{"id":470,"date":"2025-02-13T19:45:16","date_gmt":"2025-02-13T19:45:16","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/physical-applications-fresh-take\/"},"modified":"2025-02-13T19:45:16","modified_gmt":"2025-02-13T19:45:16","slug":"physical-applications-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus2\/chapter\/physical-applications-fresh-take\/","title":{"raw":"Physical Applications: Fresh Take","rendered":"Physical Applications: Fresh Take"},"content":{"raw":"\n<section class=\"textbox learningGoals\">\n<ul>\n\t<li>Calculate the mass of linear and circular objects using their density distributions<\/li>\n\t<li>Compute the work required in various situations, such as pumping fluids or moving objects along a path<\/li>\n\t<li>Determine the force exerted by water on a vertical surface underwater<\/li>\n<\/ul>\n<\/section>\n<h2>Mass and Density<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Mass-Density Formula for One-Dimensional Objects:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">For a thin rod along the [latex]x[\/latex]-axis from [latex]a[\/latex] to [latex]b[\/latex]: [latex]m = \\int_a^b \\rho(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Mass-Density Formula for Two-Dimensional Disks:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">For a disk of radius [latex]r[\/latex] with radial density: [latex]m = \\int_0^r 2\\pi x \\rho(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Concept of Density Functions:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Linear density: mass per unit length<\/li>\n\t<li class=\"whitespace-normal break-words\">Area density: mass per unit area<\/li>\n\t<li class=\"whitespace-normal break-words\">Radial density: density varying along the radius<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><strong>Problem-Solving Strategies&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">One-Dimensional Case:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Partition the rod into segments<\/li>\n\t<li class=\"whitespace-normal break-words\">Approximate mass of each segment: [latex]m_i \\approx \\rho(x_i^*) \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Sum and take the limit as [latex]n[\/latex] approaches infinity<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Two-Dimensional Case:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Partition the disk into concentric washers<\/li>\n\t<li class=\"whitespace-normal break-words\">Approximate area of each washer: [latex]A_i \\approx 2\\pi x_i^* \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Approximate mass of each washer: [latex]m_i \\approx 2\\pi x_i^* \\rho(x_i^*) \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Sum and take the limit as [latex]n[\/latex] approaches infinity<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the mass of a rod with density [latex]\\rho(x) = \\sin x[\/latex] over the interval [latex][\\frac{\\pi}{2}, \\pi][\/latex].<\/p>\n<p><br>\n[reveal-answer q=\"344543\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"344543\"]<\/p>\n<p style=\"text-align: center;\">[latex]<br>\n\\begin{array}{rcl}<br>\nm &amp;=&amp; \\int_{\\frac{\\pi}{2}}^{\\pi} \\rho(x) dx \\\\<br>\n&amp;=&amp; \\int_{\\frac{\\pi}{2}}^{\\pi} \\sin x dx \\\\<br>\n&amp;=&amp; [-\\cos x]_{\\frac{\\pi}{2}}^{\\pi} \\\\<br>\n&amp;=&amp; (-\\cos \\pi) - (-\\cos \\frac{\\pi}{2}) \\\\<br>\n&amp;=&amp; 1 - 0 = 1<br>\n\\end{array}<br>\n[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider a thin rod oriented on the [latex]x[\/latex]-axis over the interval [latex]\\left[1,3\\right].[\/latex] If the density of the rod is given by [latex]\\rho (x)=2{x}^{2}+3,[\/latex] what is the mass of the rod?<\/p>\n<p>[reveal-answer q=\"88003546\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"88003546\"]<\/p>\n<p id=\"fs-id1167794296202\">[latex]\\frac{70}{3}[\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?si=mVrmQDI1wQ35XITX?controls=0&amp;start=134\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\n<p>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\/2.5PhysicalApplications134to181_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Calculate the mass of a disk with radius [latex]4[\/latex] and radial density [latex]\\rho(x) = \\sqrt{x}[\/latex].