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Q 70.

Expert-verifiedFound in: Page 655

Book edition
6th

Author(s)
Sullivan

Pages
1200 pages

ISBN
9780321795465

Semi elliptical Arch Bridge The arch of a bridge is a semi ellipse with a horizontal major axis. The span is $30$ feet, and the top of the arch is $10$ feet above the major axis. The roadway is horizontal and is $2$ feet above the top of the arch. Find the vertical distance from the roadway to the arch at $5$-foot intervals along the roadway.

The vertical distance from the roadway to arch at $5$ foot intervals are $2.57$feet ,$4.54$ feet and $12$ feet.

The major axis of the arch is parallel to the x-axis. since the span of the arch is $30$feet, this means

$a=\frac{30}{2}\phantom{\rule{0ex}{0ex}}a=15$

The top of the arch is $10$ feet above the major axis.This means $b=10$.

The road ways lies $2$ feet above top of the arch.Thus, the coordinates of the center are $(0,-12)$.

The equation of the ellipse is $\frac{{(x-0)}^{2}}{225}+\frac{{(y+12)}^{2}}{100}=1$

The graph of the arch is

Now, putting $x=5$ in the equation of ellipse

role="math" localid="1646806151750" $\frac{25}{225}+\frac{{(y+12)}^{2}}{100}=1\phantom{\rule{0ex}{0ex}}{(y+12)}^{2}=\frac{800}{9}y=\pm \frac{20\sqrt{2}}{3}-12\phantom{\rule{0ex}{0ex}}y=-2.57,-21.42$We have considered only the upper semi- ellipse constituting the arch.

Putting $x=10$ in the equation we get.

role="math" localid="1646806607812" $\frac{100}{225}+\frac{{(y+12)}^{2}}{100}=1\phantom{\rule{0ex}{0ex}}{(y+12)}^{2}=\frac{500}{9}\phantom{\rule{0ex}{0ex}}y=\pm \frac{10\sqrt{5}}{3}-12\phantom{\rule{0ex}{0ex}}y=-4.54$

For $x=-15$ and $x=15$ $y=-12$

Thus, the vertical distances from the roadway to arch at $5$ foot intervals are $2.57feet,4.54feet$and $12feet$.

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