Growth factors

Section 5.3 Growth factors

If someone gives us two points on a graph, we can figure out a linear equation that goes through both of them: we just use the Rate of Change Formula to figure out the slope, and then figure out the intercept. Turns out we can do a similar process to figure out an exponential equation that goes through any two points.

🔗

An important part of this process will be to solve a power equation, so let’s remind ourselves how this works.

🔗

Opening Activity 5.3.1.

Power equations and exponential equations look very similar, but they are solved differently.

🔗

(a)

When you are solving a power equation, do you use the Root Formula or the Log-Divides Formula?

🔗

(b)

When you are solving an exponential equation, do you use the Root Formula or the Log-Divides Formula?

🔗

(c)

Which of these equations should be solved using the Root Formula, and which should be solved using the Log-Divides Formula?

🔗

$75=100\ast 0.92^W$
$12.3=3.04 \ast 2^G$
$1{,}500=1{,}290\ast 1.03^Y$
$85=78\ast 1.016^T$
Log-Divides Formula
$72 = 10 \ast G^9$
$136 = 0.04\, S^2$
$-2=-0.0021\, T^3$
Root Formula

🔗

Subsection Childhood obesity

Obesity among children ages 6-11 continues to increase. From 1994 to 2010, the proportion of children classified as obese rose from an average of 1.1 out of every ten children in 1994 to around 2 out of every ten children in 2010.

Aside

🔗

Assuming that the prevalence of childhood obesity increases exponentially, what is the annual percent increase and what does the equation project for the year 2020? Well, unless we are able to make drastic improvements in how children eat and how much they exercise.

🔗

Because we are told obesity is increasing exponentially we can use the template for an exponential equation.

\begin{equation*} \text{dep }=\text{ start } \ast \text{growth factor}^{\text{indep}} \end{equation*}

The variables are

\begin{align*} C \amp= \text{ obese children (out of every ten) } \sim \text{ dep} \\ Y \amp= \text{ time (years since 1994) } \sim \text{ indep} \end{align*}

Why did we choose to measure time in years since 1994? Since we were given a value of our function in the year 1994, it’s helpful for us to choose 1994 as $Y = 0\text{,}$ so that the value we are given is the starting value -- that is, “start” in our exponential equation template.

🔗

With this choice of variables, we can organize the information we know into a table. Do you see why 2010 corresponds to $Y = 16\text{?}$

time	1994	2010
$Y$	0	16
$C$	1.1	2

🔗

Activity 5.3.2.

Making a little table like the one in this section is a really helpful way to set up problems like this. Let’s practice. In each story below, name the variables, and then summarize the given values in a table.

🔗

In one town, people picking up food at the food shelf has increased exponentially, from 120 per week in 2005 to 630 per week in 2011.
🔗

letter = everyday words (units) ~ dep or indep

= $\qquad\qquad\qquad\qquad\qquad$ $\displaystyle\Big(\qquad\Big)$ ~

= $\displaystyle\Big(\qquad\Big)$ ~

🔗

(indep) $\fillinmath{\displaystyle\int\int}$ $\fillinmath{\displaystyle\int\int}$

(dep) $\fillinmath{\displaystyle\int\int}$ $\fillinmath{\displaystyle\int\int}$

🔗
In another town, attendance at parent volunteer night in 2016 was 132, and in 2019 there were twice as many parent volunteers.
🔗

letter = everyday words (units) ~ dep or indep

= $\qquad\qquad\qquad\qquad\qquad$ $\displaystyle\Big(\qquad\Big)$ ~

= $\displaystyle\Big(\qquad\Big)$ ~

🔗

(indep) $\fillinmath{\displaystyle\int\int}$ $\fillinmath{\displaystyle\int\int}$

(dep) $\fillinmath{\displaystyle\int\int}$ $\fillinmath{\displaystyle\int\int}$

🔗
In a third town, the number of high school students arrested for driving under the influence is half what it was 5 years ago. (Hint: how might you represent the unknown starting value?)
🔗

letter = everyday words (units) ~ dep or indep

= $\qquad\qquad\qquad\qquad\qquad$ $\displaystyle\Big(\qquad\Big)$ ~

= $\displaystyle\Big(\qquad\Big)$ ~

🔗

(indep) $\fillinmath{\displaystyle\int\int}$ $\fillinmath{\displaystyle\int\int}$

(dep) $\fillinmath{\displaystyle\int\int}$ $\fillinmath{\displaystyle\int\int}$

🔗

letter	=	everyday words	(units)	~	dep or indep
	=	\(\qquad\qquad\qquad\qquad\qquad\)	\(\displaystyle\Big(\qquad\Big)\)	~
	=		\(\displaystyle\Big(\qquad\Big)\)	~

