Math 141, Sem. I, 1998-99 (R.A. Brualdi)
Handout 15 (October 28, 1998)
Topic: Exponential decay


Bacteria population, human population, bank accounts, often
*grow* at a rate proportional to the amount present. If P is
the initial amount and r (expressed as a decimal) is the rate
of growth per time unit, then the population, amount, ... 
present after t time units is: 

   P (1 + r)^t.

Radioactive materials, cropland (often), ... *decay* at a rate
proportional to the amount present If P is the initial amount
and r (expressed as a decimal) is the rate of decay per time
unit, then the amount, ... present after t time units is:

   P (1 - r)^t.

The rate of decay per year of radioactive materials is often
very small. So the convention is to change the time unit in
such a way that the rate is 50% (i.e. 0.50). The time unit
thus varies and depends on the particular  radioactive
material. This time unit for such a material is called its
*half-life*:

               the time it takes for 50% (i.e. half) of the
               material to be dissipate.

Using the time unit of half-life, the above formula becomes

  P(1/2)^T

where T is the time expressed in terms of half-lives.

Example: A radioactive substance decays 1% per year. Thus if
one starts with an amount P then after t years

        P(1-0.01)^t = P(.99)^t

remains. To say that half remains after time t means that


         P(.99)^t = 1/2P = 0.50P,

that is, 

         (.99)^t = 0.50.

Our usual "trial and error" method shows that t is about 69
years, that is, the half-life is 69 years. Thus in 138 years,
i.e. two half-lives, the amount remaining is

        P(1/2)^2= P(1/4).

The amount remaining after 10 half-lives (690 years) is

        P(1/2)^10 = P(1/1024).


Example: Radium 226 has a half-life of 1620 years. So if we
start with 6 grams, then after 1620 years, 3 grams remain,
after 3240 years, 1.5 grams remain, after 4860 years, 0.75
grams remain. What is the rate of decrease per year? This is
the number r such that

    (1-r)^1620 = 0.50.

    (1-r)= 1620th root of 0.50

    (1-r)= .99957, r = .00043 (.043% each year)


Exercises

1. Page 80-1: 2, 4

2. Page 85-6: 3, 4, 7

3. Problem. (a) If Marie Curie had put 5 grams of ordinary
radium in a safe deposit box in 1898, how much would be left
today, 100 years later? 
             (b) In what year would the amount remaining be
1/100 of a gram?
             (c) How much would Marie Curie have had to put in
the safe deposit box in order that there be 3 grams remaining
in the year 2000 (102 years later).

Hand in Friday, October 30, page 85, number 4 and Exercise 3
above.