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:

$p(5) = S e^{0.1 \cdot 5} \times 10^6 = 1 \cdot e^{0.5} \times 10^6.$

Using the approximate value of $e^{0.5} \approx 1.6487$:

$p(5) \approx 1.6487 \times 10^6 \approx 1,813,593.$

To find the change during 2015, we first find:

$p(6) = S e^{0.1 \cdot 6} \times 10^6 = 1 \cdot e^{0.6} \times 10^6.$

Using $e^{0.6} \approx 1.8221$, we have:

$p(6) \approx 1.8221 \times 10^6.$
Now, calculate the population change:

$\Delta p = p(6) - p(5) = (1.8221 - 1.6487) \times 10^6 \approx 190,737.$

The population in Avalon at the beginning of 2011:

$q(1) = 3 e^{-k \cdot 1} \times 10^6 = 3 e^{-k} \times 10^6 = 3,709,795.$
Dividing by $3 \times 10^6$ gives:

\log(e^{-k}) = \log(1.236).$$ This simplifies to: $$-k = \log(1.236) \Rightarrow k = -\log(1.236) \approx -0.05.$$

To find when $p(t) = q(t)$:

$S e^{0.1t} \times 10^6 = 3 e^{-0.05t} \times 10^6.$
Cancel $10^6$ and $S = 1$:

$e^{0.1t} = 3 e^{-0.05t}.$
Rearranging gives:

$e^{0.15t} = 3.$
Taking logs:

$$$$

Thus, the year is:

2010 + 7 = 2018.$$

To find the average population:

$\text{Average} = \frac{1}{15} \int_0^{15} q(t) \,dt = \frac{1}{15} \int_0^{15} 3 e^{-0.05t} \times 10^6 \,dt.$
Using integration, this yields:

$= \frac{10^6}{15} \left[-\frac{3}{0.05} e^{-0.05t}\right]_{0}^{15}.$
Calculating gives:

$= \frac{10^6}{15} \left[-60 e^{-0.75} + 60 \right] \approx 2,743,694.$

To find the rate of change, we calculate:

$q(t) = 3 e^{-0.05t} \times 10^6.$
The derivative is:

$q'(t) = -0.05 \cdot 3 e^{-0.05t} \times 10^6.$
At $t = 8$ (the beginning of 2018):

$q'(8) = -0.15 e^{-0.4} \times 10^6 \approx -130,712.$