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March 31, 2015

March 31, 2015

Posted by **reneet** on Monday, June 2, 2014 at 7:14pm.

On the figure below, make a rough sketch of what you think the arch with the ramp would look like.

- algebra -
**Steve**, Monday, June 2, 2014 at 10:21pmIf the vertex of the parabola is at (0,630), then the arch is modeled by

y = (630/315^2)(315^2-x^2)

= 630 - 2/315 x^2

If the work is being done at a position c ft to the right of the vertex, it will be at the point (c,y(c)), on a line with slope 1/5.

Unfortunately, this makes for some very long ramps. If they're working at a point at height 376 ft (x=200), the ramp has to start out 1365 ft to the left of the arch. That's about 1/4 mile away!

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