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,
\(P(t)\text{,}\) of
Drosophila cells, in millions, as a function of time in days.
|
\(t\) (days) |
\(P(t)\) (millions) |
| \(0\) |
\(1\) |
| \(1\) |
\(2\) |
| \(2\) |
\(4\) |
| \(3\) |
\(8\) |
| \(4\) |
\(16\) |
| \(5\) |
\(32\) |
| \(6\) |
\(64\) |
Looking at the table, we see that we multiply the fruit fly population by 2 every day, so that after
\(t\) days, the initial population is multiplied by
\(2^t\text{.}\) Because the population grows by a factor of 2 each day, the function
\(P(t)\) describes exponential growth. We can express functions that describe exponential growth in a standard form.
Exponential Growth.
The function
\begin{gather*}
\blert{P(t)=P_0 b^t}
\end{gather*}
describes exponential growth, where \(P_0=P(0)\) is the initial value of the function and the positive constant \(b\) is the growth factor.
For the Drosophila cell population, the growth factor is \(b=2\text{,}\) and the initial value is \(P_0 = 1\) million cells, so we have
\begin{gather*}
P(t)=1 \cdot 2^t
\end{gather*}
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 \(C(t)\) be the number of cell phone users \(t\) years after 1985 according to this model. Make a table of values and graph \(C(t)\text{.}\)
-
Write a formula for \(C(t).\)
-
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
\(t=0\) in 1985, so
\(C(0)=1.2\text{,}\) in millions. Each value of
\(C(t)\) can be obtained by multiplying the previous value by the growth factor, 1.5.
| \(\hphantom{0}t\hphantom{0}\) |
\(C(t)\) |
\(\) |
| \(0\) |
\(1.2\) |
\(\) |
| \(1\) |
\(1.8\) |
\(\blert{1.2\times 1.5 = 1.8}\) |
| \(2\) |
\(2.7\) |
\(\blert{1.8\times 1.5 = 2.7}\) |
| \(3\) |
\(4.05\) |
\(\blert{2.7\times 1.5 = 4.05}\) |
| \(4\) |
\(6.08\) |
\(\blert{4.05\times 1.5 = 6.08}\) |
| \(5\) |
\(9.1\) |
\(\blert{6.08\times 1.5 = 9.1}\) |
| \(6\) |
\(13.7\) |
\(\blert{9.1\times 1.5 = 13.7}\) |
| \(7\) |
\(20.5\) |
\(\blert{13.7\times 1.5 = 20.5}\) |
-
The initial value of the function is
\(C_0 = C(0) = 1.2\) million. The annual growth factor is
\(b=1.5\text{,}\) so the formula is
\begin{equation*}
C(t) = 1.2(1.5)^t
\end{equation*}
-
The year 2000 is 15 years after 1985, so we evaluate the function for
\(t = \alert{15}\text{.}\)
\begin{equation*}
C(\alert{15}) = 1.2(1.5)^{\alert{15}} = 525.47
\end{equation*}
The formula predicts that over 525 million people used cell phones in 2000.
-
It is unlikely that the model will be valid indefinitely, because \(C(t)\) 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
\(P(t) = 800(1.06)^t\text{,}\) where
\(t\) 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 \(800(1.06)^5\text{:}\)
-
-
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
\(P(t)\) after
\(t\) months, and graph the function.
-
How is the exponent on the base 2 related to
\(t\text{?}\) Write a formula for the function
\(P(t)\text{.}\)
-
How many rabbits will there be after 1 year?
| \(\hphantom{0}t\hphantom{0}\) |
\(~P(t)~\) |
\(\) |
| \(0\) |
\(\hphantom{000}\) |
\(\blert{P_0 = 20}\) |
| \(3\) |
\(\hphantom{000}\) |
\(\blert{20 \cdot 2}\) |
| \(6\) |
\(\hphantom{000}\) |
\(\blert{20 \cdot 2^2}\) |
| \(9\) |
\(\hphantom{000}\) |
\(\blert{20 \cdot 2^3}\) |
| \(12\) |
\(\hphantom{000}\) |
\(\blert{20 \cdot 2^4}\) |
| \(15\) |
\(\hphantom{000}\) |
\(\blert{20 \cdot 2^5}\) |
Answer.
-
| \(\hphantom{0}t\hphantom{0}\) |
\(P(t)\) |
| \(0\) |
\(20\) |
| \(3\) |
\(40\) |
| \(6\) |
\(80\) |
| \(9\) |
\(160\) |
| \(12\) |
\(320\) |
| \(15\) |
\(640\) |
-
The exponent is the value of
\(t\) divided by 3.
\(~P(t)=20(2)^{t/3}\)
-
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:
\begin{equation*}
(a^m)^n = a^{mn}
\end{equation*}
For example,
\begin{equation*}
(x^3)^2 = x^{2\cdot 3} = x^6
\end{equation*}
Using the same idea for the rabbit population, we can see that
\begin{equation*}
(2^{1/3})^t = 2^{t(1/3)} = 2^{t/3}
\end{equation*}
So the growth factor for the rabbit population is \(2^{1/3}\text{,}\) or about 1.26. The rabbit population grows by a factor of 1.26 every month.
