Question

Use calculus to prove Archimedes' result that the area of a parabolic segment is four-thirds of the area of the inscribed triangle.

Short answer

Question: Prove that the area of a parabolic segment is four-thirds (4/3) of the area of an inscribed triangle. Answer: To prove Archimedes' result, we first represent the parabolic curve as y = ax^2 and the inscribed triangle with vertices A, B, and C. Using calculus, we find the area of the parabolic segment by integrating y = ax^2 over the interval [0, b], obtaining \(\frac{a}{3}b^3\). Then, we calculate the area of the inscribed triangle as \(\frac{1}{2}ab^3\). By comparing the area of the parabolic segment with that of the inscribed triangle, we find that their ratio is \(\frac{4}{3}\), thus proving Archimedes' result.

Step-by-step solution

Define the parabolic function and its domain

Firstly, we represent the parabolic curve as y = ax^2, where a is a constant that determines the shape of the parabola. Next, we define the interval for x-values where the inscribed triangle lies. Let the interval be [0, b].

Calculate the area of the parabolic segment

To calculate the area under the curve of y = ax^2, we integrate the function with respect to x in the interval [0, b]: Area of parabolic segment = \(\int_0^b {ax^2 dx}\). Now, integrate ax^2 with respect to x: \(\int ax^2 dx = \frac{a}{3}x^3 + C\). To find the definite integral over the interval [0, b], we evaluate this: \(\frac{a}{3}b^3 - \frac{a}{3}(0)^3 = \frac{a}{3}b^3\). So, the area of the parabolic segment is \(\frac{a}{3}b^3\).

Calculate the area of the inscribed triangle

We have coordinates A(0, 0), B(b, 0), and C(b, ab^2). The base of the triangle is b, and the height of the triangle is ab^2. Area of the inscribed triangle = \(\frac{1}{2}(\text{base})(\text{height}) = \frac{1}{2}(b)(ab^2) = \frac{1}{2}ab^3\).

Compare the area of the parabolic segment with that of the inscribed triangle

We want to show that the area of the parabolic segment is four-thirds (4/3) of the area of the inscribed triangle: \(\frac{\text{area of the parabolic segment}}{\text{area of the inscribed triangle}} = \frac{\frac{a}{3}b^3}{\frac{1}{2}ab^3}\). By simplifying this expression, we get: \(\frac{\frac{a}{3}b^3}{\frac{1}{2}ab^3} = \frac{\frac{a}{3}}{\frac{1}{2}} = \frac{2a}{3a} = \frac{2}{3}\). This means that the area of the parabolic segment is \(\frac{4}{3}\) times the area of the inscribed triangle, thus proving Archimedes' result.

Key concepts

Parabolic Segment

In the realm of calculus, a parabolic segment is an intriguing curve section. Specifically, it's the region bounded by a parabola and a line segment called the chord. For our exercise, imagine the curve described by the function \( y = ax^2 \), which is a standard equation of a parabola. Here, \( a \) is a constant that reveals the curvature's steepness. The parabolic segment's location on the x-axis is within an interval from 0 to \( b \), encapsulating the space beneath the parabola within these boundaries.

Understanding the parabolic segment is essential as it involves appreciating how calculus can be used to find precise areas beneath curves, something that our ancients, like Archimedes, estimated visually and through reasoning. The parabolic segment isn't just an abstract concept; it represents real-world applications where curved surfaces are common, such as construction arches or bridge designs.

To sum up, grasping parabolic segments offers a window into the transition from geometric intuition to analytical rigor, made possible by calculus.

Archimedes' Theorem

Archimedes' Theorem is a marvelous ancient breakthrough in geometry attributed to the iconic Greek mathematician, Archimedes. It pertains to the geometric concept of the parabolic segment, establishing a relationship between this segment and a simpler shape—an inscribed triangle.

The essence of Archimedes' Theorem is that the area of a parabolic segment is exactly \( \frac{4}{3} \) times the area of the triangle inscribed within it. Such insight was remarkable for its time, long before the invention of integral calculus. By visual reasoning and logical deduction, Archimedes was able to arrive at this conclusion, demonstrating that even with minimal tools, significant mathematical discoveries could be made.

The theorem remains relevant today, primarily because it laid foundational principles that calculus would later rigorize. It provides a beautiful example of geometrical thinking being aligned with calculus' ability to generalize and validate results quantitatively.

Definite Integrals

Definite integrals form the backbone of calculus concepts when it comes to finding the area underneath curves. They allow for the precise calculation of areas, which is crucial for understanding shapes that don't have simple geometric formulas.

A definite integral is expressed as \( \int_a^b f(x) \, dx \), where \( f(x) \) is the function being integrated over an interval [a, b]. By evaluating this integral, you can determine the area enclosed by the function \( f(x) \) and the x-axis from \( a \) to \( b \).

For the parabolic segment in our exercise, using definite integrals, we calculated \( \int_0^b ax^2 \ dx \), revealing the area under the curve \( y = ax^2 \) over [0, b]. The integration process involves finding an antiderivative and then applying the limits of the interval to obtain a numeric value.

In essence, definite integrals extend beyond abstract calculations; they translate into tangible applications in physics, engineering, economics, and beyond, where such precision is indispensable. They offer a systematic approach to measure and analyze phenomena with mathematical accuracy.