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.
| \(t\) |
\(P(t)\) |
| \(0\) |
\(1\) |
| \(1\) |
\(2\) |
| \(2\) |
\(4\) |
| \(3\) |
\(8\) |
| \(4\) |
\(16\) |
| \(5\) |
\(32\) |
| \(6\) |
\(64\) |
Because the fruit fly population grows by a factor of 2 every day, the function \(P(t)\) describes exponential growth. Functions that describe exponential growth can be expressed 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 \(b\) is the growth factor.
For the Drosophila cell population, we have
\begin{gather*}
P(t)=1 \cdot 2^t
\end{gather*}
so \(P_0 = 1\) and \(b=2\text{.}\) You can see that the graph of the function is not linear. In fact, the population grows slowly at first, but eventually grows faster and faster.
Example 7.1.1.
In 1985, there were about 1.2 million cell phone users world-wide. Since that time, the number has grown by a factor of 1.5 each year.
Make a table of values and graph the function.
Write a formula for the number, \(C(t)\text{,}\) of cell phone users \(t\) years after 1985.
How many cell phone users does the formula predict for the year 2000?
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.
| \(t\) |
\(C(t)\) |
\(\) |
| \(0\) |
\(1.2\) |
\(\) |
| \(1\) |
\(1.8\) |
\(\blert{1.2\times 1.5 = 1.8}\) |
| \(2\) |
\(2.7\) |
\(\blert{1.8\times 1.5 = 2.7}\) |
| \(3\) |
\(4.05\) |
\(\blert{2.7\times 1.5 = 4.05}\) |
| \(4\) |
\(6.08\) |
\(\blert{4.05\times 1.5 = 6.08}\) |
| \(5\) |
\(9.1\) |
\(\blert{6.08\times 1.5 = 9.1}\) |
| \(6\) |
\(13.7\) |
\(\blert{9.1\times 1.5 = 13.7}\) |
| \(7\) |
\(20.5\) |
\(\blert{13.7\times 1.5 = 20.5}\) |
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.
In the examples above, you can see that the graphs of the functions are not linear. In each case, the function grows slowly at first, but eventually grows faster and faster.
Caution 7.1.2.
Be careful when evaluating exponential growth functions. In part (c) of the previous
Example, note that
\begin{gather*}
1.2\blert{(1.5)^{15}} = 1.2 \times \blert{437.89} = 525.47~~~~~~~\blert{\text{Compute the power first.}}
\end{gather*}
According to the order of operations, we compute the power \(1.5^{15}\) first, and then multiply the result by 1.2.
Checkpoint 7.1.3. QuickCheck 1.
Checkpoint 7.1.4. Practice 1.
Note 7.1.5.
In Practice 1, the rabbit population doubled every 3 months, leading to the growth law
\begin{equation*}
P(t)=20(2)^{t/3}
\end{equation*}
To see the growth factor for the rabbit population, we use the third law of exponents to rewrite \(2^{t/3}\text{.}\)
\begin{equation*}
2^{t/3} = 2^{t(1/3)} = (2^{1/3})^t
\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.
| \(t\) |
\(L(t)\) |
| \(0\) |
\(5\) |
| \(1\) |
\(7\) |
| \(2\) |
\(9\) |
| \(3\) |
\(11\) |
| \(4\) |
\(13\) |
Slope \(m=2\)
| \(t\) |
\(E(t)\) |
| \(0\) |
\(5\) |
| \(1\) |
\(10\) |
| \(2\) |
\(20\) |
| \(3\) |
\(40\) |
| \(4\) |
\(80\) |
Growth factor \(b=2\)
\(L\) 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.
| \(t\) |
\(L(t)\) |
\(E(t)\) |
| \(0\) |
\(80,000\) |
\(80,000\) |
| \(1\) |
\(88,000\) |
\(88,000\) |
| \(2\) |
\(96,000\) |
\(96,800\) |
| \(3\) |
\(104,000\) |
\(106,480\) |
| \(4\) |
\(112,000\) |
\(117,128\) |
Subsection 7.1.4 Exponential Decay
In the examples above, exponential growth was modeled by increasing functions of the form
\begin{equation*}
P(t) = P_0 b^t
\end{equation*}
where the growth factor, \(b\text{,}\) is a number greater than 1. If we multiply the function value by a number smaller than 1, the function values will decrease. Thus, if \(0 \lt b \lt 1\text{,}\) then \(P(t) = P_0b^t\) is a decreasing function. In this case, we say that the function describes exponential decay, and the constant \(b\) is called the decay factor.
