Question

The profile of the cables on a suspension bridge may be modeled by a parabola. The central span of the bridge is 1210 m long and 128 m high. The parabola y equals 0.00035 x squared gives a good fit to the shape of the​ cables, where StartAbsoluteValue x EndAbsoluteValue less than or equals 605​, and x and y are measured in meters. Approximate the length of the cables that stretch between the tops of the two towers.

Answer 1

The approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.

Explanation:

The equation of the parabola is:

`y=0.00035x^(2)`

Compute the first order derivative of y as follows:

`y=0.00035x^(2)`

`\frac{\text{d}y}{\text{dx}}=\frac{\text{d}}{\text{dx}}[0.00035x^(2)]`

`=2\cdot 0.00035x\\\\=0.0007x`

Now, it is provided that |x | ≤ 605.

⇒ -605 ≤ x ≤ 605

Compute the arc length as follows:

`\text{Arc Length}=\int\limits^(x)_(-x) {1+(\frac{\text{dy}}{\text{dx}})^(2)} \, dx`

`=\int\limits^(605)_(-605) {\sqrt{1+(0.0007x)^(2)}} \, dx \\\\={\displaystyle\int\limits^(605)_(-605)}\sqrt{(49x^2)/(100000000)+1}\,\mathrm{d}x\\\\={(1)/(10000)}}{\displaystyle\int\limits^(605)_(-605)}√(49x^2+100000000)\,\mathrm{d}x\\\\`

Now, let

`x=(10000\tan\left(u\right))/(7)\\\\\Rightarrow u=\arctan\left((7x)/(10000)\right)\\\\\Rightarrow \mathrm{d}x=(10000\sec^2\left(u\right))/(7)\,\mathrm{d}u`

`\int dx={\displaystyle\int\limits}(10000\sec^2\left(u\right)√(100000000\tan^2\left(u\right)+100000000))/(7)\,\mathrm{d}u`

`={(100000000)/(7)}}{\displaystyle\int}\sec^3\left(u\right)\,\mathrm{d}u\\\\=(50000000\ln\left(\tan\left(u\right)+\sec\left(u\right)\right))/(7)+(50000000\sec\left(u\right)\tan\left(u\right))/(7)\\\\=\frac{50000000\ln\left(\sqrt{(49x^2)/(100000000)+1}+(7x)/(10000)\right)}{7}+5000x\sqrt{(49x^2)/(100000000)+1}`

Plug in the solved integrals in Arc Length and solve as follows:

`\text{Arc Length}=\frac{5000\ln\left(\sqrt{(49x^2)/(100000000)+1}+(7x)/(10000)\right)}{7}+\frac{x\sqrt{(49x^2)/(100000000)+1}}{2}|_{limits^(605)_(-605)}\\\\`

`=1245.253707795227\\\\\approx 1245.25`

Thus, the approximated length of the cables that stretch between the tops of the two towers is 1245.25 meters.