Let's solve the problem step by step!

The population of bears grows according to the logistic differential equation:

$\frac{dP}{dt} = 5P - 0.002P^2,$

where $P$ is the number of bears at time $t$ in years.

Part (a): $P(0) = 100$

Step 1: Find $\lim_{t \to \infty} P(t)$

The logistic equation has equilibrium points when $\frac{dP}{dt} = 0$. Solving for $P$: $\frac{dP}{dt} = 5P - 0.002P^2 = 0 \implies P(5 - 0.002P) = 0.$ This gives two equilibrium points: $P = 0 \quad \text{and} \quad P = \frac{5}{0.002} = 2500.$

The equilibrium points $P = 0$ and $P = 2500$ represent the population levels where the growth stops. To determine stability, analyze the sign of $\frac{dP}{dt}$ near these points:

• If $0 < P < 2500$, then $\frac{dP}{dt} > 0$, so the population increases.
• If $P > 2500$, then $\frac{dP}{dt} < 0$, so the population decreases.

Thus, $P = 2500$ is a stable equilibrium, and $\lim_{t \to \infty} P(t) = 2500$ for any $P(0) > 0$.

For $P(0) = 100$, the solution curve will increase towards 2500.

Step 2: Sketch the graph

For $P(0) = 100$, the population starts small and grows towards the carrying capacity of 2500. The graph is an increasing curve approaching $P = 2500$ asymptotically.

Part (b): $P(0) = 1500$

Step 1: Find $\lim_{t \to \infty} P(t)$

As established, $\lim_{t \to \infty} P(t) = 2500$ for any $P(0) > 0$.

For $P(0) = 1500$, the population is below the carrying capacity, so the solution curve will increase towards 2500.

Step 2: Sketch the graph

For $P(0) = 1500$, the population grows towards 2500, but at a slower rate compared to when $P(0) = 100$ because it starts closer to the carrying capacity. The graph is again an increasing curve approaching $P = 2500$ asymptotically.

Part (c): $P(0) = 3000$

Step 1: Find $\lim_{t \to \infty} P(t)$

For $P(0) = 3000$, the initial population is above the carrying capacity ($P = 2500$). The population will decrease towards the equilibrium point at $P = 2500$.

Step 2: Sketch the graph

For $P(0) = 3000$, the population decreases over time and approaches 2500 asymptotically. The graph is a decreasing curve approaching $P = 2500$.

Part (d): Population at maximum growth rate

The logistic growth equation has the form: $\frac{dP}{dt} = rP \left(1 - \frac{P}{K}\right),$ where $r = 5$ (growth rate) and $K = 2500$ (carrying capacity). The population grows the fastest when $P = \frac{K}{2}$.

Thus, the population of bears grows the fastest when: $P = \frac{2500}{2} = 1250.$

Summary of Results

1. (a) For $P(0) = 100$, $\lim_{t \to \infty} P(t) = 2500$, and the solution curve is increasing.
2. (b) For $P(0) = 1500$, $\lim_{t \to \infty} P(t) = 2500$, and the solution curve is increasing.
3. (c) For $P(0) = 3000$, $\lim_{t \to \infty} P(t) = 2500$, and the solution curve is decreasing.
4. (d) The population grows the fastest when $P = 1250$.

Do you need further clarification or detailed steps for any part?
Here are some related questions you might consider:

1. What is the general solution to the given logistic differential equation?
2. How do equilibrium points affect population models in general?
3. Can you explain the concept of carrying capacity in more detail?
4. How does the initial population affect the time it takes to approach equilibrium?
5. What happens if the growth rate $r$ or carrying capacity $K$ changes?

Tip: In logistic models, the population growth is fastest at half the carrying capacity because resources are optimally utilized!