Subsection 7.1.3 Comparing Linear Growth and Exponential Growth
It may be helpful to compare linear growth and exponential growth. Consider the two functions
\begin{equation*}
L(t) = 5 + 2t ~~~~~\text{ and } ~~~~~ E(t) = 5 \cdot 2^t ~~~ (t \ge 0)
\end{equation*}
whose graphs are shown below.
| \(\hphantom{0}t\hphantom{0}\) |
\(L(t)\) |
| \(0\) |
\(5\) |
| \(1\) |
\(7\) |
| \(2\) |
\(9\) |
| \(3\) |
\(11\) |
| \(4\) |
\(13\) |
| \(\hphantom{0}t\hphantom{0}\) |
\(E(t)\) |
| \(0\) |
\(5\) |
| \(1\) |
\(10\) |
| \(2\) |
\(20\) |
| \(3\) |
\(40\) |
| \(4\) |
\(80\) |
\(L\) is a linear function with
\(y\)-intercept 5 and slope 2;
\(E\) is an exponential function with initial value 5 and growth factor 2. In a way, the growth factor of an exponential function is analogous to the slope of a linear function: Each measures how quickly the function is increasing.
However, for each unit increase in
\(t\text{,}\) 2 units are
added to the value of
\(L(t)\text{,}\) whereas the value of
\(E(t)\) is
multiplied by 2. An exponential function with growth factor 2 eventually grows much more rapidly than a linear function with slope 2, as you can see by comparing the graphs or the function values in the tables.
Example 7.1.6.
A solar energy company sold $80,000 worth of solar collectors last year, its first year of operation. This year its sales rose to $88,000. The marketing department must estimate its projected sales for the next 3 years.
-
If the marketing department predicts that sales will grow linearly, what sales total should it expect next year? Graph the projected sales figures over the next 3 years, assuming that sales will grow linearly.
-
If the marketing department predicts that sales will grow exponentially, what sales total should it expect next year? Graph the projected sales figures over the next 3 years, assuming that sales will grow exponentially.
Solution.
-
Let \(L(t)\) represent the company’s total sales \(t\) years after starting business, where \(t = 0\) is the first year of operation. If sales grow linearly, then \(L(t)\) has the form \(L(t) = mt + b\text{.}\) Because \(L(0) = 80,000\text{,}\) the intercept \(b\) is 80,000. The slope \(m\) of the graph is
\begin{equation*}
\dfrac{\Delta S}{\Delta t}= \dfrac{8000 \text{ dollars}}{1\text{ year}}= 8000 \text{ dollars/year}
\end{equation*}
where \(\Delta S = 8000\) is the increase in sales during the first year. Thus, \(L(t) = 8000t + 80,000\text{,}\) and sales grow by adding $8000 each year. The expected sales total for the next year is
\begin{equation*}
L(2) = 8000(2) + 80,000 = 96,000
\end{equation*}
-
Let \(E(t)\) represent the company’s sales assuming that sales will grow exponentially. Then \(E(t)\) has the form \(E(t) = E_0 b^t\text{,}\) and the initial value is \(E_0 = 80,000\text{.}\) We find the growth factor in sales over the first year by dividing \(E(1)\) by \(E_0\text{:}\)
\begin{align*}
E(1) \amp = E_0 b^1\\
\text{so}~~b \amp = \dfrac{E(1)}{E_0} = \dfrac{88,000}{80,000} = 1.1
\end{align*}
Thus, \(E(t) = 80,000(1.1)^t\text{,}\) and the expected sales total for the next year is
\begin{equation*}
E(2) = 80,000(1.10)^2= 96,800
\end{equation*}
We evaluate each function at several points to obtain the graphs shown in the figure.
| \(\hphantom{0}t\hphantom{0}\) |
\(L(t)\) |
\(E(t)\) |
| \(0\) |
\(80,000\) |
\(80,000\) |
| \(1\) |
\(88,000\) |
\(88,000\) |
| \(2\) |
\(96,000\) |
\(96,800\) |
| \(3\) |
\(104,000\) |
\(106,480\) |
| \(4\) |
\(112,000\) |
\(117,128\) |
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, \(P(t)\text{.}\) Thus
\begin{align*}
\text{In 2000, } P(0) \amp = 4000\\
\text{In 2001, } P(1) \amp = 4000 + \blert{0.05(4000)} = 4200 \amp\amp \blert{\text{Add 5}\% \text{ of } P(0).}\\
\text{In 2002, } P(2) \amp = 4200 + \blert{0.05(4200)} = 4410 \amp\amp \blert{\text{Add 5}\% \text{ of } P(1).}
\end{align*}
and so on. Now here is the important observation about percent increase:
\begin{equation*}
\bf{\text{Adding 5% of the old population is the same as multiplying the old population by 1.05.}}
\end{equation*}
\begin{align*}
4000 + 0.05(4000) \amp = 4000(1+0.05) = 4000(1.05) \amp \amp \blert{\text{Factor out 4000.}}\\
4200 + 0.05(4200) \amp = 4200(1+0.05) = 4200(1.05) \amp \amp \blert{\text{Factor out 4000.}}
\end{align*}
Thus, we can find the current population by multiplying the old population by 1.05. In other words,
\begin{equation*}
\bf{\text{Growing by 5% is the same as growing by a factor of 1.05.}}
\end{equation*}
A formula for the population of Lakeview \(t\) years after 2000 is
\begin{equation*}
P(t) = 4000(1.05)^t
\end{equation*}
This formula describes exponential growth with a growth factor of \(b=1.05\text{.}\) In general, a function that grows at a percent rate \(r\text{,}\) where \(r\) is expressed as a decimal, has a growth factor of \(b=1+r\text{.}\)
Growth by a Constant Percent.
