At the marked point of the solution that starts at 80, which statement is true?

The population increases, and the growth rate decreases.

The population increases, and the growth rate increases.

The population decreases, and the growth rate decreases.

As t increases, what does the solution that starts at 1200 do?

It decreases and approaches 800.

It increases, because it starts with a large population.

It crosses 800 and continues to 400.

A population satisfies \dfrac{dP}{dt} = 0.1P\left(1 - \dfrac{P}{50}\right). At P = 40, is the graph of P concave up or concave down?

Concave up, because the population is increasing.

Concave up, because 40 is below the capacity.

It cannot be decided without P(0).

Topic 7.9 · BC

Describe concavity and long-term behavior

Builds on finding the population of fastest growth. 7 minutes.

A rising population with a falling growth rate

The upper graph shows the number of frogs \(P\) in a pond against time \(t\). The dashed line at 900 is the carrying capacity.

The orange line through the marker is the tangent. Its slope is the growth rate \(\cy{\frac{dP}{dt}}\), which the lower graph plots against \(t\).

Drag the marker from \(t = 0\) to the right. The population rises all the time, and the tangent becomes steeper until the population reaches the dashed orange line at 450, at about \(t = 3.5\) years.

After that point the population still rises, but the tangent becomes flatter. The lower graph falls, so the growth rate decreases while the population increases.

Before 450 the curve bends upward, so it is concave up. After 450 it is concave down, and the change happens at the inflection point.

Every start ends at the capacity

The graph has no numbers. You set the start \(P_0\) of the chalk curve, and the faint curves show other starts.

If the start is below \(K\), the rate \(\frac{dP}{dt}\) is positive, green in the lower graph, and the curve rises toward \(K\).

If the start is above \(K\), the rate is negative, shown red, and the curve falls toward \(K\).

A curve that starts below \(\frac{K}{2}\) bends upward and then downward. The bend changes at the open orange dot, where the curve crosses the dashed line at \(\frac{K}{2}\).

Every positive start ends at \(K\), but the bend depends on the start.

Notation

The first derivative \(\frac{dP}{dt}\) tells whether the population increases or decreases. Because \(k\) and \(P\) are positive, its sign is the sign of the bracket.
\[\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)\]
To find the concavity, expand the rate to \(kP - \frac{k}{K}P^2\) and differentiate with respect to \(t\). Each term gives a factor \(\frac{dP}{dt}\) by the chain rule, because \(P\) depends on \(t\).
\[\begin{aligned} \frac{d^2P}{dt^2} &= k\frac{dP}{dt} - \frac{2kP}{K}\frac{dP}{dt} \\ &= k\left(1 - \frac{2P}{K}\right)\frac{dP}{dt} \end{aligned}\]
If \(0 < P < \frac{K}{2}\), both factors are positive. The graph is concave up, so the growth rate increases.
\[\cg{+}\cdot\cg{+} > 0\]
If \(\frac{K}{2} < P < K\), the first factor is negative and \(\frac{dP}{dt}\) is positive. The graph is concave down, so the population increases but the growth rate decreases.
\[\cr{-}\cdot\cg{+} < 0\]
If \(P > K\), both factors are negative. The graph is concave up, so the population decreases and the decline slows.
\[\cr{-}\cdot\cr{-} > 0\]
For every positive start, the sign of \(\frac{dP}{dt}\) moves \(P\) toward \(K\).
\[\lim_{t\to\infty} P(t) = K\]

When it fails

A curve has an inflection point only if it crosses \(\frac{K}{2}\). From a start between \(\frac{K}{2}\) and \(K\), the graph is concave down throughout. From a start above \(K\), it is concave up throughout.

The limit \(K\) does not depend on the start, but the inflection point does.

Example 1: A start below the midpoint

The mussels in a bay satisfy \(\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{250}\right)\) with \(P(0) = 40\). Find \(\displaystyle\lim_{t\to\infty} P(t)\) and the population at which the graph changes concavity.

