Logistic Models — AP® Calculus BC Lesson

Part 1 · Hook

Why this matters

An exponential model says a population grows faster and faster forever. Real populations run out of food and space: growth is fast at first, then slows, and the population levels off. The logistic model builds that limit into the differential equation, and you can read the limit, and the moment of fastest growth, straight from the equation.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. If $\frac{dP}{dt} = 0.3P$ , then as $t$ grows, $P$ (starting at a positive value)

grows without bound
levels off at 0.3
decreases to 0
stays the same

Show the answer

An exponential model $P = P_0 e^{0.3t}$ grows without bound.

Correct: grows without bound:
levels off at 0.3:
decreases to 0:
stays the same:

2. What are the equilibrium solutions of $\frac{dy}{dt} = y(4 - y)$ ?

$y = 0$ and $y = 4$
$y = 4$ only
$y = 2$
$y = -4$ and $y = 0$

Show the answer

The rate is 0 when $y = 0$ or $4 - y = 0$ .

Correct: $y = 0$ and $y = 4$ :
$y = 4$ only:
$y = 2$ :
$y = -4$ and $y = 0$ :

Part 4 · See it

See it first

Left: the population P against time t for dP/dt = 0.0008P(1000 − P) with P(0) = 100. It rises slowly, then steeply, then levels off below the dashed carrying capacity line P = 1000, and is steepest where it crosses P = 500, near t = 2.75, where its concavity changes from up to down. Right: the rate dP/dt against P, a downward parabola that is 0 at P = 0 and P = 1000, with its largest value 200 at P = 500. — For dP/dt = 0.0008P(1000 − P) with P(0) = 100: the population rises toward the carrying capacity 1000 and is steepest at P = 500 (left). The rate as a function of P is a downward parabola with its peak, 200, at P = 500 (right). LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

The growth rate is jointly proportional to $P$ and to the room left, $K - P$ $\frac{dP}{dt} = kP(K - P)$ , a logistic differential equation
When $P$ is small, $K - P \approx K$ $\frac{dP}{dt} \approx kKP$ : growth is nearly exponential
As $P$ approaches $K$ , the factor $K - P$ approaches 0growth slows and $P$ levels off: $\lim_{t \to \infty} P(t) = K$ , the carrying capacity
The rate $kP(K - P)$ is a downward parabola in $P$ it is largest at $P = \frac{K}{2}$ , where the graph of $P(t)$ has a point of inflection

Part 6 · Key ideas

Key ideas

Logistic: $\frac{dP}{dt} = kP(K - P)$ or $\frac{dP}{dt} = rP\left(1 - \frac{P}{K}\right)$ . The carrying capacity $K$ is the positive value of $P$ that makes the rate 0.
For $0 < P(0) < K$ , $P$ increases toward $K$ ; for $P(0) > K$ , $P$ decreases toward $K$ . Either way $\lim_{t \to \infty} P(t) = K$ .
Growth is fastest at $P = \frac{K}{2}$ : below it the graph of $P$ is concave up, above it concave down.
Read these from the equation; no formula for $P(t)$ is needed.

Part 7 · Misconception

A common mistake

The wrong idea: For $\frac{dP}{dt}$ $= 0.0006P(1200 - P)$ , the population grows fastest when it is largest, near 1200.

What actually happens: Near 1200 the factor $1200 - P$ is almost 0, so growth has nearly stopped. The rate $0.0006P(1200 - P)$ is largest halfway, at $P = 600$ .

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Fish in a lake

The number of fish

P(t)

in a lake,

t

years after 200 fish were released, satisfies

\frac{dP}{dt}

= 0.0006P(1200 - P)

with

P(0) = 200

. Values of

P(t)

, rounded to the nearest fish, are shown.

Number of fish P(t) at selected times t in years
$t$ (years)	0	1	2	3	4	6	8
$P(t)$ (fish)	200	349	549	761	937	1125	1181

1. What is the carrying capacity of the lake for these fish?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in fish.

Show the answer

$\frac{dP}{dt} = 0$ when $P = 1200$ , and $P$ rises toward it: the carrying capacity is 1200 fish.

Answer: 1200 fish

2. For what number of fish is the population growing fastest?

No calculator. Type the exact value (a fraction, π, e or a root is fine), in fish.