<\/p>\n<p><br>\n[reveal-answer q=\"670779\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"670779\"]<\/p>\n<p style=\"text-align: center;\">[latex]<br>\n\\begin{array}{rcl}<br>\nm &amp;=&amp; \\int_0^4 2\\pi x \\rho(x) dx \\\\<br>\n&amp;=&amp; \\int_0^4 2\\pi x \\sqrt{x} dx \\\\<br>\n&amp;=&amp; 2\\pi \\int_0^4 x^{3\/2} dx \\\\<br>\n&amp;=&amp; 2\\pi [\\frac{2}{5}x^{5\/2}]_0^4 \\\\<br>\n&amp;=&amp; 2\\pi \\cdot \\frac{2}{5} \\cdot 32 \\\\<br>\n&amp;=&amp; \\frac{128\\pi}{5}<br>\n\\end{array}<br>\n[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]\\rho (x)=3x+2[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius [latex]2[\/latex].<\/p>\n\n[reveal-answer q=\"fs-id1167793547373\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793547373\"]\n\n<p id=\"fs-id1167793547373\">[latex]24\\pi [\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=397&amp;end=477&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>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\/2.5PhysicalApplications397to477_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<h2>Work Done by a Force<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Definition of Work:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Work is the energy transfer when a force moves an object<\/li>\n\t<li class=\"whitespace-normal break-words\">Units: Joules (J); Force in Newtons (N), Distance in meters (m)<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Work Formula for Constant Force: [latex]W = F \\cdot d[\/latex] (Force \u00d7 Distance)<\/li>\n\t<li class=\"whitespace-normal break-words\">Work Formula for Variable Force: [latex]W = \\int_a^b F(x) dx[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Application to Springs:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Hooke's Law: [latex]F(x) = kx[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]k[\/latex] is the spring constant (N\/m)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><strong>Problem-Solving Strategy<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Partition the interval [latex][a,b][\/latex] into [latex]n[\/latex] segments<\/li>\n\t<li class=\"whitespace-normal break-words\">Approximate work in each segment: [latex]W_i \\approx F(x_i^*) \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Sum the work over all segments: [latex]W \\approx \\sum_{i=1}^n F(x_i^*) \\Delta x[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">Take the limit as [latex]n \\to \\infty[\/latex] to get the exact work: [latex]W = \\lim_{n \\to \\infty} \\sum_{i=1}^n F(x_i^*) \\Delta x = \\int_a^b F(x) dx[\/latex]<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Suppose it takes a force of [latex]8[\/latex] lb to stretch a spring [latex]6[\/latex] in. from the equilibrium position. How much work is done to stretch the spring [latex]1[\/latex] ft from the equilibrium position?<\/p>\n\n[reveal-answer q=\"fs-id1167793281283\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793281283\"]\n\n<p id=\"fs-id1167793281283\">[latex]8[\/latex] ft-lb<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=811&amp;end=903&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\n<p>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\/2.5PhysicalApplications811to903_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"2.5 Physical Applications\" here (opens in new window)<\/a>.[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A spring requires a [latex]10[\/latex] N force to compress it [latex]0.2[\/latex] m. How much work is done to stretch it [latex]0.5[\/latex] m?<\/p>\n<p><br>\n[reveal-answer q=\"430870\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"430870\"]<\/p>\n<p>Find spring constant [latex]k[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]10 = k(0.2)[\/latex], so [latex]k = 50[\/latex] N\/m<\/p>\n<p>Set up the work integral:<\/p>\n<p style=\"text-align: center;\">[latex]W = \\int_0^{0.5} kx dx = \\int_0^{0.5} 50x dx[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]<br>\n\\begin{array}{rcl}<br>\nW &amp;=&amp; 50 \\int_0^{0.5} x dx \\\\<br>\n&amp;=&amp; 50 [\\frac{1}{2}x^2]_0^{0.5} \\\\<br>\n&amp;=&amp; 25 [(0.5)^2 - 0] \\\\<br>\n&amp;=&amp; 6.25 \\text{ J}<br>\n\\end{array}<br>\n[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h2>Work Done in Pumping<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Work in Pumping:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Calculating work required to pump water out of tanks<\/li>\n\t<li class=\"whitespace-normal break-words\">Varies based on tank shape and water level<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Key