Problem 59 - Precalculus: Functions and Graphs

Question

Dimensions of an arch An arch of a bridge is semielliptical, with major axis horizontal. The base of the arch is 30 feet across, and the highest part of the arch is 10 feet above the horizontal roadway, as shown in the figure. Find the height of the arch 6 feet from the center of the base. (GRAPH CANT COPY)

Step-by-Step Solution

Answer

The height of the arch 6 feet from the center is approximately 9.17 feet.

Step 1: Identify the Semielliptical Equation

The general equation for a horizontal ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), where \(2a\) is the length of the major axis and \(2b\) is the length of the minor axis. Given that the base of the arch is 30 feet, the major axis has length 30 feet, so \(a = 15\) feet. The maximum height of 10 feet makes this the minor axis, so \(b = 10\) feet.

Step 2: Substitute Known Values into Equation

Substitute \(a = 15\) and \(b = 10\) into the equation \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\) to get the equation of the semiellipse: \(\frac{x^2}{15^2} + \frac{y^2}{10^2} = 1\). Simplifying, we have \(\frac{x^2}{225} + \frac{y^2}{100} = 1\).

Step 3: Find Height at Specified Distance from Center

To find the height of the arch 6 feet from the center, substitute \(x = 6\) into the equation \(\frac{x^2}{225} + \frac{y^2}{100} = 1\). This gives \(\frac{6^2}{225} + \frac{y^2}{100} = 1\). Simplify \(\frac{36}{225}\) to get \(0.16\), so the equation becomes \(0.16 + \frac{y^2}{100} = 1\).

Step 4: Solve for y

Rearrange the equation \(0.16 + \frac{y^2}{100} = 1\) to get \(\frac{y^2}{100} = 0.84\). Multiply both sides by 100 to clear the fraction: \(y^2 = 84\). To find \(y\), take the square root of both sides: \(y = \sqrt{84}\), which simplifies to approximately 9.17 feet.

Key Concepts

Semielliptical Equation

Understanding a semielliptical equation is key to solving problems related to elliptical arches, like the one in our exercise. Essentially, a semiellipse is half of an ellipse, and its equation can be derived from the general equation of a horizontal ellipse centered at the origin:

• \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

Here, \( a \) corresponds to half of the total length of the major axis, and \( b \) to half of the total height, which represents the minor axis in this context. Since an arch is only the upper half of this shape, it relies on this equation to describe its curvature effectively. Substituting the values of \( a \) and \( b \) into this equation helps to hone in on the dimensions of the arch in practical problems. In our example, values for \( a \) and \( b \) were known, making it possible to establish the precise shape of the semielliptical arch.

Major and Minor Axes

In the context of elliptical arches, comprehending the major and minor axes is crucial as these define the fundamental dimensions of the arch.

• The **major axis** is the longest diameter of the ellipse; for our semielliptical arch, it runs horizontally along the base. Our example shows a base of 30 feet, thus the major axis **length** is 30 feet, giving \( a = 15 \) feet.
• The **minor axis** is the shortest diameter, vertical in this context and denoting the maximum height of the arch. In the exercise, the poorly connected height is 10 feet, providing \( b = 10 \) feet.

These axes are critical because they provide the parameters for the ellipse equation, enabling further calculations to ascertain specific dimensions, like the arch's height at different points.

Height Calculation

Calculating the height of a semielliptical arch at a specific point from its center involves substituting into the semiellipse equation. For our problem, the goal was to ascertain the height 6 feet away from the center of the base.Here are the steps we followed:

• First, substitute \( x = 6 \) in the equation \( \frac{x^2}{225} + \frac{y^2}{100} = 1 \).
• This manages to isolate the \( y \) terms once the \( x \) contributions have been incorporated.
• We calculated \( \frac{6^2}{225} = 0.16 \), which transformed the equation into \( 0.16 + \frac{y^2}{100} = 1 \).
• Simplify further by rearranging and solving for \( y^2: \frac{y^2}{100} = 0.84 \).
• Multiplying through by 100 yields: \( y^2 = 84 \).
• The final step is taking the square root: \( y = \sqrt{84} \approx 9.17 \). This numeric answer gives the height of the arch 6 feet from its center.

This practical application of the semielliptical equation allows us to determine structural properties such as height at various distances, which can be foundational for those learning about architectural design or structural engineering.