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The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of a bridge is 1270 ft long and 157 ft high. The parabola y=0.00039x^2 gives a good fit to the shape of the cables, where IxI less than of equal to 635, and x and y are measured in feet. Approximate the length of the cables that stretch between the tops of the two towers.

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Answer:

Step-by-step explanation:

Given

span of bridge [tex]L=1270\ ft[/tex]

height of span [tex]h=157\ ft[/tex]

Equation of Parabola

[tex]y=0.00039x^2[/tex]

[tex]|x|<635[/tex] i.e.

[tex]-635<x<635[/tex]

[tex]\frac{dy}{dx}=2\times 0.00039[/tex]

length of Arc[tex]=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(\frac{dy}{dx})^2}[/tex]

[tex]=\int_{-635}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times \int_{0}^{635}\sqrt{1+(0.00078x)^2}[/tex]

[tex]=2\times (660.08)[/tex]

[tex]=1320.16\ ft[/tex]

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