Investigation 7.1.2. Exponential Decay.
A small coal-mining town has been losing population since 1940, when 5000 people lived there. At each census thereafter (taken at 10-year intervals), the population declined to approximately 0.90 of its earlier figure.
| \(t\) |
\(P(t)\) |
|
|
| \(0\) |
\(5000\) |
|
\(\blert{P(0)=5000}\) |
| \(10\) |
\(\) |
|
\(\blert{P(10)=5000\cdot 0.90=}\) |
| \(20\) |
\(\) |
|
\(\blert{P(20)=[5000\cdot 0.90]\cdot 0.90=}\) |
| \(30\) |
\(\) |
|
\(\blert{P(3)=}\) |
| \(40\) |
\(\) |
|
\(\blert{P(4)=}\) |
| \(50\) |
\(\) |
|
\(\blert{P(5)=}\) |
Fill in the table showing the population \(P(t)\) of the town \(t\) years after 1940.
Plot the data points and connect them with a smooth curve.
-
Write a function that gives the population of the town at any time \(t\) in years after 1940.
Hint: Express the values you calculated in part (1) using powers of 0.90. Do you see a connection between the value of \(t\) and the exponent on 0.90?
Graph your function from part (3) using a calculator. (Use the table to choose an appropriate window.) The graph should resemble your hand-drawn graph from part (2).
Evaluate your function to find the population of the town in 1995. What was the population in 2000?
Example 7.1.7.
Before the introduction of disposable containers, soft drinks and draught beer were sold in refillable glass botles. During the second half of the last century, the percent of beer volume sold in refillable glass bottles declined by a factor of 0.942 each year.
In 1944, 98% of beer was sold in refillable bottles. Write a formula for the percent of beer sold in refillable bottles as a function of \(t\text{,}\) the number of years after 1944.
Graph the function from 1944 to 2000.
In 1998, only 3.3% of beer was sold in refillable bottles. How well does the model predict this number?
Solution.
We let
\(t=0\) in 1944, so that
\(P_0 = 98\text{.}\) The formula is
\begin{equation*}
P(t)=P_0 b^t = 90(0.942)^t
\end{equation*}
-
We evaluate the formula for several values of \(t\text{,}\) and plot the data points.
| Year |
1950 |
1965 |
1980 |
1995 |
| \(t\) |
\(6\) |
\(21\) |
\(36\) |
\(51\) |
| \(P(t)\) |
68.5% |
27.9% |
11.4% |
4.7% |
In 1998,
\(t=54\text{,}\) and
\begin{equation*}
P(54) = 98(0.942)^{54} = 3.89
\end{equation*}
The model predicts that 3.89% of beer was sold in refillable bottles in 1998, just slightly above the actual figure.
Checkpoint 7.1.8. Practice 2.
Checkpoint 7.1.9. QuickCheck 2.
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{gather*}
\glert{\text{Adding }5\% \text{ of the old population is the same as multiplying the old population by 1.05.}}
\end{gather*}
\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*}
\glert{\text{Growing by } 5\% \text{ 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 7.1.10.
