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