Principle:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Work = Force (Weight of water) \u00d7 Distance (Height lifted)<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">General Work Formula: [latex]W = \\int_a^b F(x) dx[\/latex], where [latex]F(x)[\/latex] depends on tank shape<\/li>\n\t<li class=\"whitespace-normal break-words\">Weight-Density of Water:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]9800[\/latex] N\/m\u00b3 (metric) or [latex]62.4[\/latex] lb\/ft\u00b3 (imperial)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Sketch the tank and choose a coordinate system<\/li>\n\t<li class=\"whitespace-normal break-words\">Calculate volume of a representative water layer<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine force using weight-density<\/li>\n\t<li class=\"whitespace-normal break-words\">Calculate lifting distance for the layer<\/li>\n\t<li class=\"whitespace-normal break-words\">Estimate work for the layer: Force \u00d7 Distance<\/li>\n\t<li class=\"whitespace-normal break-words\">Sum work for all layers (Riemann sum)<\/li>\n\t<li class=\"whitespace-normal break-words\">Take limit and evaluate integral for exact work<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Calculate work to pump water from a full inverted conical tank (height [latex]12[\/latex] ft, base radius [latex]4[\/latex] ft) until [latex]4[\/latex] ft of water remains.<\/p>\n<p><br>\n[reveal-answer q=\"56634\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"56634\"]<\/p>\n<p>Set up coordinate system: [latex]x[\/latex]-axis vertical, origin at top<\/p>\n<p>Partition interval [latex][0, 8][\/latex] ([latex]8[\/latex] ft of water pumped)<\/p>\n<p>Use similar triangles to find radius at each layer:<\/p>\n<p style=\"text-align: center;\">[latex]r_i = 4 - \\frac{x_i^*}{3}[\/latex]<\/p>\n<p>Volume of layer: [latex]V_i = \\pi (4-\\frac{x_i^*}{3})^2 \\Delta x[\/latex]<\/p>\n<p>Force: [latex]F_i \\approx 62.4\\pi (4-\\frac{x_i^*}{3})^2 \\Delta x[\/latex]<\/p>\n<p>Distance lifted: approximately [latex]x_i^*[\/latex]<\/p>\n<p>Work integral:<\/p>\n<p style=\"text-align: center;\">[latex]W = \\int_0^8 62.4\\pi x(4-\\frac{x}{3})^2 dx[\/latex]<\/p>\n<p>Evaluate integral:<\/p>\n<p style=\"text-align: center;\">[latex]W \\approx 33,450[\/latex] ft-lb<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>A tank is in the shape of an inverted cone, with height [latex]10[\/latex] ft and base radius [latex]6[\/latex] ft. The tank is filled to a depth of [latex]8[\/latex] ft to start with, and water is pumped over the upper edge of the tank until [latex]3[\/latex] ft of water remain in the tank. How much work is required to pump out that amount of water?<\/p>\n\n[reveal-answer q=\"fs-id1167793553760\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167793553760\"]\n\n<p id=\"fs-id1167793553760\">Approximately [latex]43,255.2[\/latex] ft-lb<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<h2>Hydrostatic Force and Pressure<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n\t<li class=\"whitespace-normal break-words\">Hydrostatic Pressure:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Pressure exerted by a fluid at rest due to its weight<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]p = \\rho s[\/latex], where [latex]\\rho[\/latex] is weight density and [latex]s[\/latex] is depth<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Force on Horizontal Surface:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]F = \\rho As[\/latex], where [latex]A[\/latex] is area<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Force on Vertical or Angled Surface:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">[latex]F = \\int_a^b \\rho w(x)s(x)dx[\/latex]<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]w(x)[\/latex]: width function<\/li>\n\t<li class=\"whitespace-normal break-words\">[latex]s(x)[\/latex]: depth function<\/li>\n<\/ul>\n<\/li>\n\t<li class=\"whitespace-normal break-words\">Pascal's Principle:\n\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Pressure at a given depth is the same in all directions<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n\t<li class=\"whitespace-normal break-words\">Sketch the situation and choose a coordinate system<\/li>\n\t<li class=\"whitespace-normal break-words\">Determine depth [latex]s(x)[\/latex] and width [latex]w(x)[\/latex] functions<\/li>\n\t<li class=\"whitespace-normal break-words\">Identify the weight density [latex]\\rho[\/latex] of the liquid<\/li>\n\t<li class=\"whitespace-normal break-words\">Set up and evaluate the force integral: [latex]F = \\int_a^b \\rho w(x)s(x)dx[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Calculate the force on one end of a water trough [latex]15[\/latex] ft long, with ends shaped like inverted isosceles triangles (base [latex]8[\/latex] ft, height [latex]3[\/latex] ft).