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 \(1+0.12 = 1.12\) 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.
| \(t\) |
\(P(t)\) |
\(\) |
| \(0\) |
\(P(0)=2.00\) |
\(\blert{2}\) |
| \(1\) |
\(P(1)=2.24\) |
\(\blert{2(1.12)}\) |
| \(2\) |
\(P(2)=2.51\) |
\(\blert{2(1.12)^2}\) |
| \(3\) |
\(P(3)=2.81\) |
\(\blert{2(1.12)^3}\) |
| \(4\) |
\(P(4)=3.15\) |
\(\blert{2(1.12)^4}\) |
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 7.1.11. QuickCheck 3.
Checkpoint 7.1.12. Practice 3.
Compound interest is another example of exponential growth. Suppose you deposit a sum of money into an account that pays 5% interest compounded annually. Each year, 5% of your current balance is added to your account as interest, so your balance grows by a factor of 1.05. 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*}
Subsection 7.1.6 Percent Decrease
We have seen that a percent increase of \(r\) (in decimal form) corresponds to a growth factor of \(b=1+r\text{.}\) A percent decrease of \(r\) corresponds to a decay factor of \(b=1-r\text{.}\) For example, if a population declines by 25% each year, then each year the new population is 75% of its previous value. So
\begin{equation*}
b = 1-0.25 = 0.75
\end{equation*}
and \(P(t) = P_0 (0.75)^t\text{.}\) Remember that multiplying by \(b\) gives us the population remaining, not the amount of decline.
Example 7.1.13.
According to Context magazine: "Computing prices have been falling exponentially for the past 30 years and will probably stay on that curve for another couple of decades." In fact, prices have been falling at a rate of 37% every year. Suppose an accounting firm invests $50,000 in new computer equipment.
Write a formula for the value \(V(t)\) of the equipment \(t\) years from now.
What will the equipment be worth in 5 years?
Graph the function \(V(t)\) for \(0 \le t \le 20\text{.}\)
Solution.
The initial value of the equipment is \(V_0 = 50,000\text{.}\) Every year, the value of the equipment is multiplied by
\begin{equation*}
b=1-r=1-0.37 = 0.63
\end{equation*}
After \(t\) years, the value of the equipment is
\begin{equation*}
V(t)=50,000(0.63)^t
\end{equation*}
After 5 years, we have
\begin{equation*}
V(\alert{5}) = 50,000(0.63)^{\alert{5}} = 4962.18
\end{equation*}
The value of the equipment after 5 years is $4962.18.
We evaluate the function \(V(t)\) for several values of \(t\text{,}\) and plot the points to obtain the graph shown.
| \(t\) |
\(2\) |
\(4\) |
\(6\) |
\(10\) |
\(12\) |
| \(V(t)\) |
\(19,845\) |
\(7876\) |
\(3126\) |
\(492\) |
\(195\) |
Note 7.1.14.
In the preceding
Example, the value of the computer equipment decreases by 37% each year, so 63% of the value remains, and the decay factor for the value function is 0.63, not 0.37. The function
\(V(t)\) gives the value remaining, not the amount that has depreciated.
Checkpoint 7.1.15. Practice 4.
Caution 7.1.16.
Compare these two descriptions of exponential decay:
Each year, the population decreases by 25% of its previous value.
Each year, the population decreases to 25% of its previous value.
In the first description, \(r=0.25\) and \(b=1-r = 0.75\text{.}\) If the population this year is 100, next year it will be 75. If 25% of the population is gone, 75% remains.
In the second description, only 25% of the population remains, so \(b = 0.25\text{.}\) If the population this year is 100, next year it will be 25.
Checkpoint 7.1.17. QuickCheck 4.
We summarize our observations about exponential growth and decay functions as follows.
Exponential Growth and Decay.
The function
\begin{gather*}
\blert{P(t) = P_0 b^t}
\end{gather*}
models exponential growth and decay.
If \(b \gt 1\text{,}\) then \(P(t)\) is increasing, and \(b = 1 + r\text{,}\) where \(r\) represents percent increase.
If \(0 \lt b \lt 1\text{,}\) then \(P(t)\) is decreasing, and \(b = 1 - r\text{,}\) where \(r\) represents percent decrease.
Checkpoint 7.1.18. Practice 5.
