# Logistic models with differential equations (BC only) > AP Calculus BC · AP Calculus BC CED Unit 7 > Source: https://www.owlsprep.com/study/ap-calculus-bc-u7-logistic-models-with-differential-equations/ This exclusive AP Calculus BC subtopic covers bounded growth modeling with logistic differential equations. You will learn to identify model parameters, solve separable logistic equations, analyze growth rate, concavity, inflection points, and interpret results for AP exam questions. **Prerequisites:** [Separation of variables for first-order differential equations](https://www.owlsprep.com/study/ap-calculus-bc-u7-separation-of-variables/); Exponential growth modeling; Logarithm rules and partial fraction decomposition ## Learning objectives - Identify carrying capacity and intrinsic growth rate from a logistic differential equation - Find and classify equilibrium solutions of the logistic DE - Solve for general and particular solutions of the logistic differential equation - Analyze concavity and locate inflection points of logistic solutions - Interpret logistic models in real-world contextual problems ## The Logistic Differential Equation and Equilibrium Solutions 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. **Standard Logistic Differential Equation** — Where $P(t)$ is the quantity of interest at time $t$, $k>0$ is the intrinsic growth rate (per-capita growth rate for very small populations), and $K>0$ is the carrying capacity (maximum sustainable size of $P$). *Notation:* \frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right) *Example:* When $P$ is very small, $\frac{dP}{dt} \approx kP$, matching exponential growth. When $P=K$, growth stops entirely; when $P>K$, $P$ decreases toward $K$. 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. **Worked example:** Given the differential equation $\frac{dP}{dt} = 0.3P\left(1 - \frac{P}{1800}\right)$, (a) identify the carrying capacity and intrinsic growth rate, (b) find all equilibrium solutions and classify their stability, (c) find the value of $P$ that maximizes the growth rate. 1. Compare the given equation directly to standard form $\frac{dP}{dt}=kP\left(1-\frac{P}{K}\right)$ to get parameters: 2. $$k=0.3 \quad (\text{intrinsic growth rate}), \quad K=1800 \quad (\text{carrying capacity})$$ 3. Set $\frac{dP}{dt}=0$ to find equilibria: 4. $$0.3P\left(1-\frac{P}{1800}\right)=0 \implies P=0 \text{ or } P=1800$$ 5. For $00$, so $P$ moves away from $0$ toward $1800$. For $P>1800$, $\frac{dP}{dt}<0$, so $P$ moves toward $1800$. Thus, $P=0$ is unstable, and $P=1800$ is stable. 6. The maximum of a quadratic with roots at $P=0$ and $P=1800$ occurs at the midpoint of the roots: 7. $$P = \frac{0 + 1800}{2} = 900$$ > **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. ## Solving the Logistic Differential Equation The logistic differential equation is separable, so we can use separation of variables and partial fraction decomposition to find an explicit solution for $P(t)$. **Derivation:** Derive the general and particular solution for the logistic differential equation *Starting from:* Standard logistic DE: $\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)$ 1. Separate variables: 2. $$\frac{dP}{P\left(1-\frac{P}{K}\right)} = k \, dt$$ 3. Use partial fraction decomposition to rewrite the left-hand side: 4. $$\left(\frac{1}{P} + \frac{1}{K-P}\right)dP = k \, dt$$ 5. Integrate both sides and