Logistic Models with Differential Equations: Study Guide (Calculus BC 7.9)

BC-only material. Logistic models are part of Calculus BC only. Calculus AB students do not need this topic.

Before you start: prerequisites

Prerequisite	Topic	What you use it for here
Differential equations from words	7.1	Turning “jointly proportional” into an equation
Constant solutions and slope fields	7.4	Seeing why solutions level off
Exponential models	7.8	The model that logistic growth improves on
Maximum of a quadratic; concavity	Unit 5	Finding where growth is fastest

Revisit any of these from the Calculus BC roadmap.

A note on notation

y (or P, N, B…) is the size of a quantity at time t. dy/dt is its rate of change. k and a are positive constants. a is called the carrying capacity or limiting value. “lim y(t) as t → ∞” means the value the solution approaches in the long run.

From exponential to logistic growth

In Topic 7.8 the rate of growth was proportional to the amount: dy/dt = ky. Growth then never slows down. Real populations run out of food or space, so growth must slow as the population gets close to some maximum, a.

A logistic model builds this in. The rate is jointly proportional to two things at once:

• the size of the quantity, y (more members, more growth), and
• the room left before the maximum, a − y (less room, less growth).

“Jointly proportional” means proportional to the product. So

dy/dt = k y (a − y), with k > 0 and a > 0.

Other forms you will meet. Expanding or rearranging gives the same model:

• dy/dt = r y (1 − y/a) where r = ka. Here r is the relative growth rate when y is very small.
• dy/dt = by − cy² (b, c > 0). Factor out cy: cy(b/c − y). So the carrying capacity is a = b/c.

Always rewrite the equation as k y (a − y) before you read off a.

Reading the equation without solving it

Think of the right-hand side as a function of y alone: dy/dt = ky(a − y). Its graph is a downward parabola with zeros at y = 0 and y = a.

Read four facts from this picture:

1. Constant solutions. dy/dt = 0 when y = 0 or y = a. If the quantity starts at one of these values, it stays there.
2. Direction. If 0 < y < a, dy/dt > 0, so y increases. If y > a, dy/dt < 0, so y decreases.
3. Long-run limit. A solution that starts with y(0) > 0 moves towards a and levels off, because the rate shrinks to 0 as y nears a. So lim y(t) as t → ∞ = a. If y(0) = 0, the limit is 0.
4. Fastest change. The parabola peaks at y = a/2 (halfway between its zeros). The greatest rate is k(a/2)(a/2) = ka²/4.

Concavity. Differentiate dy/dt = k(ay − y²) with respect to t, using the chain rule:

d²y/dt² = k(a − 2y) · dy/dt

For 0 < y < a, dy/dt > 0, so the sign of d²y/dt² is the sign of a − 2y. The graph is concave up below a/2 and concave down above a/2: a point of inflection at y = a/2. For y > a, both factors are negative, so the graph is concave up as it falls towards a.

When does “fastest” happen? If 0 < y(0) < a/2, the solution passes through a/2, and that is when it grows fastest. If a/2 ≤ y(0) < a, the rate only falls from the start, so for t ≥ 0 the greatest rate is at t = 0. Questions usually give a starting value below a/2.

The graph of a logistic solution

Worked example 1: a deer population

Question (no calculator). The number of deer P in a fenced nature reserve (fictional data) satisfies dP/dt = 0.0004P(800 − P), where t is in years, and P(0) = 100.

(a) Find lim P(t) as t → ∞ and explain what it means. (b) Find the population when it is growing fastest, and the rate of growth then. (c) Is the graph of P concave up or concave down at t = 0? Justify. (d) Rewrite the equation in the form rP(1 − P/800) and explain what r means. (e) How would the answers to (a) and (b) change if instead P(0) = 1000?

(a) The equation has the form kP(a − P) with k = 0.0004 and a = 800. Since P(0) = 100 > 0, lim P(t) = 800. In the long run the reserve supports about 800 deer: the carrying capacity.

(b) Growth is fastest at P = a/2 = 400 deer. The rate there is 0.0004 × 400 × 400 = 64 deer per year. Because 100 < 400, the population does pass through 400.

(c) At t = 0, dP/dt = 0.0004 × 100 × 700 = 28 deer per year. Then d²P/dt² = 0.0004(800 − 2P) · dP/dt = 0.0004 × 600 × 28 = 6.72 deer per year² > 0. So the graph is concave up at t = 0: the growth rate is still increasing.

