AP Calculus AB/BC · Unit 7 · Topic 7.9 · BC Only
Logistic Models with Differential Equations
Analyze growth limited by a carrying capacity and identify equilibria and maximum growth.
1. Topic Focus
Model rates with differential equations, read slope fields, approximate solutions, solve separable equations, and interpret exponential or logistic models.
This topic: Analyze growth limited by a carrying capacity and identify equilibria and maximum growth.
2. Key Relationship
Read every symbol with its domain, direction, units, and hypotheses before applying the relationship.
3. Visual Connection
4. Worked Example
Equilibria occur at P=0 and P=L; growth is fastest at P=L/2.
Write the governing relationship first, carry out the algebra cleanly, and finish with a sentence that answers the mathematical question.
5. Concept Development
The logistic model
A standard logistic initial-value problem is
| Symbol | Meaning |
|---|---|
| \(P(t)\) | Quantity or population at time \(t\) |
| \(K\) | Carrying capacity, the long-run sustainable level |
| \(r\) | Intrinsic relative growth constant, with units of reciprocal time |
| \(P_0\) | Initial quantity |
The factor \(rP\) produces exponential-like growth. The crowding factor \(1-P/K\) reduces that growth as \(P\) increases:
Thus the relative growth rate decreases linearly with population. When \(P\ll K\), the crowding factor is near \(1\), so the model initially behaves approximately like \(P'=rP\).
Equilibria and qualitative behavior
Setting \(P'=0\) gives equilibrium solutions \(P=0\) and \(P=K\). A sign analysis determines the direction of every nonnegative solution.
| Population | Sign of \(P'\) | Behavior |
|---|---|---|
| \(P=0\) | \(0\) | Constant equilibrium |
\(0| Positive | Increases toward \(K\) | |
| \(P=K\) | \(0\) | Constant equilibrium |
| \(P>K\) | Negative | Decreases toward \(K\) |
In the nonnegative model, \(K\) is a stable equilibrium: nearby solutions move toward it. The equilibrium \(0\) is unstable because a small positive population grows away from it. Carrying capacity is therefore a limiting level, not necessarily a hard upper bound; a population that begins above \(K\) decreases toward it.
Solving by separation of variables
For a non-equilibrium positive solution, separate and use partial fractions:
Integration gives
Solving for \(P\) and applying \(P(0)=P_0>0\) produces
The solution \(P=0\) must be listed separately because the separation step divides by \(P\). If \(P_0=K\), then \(A=0\) and the formula correctly gives the constant solution \(P=K\).
Concavity and fastest growth
Differentiate the differential equation with respect to time:
For \(0 This downward-opening parabola has its vertex at \(P=K/2\), so If \(0 If \(P_0>K/2\), that inflection occurred before \(t=0\), so the future solution is already concave down. A solution starting above \(K\) decreases and is concave up as it approaches \(K\). If \(K\), \(P_0\), and a later value \(P(t_1)=P_1\) are known with \(0 Always check units and assumptions. The basic model treats \(r\) and \(K\) as constant and ignores migration, delays, seasonal effects, and sudden environmental changes. A good AP response interprets conclusions within the model rather than claiming that a real population must follow the curve forever.Estimating a parameter from data
6. Detailed Worked Example and Error Check
Example 1: Read the model
For
the intrinsic growth constant is \(r=0.6\), the carrying capacity is \(K=800\), and the equilibria are \(0\) and \(800\). Growth is fastest at \(P=400\), with rate
individuals per time unit.
Example 2: Solve an initial-value problem
Suppose \(P'=0.4P(1-P/1000)\) and \(P(0)=100\). Then
After \(10\) time units,
The value remains below \(1000\) and approaches it over time.
Example 3: Verify the explicit solution
Let \(P=K(1+Ae^{-rt})^{-1}\). Differentiating gives
Also,
The two expressions agree, so the formula satisfies the logistic differential equation wherever it is defined.
Example 4: Begin above carrying capacity
For \(P'=0.2P(1-P/500)\) and \(P(0)=700\),
Because \(P>500\), \(P'<0\). Also \(P''=0.2P'(1-2P/500)>0\), so the population decreases, is concave up, and approaches \(500\).