(indep)	\(\fillinmath{\displaystyle\int\int}\)	\(\fillinmath{\displaystyle\int\int}\)
(dep)	\(\fillinmath{\displaystyle\int\int}\)	\(\fillinmath{\displaystyle\int\int}\)

letter	=	everyday words	(units)	~	dep or indep
	=	\(\qquad\qquad\qquad\qquad\qquad\)	\(\displaystyle\Big(\qquad\Big)\)	~
	=		\(\displaystyle\Big(\qquad\Big)\)	~

(indep)	\(\fillinmath{\displaystyle\int\int}\)	\(\fillinmath{\displaystyle\int\int}\)
(dep)	\(\fillinmath{\displaystyle\int\int}\)	\(\fillinmath{\displaystyle\int\int}\)

letter	=	everyday words	(units)	~	dep or indep
	=	\(\qquad\qquad\qquad\qquad\qquad\)	\(\displaystyle\Big(\qquad\Big)\)	~
	=		\(\displaystyle\Big(\qquad\Big)\)	~

(indep)	\(\fillinmath{\displaystyle\int\int}\)	\(\fillinmath{\displaystyle\int\int}\)
(dep)	\(\fillinmath{\displaystyle\int\int}\)	\(\fillinmath{\displaystyle\int\int}\)

🔗

Subsection Finding the growth factor

Back to our childhood obesity story. Let’s figure out an equation.

🔗

Because of our clever setup that the variable $Y$ is measured in years since 1994, the starting amount is 1.1 children out of every ten. So, our equation will look like

\begin{equation*} C = 1.1 \ast g^Y\text{.} \end{equation*}

Trouble is we don’t actually know what the growth factor $g$ is. To figure it out, we can use the second piece of information in our table: when $Y = 16\text{,}$ we know that $C = 2\text{.}$ We can put those values into our equation to get

\begin{equation*} 1.1 \ast g^{16}=2\text{.} \end{equation*}

No particular reason for switching sides, just wanted to have the variable on the left. That’s supposed to be true but we don’t know what number $g$ is so we can’t check. Argh.

🔗

Oh, wait a minute. The only unknown in that equation is the growth factor $g\text{.}$ What if we solve for $g\text{?}$ First, divide each side by 1.1 to get

\begin{equation*} \frac{\cancel{1.1}\ast g^{16}}{\cancel{1.1}} = \frac{2}{1.1}\text{,} \end{equation*}

which simplifies to

\begin{equation*} g^{16} = \frac{2}{1.1} = 2 \div 1.1= 1.818181818\ldots \end{equation*}

What kind of equation is this? Since the thing we want to solve for is the base (not the exponent), we have a power equation. We use the Root Formula with power $n=16$ and value $v=1.818181818$ to get

\begin{equation*} g = \sqrt[n]{v} = \sqrt[16]{1.818181818} = 16 \sqrt[x]{~\text{ }} 1.818181818 = 1.038071653 \approx 1.0381\text{.} \end{equation*}

🔗

Want a quicker way to find the growth factor? The entire calculation we just did all boils down to two steps:

\begin{equation*} \frac{2}{1.1} = 2 \div 1.1 = 1.818181818\ldots \end{equation*}

and then

\begin{equation*} g = \sqrt[16]{1.818181818} = 16 \sqrt[x]{~\text{ }} 1.818181818 = 1.038071653 \approx 1.0381\text{.} \end{equation*}

We can even do this calculation all at once:

\begin{equation*} g = \sqrt[16]{\frac{2}{1.1}} = 16 \sqrt[x]{~\text{ }} (2 \div 1.1)= 1.038071653 \approx 1.0381\text{.} \end{equation*}

Notice we added parentheses because the normal order of operations would do the root first and division second. We wanted the division calculated before the root.

🔗

Here’s the shortcut version in a formula.

🔗

Growth Factor Formula

If a quantity is growing (or decaying) exponentially, then the growth (or decay) factor is

\begin{equation*} \displaystyle g = \sqrt[t]{\frac{a}{s}}\text{,} \end{equation*}

where $s$ is the starting amount and $a$ is the amount after $t$ time periods.

🔗

Activity 5.3.3.

Use the tables you created in Activity 5.3.2, together with the Growth Factor Formula, to figure out the growth factor $g$ in each story below.

