Chapter 16  Applications of the Exponential Function
Chapter 16  Applications of the Exponential Function
" The most powerful force in the universe is compound interest." Albert Einstein
In this
chapter we will take a look at just a few of the many interesting
situations
where the exponential function arises, i.e population growth and compound interest.
The emphasis on the exponential
function
was that its base was multiplication, or it can be thought of as a
repetitive multiplication
function. We
proved its derivative, but now we need to explain how to integrate it.
You
may
recall from Part I, that in integrating a function over any interval, we
just
added a constant to our integral. The derivative of the function with or
without an added constant was still the same. The important point to
realize
here is that if we were calculating the integral over a definite interval
from
a to b, then the constants are
useless
and take no part in the calculation. For example if the integral of f(x)
is
F(x) + C, then the integral from a to b is:
This tells us that the change in
a function over an interval is independent of the constants that are added
on.
The constant are there to define the initial condition, such that for any,
x,
F(x) is simply plus or minus a constant. This will become clearer later on
as
we look at examples.
If
constants are added on to polynomial functions, then they are multiplied
on to
exponential functions. What do we mean by this? When we are integrating
the
Logarithmic and exponential functions we can multiply constants that have
no
effect on the derivative, they are there only to satisfy initial
condition. For
example, the integral of 1/x is:
What is then the derivative of ln
of a constant times x?
Note that we have not attached
any limits to the integral, this is called the indefinite integral as it
simply
refers to any x over any interval. Suppose we were to integrate over a
definite
interval from a to b, then just as the constants in the polynomial
functions
canceled out, so will the constant in ln cx:
Therefore the constant plays no
role in the change of function over an interval and is only there to
define the
initial condition..
Now
let us move on to some applications, first a look at populations.
Population
growth is generally expressed as a percentage of the population per year.
Such
that a 3% growth rate means a population of 1000 people will see 30 new babies for the first year, 3% of
1030
for the next year and so on. No matter how large the population, there
will
always be 3% of the population wanting to have babies every year. We can
express this relationship as follows:
This can be written more
generally as:
The only function whose rate of
change or derivative is proportional to itself is the exponential function
or:
A more accurate way of solving
this, is separating the variables and integrating the change in population
from
some initial to some final population, and integrating time from 0 to
t.
We can now integrate both sides
over the limits just defined:
From the definition of the
Logarithm we get:
Note how the initial condition of
a given population at t=0 is multiplied on to the exponential function and
not
added on, such that at t=0 we have:
Consider
the population of Pakistan, 120,000,000 people. If the birth rate is 3%
per
year, then what will be the population in 10 years?
We
do not take the death rate into consideration is because the birth rate,
takes
into account both the birth and death rate such that it reflects the net
or
true growth in population per year. Our equation
becomes:
The population will therefore be
over 148,484,468 people which represents an increase of 38,484,468 people
in
ten years. Note that this answer was not
found by taking 3% of 110 million and multiplying by ten. This fails to
work
since it assumes that every year the population is 110 million, but this
is
clearly not the case, since the population is constantly growing and
growing and
3% of the population is a larger and larger number. That is why 38,484,468
>
33,000,000.
If
you are a skeptical person, like me, you will probably doubt how accurate
our
answer is. One argument might be that since the population is increasing
every
second and not only once a year, then 38,484,468 is not the exact increase
as
it is based on a yearly increase in population growth. In fact our answer
does
indeed take into consideration that babies are born every second and not
only
one a year. To understand this, go back to our proof of the derivative of
the exponential
function:
We are taking the limit as time
goes to zero and similarly the integral:
Is also based on ∆t going
to zero. Furthermore the growth rate is a constant value and will always
be 3%,
in this case, regardless of how much time has passed in years, months or
seconds. The fact remains that there will always be 3% of the population
desiring to have babies.
is based on time
going to zero, how can we be sure that when we solve for t =10, it is
referring
to ten years and not ten months or ten seconds later.
