08} \approx 11.45$ days.
For $P^{'} = 0.3 P (1 - \frac{P}{800})$ and $P (0) = 100$ , find $P (t)$ .
2 marks
$A = \frac{800 - 100}{100} = 7$ .
Substitute $r = 0.3$ and $K = 800$ .
$P = \frac{800}{1 + 7 e^{- 0.3 t}}$
For the preceding logistic model, find the maximum growth rate.
2 marks
The maximum occurs at half the carrying capacity.
$0.3 (400) (1 - \frac{400}{800}) = 60$ .
$60$ units/time at $P = 400$ .
$P = \frac{500}{1 + 9 e^{- 0.2 t}}$ . Find the initial amount and long-term limit.
2 marks
At zero the denominator is $10$ .
As $t$ increases, the exponential term tends to zero.
$P (0) = 50$ ; limit $500$ .
Under $P^{'} = 0.4 P (1 - \frac{P}{1000})$ , state the predicted direction of change at $P = 1200$ .
2 marks
$P^{'} = 0.4 (1200) (1 - 1.2) = - 96$ .
The negative sign indicates decrease.
Decreasing, at $96$ units/time.

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 2 Specialist Mathematics topic 6.1–6.2, with the approved school-level rates, differential-equation and motion applications.

Official curriculum source ↗

In this lesson

Understand the idea.

Proportional growth and decay

Growth with a carrying capacity

See the method in action.

Worked example

Fit the exponential model.

Solve for the rate constant.

Solve the doubling condition.

Worked example

Identify the logistic parameters.

Write and check the population model.

Maximise the rate as a quadratic.

Evaluate and confirm that the trajectory reaches it.

Try it yourself.

Check your understanding

Keep practising