SOLUTION: A cable wire was tied to both ends of a 240 meters long bridge such that the cable hangs in the form of a parabola. The lowest point of the cable is 60 meters below the bridge.

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Question 1043488: A cable wire was tied to both ends of a 240 meters long bridge such that the cable hangs in the form of a parabola. The lowest point of the cable is 60 meters below the bridge.
a.) Find the equation of the cable wire, assuming that the origin is the lowest point of the cable.
b.) Using the equation, find the width of the rope at a point 38 meters above its lowest point.
c.) suppose a 180 neters rope will be tied horizontally to the cable wire. How high above the lower point of the cable wire should the rope be?
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
A cable wire was tied to both ends of a 240 meters long bridge such that the cable hangs in the form of a parabola. The lowest point of the cable is 60 meters below the bridge.
a.) Find the equation of the cable wire, assuming that the origin is the lowest point of the cable.
Use the form ax^2 + bx + c = y; lowest point is at origin; c = 0
Two given points: x= -120; y = 60 and x = +120; y = 60; two equations
-120^2a - 120b + 0 = 60
+120x^2 + 120b + 0 = 60
Use elimination
14400a - 120b = 60
14400a + 120b = 60
-------------------Addition eliminates b, find a
28800a = 120
a = 120/28800
a = .004167
:
Find b using 120^2(.001467) + 120b = 60
60 + 120b = 60
b = 0
The equation: y = .004167x^2
looks like this, green line is the bridge
+graph%28+300%2C+200%2C+-150%2C+150%2C+-20%2C+80%2C+.004167x%5E2%2C+60%29+
:
b.) Using the equation, find the width of the rope at a point 38 meters above its lowest point.
+graph%28+300%2C+200%2C+-150%2C+150%2C+-20%2C+80%2C+.004167x%5E2%2C+60%2C+38%29+, blue line is 38 ft above the min
//////////////////Sent this before I was done, the rest of the problem://///////
Calculate, find x when y=38
.004167x^2 = 38
x^2 = 38/.004167
x^2 = 9119.3
x = sqrt%289119.3%29
x = +95.5
and
x = -91.5
the length of the rope then, 2(91.5) = 191 ft
:
c.) suppose a 180 meters rope will be tied horizontally to the cable wire. How high above the lower point of the cable wire should the rope be?
x will be half of 180, +/-90
y = .004167(90^2)
y = 33.75
+graph%28+300%2C+200%2C+-150%2C+150%2C+-20%2C+80%2C+.004167x%5E2%2C+60%2C+33.75%29+; blue line is 33.75 ft above min