Problem 102 - Algebra and Trigonometry Real Mathematics, Real People

Question

Roads are often designed with parabolic surfaces to allow rain to drain off. A particular road that is 32 feet wide is 0.4 foot higher in the center than it is on the sides. (See figure.) (a) Find an equation of the parabola with its vertex at the origin that models the road surface. (b) How far from the center of the road is the road surface 0.1 foot lower than in the middle?

Step-by-Step Solution

Answer

The equation of the parabola that models the road surface is \( y = -0.0015625x^2 \) and the road surface is 0.1 foot lower than in the middle approximately 13.86 feet from the center.

Step 1: Define the parabola equation.

Since the vertex of the parabola is at the origin and the parabola opens downwards, the parabola equation would take the form \( y = ax^2 \). The goal on this step is to find the value of the constant \( a \). Using the information given, at the edge of the road (\( x = 16 \) feet), the height is lower by 0.4 feet (\( y = -0.4 \) foot). Substitute these values into the equation and solve for \( a \).

Step 2: Calculate the constant for the parabola equation

Substitute \( x = 16 \) feet and \( y = -0.4 \) foot into the equation \( y = ax^2 \), which lead to: \( -0.4 = a(16^2) \). Solving this equation gives \( a = -0.4 / 256 = -0.0015625 \). So, the equation of the parabola that models the road surface is \( y = -0.0015625x^2 \).

Step 3: Find the distance from the center where the road surface is 0.1 foot lower

Since the center has a 0.4 foot rise, a 0.1 foot lower position means \( y = 0.4 - 0.1 = 0.3 \) foot. Substitute this into the previously determined parabola equation \( 0.3 = -0.0015625x^2 \) and solve for \( x \), which results in \( x = sqrt(0.3/-a) = sqrt(0.3/0.0015625) = 13.86 \) feet (approx.). This indicates that the road surface is 0.1 foot lower than the center about 13.86 feet from the center.

Key Concepts

Vertex at the origin

When dealing with parabolas, it's important to locate the position of the vertex, which is the highest or lowest point of a parabola depending on its orientation. In this problem, we see that the vertex of our parabola is at the origin, meaning the very center of the road is located at point (0,0) on the coordinate plane.

The vertex at the origin simplifies the equation we need to use because it eliminates any horizontal or vertical shifts that usually come with a vertex not at the origin. This particular parabola represents a cross-section of the road, with rain naturally flowing away from the center. In such cases, symmetry plays a key role, suggesting that the heights on either side of the vertex (center) will decrease evenly as one moves away from the middle.

Parabola equation

A parabola is described by a specific kind of quadratic algebraic equation. When the vertex is at the origin and the parabola opens downwards, it takes the form: \( y = ax^2 \). Here, "a" is a constant that dictates the curvature of the parabola, and it can be positive or negative depending on whether the parabola opens upwards or downwards.

In this exercise, the parabola opens downwards to let rainwater drain towards the sides, so "a" will be negative. The characteristics of the parabola, such as its width and steepness, are entirely determined by the constant "a". The greater the absolute value of "a," the steeper the parabola. Thus, determining "a" is crucial to fully describing the shape of the parabola for the task at hand.

Constant calculation

The calculation of the constant "a" is crucial as it affects the exact shape and position of the parabola. In the exercise, we know that the height of the road at the edge, 16 feet from the center, is 0.4 foot lower than at the vertex.

By substituting these given values (\( x = 16, y = -0.4 \)) into the parabola equation \( y = ax^2 \), we can solve for "a": \(-0.4 = a(16^2)\).

Solving for "a", we get \( a = -0.4 / 256 = -0.0015625 \). This determines the curvature of our parabola that models the road surface correctly. Thus, the complete equation becomes \( y = -0.0015625x^2 \).

Solving quadratic equations

A quadratic equation can reveal many practical and useful insights, such as the distance along the road where certain height conditions are met. In this case, we were asked to find the distance from the center where the road surface is 0.1 foot lower than in the middle.

To solve the equation, consider that the middle of the road is 0.4 foot at the vertex. We need to find where the surface is 0.3 foot (\( y = 0.3 \) foot) using our identified parabola equation \( y = -0.0015625x^2 \).

Placing this value in the equation and isolating "x" gives us \( 0.3 = -0.0015625x^2 \). Solving for "x" provides the value \( x \approx 13.86 \) feet. This solution means at approximately 13.86 feet from the center, the road is exactly 0.1 foot lower than the vertex height. Thus, quadratic equations serve as valuable tools for solving real-world geometry problems related to measurements and optics.