2025 YIJC P1 Q13

2025 YIJC P1 Q13

Junior College 2
12 marks

Around November \(2019\), there was an outbreak of COVID-\(19\) in Wuhan, China, caused by the coronavirus SARS-CoV-\(2\) virus, which subsequently spread to the rest of the world. On \(23\) January \(2020\), there was \(1\) confirmed case in Singapore. \(70\) days later, there were \(1049\) confirmed cases.

A student is interested to study the spread of the virus in Singapore and uses the model \(\frac{\mathrm{d}N}{\mathrm{d}t} = rN\), where \(r\) is a positive constant, \(N\) denotes the total number of confirmed cases, and \(t\) is the number of days after \(23\) January \(2020\).

  1. Solve the differential equation to express \(N\) in terms of \(t\).[4]
  2. Use the model to estimate the number of days it will take to reach \(1000000\) cases.[1]

To contain the spread of the virus, Singapore implemented strict circuit breaker lockdown measures starting on \(7\) April \(2020\), when the number of confirmed cases stood at \(1481\). By the time the measures were lifted \(55\) days later, the number of confirmed cases had surged to \(35292\).

To study the spread of the virus during this period, the student uses a second model \(\frac{\mathrm{d}N}{\mathrm{d}u} = sN\), where \(s\) is a positive constant, and \(u\) is the number of days after \(7\) April \(2020\).

  1. Find the value of \(s\).[2]
  2. By comparing the values of \(r\) and \(s\), comment on whether the lockdown measures were effective in containing the spread of the virus.[1]
  3. Give a reason why neither model can estimate the actual number of confirmed cases accurately.[1]

On \(30\) December \(2020\), Singapore became the first country in Asia to start its COVID-\(19\) vaccination campaign. The student now uses a third model \(\frac{\mathrm{d}N}{\mathrm{d}v} = aN \left( 1 - \frac{N}{b} \right)\), where \(a\) and \(b\) are positive constants, and \(v\) is the number of days after \(30\) December \(2020\).

  1. Find the maximum rate of change of \(N\) in terms of \(a\) and \(b\).[3]

Solution:

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Answer:(a) \(N=\mathrm e^{\frac{\ln1049}{70}t}\), \(r\approx0.0994\). (b) Continuous threshold \(139.037\ldots\) days; first whole day \(140\). (c) \(s\approx0.0577\). (d) \(s<r\): reduced fitted growth rate. (e) Constant-rate, unbounded growth ignores changing conditions and finite population. (f) Maximum rate \(\frac{ab}{4}\), at \(N=\frac{b}{2}\).

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