🔗

In one town, people picking up food at the food shelf has increased exponentially, from 120 per week in 2005 to 630 per week in 2011.
🔗

$g = $
🔗

🔗
In another town, attendance at parent volunteer night in 2016 was 132, and in 2019 there were twice as many parent volunteers.
🔗

$g = $
🔗

🔗
In a third town, the number of high school students arrested for driving under the influence is half what it was 5 years ago. (Hint: Do you actually need to know the unknown starting value?)
🔗

$g = $
🔗

🔗

🔗

Subsection Checking our formula (and rounding)

In our childhood obesity story, we knew from the beginning that our equation was in the form $C = 1.1 \ast g^Y\text{.}$ Now that we found the growth factor $g \approx 1.0381$ we get our final equation

\begin{equation*} C = 1.1 \ast 1.0381^Y\text{.} \end{equation*}

For example, we can check that in 2010, we have $Y=16$ still and so

\begin{equation*} C = 1.1 \ast 1.0381^{16} = 1.1 \times 1.0381 \wedge \underline{16} = 2.000874004 \approx 2 \quad \checkmark \end{equation*}

🔗

You might wonder why we didn’t just round off and use the equation

\begin{equation*} C = 1.1 \ast 1.04^Y\text{.} \end{equation*}

Look what happens when we evaluate at $Y=16$ then. We would get

\begin{equation*} C = 1.1 \ast1.04^{16} = 1.1 \times 1.04 \wedge \underline{16} = 2.06027937 \approx 2.1\text{.} \end{equation*}

Not a big difference (2.1 vs. 2.0) but enough to encourage us to keep extra digits in the growth factor in our equation. Lesson here is: don’t round off the growth factor too much.

🔗

Activity 5.3.4.

Let’s explore a situation in which rounding too much really does cause problems. In one town, attendance at parent volunteer night in 2016 was 132, and in 2019 there were twice as many parent volunteers.

🔗

How many volunteers were there in 2019?
🔗

🔗
Use the Growth Factor Formula to find the value of $g$ for this story.
🔗

🔗
Round the value of $g$ to two decimal digits. How many volunteers does your equation predict for 2019?
🔗

🔗
Now try rounding the value of $g$ to one decimal digit. (This is too much rounding! Don’t do this in real life!) Now how many volunteers does your equation predict for 2019? How far off is your equation from the real data?
🔗

🔗

🔗

Subsection Finding percent change from growth factor

Back to the less harshly rounded equation

\begin{equation*} C = 1.1 \ast 1.0381^Y\text{.} \end{equation*}

We can now answer the two questions. First, in 2020 we have $Y = 2020-1994 = 26$ and so

\begin{equation*} C = 1.1 \ast 1.0381^{26} = 1.1 \times 1.0381 \wedge \underline{26} = 2.908115507 \approx 2.9\text{.} \end{equation*}

According to our equation, by 2020 there would be approximately 2.9 obese children for every ten children.

🔗

The other question was what the annual percent increase is. Think back to an earlier example. Remember that Jocelyn was analyzing health care costs in Section 2.2? They began at $2.26 million and grew 6.7% per year. She had the equation

\begin{equation*} H=2.26\ast1.067^Y\text{.} \end{equation*}

So the growth factor $g=1.067$ in the equation came from the growth rate $r=6.7\%=0.067\text{.}$ Our equation modeling childhood obesity is

\begin{equation*} C = 1.1 \ast 1.0381^Y\text{.} \end{equation*}

The growth factor of $g=1.0381$ in our equation must come from the growth rate $r= 0.0381=3.81\%\text{.}$ Think of it as converting to percent $1.0381 = 103.81 \%$ and then ignoring the 100% to see the 3.81% increase. Childhood obesity has increased around 3.81% each year. Well, on average.

🔗

Here’s the general formula relating the growth rate and growth factor.

🔗

Percent Change Formula

If a quantity changes by a percentage corresponding to growth rate $r\text{,}$ then the growth factor is

\begin{equation*} \displaystyle g=1+r \end{equation*}

🔗

🔗
If the growth factor is $g\text{,}$ then the growth rate is

\begin{equation*} r = g-1 \end{equation*}

🔗

🔗

🔗

Let’s check. We have $g=1.0381$ and so the growth rate is

\begin{equation*} r=g-1 = 1.0381-1 = 0.0381= 3.81\%\text{.} \end{equation*}

Not sure we really need these formulas, but there you have it.

🔗

By the way, formula works just fine if a quantity decreases by a fixed percent. One example we saw was Joe, who drank too much coffee. The growth (or should I say decay) factor was $g=0.87\text{.}$ That corresponds to a growth (decay) rate of

\begin{equation*} r=g-1=0.87-1=-0.83=-13\%\text{.} \end{equation*}

Again, the negative means that we have a percent decrease.