The
way we are sure that 10 refers to years and not anything else, is by
remembering that 3% growth rate is a per year figure.
So.
means that even though the growth
rate is always 3%, it doesn't take into consideration that it takes around
a
year for a baby to be born, so that 3% per year means that the there will
be
.03P new babies per year, not per month.
Another
fault with our equation is that it assumes that the 3 percent of the
entire
population at any given time t, has an equal desire to have a baby in one
year.
This is obviously untrue. Therefore our population equation is clearly
flawed
but the general idea is correct.
The
continent of South Asia has an average growth rate of 2.5% per year. If
the
current population is 1.1 billion, in how many years will the population
double
or be 2.2 billion. Our equation:
Taking the natural log of both
sides gives us:
In 27 years the Population of
South Asia will double.
In
this next example we will take a closer look at the exponential function's
role
as a solution to the equation:
. Hopefully all doubts will be cleared up now.
When
one deposits money in a savings account, it is generally for letting the
money
earn interest. Interest is just money paid for the use of money. Interest
rates
are quoted as per year or 6% per year or 9% per annum( per year). So a 6%
rate
would mean that $1000 would earn
. This is called simple interest since the money only earn
interest once a year.
Compound
interest on the other hand, is where the interest earns interest. So
$10,000
put into an account that earns 6% interest, compounded quarterly or every
three
months, undergoes a series of calculation. In each quarter we add the
interest
earned from the previous quarter to the amount of money in the bank to get
our
new balance for the next quarter. Therefore the interest earned in the
previous
quarters adds to the interest earned in the following
quarters
Quarter 1:
Quarter 2:
Quarter 3:
End of 1 Year:
The difference is not much since
the amount of interest earning interest is small. The balance in the bank
after
one period for money compounded n times a year can be found
by:
This last formula gives us the
new balance after the first time it was compounded. To find the amount for
the
second compounding period we multiply by:
Factoring out a
gives us:
The money in the bank after the 3.rd
and 4.th it has been compounded is given by
. By observing the pattern we can conclude that the balance
at the end of the year, when the money has been compounded n times
is:
If
$10,000 were put into a savings account that gives an interest rate of 6%.
compounded quarterly,
If this new balance were put into
the same account for another year we would have:
The formula then for calculating
the new balance, or how much money is in the bank after t years compounded
n
times is:
The exponent (nt) refers to
how
many times the money has been compounded. If in one year it is compounded
n
times, then in t years it is compounded nt times over the t
years.
If the same money were compounded infinite times this would mean that the
interest
is constantly earning interest. The interest earned over a
small change in time
is
added to the previous balance, creating a new balance on which the
interest
earned over the next small time interval is
based on. Therefore the rate at which the money is changing is
proportional to the
amount of money in the bank times the interest rate.
Let us examine this familiar
looking limit in the brackets,
. It looks similar to the definition for e, except for the
r, interest rate which is constant throughout the time interval. To
evaluate
this recall that the limit as
Δx
goes to zero of x/Δx
and
Δx/x
were infinity and
zero, respectively. This is true regardless of the value of x, since its
value
was fixed and did not change. Therefore by raising
to the
1/r power gives us:
Now as n tends to infinity, r/n
goes to zero, and n/r goes to infinity. This gives us the familiar answer,
e.
Raising both sides to the, r,
power we get:
This gives us the limit of what
we were initially looking for:
Or more generally for any x,
Returning back to our initial
problem of calculating the Principal in an account that compounds interest
infinite times per year:
This important answer could also
have been arrived at by realizing that money that is put into an account
that compounds
interest infinitely is changing at a rate proportional to the amount of
money
in the bank, times the interest rate. For example if at t=4, the amount of
money in the bank was $10000, and the interest rate was 6%, then at that
instant the money would be changing at the rate of $600 a year. As the
money
changes then so does the amount of money in the bank, such that our equation
becomes:
We can call Money, Principal or
just P, and separate variables:
We can integrate this equation to
solve for P in terms of r and t. We integrate time from 0 to t, and
Principal,
from P at t=0, to some final value of P.
