5.7: Problem Solving Strategies: Use a Linear Model
Learning Objectives
- Read and understand given problem situations.
- Develop and apply the strategy: use a linear model.
- Plan and compare alternative approaches to solving problems.
- Solve real-world problems using selected strategies as part of a plan.
Introduction
In this chapter, we have been estimating values using straight lines. When we use linear interpolation, linear extrapolation or predicting results using a line of best fit, it is called linear modeling. In this section, we will look at a few examples where data sets occurring in real-world problems can be modeled using linear relationships. From previous sections remember our problem solving plan:.
Step 1
Understand the problem
Read the problem carefully. Once the problem is read, list all the components and data that are involved. This is where you will be assigning your variables
Step 2
Devise a plan – Translate
Come up with a way to solve the problem. Set up an equation, draw a diagram, make a chart or construct a table as a start to solving your problem.
Step 3
Carry out the plan – Solve
This is where you solve the equation you came up with in Step 2.
Step 4
Look – Check and Interpret
Check to see if you used all your information and that the answer makes sense.
Example 1
Dana heard something very interesting at school. Her teacher told her that if you divide the circumference of a circle by its diameter you always get the same number. She tested this statement by measuring the circumference and diameter of several circular objects. The following table shows her results.
From this data, estimate the circumference of a circle whose diameter is 12 inches. What about 25 inches? 60 inches?
Object | Diameter (inches) | Circumference (inches) |
---|---|---|
Table | 53 | 170 |
Soda can | 2.25 | 7.1 |
Cocoa tin | 4.2 | 12.6 |
Plate | 8 | 25.5 |
Straw | .25 | 1.2 |
Propane tank | 13.3 | 39.6 |
Hula Hoop | 34.25 | 115 |
Solution
Let’s use the problem solving plan.
Step 1
We define our variables.
\begin{align*}x =\end{align*} diameter of the circle in inches
\begin{align*}y =\end{align*} circumference of the circle in inches
We want to know the circumference when the diameter is 12, 25 or 60 inches.
Step 2 We can find the answers either by using the line of best fit or by using linear interpolation or extrapolation. We start by drawing the scatter plot of the data.
Step 3 Line of best fit
Estimate a line of best fit on the scatter plot.
Find the equation using points (.25, 1.2) and (8, 25.5).
\begin{align*}\text{Slope} \ m & = \frac{25.5 - 1.2} {8 - .25} = \frac{24.3} {7.75} = 3.14\\ y & = 3.14x + b\\ 1.2 & = 3.14 (.25) + b \Rightarrow b = 0.42\\ \text{Equation} \ y & = 3.14x + 0.42\end{align*}
\begin{align*}\text{Diameter} & = 12 \ inches \Rightarrow y = 3.14 (12) + 0.42 = \underline{38.1\ inches}\\ \text{Diameter} & = 25 \ inches \Rightarrow y = 3.14 (25) + 0.42 = \underline{78.92\ inches}\\ \text{Diameter} & = 60 \ inches \Rightarrow y = 3.14 (60) + 0.42 = \underline{188.82\ inches}\end{align*}
In this problem the \begin{align*}\text{slope} = 3.14\end{align*}. This number should be very familiar to you – it is the number rounded to the hundredths place. Theoretically, the circumference of a circle divided by its diameter is always the same and it equals 3.14 or \begin{align*}\pi\end{align*}.
You are probably more familiar with the formula \begin{align*}C = \pi \cdot d\end{align*}.
Note: The calculator gives the line of best fit as \begin{align*}y = 3.25x- 0.57\end{align*}, so we can conclude that we luckily picked two values that gave the correct slope of 3.14. Our line of best fit shows that there was more measurement error in other points.
Step 4 Check and Interpret
The circumference of a circle is \begin{align*}\pi d\end{align*} and the diameter is simply \begin{align*}d\end{align*}. If we divide the circumference by the diameter we will get \begin{align*}\pi\end{align*}. The slope of the line is 3.14, which is very close to the exact value of \begin{align*}\pi\end{align*}. There is some error in the estimation because we expect the \begin{align*}y-\end{align*}intercept to be zero and it is not.
