the lions gate bridge in vancuver BC, is a suspension bridge that spans 1516m. Large cables are attached to the tops of the towers, 50m above the road. the road is suspended from the large cables with small vertical cables, the smallest one being 2m. find a quadratic equation to model the large cable shape.

Well, you've come to the right place if you're looking for some mathematical amusement! Now, let's get our equation-creating hats on and tackle this cable conundrum.

To model the shape of the large cable, we can start by assuming it follows a parabolic curve. We'll consider the cable to be symmetric, meaning the lowest point of the cable would be in the middle, right at the center of the bridge span.

Let's break down what we know:
- The length of the bridge span is 1516m.
- The towers have cables attached to their tops, which are 50m above the road.
- The lowest point of the cable, the center, is located at a height of 50m.
- We're looking for a quadratic equation to describe this cable shape.

Now, let's assume that the lowest point of the cable is at x = 0, and the road extends on either side symmetrically. In this case, the x-coordinate of the center of the bridge span would be x = 758m, 758m to the right and 758m to the left (1516m / 2).

We can now write the quadratic equation as follows:

y = ax^2 + 50,

where y represents the height of the large cable (above the road) and x represents the distance from the center (in meters). The constant 50 is added to ensure that the lowest point is 50m above the road.

Now, all we need to figure out is the value of a. We can substitute in the coordinates of either tower to solve for a. Let's use the coordinates of one of the towers:

Tower coordinate: (758, 50)

Substituting these values into the equation:

50 = a(758)^2 + 50.

Simplifying:

a(758)^2 = 0,
a = 0 / (758)^2,
a = 0.

Therefore, the quadratic equation to model the large cable shape is y = 0x^2 + 50. But hold on a second, something's off! It turns out that the shape of the cable is not a parabola after all! The cable shape is actually best approximated by a catenary curve, which is a completely different type of curve that arises naturally under the force of gravity.

So, my friend, while we started with a quadratic equation, we discovered that it was quite the "suspense"ion. And now, we have learned that the cable shape is not a parabola, but rather a catenary curve. Mathematics can be full of surprises!

To model the shape of the large cable, we can assume that it follows a quadratic equation form. Let's denote the horizontal distance from one of the tower bases as x (0 ≤ x ≤ 1516), and the vertical distance from the road to the large cable as y.

Given that the large cables are attached to the tops of the towers 50m above the road, we can determine that the cable's highest point will occur at (x, y) = (758, 50).

We can also find the coordinates of another point along the cable. Since the road is suspended 50m above the base of the towers, the vertical distance from the road to the cable is zero at the tower's base. Therefore, we have another point at (x, y) = (0, 0).

Considering these two points, we can calculate the equation of the quadratic:

y = a(x - h)^2 + k

Using the point (758, 50):

50 = a(758 - h)^2 + k ........(1)

Using the point (0, 0):

0 = a(0 - h)^2 + k ............(2)

Simplifying equation (2) gives us:

0 = ah^2 + k

Rearranging equation (1) gives us:

50 - a(758 - h)^2 = k

Substituting equation (2) into (1), we get:

50 - a(758 - h)^2 = ah^2

Expanding both sides of the equation:

50 - a(574564 - 1516h + h^2) = ah^2

Rearranging:

50 - 574564a + 1516ah - ah^2 = ah^2

Simplifying:

2ah^2 - 1516ah + 574564a - 50 = 0

Dividing the equation by 2a gives:

h^2 - (758/a)h + (287282/a - 25/a) = 0

This is a quadratic equation that models the shape of the large cable in terms of the variable h and the constant a.

To model the shape of the large cable, we can use a quadratic equation in the form y = ax^2 + bx + c, where x represents the horizontal distance from one tower to another, and y represents the vertical height of the cable at that point.

Let's break down the problem and find the coefficients for the quadratic equation step by step:

1. The cable is attached to the tops of the towers, which are 50m above the road level. Therefore, we can set the vertex of the parabola at (0, 50), where x is the distance from one tower to another, and y is the height of the cable.

2. The span of the bridge is 1516m, so the distance from one tower to another is half of that, i.e., 1516/2 = 758m.

3. At the midpoint of the bridge (x = 758/2 = 379m), the cable should be at its lowest point. Since the vertex is at (0, 50), the x-coordinate of the lowest point can be calculated as -b/2a. In this case, -b/2a = 379. Therefore, this gives us the first equation: -b/2a = 379.

4. At the towers, the height of the cable is 50m, so we can use the coordinates (379, 0) and (-379, 0) to calculate the additional two points on the parabola.

5. Using the general form of the quadratic equation, we can substitute the known x and y values to form two additional equations. For the point (379, 0), we have the equation 0 = a(379^2) + b(379) + c. For the point (-379, 0), we have the equation 0 = a(-379^2) + b(-379) + c.

6. Now we have a system of three equations with three unknowns (a, b, c):

1. -b/2a = 379
2. 0 = a(379^2) + b(379) + c
3. 0 = a(-379^2) + b(-379) + c

Solving this system of equations should give us the quadratic equation to model the shape of the large cable.

The cables of a suspension bridge do not follow the path of a parabola, but since that is what the question asks for ...

centre is (2,0), one other point is (758,50)
y = a(x-2)^2 + 0 , using the centre
but (758,50) lies on it
50 = a(756)^2
a = 50/574564

y = 25/287282(x-2)^2

The cables of most suspension bridges form a
"catenary" (from the Latin 'catena' for chain)

the general equation is
y = a/2(e^(x/a) + e^(-x/a))