

Interest
Calculations 
Exponential
growth and decay, application of the natural exponential
function 





Exponential
growth and decay, application of the natural exponential
function 
We
use the same formula as for continuous compound interest in many natural processes
where the rate of change of a
quantity through time is proportional to the current amount of the quantity,
represented by the differential equation 

where, y
is a function of time t,
whose value y(0)
= y_{0 }at
t
= 0, and k
is a constant. 
We solve
the given differential equation
with initial condition y(0)
= y_{0 } to find y
as function of t,
by separating the variables 


A
quantity is said to be subject to exponential growth if it
increases at a rate proportional to its current value. 
The
constant k
is called the growth rate and in
exponential growth k
> 0. The rate constant k
depends only on the process and the conditions under which it is
carried out. 


A
quantity is said to be subject to exponential decay if it
decreases at a rate proportional to its current value. 
The
constant k
is called the decay
rate and in exponential decay k
< 0. 


Example:
Suppose that
microorganisms in a culture dish grow exponentially. At the
start of an experiment there are 8,000 of bacteria, and two
hours later the population has increased to 8,600. How long will
it take for the population to reach 20,000? 
Solution: Given the initial
population N_{0}
= 8,000 and for t
=
2 hours
the population increased to N
= 8,600. 
We
first find the growth rate k
and then the time needed the population increases to 20,000. 


Radioactive decay is a
typical example to which the exponential decay model can be
applied. 
Example:
After 800
years a sample of radioisotope radium226 has decayed to 70.71%
of its initial mass, find the halflife of radium226. 
Solution:
The halflife of a radioactive material is the amount of time
required for half of a given sample to 
decay.
First
we plug the given information into the formula for exponential
decay 
N(t)
= N_{0 }e^{kt }to
find the decay constant k. 
Since
N_{0}
is the initial quantity, N(t)
is the quantity after time t
and k
is the decay constant then, 
by
substituting
N_{(t = 800)} = 0.7071·N_{0
}into
the formula 
0.7071N_{0
}= N_{0 }e ^{k}^{ ·}^{
}^{800} 
and
solving for k,
0.7071_{ }= _{
}e ^{k}^{ ·}^{ }^{800}
 ln 
ln 0.7071_{ }= 800 · k 
k_{ }= ln (0.7071) / 800 
k_{ }= 
0.0004332217. 
Thus,
the formula for the amount of radium226 present at a time t
is 
N(t)
= N_{0 }e ^{
0.0004332217 · t} 
As
we want determine the halflife or the time for half of a substance to decay,
we substitute N(t)
= (1/2) N_{0} 
into
the formula 









Intermediate
algebra contents 



Copyright
© 2004  2020, Nabla Ltd. All rights reserved. 