Problem 32 - Algebra 2

Question

STRUCTURAL DESIGN An architect's design for a building includes some large pillars with cross sections in the shape of hyperbolas. The curves can be modeled by the equation \(\frac{x^{2}}{0.25}-\frac{y^{2}}{9}=1,\) where the units are in meters. If the pillars are 4 meters tall, find the width of the top of each pillar and the width of each pillar at the narrowest point in the middle. Round to the nearest centimeter.

Step-by-Step Solution

Answer

Top width: 224 cm, middle width: 100 cm.

Step 1: Understand the Hyperbola Equation

The given hyperbola equation is \( \frac{x^2}{0.25} - \frac{y^2}{9} = 1 \). This is in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) for a hyperbola centered at the origin with horizontal transverse axis. Here, \(a^2 = 0.25\) and \(b^2 = 9\). Thus, \(a = 0.5\) meters and \(b = 3\) meters.

Step 2: Find Width at the Top of the Pillar

To find the width of the pillar at the top, calculate \(x\) when \(y = 2\) (since the full height is 4 meters, half-height is 2 meters). Use the equation: \( \frac{x^2}{0.25} - \frac{2^2}{9} = 1 \). Calculate to get \(x^2 = 1.25\), then \(x = \sqrt{1.25}\). This gives \(x \approx 1.118\), meaning the total width is about \(2 \times 1.118 = 2.236\) meters.

Step 3: Find Width at the Middle of the Pillar

At the narrowest point (middle, where \( y = 0 \)), plug \( y = 0 \) into the equation: \( \frac{x^2}{0.25} - \frac{0^2}{9} = 1 \), simplifying to \( \frac{x^2}{0.25} = 1 \). Hence, \( x^2 = 0.25 \) and \( x = 0.5 \). The width is \(2 \times 0.5 = 1\) meter.

Step 4: Convert Widths to Centimeters

Convert the widths obtained to centimeters by multiplying by 100. For the top of the pillar, \(2.236\) meters is approximately \(224\) cm. For the narrowest point, \(1\) meter is \(100\) cm.

Key Concepts

Architectural Design

Hyperbolas play a fascinating role in architectural design, particularly in structures like the large pillars mentioned in the exercise. Architects often favor hyperbolic shapes for their unique aesthetic appeal and strength. The curvature of hyperbolas provides efficient load distribution, which means the structure can withstand substantial vertical pressure.
Beautiful architectural designs may use hyperbolic forms in pillars and arches to create striking visual effects while incorporating the beneficial structural qualities of hyperbolas. Whether found in ancient arches or modern buildings, the hyperbola helps achieve both beauty and stability in architecture.

Hyperbola Equation

The hyperbola equation given is \ \( \frac{x^2}{0.25} - \frac{y^2}{9} = 1 \), which is already in its standard form.
This form is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), where \(a\) and \(b\) determine the shape and orientation of the hyperbola. In this equation, \(a^2 = 0.25\) and \(b^2 = 9\), leading to \(a = 0.5\) meters and \(b = 3\) meters.
This particular arrangement indicates that the hyperbola is centered at the origin with its transverse axis aligned horizontally along the x-axis, which significantly affects its geometry and orientation. Such features are critical in architectural applications where precise calculations define the functionality and stability of designs.

Transverse Axis

The transverse axis is the line segment that extends through the vertices of a hyperbola. It's a crucial aspect of hyperbolic structures, as it defines the widest point of the hyperbola.
In the given problem, the transverse axis is horizontal because the equation's structure \ \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) prioritizes the \(x\)-variable.
This axis influences how the pillar extends horizontally at its broadest section. Understanding the transverse axis helps architects determine how much space a hyperbolic column or entrance might occupy, providing a blueprint for the overall architectural design.

Width Calculation

Calculating the width of a hyperbolic shape involves understanding how the distance between points on the hyperbola changes at various heights. For instance:

• The top of the pillar is 4 meters high, so calculating the width at half this height \((y = 2)\) involves substituting \(y = 2\) into the hyperbola equation.
• Solving \( \frac{x^2}{0.25} - \frac{2^2}{9} = 1 \) results in \(x = \sqrt{1.25}\), leading to a total width around 2.236 meters.
• At the middle of the pillar, where \(y = 0\), the calculation \(x = 0.5\) provides a width of 1 meter.

These calculations ensure that each part of the pillar aligns with the design requirements set by architects. Converting these dimensions to centimeters finalizes the precision needed in construction, making sure all elements fit the intended specifications.