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Rates of change and differential equations — Question 12

QCAA 2020, Paper 2 · 9 marks

Q12 · 2020 · Technology-activeSimple familiar9 marks

QUESTION 12 (9 marks)

For a certain experiment, the number of yeast cells, NN, after tt hours in a test tube can be modelled by the differential equation dNdt=11000N(1000−N),t≥0.\frac{dN}{dt}=\frac1{1000}N(1000-N),\qquad t\ge0.
a)
Given 1000N(1000−N)=1N+11000−N\dfrac{1000}{N(1000-N)}=\dfrac1N+\dfrac1{1000-N}, show that the general solution of the differential equation can be expressed as ln⁡∣N1000−N∣=t+c\ln\left|\dfrac{N}{1000-N}\right|=t+c.
[2 marks]
b)
A scientist starts the experiment at 9:00 am with 100 yeast cells. Show that the solution of the differential equation can be expressed as N=10001+9e−tN=\dfrac{1000}{1+9e^{-t}}.
[3 marks]
c)
Determine the time of day when 900 yeast cells were present.
[2 marks]
d)
The scientist predicted that the number of yeast cells would eventually exceed 1200. Evaluate the reasonableness of the scientist’s prediction.
[2 marks]
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