(d) 0.0004P(800 − P) = 0.0004 × 800 × P(1 − P/800) = 0.32P(1 − P/800). So r = 0.32. When P is small compared with 800, the factor (1 − P/800) is close to 1, and the population grows at about 32% of its size per year, just like an exponential model with k = 0.32. As P grows, the factor shrinks and growth slows.

(e) The limit is still 800: the reserve holds more deer than it can support, so dP/dt = 0.0004 × 1000 × (−200) = −80 < 0 and P decreases towards 800. P never passes through 400, so there is no “fastest growth at 400”; the population falls fastest at the start, then more and more slowly.

Checks. Units: k has units of 1/(deer · year), so k × deer × deer gives deer per year. ✓ The answer to (b) is less than 800 and equals a/2. ✓

Worked example 2: from a sentence to the model

Question. In a school of 1500 students (fictional), a rumour spreads at a rate jointly proportional to the number N of students who have heard it and the number who have not. When 300 students have heard it, it is spreading at 72 students per day. At t = 0 days, 30 students have heard it.

(a) Write a differential equation for N and find the constant of proportionality. (b) Find lim N(t) as t → ∞ and interpret it. (c) How many students have heard the rumour when it is spreading fastest? What is that greatest rate?

(a) “Have not heard” is 1500 − N. Jointly proportional means the rate is k times the product:

dN/dt = kN(1500 − N).

At N = 300: 72 = k × 300 × 1200 = 360 000k, so k = 72/360 000 = 0.0002 (per student per day).

(b) N(0) = 30 > 0, so lim N(t) = 1500. In the model, eventually every student hears the rumour.

(c) Fastest when N = 1500/2 = 750 students. The greatest rate is 0.0002 × 750 × 750 = 112.5 students per day.

Interpretation. Early on the rumour spreads slowly because few people know it. Late on it spreads slowly because few people are left to tell. It spreads fastest when half the school knows.

Going further (optional): the solution formula

You are not required to solve the logistic equation in this topic. For interest only: separating variables and using partial fractions (Topic 6.12) gives

y = a / (1 + C e^(−akt)), with C = a/y(0) − 1.

You can check this by differentiating (Topic 7.2). For Worked example 2, ak = 1500 × 0.0002 = 0.3 and C = 1500/30 − 1 = 49, so N(t) = 1500/(1 + 49e^(−0.3t)). It gives N(0) = 30 ✓ and reaches 750 when 49e^(−0.3t) = 1, at t = ln 49 / 0.3 ≈ 12.973 days (calculator). For the deer, P = 800/(1 + 7e^(−0.32t)) reaches 400 at t = ln 7 / 0.32 ≈ 6.081 years, the inflection point in Figure 2.

Common misconceptions

• “The carrying capacity is the number outside the bracket.” In 0.0004P(800 − P), 0.0004 is k and 800 is a. In 0.32P(1 − P/800), 0.32 is r, not a.
• Not factoring first. In dP/dt = 0.3P − 0.0006P², the capacity is 0.3/0.0006 = 500, not 0.3 and not 0.0006.
• “Growth is fastest at the carrying capacity.” At y = a the rate is 0. The fastest rate is at a/2.
• Giving the time instead of the value. “Fastest when y = a/2” answers the question asked in this topic. The time needs the solution formula.
• “The limit depends on where you start.” Any positive start gives limit a. Only y(0) = 0 gives 0.
• “A population above a can’t happen.” It can (for example after animals are moved in). The model says it then decreases towards a.
• Thinking k is a percentage growth rate. In ky(a − y), the relative rate is k(a − y), which changes as y changes.
• Treating logistic and exponential models as the same. dy/dt = ky never levels off; dy/dt = ky(a − y) always does.

Where this leads

Logistic models close Unit 7. Unit 8 turns to applications of integration, starting with the average value of a function. Logistic equations also make good Euler’s method exercises, since Topic 7.5 needs only the equation and a starting point.

Next topic: Finding the Average Value of a Function on an Interval. Previous topic: Exponential Models with Differential Equations. See also Euler’s method. Return to the Calculus BC roadmap to see the order.

Try the practice questions now, then use the revision notes and the checklist to consolidate.