Re: Parabola -- Mathematics Help

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Subject: Re: Parabola

Author:
QUITTNER

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Date Posted: 08:17:07 06/10/05 Fri
In reply to: FB 's message, "Parabola" on 15:39:19 06/06/05 Mon

>>> A masonry bridge over a river has an arch in the form of a parabola with a vertical axis. The widths of the arch at water level are 5m above water level and 12m and 8m respectively. Find the height of the arch above the point at water level which is 2m from the middle. <<<
..... A shortened form of the equation with the origin (x=0 and y=0) at the top of the arch is
y = p * x^2 with all y being negative.
..... Put y1 at x=2, put y2 at x=4, and y3 at x=6. The x of y2 and y3 are half the total widths.
..... We also know that (y2 - y3) = 5 (IS THAT RIGHT?) where y3 is the water level.
..... Insert into the equations y=p*x^2 where * means times, and ^2 means squared:
y3=6*6*p=36p, y2=16p, and y2-y3=5. Solve for p and you then can get y1, y2 and y3. The required height above water level is y1-y3 where x=2.

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