The population of a town is being studied - Edexcel - A-Level Maths Pure - Question 22 - 2013 - Paper 1

The expected upper limit of the population can be found as t approaches infinity. As $t \to \infty$, the term $e^{-kt} \to 0$.

Thus, we have:

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

To find k, we use the given data that at t = 3, P = 2500:

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

$1 + 7e^{-3k} = \frac{8000}{2500} = 3.2.$
This simplifies to:

$7e^{-3k} = 3.2 - 1 = 2.2.$
Dividing by 7:

$e^{-3k} = \frac{2.2}{7}.$
Taking the natural logarithm of both sides:

$-3k = \ln\left(\frac{2.2}{7}\right),$
which gives us:

$k = -\frac{1}{3} \ln\left(\frac{2.2}{7}\right) \approx 0.386.$
Thus, k to 3 decimal places is 0.386.

Using k = 0.386, we calculate:

$P(10) = \frac{8000}{1 + 7e^{-0.386 \cdot 10}}.$
Calculating the exponent:

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

$P(10) = \frac{8000}{1 + 7 \cdot 0.021} = \frac{8000}{1 + 0.147} = \frac{8000}{1.147} \approx 6970.$
Therefore, the population at 10 years from the start of the study, to 3 significant figures, is approximately 6970.

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

Using the formula,

$\frac{dP}{dt} = \frac{C}{(1 + 7e^{-kt})^2} \cdot (7ke^{-kt}).$
Substituting k = 0.386 and t = 10:

$\frac{dP}{dt} = \frac{8000 \cdot 7 \cdot 0.386 \cdot e^{-3.86}}{(1 + 7 \cdot e^{-3.86})^2}.$
Carry out the calculations to give the rate of growth, which evaluates to approximately 346 at t = 10.