Question

(b) suppose two telephone poles are 40 ft apart and the length of the wire between the poles is 41 ft. if the lowest point of the wire must be 19 ft above the ground, how high up on each pole should the wire be attached? (round your answer to two decimal places.)

Answer 1

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
$\int _{0}^{20} \sqrt{1+[y'(x)]^{2}} \, dx = \int_{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx =20.5$

Because we do not know the value of a, we shall find it numerically.
Define the function
$f(a) = \int _{0}^{20} \sqrt{1+4a^{2}x^{2}} \, dx - 20.5 = 0$
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