Question

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.

Short answer

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

Step-by-step solution

Step 1. Given information.

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

$$\begin{gathered} a=\frac{30}{2} \\ a=15 \end{gathered}$$

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 $(0,-12)$.

Step 2. The equation of the ellipse and graph of the arch.

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

The graph of the arch is

Step 3. Vertical distance from roadway to arch.

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

role="math" localid="1646806151750" $$\begin{gathered} \frac{25}{225}+\frac{{(y+12)}^2}{100}=1 \\ {(y+12)}^2=\frac{800}{9}y=\pm \frac{20\sqrt{2}}{3}-12 \\ y=-2.57,-21.42 \end{gathered}$$

We have considered only the upper semi- ellipse constituting the arch.

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

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

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$12feet$.