Chapter 1 Linear Equations and Graphs

Transcription

1 Chapter 1 Linear Equations and Graphs Section 3 Linear Regression

2 Table of Content Slope as a Rate of Change Linear Regression 2

3 Learning Objectives for Section 1.3 Linear Regression The student will be able to calculate slope as a rate of change. The student will be able to calculate linear regression using a calculator. 3

4 Terms mathematical model rate of change linear regression regression analysis interpolation extrapolation predictions 4

5 Mathematical Modeling Mathematical modeling is the process of using mathematics to solve real-world problems. This process can be broken down into three steps: 1. Construct the mathematical model, a problem whose solution will provide information about the real-world problem. 2. Solve the mathematical model. 3. Interpret the solution to the mathematical model in terms of the original real-world problem. In this section we will discuss one of the simplest mathematical models, a linear equation. 5

6 Mathematical Modeling Real-world problem Mathemathical solution 2. Solve Mathemathical model In this section we will discuss one of the simplest mathematical models, a linear equation. 6

7 Slope as a Rate of Change If x and y are related by the equation y = mx + b, where m and b are constants with m not equal to zero, then x and y are linearly related. If (x 1, y 1 ) and (x 2, y 2 ) are two distinct points on this line, then the slope of the line is y y y 2 1 m x 2 x 1 x This ratio is called the rate of change of y with respect to x. Since the slope of a line is unique, the rate of change of two linearly related variables is constant. Some examples of familiar rates of change are miles per hour, price per pound, and revolutions per minute. 7

8 Example 1 Estimating Body Surface Area. Appropiate doses of medicine for both animals and humans are based on body surface area (BSA). Since weight is much easier to determine than BSA, veterinarians use the weight of an animal to estimate BSA. The following linear equation expresses BSA for canines in terms of weights: a = 16.12w where a is BSA in square inches and w is weight in pounds. A) Interpret the slope of the BSA equation. B) What is the effect of one-pound increase in weight? 8

9 Example 1 Solution. A) The rate-of-change BSA with respect to weight is 16.2 square inches per pound. B) Since slope is the ratio of rise to run, increasing w by 1 pound (run) increases a by square inches (rise). 9

10 Example of Rate of Change: Rate of Descent Parachutes are used to deliver cargo to areas that cannot be reached by other means of conveyance. The rate of descent of the cargo is the rate of change of altitude with respect to time. The absolute value of the rate of descent is called the speed of the cargo. At low altitudes, the altitude of the cargo and the time in the air are linearly related. If a linear model relating altitude a (in feet) and time in the air t (in seconds) is given by a = 14.1t +2,880, how fast is the cargo moving when it lands? 10

11 Example of Rate of Change: Rate of Descent Parachutes are used to deliver cargo to areas that cannot be reached by other means of conveyance. The rate of descent of the cargo is the rate of change of altitude with respect to time. The absolute value of the rate of descent is called the speed of the cargo. At low altitudes, the altitude of the cargo and the time in the air are linearly related. If a linear model relating altitude a (in feet) and time in the air t (in seconds) is given by a = 14.1t +2,880, how fast is the cargo moving when it lands? Answer: The rate of descent is the slope m = 14.1, so the speed of the cargo at landing is 14.1 = 14.1 ft/sec. 11

12 Example 2 Finding the Rate of Descent. A 100-pound cargo of delicade electronic equipment is dropped from an altitude of 2,880 feet and lands 200 second later. A) Find a linear model relating altitude a (in feet) and time in the air t (in seconds). B) How fast is the cargo moving when it lands? 12

13 Example 2 Solution. A) If a = mt + b is the linear equation relating altitude a and time in the air t, then the graph of this equation must pass through the following points: (t1, a1) = (0, 2,880) cargo drops The slope of the line is: (t2, a2) = (0, 2,880) cargo lands And the equation of the line is: a 0 = -14.4(t 200) a = -14.4t + 2,880 13

14 Example 2 Solution (cont.) B) The rate of descent is the slope m = -14.4, so the speed of the cargo at landing is = 14,4 ft/sec. 14

15 Linear Regression In real world applications we often encounter numerical data in the form of a table. The powerful mathematical tool, regression analysis, can be used to analyze numerical data. In general, regression analysis is a process for finding a function that best fits a set of data points. In the next example, we use a linear model obtained by using linear regression on a graphing calculator. 15

16 Example 3 [Linear Regression] Diamond Prices. Prices for round-shaped diamonds taken from an on-line trader are given in the following table. Weight (carats) Price 0.5 $2, $3, $3, $4, $5, $5,898 16

17 Example 3 [Linear Regression] (cont.) A) A linear model of the data is giving by P = 6,140c 480 where p is the price of the diamond weighting c carats. Plot the points in the table on a Cartesian coordinate system, producing a scatter plot, and graph the model on the same axes. B) Interpret the slope of the model of the data above. 17

18 Example 3 [Linear Regression] (cont.) C) Use the model to estimate the cost of a 0.85-carat diamond and the cost of a 1.2-carat diamond. Round to the nearest dollar. D) Use the model to estimate the weight of a diamond (to 2 decimals places) that sells for $4,

19 Solution Example 3 A) Scatter Plots. We can plot the data points in the previous example on a Cartesian coordinate plane by hand. We obtain the following plot: Price of emerald (thousands) We can plot the graph of our line of best fit on top of the scatter plot: Weight (tenths of a carat) 19

20 Solution Example 3 [cont] B) The rate of change [price with respect to weight] is 6,140. Increasing the weight by one carat will increase the price by about $6,140. C) We can use the equation to approximate point not in the table. c = 0.85 p = 6,140(0.85) 480 = $4,739 c = 1.2 p = 6,140(1.2) 480 = $6,888 20

21 Solution Example 3 [cont] D) To find the weight of a $4,000 diamond, we solve the equation for c: 6,140 c 480 = 4,000 [add 480] 6,140 c = 4,480 [divide by 6,140] c = 4,480/6,140 [divide by 6,140] c = 0.73 [rounded] The model used in this example was obtained using a technique called linear regression, and the model is called the regression line. 21

22 Interpolation, Extrapolation & Predictions The regression model in the Example 3 was used to approximate points that were not given in the table but would fit between points in the table. This process is called interpolation. In the next example (Example 4) we use a regression model to approximate points outside the given data set. This process is called extrapolation, and the approximations are often referred to as predictions. 22

23 Example 4 [Linear Regression] Carbon Monoxide Emissions. The table in next display contains information about U.S. carbon monoxide emissions. The linear regression model for carbon monoxide emissions from all sources is: C = c where C is the carbon monoxide emission (in millions of short tons) and t is time in years with t = 0 corresponding to A) Interpret the slope of the regression line as a rate of change. B) Use the regression model to predict carbon monoxide emissions from all sources in

24 Example 4 [Linear Regression] (cont.) TABLE U.S. Carbon Monoxide Emissions (millions of short tons) Year All Sources Highway Vehicles

25 Solution Example 4 A) The slope m = -3.6 is the rate of change of emissions with respect to time. Since the slope is negative and the slope is given in millions of tons, the emissions are decreasing at a rate of 3.6(1, ) = 3,600,000 tons per year. B) If t = 36 then, C = (38) = 51.2 or 51,200,000,000 tons so approximately 51,200,000,000 tons of carbon monoxide will be emitted in

26 Chapter 1 Linear Equations and Graphs Section 3 Linear Regression END Last Update: Juanuary 2013