<\/p>\n<p><br>\n[reveal-answer q=\"334412\"]Show Answer[\/reveal-answer]<br>\n[hidden-answer a=\"334412\"]<\/p>\n<p>Set coordinate system: [latex]x[\/latex]-axis vertical, [latex]x = 0[\/latex] at top<\/p>\n<p>Depth function: [latex]s(x) = x[\/latex]<\/p>\n<p>Width function (similar triangles): [latex]w(x) = 8 - \\frac{8}{3}x[\/latex]<\/p>\n<p>Force integral:<\/p>\n<p style=\"text-align: center;\">[latex]<br>\n\\begin{array}{rcl}<br>\nF &amp;=&amp; \\int_0^3 62.4(8 - \\frac{8}{3}x)x dx \\\\<br>\n&amp;=&amp; 62.4 \\int_0^3 (8x - \\frac{8}{3}x^2) dx \\\\<br>\n&amp;=&amp; 62.4 [4x^2 - \\frac{8}{9}x^3]_0^3 \\\\<br>\n&amp;=&amp; 748.8 \\text{ lb}<br>\n\\end{array}<br>\n[\/latex]<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n<section class=\"textbox example\">\n<p>When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?<\/p>\n\n[reveal-answer q=\"fs-id1167794138229\"]Show Solution[\/reveal-answer]<br>\n[hidden-answer a=\"fs-id1167794138229\"]\n\n<p id=\"fs-id1167794138229\">Approximately [latex]7,164,520,000[\/latex] lb or [latex]3,582,260[\/latex] t<\/p>\n<p>[\/hidden-answer]<\/p>\n<\/section>\n","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate the mass of linear and circular objects using their density distributions<\/li>\n<li>Compute the work required in various situations, such as pumping fluids or moving objects along a path<\/li>\n<li>Determine the force exerted by water on a vertical surface underwater<\/li>\n<\/ul>\n<\/section>\n<h2>Mass and Density<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Mass-Density Formula for One-Dimensional Objects:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For a thin rod along the [latex]x[\/latex]-axis from [latex]a[\/latex] to [latex]b[\/latex]: [latex]m = \\int_a^b \\rho(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Mass-Density Formula for Two-Dimensional Disks:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">For a disk of radius [latex]r[\/latex] with radial density: [latex]m = \\int_0^r 2\\pi x \\rho(x) dx[\/latex]<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Concept of Density Functions:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Linear density: mass per unit length<\/li>\n<li class=\"whitespace-normal break-words\">Area density: mass per unit area<\/li>\n<li class=\"whitespace-normal break-words\">Radial density: density varying along the radius<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><strong>Problem-Solving Strategies&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">One-Dimensional Case:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Partition the rod into segments<\/li>\n<li class=\"whitespace-normal break-words\">Approximate mass of each segment: [latex]m_i \\approx \\rho(x_i^*) \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Sum and take the limit as [latex]n[\/latex] approaches infinity<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Two-Dimensional Case:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Partition the disk into concentric washers<\/li>\n<li class=\"whitespace-normal break-words\">Approximate area of each washer: [latex]A_i \\approx 2\\pi x_i^* \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Approximate mass of each washer: [latex]m_i \\approx 2\\pi x_i^* \\rho(x_i^*) \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Sum and take the limit as [latex]n[\/latex] approaches infinity<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Find the mass of a rod with density [latex]\\rho(x) = \\sin x[\/latex] over the interval [latex][\\frac{\\pi}{2}, \\pi][\/latex].<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q344543\">Show Answer<\/button><\/p>\n<div id=\"q344543\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]<br \/> \\begin{array}{rcl}<br \/> m &=& \\int_{\\frac{\\pi}{2}}^{\\pi} \\rho(x) dx \\\\<br \/> &=& \\int_{\\frac{\\pi}{2}}^{\\pi} \\sin x dx \\\\<br \/> &=& [-\\cos x]_{\\frac{\\pi}{2}}^{\\pi} \\\\<br \/> &=& (-\\cos \\pi) - (-\\cos \\frac{\\pi}{2}) \\\\<br \/> &=& 1 - 0 = 1<br \/> \\end{array}<br \/>[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Consider a thin rod oriented on the [latex]x[\/latex]-axis over the interval [latex]\\left[1,3\\right].