The population of mice on an isolated island can be modelled by the function $$m(t) = a \sin \left( \frac{\pi}{26} t \right) + b,$$ where $t$ is the time in weeks and $0 \leq t \leq 52$ - HSC - SSCE Mathematics Advanced - Question 31 - 2020 - Paper 1

To determine when both populations are increasing, we need to find the intervals where the derivatives of both functions are positive.

1. For the mice population $m(t)$:

   • Find the derivative:
     $m'(t) = a \cdot \frac{\pi}{26} \cos \left( \frac{\pi}{26} t \right).$
   • Set $m'(t) > 0$. Since $a = 15000$, we get: $15000 \cdot \frac{\pi}{26} \cos \left( \frac{\pi}{26} t \right) > 0,$
     which implies that $\cos \left( \frac{\pi}{26} t \right) > 0.$
     This occurs in the ranges $0 < t < 13$ and $39 < t < 52$.

2. For the cat population $c(t)$:

   • Find the derivative:
     $c'(t) = 80 \cdot \frac{\pi}{26} \sin \left( \frac{\pi}{26} (t - 10) \right).$
   • Set $c'(t) > 0$. Since the factor is positive, we have: $\sin \left( \frac{\pi}{26} (t - 10) \right) > 0.$
     This occurs in the ranges $10 < t < 36$.

Combining both intervals, we find that:

• Both populations are increasing for $10 < t < 13$.

The cat population reaches its maximum at $t = 36$. To find the rate of change of the mice population at this time, we need to find the derivative of $m(t)$:

1. Differentiate $m(t)$: $m'(t) = 15000 \cdot \frac{\pi}{26} \cos \left( \frac{\pi}{26} t \right).$

2. Substitute $t = 36$ into the derivative:

   • First calculate: $m'(36) = 15000 \cdot \frac{\pi}{26} \cos \left( \frac{\pi}{26} \cdot 36 \right).$
   • Calculate the cosine value: $\frac{\pi}{26} \cdot 36 = 4.3633 \implies \cos(4.3633) \approx -0.6427.$
   • Thus: $m'(36) \approx 15000 \cdot \frac{\pi}{26} \cdot (-0.6427) \approx -643.$

Therefore, the rate of change of the mice population when the cat population reaches a maximum is approximately $-643$ mice per week, indicating that the mice population is decreasing.