Asked by Marky
An arc in the shape of parabola measures 6 m across the base and its vertex is 2.5 m above the base. determine he lenght of the beam parallel to the base and 2m above it.
Answers
Answered by
Steve
if we set the center of the base at (0,0) then the parabola is modeled by the equation
y = 5/2 - 5/18 x^2
Solve for x when y=2, and the beam has length 2x.
y = 5/2 - 5/18 x^2
Solve for x when y=2, and the beam has length 2x.
Answered by
Anonymous
use this formula to solve the x when y=h(1-x^2/a^2)
answer = 2.68 m
answer = 2.68 m
Answered by
Je
Use squared property of parabola
(x sub 1)^2 / (y sub 1)^2 = (x sub 2)^2 / (y sub 2)^2
where xsub1 = 3, ysub1 = 2.5, ysub2 = 2
Not sure if xsub2 = length of beam
or xsub2 = length of beam ÷ 2
But if xsub2 = length of beam, xsub2 = 2.68m (which is the answer in the answer key)
(x sub 1)^2 / (y sub 1)^2 = (x sub 2)^2 / (y sub 2)^2
where xsub1 = 3, ysub1 = 2.5, ysub2 = 2
Not sure if xsub2 = length of beam
or xsub2 = length of beam ÷ 2
But if xsub2 = length of beam, xsub2 = 2.68m (which is the answer in the answer key)
Answered by
Je
CORRECTION!! Let vertex be at (0,0)
Using squared property of parabola
(x sub 1)^2 / (y sub 1)^2 = (x sub 2)^2 / (y sub 2)^2
where x and y are measured from the vertex
xsub1 = 6m, ysub1 = 2.5m, ysub2 = 0.5m, and
xsub2 = length of beam 2 m from the base
(6)^2 / 2.5 = (x sub 2)^2 / 0.5
xsub2 = 2.68
Therefore, length of beam 2 m from the base = 2.68m
Using squared property of parabola
(x sub 1)^2 / (y sub 1)^2 = (x sub 2)^2 / (y sub 2)^2
where x and y are measured from the vertex
xsub1 = 6m, ysub1 = 2.5m, ysub2 = 0.5m, and
xsub2 = length of beam 2 m from the base
(6)^2 / 2.5 = (x sub 2)^2 / 0.5
xsub2 = 2.68
Therefore, length of beam 2 m from the base = 2.68m
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