The population of a town is being studied - Edexcel - A-Level Maths Pure - Question 3 - 2013 - Paper 8

As $t$ approaches infinity, $e^{-kt}$ approaches 0. Therefore, the expected upper limit for the population is: $P(t \to \infty) = \frac{8000}{1 + 0} = 8000.$ The expected upper limit of the population is 8000.

Given that $P(3) = 2500$, we set up the equation: $2500 = \frac{8000}{1 + 7e^{-3k}}.$ Rearranging gives: $1 + 7e^{-3k} = \frac{8000}{2500} = 3.2,$ which leads to: $7e^{-3k} = 3.2 - 1 = 2.2.$ This results in: $e^{-3k} = \frac{2.2}{7}.$ Taking the natural logarithm:

Using the value of $k = 0.386$, we can find $P(10)$: $P(10) = \frac{8000}{1 + 7e^{-10 \cdot 0.386}}.$ Calculating: $e^{-3.86} \approx 0.021.$ Thus: $P(10) \approx \frac{8000}{1 + 7 \cdot 0.021} \approx \frac{8000}{1 + 0.147} \approx \frac{8000}{1.147} \approx 6967.82.$ Rounding to 3 significant figures gives 6970.

To find rac{dP}{dt}, we differentiate $P$ with respect to $t$: $\frac{dP}{dt} = \frac{8000 \cdot \left(-7 \cdot e^{-kt}\right) \cdot (-k)}{(1 + 7e^{-kt})^2}.$ Substituting $t = 10$: $\frac{dP}{dt} = \frac{8000 \cdot k \cdot 7e^{-10k}}{(1 + 7e^{-10k})^2}.$ Calculating at $t = 10$ and substituting $k = 0.386$: $\frac{dP}{dt} \approx 346.$ The rate at which the population is growing at 10 years is approximately 346.