Logistic Models with Differential Equations

AP Calculus BC · Differential Equations

Logistic Models with Differential Equations

Exponential growth with a brake: the rate fades as the population nears its ceiling.

key idea$\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)\ \Rightarrow\ P\to K,\ \ \text{fastest at }P=\tfrac{K}{2}$

Setting up the model

Exponential growth has no ceiling; real populations do. The logistic model multiplies the exponential rate by a braking factor that shrinks to zero as the population nears its carrying capacity $K$. Growth starts nearly exponential, peaks halfway up, then tails off.

Reading a logistic equation

Get it into the standard shape $\dfrac{dP}{dt}=kP\left(1-\dfrac{P}{K}\right)$. Anything of the form $aP-bP^{2}$ can be factored into it.
Read the carrying capacity $K$ off directly — it is the non-zero value of $P$ that makes the right-hand side zero.
Read $k$ as the growth rate that would apply if $P$ were tiny, when the braking factor is close to $1$.
The equilibrium solutions are $P=0$ and $P=K$. Neither can be crossed, so the initial value decides which side the whole story plays out on.
For anything asking about fastest growth, go straight to $P=\dfrac{K}{2}$ — and note that this is a statement about the population, not about a specific time.

every solution is drawn to $P=K$

fastest growth at $P=\dfrac{K}{2}$

With $K=12$: a population starting below the capacity rises in an S-curve, and one starting above it falls — both settle onto $P=12$ without ever touching it. The dot marks $P=6$, exactly half the capacity, where the curve is steepest and its concavity flips.

what $P$ does in each band

	$0\lt P\lt\dfrac{K}{2}$	$P=\dfrac{K}{2}$	$\dfrac{K}{2}\lt P\lt K$	$P\gt K$
$\dfrac{dP}{dt}$	+	max	+	−
$\dfrac{d^{2}P}{dt^{2}}$	+	0	−	+
verdict	rising, cup	steepest point	rising, cap	falling, cup

Worked examples

EX 1

Read $K$ and $k$ off the equation

factor to standard form

For $\dfrac{dP}{dt}=0.06P-0.0002P^{2}$, find the carrying capacity and the growth constant.

Factor out $P$

$\dfrac{dP}{dt}=P\left(0.06-0.0002P\right)$.

Pull out $k$

$=0.06P\left(1-\dfrac{0.0002}{0.06}P\right)$.

Simplify the fraction

$\dfrac{0.0002}{0.06}=\dfrac{1}{300}$, so the equation is $0.06P\left(1-\dfrac{P}{300}\right)$.

Read the constants

$k=0.06$ and $K=300$.

Shortcut check

Setting the right side to zero gives $P=0$ or $P=300$ immediately — the fastest route to $K$ when only the capacity is wanted.

EX 2

Long-run behaviour

which equilibrium wins

For $\dfrac{dP}{dt}=0.5P\left(1-\dfrac{P}{800}\right)$ with $P(0)=1200$, find $\displaystyle\lim_{t\to\infty}P(t)$.

Identify the capacity

$K=800$.

Locate the starting value

$P(0)=1200\gt 800$, so the population starts above capacity.

Sign the rate

With $P\gt K$ the factor $1-\dfrac{P}{K}$ is negative, so $\dfrac{dP}{dt}\lt 0$ and $P$ falls.

Apply the barrier

$P=800$ is an equilibrium solution and cannot be crossed.

Conclude

$\displaystyle\lim_{t\to\infty}P(t)=800$. The limit is $K$ from either side — only the direction of approach differs.

EX 3

When is growth fastest?

$P=K/2$, not $t=0$

A fish population follows $\dfrac{dP}{dt}=0.3P\left(1-\dfrac{P}{5000}\right)$ with $P(0)=400$. At what population is it growing fastest, and how fast is that?

Use the halfway rule

Growth is fastest at $P=\dfrac{K}{2}=2500$.

Compute the rate there

$\dfrac{dP}{dt}=0.3(2500)\left(1-\dfrac{2500}{5000}\right)=0.3(2500)(0.5)$.

Evaluate

$375$ fish per unit time.

Confirm it is a maximum

$\dfrac{dP}{dt}$ as a function of $P$ is a downward parabola with roots $0$ and $5000$; its vertex sits at $P=2500$.

Answer the question asked

The population is $2500$ at that moment. Nothing here requires knowing the time at which it happens.

EX 4

Concavity and the inflection point

the second derivative

Explain why a logistic solution starting below $\dfrac{K}{2}$ has exactly one inflection point.

Differentiate the equation

$\dfrac{d^{2}P}{dt^{2}}=k\dfrac{dP}{dt}\left(1-\dfrac{2P}{K}\right)$, by the product rule on $kP\left(1-\dfrac{P}{K}\right)$.

Find where it vanishes

$\dfrac{dP}{dt}\neq0$ strictly between the equilibria, so $P''=0$ requires $1-\dfrac{2P}{K}=0$, that is $P=\dfrac{K}{2}$.

Check the sign change

Below $\dfrac{K}{2}$ the bracket is positive, above it negative — a genuine change in concavity.

Count the crossings

$P$ increases steadily from below $\dfrac{K}{2}$ toward $K$, so it passes that level exactly once.

Conclude

One inflection point, at $P=\dfrac{K}{2}$: growth accelerates up to it and decelerates after.

Units & interpretation

Quantity	Where it comes from	What it means
$K$	the non-zero root of the right side	the ceiling the population settles at
$k$	the coefficient of $P$ in standard form	the growth rate when $P$ is small
$P=\dfrac{K}{2}$	$P''=0$	the population of fastest growth
$P=0$ and $P=K$	roots of the right side	equilibrium solutions — barriers
$\dfrac{dP}{dt}$	the equation itself	individuals per unit time

A logistic answer is almost never obtained by solving the differential equation. Every question above is answered from the equation directly, which is exactly how the exam expects it to be handled.

Watch out

Fastest growth is a population, not a time

The answer to ‘when is it growing fastest?’ is $P=\dfrac{K}{2}$. Finding the corresponding $t$ requires the explicit solution and is rarely asked.

The limit is $K$ from either side

Starting above the capacity gives a decreasing solution that still tends to $K$. Do not assume logistic means increasing.

Factor before reading $K$

In $aP-bP^{2}$ the capacity is $\dfrac{a}{b}$, not $b$. Put the equation into standard form first.

Do not solve unless you are asked to

Carrying capacity, long-run limit, fastest growth and concavity all come straight from the equation. Separating variables here wastes several minutes.

Course map. Topic 7.9 · LO FUN-7.G, FUN-7.H — interpret the meaning of a logistic differential equation and identify the carrying capacity (EK FUN-7.G.1, FUN-7.H.1). BC-only.