The reason the line of best fit method works the best is that the data is very linear. All the points are close to the straight line but there is some slight measurement error. The line of best fit averages the error and gives a good estimate of the general trend.
Note: The linear interpolation and extrapolation methods give estimates that aren't as accurate because they use only two points in the data set. If there are measurement errors in the points that are being used, then the estimates will lose accuracy. Normally, it is better to compute the line of best fit with a calculator or computer.
Example 2
A cylinder is filled with water to a height of 73 centimeters. The water is drained through a hole in the bottom of the cylinder and measurements are taken at two second intervals. The table below shows the height of the water level in the cylinder at different times.
a) Find the water level at 15 seconds.
b) Find the water level at 27 seconds.
Water Level in Cylinder at Various Times
Time (seconds) | Water level (cm) |
---|---|
0.0 | 73 |
2.0 | 63.9 |
4.0 | 55.5 |
6.0 | 47.2 |
8.0 | 40.0 |
10.0 | 33.4 |
12.0 | 27.4 |
14.0 | 21.9 |
16.0 | 17.1 |
18.0 | 12.9 |
20.0 | 9.4 |
22.0 | 6.3 |
24.0 | 3.9 |
26.0 | 2.0 |
28.0 | 0.7 |
30.0 | 0.1 |
Solution
Let’s use the problem solving plan.
Step 1
Define our variables
\begin{align*}x=\end{align*} time in seconds
\begin{align*}y =\end{align*} water level in centimeters
We want to know the water level at time 15, 27 and -5 seconds.
Step 2 We can find the answers either by using the line of best fit or by using linear interpolation or extrapolation. We start by drawing the scatter plot of the data.
Step 3 Method 1 Line of best fit
Draw an estimate of the line of best fit on the scatter plot. Find the equation using points (6, 47.2) and (24, 3.9).
\begin{align*}\text{Slope} & & m & = \frac{3.9 - 47.2} {24 - 6} = \frac{-43.3} {18} = - 2.4\\ & & y & = -2.4x + b\\ & & 47.2 & = -2.4(6) + b \Rightarrow b = 61.6\\ \text{Equation} & & y & = -2.4 x + 61.6\end{align*}
\begin{align*}\text{Time} & = 15 \ seconds \Rightarrow y = -2.4 (15) + 61.6 = \underline{25.6 \ cm}\\ \text{Time} & = 27 \ seconds \Rightarrow y = -2.4 (27) + 61.6 = \underline{-3.2 \ cm}\end{align*}
The line of best fit does not show us accurate estimates for the height. The data points do not appear to fit a linear trend so the line of best fit is close to very few data points.
Method 2: Linear interpolation or linear extrapolation.
We use linear interpolation to find the water level for the times 15 and 27 seconds, because these points are between the points we know.
\begin{align*}\text{Time} = 15 \ seconds \end{align*}
Connect points (14, 21.9) and (16, 17.1) and find the equation of the straight line.
\begin{align*} m = \frac{17.1 - 21.9} {16 - 14} = \frac{-4.8} {2} = -2.4 \ y = -2.4x + b \Rightarrow 21.9 = -2.4 (14) + b \Rightarrow b = 55.5\end{align*}
Equation \begin{align*}y = -2.4x + 55.5\end{align*}
Plug in \begin{align*}x=15\end{align*} and obtain \begin{align*}y = -2.4(15) + 55.5 = 19.5 \ cm\end{align*}
\begin{align*}\text{Time} = 27 \ seconds\end{align*}
Connect points (26, 2) and (28, 0.7) and find the equation of the straight line.
\begin{align*} m & = \frac{0.7 - 2} {28 - 26} = \frac{-1.3} {2} = -.65\\ y & = -.65x + b \Rightarrow 2 = - .65 (26) + b \Rightarrow b = 18.9\end{align*}
Equation \begin{align*}y= -.65x = 18.9\end{align*}
Plug in \begin{align*}x = 27\end{align*} and obtain \begin{align*} y = -.65 (27) + 18.9 =1.35 \ cm\end{align*}
We use linear extrapolation to find the water level for time -5 seconds because this point is smaller than the points in our data set.
