4.1: Linear Growth
- Page ID
- 22324
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Starting at the age of 25, imagine if you could save $20 per week, every week, until you retire, how much money would you have stuffed under your mattress at age 65? To solve this problem, we could use a linear growth model. Linear growth has the characteristic of growing by the same amount in each unit of time. In this example, there is an increase of $20 per week; a constant amount is placed under the mattress in the same unit of time.
If we start with $0 under the mattress, then at the end of the first year we would have \($20 \cdot 52 = $1040\). So, this means you could add $1040 under your mattress every year. At the end of 40 years, you would have \($1040 \cdot 40 = $41,600\) for retirement. This is not the best way to save money, but we can see that it is calculated in a systematic way.
A quantity grows linearly if it grows by a constant amount for each unit of time.
Suppose in Flagstaff Arizona, the number of residents increased by 1000 people per year. If the initial population was 46,080 in 1990, can you predict the population in 2013? This is an example of linear growth because the population grows by a constant amount. We list the population in future years below by adding 1000 people for each passing year.
| 1990 | 1991 | 1992 | 1993 | 1994 | 1995 | 1996 | |
|---|---|---|---|---|---|---|---|
| Year | 0 | 1 | 2 | 3 | 4 | 5 | 6 |
| Population | 46,080 | 47,080 | 48,080 | 49,080 | 50,080 | 51,080 | 52,080 |
The population growth can be modeled with a linear equation. The initial population P0 is 48,080. The future population depends on the number of years, t, after the initial year. The model is P(t) = 46,080 + 1000t
To predict the population in 2013, we identify how many years it has been from 1990, which is year zero. So n = 23 for the year 2013.
\[P(23)=46,080+1000(23)=69,080 \nonumber \]
The population of Flagstaff in 2013 would be 69,080 people.
|
Linear Growth Model: Linear growth begins with an initial population called \(P_{0}\). In each time period or generation t, the population changes by a constant amount called the common difference d. The basic model is: \[P(t) = P_{0} + td \nonumber \] |
Dora has inherited a collection of 30 antique frogs. Each year she vows to buy two frogs a month to grow the collection. This is an additional 24 frogs per year. How many frogs will she have is six years? How long will it take her to reach 510 frogs?
Solution
The initial population is \(P_{0} = 30\) and the common difference is \(d = 24\). The linear growth model for this problem is:
\[P(t) = 30 + 24t \nonumber \]
The first question asks how many frogs Dora will have in six years so, t = 6.
\[P(6) = 30+24(6) = 30 + 144 = 174 \nonumber \] frogs.
The second question asks for the time it will take for Dora to collect 510 frogs. So, \(P(t) = 510\)and we will solve for t.
\[\begin{align*}510 &= 30 + 24t \\ 480 &= 24t \\ 20 &= t \end{align*} \nonumber \]
It will take 20 years to collect 510 antique frogs.
Note: The graph of the number of antique frogs Dora accumulates over time follows a straight line.
Assume a car depreciates by the same amount each year. Joe purchased a car in 2010 for $16,800. In 2014 it is worth $12,000. Find the linear growth model. Predict how much the car will be worth in 2020.
\(P_{0} = 16,800\) and \(P_{4} = 12,000\)
To find the linear growth model for this problem, we need to find the common difference d.
\[\begin{align*} P(t) &= P_{0} + td \\ 12,000 &= 16,800 + 4d \\ -4800 &= 4d \\ -1200 &= d \end{align*} \nonumber \]
The common difference of depreciation each year is \(d = $-1200\). Thus, the linear growth model for this problem is: \(P(t) = 16,800 - 1200t\)
Now, to find out how much the car will be worth in 2020, we need to know how many years that is from the purchase year. Since it is ten years later, \(t=10\).
\[P(10)=16,800-1200(10)=16,800-12,000=4,800 \nonumber \]
The car is worth $4800 in 2020.
Note: The value of the car over time follows a decreasing straight line.