The function
\begin{gather*}
\blert{P(t) =P_0 (1+r)^t}
\end{gather*}
describes exponential growth at a constant percent rate of growth, \(r\text{.}\)
The
initial value of the function is
\(P_0 = P(0)\text{,}\) and
\(b=1+r\) 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 \(t\) years after inflation began.
-
How much did a loaf of bread cost after 6 years? After 30 months? \(~\alert{\text{[TK]}}\)
-
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
\(1+0.12 = 1.12\) every year. If
\(P(t)\) represents the price of bread after
\(t\) years, then
\(P(0)=2\text{,}\) and we multiply the price by 1.12 every year, as shown in the table.
| \(\hphantom{0}t\hphantom{0}\) |
\(\hphantom{0}P(t)\hphantom{0}\) |
\(\) |
\(\) |
| \(0\) |
\(2.00\) |
|
\(\blert{P(0)=2}\) |
| \(1\) |
\(2.24\) |
|
\(\blert{P(1)=2(1.12)}\) |
| \(2\) |
\(2.51\) |
|
\(\blert{P(2)=2(1.12)^2}\) |
| \(3\) |
\(2.81\) |
|
\(\blert{P(3)=2(1.12)^3}\) |
| \(4\) |
\(3.15\) |
|
\(\blert{P(4)=2(1.12)^4}\) |
-
After
\(t\) years of inflation the original price of $2 has been multiplied
\(t\) times by a factor of 1.12. Thus,
\begin{equation*}
P(t)=2(1.12)^t
\end{equation*}
-
To find the price of bread at any time after inflation began, we evaluate the function at the appropriate value of
\(t\text{.}\)
\begin{equation*}
P(\alert{6}) = 2(1.12)^{\alert{6}} \approx 3.95
\end{equation*}
After 6 years the price was $3.95. Thirty months is 2.5 years, so we evaluate
\(P(2.5)\text{.}\)
\begin{equation*}
P(\alert{2.5}) = 2(1.12)^{\alert{2.5}} \approx 2.66
\end{equation*}
After 30 months the price was $2.66.
-
To graph the function
\begin{gather*}
P(t) = 2(1.12)^t
\end{gather*}
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.
Fill in the blanks.
-
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
\(P(t)\text{,}\) the population of Tombstone
\(t\) months after its founding.
-
Complete the table and sketch a graph of
\(P(t)\text{.}\)
| \(t\) |
\(0\) |
\(5\) |
\(10\) |
\(15\) |
\(20\) |
\(25\) |
| \(P(t)\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
\(\hphantom{000}\) |
-
Tombstone’s peak population was about 10,000 people. Use your graph to estimate the time it took to reach that figure.
Answer.
-
806
-
\(\displaystyle P(t)=100(1.19)^t\)
-
| \(t\) |
\(0\) |
\(5\) |
\(10\) |
\(15\) |
\(20\) |
\(25\) |
| \(P(t)\) |
100 |
239 |
569 |
1359 |
3243 |
7739 |
-
About 26 months
Compound interest is another example of exponential growth. Suppose you deposit a sum of money,
\(P\text{,}\) 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,
\(B\text{,}\) grows by a factor of 1.05.
You can see this more clearly with a little calculation. At interest rate \(r\text{,}\) after 1 year your balance is
\begin{equation*}
B = \text{Principal} + \text{Interest} = P + Pr = P(1+r)
\end{equation*}
So the balance grew by a factor of \((1+r)\text{.}\) After the second year, the new balance of \(B = P(1+r)\) grows by another factor of \((1+r)\text{,}\) giving
\begin{equation*}
B = P(1+r)(1+r) = P(1+r)^2
\end{equation*}
In general, we have the following formula.
Compound Interest.
If a principal of \(P\) dollars is invested in an account that pays an interest rate \(r\) compounded annually, the balance \(B\) after \(t\) years is given by
\begin{equation*}
\blert{B=P(1+r)^t}
\end{equation*}
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.
| \(t\) |
\(P(1+r)^t\) |
\(B(t)\) |
| \(0\) |
\(100\) |
\(100\) |
| \(1\) |
\(100(1.05)\) |
\(105\) |
| \(2\) |
\(100(1.05)^2\) |
\(1110.25\) |
| \(3\) |
\(100(1.05)^3\) |
\(115.76\) |
