Question

A parabolic arch has a span of 120 feet and a maximum height of 25 feet. Choose suitable rectangular coordinate axes and find the equation of the parabola. Then calculate the height of the arch at points 10 feet, 20 feet, and 40 feet from the center.

Short answer

The equation of the parabola is \(y = -\frac{1}{144}x^2 + 25\). Heights: 10 ft - 24.31, 20 ft - 22.22, 40 ft - 13.89.

Step-by-step solution

- Setting Up the Coordinate System

Place the origin of the coordinate system at the vertex of the parabola. Therefore, the vertex is at (0, 25). The span of the arch is 120 feet, so the distance from the center to either end is 60 feet. Thus, the points (-60, 0) and (60, 0) lie on the parabola.

- Using the Standard Form of a Parabola

The standard form of a parabolic equation is \(y = ax^2 + bx + c\). Substituting the vertex coordinates, (0, 25), we get \(25 = a(0)^2 + b(0) + c\), so \(c = 25\). The equation simplifies to \(y = ax^2 + bx + 25\). However, since the parabola is symmetric and opens downward, we use the vertex form \(y = a(x-h)^2 + k\) where (h, k) = (0, 25), resulting in \(y = a(x)^2 + 25\).

- Finding 'a' Using Given Points

Substitute (60, 0) into the vertex form: \(0 = a(60)^2 + 25\). Solving for 'a' gives \(0 = 3600a + 25\), or \(3600a = -25\). Thus, \(a = -\frac{25}{3600} = -\frac{1}{144}\). The equation of the parabola is \(y = -\frac{1}{144}x^2 + 25\).

- Calculating the Height at 10 Feet from the Center

Substitute |10| for x in the equation \(y = -\frac{1}{144}x^2 + 25\): \(y = -\frac{1}{144}(10)^2 + 25\). This simplifies to \(y = -\frac{1}{144}(100) + 25 = -\frac{100}{144} + 25\). Converting \(-\frac{100}{144}\) to decimal, we get \(y ≈ -0.6944 + 25 ≈ 24.3056\).

- Calculating the Height at 20 Feet from the Center

Substitute |20| for x in the equation: \(y = -\frac{1}{144}(20)^2 + 25\), simplifying to \(y = -\frac{1}{144}(400) + 25 = -\frac{400}{144} + 25\). Converting \(-\frac{400}{144}\) to decimal, we get \(y ≈ -2.7778 + 25 ≈ 22.2222\).

- Calculating the Height at 40 Feet from the Center

Substitute |40| for x in the equation: \(y = -\frac{1}{144}(40)^2 + 25\), simplifying to \(y = -\frac{1}{144}(1600) + 25 = -\frac{1600}{144} + 25\). Converting \(-\frac{1600}{144}\) to decimal, we get \(y ≈ -11.1111 + 25 ≈ 13.8889\).

Key concepts

Coordinate System

Understanding how to set up a coordinate system is crucial when working with parabolic equations. For this exercise, the origin of the coordinate system is placed at the highest point of the parabolic arch, which is its vertex. By doing this, the vertex's coordinates become \(0, 25\), where 25 is the maximum height of the arch. The span, which is the distance across the arch, is 120 feet. This makes the ends of the arch fall at \(-60, 0\) and \(60, 0\) on the coordinate plane.

• Place the origin at the vertex \(0, 25\)
• Span of 120 feet means ends are at \(-60, 0\) and \(60, 0\)

This setup simplifies the problem, making it easier to find the equation of the parabola.

Vertex Form

The vertex form of a parabolic equation is essential for solving this type of problem because it provides a direct way to incorporate the coordinates of the vertex. The vertex form equation is given by: \ y = a(x-h)^2 + k \ where \(h, k\) represents the vertex. With our vertex at \(0, 25\), we substitute these values into the equation to get: \ y = a(x-0)^2 + 25 \ which simplifies to: \ y = ax^2 + 25 \/ In this form, it becomes simpler to find the value of 'a', which will adjust the parabola to meet the specific conditions of the problem.

Finding 'a'

To find the 'a' value in our parabolic equation, we use the given point where the parabola intersects the x-axis. In this problem, that point is \(60, 0\). To find 'a,' we substitute \(60, 0\) into the vertex form equation:
\(0 = a(60)^2 + 25\)
This simplifies to: \(0 = 3600a + 25\),
from which we solve for 'a': \(3600a = -25\), \ a = -\frac{25}{3600} = -\frac{1}{144}.\
Therefore, the equation of our parabola becomes: \ y = -\frac{1}{144}x^2 + 25. \/ This 'a' value tells us the parabola opens downwards and helps define the shape of the arch for further height calculations at various points.