SOLUTION: A suspension bridge with weight uniformly distributed along its length has twin towers that extend 50 meters above the road surface and are 400 meters apart. The cables are parabol

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Question 1069328: A suspension bridge with weight uniformly distributed along its length has twin towers that extend 50 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 the road is level)
Answer by ankor@dixie-net.com(22740) About Me  (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 50 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 the road is level)
:
find the quadratic equation y = ax^2 + bx + c of the bridge.
x=0 is the center of the bridge,
the towers 400ft apart, will be x=-200 and x=+200
Minimum occurs at x=0, y=0 therefore c = 0
x=-200, y = 50
(-200^2)a - 200b = 50
40000a - 200b = 50
and
x=+200, y = 50
(200^2)a + 200b = 50
40000a + 200b = 50
Use elimination
40000a - 200b = 50
40000a + 200b = 50
---------------------addition eliminates b, find a
80000a = 100
a = 100/80000
a = .00125
the equation: y = .00125x^2, looks like this, green line is 50 ft, x=+/-200
+graph%28+300%2C+200%2C+-250%2C+250%2C+-10%2C+70%2C+.00125x%5E2%2C+50%29+
:
"Find the height of The cables at a point 100 meters from the center."
x = 100
y = .0125(100^2)
y = 12.5 meters (green line)
+graph%28+300%2C+200%2C+-250%2C+250%2C+-10%2C+70%2C+.00125x%5E2%2C+12.5%29+