Back to the on-screen lesson ·
Exponential growth with a brake: $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{L}\right)$, the carrying capacity it settles at, the half-capacity where growth is fastest, and the S shape that separates a rising total from a falling rate.
Paper packet. Every task here also exists on screen, where it is checked automatically; answers written on paper are not assessed by Nydus. When you are back at a device, enter your answers there.
By the end of this lesson you can read a logistic model directly off its differential equation: the carrying capacity where the bracket vanishes, the population at which growth is fastest, the value of that fastest rate, and what happens to a population above the capacity as well as below it. You can also say what the inflection of an S-shaped curve means about a quantity in the world, and why a total that is still rising and a rate that has begun to fall are compatible.
You can separate a differential equation, and you know that $\dfrac{dP}{dt} = kP$ gives exponential growth — a population that doubles for ever. Nothing does. This lesson puts a brake on that model, and the result describes populations, epidemics and the spread of almost anything.
Logistic equation: $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{L}\right)$.
Carrying capacity $L$: the population the model levels off at.
Equilibrium: a population that does not change — here $P = 0$ and $P = L$.
Inflection: where the population curve changes concavity, at $P = \tfrac{L}{2}$; it is where the growth rate is largest.
Start from $\dfrac{dP}{dt} = kP$ and multiply by a factor that switches the growth off as the population approaches $L$:
$$\frac{dP}{dt} = kP\left(1 - \frac{P}{L}\right)$$
When $P$ is small the bracket is nearly $1$ and the model is indistinguishable from exponential. As $P$ climbs towards $L$ the bracket closes towards zero and so does the growth.
The capacity is an equilibrium, and a stable one. Above $L$ the bracket is negative, so the population falls back; below it, the population rises. $P = 0$ is an equilibrium too, but an unstable one — any population at all grows away from it.
Growth is fastest at half the capacity. Read the right-hand side as a function of $P$: it is a downward parabola with zeros at $0$ and $L$, so it peaks halfway between them, at $P = \tfrac{L}{2}$, where it equals $\tfrac{kL}{4}$.
That point is the inflection of the population curve. Before it the population is growing faster every year; after it the growth eases, while the total goes on rising. Those two sentences are easy to confuse and the difference between them is a second derivative — which is the single most useful thing this model teaches, because it is the thing most often got wrong in public.
Notice what was never needed. Every statement above came off the differential equation directly. The formula for $P(t)$ is not required for any of them.
Another way: picture
Picture a jar of yeast. At first there is food everywhere and the population doubles and doubles; in the middle it is growing as fast as it ever will; towards the end the jar is crowded and each new cell has to wait for space. The curve is a stretched S, steepest in the middle, and flattening towards a ceiling the jar itself sets.
Another way: steps
| Exponential | Logistic | |
|---|---|---|
| Equation | $\dfrac{dP}{dt} = kP$ | $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{L}\right)$ |
| Shape | ever-steepening | a stretched S |
| Fastest growth | at the end, for ever | at $P = \tfrac{L}{2}$ |
| Long run | unbounded | settles at $L$ |
For small $P$ the two are the same model, which is why early data in an epidemic looks exponential and why extrapolating it is so tempting and so wrong. What the logistic model adds is a ceiling, and everything interesting happens on the way to it.
Thinking growth is fastest at the capacity. It is zero there. That is what makes it a capacity.
Reading the inflection as the peak of the population. It is the peak of the growth rate. The population is only half way, and the second half is still to come.
Confusing 'cases are falling' with 'new cases are falling'. The first is about the total, the second about its derivative. A quantity can be rising, and rising more slowly, at the same moment — and in the middle of a logistic curve it always is.
$\dfrac{dP}{dt} = 0.2P\left(1 - \dfrac{P}{800}\right)$: the bracket vanishes at $P = 800$, so that is the capacity.
Read $L$ off the bracket.
Growth is fastest at $P = 400$, half of it.
Halfway between the zeros.
The rate there is $\dfrac{kL}{4} = \dfrac{0.2 \times 800}{4} = 40$ per unit time — the steepest the curve ever gets.
The same model with $P = 1000$, which is above the capacity of $800$.
The bracket is $1 - \dfrac{1000}{800} = -0.25$, so $\dfrac{dP}{dt} = 0.2 \times 1000 \times (-0.25) = -50$.
A negative growth rate.
The population falls back towards $800$. The capacity attracts from both sides, which is what makes it stable.
The bracket vanishes at $P = 600$, so the carrying capacity is $600$.
Read it off the denominator.
Growth is therefore fastest at $P = 300$, half the capacity, and that is where the population curve has its inflection.
Halfway between the two zeros.
The peak rate is $\dfrac{0.5 \times 600}{4} = 75$ — and a population of $300$ that is growing at $75$ still has half its eventual size ahead of it.
For $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{L}\right)$, match each feature to where it sits.
| $P = L$, where the bracket vanishes | $P = \dfrac{L}{2}$, halfway between the zeros | $\dfrac{kL}{4}$ | $P = 0$ | |
|---|---|---|---|---|
| The carrying capacity | ||||
| The population where growth is fastest | ||||
| The fastest growth rate | ||||
| The equilibrium a population grows away from |
A population satisfies $\dfrac{dP}{dt} = \dfrac{1}{2}P\left(1 - \dfrac{P}{600}\right)$. Complete the table.
| Value | |
|---|---|
| Carrying capacity | |
| Population where growth is fastest | |
| Fastest growth rate |
For $\dfrac{dP}{dt} = \dfrac{1}{5}P\left(1 - \dfrac{P}{1000}\right)$, what is the largest the growth rate ever gets?
Answer:
For $\dfrac{dP}{dt} = \dfrac{1}{10}P\left(1 - \dfrac{P}{1000}\right)$, at what population is the growth fastest?
Put the steps of reading everything off $\dfrac{dP}{dt} = kP\left(1 - \dfrac{P}{500}\right)$ into the order you do them.
Number the steps in order (write the number in the box):
An epidemic follows a logistic curve that levels off at $1200$ cases. The total has just reached $600$, and health officials announce that the number of **new** cases each day has begun to fall for the first time. Select every statement that follows.
This task has no paper form; do it on a device.
Lesson test: one question per skill, one attempt each, no hints. Your answers are checked when you submit.
For $\dfrac{dP}{dt} = \dfrac{1}{2}P\left(1 - \dfrac{P}{800}\right)$, what is the largest the growth rate ever gets?
Answer:
You can read a logistic model and say what its shape implies. Without looking: where does the growth rate reach zero, at what population is it largest, and what is still true of the total once the growth rate has begun to fall?
9. Your turn: $\dfrac{dP}{dt} = 0.5P\left(1 - \dfrac{P}{600}\right)$, step 3