{"id":495,"date":"2025-02-13T19:45:28","date_gmt":"2025-02-13T19:45:28","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-of-the-hyperbolic-functions-learn-it-3\/"},"modified":"2025-02-13T19:45:28","modified_gmt":"2025-02-13T19:45:28","slug":"calculus-of-the-hyperbolic-functions-learn-it-3","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/calculus-of-the-hyperbolic-functions-learn-it-3\/","title":{"raw":"Calculus of the Hyperbolic Functions: Learn It 3","rendered":"Calculus of the Hyperbolic Functions: Learn It 3"},"content":{"raw":"\n

Applications of Hyperbolic Functions<\/h2>\n

Hyperbolic functions have practical applications, particularly in the modeling of hanging cables. When a cable of uniform density hangs between two supports, it forms a curve known as a catenary<\/strong>. Examples of catenaries include high-voltage power lines, chains hanging between posts, and strands of a spider's web.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"651\"]\"An Figure 3. Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)[\/caption]\n\n

Mathematically, catenaries can be modeled using hyperbolic functions. Specifically, functions of the form [latex]y=a\\text{cosh}\\left(\\frac{x}{a}\\right)[\/latex] represent catenaries. For instance, the graph of [latex]y=2\\text{cosh}\\left(\\frac{x}{2}\\right)[\/latex] demonstrates this shape effectively.<\/p>\n\n[caption id=\"\" align=\"aligncenter\" width=\"342\"]\"This Figure 4. A hyperbolic cosine function forms the shape of a catenary.[\/caption]\n\n

This visualization helps in understanding how hyperbolic functions apply to real-world structures.<\/p>\n

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When solving problems related to catenaries and their lengths, we use the arc length formula. This formula helps us determine the length of the hanging cable modeled by a hyperbolic function.<\/p>\n

Recall that the formula for arc length is<\/p>\n

[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\n<\/section>\n
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Assume a hanging cable has the shape [latex]10\\text{cosh}\\left(\\frac{x}{10}\\right)[\/latex] for [latex]-15\\le x\\le 15,[\/latex] where [latex]x[\/latex] is measured in feet. Determine the length of the cable (in feet).<\/p>\n\n[reveal-answer q=\"fs-id1167793307481\"]Show Solution[\/reveal-answer]
\n[hidden-answer a=\"fs-id1167793307481\"]\n\n

We have [latex]f(x)=10\\text{cosh}(x\\text{\/}10),[\/latex] so [latex]{f}^{\\prime }(x)=\\text{sinh}(x\\text{\/}10).[\/latex] Then<\/p>\n

[latex]\\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}[\/latex]<\/div>\n

Now recall that [latex]1+{\\text{sinh}}^{2}x={\\text{cosh}}^{2}x,[\/latex] so we have<\/p>\n

[latex]\\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}[\/latex]<\/div>\n

Watch the following video to see the worked solution to this example.<\/p>\n