At P = 150, what happens to the culture?

It grows, because 150 is a large population.

Determine growth or decline

Builds on identifying the carrying capacity and the equilibria. 6 minutes.

Does the herd grow or shrink?

The curve shows how fast a herd of elk changes. The horizontal axis is the number of elk \(P\), and the vertical axis is the rate \(\cy{\frac{dP}{dt}}\), in elk per year.

The curve is green where the rate is positive, because the herd gains elk. It is red where the rate is negative, because the herd loses elk.

The orange marker starts at 300 elk. The curve is green there and the rate is 90 elk per year, so the herd grows.

Move the marker to 1500 elk. The curve is red and the rate is \(-150\) elk per year, so the herd shrinks even though it is large.

At 1200 elk the curve crosses the axis, so the rate is 0 and the herd stays the same size.

The side of K decides the sign

The graph now has no numbers. It shows any logistic population, and the curve crosses the axis at 0 and at the carrying capacity \(K\).

Below the graph, a second line repeats the population axis, and its arrows point the way the population moves.

Between 0 and \(K\) the curve is green, so the population increases, and the arrow on the line points to the right.

To the right of \(K\) the curve is red, so the population decreases, and the arrow points to the left.

The arrows on both sides of \(K\) point toward \(K\). Which side of \(K\) the population is on decides the sign of the rate.

Notation

The rate is a product of three factors: the constant \(k\), the population \(P\), and the bracket \(1 - \frac{P}{K}\). Both \(k\) and \(P\) are positive.
\[\frac{dP}{dt} = k\cdot P\cdot\left(1 - \frac{P}{K}\right)\]
Two factors are positive, so the bracket alone decides the sign. The curve turns green where the rate is positive and red where it is negative.
If \(P < K\), then \(\frac{P}{K}\) is less than 1 and the bracket is positive, so the rate is positive and the population increases.
\[\begin{aligned} 0 0 \\ &\Rightarrow \cg{\frac{dP}{dt} > 0} \end{aligned}\]
If \(P > K\), then \(\frac{P}{K}\) is greater than 1 and the bracket is negative, so the rate is negative and the population decreases.
\[\begin{aligned} P > K &\Rightarrow 1 - \frac{P}{K} < 0 \\ &\Rightarrow \cr{\frac{dP}{dt} < 0} \end{aligned}\]
If \(P = K\), the bracket is 0, so the rate is 0 and the population does not change.

When it fails

The claim that a larger population grows faster fails above \(K\). Press the button to move the marker to the far right, where the population is largest.

Above \(K\) the bracket is negative, and it becomes more negative as \(P\) grows, so the largest population has the most negative rate. The sign of the rate depends on the side of \(K\), not on the size of \(P\).

Example 1: One population below K and one above

A trout population satisfies \(\dfrac{dP}{dt} = 0.06P\left(1 - \dfrac{P}{3000}\right)\). Decide whether the population grows or shrinks when \(P = 1000\) and when \(P = 4000\).

The capacity is \(K = 3000\). The population 1000 is below it, and 4000 is above it.
\[1000 < 3000 < 4000\]
At \(P = 1000\) the bracket is positive, so the rate is positive and the population grows by 40 trout per year.
\[0.06(1000)\left(1 - \frac{1000}{3000}\right) = \cg{40}\]
At \(P = 4000\) the bracket is negative, so the rate is negative and the population shrinks by 80 trout per year.
\[0.06(4000)\left(1 - \frac{4000}{3000}\right) = \cr{-80}\]

Example 2: A flock above the capacity

A flock of pigeons satisfies \(\dfrac{dP}{dt} = 0.002P(90 - P)\) with \(P(0) = 120\). Does the flock grow or shrink at time 0?

The rate has three factors. The number 0.002 is positive, \(P = 120\) is positive, and the third factor \(90 - P\) is 0 at the capacity 90.
A tempting wrong move: Two of the factors are positive, so the rate is positive and the flock grows.
\[\frac{dP}{dt} > 0\]
This step ignores the third factor. The curve at \(P = 120\) lies below the axis and is red, so the rate there is negative.
Include the third factor. It is negative at \(P = 120\).
\[90 - 120 = -30\]
One negative factor makes the product negative. The flock shrinks by 7.2 birds per year at time 0, because 120 is above the capacity 90.
\[0.002(120)(-30) = \cr{-7.2}\]

Practice

A population satisfies \(\dfrac{dP}{dt} = 0.8P - 0.004P^2\). For which populations \(P > 0\) does the population grow?

Factor the rate until the bracket has the form \(1 - \frac{P}{K}\). The population grows where the curve is green.

Solution

Factor out \(0.8P\). The second term divided by \(0.8P\) is \(\frac{0.004}{0.8}P = \frac{P}{200}\).
\[\begin{gathered} 0.8P - 0.004P^2 \\ = 0.8P\left(1 - \frac{P}{200}\right) \end{gathered}\]
The capacity is \(K = 200\), and the factors 0.8 and \(P\) are positive.
The rate is positive when the bracket is positive, which means \(P < 200\).
\[\cg{0 < P < 200}\]
The population grows for \(0 200\).

Questions

The graph shows the rate \(\dfrac{dP}{dt}\) of a cell culture against its population \(P\), in thousands of cells. The equation is \(\dfrac{dP}{dt} = 0.05P\left(1 - \dfrac{P}{100}\right)\).

For which populations does the culture grow?
- \(0 100\)
- \(0 0\)
Answer
\(0 < P < 100\)
At \(P = 150\), what happens to the culture?
- It shrinks toward 100.
- It grows, because 150 is a large population.
- It stays at 150.
- It shrinks to 0.
Answer
It shrinks toward 100.
A population satisfies \(\dfrac{dP}{dt} = 0.9P - 0.003P^2\). For which \(P > 0\) does the population shrink?
- \(P > 300\)
- \(0 333\)
- \(P > 150\)
Answer
\(P > 300\)

Summary

You factor the rate as \(kP\left(1 - \frac{P}{K}\right)\) and compare the population with \(K\), which gives the sign of \(\frac{dP}{dt}\).

This works because \(k\) and \(P\) are positive, so only the bracket changes sign, and it changes sign at \(P = K\).

You use it to decide whether a logistic population grows or shrinks at a given size, even one above the capacity.

This is one of five skills in topic 7.9 of Which & Why: AP Calculus.