{"id":1881,"date":"2015-11-12T18:30:43","date_gmt":"2015-11-12T18:30:43","guid":{"rendered":"https:\/\/courses.candelalearning.com\/collegealgebra1xmaster\/?post_type=chapter&p=1881"},"modified":"2015-11-12T18:30:43","modified_gmt":"2015-11-12T18:30:43","slug":"solving-applied-problems-involving-hyperbolas","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/ivytech-collegealgebra\/chapter\/solving-applied-problems-involving-hyperbolas\/","title":{"raw":"Solving Applied Problems Involving Hyperbolas","rendered":"Solving Applied Problems Involving Hyperbolas"},"content":{"raw":"

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!\n\n[caption id=\"\" align=\"aligncenter\" width=\"488\"]\"descriptionFigure 10.<\/b> Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)[\/caption]\n\nThe first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In\u00a0Example 6\u00a0we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.\n<\/p>

\n

Example 6: Solving Applied Problems Involving Hyperbolas<\/h3>\nThe design layout of a cooling tower is shown in Figure 11. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 11.<\/b> Project design for a natural draft cooling tower[\/caption]\n\nFind the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola<\/strong>\u2014indicated by the intersection of dashed perpendicular lines in the figure\u2014is the origin of the coordinate plane. Round final values to four decimal places.\n\n<\/div>\n
\n

Solution<\/h3>\nWe are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], where the branches of the hyperbola form the sides of the cooling tower. We must find the values of [latex]{a}^{2}[\/latex] and [latex]{b}^{2}[\/latex] to complete the model.\n\nFirst, we find [latex]{a}^{2}[\/latex]. Recall that the length of the transverse axis of a hyperbola is [latex]2a[\/latex]. This length is represented by the distance where the sides are closest, which is given as [latex]\\text{ }65.3\\text{ }[\/latex] meters. So, [latex]2a=60[\/latex]. Therefore, [latex]a=30[\/latex] and [latex]{a}^{2}=900[\/latex].\n\nTo solve for [latex]{b}^{2}[\/latex], we need to substitute for [latex]x[\/latex] and [latex]y[\/latex] in our equation using a known point. To do this, we can use the dimensions of the tower to find some point [latex]\\left(x,y\\right)[\/latex] that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y<\/em>-axis bisects the tower, our x<\/em>-value can be represented by the radius of the top, or 36 meters. The y<\/em>-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,\n
[latex]\\begin{array}{ll}\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Standard form of horizontal hyperbola}.\\hfill \\\\ \\text{ }{b}^{2}=\\frac{{y}^{2}}{\\frac{{x}^{2}}{{a}^{2}}-1}\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Isolate }{b}^{2}\\hfill \\\\ \\text{ }=\\frac{{\\left(79.6\\right)}^{2}}{\\frac{{\\left(36\\right)}^{2}}{900}-1}\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Substitute for }{a}^{2},x,\\text{ and }y\\hfill \\\\ \\text{ }\\approx 14400.3636\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Round to four decimal places}\\hfill \\end{array}[\/latex]<\/div>\nThe sides of the tower can be modeled by the hyperbolic equation\n
[latex]\\frac{{x}^{2}}{900}-\\frac{{y}^{2}}{14400.3636 }=1,\\text{or}\\frac{{x}^{2}}{{30}^{2}}-\\frac{{y}^{2}}{{120.0015}^{2} }=1[\/latex]<\/div>\n<\/div>\n
\n

Try It 6<\/h3>\nA design for a cooling tower project is shown in Figure 12. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola\u2014indicated by the intersection of dashed perpendicular lines in the figure\u2014is the origin of the coordinate plane. Round final values to four decimal places.\n\n[caption id=\"\" align=\"aligncenter\" width=\"487\"]\"\"Figure 12<\/b>[\/caption]\n\nSolution<\/a>\n\n<\/div>","rendered":"

As we discussed at the beginning of this section, hyperbolas have real-world applications in many fields, such as astronomy, physics, engineering, and architecture. The design efficiency of hyperbolic cooling towers is particularly interesting. Cooling towers are used to transfer waste heat to the atmosphere and are often touted for their ability to generate power efficiently. Because of their hyperbolic form, these structures are able to withstand extreme winds while requiring less material than any other forms of their size and strength. For example, a 500-foot tower can be made of a reinforced concrete shell only 6 or 8 inches wide!<\/p>\n

