The Logistic Equation
A petri dish starts with 100 bacteria. They double every hour, but the dish can only hold 10 000. Does growth just stop at capacity? What does the population curve look like? The logistic equation $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$ captures this S-shaped reality. In this lesson you'll solve it using partial fractions, sketch the solution curve, and answer HSC-style questions on carrying capacity and inflection.
A population starts at $P_0 = 200$ with carrying capacity $M = 1000$ and growth rate $k = 0.4$. Without using a formulasketch or describe what you expect the population-vs-time curve to look like. Does growth speed up, slow down, or both?
Exponential growth ($\tfrac{dP}{dt} = kP$) predicts unlimited growth, unrealistic for any real population. The logistic model adds a braking term $\bigl(1 - \tfrac{P}{M}\bigr)$ that shrinks to zero as $P \to M$:
When $P \ll M$ the factor $\approx 1$ and growth is nearly exponential. When $P \approx M$ the factor $\approx 0$ and growth stalls. The inflection point fastest growth, occurs at exactly $P = M/2$.
$\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$
Key facts
- Logistic DE: $\tfrac{dP}{dt} = kP(1 - P/M)$; $M$ is carrying capacity
- Inflection point at $P = M/2$; solution is an S-curve
- General solution: $P(t) = \dfrac{M}{1 + Ae^{-kt}}$ where $A = \dfrac{M - P_0}{P_0}$
Concepts
- Why the braking factor $(1 - P/M)$ produces bounded growth
- How partial fractions decompose $\tfrac{1}{P(M-P)}$
- The biological meaning of $k$, $M$, and the inflection point
Skills
- Solve the logistic DE by separation of variables and partial fractions
- Apply initial conditions to find $A$ and interpret the solution
- Identify $k$, $M$, $P_0$ from a context and find $P(t)$
Start with $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$. Rewrite as $\dfrac{dP}{dt} = \dfrac{k}{M} P(M - P)$ and separate variables:
Partial fractions: Write $\dfrac{1}{P(M-P)} = \dfrac{A}{P} + \dfrac{B}{M-P}$. Multiplying both sides by $P(M-P)$ and comparing coefficients gives $A = \tfrac{1}{M}$, $B = \tfrac{1}{M}$. So:
Multiplying both sides by $M$ and exponentiating:
where $A = \dfrac{M - P_0}{P_0}$ is determined by the initial condition $P(0) = P_0$.
Start with $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$. Rewrite as $\dfrac{dP}{dt} = \dfrac{k}{M} P(M - P)$ and separate variables:
Pause, copy the logistic solution $P(t)=\frac{M}{1+Ae^{-kt}}$ with $A=(M-P_0)/P_0$, and the partial-fraction step that generates it into your book.
Quick check: A population satisfies $\dfrac{dP}{dt} = 0.3P\!\left(1 - \dfrac{P}{500}\right)$ with $P(0) = 50$. What is the value of $A$ in the general solution $P(t) = \dfrac{M}{1+Ae^{-kt}}$?
We just saw that separating $\frac{dP}{dt}=kP(1-P/M)$ and using partial fractions gives $P(t)=\frac{M}{1+Ae^{-kt}}$ where $A=(M-P_0)/P_0$. That raises a question: the growth rate $\frac{dP}{dt}$ starts small, rises, then falls back to zero, at what exact value of $P$ is it maximised, and what is that maximum rate? This card answers it → the inflection occurs at $P=M/2$, and the maximum growth rate is $\frac{dP}{dt}=kM/4$.
The inflection point of $P(t)$ is where $\dfrac{d^2P}{dt^2} = 0$, i.e.\ where the growth rate $\dfrac{dP}{dt}$ is a maximum.
Define $f(P) = kP(1-P/M)$. Differentiate with respect to $P$:
$f'(P) = k - \dfrac{2kP}{M} = 0 \quad \Rightarrow \quad P = \dfrac{M}{2}$
At this population the growth rate equals $\dfrac{dP}{dt}\Big|_{P=M/2} = k \cdot \dfrac{M}{2} \cdot \dfrac{1}{2} = \dfrac{kM}{4}$.
The inflection point of $P(t)$ is where $\dfrac{d^2P}{dt^2} = 0$, i.e.\ where the growth rate $\dfrac{dP}{dt}$ is a maximum.
Pause, copy the inflection-point result: $P=M/2$ gives maximum growth rate $kM/4$; above $M/2$ the growth decelerates; $P=M$ is the asymptote into your book.
Did you get this? True or false: for the logistic equation with $M = 800$, the population growth rate is greatest when $P = 400$.
Worked examples · 3 in a row, reveal as you go
A fish population satisfies $\dfrac{dP}{dt} = 0.2P\!\left(1 - \dfrac{P}{1000}\right)$ with $P(0) = 100$. Find $P(t)$.
