SOLUTION: The roadbed of one section of a suspension bridge is hanging from a large cable suspended between two towers that are 200 feet apart. The cable forms a parabola that is 60 feet ab

Question 237556: The roadbed of one section of a suspension bridge is hanging from a large cable suspended between two towers that are 200 feet apart. The cable forms a parabola that is 60 feet above the roadbed at the towers and 10 feet above the roadbed at the lowest point.

A)Express the vertical distance D(x) (in feet) from the roadbed in to the suspension cable in terms of x and state the domain of D.
B)The roadbed is supported by seven equally spaced vertical cables. Find the combined total length of these supporting cables?
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!

The graph shows the cable between the posts at x = -100 and x = 100

The lines are at y = 10 and y = 60 to show you that the cable meets the specifications at the tower and at the minimum point.

The equation of the graph is y = .005x^2 + 10

The minimum point is at x = 0

The x value extends from x = -100 to x = +100

The road bed is at y = 0.

y = f(x) is equal to the height of the cable from the roadbed.

D(x) = f(x), so:

D(x) = .005x^2 + 10

The domain of D(x) is the value of x from x = -100 to x = +100

The range of D(x) is from 10 to 60.

Values of x and values of y are expressed in feet.

7 equally spaced cables supporting the roadbed would extend from the suspension cable to the roadbed.

These would be equally spaced 25 feet apart.

They would be at:

x = -75
x = -50
x = -25
x = 0
x = 25
x = 50
x = 75

To find the height at these points from the roadbed, solve for D(x) at each of these points.

D(-75) = (.005)*(-75)^2 + 10 = 38.125 feet from the roadbed.
D(-50) = (.005)*(-50)^2 + 10 = 22.5 feet from the roadbed.
D(-25) = (.005)*(-25)^2 + 10 = 13.125 feet from the roadbed.
D(0) = (.005)*(0)^2 + 10 = 10 feet from the roadbed.
D(25) = (.005)*(25)^2 + 10 = 13.125 feet from the roadbed.
D(50) = (.005)*(50)^2 + 10 = 22.5 feet from the roadbed.
D(75) = (.005)*(75)^2 + 10 = 38.125 feet from the roadbed.

Each Tower will be at x = -100 and at x = 100 feet.

D(-100) = (.005)*(-100)^2 + 10 = 60 feet from the roadbed.
D(100) = (.005)*(100)^2 + 10 = 60 feet from the roadbed.

The formula was created as follows:

Standard formula for a quadratic equation is ax^2 + bx + c = 0

The minimum point of a quadratic equation is at x = -b/2a.

Since that had to be at x = 0, that meant that -b/2a was equal to 0 which meant that b had to be equal to 0.

When x = 0, the quadratic equation becomes c = 0 because the a factor drops out since it is being multiplied by x = 0. The b factor was already canceled out because it was equal to 0.

Since the minimum point of the cable had to be 10, that meant that c = 10.

The equation became:

ax^2 + 10 = 0

At x = -100, the height of the cable had to be 60.

That meant that ax^2 + 10 = 60 when x = -100.

equation became:

a*(-100)^2 + 10 = 60

that became:

10,000*a + 10 = 60

Solve for a to get a = 50/10000 = .005

This same value was also applicable at x = 100.

Equation became:

y = .005x^2 + 10

Let D(x) = y and equation became:

D(x) = .005x^2 + 10

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