Parabolic problem?

Question

The cables of a suspension bridge are 50 ft above the roadbed near the towers of the bridge and 10 ft above it in the center of the bridge. The roadbed is 200 ft longand vertical cables are spaced every 20 ft along it. Calculate the lengths of these vertical cables.

Answer 1

We want the equation for a parabola that does what you describe . . . put the middle of the bridge at the origin and let the length of the bridge be from x=-100ft to x=100ft. At these points there are towers that are 50ft high. In the middle, where x=0, h=10ft, then our equation is of the form

h=(something)*x^2 + 10ft (if we're saying it's a parabola)

we know 50ft=(something)(100ft)^2 +10ft

so .... something = 40/(100^2ft) = .004/ft so the height along the bridge is

h = .004x^2/ft +10ft

Too bad you let me do the fun part. Now go along the bridge every 20ft, from -100, -80, . . . 100 calculate the heights, add them, and then multiply by two to account for both sides of the bridge.

Answer 2

Assuming the cables form a parabola (which is dubious - I would think closer to a catenary)
f(x) = ax^2 + b is the equation of a parabola (symmetric about the y-axis for convenience.)
Since f(0) = 10, we know b=10
f(x) = ax^2 + 10
Since f(100) = 10000a + 10 = 50
10000a = 40
a = 1/250

f(x) = x^2 / 250 + 10

There are 11 vertical cables:
f(0) = 10
f(20) = f(-20) = 400/250 + 10 = 11.6
f(±40) = 1600/250 + 10 = 16.4
f(±60) = 3600/250 + 10 = 24.4
f(±80) = 6400/250 + 10 = 35.6
f(±100) = 50