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Many real world phenomena can be modeled by functions that describe how things grow or decay as time passes. Examples of such phenomena include population growth or decline, bacteria, viruses, radioactive substances, investments and mortgages.

Under ideal conditions a bacterial culture will grow according to the following law.
                y = y₀e^kt   where

• y is the size of the population at time t seconds
• y₀ is the initial size of the population (at time t = 0)
• k is a positive constant related to the growth rate of the population

  Note:  If k > 0,  we have exponential growth and if k < 0 we have exponential decay.

Example 1:  A bacteria culture starts with 200 bacteria.  After 4 hours the population is 700 bacteria.

• a)     Find the growth constant k.
• b)     Find the number of bacteria after t hours.
• c)      Find the number of bacteria after 2 hours.
• d)     When will the bacteria culture reach a population of 2800?

Solution
a)     The initial population is  y₀ = 200 with the result that:
y = y₀e^kt  
y = 200e^kt


  We also know that y = 700 when t = 4, yielding

            700 = 200e^4k              **  Dividing both sides by 200
               e^4k=3.5                    **  Taking lns of both sides
        ln(e^4k) = ln(3.5)
               4k = ln3.5
                 k = ln3.5/4 = 0.3132

                  b)  The number of bacteria after “t” hours is obtained by substituting k = 0.3132 into the formula:
y = 200e^kt
                        y = 200e^0.3132t

  c)  Let  t = 2  to find the number of bacteria after 2 hours.
y = 200e ^0.3132(2)
    = 374 bacteria

d)     Let  y = 2800 in the above formula to find when the bacteria count will reach 2800.
2800 = 200e^0.3132t                        **  Divide both sides by 200
                        14 = e^0.3132t                **  Take lns of both sides
                  ln(14) = ln(e^0.3132t)
                   ln(14) = 0.3132t              ** Divide both sides by 0.3132  
                                  t = ln14/0.3132
                                    = 8.4
Therefore the bacteria count will reach 2800 after 8.4 hours.

The half-life of a radioactive substance is the period of time a given amount will decay to half of it’s original amount.

The rate of decay of a radioactive substance is also given by the above equation  y = y₀ e ^kt  where  k < 0  and 

• y  is the mass of the sample at any time t
• y₀  is the initial mass of the sample  (at time  t = 0)
• k  is a negative constant related to the decay rate of the sample

  Example 2:  Iridium – 192 is a substance used in breast cancer therapy.  The half life of iridium – 192 is 74 days.  If a hospital has a sample of 800 mg.

• a)      Determine the decay constant constant  k.
• b)      Hoe much will remain after 1 year?
• c)      How long will it take to decay to 10 mg?

Solution:
a)     The initial mass is  y₀ = 800 with the result that:
y = y₀ e ^kt  
y = 800 e ^kt
 
We also know that after 74 days half of the original sample of 800 mg will remain.
Hence  y = 400 when t = 74, yielding
            400 = 800 e ^74k                      **  Divide both sides by 800
               e^74k = 0.5                             **  Take lns of both sides
        ln(e^74k) = ln(0.5)
               74k = ln0.5
                 k = ln0.5/74 = - 0. 0093669                  
        The number of bacteria after “t” hours is obtained by substituting k = - 0. 0093669 into the formula
y = 800e^kt     
                    y = 800e^-0.0093669t

  b)  Let  t = 365  to find the mass after 1 year or 365 days.
            y = 200e ^0.3132(365)
               = 26.2 mg
         Hence there are 26.2 mg left after I year.

           c)  Let  y = 10 in the above formula to find when the mass will decay to 10 mg.
                          10 = 800e^-0.0093669t                           **  Divide both sides by 800
                 0.0125 = e^{-0.0093669t t}                                 **  Take lns of both sides
           ln(0.0125) = ln(e^-0.0093669t)
           ln(0.0125) = - 0.0093669t                              ** Divide both sides by 0.3132  
                            t = ln0.0125/(- 0.0093669)
                              = 468
Therefore the original mass of 800 mg will decay to 10 mg  after 468 days.

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