To model the population of deer over time, we can use the logistic growth model since the population has a carrying capacity and a growth rate. The logistic growth equation is:

$P(t) = \frac{K}{1 + \left( \frac{K - P_0}{P_0} \right) e^{-rt}}$

where:

• $P(t)$ is the population at time $t$,
• $K$ is the carrying capacity,
• $P_0$ is the initial population,
• $r$ is the growth rate,
• $e$ is Euler's number (approximately 2.718).

Given values:

• $K = 200200$ (carrying capacity),
• $r = 0.02$ (growth rate of 2% per unit of time),
• $P_0 = 5050$ (initial population),
• $t$ is time (the variable we're interested in).

Substitute these values into the equation:

$P(t) = \frac{200200}{1 + \left( \frac{200200 - 5050}{5050} \right) e^{-0.02t}}$

Now, simplify the fraction inside the equation:

$\frac{200200 - 5050}{5050} = \frac{195150}{5050} \approx 38.64$

So, the population at time $t$ becomes:

$P(t) = \frac{200200}{1 + 38.64 e^{-0.02t}}$

This equation describes the population of deer at any given time $t$.

Would you like more details on how to interpret this function or how to compute the population at specific time points?

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5 Related Questions:

1. How does the growth rate affect the time it takes for the population to reach half of the carrying capacity?
2. What happens to the population if the initial population is closer to the carrying capacity?
3. How can we estimate the time when the population reaches 90% of the carrying capacity?
4. What is the significance of the carrying capacity in logistic growth?
5. How would changing the growth rate to 4% impact the population growth?

Tip: Logistic growth models are ideal for populations in environments with limited resources, as they account for the saturation of growth at the carrying capacity.