This yields the solution:
This
last answer is the same answer we derived using the definition of compound
interest, which gives us the amount of money in a bank after t years and
infinite compounding with an initial deposit of P_{0}.
Differentiating
this equation with respect to time:
Despite how difficult it may seem
to explain this change the fact remains that the money is changing at a
rate proportional
to the amount of money itself, such that as the money begins to change
over a
small change in time, the amount of money in the bank changes by a small
amount
creating a new balance that is only accurate for the next
Δt . It is really this
simple.
In
this example we will compare simple interest to compound interest.
Consider a
poor college student who decides to put his summer earnings of $2000 into
a
savings account for 3 years. His choices are to City Bank that offers
a
simply interest rate of 3.2% per year or Carolina Bank that offer an
interest
rate of 2.9% compounded continuously. Where should he put his money? Also
note
that City Bank offers a free Tshirt for opening a savings account.
Lets
first consider City Bank. The rate at which his money will be changing is
the
rate of interest time the amount of money in the bank or (3.2/100)(2000) =
$64/year. Remember this is the definition of simple interest, the rate at
which
your money is changing is only dependent on how much Money you initially
open
the savings account with.
With Carolina Bank, the interest
rate is 2.9% compounded continuously. We can just go directly to our
equation
for continuous compounding:
This tells us that the interest
earning interest only contributed to an extra $9.52, barely more than the
cost
of the shirt. This is because the interest earned was small, and the
interest
it earns is even smaller. Now if we were to put the money on for 30 years,
then
the interest earned will have enough time to earn some substantial
interest.
In simple interest the amount of
money in the bank remains constant, while in compound interest the balance
is
increasing constantly.
Just
as exponential functions increase with rates proportional to the function
itself, they can also decrease. This happens when:
Whose solution is simply:
The evolution of living beings is also based on the priniciples of exponential growth. Most people can not understand
how small changes to populations per generation can lead to large differences in species over long periods of time.
This works exactly the same way money grows to a large amount when interest is compounding.
Sean B. Carroll writes in his book, The Making of the Fittest:
In the case of evolution, the amount of the initial investment is analagous to the number of
individuals in a population with a trait. And the interest rate is analogous to the small selected advantage that
trait confers on those individuals that carry it.
While the time required for one trait to become prevalant in a
population is greater than one lifetime, it is often no more than a few hundred generations. This is a blink of an eye
in geologic time. This tells us that small differences among individuals when compounded by natural selection over
time really do add up to the large differences we see among species.
This
concludes our study of the exponential and logarithmic function.
It is important to understand the differences and similarities
to
the polynomial function. The base of the exponential function is multiplication and
exponential growth can be thought of as repetitive multiplication. In
this
chapter we saw how this applies to multiplying populations and multiplying
money.
Questions

The power of compound interest lets you amass large wealth as long
as you start investing early. In the early years your investment will
grow linearly , but after some years you will see the investment grow
very fast. Make a bar graph in excel for investing $10,000 once a year
for 30 years. If you do not have excel, you can download either Open Office or Gnumeric
, both are free spreadsheet softwares. Assume the interest is compounded
continuously at a rate of 6%.
Write a formula to calculate the value of
the $10,000 at the end of the first year. Then in the next row add $10,000
to the new principal and use the same formula to find the value of the new principal
afer that year ends. The graph you make should illustrate the power of investing early
in life and reaping large rewards towards the end .

Exponential decay occurs whenever the rate at which the function is
decreasing
is proportional to the function itself.
Consider
the element Uranium that undergoes radioactive decay at the rate of 12% of
the
material per year. This decay rate, 12%, is independent of the amount of
material present. Find the time taken for half of Uranium to disappear, or
the
half life.
This half life is true for any amount
of material. A 1000 kg sample and a 1 kg sample will both take the same
time to
reduce to half their original mass.