Example 5: Locate the inflection point
For the model in Example 2, fastest growth occurs at \(P=500\). Solve
to obtain
At that time the growth rate is \(rK/4=(0.4)(1000)/4=100\). The population is not largest there; its rate of increase is largest there.
Example 6: Estimate the growth constant
A population has \(K=1000\), \(P(0)=100\), and \(P(5)=300\). Since \(A=9\),
Therefore
per time unit.
Example 7: Compare relative and absolute growth
For \(P'=0.5P(1-P/10000)\), at \(P=100\) the relative growth rate is
which is close to the exponential rate \(0.5\). At \(P=5000\), the relative rate has fallen to \(0.25\), although the absolute rate is maximal there.
Example 8: Read the rate as a function of population
For \(P'=0.4P(1-P/1000)\), the rates at \(P=0,250,500,750,1000\) are respectively
The equal rates at \(250\) and \(750\) reflect the symmetry of the quadratic \(G(P)\), and its vertex confirms the maximum at \(P=500\).
Common errors
- Calling \(K/2\) the maximum population instead of the population at maximum growth.
- Treating \(r\) as the actual relative rate at every population; the relative rate is \(r(1-P/K)\).
- Forgetting the equilibrium solution \(P=0\) after dividing by \(P\).
- Assuming every logistic curve is S-shaped for \(t\ge0\); the visible concavity depends on the initial value.
- Using exponential growth indefinitely even when a carrying capacity is part of the context.
7. AP Reasoning Routine
Translate the context into a rate equation, verify candidate solutions by substitution, carry constants through integration, and apply initial conditions last.
- Identify the representation and requested quantity.
- State the rule or theorem and verify its conditions.
- Keep exact values until the final requested approximation.
- Interpret sign, units, interval, and context.
Analyze each logistic model. Give exact values before decimal approximations.
(a) For \(P'=0.3P(1-P/1200)\), identify \(r\), \(K\), the equilibria, and the sign of \(P'\) on the nonnegative intervals they determine.
(b) In the model from part (a), at what population is growth fastest, and what is the maximum growth rate?
(c) Solve \(P'=0.2P(1-P/500)\), \(P(0)=50\).
(d) Use the solution from part (c) to find \(P(10)\).
(e) When does the solution from part (c) reach its inflection point?
(f) Describe the direction, concavity, and long-run behavior of a solution to \(P'=0.1P(1-P/500)\) with \(P(0)=700\).
(g) For \(P'=0.4P(1-P/1000)\), find both the relative and absolute growth rates when \(P=200\).
(h) A logistic population has \(K=800\), \(P(0)=80\), and \(P(4)=200\). Find \(r\).
(i) Explain one mathematical difference between exponential and logistic relative growth rates, and state when the logistic model resembles exponential growth.
(j) State the solutions for the initial conditions \(P(0)=0\) and \(P(0)=K\). Explain why one requires special care during separation of variables.
Check the solution
(a) \(r=0.3\), \(K=1200\), and the equilibria are \(P=0,1200\). For \(0
(b) Growth is fastest at \(P=K/2=600\), and the maximum rate is \(rK/4=(0.3)(1200)/4=90\).
(c) \(A=(500-50)/50=9\), so \(P(t)=500/(1+9e^{-0.2t})\).
(d) \(P(10)=500/(1+9e^{-2})\approx225.43\).
(e) The inflection occurs at \(P=250\), so \(t=(\ln9)/0.2=5\ln9\approx10.99\).
(f) Since \(P>500\), the solution decreases and is concave up. It approaches the stable equilibrium \(P=500\) from above.
(g) \(P'/P=0.4(1-200/1000)=0.32\) per time unit, and \(P'=0.32(200)=64\) individuals per time unit.
(h) Here \(A=(800-80)/80=9\). Since \(200=800/(1+9e^{-4r})\), \(e^{-4r}=1/3\), so \(r=(\ln3)/4\approx0.275\).
(i) Exponential growth has constant relative rate \(P'/P=r\). Logistic growth has decreasing relative rate \(P'/P=r(1-P/K)\). When \(P\) is small relative to \(K\), the latter is approximately \(r\), so the models initially look similar.
(j) The solutions are \(P(t)=0\) and \(P(t)=K\), respectively. Separation divides by \(P\), so it loses the zero solution unless that equilibrium is recorded first.