Exponential growth and decay

$\newenvironment {prompt}{}{} \newcommand {\ungraded }[0]{} \newcommand {\framed }[0]{\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\shaded }[0]{\def \FrameCommand {\fboxsep =\FrameSep \colorbox {shadecolor}}\MakeFramed {\FrameRestore }} \newcommand {\shaded* }[0]{\def \FrameCommand {\fboxsep =\FrameSep \colorbox {shadecolor}}\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\leftbar }[0]{\def \FrameCommand {\vrule width 3pt \hspace {10pt}}\MakeFramed {\advance \hsize -\width \FrameRestore }} \newcommand {\titled-frame }[1]{\def \FrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame {\textbf {#1}}}\def \FirstFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame [$\blacktriangleright $]{\textbf {#1}}}\def \MidFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame [$\blacktriangleright $]{\textbf {#1\ (cont)}}}\def \LastFrameCommand {\fboxsep 8pt\fboxrule 2pt \TitleBarFrame {\textbf {#1\ (cont)}}}\MakeFramed {\advance \hsize -20pt \FrameRestore }} \newcommand {\FrameCommand }[0]{\setlength \fboxrule {\FrameRule }\setlength \fboxsep {\FrameSep }\fbox } \newcommand {\FirstFrameCommand }[0]{\FrameCommand } \newcommand {\MidFrameCommand }[0]{\FrameCommand } \newcommand {\LastFrameCommand }[0]{\FrameCommand } \newcommand {\FrameHeightAdjust }[0]{6pt} \newcommand {\bysame }[0]{\mbox {\rule {3em}{.4pt}}\,} \newcommand {\N }[0]{\mathbb N} \newcommand {\C }[0]{\mathbb C} \newcommand {\W }[0]{\mathbb W} \newcommand {\Z }[0]{\mathbb Z} \newcommand {\Q }[0]{\mathbb Q} \newcommand {\R }[0]{\mathbb R} \newcommand {\A }[0]{\mathbb A} \newcommand {\D }[0]{\mathcal D} \newcommand {\F }[0]{\mathcal F} \newcommand {\ph }[0]{\varphi } \newcommand {\ep }[0]{\varepsilon } \newcommand {\aph }[0]{\alpha } \newcommand {\QM }[0]{\begin {center}{\huge \textbf {?}}\end {center}} \newcommand {\le }[0]{\leqslant } \newcommand {\ge }[0]{\geqslant } \newcommand {\a }[0]{\wedge } \newcommand {\v }[0]{\vee } \newcommand {\l }[0]{\ell } \newcommand {\mat }[0]{\mathsf } \newcommand {\vec }[0]{\mathbf } \newcommand {\subset }[0]{\subseteq } \newcommand {\supset }[0]{\supseteq } \newcommand {\emptyset }[0]{\varnothing } \newcommand {\xto }[0]{\xrightarrow } \newcommand {\qedsymbol }[0]{$\blacksquare $} \newcommand {\bibname }[0]{References and Further Reading} \newcommand {\bar }[0]{\protect \overline } \newcommand {\hat }[0]{\protect \widehat } \newcommand {\tilde }[0]{\widetilde } \newcommand {\tri }[0]{\triangle } \newcommand {\minipad }[0]{\vspace {1ex}} \newcommand {\leftexp }[2]{{\vphantom {#2}}^{#1}{#2}} \newcommand {\epsilon }[0]{\varepsilon } \newcommand {\theta }[0]{\vartheta } \newcommand {\l }[0]{\ell } \newcommand {\d }[0]{\,d} \newcommand {\sectionOutcomes }[0]{} \newcommand {\HyperFirstAtBeginDocument }[0]{\AtBeginDocument }$

Percent increase and decrease; exponential growth and decay.

The population of Metroville is $380,\!000$ in January $2024$ , and it has been growing at $3\%$ per year. We say the growth rate has been $3\%$ .

What will the population be January $2025$ ? $\answer {380000\cdot 1.03}$ .

Correct!

Some people make the computations as follows: First, compute $3\%$ of the population, which is $\answer {0.03\cdot 380000}$ . Then add that to the original population.

Other people use a more efficient method: Just multiply $380,\!000$ by $\answer {1.03}$ .

Question: Why are these methods both correct? In other words, why is $0.03\cdot 380,\!000 + 380,\!000 = 1.03 \cdot 380,\!000?$ Answer: By the $\answer [format=string]{distributive}$ property, thinking of $380,\!000$ as a common $\answer [format=string]{factor}$ .

When thinking of growth in percentage terms, it helps to distinguish the growth rate, in this case $3\%$ , from the growth factor, which is $1.03$ . And it often helps to think of the growth factor in percentage terms, for after 1 year, the population will be $103\%$ of the starting population.

