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 17469: 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 the road is level.? Please help.
Answer by rapaljer(4671) About Me  (Show Source):
You can put this solution on YOUR website!
If the road is suspended from towers whose cables are parabolic in shape, then picture a graph in which the road surface is the x-axis, and the point (0,0) represents the point that is on the road surface midway between the two towers. Draw a parabola whose vertex is at (0,0), and it curves upward to two points, one on either side of vertex at a distance x= 200 or x= -200, and y for each of these points is 75.

In other words we need a parabola whose equation is in the form y+=a+x%5E2, that passes through the point (200, 75). Substitute the values of x and y into the given formula and solve for a.

y+=+ax%5E2
75+=+a%2A200%5E2
75%2F40000+=+a+
3%2F1600+=+a

The equation for the parabolic cable is now y+=+%283x%5E2%29%2F1600

The question is to find the height of the cable when x= 100. In other words find y when x=100.
y=+%283%2A100%5E2%29%2F1600= 75%2F4 or 18.75 meters.

The graph may serve as a partial check to see if it looks reasonable! Even better than this, I also graphed this equation on a graphing calculator, and using the TRACE function, I was able to check the value of x = 200, and sure enough, I got y = 75. Then I also checked the value of x= 100, and I got 18.75 EXACTLY!!! How about that for a check??!!
graph+%28600%2C300%2C+-250%2C+250%2C+-10%2C+75%2C+%283x%5E2%29%2F1600%29+

R^2 at SCC