\"description<\/p>\n

Figure 10.<\/b> Cooling towers at the Drax power station in North Yorkshire, United Kingdom (credit: Les Haines, Flickr)<\/p>\n<\/div>\n

The first hyperbolic towers were designed in 1914 and were 35 meters high. Today, the tallest cooling towers are in France, standing a remarkable 170 meters tall. In\u00a0Example 6\u00a0we will use the design layout of a cooling tower to find a hyperbolic equation that models its sides.\n<\/p>\n

\n

Example 6: Solving Applied Problems Involving Hyperbolas<\/h3>\n

The design layout of a cooling tower is shown in Figure 11. The tower stands 179.6 meters tall. The diameter of the top is 72 meters. At their closest, the sides of the tower are 60 meters apart.<\/p>\n

\"\"<\/p>\n

Figure 11.<\/b> Project design for a natural draft cooling tower<\/p>\n<\/div>\n

Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola<\/strong>\u2014indicated by the intersection of dashed perpendicular lines in the figure\u2014is the origin of the coordinate plane. Round final values to four decimal places.<\/p>\n<\/div>\n

\n

Solution<\/h3>\n

We are assuming the center of the tower is at the origin, so we can use the standard form of a horizontal hyperbola centered at the origin: [latex]\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1[\/latex], where the branches of the hyperbola form the sides of the cooling tower. We must find the values of [latex]{a}^{2}[\/latex] and [latex]{b}^{2}[\/latex] to complete the model.<\/p>\n

First, we find [latex]{a}^{2}[\/latex]. Recall that the length of the transverse axis of a hyperbola is [latex]2a[\/latex]. This length is represented by the distance where the sides are closest, which is given as [latex]\\text{ }65.3\\text{ }[\/latex] meters. So, [latex]2a=60[\/latex]. Therefore, [latex]a=30[\/latex] and [latex]{a}^{2}=900[\/latex].<\/p>\n

To solve for [latex]{b}^{2}[\/latex], we need to substitute for [latex]x[\/latex] and [latex]y[\/latex] in our equation using a known point. To do this, we can use the dimensions of the tower to find some point [latex]\\left(x,y\\right)[\/latex] that lies on the hyperbola. We will use the top right corner of the tower to represent that point. Since the y<\/em>-axis bisects the tower, our x<\/em>-value can be represented by the radius of the top, or 36 meters. The y<\/em>-value is represented by the distance from the origin to the top, which is given as 79.6 meters. Therefore,<\/p>\n

[latex]\\begin{array}{ll}\\frac{{x}^{2}}{{a}^{2}}-\\frac{{y}^{2}}{{b}^{2}}=1\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Standard form of horizontal hyperbola}.\\hfill \\\\ \\text{ }{b}^{2}=\\frac{{y}^{2}}{\\frac{{x}^{2}}{{a}^{2}}-1}\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Isolate }{b}^{2}\\hfill \\\\ \\text{ }=\\frac{{\\left(79.6\\right)}^{2}}{\\frac{{\\left(36\\right)}^{2}}{900}-1}\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Substitute for }{a}^{2},x,\\text{ and }y\\hfill \\\\ \\text{ }\\approx 14400.3636\\hfill & \\begin{array}{cccc}& & & \\end{array}\\text{Round to four decimal places}\\hfill \\end{array}[\/latex]<\/div>\n

The sides of the tower can be modeled by the hyperbolic equation<\/p>\n

[latex]\\frac{{x}^{2}}{900}-\\frac{{y}^{2}}{14400.3636 }=1,\\text{or}\\frac{{x}^{2}}{{30}^{2}}-\\frac{{y}^{2}}{{120.0015}^{2} }=1[\/latex]<\/div>\n<\/div>\n
\n

Try It 6<\/h3>\n

A design for a cooling tower project is shown in Figure 12. Find the equation of the hyperbola that models the sides of the cooling tower. Assume that the center of the hyperbola\u2014indicated by the intersection of dashed perpendicular lines in the figure\u2014is the origin of the coordinate plane. Round final values to four decimal places.<\/p>\n

\"\"<\/p>\n

Figure 12<\/b><\/p>\n<\/div>\n

Solution<\/a><\/p>\n<\/div>\n\n\t\t\t

\n\t\t\t

Candela Citations<\/h3>\n\t\t\t\t\t
\n\t\t\t\t\t\t
\n\t\t\t\t\t\t\t
CC licensed content, Specific attribution<\/div>