If the clearance is 15 ft at the edges of the pavement, then we have
y = 25-ax^2
y(30) = 15
so now we know that
25 - 900a = 15
a = 1/90
now, knowing that
y = 25 - 1/90 x^2, we have
y=0 at x = 15√10 ≈ 47.43
so the arch is 94.86 ft wide at its base
I assume you are not really after the actual length of the curve, since that requires calculus.
if the parkway is 60 feet wide, the minimum clearance is 15 feet and the parabolic arch is 25 feet high, how long is the arch?
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