Step 4 Check and Interpret
In this example, the linear interpolation and extrapolation method gives better estimates of the values that we need to solve the problem. Since the data is not linear, the line of best fit is not close to many of the points in our data set. The linear interpolation and extrapolation methods give better estimates because we do not expect the data to change greatly between the points that are known.
Lesson Summary
- Using linear interpolation, linear extrapolation or prediction using a line of best fit is called linear modeling.
- The four steps of the problem solving plan are:
- Understand the problem
- Devise a plan – Translate
- Carry out the plan – Solve
- Look – Check and Interpret
Review Questions
The table below lists the predicted life expectancy based on year of birth (US Census Bureau).
Use this table to answer the following questions.
Birth Year | Life expectancy in years |
---|---|
1930 | 59.7 |
1940 | 62.9 |
1950 | 68.2 |
1960 | 69.7 |
1970 | 70.8 |
1980 | 73.7 |
1990 | 75.4 |
2000 | 77 |
- Make a scatter plot of the data
- Use a line of best fit to estimate the life expectancy of a person born in 1955.
- Use linear interpolation to estimate the life expectancy of a person born in 1955.
- Use a line of best fit to estimate the life expectancy of a person born in 1976.
- Use linear interpolation to estimate the life expectancy of a person born in 1976.
- Use a line of best fit to estimate the life expectancy of a person born in 2012.
- Use linear extrapolation to estimate the life expectancy of a person born in 2012.
- Which method gives better estimates for this data set? Why?
The table below lists the high temperature for the first day of the month for year 2006 in San Diego, California (Weather Underground).
Use this table to answer the following questions.
Month number | Temperature (F) |
---|---|
1 | 63 |
2 | 66 |
3 | 61 |
4 | 64 |
5 | 71 |
6 | 78 |
7 | 88 |
8 | 78 |
9 | 81 |
10 | 75 |
11 | 68 |
12 | 69 |
- Draw a scatter plot of the data
- Use a line of best fit to estimate the temperature in the middle of the \begin{align*}4^{th}\end{align*} month (month 4.5).
- Use linear interpolation to estimate the temperature in the middle of the \begin{align*}4^{th}\end{align*} month (month 4.5).
- Use a line of best fit to estimate the temperature for month 13 (January 2007).
- Use linear extrapolation to estimate the temperature for month 13 (January 2007).
- Which method gives better estimates for this data set? Why?
Review Answers
- Equation of line of best fit using points (1940, 62.9) and (1990, 75.4) \begin{align*}y = .25x- 422.1\end{align*}.
- 66.7 years
- \begin{align*}y = .15x- 224.3\end{align*}, 69.0 years
- 71.9 years
- \begin{align*}y = .29x- 500.5\end{align*}, 72.5 years
- 80.9 years
- \begin{align*}y= .16x- 243\end{align*}, 78.9 years
- A line of best fit gives better estimates because data is linear.
- Equation of line of best fit using points (2, 66) and (10, 75) \begin{align*}y = 1.125x + 63.75\end{align*}
- 68.8 F
- \begin{align*}y = 7x + 36\end{align*}, 67.5 F
- 78.4 F
- \begin{align*}y = x + 57\end{align*}, 70 F
- Linear interpolation and extrapolation give better estimates because data is not linear.
Texas Instruments Resources
In the CK-12 Texas Instruments Algebra I FlexBook, there are graphing calculator activities designed to supplement the objectives for some of the lessons in this chapter. See http://www.ck12.org/flexr/chapter/9615.
Notes/Highlights Having trouble? Report an issue.
Color | Highlighted Text | Notes | |
---|---|---|---|
Show More |