Chapter 8.1, Problem 63E

Solution

To Find: The length of the vertical cables supporting the specified suspension bridge.

The support cable at the lowest point is $10\text{ feet}$ long. On both sides of it the support cables have the lengths $12.77,21.11,35,54.44,\text{ and }79.44\text{ feet}$ in order of increasing distance from the lowest point.

Given:

The parabolic main cables of the suspension bridge are attached to towers that are $600\text{ feet}$ apart. The main cables are attached to the towers at the height of $110\text{ feet}$ above the roadway. The main cables are $10\text{ feet}$ above the roadway at their lowest points. The vertical support cables are attached $50\text{ feet}$ apart along the level roadway.

  

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 level roadway is represented by the $x$ -axis.

  

The line segment $AB$ represents the $600\text{ feet}$ distance between the towers. The mid-point of this line segment is the origin. The point $P\left(0,10\right)$ represents the lowest point of the parabolic main cable which is given to be $10\text{ feet}$ above the roadway (the $x$ -axis). The points $F\left(300,110\right)$ and $F^{\prime }\left(-300,110\right)$ represent the points where the main cable is attached to the towers given to be $110\text{ feet}$ above the roadway.

Note that the parabola appearing in the figure has vertex at $\left(0,10\right)$ . Also, the parabola passes through the $F\left(300,110\right)$ so the coordinates of this point must satisfy the equation of the parabola.

Write the equation of the given concave up parabola having vertex $\left(h,k\right)=\left(0,10\right)$ and axis along the $y$ -axis, using the standard equation ${\left(x-h\right)}^2=4p\left(y-k\right)$ :

  $\begin{array}{c}{\left(x-h\right)}^2=4p\left(y-k\right)\\ {\left(x-0\right)}^2=4p\left(y-10\right)\\ x^2=4p\left(y-10\right)\end{array}$

Since the parabola also passes through $F\left(300,110\right)$ , determine $p$ by substituting $x=300$ , and, $y=110$ in the equation $x^2=4p\left(y-10\right)$ .

  $\begin{array}{c}{x}^2=4p\left(y-10\right)\\ {\left(300\right)}^2=4p\left(110-10\right)\\ 90000=4p\left(100\right)\\ \frac{90000}{400}=p\\ 225=p\end{array}$

Substitute $p=225$ in the equation $x^2=4p\left(y-10\right)$ and then isolate $y$ :

  $\begin{array}{c}{x}^2=4p\left(y-10\right)\\ x^2=4\left(225\right)\left(y-10\right)\\ x^2=900\left(y-10\right)\\ \frac{x^2}{900}=y-10\\ \frac{x^2}{900}+10=y\end{array}$

The relation $y=\frac{x^2}{900}+10$ give the height of the parabolic cable for a particular $x$ -coordinate.

Since it is given that the vertical support cables are $50\text{ feet}$ apart. In the above figure, they are represented as vertical line segments from the roadway up to the main cable at the $x$ -coordinates $-250,-200,-150,-100,\text{ and},-50$ towards the left of the lowest point at $x=0$ .

Also, such support cables are at symmetrical positions towards the right of the lowest point namely at the $x$ -coordinates $250,200,150,100,\text{ and},50$ .

Calculate the lengths of vertical support cables by tabulating the value of $y=\frac{x^2}{900}+10$ for the $x$ -coordinates $\pm 250,\pm 200,\pm 150,\pm 100,\text{ and},\pm 50$ .

$x$	$y=\frac{x^2}{900}+10$
$\pm 250$	$\frac{{\left(\pm 250\right)}^2}{900}+10\approx 79.44$
$\pm 200$	$\frac{{\left(\pm 200\right)}^2}{900}+10\approx 54.44$
$\pm 150$	$\frac{{\left(\pm 150\right)}^2}{900}+10=35$
$\pm 100$	$\frac{{\left(\pm 100\right)}^2}{900}+10\approx 21.11$
$\pm 50$	$\frac{{\left(\pm 50\right)}^2}{900}+10\approx 12.77$

Conclusion:

The support cable at the lowest point $x=0$ is $10\text{ feet}$ long. On both sides of it the support cables have the lengths $12.77,21.11,35,54.44,\text{ and }79.44\text{ feet}$ in order of increasing distance from the lowest point.