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Longhand

Before the final

Misreading the arrows on a phase line

The habit: Draw the arrows from the sign of f between the steady states: a steady state is stable only when both arrows point at it, and a solution follows the arrow where it starts.

3 practice questions are built to catch it, in 1 topic.

See it fail in a figure: Not the nearest, in the logistic growth and phase lines lesson

Practise against it: 2 questions

What it looks like

A question from logistic growth and phase lines, with the answer this slip gives beside the right one.

Introductory2 marks · 3 min

The differential equation below has steady states y=1y = 1 and y=4y = 4. Which statement describes their stability?

dydt=(y−1)(y−4)\frac{dy}{dt} = (y - 1)(y - 4)
With the slip
y = 1 is unstable and y = 4 is stable.
Worked through
y = 1 is stable and y = 4 is unstable.

Check the sign of (y − 1)(y − 4) below 1, between 1 and 4, and above 4: it says which way solutions move near each steady state.

See the worked solution

Where it comes up

Practise against it

Each of these has an answer this slip gives. Work them on paper: if you fall into it, the verdict names it and the question comes back in review.

  1. 01
    A fish stock of N(t)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? dNdt=0.4N(1−N500)−32\displaystyle \frac{dN}{dt} = 0.4N\left(1 - \frac{N}{500}\right) - 32

    Logistic growth and phase linesStandard

  2. 02
    A fish population of N(t)N(t) thousand is harvested at 8 thousand fish per year, so that the equation below holds, and N(0)=15N(0) = 15. A student predicts the long-run population as shown. Find the first line that is wrong. dNdt=0.5N(1−N100)−8\displaystyle \frac{dN}{dt} = 0.5N\left(1 - \frac{N}{100}\right) - 8

    Logistic growth and phase linesStandard

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