Logistic growth and phase lines, question 11

Topic: Logistic growth and phase lines
Difficulty: Standard
Suggested time: 4 min
Marks: 2
Theory: UBC Math 100 §13.1Key results

A fish population of $N(t)$ thousand is harvested at 8 thousand fish per year, so that the equation below holds, and $N(0) = 15$ . A student predicts the long-run population as shown. Find the first line that is wrong.

\frac{dN}{dt} = 0.5N\left(1 - \frac{N}{100}\right) - 8

$0.5N\left(1 - \frac{N}{100}\right) - 8 = 0 \iff N^{2} - 100N + 1600 = 0$
Steady states: set dN/dt = 0 and multiply through by −200.
$(N - 20)(N - 80) = 0 \quad\Longrightarrow\quad N = 20 \text{ or } N = 80$
Factor.
$N = 10:\ \frac{dN}{dt} = -3.5 \qquad N = 50:\ \frac{dN}{dt} = 4.5 \qquad N = 100:\ \frac{dN}{dt} = -8$
One test value in each interval: 80 is stable and 20 is unstable.
$N(t) \to 80$
Solutions approach the stable steady state, so the population settles at 80 thousand.

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Theory UBC Math 100 §13.1Key results

Corrected working

Strategy

Find the steady states and the sign of dN/dt between them, then follow the arrow from the starting value rather than heading for the stable steady state.

Working

$N^{2} - 100N + 1600 = 0 \iff N = 20 \text{ or } N = 80$
The steady states, as the student found.
$N < 20:\ \frac{dN}{dt} < 0, \qquad 20 < N < 80:\ \frac{dN}{dt} > 0, \qquad N > 80:\ \frac{dN}{dt} < 0$
So 80 is stable and 20 is unstable.
$N = 15:\quad \frac{dN}{dt} = 0.5(15)(0.85) - 8 = -1.625 < 0$
The start is below 20, so the population decreases.
$\frac{dN}{dt} \le -1.625 \text{ for } 0 \le N \le 15:\quad N = 0 \text{ within } \frac{15}{1.625} \approx 9.2 \text{ years}$
It cannot cross 20, and dN/dt only grows more negative as N falls. 0 is not a steady state (dN/dt = −8 there), so N does not level off: it reaches 0, and the population is fished out.

Answer

N(t) \text{ falls to } 0: \quad \text{the population is fished out}

Check

Below 20 thousand the natural growth 0.5N(1 − N/100) stays under the catch of 8 thousand a year, at 15 only about 6.4, so the fish run out.

Where this goes wrong

Sending every solution to the stable steady state. A solution follows the arrow where it starts, and below the unstable steady state 20 that arrow points down, towards extinction.

Leans on

Reading solutions from the sign of dy/dt, from Slope fields and Euler’s method

Try a similar question

A fish stock of

N(t)

tonnes grows logistically and is harvested at a constant 32 tonnes per year, so that the equation below holds. Find the steady states. If the stock is 150 tonnes when harvesting begins, what does it approach in the long run?

\frac{dN}{dt} = 0.4N\left(1 - \frac{N}{500}\right) - 32

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Logistic growth and phase lines practice question 11