Asked by Anonymous
A manhole is 12 ft from the centerline of a 30-ft wide street that has a 6-in. parabolic crown. The street center at the station of the manhole is at elevation 612.58 ft. What is the elevation of the manhole cover?
Answers
Answered by
MathMate
We have a parabola, and the crown is 6 inches over a 30-foot width, i.e. 15' on each side of the crown.
The elevation of the crown is 612.58, and on each side of the road the elevation is 612.58-0.5'=612.08.
Taking x=0 at the crown, the equation of a parabola with the crown (vertex) at (0,612.58) is
y=f(x)=-(0.5/225)x^2+612.58
by fitting a parabola through these three points.
(0,612.58), (-15,612.08), (15,612.08)
So the elevation (of the centre) of the manhole can be evaluated at x=12, or f(12).
The elevation of the crown is 612.58, and on each side of the road the elevation is 612.58-0.5'=612.08.
Taking x=0 at the crown, the equation of a parabola with the crown (vertex) at (0,612.58) is
y=f(x)=-(0.5/225)x^2+612.58
by fitting a parabola through these three points.
(0,612.58), (-15,612.08), (15,612.08)
So the elevation (of the centre) of the manhole can be evaluated at x=12, or f(12).
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