We can generalize these ideas as follows: If a quantity $P$ is growing at a growth rate of $r$ (written as a decimal, interpreted as a percent), then after one unit of time, the new quantity will be $rP + P$ , which can be factored as $\left (\answer {1+r}\right )P$ . In other words, when the growth rate is $r$ , the growth factor is $\answer {1+r}$ .

Kayla purchased a new car for $\$16,\!000$ in January $2024$ . But once a car has been driven off the lot, it depreciates, which is to say it loses value. Kayla’s accountant tells here that in normal economic times and with typical annual driving, her car will depreciate about $8\%$ per year.

Using this rate of depreciation, what will the value of the car be in January $2025$ ? $\answer {16000\cdot 0.92}$ dollars.

Correct!

Some people make the computations as follows: First, compute $8\%$ of the purchase price, which is $\answer {0.08\cdot 16000}$ dollars. Then subtract that from the purchase price.

Other people use a more efficient method: Just multiply $\$16,\!000$ by $\answer {0.92}$ .

Question: Why are these methods both correct? In other words, why is $16,\!000 - 0.08\cdot 16,\!000 = 0.92 \cdot 16,\!000?$ Answer: By the $\answer [format=string]{distributive}$ property, thinking of $16,\!000$ as a common $\answer [format=string]{factor}$ .

In finance, the opposite of depreciation is appreciation, which means an increase in value. In mathematics, appreciation and depreciation are examples of growth and decay.

When thinking of decay in percentage terms, it helps to distinguish the decay rate, in this case $8\%$ , from the decay factor, which is $0.92$ . And it often helps to think of the decay factor in percentage terms, for after 1 year, the value of the car will be $92\%$ of the purchase price.

We can generalize these ideas as follows: If a quantity $P$ is decaying at a rate of $r$ (written as a decimal, interpreted as a percent), then after one unit of time, the new quantity will be $P - rP$ , which can be factored as $\left (\answer {1-r}\right )P$ . In other words, when the decay rate is $r$ , the decay factor is $\answer {1-r}$ .

Back to Metroville, where the population was $380,\!000$ in January $2024$ , and it has been growing at $3\%$ per year.

Write a recursive formula for this situation:

$f(n+1) =$ $\answer {1.03}$ .

Using January 2024 as “year 0,” so that $f(0)=380000$ , write an explicit formula (without commas) to model this situation: $f(t) = \answer {380000\cdot 1.03^t}$ .

Use your explicit formula to compute the following values when the input value makes sense. If the input value does not make sense in the domain of the value, enter ‘NA’ without quotes.

(a): $f(-104) = \answer {NA}$
(b): $f(-3) = \answer [tolerance=100]{380000\cdot 1.03^{-3}}$
(c): $f(2.5) = \answer [tolerance=100]{380000\cdot 1.03^{2.5}}$
(d): $f(4) = \answer [tolerance=100]{380000\cdot 1.03^4}$
(e): $f(220) = \answer {NA}$

Correct! Some of these make sense in the context. For example, $f(-3)$ would approximate the population in January $\answer {2021}$ , $f(4)$ would approximate the population in January $\answer {2028}$ , and $f(2.5)$ would approximate the population in $\answer [format=string]{July}$ (month) $\answer {2026}$ (year).

But $f(-104)$ would mean the population in $\answer {1920}$ , before major world-changing events like World War II and the Great Depression. Similarly, $f(220)$ would mean the population in $\answer {2444}$ , and there are likely to be several world-changing events before then.

Back to Kayla’s new car, which she purchased for $\$16,\!000$ in January $2024$ , and is estimated to depreciate $8\%$ per year.

Write a recursive formula for this situation:

$f(n+1) =$ $\answer {0.92}$ .

Using January 2024 as “year 0,” so that $f(0)=16000$ , write an explicit formula (without commas) to model this situation: $f(t) = \answer {16000\cdot 0.92^t}$ .

Use your explicit formula to compute the following values when the input value makes sense. If the input value does not make sense in the domain of the value, enter ‘NA’ without quotes.

(a): $f(-3) = \answer {NA}$
(b): $f(1.75) = \answer [tolerance=10]{16000\cdot 0.92^{1.75}}$
(c): $f(6) = \answer [tolerance=10]{16000\cdot 0.92^6}$
(d): $f(30) = \answer {NA}$

Correct! Some of these make sense in the context. For example, $f(6)$ would approximate the car’s value in January $\answer {2030}$ , and $f(1.75)$ would approximate the car’s value in $\answer [format=string]{October}$ (month) $\answer {2025}$ (year).

But $f(-3)$ would mean the car’s value in $\answer {2021}$ , before it was built. Also, $f(30)$ would mean the car’s value in $\answer {2054}$ , when it is unlikely to be still road-worthy. If it is road-worthy, it will be an antique.