Problem 101
Question
A cable of the Golden Gate Bridge is suspended (in the shape of a parabola) between two towers that are 1280 meters apart. The top of each tower is 152 meters above the roadway. The cable touches the roadway midway between the towers. (a) Draw a sketch of the cable. Locate the origin of a rectangular coordinate system at the center of the roadway. Label the coordinates of the known points. (b) Write an equation that models the cable. (c) Complete the table by finding the height \(y\) of the suspension cable over the roadway at a distance of \(x\) meters from the center of the bridge. $$\begin{array}{|c|c|c|c|c|c|} \hline x & 0 & 200 & 400 & 500 & 600 \\ \hline y & & & & & \\ \hline \end{array}$$
Step-by-Step Solution
Verified Answer
The equation of the parabola is \( y = -0.00037x^2 \). The heights of the cable at different x-values are \(y=0,-14.8,-59.2,-92.5,-133.2\) respectively for values \(x=0,200,400,500,600\).
1Step 1: Drawing the Sketch and Labelling the Coordinates
Center the origin \(O\) on the roadway between the two towers. Label the highest point of the two towers as \(A\) and \(B\) at \((-640,152)\) and \( (640,152)\) respectively and the lowest point \(C\) where the cable touches the roadway as \((0,0)\).
2Step 2: Derive the Equation of the Parabola
Since the cable is a parabola opening downwards, its equation is of the form \(y = ax^2+ bx + c\). Using that the vertex is at \( C(0,0) \), the equation simplifies to \( y = ax^2 \). Additionally, since point A at \((-640,152)\) lies on the curve, substitute that into the equation, leading to \( 152 = a(-640)^2 \) , allows you to solve for \(a\), which gives \(a = \frac{152}{(-640)^2} = -0.00037\). Thus, the equation of the parabola is \(y = -0.00037x^2 \).
3Step 3: Calculating the Heights
Now that the equation of the parabola is known, substitute the given x-values into the equation to get the corresponding y-values (heights). This gives for \(x=0,200,400,500,600\) the heights \(y=0,-14.8,-59.2,-92.5,-133.2\) respectively.
Key Concepts
Quadratic FunctionsCoordinate GeometryEquations of Parabolas
Quadratic Functions
Quadratic functions are a type of polynomial usually represented in the form \( y = ax^2 + bx + c \).
This function creates a parabola when graphed on a coordinate plane.
The key features of a parabola include its vertex, axis of symmetry, and direction of opening.
In the given problem, the equation simplifies to \( y = ax^2 \) because both the vertex
The vertex being at the origin simplifies the analysis significantly, as it means that the highest or lowest point is directly at \((0,0)\).
The coefficient \( a \) controls the width and direction of the parabola. In this example, the negative \( a \) value shows that the parabola opens downward, which represents the suspension cable dipping downwards.
This function creates a parabola when graphed on a coordinate plane.
The key features of a parabola include its vertex, axis of symmetry, and direction of opening.
In the given problem, the equation simplifies to \( y = ax^2 \) because both the vertex
- (point \( C(0,0) \))
- the axis of symmetry being vertical
The vertex being at the origin simplifies the analysis significantly, as it means that the highest or lowest point is directly at \((0,0)\).
The coefficient \( a \) controls the width and direction of the parabola. In this example, the negative \( a \) value shows that the parabola opens downward, which represents the suspension cable dipping downwards.
Coordinate Geometry
Coordinate geometry, also known as analytic geometry, involves using a coordinate plane to analyze geometric shapes using algebra.
In this problem, the cable of the Golden Gate Bridge is modeled using a parabola, with points marked strategically on a rectangle coordinate system.
The coordinates given are
Understanding these points' locations helps us set up equations and solve for unknowns, simplifying complex structural concepts into more manageable mathematical representations.
Coordinate geometry thus provides a powerful tool for visualizing and solving real-world problems.
In this problem, the cable of the Golden Gate Bridge is modeled using a parabola, with points marked strategically on a rectangle coordinate system.
The coordinates given are
- \((-640, 152)\) and \((640, 152)\) for the tops of the towers.
- \((0, 0)\) where the cable touches the roadway.
Understanding these points' locations helps us set up equations and solve for unknowns, simplifying complex structural concepts into more manageable mathematical representations.
Coordinate geometry thus provides a powerful tool for visualizing and solving real-world problems.
Equations of Parabolas
To form the equation of a parabola, we use points on the parabola to establish relationships between the x and y coordinates.
In our case, the standard form of the equation based on the given information is \( y = ax^2 \).
The equation is derived using the point \((0, 0)\) as the vertex and substituting the point \((-640, 152)\) to solve for \(a\).
Calculation shows that \(a = \frac{152}{(-640)^2} = -0.00037\), illustrating the parabolic path precisely.
This gives the equation \( y = -0.00037x^2 \). Each point on this curve is a solution to this equation, showing the cable's height above the road.
Using such techniques enables precise modeling of physical phenomena, revealing how theoretical math principles manifest in structures like bridges.
In our case, the standard form of the equation based on the given information is \( y = ax^2 \).
The equation is derived using the point \((0, 0)\) as the vertex and substituting the point \((-640, 152)\) to solve for \(a\).
Calculation shows that \(a = \frac{152}{(-640)^2} = -0.00037\), illustrating the parabolic path precisely.
This gives the equation \( y = -0.00037x^2 \). Each point on this curve is a solution to this equation, showing the cable's height above the road.
Using such techniques enables precise modeling of physical phenomena, revealing how theoretical math principles manifest in structures like bridges.
Other exercises in this chapter
Problem 99
Use the Law of sines or the Law of cosines to solve the triangle. $$B=71^{\circ}, a=21, c=29$$
View solution Problem 99
Water is flowing from a horizontal pipe 48 feet above the ground. The falling stream of water has the shape of a parabola whose vertex (0,48) is at the end of t
View solution Problem 102
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
View solution Problem 104
A satellite in a 100 -mile-high circular orbit around Earth has a velocity of approximately 17,500 miles per hour. When this velocity is multiplied by \(\sqrt{2
View solution