Applications of Hyperbolic Functions

Learning Outcomes

Describe the common applied conditions of a catenary curve

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 3. 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 $y=a\text{cosh}\left(\frac{x}{a}\right)$ are catenaries. Figure 4 shows the graph of $y=2\text{cosh}\left(\frac{x}{2}\right).$

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 4. A hyperbolic cosine function forms the shape of a catenary.

example: Using a Catenary to Find the Length of a Cable

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

Show Solution

Recall from Section 6.4 that the formula for arc length is

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

$\text{Arc Length}={\displaystyle\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}dx.$

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

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

$\begin{array}{cc}\hfill \text{Arc Length}& ={\displaystyle\int }_{a}^{b}\sqrt{1+{\left[{f}^{\prime }(x)\right]}^{2}}dx\hfill \\ & ={\displaystyle\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}dx.\hfill \end{array}$

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

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

$\begin{array}{cc}\hfill \text{Arc Length}& ={\displaystyle\int }_{-15}^{15}\sqrt{1+{\text{sinh}}^{2}(\frac{x}{10})}dx\hfill \\ & ={\displaystyle\int }_{-15}^{15}\text{cosh}(\frac{x}{10})dx\hfill \\ & =10\text{sinh}{(\frac{x}{10})|}_{-15}^{15}=10\left[\text{sinh}(\frac{3}{2})-\text{sinh}(-\frac{3}{2})\right]=20\text{sinh}\left(\frac{3}{2}\right)\hfill \\ & \approx 42.586\text{ft}\text{.}\hfill \end{array}$

Watch the following video to see the worked solution to Example: Using a Catenary to Find the Length of a Cable.

Closed Captioning and Transcript Information for Video

Try It

Assume a hanging cable has the shape $15\text{cosh}\left(\frac{x}{15}\right)$ for $-20\le x\le 20.$ Determine the length of the cable (in feet).

Show Solution

52.95 ft

$52.95\text{ft}$

Hint

Use the procedure from the previous example.

Module 6: Applications of Integration