Using the solution from Problem 1, at what time $t$ does the population reach the inflection point?
Show that $P(t) = \dfrac{M}{1 + Ae^{-kt}}$ satisfies $\dfrac{dP}{dt} = kP\!\left(1 - \dfrac{P}{M}\right)$.
Fill the gap: For the logistic equation with $M = 600$ and $k = 0.5$, the maximum growth rate equals $\dfrac{kM}{4} = $ .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for a logistic population with $P_0 = M/2$, the constant $A$ in $P(t) = M/(1+Ae^{-kt})$ equals 1.
Activities · practice with the ideas
A population satisfies $\tfrac{dP}{dt} = 0.1P(1 - P/2000)$ with $P(0) = 200$. Write down the values of $k$, $M$, and $A$, then state $P(t)$.
For the DE in Activity 1, at what population value is the growth rate a maximum? What is that maximum growth rate?
Use partial fractions to show that $\displaystyle\int \frac{dP}{P(1000-P)} = \frac{1}{1000}\ln\left|\frac{P}{1000-P}\right| + C$.
A population with $M = 500$, $k = 0.6$, $P(0) = 50$ satisfies the logistic model. Find the time when $P = 400$. Give your answer in exact form.
Explain why a logistic population with $P_0 > M$ (population starts above carrying capacity) would decrease towards $M$. What is the sign of $\tfrac{dP}{dt}$ in this case?
Odd one out: Three of these statements about the logistic equation are correct. Which one is NOT?
Earlier you sketched or described the population curve for $P_0 = 200$, $M = 1000$, $k = 0.4$.
The exact solution is $P(t) = \dfrac{1000}{1 + 4e^{-0.4t}}$ (here $A = \tfrac{1000-200}{200} = 4$). Growth accelerates up to $P = 500$ (inflection at $t = \tfrac{\ln 4}{0.4} \approx 3.47$), then decelerates, asymptoting to 1000. Did your sketch capture the S-shape with deceleration near $M$?
Pick your answer, then rate your confidencethat tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. Write down the general solution of $\dfrac{dP}{dt} = 0.4P\!\left(1 - \dfrac{P}{800}\right)$ with $P(0) = 160$. (2 marks)
Q2. For the logistic equation in Q1, find the time at which the growth rate is a maximum. Give your answer in exact form. (3 marks)
Q3. Show that if $P(t) = \dfrac{M}{1+Ae^{-kt}}$, then $\dfrac{dP}{dt} = \dfrac{k}{M}P(M-P)$. (3 marks)
Comprehensive answers (click to reveal)
Activity answers: 1. $k=0.1$, $M=2000$, $A=(2000-200)/200=9$, $P(t)=2000/(1+9e^{-0.1t})$ · 2. Inflection at $P=1000$; max rate $=0.1\times2000/4=50$ per unit time · 3. Partial fractions: $A=B=1/1000$; integrate to get $\tfrac{1}{1000}\ln|P/(1000-P)|+C$ · 4. $A=(500-50)/50=9$; set $400=500/(1+9e^{-0.6t})$, solve $e^{-0.6t}=1/36$, $t=\tfrac{\ln 36}{0.6}=\tfrac{5\ln 36}{3}$ · 5. When $P>M$, $(1-P/M)<0$ so $dP/dt<0$, population decreases towards $M$.
Q1 (2 marks): $A=(800-160)/160=4$ [1]. $P(t)=\dfrac{800}{1+4e^{-0.4t}}$ [1].
Q2 (3 marks): Inflection at $P=M/2=400$ [1]. Set $\dfrac{800}{1+4e^{-0.4t}}=400$: $1+4e^{-0.4t}=2$, $e^{-0.4t}=\tfrac{1}{4}$ [1]. $t=\dfrac{\ln 4}{0.4}=\dfrac{5\ln 4}{2}$ [1].
Q3 (3 marks): Differentiate $P=M/(1+Ae^{-kt})$: $dP/dt = MkAe^{-kt}/(1+Ae^{-kt})^2$ [1]. Note $P = M/(1+Ae^{-kt})$, so $P/M = 1/(1+Ae^{-kt})$, hence $1-P/M = Ae^{-kt}/(1+Ae^{-kt})$ [1]. Multiply: $\tfrac{k}{M}P(M-P) = k \cdot \tfrac{M}{1+Ae^{-kt}} \cdot \tfrac{MAe^{-kt}}{1+Ae^{-kt}} \cdot \tfrac{1}{M} = \tfrac{MkAe^{-kt}}{(1+Ae^{-kt})^2} = dP/dt$ ✓ [1].
Five timed questions. Beat the boss to bank a tier, gold (90% + speed), silver (75%), or bronze (50%). Replays welcome.
⚔ Enter the arenaClimb platforms by answering logistic equation questions. Lighter alternative to the boss.
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