Opening the book…
- Apply a model to the linear relationship between two variables.
Introduction
Marco is a collector of antique soda bottles. His collection currently contains \(437\) bottles. Every year, he budgets enough money to buy \(32\) new bottles. Can we determine how many bottles he will have in \(5\) years and how long it will take for his collection to reach \(1000\) bottles?
While both of these questions you could probably solve without an equation or formal mathematics, we are going to formalize our approach to this problem to provide a means to answer more complicated questions.
Suppose that \(P_{n}\) represents the number, or population, of bottles Marco has after \(n\) years. So \(P_{0}\) would represent the number of bottles now, \(P_{1}\) would represent the number of bottles after \(1\) year, \(P_{2}\) would represent the number of bottles after \(2\) years, and so on. We could describe how Marco’s bottle collection is changing using:
\begin{align*}
P_{0}&=437\\
P_{n}&=P_{n-1}+32\\
\end{align*}
This is called a recursive relationship. A recursive relationship is a formula that relates the next value in a sequence to the previous values. Here, the number of bottles in year \(n\) can be found by adding \(32\) to the number of bottles in the previous year, \(P_{n-1}\). Using this relationship, we could calculate:
\begin{align*}
P_{1}&=P_{0}+32=437+32=469\\
P_{2}&=P_{1}+32=469+32=501\\
P_{3}&=P_{2}+32=501+32=533\\
P_{4}&=P_{3}+32=533+32=565\\
P_{5}&=P_{4}+32=565+32=597\\
\end{align*}
We have answered the question of how many bottles Marco will have in \(5\) years. However, solving how long it will take for his collection to reach \(1000\) bottles would require a lot more calculations.
While recursive relationships are excellent for describing simply and cleanly how a quantity is changing, they are not convenient for making predictions or solving problems that stretch far into the future. For that, a closed or explicit form of the relationship is preferred.
An explicit equation allows us to calculate Pn directly without needing to know Pn-1. While you may already be able to guess the explicit equation, let us derive it from the recursive formula. We can do so by selectively not simplifying as we go:
\begin{array}{lr}P_{I}=437+32 & =437+1(32) \\ P_{2}=P_{1}+32=437+32+32 & =437+2(32) \\ P_{3}=P_{2}+32=(437+2(32))+32 & =437+3(32) \\ P_{4}=P_{3}+32=(437+3(32))+32 & =437+4(32)\end{array}
You can probably see the pattern now and generalize that
\[P_{n}=437+n(32)=437+32 n\nonumber\]
Using this equation, we can calculate how many bottles he’ll have after \(5\) years:
\[P_{5}=437+32(5)=437+160=597\nonumber\]
We can now also solve for when the collection reaches \(1000\) bottles by substituting in 1000 for \(P_{n}\) and solving for \(n\).
\begin{align*}
1000&=437+32 n\\
563&=32 n\\
n&=\frac{563}{32}\\
n&=17.59\\
\end{align*}
So Marco will reach \(1000\) bottles in \(18\) years.
In the previous example, Marco’s collection grew by the same number of bottles every year. This constant change is the defining characteristic of linear growth. Plotting the values we calculated for Marco’s collection, we can see they form a straight line, indicating linear growth.
Linear Growth and Depreciation:
If a quantity starts at size \(P_{0}\) and grows (increases) by \(m\) every time period, then the quantity after \(n\) time periods can be determined using the following model (relation). Where \(P_{n}\) is the quantity after \(n\) steps.
\[P_{n}=P_{0}+nm\nonumber\]
This equation represents the common difference \(m\), which is the constant change in each time period. If \(m\) is negative, the model represents linear depreciation
\[P_{n}=P_{0} - m n\nonumber\]
Note: \(P_{n}\) is the same as \(P(n),\) amount after time \(n\).
This equation should look familiar – it is the same linear equation you learned in algebra, probably stated as
\begin{align*}
f(x)=y&=mx+b\text{~~~~~~~~Linear Equation in Slope Intercept Form}\\
\end{align*}
Recall that \(m\) was the slope and \(b\) is an initial amount.
