Clearance Through a Parabolic Road Arch
A road passes through a parabolic arch. The arch is 40 m wide at road level and rises to a maximum height of 12 m at its centre. An engineer models the arch in a coordinate system with the origin at the centre of the road, the x-axis along the road, and the y-axis vertical. The arch is symmetric about the y-axis. A transport company wants to move a rectangular load 8 m high through the arch. A second company proposes a load 10 m high and 18 m wide. The engineer must find the equation of the parabola, determine the maximum width of the 8 m high load that can pass centrally, and check the second proposal. The road is level, and the load must clear the arch at both top corners.
1. Derive the equation of the parabolic arch in the given coordinate system.
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The vertex is at (0,12), so the equation has the form y=a x²+12. At road level the arch meets the road at x=±20, where y=0. Using x=20 gives 0=a(400)+12, so a=−12/400=−3/100. Therefore y=12−(3/100)x². Check: x=0 gives y=12 m, and x=20 gives y=0 m.
2. Find the maximum width of a rectangular load 8 m high that can pass centrally under the arch.
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The load's top corners are at y=8 and x=±w/2. Set 8=12−(3/100)x², so (3/100)x²=4, x²=400/3, and x≈11.55 m. The width is 2x≈23.1 m. A wider load would have top corners above the arch; a narrower one clears. Thus the maximum central width is about 23.1 m, or about 23 m in practice.
3. A second company proposes a load 10 m high and 18 m wide. Check whether it clears the arch, and state the limiting point.
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The half-width is 9 m, so the critical top corners are at x=±9. The arch height there is y=12−(3/100)(81)=12−2.43=9.57 m. The load is 10 m high, so it exceeds the arch by 10−9.57=0.43 m at each top corner. It cannot pass centrally. The limiting points are the outer top corners, not the centre, because the arch is highest at the centre.