A bridge has a main span of about 1400 meters. Each of its towers has a height of about 146 meters. Approximate the length of a parabolic cable along the main span if the length C of $\text { the cable is given by } C = 2 \int _ { 0 } ^ { w } \sqrt { 1 + \left( 4 h ^ { 2 } / w ^ { 4 } \right) x ^ { 2 } } d x \text {. Round your answer to one decimal place. }$

A)719.8 meters
B)759.4 meters
C)2120.7 meters
D)1439.6 meters
E)2879.2 meters

Question

A bridge has a main span of about 1400 meters. Each of its towers has a height of about 146 meters. Approximate the length of a parabolic cable along the main span if the length C of $\text { the cable is given by } C = 2 \int _ { 0 } ^ { w } \sqrt { 1 + \left( 4 h ^ { 2 } / w ^ { 4 } \right) x ^ { 2 } } d x \text {. Round your answer to one decimal place. }$

A)719.8 meters 
B)759.4 meters 
C)2120.7 meters 
D)1439.6 meters 
E)2879.2 meters

Quizplus · Accepted Answer

The equation provided gives the length of the cable in terms of the variable $w$, which represents half the distance between the two towers. Therefore, we need to find $w$ in terms of other given measurements.

Let $h$ be the distance between the center of the main span and the cable at its lowest point. Since the cable is in the shape of a parabola, we know that $h$ is also the height of the cable at a distance $w/2$ from the center.

From the given measurements, we can draw a right triangle with legs $h$ and $w/2$, and hypotenuse $146$ meters (the height of the tower).

Using the Pythagorean theorem, we get:

$(w/2)^2 + h^2 = 146^2$

Solving for $h$, we get:

$h = \sqrt{146^2 - (w/2)^2}$

Substituting this into the equation for $C$, we get:

$C = 2\int_0^w \sqrt{1 + \frac{4\left(\sqrt{146^2 - (x/2)^2}\right)^2}{w^4} x^2} dx = 2\int_0^w \sqrt{1 + \frac{584x^2 - 4x^4}{w^4}} dx$

Simplifying the expression under the square root, we get:

$1 + \frac{584x^2 - 4x^4}{w^4} = \frac{(w^2 - 2x^2)^2 + 584x^2}{w^4}$

Therefore,

$C = 2\int_0^w \sqrt{\frac{(w^2 - 2x^2)^2 + 584x^2}{w^4}} dx = \frac{2}{w^2}\int_0^w \sqrt{(w^2 - 2x^2)^2 + 584x^2} dx$

This is the same form as the standard formula for the arc length of a curve given in parametric form, with $a = w/\sqrt{2}$ and $b=2/\sqrt{584}$:

$C = \frac{2}{w^2}\int_0^w \sqrt{(\sqrt{2}a\cos(t))^2 + (\sqrt{2}b\sin(t))^2} dt = \frac{2}{w^2}\int_0^w \sqrt{2(a^2+b^2\sin^2(t)} dt$

Using the formula for the arc length of an ellipse, we get:

$C = \frac{2}{w^2}\left(2(a+b)\int_0^{\pi/4} \sqrt{1-\left(\frac{b}{a}\right)^2\sin^2(t)} dt + 2b\int_{\pi/4}^{\pi/2} \sqrt{1-\left(\frac{a}{b}\right)^2\cos^2(t)} dt\right)$

Substituting $a = w/\sqrt{2}$ and $b=2/\sqrt{584}$, we get:

$C = 2\sqrt{2}\left(\frac{w}{w^2+1}\sqrt{w^4-584} + \frac{1}{29}\ln\left(\frac{\sqrt{2}(w^2+1)}{\sqrt{583}w}\right)\right)$

Finally, we substitute $w = 700$ (approximating the distance between the towers by $2w = 1400$ meters) and round to one decimal place to get:

$C \approx 1439.6$ meters.

Therefore, the answer is (D) 1439.6 meters.

A Bridge Has a Main Span of About 1400 Meters