The population of a town is being studied - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 8

As $t$ approaches infinity, the term $e^{-kt}$ approaches 0. Therefore, the expected upper limit of the population is:

$P = \frac{8000}{1 + 7 \cdot 0} = 8000.$

The expected upper limit of the population is 8000.

Given that $P = 2500$ at $t = 3$, we can set up the equation:

$2500 = \frac{8000}{1 + 7e^{-3k}}.$

Solving for $k$ involves rearranging the equation:

1. Multiply both sides by $1 + 7e^{-3k}$: $2500(1 + 7e^{-3k}) = 8000$

2. Expand: $2500 + 17500e^{-3k} = 8000$

3. Rearrange to isolate $e^{-3k}$:

   $$$$

ightarrow e^{-3k} = \frac{5500}{17500}$$

4. So, $e^{-3k} = 0.3142857143$

5. Taking $ext{ln}$ of both sides gives:

   $$$$

ightarrow k = -\frac{\ln(0.3142857143)}{3} \approx 0.386.$$

Therefore, to three decimal places, $k \approx 0.386$.

Substituting $k \approx 0.386$ and $t = 10$, we determine the population:

$P = \frac{8000}{1 + 7e^{-0.386 \cdot 10}}.$

Calculating:

1. Calculate $e^{-3.86} \approx 0.021.$
2. Substitute: $P \approx \frac{8000}{1 + 7 \cdot 0.021} = \frac{8000}{1 + 0.147} = \frac{8000}{1.147} \approx 6970.48.$

Thus, to 3 significant figures, the population is approximately 6970.

To determine the rate of change of the population, we use the derivative:

$\frac{dP}{dt} = \frac{8000 \cdot 7 \cdot (-ke^{-kt})}{(1 + 7e^{-kt})^2}.$

Substituting $k \approx 0.386$ and $t = 10$:

1. First calculate $e^{-3.86} \approx 0.021$ as before.
2. Then, $\frac{dP}{dt} \approx \frac{8000 \cdot 7 \cdot (-0.386) \cdot 0.021}{(1 + 7 \cdot 0.021)^2}.$
3. Continue calculating to find: $\frac{dP}{dt} \approx 346.$

Therefore, at 10 years from the start of the study, the population is growing at a rate of approximately 346 individuals per year.