The population of a town is being studied - Edexcel - A-Level Maths Pure - Question 1 - 2013 - Paper 8

As $t$ approaches infinity, the term $7e^{-kt}$ approaches 0. Thus, we can find the expected upper limit:

$P = \frac{8000}{1 + 0} = 8000.$
Hence, the expected upper limit of the population is 8000.

Given $P = 2500$ at $t = 3$, we substitute into the equation:

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

Rearranging gives:

$1 + 7e^{-3k} = \frac{8000}{2500} \implies 1 + 7e^{-3k} = 3.2.$

This leads to:

$7e^{-3k} = 3.2 - 1 = 2.2 \implies e^{-3k} = \frac{2.2}{7} \implies -3k = \ln\left(\frac{2.2}{7}\right) \implies k = -\frac{1}{3}\ln\left(\frac{2.2}{7}\right).$

Calculating this gives:
$k \approx 0.386.$
Therefore, the value of $k$ to 3 decimal places is 0.386.

Using $k \approx 0.386$, we substitute $t = 10$ into the population equation:

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

Calculating the exponent:

$e^{-3.86} \approx 0.021.$

Thus,

$P \approx \frac{8000}{1 + 7 \cdot 0.021} \approx \frac{8000}{1 + 0.147} \approx \frac{8000}{1.147} \approx 6970.3.$
Hence, the population at 10 years is approximately 6970.

To find the rate of population growth, we first differentiate the population function:

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

Substituting $t = 10$ into this expression gives:

$\frac{dP}{dt} = \frac{8000 \cdot (-7) e^{-3.86}}{(1 + 7e^{-3.86})^2}.$

Calculating $e^{-3.86} \approx 0.021$ from earlier, we get:

$\frac{dP}{dt} \approx \frac{8000 \cdot (-7) \cdot 0.021}{(1 + 7 \cdot 0.021)^2} \approx \frac{-1176}{1.147^2} \approx -346.$
Thus, the rate at which the population is growing at 10 years is approximately 346.