Refer to the diagram shown below.
The suspended wire is in the shape of a parabola defined by the equation
y = ax²
where a = a positive constant.
The derivative of y with respect to x is y' = 2ax.
The vertex is at (0,0) and the line of symmetry is x = 0.
The suspended length is 41 ft, therefore half the suspended length is 20.5 ft.
The length between x = 0 and x = 20 is given by
[tex]\int _{0}^{20} \sqrt{1+[y'(x)]^{2}} \, dx = \int_{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx =20.5[/tex]
Because we do not know the value of a, we shall find it numerically.
Define the function
[tex]f(a) = \int _{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx - 20.5 = 0 [/tex]
The plot for f(a) versus a yields an approximate solution (from Matlab) of a = 0.01 (shown in the figure).
Therefore
y = 0.01x²
When x = 20 ft, h = 0.01(400) = 4 ft
Because the vertex of the parabola is 19 ft above ground, the support points for the wire are 19 + h = 23 ft above ground.
Answer: 23.00 ft
