Bridges

Calculus Level 5

You have been commissioned to build a freely suspended bridge of uniform density across a river of width 20 metres, normal to the edge. Oddly enough, the commissioner asks you to maximise the amount of material used without making it too steep.

Given that the maximum gradient that people can walk on is 0.5, find the required length of material in metres.

Model the bridge as a catenary.

Image credit: Flickr Tim O'Shea

The answer is 20.78.

2 solutions

Jake Lai
Dec 5, 2014

Let the middle of the bridge be centered above the origin.

The bridge is a catenary, described by the equation $y = a \cosh(x/a)$ . The maximum gradient $0.5$ of the bridge occurs at the end of the bridge $x = 10$ .

$\frac{dy}{dx}|_{x = 10} = \sinh(10/a) = 0.5$

Solving for $a$ , we get

$a = \frac{10}{\sinh^{-1}(0.5)}$

Using the Whewell equation for the catenary $s = a \tan \varphi\ = a \frac{dy}{dx}$ where $s$ is arclength and $\varphi$ is angle of inclination, we obtain

$s = \frac{5}{\sinh^{-1}(0.5)} \approx 10.3904$

Of course, the $s$ above is only half the bridge, so twice $s$ is approximately $\boxed{20.78}$ .

(Thanks a lot to Jon Haussman for informing me that my solution was incorrect.)

I wish I knew the bridge is a catenary :'(

Agnishom Chattopadhyay - 6 years, 6 months ago

Hehe, this is one of those more esoteric questions.

Jake Lai - 6 years, 6 months ago

I get the same value of $a$ , which is about 20.78. But if I understand the Whewell equation correctly, then shouldn't $\varphi$ satisfy $\tan \varphi = 0.5$ ? This makes $s = a \tan \varphi = \frac{a}{2}.$ This is half the arc length, so the total arc length is just $a \approx 20.78$ .

We can also use the formula for arc length: $\begin{aligned} f(x) &= \frac{a}{2} (e^{x/a} + e^{-x/a}), \\ f'(x) &= \frac{1}{2} (e^{x/a} - e^{-x/a}), \end{aligned}$ so the arc length from $x = 0$ to $x = 10$ is $\begin{aligned} \int_0^{10} \sqrt{1 + [f'(x)]^2} \: dx &= \int_0^{10} \sqrt{1 + \frac{1}{4} e^{2x/a} - \frac{1}{2} + \frac{1}{4} e^{-2x/a}} \: dx \\ &= \int_0^{10} \sqrt{\frac{1}{4} e^{2x/a} + \frac{1}{2} + \frac{1}{4} e^{-2x/a}} \: dx \\ &= \int_0^{10} \frac{1}{2} (e^{x/a} + e^{-x/a}) \: dx \\ &= \frac{a}{2} (e^{x/a} - e^{-x/a}) \big|_0^{10} \\ &= \frac{a}{2} (e^{10/a} - e^{-10/a}). \end{aligned}$

But $f'(10) = \frac{1}{2} (e^{10/a} - e^{-10/a}) = 0.5,$ so $e^{10/a} - e^{-10/a} = 1$ , and the above arc length is $a/2$ . Again, we double this to get the total arc length of $a$ .

Jon Haussmann - 6 years, 6 months ago

At the very least, the length of the bridge has to be longer than the width of the river. Hence 17.2 is not the correct answer.

Calvin Lin Staff - 6 years, 6 months ago

Yeah, I misinterpreted the Whewell equation. Thanks so much for bringing this up; I'll change the answer.

Jake Lai - 6 years, 6 months ago

Thanks. I have updated the answer to 20.78.

Thanks for updating the solution :)

Calvin Lin Staff - 6 years, 6 months ago

Rohit Ner
May 16, 2016

Since I am not aware of the term catenary, I assumed the bridge to be parabolic.

Let ${x}^2=4ay$ be the eqn of the curve.

Interpreting the given data, the slope at $x=10$ is $0.5$ .

$\begin{aligned}\dfrac{dy}{dx}&=\dfrac{x}{2a}\\\dfrac{1}{2}&=\dfrac{10}{2a}\\\Rightarrow a&=10\\\Rightarrow y&=\dfrac{{x}^2}{40}\end{aligned}$

Arc length of a curve $y=f(x)$ lying between $a$ and $b$ is given by

$\displaystyle\int_{a}^{b}{\sqrt{1+{\left(\dfrac{dy}{dx}\right)}^2}dx}$

So the required length of material is

$\begin{aligned}\displaystyle\int_{-10}^{10}{\sqrt{1+{\left(\dfrac{x}{20}\right)}^2}dx}&=2\displaystyle\int_{0}^{10}{\sqrt{1+{\left(\dfrac{x}{20}\right)}^2}dx}\\&=\left[ x\sqrt { 1+{ \left( \frac { x }{ 20 } \right) }^{ 2 } } +20\ln { \left| \frac { x }{ 20 } +\sqrt { 1+{ \left( \frac { x }{ 20 } \right) }^{ 2 } } \right| } \right] \begin{matrix} 10 \\ \\ 0 \end{matrix}\\&\large\color{#3D99F6}{=\boxed{10\sqrt{\dfrac{5}{4}}+20\ln {\left| \dfrac{1}{2} +\sqrt{\dfrac{5}{4}} \right|}}} \end{aligned}$

This is a good approximation of the required value $\approx 20.80$ .

Bridges

Image credit: Flickr Tim O'Shea

The answer is 20.78.

2 solutions

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