Problem 84 - Calculus Early Transcendentals

Question

Completed in 1937, San Francisco's Golden Gate Bridge is \(2.7 \mathrm{km}\) long and weighs about 890,000 tons. The length of the span between the two central towers is \(1280 \mathrm{m}\) the towers themselves extend \(152 \mathrm{m}\) above the roadway. The cables that support the deck of the bridge between the two towers hang in a parabola (see figure). Assuming the origin is midway between the towers on the deck of the bridge, find an equation that describes the cables. How long is a guy wire that hangs vertically from the cables to the roadway \(500 \mathrm{m}\) from the center of the bridge?

Step-by-Step Solution

Answer

Answer: The length of the guy wire is approximately 109.2 meters.

Step 1: 1. Set up given information and coordinates on the coordinate plane

Let's use a coordinate plane to represent the given information. Set the origin to be the midpoint between the two central towers. Let's place the two central towers at \((-640,0)\) and \((640,0)\). Notice that the distance between the 2 towers is \(1280 \mathrm{m}\), exactly as given in the problem.

Step 2: 2. Identify key points on the parabola

Let's denote \(P=(-640,0)\) and \(Q=(640,0)\), as these are the two points where the cables meet the bridge deck on either side of the central towers. There's also the highest point of the cable above the mid-point of the bridge. Let's call this the vertex and denote it as \(R=(0,152)\).

Step 3: 3. Determine the equation for the parabola

Since the parabola is symmetric around the x-axis and has a vertex at \((0,152)\), we can use the vertex form of a parabola equation: \(y=a(x-h)^{2}+k\), where \((h,k)\) is the vertex. In this case, \(h=0\) and \(k=152\), so the equation can be simplified to \(y=ax^{2}+152\). We now need to find the value of \(a\). We can use either point \(P\) or \(Q\). Let's use point \(P=(-640,0)\). Plug this into the equation: \(0=(-640)^{2}a+152\).

Step 4: 4. Solve for the value of a

From the equation in step 3, we get \(0=640^{2}a+152\). Solving for \(a\), we have \(a=\frac{-152}{640^{2}}=-\frac{19}{51200}\).

Step 5: 5. The equation of the parabola representing the cables

Revising the equation from step 3 using the value of \(a\) found in step 4, we obtain the final equation: \(y=-\frac{19}{51200}x^{2}+152\).

Step 6: 6. Determine the length of the guy wire

At \(x=500 \mathrm{m}\) from the center of the bridge, we can find the height of the cable above the roadway by plugging \(x\) into the equation: \(y= -\frac{19}{51200}(500)^{2}+152 \approx 109.2 \mathrm{m}\) Now we can use the distance formula to find the length of the guy wire that hangs vertically from the cables to the roadway. Let the point where the guy wire meets the cable be denoted as \(S\), and the point where it meets the roadway be denoted as \(T\). Coordinates for points \(S\) and \(T\) are \((500,109.2)\) and \((500,0)\), respectively. The distance formula is given by \(d=\sqrt{(x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2}}\). Plugging in the coordinates of points \(S\) and \(T\) into the formula, we have: \(d = \sqrt{(500-500)^{2} + (109.2-0)^{2}}=\sqrt{(0)^{2} + (109.2)^{2}}=109.2 \mathrm{m}\). The length of the guy wire that hangs vertically from the cables to the roadway \(500 \mathrm{m}\) from the center of the bridge is approximately \(109.2 \mathrm{m}\).

Key Concepts

Vertex Form

The vertex form of a parabola is very handy when dealing with equations related to parabolic structures, such as the cables of a suspension bridge. This form is represented as:

• \( y = a(x-h)^2 + k \)

Here,

• \((h, k)\) represents the vertex of the parabola, which is its highest or lowest point depending on the direction in which the parabola opens.
• \(a\) determines the "width" and "direction" of the parabola. Positive \(a\) opens upwards, while negative \(a\) opens downwards.

In our problem, the vertex of the parabola is at the highest point of the cable, dramatically aiding in constructing the equation. The vertex is at \((0, 152)\) given by the problem, leading us to the equation \(y = ax^2 + 152\), tasked with finding \(a\). Thus, through the vertex form, we efficiently describe how the cable's height varies along the length of the bridge.

Symmetry

Parabolas are typically symmetric around their vertical axis, termed the axis of symmetry. This line runs through the vertex, acting as a mirror that divides the parabola into two identical halves.

• This symmetry helps immensely when simplifying calculations and graph construction, as only one side needs to be analyzed before reflecting across the axis.
• In our case, setting the coordinate system with the origin halfway between the towers helps utilize this symmetry to find key points and eventually derive the equation.

The origin, being at \(x = 0\), ensures the equation \(y = ax^2 + 152\) maintains this symmetry by having \((0, 152)\) as its peak at the middle of the bridge. The computed parabola naturally mirrors itself around this axis, signifying a precise calculation tool for both mathematical and engineering solutions.

Coordinate Geometry

Coordinate Geometry empowers us to analyze geometric figures algebraically by representing them in a coordinate system.

• Here, the towers are on the x-axis at \((-640, 0)\) and \((640, 0)\), and the parabolic cable arc lies between them, symmetrically unfolding around the origin.
• This placement illustrates not just the position but also integrates geometric features with numeric information, crucial for solving such engineering-based problems.

Connecting geometry with algebra, we can say that the coordinates of key points like the towers and vertex become the bedrock for constructing the parabola’s equation. The knowledge of this union of geometry and algebra permits one to visually grasp the solution by plotting the points and understanding the symmetrical nature of the resulting parabola.

Distance Formula

The distance formula helps find the length of a straight line segment between two points in a coordinate plane. It's expressed as:

• \( d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2} \)

Applying it within our problem,

• The guy wire descends from the cable down to the roadway, with endpoints at \((500, 109.2)\) and \((500, 0)\).
• We calculate the distance or length of this guy wire, which reflects directly between these points as \(109.2\) meters.

The formula, simple yet powerful, reveals not just the functional attributes of geometry on paper but helps in physical calculations essential to engineering tasks. It confirms the importance of mastering basic mathematical tools for solving real-world problems proficiently.