Logistic Models with Differential Equations - AP Calculus BC

1. What are the units of $r$ in the logistic model $\frac{dP}{dt} = rP(1 - \frac{P}{K})$?

   Inverse time. Since $\frac{dP}{dt}$ has units of population per time.

2. What is logistic growth's behavior at $t \to \text{infinity}$?

   $P(t) \to K$. As $t \to \infty$, the exponential term vanishes leaving $P = K$.

3. Describe the asymptotic behavior of $P(t)$ in logistic growth.

   Approaches $K$ as $t \to \text{infinity}$. The exponential term vanishes as $t$ increases without bound.

4. What is the significance of the point of inflection in a logistic growth model?

   It is where growth rate is maximum. The inflection point occurs where the second derivative equals zero.

5. What does $P(0)$ represent in the context of logistic growth?

   Initial population size. $P(0)$ is the starting population at time $t = 0$.

6. What characterizes the initial phase of logistic growth?

   Exponential-like growth. Early phase resembles exponential growth when $P << K$.

7. What characterizes a stable equilibrium in logistic growth?

   Population returns to $K$. Small perturbations from $K$ decay back to equilibrium.

8. State the solution for $P(t)$ when $P(0) = \frac{K}{2}$ in logistic growth.

   $P(t) = \frac{K}{1 + e^{-rt}}$. When $P(0) = \frac{K}{2}$, then $A = 1$ in the solution.

9. Find $P(t)$ when $r = 0.5$, $K = 100$, and $P(0) = 10$.

   $P(t) = \frac{100}{1 + 9e^{-0.5t}}$. Using $A = \frac{K - P(0)}{P(0)} = \frac{90}{10} = 9$.

10. Explain the impact of $K$ in the logistic growth equation.

    Sets upper limit for population. $K$ represents the environmental limit on population size.

11. What does $r$ represent in the logistic differential equation?

    Growth rate. $r$ is the intrinsic growth rate when population is small.

12. What happens to growth rate as $P$ approaches $K$ in logistic growth?

    Growth rate approaches zero. The term $(1 - \frac{P}{K})$ approaches zero as $P \to K$.

13. Describe the role of $e^{-rt}$ in logistic growth solution.

    Determines rate of approach to $K$. The exponential decay controls how quickly $P$ approaches $K$.

14. Calculate $P(t)$ when $K = 200$, $r = 0.2$, and $P(0) = 100$. Use logistic solution.

    $P(t) = \frac{200}{1 + e^{-0.2t}}$. With $P(0) = 100 = \frac{K}{2}$, we get $A = 1$.

15. How do you find the constant $A$ in the logistic solution $P(t) = \frac{K}{1 + Ae^{-rt}}$?

    Use initial condition $P(0)$. Substitute $P(0)$ into the solution to solve for $A$.

16. Find the value of $A$ if $P(0) = 5$, $K = 10$, and $r = 0.3$. Use $P(t) = \frac{K}{1 + Ae^{-rt}}$.

    $A = 1$. Using $A = \frac{K - P(0)}{P(0)} = \frac{10 - 5}{5} = 1$.

17. In logistic growth, what phase follows the inflection point?

    Decelerating growth phase. After inflection, growth rate decreases as $P$ approaches $K$.

18. Compute $P(t)$ for $r = 1$, $K = 50$, and $P(0) = 25$. Use logistic solution.

    $P(t) = \frac{50}{1 + e^{-t}}$. With $P(0) = 25 = \frac{K}{2}$, we have $A = 1$.

19. What property of logistic growth makes it realistic for modeling populations?

    Incorporates carrying capacity. Unlike exponential growth, logistic models have a maximum capacity.

20. What is the logistic growth rate at $P = \frac{K}{2}$?

    Maximum growth rate. At the inflection point, $\frac{dP}{dt}$ reaches its maximum value.

21. In logistic growth, what happens as $P$ approaches $K$?

    Growth rate decreases. Factor $(1 - \frac{P}{K})$ approaches zero as $P$ nears $K$.

22. Identify the inflection point time $t$ for logistic model $P(t) = \frac{K}{1 + Ae^{-rt}}$.

    $t = \frac{\text{ln}(A)}{r}$. Inflection occurs when the exponential term $Ae^{-rt} = 1$.

23. Determine the point of inflection for $P(t) = \frac{K}{1 + Ae^{-rt}}$.

    $P = \frac{K}{2}$. Inflection occurs at half the carrying capacity in logistic growth.

24. What happens to $P(t)$ when $r$ is zero in the logistic model?

    Population remains constant. Zero growth rate means $\frac{dP}{dt} = 0$ for all $t$.

25. What condition leads to logistic growth equilibrium?

    $P = K$. At equilibrium, $\frac{dP}{dt} = 0$, which occurs when $P = K$.

26. State the logistic growth model in terms of $y(t)$ if $y(0) = 1$ and $K = 10$.

    $y(t) = \frac{10}{1 + 9e^{-rt}}$. With $y(0) = 1$ and $K = 10$, we get $A = 9$.

27. What is the impact of a negative $r$ in the logistic model?

    Population declines. Negative $r$ reverses the direction of population change.

28. What is the behavior of the logistic model when $P > K$?

    Population decreases. When $P > K$, the factor $(1 - \frac{P}{K})$ becomes negative.

29. What is the effect of a larger $r$ on the logistic growth curve?

    Faster growth towards $K$. Higher $r$ means the population reaches $K$ more quickly.

30. What is the solution form of a logistic differential equation?

    $P(t) = \frac{K}{1 + Ae^{-rt}}$. General solution with constant $A$ determined by initial conditions.

31. What is the general form of a logistic differential equation?

    $\frac{dP}{dt} = rP(1 - \frac{P}{K})$. Standard form with growth rate $r$, population $P$, and carrying capacity $K$.

32. Find the equilibrium solutions of $\frac{dP}{dt} = rP(1 - \frac{P}{K})$.

    $P = 0$ and $P = K$. Set $\frac{dP}{dt} = 0$ and solve for $P$.

33. Identify the phase where logistic growth is approximately linear.

    Near point of inflection. Around the inflection point, growth rate is roughly constant.

34. Identify the carrying capacity in the logistic equation $\frac{dP}{dt} = rP(1 - \frac{P}{K})$.

    Carrying capacity $K$. $K$ is the maximum sustainable population in the logistic model.

35. How is the constant $A$ related to initial population in logistic growth?

    $A = \frac{K - P(0)}{P(0)}$. Derived from solving $P(0) = \frac{K}{1 + A}$ for $A$.

36. Identify the type of growth when $P$ is much less than $K$ in the logistic model.

    Approximately exponential growth. When $P << K$, the factor $(1 - \frac{P}{K}) \approx 1$.

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