Transcript RecRltns02

GROWTH & DECAY
• NB
• Removing 15% leaves behind 85% or 0.85 which is
called the DECAY factor.
• Adding on 21% gives us 121% or 1.21 and this is
called the GROWTH factor.
• Growth and decay factors allow us a quick method
of tackling repeated % changes.
Ex1 An agar plate contains 10000 bacteria which
are being killed off at a rate of 17% per hour by
a particular disinfectant.
(a) How many bacteria are left after 3 hours?
(b) How many full hours are needed so that there
are fewer than 4000 bacteria?
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Suppose that un represents the number of bacteria
remaining after n hours.
Removing 17% leaves behind 83% so the DECAY factor is
0.83
and
un+1 = 0.83 un
(a)
u0 = 10000
u1 = 0.83u0 = 0.83 X 10000 = 8300
u2 = 0.83u1 = 0.83 X 8300 = 6889
u3 = 0.83u2 = 0.83 X 6889 = 5718
So there are 5718 bacteria after 3 hours.
(b)
u4 = 0.83u3 = 0.83 X 5718 = 4746
u5 = 0.83u4 = 0.83 X 4746 = 3939
This is less than 4000 so it takes 5 full hours to
fall below 4000.
Ex2
The population of a town is growing at a
rate of 14% per annum.
If P0 is the initial population and Pn is the
population after n years …..
then find a formula for Pn in terms of P0.
Find roughly how long it takes the population
to treble.
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Adding on 14% gives us 114% so the GROWTH
factor is 1.14
and
Pn+1 = 1.14 Pn
P1 = 1.14 P0
P2 = 1.14 P1 = 1.14 X 1.14 P0 = (1.14)2 P0
P3 = 1.14 P2 = 1.14 X (1.14)2 P0 = (1.14)3 P0
etc
So in general we have
Pn = (1.14)n P0
If the population trebles then we need to have
Pn > 3 P0
or
we get
(1.14)n P0 > 3 P0
(1.14)n > 3
Dividing by P0
We now use a bit of trial and error along with the
^
or
If n = 5
xy
buttons on the calculator.
then (1.14)5 = 1.92…
too small
If n = 9 then (1.14)9 = 3.25…
too big
If n = 7 then (1.14)7 = 2.50…
too small
If n = 8 then (1.14)8 = 2.85…
too small but
closest to 3.
From the above we can say it takes just over 8 years
for the population to treble.