

For "O" level Add maths students to practise
Question 5(a) (from Pan Pacific Additional Mathematics Page 421)
In an experiment, the growth rate of bacteria in a liquid at any time is proportional to N, the number of bacteria after t seconds. Given that the initial number of bacteria is A, show that N = Ae^(kt) , where k is a constant.
Steps for Solution for Q5(a)
(1) Given that the initial number of bacteria is A, show that N = Ae^(kt) ,
where k is a constant.
(2) Since the growth rate of bacteria (ie dN/dt) in a liquid at any time
is proportional to N (ie kN), the number of bacteria after t seconds.
So, dN/dt = kN
(3) dN/dt = kN
dN/N = kdt
(1/N) dN = k dt
Integrate on both sides
Integrate (1/N) dN = Integrate k dt
ln N + C1 = kt + C2
ln N = kt + C2  C1
N = e ^ ( kt + C2  C1)
N = e ^ ( C2  C1 ) e ^ (kt)
(4) Initial number of bacteria is A ie when t = 0 , N = A
Substitute t = 0 , N = A into N = e ^ ( C2  C1 ) e ^ (kt)
A = e ^ ( C2  C1 ) e^ (k x 0)
A = e ^ ( C2  C1 ) e^0
A = e ^ ( C2  C1 ) x 1 since e^0 = 1
A = e ^ ( C2  C1 )
(5) Substitute A = e ^ ( C2  C1 ) into N = e ^ ( C2  C1 ) e ^ (kt)
N = Ae^(kt) (Shown)
Thank you for your kind attention.
Regards,
ahm97sic
