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## Getting a Little "Lift" out of Calculus Part I: Answers

1. When I printed out the enlarged picture of the plotter view panel, I found the curve to cover 6.5 cm. Because of this, I divided it into 12 rectangles with width .5 cm each (see Figure 1). I then took the height measurements of each rectangle and recorded them in a spreadsheet (see Table 1).
2. The program displays the plot from 0 to 1 square foot, so I converted the X-axis to (0 to 144) square inches. I then divided the displayed 144 inches by the measured 6.5 cm to find an X-scaling factor of 22.15. On the Y-axis, 1.0 psi was equivalent to 6.0 cm, which gives the Y-scaling factor of .1666. On the spread sheet, I have multiplied by the approtiate scaling factors.
3. I then multiplied the scaled values to find the total area under the curve. This gave a value of 68.4 lbs which represented the lift force.
4. I found the following functions to represent the curves (shown in Figure 2):

f1(x) = 14.81

f2(x) = .0065 x + 13.87

5. I finished by solving the following integral:
 Integral (X=0 to X=144) [f1(x) - f2(x)] dx 135.3 - 66.3 = 69.0 lbs
6. The value calculated by FoilSim is 66 lbs, so both answers compare favorably.

Figure 1

Figure 2

Table 1

 Measured Scaled Rectangle Width (cm) Height (cm) Width (sq. in.) Height (lbs/sq. in.) Area (lbs) 1 0.50 4.2 11.07 .7 7.7 2 0.50 5.0 11.07 .83 9.2 3 0.50 4.6 11.07 .766 8.5 4 0.50 4.4 11.07 .733 8.1 5 0.50 4.1 11.07 .683 7.5 6 0.50 3.6 11.07 .6 6.6 7 0.50 3.1 11.07 .516 5.7 8 0.50 2.6 11.07 .433 4.8 9 0.50 2.1 11.07 .35 3.9 10 0.50 1.7 11.07 .283 3.1 11 0.50 1.1 11.07 .183 2.0 12 0.50 .7 11.07 .117 1.3 Total (in lbs) 68.4

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