The shape of the main cable that supports the deck of a brid

The shape of the main cable that supports the deck of a bridge is a catenary, a unction of the type y = e^ax - e^ax/2. (Also know as a hyperbolic cosine function (cosh x)). Below is the graph of this function when a = 2. The graph looks like a parabola, so try to fit a quadratic function to this graph by using three points on the graph with the x-coordinates of 0, + or -.5. Using three other values for x, find the difference between the correct function values and those given by the quadratic approximation An side: The picture is the Ben Franklin bridge at dusk. This is a suspension bridge that for a long time had the distinction of being the longest suspension bridge in the world.

Solution

We have y = cosh x = ( e^ax + e^-ax)/ 2 as the equation of the catenary representing the main cable that supports the deck of a bridge. We have been provided with the graph of this function when a = 2 i.e. the graph of the function y = ( e^2x + e^-2x )/ 2 . When x = 0, y = 1 ( as e⁰ = 1), when x = 0.5, y = ( e + 1/e)/2 = 1.543 ( approximately), when x = - 0.5, y = 1.543 ( approximately). Also when x = 1 or -1, we have y = 3.762 , when x = 1.5 0r, - 1.5, we have y = 10.068 and when x = 2 0r -2 , we have y = 27.308. The graph of the function resembles a parabola with vertex at ( 0, 1). The graph is symmetric about Y – Axis. The parabola opens upwards. Let the equation of the parabola be f (x)= px² + 1 where p is a constant ( this parabola has vertex at ( 0, 1). Since the graph passes through ( 0.5, 1.543), we have 1.543 = p*( 0.5)² + 1 or, 0.25p = 1.543 – 1 or, p = 0.543/ 0.25 = 2.172. Then, the equation of the required quadratic function is = 2.172 x² + 1. When x = 1 or, - 1, we have f (x) = 2.172 + 1 = 3.172; when x = 1.5 or – 1.5, we have f (x) = 2.172(2.25) + 1 = 6.112. When x = 2 or, -2, we have f (x) = 2.172(4) + 1 = 9.688.

Value of x

Value y = (e^2x + e^-2x )/2

Value of f (x) = 2.172 x² + 1

Value of y – f(x)

0

1

1

0

0.5 or, - 0.5

1.543

1.543

0

1 or, -1

3.762

3.172

0.59

1.5 or, - 1.5

10.068

6.112

3.956

2 or, -2

27.308

9.688

17.62

Note: The equation given in the question is not that of a cosh function. We have taken the correct equation of a cosh function as the graph is that of the cosh function.

Value of x	Value y = (e2x + e-2x )/2	Value of f (x) = 2.172 x2 + 1	Value of y – f(x)
0	1	1	0
0.5 or, - 0.5	1.543	1.543	0
1 or, -1	3.762	3.172	0.59
1.5 or, - 1.5	10.068	6.112	3.956
2 or, -2	27.308	9.688	17.62