```Curve Fitting--3

S. Arlinghaus

APPLICATIONS OF EXPONENTIAL FUNCTIONS

1.  Exponential growth and decay

Basic definition

Log_a y = x if and only if a^x=y

Assumption:
There are a number of natural quantities, ranging from population to
amount of
radioactivity, whose rate of growth or decay at any time is viewed to be
proportional
to the amount of that quantity present.

Laws of exponential growth and decay follow from this assumption

Suppose N is the size of a population at time t, and that N_0 is the
starter set:

N = N_0 e^(kt)

represents exponential growth when k>0 and exponential decay when k<0.

To see that in fact N_0 is the starter set, consider the equation when
t=0--at the start.  Then, N=N_0.

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The assumption above, when stated more formally, leads directly to the
notational representation
of the growth/decay formula.  Consider the assumption restated as:
the rate of growth is equal to a constant of proportionality times
the amount of the quantity at time t.
Then, this latter statement is written symbolically as:
dN / dt = kN, where k is the constant of proportionality.

Separate the variables of this differential equation:

1/N dN = k dt

Integrate:

INT 1/N dN = INT k dt

ln |N| = kt + C_1
here N > 0
so ln N = kt + C_1

Therefore,
N = e^(kt + C_1) = e^(kt)e^(C_1)=e^(C_1)e^(kt)

Let N_0 = e^(C_1)

Thus, N = N_0 e^(kt)              ----  exponential growth is unbounded as
t approaches infinity.
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Example of use of this equation:

Suppose in a city that the rate at which the population grows at any
time is proportional to the
size of the population.
If the population of a city was 125, 000 in 1950, and 140,000 in
1970, what is expected in
1990 and in 2000 , assuming use of growth equation?

Assume use of N = N_0 e^(kt)

Find values for the constants N_0 and k.

1950:  t=0, the starting time.  At the starting time, N_0 = 125,000.
Thus, N = 125,000e^(kt)

In 1970, t=20.  At this point, N=140,000.
So, 140,000=125,000e^(20k)
Solve for k:
e^(20k) = 140,000 / 125,000 = 1.12
Therefore,
20 k = ln (1.12)
Thus,
k = 1/20 * ln (1.12) = 0.0056664, a growth rate of 0.567 % per year.

Now, obtain the general equation--we have found N_0 and k:

N = 125,000 e^((t/20)ln(1.12))
= 125,000 [e^(ln(1.12))]^(t/20)
= 125,000(1.12)^(t/20)

So, this general equation may now easily be used to forecast, based on
previous values:
In 1990, t = 40, so the equation yields:
N=125,000(1.12)^(40/20) = 156,800

In 2000, t = 50, so the equation yields:
N=125,000(1.12)^(50/20) = 165,942.

2.  Population Doubling Time

Use the growth equation to understand population doubling time.
In the context of the example above, when will the 1950 population
double??
Symbolically:
2N_0 = N_0 e^(kt)  when will the total pop. on the left be twice the
starter set (N=2N_0)?

Solve for t:
2N_0 = N_0e^(Kt)
2=e^(kt)
ln 2 = ln e^(kt)
ln 2 = kt
therefore,
t=(ln 2)/k = 0.6931472/k
Multiply numerator and denominator by 100--the denominator is % growth per
year.
The numerator is about 70.  Sometimes this rule is referred to as the rule
of 72.
To use the rule, divide the %growth rate into 70 to get the doubling time.
Consider the unique prime factorization of each of 70 and 72:

70=10*7=2*5*7
72=8*9=2*2*2*3*3

Because 72 has many  more  integral divisors than does 70,
it is easier to work with in doing arithmetic in one's head.

This is the rule of 70 cited in Meadows.

Example:  Suppose a population is growing at a rate of 10% per year.  When
will it double?

It will double after 70/10=7 years.

What is the corresponding rule for population tripling?

Solve for t:
3N_0 = N_0e^(kt)
3 = e^(kt)
ln 3 = kt
t = (ln 3)/k = 1.0986123/k.
the corresponding rule is a rule of 110.

Thus, for example, a 10% per year rate of population growth

doubles the population in 70/10 = 7 years
triples the population in 110/10  = 11 years.

What is the corresponding rule for an n-fold population increase???