[\/latex] If the density of the rod is given by [latex]\\rho (x)=2{x}^{2}+3,[\/latex] what is the mass of the rod?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q88003546\">Show Solution<\/button><\/p>\n<div id=\"q88003546\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794296202\">[latex]\\frac{70}{3}[\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<p><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?si=mVrmQDI1wQ35XITX?controls=0&amp;start=134\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>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\/2.5PhysicalApplications134to181_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Calculate the mass of a disk with radius [latex]4[\/latex] and radial density [latex]\\rho(x) = \\sqrt{x}[\/latex].<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q670779\">Show Answer<\/button><\/p>\n<div id=\"q670779\" class=\"hidden-answer\" style=\"display: none\">\n<p style=\"text-align: center;\">[latex]<br \/> \\begin{array}{rcl}<br \/> m &=& \\int_0^4 2\\pi x \\rho(x) dx \\\\<br \/> &=& \\int_0^4 2\\pi x \\sqrt{x} dx \\\\<br \/> &=& 2\\pi \\int_0^4 x^{3\/2} dx \\\\<br \/> &=& 2\\pi [\\frac{2}{5}x^{5\/2}]_0^4 \\\\<br \/> &=& 2\\pi \\cdot \\frac{2}{5} \\cdot 32 \\\\<br \/> &=& \\frac{128\\pi}{5}<br \/> \\end{array}<br \/>[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>Let [latex]\\rho (x)=3x+2[\/latex] represent the radial density of a disk. Calculate the mass of a disk of radius [latex]2[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793547373\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793547373\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793547373\">[latex]24\\pi[\/latex]<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=397&amp;end=477&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/2.5PhysicalApplications397to477_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<h2>Work Done by a Force<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Definition of Work:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Work is the energy transfer when a force moves an object<\/li>\n<li class=\"whitespace-normal break-words\">Units: Joules (J); Force in Newtons (N), Distance in meters (m)<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Work Formula for Constant Force: [latex]W = F \\cdot d[\/latex] (Force \u00d7 Distance)<\/li>\n<li class=\"whitespace-normal break-words\">Work Formula for Variable Force: [latex]W = \\int_a^b F(x) dx[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Application to Springs:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Hooke&#8217;s Law: [latex]F(x) = kx[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]k[\/latex] is the spring constant (N\/m)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><strong>Problem-Solving Strategy<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Partition the interval [latex][a,b][\/latex] into [latex]n[\/latex] segments<\/li>\n<li class=\"whitespace-normal break-words\">Approximate work in each segment: [latex]W_i \\approx F(x_i^*) \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Sum the work over all segments: [latex]W \\approx \\sum_{i=1}^n F(x_i^*) \\Delta x[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">Take the limit as [latex]n \\to \\infty[\/latex] to get the exact work: [latex]W = \\lim_{n \\to \\infty} \\sum_{i=1}^n F(x_i^*) \\Delta x = \\int_a^b F(x) dx[\/latex]<\/li>\n<\/ul>\n<\/div>\n<section class=\"textbox example\">\n<p>Suppose it takes a force of [latex]8[\/latex] lb to stretch a spring [latex]6[\/latex] in. from the equilibrium position. How much work is done to stretch the spring [latex]1[\/latex] ft from the equilibrium position?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793281283\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793281283\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793281283\">[latex]8[\/latex] ft-lb<\/p>\n<p>Watch the following video to see the worked solution to this example.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/jZAEKDiWkHA?controls=0&amp;start=811&amp;end=903&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p>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\/2.5PhysicalApplications811to903_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;2.5 Physical Applications&#8221; here (opens in new window)<\/a>.