The Lions Gate bridge in Vancouver is a suspension bridge. The main span between the two towers, is 427m long. Large cables are attached to the ends of the both towers, 50m above the road. Each large cable forms a parabola. The road is suspended from the large cables by the series of shorter vertical cables. The shortest vertical cable measures 2m. Find the quadratic function that models one of the large cables?
Suppose we draw the cables on an x-y axis with the road as the x-axis.
and the posts at (-213.5,0) and (213.5,0) , with their tops at (-213.5,50) and (213.5, 50)
That would make (0,20) the vertex of the parabola
the equation using y = a(x-p)^2 + q , where (p,q) is the vertex, would be
y = a(x-0)^2 + 20
but (213.5, 0) lies on it
0 = a(213.5)^2 + 20
a = -20/213.5^2 = appr - .000439
a model equation could be
y = -.000439x^2 + 20
BTW, having crossed that bridge several times, I would have noticed that the suspended road is actually curved, and the cables are not a parabolas but rather a "Catenary"
To find the quadratic function that models one of the large cables, we need to determine the vertex form of the parabola represented by the large cable.
Given:
- Main span length = 427m
- Height of the towers above the road = 50m
- Shortest vertical cable length = 2m
Let's assume the large cable's vertex is at the origin (0, 0). The x-coordinate of the vertex represents the midpoint of the main span.
Since the main span is 427m long, the x-coordinate of the vertex is halfway, which is x = 427/2 = 213.5m.
At the vertex (x, y) = (213.5, 0), the equation of the parabola is y = a(x - h)^2 + k, where (h, k) represents the vertex.
Plugging in the values, we have:
0 = a(213.5 - 213.5)^2 + 0
Simplifying, we get:
0 = a(0) + 0
0 = 0
Since the y-coordinate of the vertex is 0, the equation simplifies to:
y = a(x - 213.5)^2
To find the value of 'a', we need another point on the parabola. Let's consider the point (0, 50), which represents the attachment point of the large cable to the tower.
Plugging in the values, we have:
50 = a(0 - 213.5)^2
Simplifying further, we get:
50 = a(213.5^2)
Divide both sides by 213.5^2:
a = 50 / (213.5^2)
Calculating this value, we find:
a ≈ 0.001162
So, the quadratic function that models one of the large cables is:
y = 0.001162(x - 213.5)^2
To find the quadratic function that models one of the large cables, let's start by determining the coordinates of the vertex of the parabolic curve. The vertex represents the highest point on the cable.
Given:
- The main span between the two towers is 427m long.
- The large cables are attached to the ends of both towers, 50m above the road.
- Each large cable forms a parabola.
Since the road is suspended from the large cables, the vertex of the parabolic curve will be directly above the midpoint of the span, which is at a distance of half the main span length.
The coordinates of the vertex are (h, k), where:
- h represents the horizontal distance from the left tower to the vertex.
- k represents the vertical distance from the road to the vertex.
We know that the main span is 427m long, so the distance from each tower to the midpoint is 427/2 = 213.5m.
Given that the large cables are attached 50m above the road, the vertical distance from the road to the vertex is k = 50m.
Therefore, the coordinates of the vertex are (h, 50).
Now let's determine the equation of the parabolic function in vertex form, which is given by:
y = a(x - h)^2 + k
Substituting the vertex coordinates into the equation, we get:
y = a(x - h)^2 + 50
Now we need to find the value of 'a', which represents the vertical stretch or compression factor of the parabola.
To do this, we can use one of the given lengths (the shortest vertical cable) as a reference.
Given that the shortest vertical cable measures 2m, we can find the value of 'a' by substituting the coordinates of any point on the parabola into the equation and solving for 'a'.
Let's choose the coordinates of the left tower (0, 0) as a reference point. Substituting these values into the equation, we have:
0 = a(0 - h)^2 + 50
0 = ah^2 + 50
Now, substitute the coordinates of the right tower (213.5, 0) into the equation:
0 = a(213.5 - h)^2 + 50
0 = a(213.5^2 - 427h + h^2) + 50
0 = a(45592.25 - 427h + h^2) + 50
Simplify the equation and combine like terms:
0 = 45592.25a - 427ah + ah^2 + 50
To solve for 'a', we have a quadratic equation in 'h'. Rearrange it in standard form:
ah^2 - 427ah + 45592.25a + 50 = 0
Now we have the quadratic function that models one of the large cables, represented by the equation:
y = a(x - h)^2 + 50
However, the specific value of 'a' depends on the height of the cable above the road and will vary. Therefore, without the height information, we cannot determine the exact quadratic function.