Ulearngo

Applications

One physical application of hyperbolic functions involves hanging cables. If a cable of uniform density is suspended between two supports without any load other than its own weight, the cable forms a curve called a catenary. High-voltage power lines, chains hanging between two posts, and strands of a spider’s web all form catenaries. The following figure shows chains hanging from a row of posts.

An image of chains hanging between posts that all take the shape of a catenary.
Figure 6.83 Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)

Hyperbolic functions can be used to model catenaries. Specifically, functions of the form \(y = a \text{cosh} \left(\right. x / a \left.\right)\) are catenaries. Figure 6.84 shows the graph of \(y = 2 \text{cosh} \left(\right. x / 2 \left.\right) .\)

This figure is a graph. It is of the function f(x)=2cosh(x/2). The curve decreases in the second quadrant to the y-axis. It intersects the y-axis at y=2. Then the curve becomes increasing.
Figure 6.84 A hyperbolic cosine function forms the shape of a catenary.

Example 6.51

Using a Catenary to Find the Length of a Cable

Assume a hanging cable has the shape \(10 \text{cosh} \left(\right. x / 10 \left.\right)\) for \(−15 \leq x \leq 15 ,\) where \(x\) is measured in feet. Determine the length of the cable (in feet).

Solution

Recall from Section \(2.4\) that the formula for arc length is

\[\text{Arc Length} = \int_{a}^{b} \sqrt{1 + \left[\right. f^{'} \left(\right. x \left.\right) \left]\right.^{2}} d x .\]

We have \(f \left(\right. x \left.\right) = 10 \text{cosh} \left(\right. x / 10 \left.\right) ,\) so \(f^{'} \left(\right. x \left.\right) = \text{sinh} \left(\right. x / 10 \left.\right) .\) Then

\[\begin{aligned} \text{Arc Length} & = \int_{a}^{b} \sqrt{1 + \left[\right. f^{'} \left(\right. x \left.\right) \left]\right.^{2}} d x \\ & = \int_{−15}^{15} \sqrt{1 + \text{sinh}^{2} \left(\right. \frac{x}{10} \left.\right)} d x . \end{aligned}\]

Now recall that \(1 + \text{sinh}^{2} x = \text{cosh}^{2} x ,\) so we have

\[\begin{aligned} \text{Arc Length} & = \int_{−15}^{15} \sqrt{1 + \text{sinh}^{2} \left(\right. \frac{x}{10} \left.\right)} d x \\ & = \int_{−15}^{15} \text{cosh} \left(\right. \frac{x}{10} \left.\right) d x \\ & = 10 \text{sinh} \left(\left(\right. \frac{x}{10} \left.\right) \left|\right.\right)_{−15}^{15} = 10 \left[\right. \text{sinh} \left(\right. \frac{3}{2} \left.\right) - \text{sinh} \left(\right. - \frac{3}{2} \left.\right) \left]\right. = 20 \text{sinh} \left(\right. \frac{3}{2} \left.\right) \\ & \approx 42.586 \text{ft} . \end{aligned}\]

Checkpoint 6.51

Assume a hanging cable has the shape \(15 \text{cosh} \left(\right. x / 15 \left.\right)\) for \(−20 \leq x \leq 20 .\) Determine the length of the cable (in feet).

This lesson is part of:

Applications of Integration

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