Logistic models with differential equations (BC only) — AP Calculus BC
AP Calculus BC · AP Calculus BC CED Unit 7 · 14 min read
1. The Logistic Differential Equation and Equilibrium Solutions★★☆☆☆⏱ 4 min
Logistic models correct the unrealistic unlimited growth assumption of exponential growth by accounting for finite resource limits. They are used to model populations, disease spread, technology adoption, and other quantities that approach a maximum sustainable size called the carrying capacity $K$. This topic makes up 2-3% of the total AP Calculus BC exam score, appearing in both multiple-choice and free-response sections.
Equilibrium solutions are constant solutions where the growth rate $\frac{dP}{dt}=0$. For the logistic equation, the two equilibria are $P=0$ (unstable: any small positive $P$ will grow away from 0) and $P=K$ (stable: any non-zero $P$ will approach $K$ as $t \to \infty$). The growth rate $\frac{dP}{dt}$ is a downward-opening quadratic function of $P$, so its maximum occurs at $P = \frac{K}{2}$, meaning growth is fastest when the quantity reaches half the carrying capacity.
Exam tip: Always rewrite the logistic equation to match the standard form where the leading term inside the parentheses is 1 before identifying $K$. Do not guess $K$ from non-standard factored forms—this is the most common MCQ mistake.
2. Solving the Logistic Differential Equation★★★☆☆⏱ 4 min
The logistic differential equation is separable, so we can use separation of variables and partial fraction decomposition to find an explicit solution for $P(t)$.
Exam tip: Always check your particular solution by plugging in $t=0$ to confirm it gives $P_0$. This catches 90% of common algebra errors from rearranging terms.
3. Concavity and Inflection Points of Logistic Solutions★★★☆☆⏱ 3 min
To analyze the shape of the logistic solution curve $P(t)$, we calculate the second derivative $\frac{d^2P}{dt^2}$ using the chain rule to identify concavity and inflection points.
For $0<P<K$, $\frac{dP}{dt}>0$, so the sign of $\frac{d^2P}{dt^2}$ matches the sign of $\left(1-\frac{2P}{K}\right)$. When $P<\frac{K}{2}$, $\frac{d^2P}{dt^2}>0$ so $P(t)$ is concave up. When $P>\frac{K}{2}$, $\frac{d^2P}{dt^2}<0$ so $P(t)$ is concave down. This means the logistic curve has an inflection point (where concavity changes) exactly at $P=\frac{K}{2}$, which is also where the growth rate is maximized. If the initial population $P_0>K$, $P(t)$ decreases toward $K$ and never crosses $P=\frac{K}{2}$, so there is no inflection point for $t>0$.
Exam tip: Remember we find concavity of $P(t)$ as a function of $t$, not of $\frac{dP}{dt}$ as a function of $P$. Always use the chain rule to get the second derivative with respect to $t$, do not stop at the derivative of $\frac{dP}{dt}$ with respect to $P$.
4. AP-Style Worked Examples★★★★☆⏱ 3 min
Common Pitfalls
Why: Students assume the constant term in the factored binomial is always $K$, without rewriting the equation to standard form.
Why: Students misremember logarithm rules when rushing through the partial fraction step.
Why: Simple fraction flip when rearranging the initial condition equation.
Why: Students memorize that inflection is always at $K/2$, but forget that if $P$ starts above $K$, it decreases toward $K$ and never crosses $K/2$.
Why: Students confuse 'maximum population size' with 'maximum growth rate of the population'.