When football-shape objects are represented by ellipses, rather than sections of two

parabolas(a tighter point at upper/lower respective limits), what is the difference in

area?A Multi-Step Calculus Question

Introduction

A football, is as most Americans know, is essentially a cylinder that curves along its length to a point at each end. The football also curves up and down from the top to the bottom. Now, the football only has significant height if it is inflated. If a football is deflated, the ball will maintain its shape/curves lengthwise. A football, with its width (minor axis) could be represented by an ellipse(roughly). The ellipse equation allows for clear vertices and axis lengths. However, the area is less, for after the co-vertices of it, the real football, approaches its maximum limit location faster. A more accurate visual representation could be made by sections of two parabolas. However, for any work being done in such a way it is inconvenient. The literature will ask the question; is this necessary? How far off it the total area as compared to an ellipse or two parabolas. (Modern architecture, with buildings similar to these two shapes, may also tend to need such literature).

Subsection of Introduction-Hypothesis

If a football shaped object is represented by an ellipse, then the area of the object will be calculated to be greater than the reality, while the amount of error can be counting the number of squares on the major axis, multiplied by the number of squares in between the two parabolas (closer representation of a football) and the ellipse.

Subsection of introduction- Variables

The independent variable is the way the football shaped object is represented in 2D. The dependent variable is the gap between the ellipse edge and the parabolic edge, which causes a difference in the area of each object.

Methods/ Finding the Equation of the Ellipse

• A vertically oriented ellipse is represented by the equation;

[(x-h)^2/b^2]+[(y-k)^2/b^2] (assuming the football is 36 units (say, cm) in length when deflated and flattened)

Vertices: h, k+a 18, 6+18 18, 24

h, k-a 18, 6-18 18, 12

Co-Vertices: h+b, k 18+6, 6 24, 6

h-b, k 18-6, 6 12, 6

Hence the equation for the ellipse is [(x-18)^2/36]+[(y-6)^2/324]

Methods/Finding the Equation of the Parabolas

• The parabolas are themselves to have vertices (respective) at the vertices of the ellipse. The points are (18, -12) and (18, 24). In order for a point on each of the parabolas to be at the covertices of (24, 6) and (12, 6), the coefficient for each equation must be 0.5 (if the coefficient was the standard “1” the rise would be 6^2 or 36. Using the distance formula the 6- - 12 the difference is 18, exactly one half of 36).

• Hence the equations are [(X-18)^2/2]-12 and [-1(X-18)^2/2]+24

• The parabola is used in its y values until y=6, then the other is used. The parabolas touch the ellipse at y=6, when x=12 and x=24.

Methods/Area of an ellipse and Area Under a Curve(Diagram next slide)

• To find the area of an ellipse the equation of

Major axis times Minor Axis x Pi equals area in units squared

Hence the area of the ellipse is 108 ח units^2.

As for the parabola the area under a curve will be found, using data from the –X^2 graph ranging from x=12 to x=24, on an online website (Tutor vista/ NCS pearson ). The result found was 216 units^2. A line vertical line was drawn at 12, meaning that when all data below y=6 needs to be removed, it is a rectangle, whose area is easy to calculate (this is because, the football is represented by the ellipse and one portion of the parabolas.) Because the rectangle is 6 by 12 units, the result is 72 units^2. Subtract that from 216 and the result is 144. Multiply by two for two parabolas. 288 units^2 vs. .units^2 ח 188

Please note, due to technological limitations the diagonal line should curveUntil it reaches the vertices and co vertices of the ellipse.

Question is; will it curve at a rate that averages out to be the same asthe part of the ellipse it is next to?

Analysis

• The difference between the areas of each was an irrational number. The gaps between each object do not equal the amount. Therefore, the gap is to be determined by its own equation, and is not the visual 18 x 4, which is 72. • In Conclusion, the hypothesis was incorrect, however,

complex equations for shapes to determine areas and their differences gave an answer of 51, which was about 30% of the estimation of 72.

Works Cited

• http://calculator.tutorvista.com/area-under-the-curve-calculator.html

• https://support.google.com/websearch/answer/3284611?p=ws_geometry&hl=en&rd=2#geometry• http://www.mathwarehouse.com/ellipse/area-of-ellipse.php