Chapter 10, Problem 79RE

To determine

Find the height of the arch.

Answer to Problem 79RE

At $5$ feet, $\boxed{19.72}$ feet

At $10$ feet, $\boxed{18.86}$ feet

At $20$ feet, $\boxed{14.91}$ feet

Explanation of Solution

Given information:

A bridge is built in the shape of a semielliptical arch. The bridge has a span of $60$ feet and a maximum height of $20$ feet. Find the height of the arch at distances of $5,$

  $10$ and $20$ feet from the center.

Calculation:

The span of the bridge is given as $60$ feet. So, the length of the major axis is $2b=60$ .

Now, divide both sides of the equation by $2$

  $\frac{2b}{2}=\frac{60}{2}$

  $b=30$

Thus, the maximum height of the bridge is $20$ feet, which is half the length of the minor axis denoted by $a$ . So, $a=20$ .

Take the center of the ellipse as $\left(0,0\right)$ . The general form of the equation of the ellipse is given by $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ .

Now, substitute $20$ for $a$ and $30$ for $b$ in the equation $\frac{x^2}{{\left(30\right)}^2}+\frac{y^2}{{\left(20\right)}^2}=1$

  $\frac{x^2}{900}+\frac{y^2}{400}=1$

To find the height of the arch at a distance of $5$ feet from the center. Substitute $5$ for $x$ in the equation $\frac{x^2}{900}+\frac{y^2}{400}=1$ and find the corresponding value of y.

  $\frac{5^2}{900}+\frac{y^2}{400}=1$

  $\frac{25}{900}+\frac{y^2}{400}=1$

Now, subtract $\frac{25}{900}$ from both sides of the equation,

  $\frac{25}{900}+\frac{y^2}{400}-\frac{25}{900}=1-\frac{25}{900}$

  $\frac{y^2}{400}=0.972$

Multiply both sides by $400$ ,

  $\frac{y^2}{400}\times 400=0.972\times 400$

  $y^2=388.8$

  $y=\sqrt{388.8}$

  $y=19.72$

Hence, the height of the arch at $5$ feet from the center is $\boxed{19.72}$ feet.

To find the height of the arch at a distance of $10$ feet from the center, substitute $10$ for $x$ in the equation $\frac{x^2}{900}+\frac{y^2}{400}=1$ and the corresponding values of $y$ .

  $\frac{{10}^2}{900}+\frac{y^2}{400}=1$

  $\frac{100}{900}+\frac{y^2}{400}=1$

Now, subtract $\frac{100}{900}$ from both sides of the equation,

  $\frac{100}{900}+\frac{y^2}{400}-\frac{100}{900}=1-\frac{100}{900}$

  $\frac{y^2}{400}=0.888$

Multiply both sides by $400$ .

  $\frac{y^2}{400}\times 400=0.888\times 400$

  $y^2=355.55$

  $y=\sqrt{355.55}$

  $y=18.86$

Hence, the height of the arch at $10$ feet from the center is $\boxed{18.86}$ feet.

To find the height of the arch at a distance of $20$ feet from the center, substitute $20$ for $x$ in the equation $\frac{x^2}{900}+\frac{y^2}{400}=1$ and find the corresponding values of $y$ .

  $\frac{{20}^2}{900}+\frac{y^2}{400}=1$

  $\frac{400}{900}+\frac{y^2}{400}=1$

Now, subtract $\frac{400}{900}$ from both sides of the equation,

  $\frac{400}{900}+\frac{y^2}{400}-\frac{400}{900}=1-\frac{400}{900}$

  $\frac{y^2}{400}=0.5555$

Multiply both sides of the equation by $400$

  $\frac{y^2}{400}\times 400=0.5555\times 400$

  $y^2=222.22$

  $y=\sqrt{222.22}$

  $y=14.91$

Hence, the height of the arch at $20$ feet from the center is $\boxed{14.91}$ feet.