combine logarithms: 6. $$\ln|P| - \ln|K-P| = kt + C \implies \ln\left|\frac{P}{K-P}\right| = kt + C$$ 7. Exponentiate both sides, let $A = e^C$: 8. $$\frac{P}{K-P} = Ae^{kt}$$ 9. Rearrange to solve for $P(t)$ to get the general solution: 10. $$P(t) = \frac{KAe^{kt}}{1 + Ae^{kt}}$$ 11. Substitute initial condition $P(0) = P_0$ to solve for $A = \frac{P_0}{K-P_0}$ *Conclusion:* The particular solution for initial condition $P(0) = P_0$ is: $P(t) = \frac{KP_0}{P_0 + (K-P_0)e^{-kt}}$. As $t \to \infty$, $P(t) \to K$, confirming the stable equilibrium at $P=K$. **Worked example:** A population grows according to $\frac{dP}{dt} = 0.2P\left(1 - \frac{P}{600}\right)$, with initial population $P(0)=150$. Find the particular solution $P(t)$. 1. Identify the parameters from the DE and initial condition: 2. $$k=0.2, \quad K=600, \quad P_0=150$$ 3. Substitute into the particular solution formula: 4. $$P(t) = \frac{KP_0}{P_0 + (K-P_0)e^{-kt}} = \frac{600 \cdot 150}{150 + (600-150)e^{-0.2t}}$$ 5. Simplify the expression by dividing numerator and denominator by 150: 6. $$P(t) = \frac{90000}{150 + 450e^{-0.2t}} = \frac{600}{1 + 3e^{-0.2t}}$$ 7. Verify the initial condition to check for errors: 8. $$P(0) = \frac{600}{1+3} = 150$$ 9. This matches the given initial value, confirming the solution is correct. > **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. ## Concavity and Inflection Points of Logistic Solutions 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. **Derivation:** Find the second derivative of $P(t)$ for the logistic DE *Starting from:* $\frac{dP}{dt} = kP\left(1 - \frac{P}{K}\right)$ 1. Differentiate both sides with respect to $t$ using the chain rule: 2. $$\frac{d^2P}{dt^2} = \frac{d}{dP}\left(kP\left(1-\frac{P}{K}\right)\right) \cdot \frac{dP}{dt}$$ 3. Compute the derivative with respect to $P$: 4. $$\frac{d}{dP}\left(kP - \frac{kP^2}{K}\right) = k\left(1 - \frac{2P}{K}\right)$$ *Conclusion:* $\frac{d^2P}{dt^2} = k\left(1-\frac{2P}{K}\right)\frac{dP}{dt} = k^2P\left(1-\frac{P}{K}\right)\left(1-\frac{2P}{K}\right)$ For $00$, 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 $t>0$. **Worked example:** For the logistic model $\frac{dP}{dt} = 0.3P\left(1-\frac{P}{900}\right)$ with $P(0)=100$, find the $t$-coordinate of the inflection point and state the intervals of concavity. 1. Inflection for solutions starting below $K$ occurs at $P = \frac{K}{2}$: 2. $$P = \frac{900}{2} = 450$$ 3. Write the particular solution for $P(t)$ using the standard formula: 4. $$P(t) = \frac{900 \cdot 100}{100 + 800e^{-0.3t}} = \frac{900}{1 + 8e^{-0.3t}}$$ 5. Set $P(t) = 450$ and solve for $t$: 6. $$450 = \frac{900}{1+8e^{-0.3t}} \implies 1 + 8e^{-0.3t} = 2 \implies e^{-0.3t} = \frac{1}{8} \implies t = \frac{\ln 8}{0.3} = 10\ln 2 \approx 6.93$$ 7. State intervals of concavity based on the sign of the second derivative: For $00$, so $P(t)$ is concave up. For $t>10\ln 2$, $P>450$ so $\frac{d^2P}{dt^2}<0$, so $P(t)$ is concave down. > **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 $P$. ## AP-Style Worked Examples **Worked example:** Multiple Choice: What is the carrying capacity of a population growing according to $\frac{dP}{dt} = 0.04P(800 - P)$? 