Why Is Alaska So Cool?Activity 3 Answer Key

Activity 3: Diameter Detective and Area Analyzer
Answer Key
Part 1

This first experiment analyzes the change in diameters of the circles of light as the distance from the source of light increases. See the Excel spreadsheet for sample data and graphs. See also the document “Behavior of Light” for a discussion about how light spreads out from its source.

  1. What type of regression does the data seem to fit?
    The type of regression is linear. The data should lie approximately in a straight line. This should seem reasonable because the diameter of a circle is a one-dimensional, linear measurement.
  1. What is the regression equation?
    The regression equation will vary depending on your data. The equation for the sample data is y = 2.2931x + 7.8438.
  1. Is the equation linear? If so, describe what the values of a (or m) and b represent.
    Yes, the equation is linear. The value of a (or m if you think of the equation in y = mx + b form), 2.2931, is the slope of the line. It represents the rate of change of the diameters from one measure of distance to the next. The value of b, 7.8438, should be the same value as the diameter of the light source at distance 0 (or close to it, depending on how closely your line fits the actual data).
  1. What is the r2 value?
    The r2 value will vary depending on your data. The value for the sample set of data is 0.97. The closer the value is to 1, the better the line approximates the data compared to other linear fits.
  1. What will happen to the circle diameters as the distance from the light source increases?
    The circle diameters will continue to increase at the same rate as the distance from the light source increases.

  1. Is the relationship of the diameters of the circles to the distance from the light source a direct variation? Explain.
    A direct variation was discussed on page 43 in the Mathematical Analysis section of Computational Science and Mathematical Modeling: Using the TI-83 Plus. A direct variation is of the form y = kx. A linear equation does represent a direct variation if it goes through the origin, because in that case the equation is simply y = mx + 0 so y = mx and the slope of the line is the same as the constant of variation. If a line does not go through the origin, then it could be “shifted” down (or up) by the value of b, the y-intercept, by subtracting the value of b from each of the data points. The new “shifted” data points would represent a direct variation of the distance, x, from the light source.
  1. Use the regression equation to predict the diameters for two new distances from the light source. Give the values and results here.
    Your predictions for two new distances will vary. Sample predictions based on the sample data and the above equation are shown. At 7 cm, the diameter of the circle would be 2.2931  7 + 7.8438 = 23.8955 cm. At 10 cm, the diameter of the circle would be 2.2931  10 + 7.8438 = 30.7748.

Print a copy of the Excel graph to submit.
Graphs will vary depending on data, but all should be appropriately titled, with axes labeled and regression lines shown on graph.

Part 2

This second experiment analyzes the change in areas of the circles of light as the distance from the source of light increases. See the Excel spreadsheet for sample data and graphs.

  1. What type of regression does the data seem to fit?
    The type of regression seems to be a quadratic regression (a polynomial of degree 2). This seems reasonable because the formula for the area of a circle includes the square of the radius, so it would be an equation of the form y = ax2 + bx + c.
    Note: Some students have found that an exponential regression is a good fit, or perhaps a power regression. Remember that there may be several types of regressions that will match a small data set fairly well. How you decide which regression is the best depends on how much information you have. If you could continue the experiment and gather more data points, which equation would be the best predictor of future values? You can read these topics in the Help in Excel to understand more about how Excel is calculating regression equations: “Choosing the best trendline for your data” and “Equations for calculating trendlines.” Some types of regressions will not be available depending on the data in your data set. For example, you cannot create a power trendline if your data contains zero or negative values.
  2. What is the regression equation?
    Answers will vary. The equation for the sample data is y = 6.9804x2 + 14.798x + 56.168.
  1. Is the equation linear? If not, describe the equation and what the coefficients might represent.
    No, the equation is not linear. It is a quadratic equation of the form y = ax2 + bx + c. The simplest quadratic equation is y = x2. When quadratic equations are graphed, they form a parabola. The coefficient of the x2 term, a, helps determine the shape of the graph. Try graphing these equations on the TI-83:
    y = x2; y = 3x2; y = 5x2; y = 0.5x2; y = 0.25x2.
    You should see that the greater the value of a, the “steeper” the parabola is. The smaller the value of a, the more “shallow” the parabola is. All of these parabolas are said to “open up.” Now try graphing y = -4x2 and y = -0.75x2. Do you see that these parabolas “open down?” So the value of a also determines the direction the parabola opens.
    The vertex is the point of the parabola where it turns. All parabolas of the form
    y = ax2 have a vertex at (0,0). If you have an x term with a coefficient b in your equation, as you probably do, it is because the vertex of the parabola is not at the origin. If you are interested, you can read a more detailed discussion of how the quadratic equations are related to their graphs in the document titled “Discussion of Quadratic Equations and their Graphs.” This topic is usually taught in Algebra II.
    The value of c in the equation represents the area of the circle at 0 distance from the light source. That would essentially be the size of the flashlight you used.
  2. What is the r2 value?
    The r2 value will vary depending on your data. For the sample data, the value is 0.9634.
  1. What will happen to the circle areas as the distance from the light source increases?
    The areas of the circle will increase as the distance from the light source increases. They will increase at a faster rate the further you get from the light source.
  1. Is the relationship of the areas of the circles to the distance from the light source a direct variation? Explain.
    A direct variation is of the form y = kxn where n is greater than 0, so y = kx2 would be a direct variation. The area varies directly as the square of the distance. As in a direct variation that is linear, a direct variation of a square would go through the origin. If the vertex of your graph does not go through the origin, the data could be “shifted” so that it did. This shift is not as simple as it is for a linear equation, but is explained in the Discussion of Quadratic Equations and
    Their Graphs.
  1. Use the regression equation to predict the areas for two new distances from the light source. Give the values and results here.
    Predictions of areas for two new distances will vary depending on your data. Sample predictions based on the sample data and the above equation are shown. At 7 cm, the area of the circle would be (6.9804  72) +(14.798  7) + 56.168 = 501.7936 square cm. At 10 cm, the area of the circle would be (6.9804  102) +(14.798  10) + 56.168 = 902.188 square cm.

Print a copy of the Excel graph to submit.
Graphs will vary depending on data, but all should be appropriately titled, with axes labeled and regression lines shown on graph.

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