🔗

Activity 5.3.5.

For each story below, translate the growth factors you computed in Activity 5.3.3 into percent changes. Use the Percent Change Formula.

🔗

In one town, people picking up food at the food shelf has increased exponentially, from 120 per week in 2005 to 630 per week in 2011.
🔗

$r = $
🔗

🔗
In another town, attendance at parent volunteer night in 2016 was 132, and in 2019 there were twice as many parent volunteers.
🔗

$r = $
🔗

🔗
In a third town, the number of high school students arrested for driving under the influence is half what it was 5 years ago. (Hint: Do you actually need to know the unknown starting value?)
🔗

$r = $
🔗

🔗

🔗

Do you know …

How to find the growth/decay factor given the starting amount and another point of information?
🔗

🔗
How to find the growth/decay factor given the doubling time or half-life?
🔗

🔗
When we use the Percent Change Formula, and when we use the Growth Factor Formula instead? Ask your instructor if you need to remember the Percent Change Formula and Growth Factor Formula or if they will be provided during the exam.
🔗

🔗
How to evaluate the Percent Change Formula and Growth Factor Formula using your calcuator?
🔗

🔗
How to read the starting amount and percent increase/decrease from the equation?
🔗

🔗

🔗

If you’re not sure, work the rest of exercises and then return to these questions. Or, ask your instructor or a classmate for help.

🔗

Exercises Exercises

Exercises 1-4 are available in a separate workbook format.

🔗

1.

In 1962, my grandfather had savings bonds that matured to $200.

Aside

He gave those to my mother to keep for me. These bonds have continued to earn interest at a fixed, guaranteed rate so I have yet to cash them in. The table lists the value at various times since then.

🔗

year	1962	1970	1980	1990	2000	2010
years since 1962	0	8	18	28	38	48
value	200.00	318.77	570.87	1,022.34	1,830.85	3,278.77

(a)

Use the Growth Factor Formula to find the annual growth factor for the time period from 1962 to 1970.

🔗

(b)

Repeat for 1970 to 1980.

🔗

(c)

What do you notice? What in the story told you that would happen?

🔗

(d)

What is the corresponding interest rate?

🔗

(e)

Write an equation for the value of bonds over time.

🔗

(f)

Use your equation to check the information for 1990, 2000, and 2010.

🔗

(g)

In what year will the bond be worth over $5,000? Set up and solve an equation to decide.

🔗

(h)

Draw a graph using the data in the table, but not your answer to part (g). Include another year that is later than your answer to part (g).

🔗

(i)

Does your answer to part (g) agree with your graph? If not, fix your work.

🔗

2.

Have you read news stories about archaeological digs where a specimen (like a bone) is found that dates back thousands of years? How do scientists know how old something is? One method uses the radioactive decay of carbon.

Aside

After an animal dies the carbon-14 in its body very slowly decays. By comparing how much carbon-14 remains in the bone to how much carbon-14 should have been in the bone when the animal was alive, scientist can estimate how long the animal has been dead. Clever, huh? Actually, it is so clever that Willard Libby won the Nobel Prize in Chemistry for it. The key information to know is that the half-life of carbon-14 (the amount of time it takes for half of the original amount of carbon-14 to decay) is about 5,730 years. For this problem, suppose a bone were found that should have contained 300 milligrams of carbon-14 when the animal was alive.

🔗

(a)

Find the annual “growth” factor.

Aside

Keep at least six digits after the decimal place for your calculations.

🔗

(b)

Name the variables and write an equation describing the dependence.

🔗

(c)

How many milligrams of carbon-14 should remain in this bone after 1,000 years? After 10,000 years? After 100,000 years?

🔗

(d)

How many milligrams of carbon-14 should remain in this bone after 1 million years? Explain the “scientific notation” answer your calculator gives you.

🔗

(e)

Draw a graph that shows up to 10,000 years.

🔗

(f)

If the bone is determined to have 100 milligrams of carbon-14, approximately how long ago did it die? Start by estimating the answer from your graph.

🔗

(g)

Now use successive approximation to revise your estimate.

🔗

(indep)
(dep)
hi / lo

🔗

(h)

Finally, solve the equation exactly.

🔗

3.

For each story, find the annual growth factor $g$ and annual growth rate $r$ as a percent.

🔗

First decide if you can use the Percent Change Formula or if you will need to use the Growth Factor Formula. Don’t forget to include the negative sign for decay rates.

🔗