Pre- Calculus. Help me please.
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Pre- Calculus. Help me please.
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Problem:
Parabolic cable of a 60m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical cables are to be spaced every 6m long along this portion of the roadbed. Calculate the lengths of first two of these vertical cables from the vertex.
Given:
The problem states that it is a parabolic cable
parabola has an equation of (x - h)^2 = 4a(y - k)
width from center to the 5th vertical cable = 60m / 2 = 30
height = 16 m
for every vertical cable, the spacing is 30/5 = 6m
Solution:
solve for a
Vertex at the center (0, 3)
coordinates for the width and the height (30, 16)
(30 - 0)^2 = 4a (16 - 3)
30^2 = 4a (16 - 3)
900 = 4a(13)
900/52 = a
solve for y1 as the first vertical cable
coordinates (6, y1) ---- from vertex to the first vertical cable
x^2 = 4a (y1 - 3)
x^2 = 4(900/52)(y1 - 3)
x^2 = (3600/52)(y1 - 3)
(6^2)(52) = 3600(y1-3)
36(52) = 3600y - 10800
1872 + 10800 = 3600y
12672 = 3600y
y1 = 3.52 m
solve for y2 as the second vertical cable
coordinates (12, y2) ---- from vertex to the 2nd vertical cable
x^2 = 4a (y2 - 3)
12^2 = 4(900/52)(y2 - 3)
144 = (3600/52)(y2 - 3)
144(52) = 3600y2 - 10800
7488 + 10800 = 3600y2
18288 = 3600y2
y2 = 5.08m
Answer:
The lengths of the first two vertical cables are 3.52m and 5.08m
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