SOLUTION: A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabo
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Question 187809: A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart. The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. Find the height of the cables at a point 100 meters from the center. (Assume that the road is level.)
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
A suspension bridge with weight uniformly distributed along its length has twin towers that extend 75 meters above the road surface and are 400 meters apart.
You have points (-200,75),(0,0),(200,75
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The cables are parabolic in shape and are suspended from the tops of the towers. The cables touch the road surface at the center of the bridge. (Assume that the road is level.)
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Use the three points in the equation f(x) = ax^2 + bx + c to generate three
equations to solve for a,b, and c.
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(-200,75): a(-200)^2 + b(-200) + c = 75
(0,0)....: a(0)^2 + b(0) + c = 0
(200,75).: a(200)^2 + b(200) + c = 75
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Rearrange after noticing that c=0 in the 2nd equation:
200^2a -200b = 75
200^2a +200b = 75
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Subtract 1st Eq. from 2nd to get:
400b = 0
b = 0
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Substitutute into 200^2a -200b = 75 to solve for "a":
200^2a = 75
a = 75/(200^2)
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Equation of the parabola: f(x) = [75/200^2]x^2
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Then f(100) = [75/200^2][100^2] = 18.75 meters
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Cheers,
Stan H.
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