Question

The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of a bridge is 1270 ft long and 157 ft high. The parabola y=0.00039x^2 gives a good fit to the shape of the cables, where IxI less than of equal to 635, and x and y are measured in feet. Approximate the length of the cables that stretch between the tops of the two towers.

Answer 1

Answer:

Step-by-step explanation:

Given

span of bridge $L=1270\ ft$

height of span $h=157\ ft$

Equation of Parabola

$y=0.00039x^2$

$|x|$ i.e.

$-635$

$\frac{dy}{dx}=2\times 0.00039$

length of Arc$=\int_{a}^{b}\sqrt{1+(\frac{dy}{dx})^2}$

$=\int_{-635}^{635}\sqrt{1+(\frac{dy}{dx})^2}$

$=\int_{-635}^{635}\sqrt{1+(0.00078x)^2}$

$=2\times \int_{0}^{635}\sqrt{1+(0.00078x)^2}$

$=2\times (660.08)$

$=1320.16\ ft$