\[\begin{aligned}
\text{slope}&=\frac{\text{rise}}{\text{run}}=\frac{\text{change in \(x\)}}{\text{change in \(y\)}}\\
\end{aligned}\nonumber\]
The linear model in the form of a word is \[\boxed{\text{Amount after time \(t\)}=\text{Initial Amount}+\text{Rate of change (slope)}\times t\nonumber}\]
Scenario |
Constant Rate of Change |
Slope:\(~m=\frac{\text{rise}}{\text{run}}\) |
|---|---|---|
Hourly Wages |
Earning \(\$15\) for every hour worked. |
\[\begin{aligned} \text{slope}&=\frac{15\text{~dollars}}{1\text{~hour}}\\ \end{aligned}\nonumber\] |
Water Filling a Tank |
A hose is pouring \(5\) gallons into a pool every minute. |
\[\begin{aligned} \text{slope}&=\frac{5\text{~gallons}}{1\text{~minute}}\\ \end{aligned}\nonumber\] |
Uniform Motion |
A car traveling at a steady \(60\) mph on a highway. |
\[\begin{aligned} \text{slope}&=\frac{60\text{~miles}}{1\text{~hour}}\\ \end{aligned}\nonumber\] |
Subscription Services |
A company gains \(100\) new users every month. |
\[\begin{aligned} \text{slope}&=\frac{100\text{~users}}{1\text{~month}}\\ \end{aligned}\nonumber\] |
| Car Depreciation | A car decrease in value \(\$2500\) per year | \[\begin{aligned} \text{slope}&=\frac{-2,500\text{~dollars}}{1\text{~year}}\\ \end{aligned}\nonumber\] |
Rather than using \(y\) or \(f(x)\) to represent a model, we can use a variety of other notations. For example, a model may be written as
\(G(w)\), \(C(x)\), \(V(x)\), \(P(n)\), or \(S(t)\) etc.
Sometimes the model is written simply as \(G\), \(C\), \(V\), \(P\), or \(S\) etc, depending on the context.
Your Turn \(\PageIndex{1} \): Find Linear Function
Example \(\PageIndex{1}\): Interpret Slope
A tour guide is planning a ski trip. The equation \(C(x)=23x+450\) models the relation between the cost in dollars, \(C(x)\), of the ski trip and the number of skiers, \(x\). The slope tells us.
- For every additional skier, the ski trip costs an additional \(\$450\).
- If no skiers show up, the ski trip costs \(\$23\).
- For every additional skier, the ski trip costs an additional \(\$23\).
- If no skiers show up, the ski trip costs \(\$450\).
- Answer
-
The slope is the ratio of the rise (change in cost in this case) and the run (change in the number of skiers). Since the slope is \(23\), then we know:
\[\begin{aligned}
23&=\frac{\text{rise}}{\text{run}}\\
\frac{23}{1}&=\frac{\text{change in cost}}{\text{change in number of skier}}\\
\end{aligned}\nonumber\]
So we might read this as follows: If the number of skiers increases by \(1\), then the cost will increase by \(23\).
Alternatively, this says that for every additional skier, the ski trip will cost an additional \(23\) dollars.
Your Turn \(\PageIndex{1} \): Find Linear Function
Example \(\PageIndex{2}\): Streetlight in the City
A city currently has \(130\) streetlights. As part of an urban renewal program, the city council has decided to install \(2\) additional streetlights at the end of each week for the next year. Write an equation that models the number of streetlights, \(P(t),\) after \(t\) weeks. How many streetlights will the city have at the end of \(35\) weeks? When will there be \(230\) streetlights?
- Answer
-
The initial streetlights are \(P_{0}=130\), and the common difference is \(d=2\), considered a slope. The linear growth model for this problem is:
\[P(t)=P_{t}=P_0+dt=130+2t\nonumber\]
The above equation models the number of streetlights after \(t\) weeks.
The next question asks how many streetlights are in the city at the end of \(35\) weeks, \(t=35\).
\[P(35)=130+2(35)=130+70=200\nonumber\]
Finally, we are asked to find the time when there is \(230\) straight light. So, \(P(t)=230\), we will solve for t.
\[\begin{aligned}
230&=130+2t\\
230-130&=2t\\
50&=t\\
\end{aligned}\nonumber\]There will be \(230\) streetlights after \(50\) weeks. We can also see the linear relationship in the following graph.
Figure \(\PageIndex{1}\): Number of Streetlight Over Time
Example \(\PageIndex{3}\): City Population
A city currently has \(1.89\) thousand residents. Each year, the city's population grows by around \(220\) persons.
- Write a function to model this situation where \(t\) is the number of years and \(P(t))\) the population after \(t\) years.
- After \(3\) years, what will the approximate population of the city be?
- Answer
-
\((a)\) The initial population is \(P_{0}=1.89\times1,000=1890\) and slope (rate), \(m,\) is \(220\) person grow per year. Note that the slope is positive here. Using a linear model that models the number of people after \(t\) years.
\[P(t)=P_{t}=P_0+mt=1890+220t \nonumber\]
\((b)\) To find the approximate population of the city after \(3\) years, put \(t=3.\)
\[P(3)=1,890+3(220)=1,890+660=2,550 \nonumber\]
So the city population after \(3\) years is \(2,550\) or \(2.55\) thousand.
If the line passes through two points (\(x_1,y_1)\) and (\(x_2,y_2),\) the slope of the line is given by the following formula. To find the slope, we need to identify two points and use the following formula:
\begin{align*}
\text{Slope}=\frac{y_2-y_1}{x_2-x_1}
\end{align*}
Now we have \(m\). The \(y\)-intercept, \(b\), is the initial amount (or value). Use \(y=mx+b\) to find the model.
