2021 DHS Promo Q12

2021 DHS Promo Q12

13 marks
Promo

A fish pond has a surface area of \(100\)m\(^{2}\). Past observations on the growth of algae in similar ponds estimates that the area, \(A\)m\(^{2}\), of algae present at time \(t\) days is such that the rate at which \(A\) is increasing is proportional to the product of the area of the pond covered by the algae and the area of the pond not covered by the algae.

The owner of the pond monitors the growth of algae in his pond. At \(t=2\), he noted that the algae had grown to cover \(11\%\) of the pond and was growing at a rate of \(11\)m\(^{2}\) per day. He then decided to clear the algae at a constant rate of \(11\)m\(^{2}\) each day, starting from \(t=3\).

  1. Show that, from \(t=3\), the differential equation relating \(A\) and \(t\) is
    \(\frac{\mathrm{d}A}{\mathrm{d}t}=-\frac{1}{89}\left( A-11 \right)\left( A-89 \right)\).[2]
  2. Assuming that \(A=20\) when \(t=3\), show that \(\frac{A-89}{A-11}=-\frac{23}{3}{{\mathrm{e}}^{q\left( 3-t \right)}}\), where \(q\) is a positive constant to be determined.[6]
  3. To maintain the balance of the ecosystem in the pond, the owner wishes to ensure that algae always cover an area less than \(80\%\) of the pond surface. Explain why this cannot be attained.[2]
  4. Find the value of \(t\) when the maximum possible rate of change in the area of algae occurs.[2]
  5. State one possible condition in which the model in part (i) is not valid.[1]

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Answer:(i) \(\frac{\mathrm{d}A}{\mathrm{d}t}=-\frac{1}{89}(A-11)(A-89)\). (ii) \(q=\frac{78}{89}\). (iii) \(A\to89\text{ m}^2\), so the area eventually exceeds \(80\text{ m}^2\). (iv) \(t=3-\frac{89}{78}\ln\left(\frac{3}{23}\right)\approx5.32\) days. (v) One possible invalidating condition is seasonal variation in the algae growth rate.

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