The Logistic Growth Model ($S$-Curve)
Unconstrained exponential growth assumes infinite resources. Real biological populations and market penetrations are bounded by carrying capacity $L$, satisfying the logistic differential equation:
Key Mathematical Properties:
- Carrying Capacity ($L$): The horizontal asymptote as $t \to \infty$: $\lim_{t \to \infty} P(t) = L$.
- Initial Value ($P(0)$): When $t = 0$: $P(0) = \frac{L}{1 + C} \implies C = \frac{L - P(0)}{P(0)}$.
- Inflection Point (Maximum Growth Rate): The point of fastest population growth occurs exactly at half carrying capacity: $$P = \frac{L}{2} \quad \text{at time } t = \frac{\ln(C)}{k}$$ $$\left(\frac{dP}{dt}\right)_{\max} = k\left(\frac{L}{2}\right)\left(1 - \frac{1}{2}\right) = \frac{kL}{4}$$
Piecewise Modelling & Spline Interpolation
When modeling roads, aircraft wings, or roller coaster tracks, smooth transition between segments requires continuous value and matching gradients:
Sinusoidal Modelling with Phase Shifts
Cyclic phenomena (tidal heights, daylight hours, temperature fluctuations) are modeled using:
A population of fish in a lake is modeled by $P(t) = \frac{4800}{1 + 15e^{-0.25t}}$, where $t$ is measured in months.
- Find the initial fish population.
- Write down the carrying capacity of the lake.
Solution:
(a) At $t = 0$: $P(0) = \frac{4800}{1 + 15e^0} = \frac{4800}{16} = 300$ fish. [M1][A1]
(b) As $t \to \infty$, $e^{-0.25t} \to 0$, so $P \to 4800$. The carrying capacity is $4800$ fish. [A1]
For the fish population in Exercise 1, find the time $t$ at which the population is growing at its maximum rate, and state the population at that moment.
Solution:
The maximum growth rate occurs at half the carrying capacity: $P = \frac{4800}{2} = 2400$ fish. [M1][A1]
Set $P(t) = 2400 \implies \frac{4800}{1 + 15e^{-0.25t}} = 2400 \implies 1 + 15e^{-0.25t} = 2$. [M1]
$15e^{-0.25t} = 1 \implies e^{-0.25t} = \frac{1}{15} \implies -0.25t = -\ln(15)$.
$t = \frac{\ln(15)}{0.25} \approx \frac{2.708}{0.25} = 10.83... \approx 10.8\text{ months}$. [A1]
Calculate the maximum rate of fish population growth in fish per month for the model in Exercises 1 and 2.
Solution:
Maximum growth rate is $\left(\frac{dP}{dt}\right)_{\max} = \frac{kL}{4}$. [M1]
Here $k = 0.25$ and $L = 4800$. [A1]
$\left(\frac{dP}{dt}\right)_{\max} = \frac{0.25 \times 4800}{4} = \frac{1200}{4} = 300\text{ fish per month}$. [A1]
A disease outbreak in a town of 10,000 residents is modeled by $I(t) = \frac{10000}{1 + 99e^{-0.4t}}$. Find the number of days until $80\%$ of the town has been infected.
Solution:
$80\%$ of $10,000 = 8000$ residents. [A1]
$\frac{10000}{1 + 99e^{-0.4t}} = 8000 \implies 1 + 99e^{-0.4t} = \frac{10000}{8000} = 1.25$. [M1]
$99e^{-0.4t} = 0.25 \implies e^{-0.4t} = \frac{0.25}{99} \approx 0.002525$. [M1]
$t = \frac{\ln(0.002525)}{-0.4} = \frac{-5.9814}{-0.4} \approx 14.95... \approx 15.0\text{ days}$. [A1]
A section of a highway curve is defined by the piecewise function:
$$f(x) = \begin{cases} 0.5x^2 + 2 & x \le 2 \\ ax + b & x > 2 \end{cases}$$Find the values of constants $a$ and $b$ such that the highway is continuous and has a smooth transition (differentiable) at $x = 2$.
Solution:
1. Continuity at $x = 2$: $0.5(2^2) + 2 = a(2) + b \implies 2 + 2 = 2a + b \implies 2a + b = 4$. [M1][A1]
2. Smooth differentiability at $x = 2$: $\frac{d}{dx}(0.5x^2 + 2) = x$. At $x = 2$, gradient $= 2$. [M1]
For $x > 2$, $\frac{d}{dx}(ax + b) = a \implies a = 2$. [A1]
Substitute $a = 2$: $2(2) + b = 4 \implies 4 + b = 4 \implies b = 0$. [A1]
Constants: $a = 2, \quad b = 0$.
The depth of water $h$ (meters) in a harbour over a 24-hour cycle is modeled by $h(t) = 4.5\cos\left(\frac{\pi}{6}t\right) + 8.0$, where $t$ is hours after midnight.
- Find the maximum depth (high tide) and minimum depth (low tide).
- Determine the period of the tidal oscillation in hours.
