dExample Solution:C=3.19d-0.464 Interpretation: See solution.
Practice makes perfect
a The following table contains the measurements of the diameter d and circumference C of some round boxes, plates, and glasses. The measurements are taken to the nearest millimeter and recorded in centimeters.
Object
d (cm)
C (cm)
1
8.0
24.9
2
9.9
31.4
3
3.8
11.7
4
14.7
46.4
5
8.7
27.6
6
7.6
23.6
7
15.8
50.1
8
13.5
42.5
9
15.0
47.5
10
18.4
58.3
You will most likely have different data, but if the diameter of your object is close to one in the table, then the circumference should also be close.
b We can use a graphing calculator to find the quotient of the circumference and the diameter for each row. First, we need to enter the data in the calculator. We can do this by pressing STAT and then choosing the first option, Edit, in the menu.
Having done this, we will see a number of columns marked L1, L2, and L3.
Using the keypad, we enter our data set into the first two lists. The first list is where we have entered the diameters and the second list has the circumferences. Note that on the screen below you see only the last seven lines of data.
Once the data is entered, we can go back to the calculator screen and use STO▶ to store the quotient of List 2 (circumference) and List 1 (diameter) in List 3. You can access the list names (L1, L2, and L3) using 2ND and the numerical keys.
Pressing STAT and choosing Edit once more will take us back to the list editing window, where now we can see the quotient of the circumference and diameter in the third column.
Note that on the screen above you see only the last seven lines of data. The table below contains all quotients rounded to the nearest hundredth as requested in the question.
Object
d (cm)
C (cm)
C/d
1
8.0
24.9
3.11
2
9.9
31.4
3.17
3
3.8
11.7
3.08
4
14.7
46.4
3.16
5
8.7
27.6
3.17
6
7.6
23.6
3.11
7
15.8
50.1
3.17
8
13.5
42.5
3.15
9
15.0
47.5
3.17
10
18.4
58.3
3.17
Note that the values of the quotients are close to each other. For our measurements, the quotients range between 3.08 and 3.17. You should get similar values for your data. If the quotient in one row is far from these values, go back and double check the corresponding measurements.
c Our largest diameter value is 18.4, and we should choose the horizontal scale of the coordinate system to be able to see the corresponding point. Similarly, the vertical scale should be chosen with the largest circumference value, 58.3, in mind. You may need to use a different scale if you measured larger objects.
Note that the points on the scatter plot are arranged very close to a straight line. You should see a similar trend. If the position of one of your points does not fit this linear pattern, go back and double check the corresponding measurements.
Checking the Scatter Plot Using the Calculator
We can use a graphing calculator to check our plot. To make a scatter plot of the data, which we already entered in the calculator for Part B, press 2ND and then Y=, and choose plot 1. In plot 1, you turn on the scatter plot, then choose L1 and L2 as your x- and y-lists, as well as an appropriate mark for the data points.
Before pressing graph, you might want to change the Window so it fits your data. To do that you press WINDOW.
Finally, by pressing GRAPH we can plot our values.
The calculator confirms our scatter plot.
d Once the values are entered, we can perform a linear regression to find the equation of the best fit line. Press STAT and then use the keypad to choose the CALC menu. Here you have all available types of regressions the calculator can carry out.
By choosing the fourth option, LinReg ax+b, the calculator calculates a linear regression on our dataset.
The linear regression the calculator gives us y=3.19x-0.464. To visually confirm that this line goes close to the data points, let's graph this line over the scatter plot.
In the equation of the regression line, the calculator uses x and y as the independent and dependent variables. In our example the diameter d is the independent variable and the circumference C is the dependent variable, so we get the best fit line by replacing x with d and y with C.
C=3.19d-0.464
We are now asked to explain what this equation, and in particular, the slope represents.
Interpreting the Equation
This equation gives a formula that approximates the circumference of a circle in terms of its diameter.
Interpreting the Slope
We can compare our equation with the formula C=π d, which is the accurate expression of the circumference of a circle in terms of its diameter.
Exact Formula
Aproximation
C=π d
C≈ 3.19d-0.464
For large d-values, - 0.464 is negligible compared to 3.19d, so comparing our approximation with the exact expression gives that the slope, 3.19, is an approximation of π. If your measurements are more accurate than ours, you may get a better approximation.
