The given differential equation modeling population growth is:

$\frac{dP}{dt} = 1.1P\left(1 - \frac{P}{4100}\right)$

This is a logistic growth model, where the growth rate depends on the current population $P$ and its carrying capacity (here, 4100). Now let's address each part:

(a) For what values of $P$ is the population increasing?

The population is increasing when $\frac{dP}{dt} > 0$. For the given equation:

$1.1P\left(1 - \frac{P}{4100}\right) > 0$

This inequality holds when:

1. $P > 0$ (since $P$ represents population, it must be non-negative)
2. $1 - \frac{P}{4100} > 0$ which implies $P < 4100$

So the population is increasing when $0 < P < 4100$.

Answer: $P \in (0, 4100)$

(b) For what values of $P$ is the population decreasing?

The population decreases when $\frac{dP}{dt} < 0$, i.e.,

$1.1P\left(1 - \frac{P}{4100}\right) < 0$

This happens if $P > 4100$, because the term $1 - \frac{P}{4100}$ will become negative. Additionally, $P = 0$ results in no population, hence $\frac{dP}{dt} = 0$, which is not increasing but stable.

Answer: $P \in (4100, \infty)$

(c) What are the equilibrium solutions?

Equilibrium occurs when $\frac{dP}{dt} = 0$. Setting the equation equal to zero:

$1.1P\left(1 - \frac{P}{4100}\right) = 0$

This implies either $P = 0$ or $1 - \frac{P}{4100} = 0$, leading to $P = 4100$.

Answer: $P = 0, 4100$

Do you want further details or have any questions?

Here are 5 related questions you might explore:

1. How does the carrying capacity influence the population dynamics in a logistic model?
2. What happens to the population when it exceeds the carrying capacity?
3. Can the population ever stabilize at a value other than 0 or the carrying capacity?
4. How would the dynamics change if the growth rate were less than 1.1?
5. Could an external factor, like predation, be modeled by modifying this equation?

Tip: In logistic growth models, the population grows fastest when it is at half of the carrying capacity.