Chapter 8.1, Problem 64E

Solution

To find:

The maximum clearance of the given parabolic bridge arch from the given road.

  $\frac{64}{3}\approx 21.33\text{ feet}$ .

Given:

The parabolic arch spans $60\text{ feet}$ wide with a $30\text{ feet}$ wide road passing underneath the bridge.

The minimum clearance of the bridge from the road is $16\text{ feet}$ .

  

Concepts Used:

The equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ represents a parabola with vertex $\left(h,k\right)$ and having an axis parallel to the $y$ -axis.

Modelling a geometrical situation on coordinate plane.

Calculations:

Draw a figure to represent the situation on coordinate plane. In the figure below the ground level is represented by the $x$ -axis.

  

The line segment $AB$ represents the $60\text{ feet}$ span of the parabolic arch. The mid-point of this span is the origin. The line segment $CD$ represents the $30\text{ feet}$ wide road. The minimum clearance of $16\text{ feet}$ between the bridge and the road underneath is represented by the height of the arch at both ends of the width of the road, namely, the line segments $CE$ and $DF$ . The line segment $OG$ represents the maximum clearance of the road from the arch and is to be determined.

Note that the parabola appearing in the figure has vertex right above the origin, so it’s $x$ -coordinate is zero. Write the vertex as $\left(0,k\right)$ . The height of the vertex above the origin (ground level) $k$ is the required maximum clearance. Also, the parabola passes through the points $B\left(30,0\right)$ and $F\left(15,16\right)$ so the coordinates of these points must satisfy the equation of the parabola.

Write the equation of the given concave down parabola having vertex $\left(h,k\right)=\left(0,k\right)$ and axis along the $y$ -axis, using the standard equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ :

  $\begin{array}{c}{\left(x-h\right)}^2=4p\left(y-k\right)\\ {\left(x-0\right)}^2=4p\left(y-k\right)\\ x^2=4p\left(y-k\right)\end{array}$

Since the parabola passes through $B\left(30,0\right)$ , find a condition on $p$ and $k$ by substituting $x=30$ and $y=0$ in the equation $x^2=4p\left(y-k\right)$ .

  $\begin{array}{c}{x}^2=4p\left(y-k\right)\\ {\left(30\right)}^2=4p\left(0-k\right)\\ 900=4p\left(-k\right)\\ 900=-4pk\\ \frac{900}{-4k}=p\\ -\frac{225}{k}=p\end{array}$

Since the parabola also passes through $F\left(15,16\right)$ , determine $k$ by substituting $x=15$ , $y=16$ , and $p=-\frac{225}{k}$ in the equation $x^2=4p\left(y-k\right)$ .

  $\begin{array}{c}{x}^2=4p\left(y-k\right)\\ {\left(15\right)}^2=4\left(-\frac{225}{k}\right)\left(16-k\right)\\ 225=4\left(-\frac{225}{k}\right)\left(16-k\right)\\ 1=4\left(-\frac{1}{k}\right)\left(16-k\right)\\ k=-4\left(16-k\right)\\ k=-64+4k\\ 64=3k\\ \frac{64}{3}=k\\ 21.33\approx k\end{array}$

Thus, $900=-4pk$ .

Conclusion:

  $k=\frac{64}{3}\approx 21.33\text{ feet}$ is the required maximum clearance of the road from the arch.