Show the answer

$\frac{dP}{dt}$ $= 0.0006P(1200 - P)$ is a downward parabola in $P$ with zeros 0 and 1200, so its largest value is at $P = 600$ , half the carrying capacity.

Answer: 600 fish

3. Between which two times in the table is the population growing fastest?

Between $t = 6$ and $t = 8$
Between $t = 0$ and $t = 1$
Between $t = 3$ and $t = 4$
Between $t = 2$ and $t = 3$

Show the answer

Growth is fastest when $P = 600$ , which happens between $t = 2$ and $t = 3$ . The table agrees: $761 - 549 = 212$ is the largest one-year change.

Between $t = 6$ and $t = 8$ : Near the carrying capacity growth slows: $P$ rises only about 56 fish over these two years.
Between $t = 0$ and $t = 1$ : At first the population is small, so growth is slow; it speeds up until $P = 600$ .
Between $t = 3$ and $t = 4$ : $P$ is already above 600 here, so the rate has started to fall: the jump is about 176, smaller than 212.
Correct: Between $t = 2$ and $t = 3$ : Right: $P$ passes 600 between $P(2) \approx 549$ and $P(3) \approx 761$ ; the table also shows its largest jump there.

4. A population satisfies $\frac{dy}{dt} = 0.5y\left(1 - \frac{y}{300}\right)$ with $y(0) = 20$ . What is $\lim_{t \to \infty} y(t)$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$1 - \frac{y}{300} = 0$ at $y = 300$ , the carrying capacity; a population starting at 20 rises toward it.

Answer: 300

5. Which differential equation is a logistic model?

$\frac{dy}{dt} = 3y^2$
$\frac{dy}{dt} = 3y(10 - y)$
$\frac{dy}{dt} = 3y$
$\frac{dy}{dt} = 3(10 - y)$

Show the answer

A logistic model has the form $\frac{dy}{dt} = ky(K - y)$ , or $ry\left(1 - \frac{y}{K}\right)$ .

$\frac{dy}{dt} = 3y^2$ : The rate grows with $y$ without limit; nothing slows it down.
Correct: $\frac{dy}{dt} = 3y(10 - y)$ : Right: the rate is jointly proportional to $y$ and to $10 - y$ .
$\frac{dy}{dt} = 3y$ : The rate is proportional to $y$ alone: an exponential model, with no carrying capacity.
$\frac{dy}{dt} = 3(10 - y)$ : The rate is proportional to $10 - y$ alone; it is largest at $y = 0$ , not halfway.

6. In a school of 5000 students, the number $R$ who have heard a rumor satisfies $\frac{dR}{dt}$ $= 0.0002R(5000 - R)$ , with $R(0) = 100$ . Which statement is true?

The rumor spreads fastest when 5000 students have heard it, and $R$ approaches 5000
The rumor spreads fastest when 2500 students have heard it, and $R$ approaches 2500
The rumor spreads fastest at the start, when 100 students have heard it
The rumor spreads fastest when 2500 students have heard it, and $R$ approaches 5000

Show the answer

$\frac{dR}{dt}$ is largest at $R = \frac{5000}{2} = 2500$ , and since $0 < R(0) < 5000$ , $\lim_{t \to \infty} R(t)$ $= 5000$ .

The rumor spreads fastest when 5000 students have heard it, and $R$ approaches 5000: At $R = 5000$ the rate is $0.0002(5000)(0) = 0$ : the spreading has stopped.
The rumor spreads fastest when 2500 students have heard it, and $R$ approaches 2500: 2500 is where the rate peaks; $R$ keeps growing past it, toward 5000.
The rumor spreads fastest at the start, when 100 students have heard it: At $R = 100$ few students are passing it on: the rate is $0.0002(100)(4900) = 98$ , below the peak $0.0002(2500)(2500)$ $= 1250$ .
Correct: The rumor spreads fastest when 2500 students have heard it, and $R$ approaches 5000: Right: the rate is largest at half the carrying capacity, and $R$ rises toward 5000.

Part 9 · Summary

Summary

A logistic model,

\frac{dP}{dt} = kP(K - P)

, has a rate jointly proportional to the quantity and to the room left. Any positive starting value approaches the carrying capacity

K

(from below it rises, from above it falls), and growth is fastest at

P = \frac{K}{2}

, the point of inflection of

P(t)

. All of this is read from the equation, without solving it.

Part 10 · Up next

What comes next

Part 11 · Connections