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) = yat  t = 0,  and k is a constant.
We solve the given differential equation with initial condition  y(0) = y0 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 N0 = 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 radium-226 has decayed to 70.71% of its initial mass, find the half-life of radium-226.
Solution:  The half-life 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) = N0 ekt    to find the decay constant k.
Since N0 is the initial quantity, N(t) is the quantity after time t and k is the decay constant then,
by substituting   N(t = 800) = 0.7071Ninto the formula
0.7071N0 = N0 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 radium-226 present at a time t is 
N(t) = N0 e - 0.0004332217 t
As we want determine the half-life or the time for half of a substance to decay, we substitute N(t) = (1/2) N0
into the formula
Intermediate algebra contents
Copyright 2004 - 2020, Nabla Ltd.  All rights reserved.