Question

The Greeks solved several calculus problems almost 2000 years before the discovery of calculus. One example is Archimedes' calculation of the area of the region \(R\) bounded by a segment of a parabola, which he did using the "method of exhaustion." As shown in the figure, the idea was to fill \(R\) with an infinite sequence of triangles. Archimedes began with an isosceles triangle inscribed in the parabola, with area \(A_{1}\), and proceeded in stages, with the number of new triangles doubling at each stage. He was able to show (the key to the solution) that at each stage, the area of a new triangle is \(\frac{1}{8}\) of the area of a triangle at the previous stage; for example, \(A_{2}=\frac{1}{8} A_{1},\) and so forth. Show, as Archimedes did, that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_{1}\).

Short answer

Question: Show that the area of the region R, filled with an infinite sequence of triangles, is 4/3 times the area of the first isosceles triangle A1. Answer: The area of the region R is 4/3 times the area of the first isosceles triangle A1 because the total area of all triangles can be represented as an infinite geometric series with first term A1 and common ratio 1/4. Using the formula for the sum of an infinite geometric series, we find that the area of region R equals (4/3)A1.

Step-by-step solution

Write the Area of All Triangles

We will start by writing the area of the triangles at each stage: 1. Area of the first triangle (\(A_1\)) is given. 2. Area of the second stage triangles is \(\frac{1}{8}A_1\). 3. Area of the third stage triangles is \(\frac{1}{8^2}A_1\), but there are \(2^2\) such triangles, thus total area for triangles at this stage is \((\frac{1}{8^2})2^2A_1\) 4. Suppose at the \(n^{th}\) stage the area of triangles is \((\frac{1}{8^{n-1}})2^{n-1} A_1\)

Compute the Total Area of All Triangles

The total area of all triangles is the sum of the areas at each stage. We will now write this as a geometric series and then calculate the sum. Total Area = \(A_1+\frac{1}{8}A_1+(\frac{1}{8^2})2^2A_1+(\frac{1}{8^3})2^3A_1+...+(\frac{1}{8^{n-1}})2^{n-1} A_1+...\) This geometric series has first term \(A_1\) and common ratio \(\frac{1}{4}\).

Find the Sum of the Geometric Series

To find the sum of an infinite geometric series, we use the formula: Sum = \(\frac{first\ term}{1 - common\ ratio}\) In our case, the first term is \(A_1\) and the common ratio is \(\frac{1}{4}\). Therefore, the sum (which represents the area of region \(R\)) is: Area of \(R\) = \(\frac{A_1}{1 - \frac{1}{4}} = \frac{A_1}{\frac{3}{4}} = \frac{4}{3}A_1\) This completes the proof that the area of \(R\) is \(\frac{4}{3}\) times the area of \(A_1\).

Key concepts

Infinite Geometric Series

An infinite geometric series is a fascinating concept that involves adding an infinite sequence of numbers. Each successive number is derived by multiplying the previous number by a common ratio. In the original problem, we explore this idea using a series of triangles within a parabola. We start with the area of the first triangle and each subsequent triangle's area is a fraction of the previous one, specifically, each area is \[\frac{1}{8} \] of the preceding stage's total area.
The formula for the sum of an infinite geometric series is essential here. It is given by \[\frac{a}{1 - r}\] where \(a\) is the first term and \(r\) is the common ratio between terms. This approach allows us to sum an infinite number of terms in a manageable way. In our case, the common ratio is \[\frac{1}{4}\]. Therefore, the total area can be calculated, allowing the result to be expressed neatly as \[\frac{4}{3} \] times the area of the first triangle.

Area of a Parabola Segment

The area of a parabola segment is the area found between a parabola and a line segment. In our example, Archimedes chose to investigate this area using the method of exhaustion, an early form of calculus. By filling the area with an infinite series of triangles, he could approximate the total area more effectively.
Starting with a single triangle inside the parabola and progressively adding smaller triangles, the process narrows down on the parabola segment's actual area. By understanding the progression of the series, getting increasingly smaller triangles allows us to approximate the area of the parabola segment with increasing accuracy.

• The first triangle sets the basis of the area to compare every other triangle.
• Each subsequent triangle's area is calculated as smaller fractions down the series, filling tighter gaps in the uncalculated space.

Initially complicated, this process becomes straightforward with practice and understanding of infinite geometric series.

Ancient Greek Mathematics

Ancient Greek mathematics is incredibly influential in the history of human thought. The Greeks, including luminaries like Archimedes, pioneered methods that paved the way for modern calculus. Archimedes' method of exhaustion employed techniques similar to integral calculus: approximating areas under curves by summing an infinite series of shapes.
In Archimedes' time, there was no formal algebraic notation or calculus. However, thinkers like him used logic and geometry to uncover truths about the world. His work with parabola segments and understanding of geometric series are remarkable examples of this.

• His method of exhaustion was a precursor to limits, a fundamental concept in calculus.
• These foundational ideas influenced mathematicians centuries later during the development of modern calculus by Newton and Leibniz.

The ability to resolve such intricate problems speaks to the depth of knowledge and insight possessed by the ancient Greeks, creating a legacy that persists in mathematical practice to this day.