{"id":1196,"date":"2021-06-30T17:02:08","date_gmt":"2021-06-30T17:02:08","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/applications-of-hyperbolic-functions\/"},"modified":"2022-03-19T03:49:23","modified_gmt":"2022-03-19T03:49:23","slug":"applications-of-hyperbolic-functions","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus2\/chapter\/applications-of-hyperbolic-functions\/","title":{"raw":"Applications of Hyperbolic Functions","rendered":"Applications of Hyperbolic Functions"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Describe the common applied conditions of a catenary curve<\/li>\r\n<\/ul>\r\n<\/div>\r\n<p id=\"fs-id1167793964561\">One physical application of hyperbolic functions involves <span class=\"no-emphasis\">hanging cables<\/span>. 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 <strong>catenary<\/strong>. High-voltage power lines, chains hanging between two posts, and strands of a spider\u2019s web all form catenaries. The following figure shows chains hanging from a row of posts.<\/p>\r\n\r\n<div id=\"CNX_Calc_Figure_06_09_003\" class=\"wp-caption aligncenter\">\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"651\"]<img class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213354\/CNX_Calc_Figure_06_09_003.jpg\" alt=\"An image of chains hanging between posts that all take the shape of a catenary.\" width=\"651\" height=\"488\" \/> Figure 3. Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)[\/caption]\r\n\r\n<\/div>\r\n<p id=\"fs-id1167793591346\">Hyperbolic functions can be used to model catenaries. Specifically, functions of the form [latex]y=a\\text{cosh}\\left(\\frac{x}{a}\\right)[\/latex] are catenaries. Figure 4 shows the graph of [latex]y=2\\text{cosh}\\left(\\frac{x}{2}\\right).[\/latex]<\/p>\r\n\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"342\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213358\/CNX_Calc_Figure_06_09_004.jpg\" alt=\"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.\" width=\"342\" height=\"347\" \/> Figure 4. A hyperbolic cosine function forms the shape of a catenary.[\/caption]\r\n\r\n<div id=\"fs-id1167793370895\" class=\"textbook exercises\">\r\n<h3>example: Using a Catenary to Find the Length of a Cable<\/h3>\r\nAssume 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).\r\n<div id=\"fs-id1167793884153\" class=\"exercise\">[reveal-answer q=\"fs-id1167793307481\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793307481\"]\r\n<p id=\"fs-id1167793307481\">Recall from Section 6.4 that the formula for arc length is<\/p>\r\n\r\n<div id=\"fs-id1167793943925\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\r\n<p id=\"fs-id1167794160035\">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>\r\n\r\n<div id=\"fs-id1167794337061\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\text{Arc Length}&amp; ={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{-15}^{15}\\sqrt{1+{\\text{sinh}}^{2}(\\frac{x}{10})}dx.\\hfill \\end{array}[\/latex]<\/div>\r\n<p id=\"fs-id1167793964837\">Now recall that [latex]1+{\\text{sinh}}^{2}x={\\text{cosh}}^{2}x,[\/latex] so we have<\/p>\r\n\r\n<div id=\"fs-id1167794333245\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\begin{array}{cc}\\hfill \\text{Arc Length}&amp; ={\\displaystyle\\int }_{-15}^{15}\\sqrt{1+{\\text{sinh}}^{2}(\\frac{x}{10})}dx\\hfill \\\\ &amp; ={\\displaystyle\\int }_{-15}^{15}\\text{cosh}(\\frac{x}{10})dx\\hfill \\\\ &amp; =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 \\\\ &amp; \\approx 42.586\\text{ft}\\text{.}\\hfill \\end{array}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Using a Catenary to Find the Length of a Cable.\r\n\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/fN1noldfkDE?controls=0&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\r\n[hidden-answer a=\"266834\"]For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.\r\n\r\nYou can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/6.9.3_73_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"6.9.3 #73 (edited)\" here (opens in new window)<\/a>.[\/hidden-answer]\r\n<div id=\"fs-id1167794326134\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\nAssume a hanging cable has the shape [latex]15\\text{cosh}\\left(\\frac{x}{15}\\right)[\/latex] for [latex]-20\\le x\\le 20.[\/latex] Determine the length of the cable (in feet).\r\n<div>[reveal-answer q=\"847806\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"847806\"]Use the procedure from the previous example.[\/hidden-answer]<\/div>\r\n<div><\/div>\r\n<div id=\"fs-id1167793414872\" class=\"exercise\">[reveal-answer q=\"fs-id1167793292342\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1167793292342\"][latex]52.95\\text{ft}[\/latex][\/hidden-answer]<\/div>\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]223450[\/ohm_question]\r\n\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Describe the common applied conditions of a catenary curve<\/li>\n<\/ul>\n<\/div>\n<p id=\"fs-id1167793964561\">One physical application of hyperbolic functions involves <span class=\"no-emphasis\">hanging cables<\/span>. 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 <strong>catenary<\/strong>. High-voltage power lines, chains hanging between two posts, and strands of a spider\u2019s web all form catenaries. The following figure shows chains hanging from a row of posts.