Exponential and logistic models | Year 12

01 / LEARN

The idea, step by step

Proportional growth and decay

The equation $P^{'} = k P$ means that the instantaneous rate is proportional to the amount present. Its solution is $P = P_{0} e^{k t}$ . Here $k$ is a continuous rate in reciprocal time units, not a fixed percentage added after each whole time period.

For growth $k > 0$ , doubling time is $ln 2 / k$ . For decay $k < 0$ , half-life is $ln 2 / (- k)$ . State the time unit and the period over which the constant-rate assumption is plausible.

Growth with a carrying capacity

A logistic model $P^{'} = r P (1 - P / K)$ has equilibrium amounts 0 and $K$ . For $0 < P < K$ its rate is positive; for $P > K$ the model predicts a decrease. The parameter $K$ is the carrying capacity, while $r$ has reciprocal-time units.

For positive initial data the solution has form $P = K / (1 + A e^{- r t})$ , where $A = (K - P_{0}) / P_{0}$ . The largest positive growth rate occurs at $P = K / 2$ . This model is an approximation: real carrying capacity and environmental conditions may change.

A common trap

Treat the carrying capacity as the initial population.

Instead: The initial value is P (0); carrying capacity is the long-term equilibrium K in the model.

02 / WORKED EXAMPLES

See the thinking

Example 1

Worked example

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

$300 = 200 e^{4 k}$ , hence $4 k = ln 1.5$ .
Doubling time solves $2 = e^{k T}$ , so $T = ln 2 / k$ .
Use the unrounded k when calculating T.

k = ln (1.5) / 4 \approx 0.1014

- 1

; doubling time

\approx 6.84

yr.

Example 2

Worked example

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

$A = (1000 - 200) / 200 = 4$ .
The growth rate is greatest at $P = 500$ .
$P^{'} = 0.4 (500) (1 - 0.5) = 100$ .

P = 1000 / (1 + 4 e^{- 0.4 t})

; maximum rate

100

units/time.

Check your understanding

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

Show answer

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

The vertex of $0.2 P - 0.2 P^{2} / 600$ is at $P = K / 2$ .

03 / WORKSHEET

Your turn

6 questions · 12 marks · Work at your own pace.

Worksheet PDF 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 $P^{'} = 0.3 P (1 - P / 800)$ and $P (0) = 100$ , find $P (t)$ .
2 marks
For the preceding logistic model, find the maximum growth rate.
2 marks
$P = 500 / (1 + 9 e^{- 0.2 t})$ . Find the initial amount and long-term limit.
2 marks
Under $P^{'} = 0.4 P (1 - P / 1000)$ , 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 $e^{5 k} = 1 / 2$ .
$k = - ln 2 / 5$ per hour.
$M = 80 e^{- (ln 2) t / 5}$
$N = 120 e^{0.08 t}$ , t in days. Find the time to reach 300.
2 marks
Divide by 120 and take logarithms.
$0.08 t = ln 2.5$ .
$t = ln (2.5) / 0.08 \approx 11.45$ days.
For $P^{'} = 0.3 P (1 - P / 800)$ and $P (0) = 100$ , find $P (t)$ .
2 marks
$A = (800 - 100) / 100 = 7$ .
Substitute r=0.3 and K=800.
$P = 800 / (1 + 7 e^{- 0.3 t})$
For the preceding logistic model, find the maximum growth rate.
2 marks
The maximum occurs at half the carrying capacity.
$0.3 (400) (1 - 400 / 800) = 60$ .
$60$ units/time at $P = 400$ .
$P = 500 / (1 + 9 e^{- 0.2 t})$ . 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 $P^{'} = 0.4 P (1 - P / 1000)$ , 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 ↗

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