Sometimes it is possible to predict the future population in a city using a function - Leaving Cert Mathematics - Question 7 - 2017

To find the population at the beginning of 2015, we calculate:

$p(5) = 1.1e^{0.1 \times 5} \times 10^6$
Calculating this, we get:

$p(5) = 1.1e^{0.5} \times 10^6 \approx 1,813,593$
Thus, the predicted population is approximately 1,813,593 people.

To find the change in population during 2015, we compute:

$p(6) - p(5)$
Using the previously calculated $p(5)$ value:

$p(6) = 1.1e^{0.1 \times 6} \times 10^6$
The change is:

$p(6) - p(5) = 1970973 - 1813593 = 190737$

Given that the population in Avalon at the beginning of 2011 is 3,709,795, we write:

$q(1) = 3 - 9e^{k \times 1} \times 10^6$
Setting this equal to 3,709,795:

$3 - 9e^{k} \times 10^6 = 3,709,795$
Solving for k:

$-9e^{k} \times 10^6 = 3,709,795 - 3$
And then:

$k = -0.05$

To find when the populations are equal, we set:

$p(t) = q(t)$
Thus,

$Se^{0.1t} \times 10^6 = 3 - 9e^{-0.05t} \times 10^6$
Solving this gives us:

$1.1e^{0.1t} = 3 - 9e^{-0.05t}$
This leads to:

$e^{0.15t} = \frac{3-3.6}{1.1}$
Finding t gives us about 8.44 years, indicating that in 2018, both populations will be equal.

The average population can be found using the integral:

$\text{Average} = \frac{1}{15} \int_{0}^{15} q(t) dt$
Calculating this will provide the average population over that time period.

The rate of change can be found by deriving the function:

$q'(t) = k imes 3 - 9e^{-0.05t} \times 10^6$
Substituting $t = 8$ (for 2018) we get:

$q'(8) = -0.05 \times 3 - 9e^{-0.4} \times 10^6$
Calculating gives us the precise rate of change.