Problem 67 - College Algebra with Corequisite Support

Question

An arch is in the shape of a parabola. It has a span of 100 feet and a maximum height of 20 feet. Find the equation of the parabola, and determine the height of the arch 40 feet from the center.

Step-by-Step Solution

Answer

The equation is \( y = -\frac{1}{125}x^2 + 20 \), and the height 40 feet from the center is 7.2 feet.

Step 1: Define the Coordinate System

Initially, place the vertex of the parabola, which is the highest point of the arch at the origin of the coordinate system. Thus, the vertex is at \((0, 20)\), because the maximum height is 20 feet.

Step 2: Use the Vertex Form of a Parabola

The equation of a parabola in vertex form is \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex. Here \(h = 0\) and \(k = 20\), so the equation becomes \( y = a(x)^2 + 20 \).

Step 3: Determine 'a' Using Given Span

The parabola spans 100 feet, meaning it extends from \(-50\) to \(50\) on the x-axis. At these points, the height \(y\) is 0. Substitute \((x, y) = (50, 0)\) into the equation: \( 0 = a(50)^2 + 20 \). Simplifying gives \( 2500a = -20 \), thus \( a = -\frac{1}{125} \).

Step 4: Write the Equation of the Parabola

Substitute \( a = -\frac{1}{125} \) into the vertex form equation: \( y = -\frac{1}{125}x^2 + 20 \). This is the equation of the parabola.

Step 5: Calculate Height 40 Feet from Center

To find the height 40 feet from the center, substitute \( x = 40 \) into the parabola's equation: \( y = -\frac{1}{125}(40)^2 + 20 \). Simplifying, \( y = -\frac{1}{125}(1600) + 20 = -12.8 + 20 = 7.2 \).

Key Concepts

Vertex Form

Understanding the vertex form of a parabola is crucial. This form is represented by the equation \( y = a(x-h)^2 + k \), where \((h, k)\) is the vertex or the highest point of the parabola. The parameter \( a \) determines the direction and width of the parabola. The vertex form is particularly helpful for graphing because it clearly shows the position of the vertex. In this problem, the arch's maximum height (vertex) is at \((0, 20)\), and since \( h = 0 \) and \( k = 20 \), the initial equation is \( y = a(x)^2 + 20 \). This setup aligns the vertex with the known maximum height of the arch, streamlining further calculations.

Coordinate System

Selecting an appropriate coordinate system simplifies solving many problems. Here, positioning the vertex at the origin \((0, 20)\) suits this problem perfectly, as it centers the coordinate system directly beneath the highest point of the arch. By placing the origin here, calculations become more straightforward, allowing us to analyze the parabola easily across the x and y axes. This decision is strategic, aligning mathematical operations with physical measurements and enhancing our ability to analyze spans, heights, and other important metrics of the parabola.

Height Calculation

Calculating the height of the parabola at specific points requires substituting x-values into the parabola's equation. The goal is to find the height \( y \) at a certain distance from the center. For example, to find the parabola's height 40 feet from the center, substitute \( x = 40 \) into the equation \( y = -\frac{1}{125}(40)^2 + 20 \). This yields \( y = -12.8 + 20 = 7.2 \) feet.

• This technique allows you to determine precise heights quickly.
• The method ensures you're computing accurate, location-specific heights.

Understanding this process equips you to apply these calculations in various practical scenarios, such as constructing curved structures.

Span of Parabola

The span of a parabola refers to the horizontal distance it covers, which in this problem is 100 feet. This span stretches from \(x = -50\) to \(x = 50\) on the x-axis. Establishing these points is vital as they outline the parabola's base width. At both endpoints, where the arch touches the ground, the height \( y \) is 0. By substituting one of these points, like \((x, y) = (50, 0)\), into the equation \( 0 = a(50)^2 + 20 \), you can solve for \( a \). Solving gives \( a = -\frac{1}{125} \), completing the specific parabolic equation.

• Recognizing the span allows effective space usage in design.
• Comprehending these bounds aids in visualizing the parabola's real-world application.

Thus, the span helps frame the big picture of the parabola's structure.