Chapter 11.1, Problem 55E

Suspension Bridge In a suspension bridge the shape of the suspension cables is parabolic. The bridge shown in the figure has towers that are 600 m apart, and the lowest point of the suspension cables is 150 m below the top of the towers. Find the equation of the parabolic part of the cables, placing the origin of the coordinate system at the vertex. [Note: This equation is used to find the length of cable needed in the construction of the bridge.]

To determine

To find: The equation of the parabolic part of the cables, by placing origin of the coordinate system as the vertex. Use it and find the length of the cable used in the construction of suspension bridge.

Answer to Problem 55E

The equation of the parabolic part of the cables is $y=\frac{1}{600}{x}^2$.

Explanation of Solution

Given:

The distance between the tower is $600$ m.

Definition used:

 “The equation of the parabola with vertex $V(0,0)$, focus $F(0,p)$ and directrix $y=-p$ is $x^2=4py$.”

Formulae used:

“The length of the curve $y=f\left(x\right)$ between the lines $x=a$ and $x=b$ is given by

${\displaystyle {\int }_a^b\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}$.”

Calculation:

Let the equation of the parabolic part of the cable be $x^2=4py$ (1)

Now assume the vertex of the parabolic part of the cable be origin.

That is, $V\left(0,0\right)$.

The distance between the tower is $600$ m.

The axis (y-axis) divides the span in to two equal halves.

Therefore, the point $C\left(0,150\right)$ lies on the axis of the parabola as shown below in Figure 1.

Now AB the diameter passing through the point $C\left(0,150\right)$ and perpendicular to the axis (y-axis).

Therefore, the end points of the diameter AB are $A\left(-300,150\right)$ and $B\left(300,150\right)$ as shown in Figure 1.

Since, the points $A\left(-300,150\right)$ and $B\left(300,150\right)$ are on the equation (1).

Substitute $B\left(300,150\right)$ in equation (1),

$\begin{array}{l}90000=4p\left(150\right)\\ p=\frac{90000}{600}\\ p=150\end{array}$

Substitute $p=150$ in equation (1), the equation becomes $x^2=600y$.

Therefore, the equation of the parabolic part of the cable is $x^2=600y$.

Find the length of the cable used in the construction of suspension bridge as follows.

Rewrite the equation as $y=\frac{1}{600}{x}^2$ and then apply the formulae for length of the curve given above.

In this problem for the parabolic part the string is used between the lines $x=-300$ and $x=300$

$\begin{array}{c}\text{The length of the string used in the parabolic part }={\displaystyle {\int }_{-300}^{300}\sqrt{1+{\left(\frac{dy}{dx}\right)}^2}dx}\\ ={\displaystyle {\int }_{-300}^{300}\sqrt{1+{\left(\frac{x}{300}\right)}^2}dx}\end{array}$

By evaluating the integral obtain the length of the string used in the construction of the suspension bridge.

Therefore, the required equation of parabolic part of the cable is $y=\frac{1}{600}{x}^2$.