The start 40 is positive, so the population approaches the carrying capacity 250.
\[\lim_{t\to\infty} P(t) = 250\]
Differentiate the rate with the chain rule. Here \(\frac{2P}{250} = \frac{P}{125}\).
\[\frac{d^2P}{dt^2} = 0.2\left(1 - \frac{P}{125}\right)\frac{dP}{dt}\]
The bracket is 0 at \(P = 125\), and \(\frac{dP}{dt}\) stays positive for \(0 < P < 250\), so the concavity changes only at 125.
\[1 - \frac{P}{125} = 0 \;\Rightarrow\; P = \cy{125}\]
The start 40 is below 125, so the graph is concave up until \(P = 125\) and concave down after that.

Example 2: A start above the capacity

A population satisfies \(\dfrac{dP}{dt} = 0.3P\left(1 - \dfrac{P}{80}\right)\) with \(P(0) = 120\). Is \(P\) increasing or decreasing, and is its graph concave up or concave down?

The start 120 is above 80, so the rate is negative and \(P\) decreases toward 80.
\[0.3(120)\left(1 - \frac{120}{80}\right) = \cr{-18}\]
The chain rule gives the second derivative. Here \(\frac{2P}{80} = \frac{P}{40}\).
\[\frac{d^2P}{dt^2} = 0.3\left(1 - \frac{P}{40}\right)\frac{dP}{dt}\]
A tempting wrong move: At \(P = 120\) the bracket \(1 - \frac{120}{40} = -2\) is negative, so the second derivative is negative and the graph is concave down.
\[\frac{d^2P}{dt^2} < 0\]
This step reads only one factor. The chalk curve flattens as it falls toward the dashed line, so it is concave up. A concave down curve, like the red dashed one, becomes steeper and falls below 80.
Include both factors. The rate \(-18\) is also negative, so the product is positive and the graph is decreasing and concave up.
\[0.3(-2)(-18) = \cg{10.8}\]

Practice

A population satisfies \(\dfrac{dP}{dt} = 0.4P\left(1 - \dfrac{P}{500}\right)\) with \(P(0) = 100\). When \(P = 300\), is the population increasing or decreasing? Is the growth rate increasing or decreasing?

Move the marker to where \(P = 300\). The population increases when \(\frac{dP}{dt} > 0\), and the growth rate increases when \(\frac{d^2P}{dt^2} > 0\).

Solution

Find the rate at \(P = 300\). It is positive, so the population is increasing.
\[0.4(300)\left(1 - \frac{300}{500}\right) = \cg{48}\]
The chain rule gives the second derivative at \(P = 300\).
\[\begin{aligned} \frac{d^2P}{dt^2} &= 0.4\left(1 - \frac{2P}{500}\right)\frac{dP}{dt} \\ &= 0.4(-0.2)(48) = \cr{-3.84} \end{aligned}\]
The second derivative is negative, so the growth rate is decreasing. The population 300 is above \(\frac{K}{2} = 250\), and the lower graph is falling there.

Questions

The graph shows a solution of \(\dfrac{dP}{dt} = 0.25P\left(1 - \dfrac{P}{800}\right)\) with \(P(0) = 80\). A faint curve shows the solution from \(P(0) = 1200\).

At the marked point of the solution that starts at 80, which statement is true?
- The population increases, and the growth rate decreases.
- The population increases, and the growth rate increases.
- The population decreases, and the growth rate decreases.
- The population is constant.
Answer
The population increases, and the growth rate decreases.
As \(t\) increases, what does the solution that starts at 1200 do?
- It decreases and approaches 800.
- It decreases to 0.
- It increases, because it starts with a large population.
- It crosses 800 and continues to 400.
Answer
It decreases and approaches 800.
A population satisfies \(\dfrac{dP}{dt} = 0.1P\left(1 - \dfrac{P}{50}\right)\). At \(P = 40\), is the graph of \(P\) concave up or concave down?
- Concave down.
- Concave up, because the population is increasing.
- Concave up, because 40 is below the capacity.
- It cannot be decided without \(P(0)\).
Answer
Concave down.

Summary

The sign of \(\frac{dP}{dt}\) tells whether the population increases, and the sign of \(\frac{d^2P}{dt^2} = k\left(1 - \frac{2P}{K}\right)\frac{dP}{dt}\) tells whether the growth rate increases.

The bracket changes sign at \(\frac{K}{2}\), where an S-shaped curve has its inflection point, and every positive start ends at \(K\).

You use this to describe a logistic graph from any start and to tell an increasing population from an increasing growth rate.

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