1. Rewrite the DE to match standard logistic form by factoring 800 out of the binomial: 2. $$\frac{dP}{dt} = 0.04P \cdot 800\left(1-\frac{P}{800}\right) = 32P\left(1-\frac{P}{800}\right)$$ 3. Matching to standard form, $K=800$. The value 32 is the intrinsic growth rate $k$, not $K$, so the correct answer is 800. **Worked example:** Real-World FRQ: A coffee shop introduces a new seasonal drink, with a maximum of 800 total regular customers that can try it. The rate of new customers follows the logistic model $\frac{dC}{dt} = 0.5C\left(1-\frac{C}{800}\right)$, with $C(0)=20$ (20 customers tried it on opening day). How many customers have tried the drink after 5 days, rounded to the nearest whole number? 1. Identify parameters: $K=800$, $k=0.5$, $P_0=20$, substitute into the particular solution formula: 2. $$C(t) = \frac{KP_0}{P_0 + (K-P_0)e^{-kt}} = \frac{800 \cdot 20}{20 + 780e^{-0.5t}} = \frac{16000}{20 + 780e^{-0.5t}}$$ 3. Substitute $t=5$ and calculate: 4. $$e^{-0.5(5)} = e^{-2.5} \approx 0.082085, \quad \text{denominator} \approx 20 + 780(0.082085) \approx 84.026$$ 5. Solve for $C(5)$ and interpret: 6. $$C(5) \approx \frac{16000}{84.026} \approx 190$$ 7. After 5 days, approximately 190 of the 800 total regular customers have tried the new seasonal drink. ## Common pitfalls - **Wrong:** For $\frac{dP}{dt} = 0.2P(1000 - 2P)$, calculate carrying capacity as $K=1000$ instead of $K=500$. - Why it fails: Students assume the constant term in the factored binomial is always $K$, without rewriting the equation to standard form. - Correct: Always factor out the coefficient of $P$ from the binomial to get a leading constant term, so the denominator under $P$ is $K$. For this example, $0.2P(1000-2P) = 200P\left(1-\frac{P}{500}\right)$, so $K=500$. - **Wrong:** After integrating, incorrectly combine $\ln|P| - \ln|K-P|$ as $\ln|P(K-P)|$. - Why it fails: Students misremember logarithm rules when rushing through the partial fraction step. - Correct: Write the logarithm rule explicitly $\ln A - \ln B = \ln\left(\frac{A}{B}\right)$ before exponentiating to confirm. - **Wrong:** Solving for the constant $A$ and getting $A=\frac{K-P_0}{P_0}$ instead of $A=\frac{P_0}{K-P_0}$. - Why it fails: Simple fraction flip when rearranging the initial condition equation. - Correct: After finding $P(t)$, plug in $t=0$ to confirm you get $P_0$, which will catch this error immediately. - **Wrong:** Claiming an inflection point at $P=K/2$ for a solution starting at $P_0 > K$. - Why it fails: Students memorize that inflection is always at $K/2$, but forget that if $P$ starts above $K$, it decreases toward $K$ and never crosses $K/2$. - Correct: Always check the range of $P(t)$ for your initial condition before stating there is an inflection point. - **Wrong:** Answering a question asking for maximum growth rate by reporting the carrying capacity $K$. - Why it fails: Students confuse 'maximum population size' with 'maximum growth rate of the population'. - Correct: If asked for maximum growth rate, calculate $\frac{kK}{4}$, the value of $\frac{dP}{dt}$ at $P=K/2$, not $K$. ## Cheatsheet | Category | Formula | Notes | | --- | --- | --- | | Standard Logistic Differential Equation | \frac{dP}{dt} = kP\left(1-\frac{P}{K}\right) | $P(t)$ = quantity at time $t$, $k$ = intrinsic growth rate, $K$ = carrying capacity | | Equilibrium Solutions | $P=0$ (unstable), $P=K$ (stable) | Occur where $\frac{dP}{dt}=0$; all non-zero solutions approach $P=K$ as $t\to\infty$ | | Maximum Growth Rate | Maximum $\frac{dP}{dt} = \frac{kK}{4}$ at $P=\frac{K}{2}$ | Applies only for $0K$ | | Long-Run Behavior | $\lim_{t\to\infty} P(t) = K$ | For all $0