Solution:
(a) Max depth $= D + A = 8.0 + 4.5 = 12.5\text{ meters}$. [A1]
Min depth $= D - A = 8.0 - 4.5 = 3.5\text{ meters}$. [A1]
(b) Period $= \frac{2\pi}{B} = \frac{2\pi}{\pi/6} = 2\pi \times \frac{6}{\pi} = 12\text{ hours}$. [M1][A1]
Using the tidal model in Exercise 6, find the first time after midnight when the water depth is exactly $10.0\text{ meters}$.
Solution:
Set $h(t) = 10.0 \implies 4.5\cos\left(\frac{\pi}{6}t\right) + 8.0 = 10.0$ [M1]
$4.5\cos\left(\frac{\pi}{6}t\right) = 2.0 \implies \cos\left(\frac{\pi}{6}t\right) = \frac{2.0}{4.5} = \frac{4}{9} \approx 0.4444$. [M1]
$\frac{\pi}{6}t = \arccos(0.4444) \approx 1.1102\text{ radians}$. [M1]
$t = 1.1102 \times \frac{6}{\pi} \approx 2.12\text{ hours}$ (approx 02:07 AM). [A1]
Explain why the second derivative $P''(t)$ of the logistic function $P(t) = \frac{L}{1 + Ce^{-kt}}$ is equal to zero at $P = \frac{L}{2}$.
Solution:
Differentiating $\frac{dP}{dt} = kP - \frac{k}{L}P^2$ with respect to $t$ gives: $\frac{d^2 P}{dt^2} = k\frac{dP}{dt} - \frac{2k}{L}P\frac{dP}{dt} = k\frac{dP}{dt}\left(1 - \frac{2P}{L}\right)$. [M1][A1]
Setting $\frac{d^2 P}{dt^2} = 0$ requires $1 - \frac{2P}{L} = 0 \implies 2P = L \implies P = \frac{L}{2}$. [R1]
This confirms that the point of inflection (where concavity changes from concave up to concave down) occurs exactly at half the carrying capacity.
A temperature model is given by $T(t) = 8\sin(0.5236(t - 3)) + 21$, where $t$ is in months ($t = 1$ is January). Find the amplitude, midline, and the month when the temperature reaches its peak.
Solution:
Amplitude $A = 8^\circ\text{C}$; Midline $D = 21^\circ\text{C}$. [A2]
The standard sine function reaches its maximum at an argument of $\frac{\pi}{2} \approx 1.5708$. [M1]
$0.5236(t - 3) = \frac{\pi}{2} \implies t - 3 = \frac{\pi/2}{2\pi/12} = 3 \implies t = 6$ (June). [A1]
Contrast the behavior of the logistic function $P(t) = \frac{L}{1 + Ce^{-kt}}$ with exponential growth $P(t) = P_0 e^{kt}$ for large values of $t$, and explain which is more appropriate for long-term ecological forecasting.
Solution:
As $t \to \infty$, exponential growth diverges to infinity ($P(t) \to \infty$), whereas the logistic function stabilizes and approaches the carrying capacity $L$. [A1]
The logistic model is far more realistic for ecological forecasting because natural ecosystems have finite space, water, and food resources that constrain unlimited growth. [R2]
Core Formulas (Topic 2 AHL)
| Logistic Function | $$P(t) = \frac{L}{1 + C e^{-kt}}$$ |
| Point of Maximum Growth | $$P = \frac{L}{2} \quad \text{at time } t = \frac{\ln(C)}{k}$$ |
| Max Growth Rate | $$\left(\frac{dP}{dt}\right)_{\max} = \frac{kL}{4}$$ |
| Sinusoidal Model Period | $$f(t) = A\sin(B(t - C)) + D \implies \text{Period} = \frac{2\pi}{B}$$ |
Top 5 IB Exam Traps to Avoid
- Trap 1: Confusing Initial Population with $C$: The initial value is $P(0) = \frac{L}{1 + C}$, NOT $C$. $C = \frac{L - P(0)}{P(0)}$.
- Trap 2: Forgetting the Factor of $\frac{1}{4}$ in Max Growth: The maximum rate is $\frac{kL}{4}$, not $kL$ or $\frac{kL}{2}$.
- Trap 3: Radians vs Degrees in Sinusoidal Models: IB DP sinusoidal models are ALWAYS evaluated in radians mode unless the prompt explicitly specifies degrees.
- Trap 4: Spline Differentiability conditions: To prove a smooth transition between pieces, you must verify BOTH $g(a) = h(a)$ (continuity) AND $g'(a) = h'(a)$ (smoothness).
- Trap 5: Horizontal Asymptote Misinterpretation: The carrying capacity $L$ is a horizontal asymptote that the curve never crosses from below.
5-Question Concept Check
Score: - / 51. In the logistic function $P(t) = \frac{1200}{1 + 3e^{-0.2t}}$, what is the carrying capacity?
2. At what population level does the maximum growth rate occur for a logistic model with carrying capacity $L$?
3. What is the initial population $P(0)$ for $P(t) = \frac{1000}{1 + 9e^{-0.5t}}$?
4. What is the period of the sinusoidal function $f(t) = 5\sin\left(\frac{\pi}{4}t\right) + 12$?
5. For two pieces $g(x)$ and $h(x)$ to meet smoothly at junction $x = c$, which two conditions must hold?