<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A spring requires a [latex]10[\/latex] N force to compress it [latex]0.2[\/latex] m. How much work is done to stretch it [latex]0.5[\/latex] m?<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q430870\">Show Answer<\/button><\/p>\n<div id=\"q430870\" class=\"hidden-answer\" style=\"display: none\">\n<p>Find spring constant [latex]k[\/latex]:<\/p>\n<p style=\"text-align: center;\">[latex]10 = k(0.2)[\/latex], so [latex]k = 50[\/latex] N\/m<\/p>\n<p>Set up the work integral:<\/p>\n<p style=\"text-align: center;\">[latex]W = \\int_0^{0.5} kx dx = \\int_0^{0.5} 50x dx[\/latex]<\/p>\n<p>Evaluate the integral:<\/p>\n<p style=\"text-align: center;\">[latex]<br \/> \\begin{array}{rcl}<br \/> W &=& 50 \\int_0^{0.5} x dx \\\\<br \/> &=& 50 [\\frac{1}{2}x^2]_0^{0.5} \\\\<br \/> &=& 25 [(0.5)^2 - 0] \\\\<br \/> &=& 6.25 \\text{ J}<br \/> \\end{array}<br \/>[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Work Done in Pumping<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Work in Pumping:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Calculating work required to pump water out of tanks<\/li>\n<li class=\"whitespace-normal break-words\">Varies based on tank shape and water level<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Key Principle:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Work = Force (Weight of water) \u00d7 Distance (Height lifted)<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">General Work Formula: [latex]W = \\int_a^b F(x) dx[\/latex], where [latex]F(x)[\/latex] depends on tank shape<\/li>\n<li class=\"whitespace-normal break-words\">Weight-Density of Water:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]9800[\/latex] N\/m\u00b3 (metric) or [latex]62.4[\/latex] lb\/ft\u00b3 (imperial)<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Sketch the tank and choose a coordinate system<\/li>\n<li class=\"whitespace-normal break-words\">Calculate volume of a representative water layer<\/li>\n<li class=\"whitespace-normal break-words\">Determine force using weight-density<\/li>\n<li class=\"whitespace-normal break-words\">Calculate lifting distance for the layer<\/li>\n<li class=\"whitespace-normal break-words\">Estimate work for the layer: Force \u00d7 Distance<\/li>\n<li class=\"whitespace-normal break-words\">Sum work for all layers (Riemann sum)<\/li>\n<li class=\"whitespace-normal break-words\">Take limit and evaluate integral for exact work<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Calculate work to pump water from a full inverted conical tank (height [latex]12[\/latex] ft, base radius [latex]4[\/latex] ft) until [latex]4[\/latex] ft of water remains.<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q56634\">Show Answer<\/button><\/p>\n<div id=\"q56634\" class=\"hidden-answer\" style=\"display: none\">\n<p>Set up coordinate system: [latex]x[\/latex]-axis vertical, origin at top<\/p>\n<p>Partition interval [latex][0, 8][\/latex] ([latex]8[\/latex] ft of water pumped)<\/p>\n<p>Use similar triangles to find radius at each layer:<\/p>\n<p style=\"text-align: center;\">[latex]r_i = 4 - \\frac{x_i^*}{3}[\/latex]<\/p>\n<p>Volume of layer: [latex]V_i = \\pi (4-\\frac{x_i^*}{3})^2 \\Delta x[\/latex]<\/p>\n<p>Force: [latex]F_i \\approx 62.4\\pi (4-\\frac{x_i^*}{3})^2 \\Delta x[\/latex]<\/p>\n<p>Distance lifted: approximately [latex]x_i^*[\/latex]<\/p>\n<p>Work integral:<\/p>\n<p style=\"text-align: center;\">[latex]W = \\int_0^8 62.4\\pi x(4-\\frac{x}{3})^2 dx[\/latex]<\/p>\n<p>Evaluate integral:<\/p>\n<p style=\"text-align: center;\">[latex]W \\approx 33,450[\/latex] ft-lb<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>A tank is in the shape of an inverted cone, with height [latex]10[\/latex] ft and base radius [latex]6[\/latex] ft. The tank is filled to a depth of [latex]8[\/latex] ft to start with, and water is pumped over the upper edge of the tank until [latex]3[\/latex] ft of water remain in the tank. How much work is required to pump out that amount of water?