Subsection 7.1.2 Growth Factors
Researchers often use cell lines from the fruit fly
Drosophila melanogaster to study protein interactions related to cancer and other diseases. From 60% to 70% of human disease genes are found in
Drosophila cells, and gene discoveries in the flies have led to parallel studies in vertebrates.
One milliliter of culture contains about 1 million
Drosophila cells, and the population doubles every 24 hours. The table shows the population,
of
Drosophila cells, in millions, as a function of time in days.
Looking at the table, we see that we multiply the fruit fly population by 2 every day, so that after
days, the initial population is multiplied by
Because the population grows by a factor of 2 each day, the function
describes exponential growth. We can express functions that describe exponential growth in a standard form.
Exponential Growth.
describes
exponential growth, where
is the
initial value of the function and the positive constant
is the
growth factor.
For the
Drosophila cell population, the growth factor is
and the initial value is
million cells, so we have
There is a sort of similarity between the formula for exponential functions and the formula for linear functions. Each has an initial value and a constant that describes change. But compare the graph of exponential growth, described by a constant growth factor, with linear growth. You can see that the graph of the fruit fly population is not a straight line with a constant slope, as a linear function would be.
Example 7.1.1.
In 1985, there were about 1.2 million cell phone users world-wide. For some years after that time, the number grew by a factor of 1.5 each year.
Let be the number of cell phone users years after 1985 according to this model. Make a table of values and graph
Write a formula for
How many cell phone users does this model predict for the year 2000?
Do you think this model will be valid indefinitely? Why or why not?
Solution.
-
We let in 1985, so in millions. Each value of can be obtained by multiplying the previous value by the growth factor, 1.5.
The initial value of the function is
million. The annual growth factor is
so the formula is
The year 2000 is 15 years after 1985, so we evaluate the function for
The formula predicts that over 525 million people used cell phones in 2000.
It is unlikely that the model will be valid indefinitely, because will eventually exceed the population of Earth.
Once again, in the examples above, you can see that the graphs of these exponential functions are not linear. In each case, the function grows slowly at first, but eventually grows faster and faster.
Checkpoint 7.1.3. QuickCheck 1.
A population grows according to the formula
where
is in years.
What was the starting value of the population?
What was the population one year later?
What does 1.06 tell you about the population?
Choose the correct first step to evaluate
Multiply 800 by 1.06 or
Raise 1.06 to the 5th power
In the next Practice, we consider a population that doubles not every month, but every three months.
Checkpoint 7.1.4. Practice 1.
A colony of rabbits started with 20 rabbits and doubles every 3 months.
Complete the table for the number of rabbits after months, and graph the function.
How is the exponent on the base 2 related to Write a formula for the function
How many rabbits will there be after 1 year?
Answer.
The exponent is the value of divided by 3.
320 rabbits
Now suppose we would like to know the
monthly growth factor for the rabbit population, that is, by what factor did the population grow every month?
Recall that when we raise a power to a power we can multiply the exponents, like this:
Using the same idea for the rabbit population, we can see that
So the growth factor for the rabbit population is
or about 1.26. The rabbit population grows by a factor of 1.26 every month.
Subsection 7.1.5 Percent Increase
Exponential growth is often described as growth by a certain percent increase. Suppose the town of Lakeview had 4000 residents in the year 2000, and grew at a rate of 5% per year. This means that each year we add 5% of last year’s population to find the current population,
Thus
and so on. Now here is the important observation about percent increase:
Thus, we can find the current population by multiplying the old population by 1.05. In other words,
A formula for the population of Lakeview
years after 2000 is
This formula describes exponential growth with a growth factor of
In general, a function that grows at a percent rate
where
is expressed as a decimal, has a growth factor of
Growth by a Constant Percent.
describes exponential growth at a constant percent rate of growth,
The
initial value of the function is
and
is the
growth factor.
Many quantities besides population can grow by a fixed percent. For example, an inflation rate gives the percent rate at which prices are rising.
Example 7.1.11.
During a period of rapid inflation, prices rose by 12% each year. At the beginning of this time, a loaf of bread cost $2.
Make a table showing the cost of bread over the next four years.
Write a function that gives the price of a loaf of bread years after inflation began.
How much did a loaf of bread cost after 6 years? After 30 months?
Graph the function found in (b).
Solution.
-
The percent increase in the cost of bread is 12% every year. Therefore, the growth factor for the cost of bread is every year. If represents the price of bread after years, then and we multiply the price by 1.12 every year, as shown in the table.
After
years of inflation the original price of $2 has been multiplied
times by a factor of 1.12. Thus,
To find the price of bread at any time after inflation began, we evaluate the function at the appropriate value of
After 6 years the price was $3.95. Thirty months is 2.5 years, so we evaluate
After 30 months the price was $2.66.
we evaluate it for several values, as shown in the table. We plot the points and connect them with a smooth curve to obtain the graph shown.
Checkpoint 7.1.12. QuickCheck 3.
Increasing by 10% is the same as multiplying by .
If a population grows by 2% annually, its growth factor is .
If a population grows by 46% annually, its growth factor is .
If a population grows by 100% annually, its growth factor is .
Checkpoint 7.1.13. Practice 3.
Tombstone, Arizona was the most famous "boomtown" during the gold rush in the American west. It was established in December, 1879, after the discovery of a large silver deposit nearby. The original town had 40 dwellings and a population of 100. Over the next two to three years, the population grew at an average rate of 19% per month.
What was the population one year later, in December, 1880?
Write a formula for the population of Tombstone months after its founding.
-
Complete the table and sketch a graph of
Tombstone’s peak population was about 10,000 people. Use your graph to estimate the time it took to reach that figure.
Answer.
806
-
|
|
|
|
|
|
|
|
100 |
239 |
569 |
1359 |
3243 |
7739 |
About 26 months
Compound interest is another example of exponential growth. Suppose you deposit a sum of money,
into an account that pays 5% interest compounded annually. "Compounded" means that each year your interest, 5% of your current balance, is added to your account, so your balance,
grows by a factor of 1.05.
You can see this more clearly with a little calculation. At interest rate
after 1 year your balance is
So the balance grew by a factor of
After the second year, the new balance of
grows by another factor of
giving
In general, we have the following formula.
Compound Interest.
If a
principal of
dollars is invested in an account that pays an interest rate
compounded annually, the
balance after
years is given by
As an example, suppose you put $100 in an account that pays 5% interest compounded annually. Using the formula above, we can make a table showing the balance in your account over the next few years.