Example \(\PageIndex{4}\): Linear Function Using Two Points
The population of elk in a national forest was measured to be \(12,000\) in \(2003\) and was measured again to be \(15,000\) in \(2007.\) If the population continues to grow linearly at this rate. Find the function that models the elk population in time \(t.\) What will the elk population be in \(2027?\)
- Answer
-
To begin, we need to define how we’re going to measure \(t\). Remember that \(P_0\) is the population when \(t = 0\), so we probably don’t want to literally use the year \(0.\) Since we already know the population in \(2003,\) let us define \(t = 0\) to be the year \(2003.\) Then
\(P_0=P(0)= 12,000,\). This is the y-intercept of the linear equation.
Next, we need to find \(d\). Remember, \(d\) is the growth per time period, in this case, growth per year (slope, \(m\)). Between the two measurements, the population grew by \(15,000-12,000 = 3,000,\) but it took \(2007-2003 = 4\) years to grow that much.
To find the annual growth, we can divide \(3000\) elk by \(4\) years, giving \(750\) elk in \(1\) year.
Alternatively, you can use the slope formula from algebra to determine the common difference or slope. Here (\(x_1,y_1)=(2003,12000)\) and (\(x_2,y_2)=(2007,15000)\)
\begin{align*}
\text{slope}&=\frac{y_2-y_1}{x_2-x_1}\\
&=\frac{15,000-12,000}{2007-2003}\\
&=\frac{3000}{4}\\&=750
\end{align*}We can now write an equation for the growth model of the elk population as follows
\[P_{t}=P(t)=12,000+750 t\nonumber\]
To answer the question, we need to first note that the year \(2027\) will be \(t = 24\), since \(2027\) is \(24\) years after \(2003.\) Use a linear model to find \(P(24)\)
\begin{align*}
P_{24}&=12,000+750(24)\\
&=30,000
\end{align*}The elk population in \(2027\) will be \(30,000\)
Example \(\PageIndex{5}\): Car Depreciation
Assume a car depreciates by the same amount each year. Max purchased a car in \(2020\) for \(\$26,800\). In \(2024\), it was worth \(\$22,000\). Find the linear model that expresses the value \(V(t)\) as a function of how many years old is, \(t\). Predict how much the car will be worth in \(2028\).
- Answer
-
Letting \(n = 0\) correspond with \(2020 \) would give \(b=\$26,800\), initial value (y-intercept) or initial amount.
Next, we need to find \(m\). Remember, \(m\) is the rate of change per year (slope, \(m\)).
We can find the slope using the two-point formula.
Here \((x_1,y_1)=(2020,\$26,800)\) and (\(x_2,y_2)=(2024,\$22,000)\)
\begin{align*}
\text{slope}&=\frac{y_2-y_1}{x_2-x_1}\\
&=\frac{\$22,000-\$26,800}{2024-2020}\\
&=\frac{-\$4,800}{4}\\&=-\$1,200\\
\end{align*}That means the car value is decreasing by \(-\$1,200\) per year. So the model equation in the form of \(y=mx+b\) is
\[V(t)=26,800-1,200t\nonumber\]
Now, to find out how much the car will be worth in \(2028\), we need to know how many years that is from the purchase year. Since it is eight years later. So put \(t=8\) in the model equation.
\begin{align*}
V(t)&=26,800-1,200t~~~~~~~\text{Equation of the model}\\
&=26,800-1,200(8)\\
&=26,800-9,600\\
&=17,200\\
\end{align*}So the value of car in \(2028\) (after \(8\) years) will be \(\$17,200\)
Your Turn \(\PageIndex{5}\): Value of Bulldozer
When Good Models go Bad?
[1] When using mathematical models to predict future behavior, it is essential to consider that very few trends will persist indefinitely. Consider the following example.
Example: Suppose a four-year-old boy is currently \(39\) inches tall, and you are told to expect him to grow \(2.5\) inches a year. How tall will he be when he is \(6\) years old? \(50\) years old?
We can set up a growth model, with \(t=0\) corresponding to \(4\) years old
\[P(t)=P_{t}=P_0+dt=39+2.5t \nonumber \]
So at 6 years old, \(t=2\) , and we expect him to be:
\[P(2)=39+2.5(2)=44~~\text{inches}\nonumber \]
Any mathematical model will eventually break down. Certainly, we shouldn’t expect this boy to continue growing at the same rate throughout his life. If he did, at age \(50\) he would be inches tall \(154\) inches or \(12.8\) feet tall as we see below.
So at \(50\) years old, \(t=46\) , and we expect him to be:\[P(46)=39+2.5(46)=154~~\text{inches}\nonumber \]
This page titled 2.1: Linear Growth was last modified on Fri, 18 Sep 2026 20:30:37 GMT and is shared under a CC BY-SA 3.0 license and was authored, remixed, and/or curated by Cheban Acharya via source content that was edited to the style and standards of the LibreTexts platform.
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