Problem 48 - Algebra 1: Concepts and Skills

Question

The Gateway Arch in St. Louis, Missouri, has the shape of a catenary (a U-shaped curve similar to a parabola). It can be approximated by the following model, where x and y are measured in feet. Gateway Arch model: \(y=-\frac{7}{1000}(x+300)(x-300)\) How far apart are the legs of the arch at the base?

Step-by-Step Solution

Answer

The legs of the Gateway Arch in St. Louis, Missouri, are 600 feet apart at the base.

Step 1: Identify the height of the arch at its base

Setting \( y = 0 \), meaning the height of the arch, the equation becomes: \( 0=-\frac{7}{1000}(x+300)(x-300) \) In order for this equation to be true, either \( (x+300) = 0 \) or \( (x-300) = 0 \), because any number multiplied by zero equals zero.

Step 2: Solve for x to find the distances from the center

Solving \( (x+300) = 0 \) for \( x \) gives: \( x = -300 \) Solving \( (x-300) = 0 \) for \( x \) gives: \( x = 300 \) So one leg of the arch is 300 feet to the left of the center (at \( x = -300 \)) and the other is 300 feet to the right of the center (at \( x = 300 \)).

Step 3: Find the total distance between the arch legs at the base

The total distance from one leg to the other is the sum of the distances from the center to either leg. \( Distance = x_{right} - x_{left} = 300 - (-300) = 600 \) feet. So the legs of the arch are 600 feet apart at the base.

Key Concepts

Quadratic Equation

A quadratic equation is an equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants, and \( a \) is not zero. It is called "quadratic" because "quad" means square, indicating the highest power of the variable \( x \) is 2. The general solution to a quadratic equation is obtained using the quadratic formula: \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]For any quadratic, the solutions are the values of \( x \) that satisfy the equation.

• The solutions can be real or complex depending on the value of the discriminant \( b^2 - 4ac \).
• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root, meaning the curve touches the x-axis at one point.
• If the discriminant is negative, there are no real roots, and the solutions are complex.

In the context of the Gateway Arch, we use a slightly adapted form of the quadratic equation, set to zero to find the 'roots', or the points where the curve intersects the x-axis. These points indicate the base distances of the arch legs, calculated by finding the solutions \(-300\) and \(300\).

Graphing Parabolas

Graphing a parabola involves plotting points that satisfy the quadratic equation, giving us a U-shaped curve. This curve can open upwards or downwards, depending on the sign of the coefficient \( a \) from the quadratic equation form \( y = ax^2 + bx + c \).

• If \( a > 0 \), the parabola opens upwards.
• If \( a < 0 \), it opens downwards.

The vertex of the parabola, a key feature, is its highest or lowest point. For this vertex, the \( x \)-coordinate can be found using the formula \( x = -\frac{b}{2a} \).
In the case of the Gateway Arch, the equation \( y=-\frac{7}{1000}(x+300)(x-300) \) represents a downward-opening parabola because the leading coefficient, \(-\frac{7}{1000}\), is negative. From the equation, we identify the x-intercepts directly as \( x = -300 \) and \( x = 300 \), marking where the parabola crosses the x-axis, or the points where the arch touches the ground.The shape of the curve very closely resembles that of a catenary, a specific type of arch structure.

Distance Calculation

Distance calculation between two points is a useful tool in geometry, often applied in various contexts. The simplest scenario on a line is calculating the distance between two points, \( x_1 \) and \( x_2 \), as \( \text{Distance} = |x_2 - x_1| \).
This key concept was used in the analysis of the Gateway Arch, where the x-intercepts of \(-300\) and \(300\) tell us the position of the arch's legs.

• The distance between these two x-intercepts represents the span of the arch at its base.
• Calculating \( 300 - (-300) \) results in a 600-foot span.

Visually, this is the length horizontally from one base point to the other, ensuring that we account correctly for the total span by summing both the positive and negative side of the central axis.
In real-world applications, understanding such distances is crucial, as accurately measuring architectural structures is necessary for both structural integrity and aesthetic value.