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167793553760\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167793553760\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167793553760\">Approximately [latex]43,255.2[\/latex] ft-lb<\/p>\n<\/div>\n<\/div>\n<\/section>\n<h2>Hydrostatic Force and Pressure<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea&nbsp;<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Hydrostatic Pressure:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Pressure exerted by a fluid at rest due to its weight<\/li>\n<li class=\"whitespace-normal break-words\">[latex]p = \\rho s[\/latex], where [latex]\\rho[\/latex] is weight density and [latex]s[\/latex] is depth<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Force on Horizontal Surface:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]F = \\rho As[\/latex], where [latex]A[\/latex] is area<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Force on Vertical or Angled Surface:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">[latex]F = \\int_a^b \\rho w(x)s(x)dx[\/latex]<\/li>\n<li class=\"whitespace-normal break-words\">[latex]w(x)[\/latex]: width function<\/li>\n<li class=\"whitespace-normal break-words\">[latex]s(x)[\/latex]: depth function<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Pascal&#8217;s Principle:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Pressure at a given depth is the same in all directions<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p class=\"font-bold\"><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol class=\"-mt-1 list-decimal space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Sketch the situation and choose a coordinate system<\/li>\n<li class=\"whitespace-normal break-words\">Determine depth [latex]s(x)[\/latex] and width [latex]w(x)[\/latex] functions<\/li>\n<li class=\"whitespace-normal break-words\">Identify the weight density [latex]\\rho[\/latex] of the liquid<\/li>\n<li class=\"whitespace-normal break-words\">Set up and evaluate the force integral: [latex]F = \\int_a^b \\rho w(x)s(x)dx[\/latex]<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>Calculate the force on one end of a water trough [latex]15[\/latex] ft long, with ends shaped like inverted isosceles triangles (base [latex]8[\/latex] ft, height [latex]3[\/latex] ft).<\/p>\n<div class=\"wp-nocaption \"><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q334412\">Show Answer<\/button><\/p>\n<div id=\"q334412\" class=\"hidden-answer\" style=\"display: none\">\n<p>Set coordinate system: [latex]x[\/latex]-axis vertical, [latex]x = 0[\/latex] at top<\/p>\n<p>Depth function: [latex]s(x) = x[\/latex]<\/p>\n<p>Width function (similar triangles): [latex]w(x) = 8 - \\frac{8}{3}x[\/latex]<\/p>\n<p>Force integral:<\/p>\n<p style=\"text-align: center;\">[latex]<br \/> \\begin{array}{rcl}<br \/> F &=& \\int_0^3 62.4(8 - \\frac{8}{3}x)x dx \\\\<br \/> &=& 62.4 \\int_0^3 (8x - \\frac{8}{3}x^2) dx \\\\<br \/> &=& 62.4 [4x^2 - \\frac{8}{9}x^3]_0^3 \\\\<br \/> &=& 748.8 \\text{ lb}<br \/> \\end{array}<br \/>[\/latex]<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>When the reservoir is at its average level, the surface of the water is about 50 ft below where it would be if the reservoir were full. What is the force on the face of the dam under these circumstances?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1167794138229\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1167794138229\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1167794138229\">Approximately [latex]7,164,520,000[\/latex] lb or [latex]3,582,260[\/latex] t<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":6,"menu_order":11,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":450,"module-header":"","content_attributions":[],"internal_book_links":[],"video_content":null,"cc_video_embed_content":{"cc_scripts":"","media_targets":[]},"try_it_collection":null,"_links":{"self":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/470"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/users\/6"}],"version-history":[{"count":0,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/470\/revisions"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/parts\/450"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapters\/470\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/media?parent=470"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/pressbooks\/v2\/chapter-type?post=470"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/contributor?post=470"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus2\/wp-json\/wp\/v2\/license?post=470"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}