Chapter 9.1, Problem 97E

Solution

(a)

To draw sketch of cable and label the coordinates of the known points

  

Required coordinates are $\left(-640,152\right)$ and $\left(640,152\right)$

Given:

Distance between two towers=1280 meters

Height of towers from roadway= 152 meters

Concept Used:

Let the origin be the midpoint of two towers such that top of the towers are at coordinates $\left(-640,152\right)$ and $\left(640,152\right)$ according to which horizontal distance between towers $=640-\left(-640\right)=1280meters$

Given height of towers = 152 meters

Calculation:

Required sketch is as follows:

  

Conclusion:

Coordinates of known points are $\left(-640,152\right)$ and $\left(640,152\right)$

(b)

To write equation of model of cable

Equation of model of cable is

  $x^2=\frac{51200}{19}y$

Given:

A cable of the Golden Gate Bridge is suspended in the shape of parabola between two towers.

Concept Used:

The standard equation of a parabola that has vertical axis of symmetry is:

  ${\left(x-h\right)}^2=4p\left(y-k\right)\left(where\left(h,k\right)\text{ is vertex of parabola}\right)$

Calculation:

From the sketch drawn in part (a), vertex is located at the coordinates $\left(0,0\right)$ and the endpoint is at coordinates $\left(640,152\right)$

From above clearly seen

  $\left(h,k\right)\to (0,0)\text{ and }\left(x,y\right)\to \left(640,152\right)$

Substituting values of vertex and endpoints in standard equation of parabola to find value of p

  $\begin{array}{l}{\left(640-0\right)}^2=4p\left(152-0\right)\\ 409600=608p\\ p=\frac{409600}{608}\\ p=\frac{128000}{19}\end{array}$

Getting equation of parabola by putting values of $\left(h,k\right)$ and $p$ in standard equation of parabola

  $\begin{array}{l}{\left(x-0\right)}^2=4\left(\frac{12800}{19}\right)\left(y-0\right)\\ x^2=\frac{51200}{19}y\end{array}$

Conclusion:

Equation of model of cable is

  $x^2=\frac{51200}{19}y$

(c)

To complete the table by finding height y of the suspension cable over the roadway at a distance of x meters from the centre of bridge

x	0	200	400	500	600
y	0	14.84	59.38	92.77	133.59

Given:

The distance between the centre of bridge and suspension cable over roadway is x and table as below:

  

Concept Used:

From part (b)

Equation of model of cable is

  $x^2=\frac{51200}{19}y$

Solve for y and then substitute values of x given in table to find value of y.

Calculation:

Equation of model of cable is

  $x^2=\frac{51200}{19}y$

  $y=\frac{19}{51200}{x}^2$

  $\begin{array}{l}forx=0\\ y=\frac{19}{51200}\times {\left(0\right)}^2=0\\ forx=200\\ y=\frac{19}{51200}\times {\left(200\right)}^2=14.84\\ forx=400\\ y=\frac{19}{51200}\times {\left(400\right)}^2=59.38\\ forx=500\\ y=\frac{19}{51200}\times {\left(500\right)}^2=92.77\\ forx=600\\ y=\frac{19}{51200}\times {\left(600\right)}^2=133.59\end{array}$

Conclusion:

Required table is given below:

x	0	200	400	500	600
y	0	14.84	59.38	92.77	133.59