<\/p>\n<div id=\"CNX_Calc_Figure_06_09_003\" class=\"wp-caption aligncenter\">\n<div style=\"width: 661px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213354\/CNX_Calc_Figure_06_09_003.jpg\" alt=\"An image of chains hanging between posts that all take the shape of a catenary.\" width=\"651\" height=\"488\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 3. Chains between these posts take the shape of a catenary. (credit: modification of work by OKFoundryCompany, Flickr)<\/p>\n<\/div>\n<\/div>\n<p id=\"fs-id1167793591346\">Hyperbolic functions can be used to model catenaries. Specifically, functions of the form [latex]y=a\\text{cosh}\\left(\\frac{x}{a}\\right)[\/latex] are catenaries. Figure 4 shows the graph of [latex]y=2\\text{cosh}\\left(\\frac{x}{2}\\right).[\/latex]<\/p>\n<div style=\"width: 352px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11213358\/CNX_Calc_Figure_06_09_004.jpg\" alt=\"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.\" width=\"342\" height=\"347\" \/><\/p>\n<p class=\"wp-caption-text\">Figure 4. A hyperbolic cosine function forms the shape of a catenary.<\/p>\n<\/div>\n<div id=\"fs-id1167793370895\" class=\"textbook exercises\">\n<h3>example: Using a Catenary to Find the Length of a Cable<\/h3>\n<p>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<div id=\"fs-id1167793884153\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793307481\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793307481\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793307481\">Recall from Section 6.4 that the formula for arc length is<\/p>\n<div id=\"fs-id1167793943925\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\text{Arc Length}={\\displaystyle\\int }_{a}^{b}\\sqrt{1+{\\left[{f}^{\\prime }(x)\\right]}^{2}}dx.[\/latex]<\/div>\n<p id=\"fs-id1167794160035\">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<div id=\"fs-id1167794337061\" class=\"equation unnumbered\" style=\"text-align: center;\">[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<p id=\"fs-id1167793964837\">Now recall that [latex]1+{\\text{sinh}}^{2}x={\\text{cosh}}^{2}x,[\/latex] so we have<\/p>\n<div id=\"fs-id1167794333245\" class=\"equation unnumbered\" style=\"text-align: center;\">[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<\/div>\n<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Using a Catenary to Find the Length of a Cable.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/fN1noldfkDE?controls=0&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/p>\n<div id=\"q266834\" class=\"hidden-answer\" style=\"display: none\">For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.<\/p>\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/6.9.3_73_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;6.9.3 #73 (edited)&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<div id=\"fs-id1167794326134\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p>Assume a hanging cable has the shape [latex]15\\text{cosh}\\left(\\frac{x}{15}\\right)[\/latex] for [latex]-20\\le x\\le 20.[\/latex] Determine the length of the cable (in feet).<\/p>\n<div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q847806\">Hint<\/span><\/p>\n<div id=\"q847806\" class=\"hidden-answer\" style=\"display: none\">Use the procedure from the previous example.<\/div>\n<\/div>\n<\/div>\n<div><\/div>\n<div id=\"fs-id1167793414872\" class=\"exercise\">\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1167793292342\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1167793292342\" class=\"hidden-answer\" style=\"display: none\">[latex]52.95\\text{ft}[\/latex]<\/div>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm223450\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=223450&theme=oea&iframe_resize_id=ohm223450&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-1196\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>6.9.3 #73 (edited). <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 2. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\">https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":36,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 2\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-2\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"6.9.3 #73 (edited)\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-1196","chapter","type-chapter","status-publish","hentry"],"part":1160,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1196","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":2,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1196\/revisions"}],"predecessor-version":[{"id":2519,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1196\/revisions\/2519"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/1160"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/1196\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=1196"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=1196"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=1196"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=1196"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}