Esta lección aún no está traducida. Por ahora se muestra en inglés; la versión en español está en camino.
Exponential growth goes on forever, but real populations run out of food or space. A logistic model starts out growing almost exponentially, then slows and levels off at a ceiling called the carrying capacity. The same S-shaped curve describes the spread of a rumor or the adoption of a new product.
🔑 Idea clave
$$P(t) = \frac{K}{1 + Ae^{-rt}}$$ where $K$ is the carrying capacity, $r > 0$ is the growth rate, and $A = \frac{K - P_0}{P_0}$ comes from the starting value $P_0$. The population grows fastest when it is at $\frac{K}{2}$, which happens at $t = \frac{\ln A}{r}$.
💡 Por qué funciona
Check the formula's behavior. At $t = 0$, $e^0 = 1$, so $P(0) = \frac{K}{1 + A}$; solving for $A$ gives $A = \frac{K - P_0}{P_0}$. As $t$ grows, $e^{-rt}$ shrinks to $0$ and $P(t)$ approaches $K$ without passing it. When $Ae^{-rt} = 1$, the denominator is $2$ and $P = \frac{K}{2}$; solving $e^{-rt} = \frac{1}{A}$ gives $t = \frac{\ln A}{r}$. The model comes from assuming the growth rate is proportional both to the population and to the room left, $1 - \frac{P}{K}$: small populations grow almost exponentially, and growth stalls near $K$.
The logistic curve $f$ and the exponential $g$ with the same start and rate. They agree at first; then the logistic curve bends toward $K$.
Worked examples
✏️ Ejemplo resuelto: Reading a model
A fish population follows $P(t) = \dfrac{1000}{1 + 9e^{-0.5t}}$, with $t$ in years. Find the starting population, the carrying capacity and when the population reaches $500$.
Reading $K$ as the starting value, or $A$ as the starting value. $K$ is the long-run ceiling; the start is $P(0) = \frac{K}{1 + A}$. Check by putting $t = 0$ into your model: you should get the initial population.
Práctica
Apruebas con 3 respuestas correctas, las 2 últimas seguidas. Cada respuesta correcta da 3 XP.