{"id":3200,"date":"2024-06-13T17:46:24","date_gmt":"2024-06-13T17:46:24","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=3200"},"modified":"2024-08-05T02:11:10","modified_gmt":"2024-08-05T02:11:10","slug":"related-rates-fresh-take","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/related-rates-fresh-take\/","title":{"raw":"Related Rates: Fresh Take","rendered":"Related Rates: Fresh Take"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Show how quantities change using derivatives and explore how these changes are connected<\/li>\r\n\t<li>Apply the chain rule to calculate how one changing quantity affects another<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Related-Rates Problem-Solving<\/h2>\r\n<div class=\"textbox shaded\">\r\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\r\n<ul>\r\n\t<li class=\"whitespace-normal break-words\">Related rates problems involve finding the rate of change of one quantity when it's related to another quantity whose rate of change is known.<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Use the chain rule to relate the rates of change of different quantities.<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Common Applications:\r\n\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Geometric problems (e.g., expanding circles, changing triangles)<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Physical scenarios (e.g., water flowing, objects moving)<\/li>\r\n<\/ul>\r\n<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Important Considerations:\r\n\r\n\r\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\r\n\t<li class=\"whitespace-normal break-words\">Draw a diagram if possible<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Don't substitute values too early (before differentiation)<\/li>\r\n\t<li class=\"whitespace-normal break-words\">Pay attention to units and sign conventions<\/li>\r\n<\/ul>\r\n<\/li>\r\n<\/ul>\r\n<p><strong>Problem-Solving Strategy<\/strong><\/p>\r\n<ol>\r\n\t<li>Assign variables to all relevant quantities<\/li>\r\n\t<li>Establish the given information and what needs to be found<\/li>\r\n\t<li>Develop an equation relating the variables<\/li>\r\n\t<li>Differentiate the equation with respect to time<\/li>\r\n\t<li>Substitute known values and solve for the unknown rate<\/li>\r\n<\/ol>\r\n<\/div>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>A conical water tank is being filled at a rate of [latex]10 \\text{ft}^3\/\\text{min}[\/latex]. The tank has a height of [latex]12 \\text{ft}[\/latex] and a base radius of [latex]6 \\text{ft}[\/latex]. At what rate is the water level rising when the water is [latex]4 \\text{ft}[\/latex] deep?<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"16696\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"16696\"]<\/p>\r\n<p>Assign variables:<\/p>\r\n<p>[latex]r [\/latex] = radius of water surface<br \/>\r\n[latex]h[\/latex] = height of water<br \/>\r\n[latex]V[\/latex] = volume of water<\/p>\r\n<p><br \/>\r\nGiven information:<\/p>\r\n<p>[latex]\\frac{dV}{dt} = 10[\/latex] ft\u00b3\/min<br \/>\r\nTank height = [latex]12[\/latex] ft, base radius = [latex]6[\/latex] ft<br \/>\r\nNeed to find [latex]\\frac{dh}{dt}[\/latex] when [latex]h = 4[\/latex] ft<\/p>\r\n<p><br \/>\r\nDevelop equation:<\/p>\r\n<p>Volume of a cone: [latex]V = \\frac{1}{3}\\pi r^2h[\/latex]<br \/>\r\nSimilar triangles: [latex]\\frac{r}{h} = \\frac{6}{12} = \\frac{1}{2}[\/latex], so [latex]r = \\frac{h}{2}[\/latex]<br \/>\r\n[latex]V = \\frac{1}{3}\\pi(\\frac{h}{2})^2h = \\frac{\\pi}{12}h^3[\/latex]<\/p>\r\n<p>Differentiate with respect to time:<\/p>\r\n<p>[latex]\\frac{dV}{dt} = \\frac{\\pi}{4}h^2\\frac{dh}{dt}[\/latex]<\/p>\r\n<p>Substitute and solve:<\/p>\r\n<p>[latex]10 = \\frac{\\pi}{4}(4^2)\\frac{dh}{dt}[\/latex]<br \/>\r\n[latex]\\frac{dh}{dt} = \\frac{10}{4\\pi} \\approx 0.796 , \\text{ft}\/\\text{min}[\/latex]<\/p>\r\n<p>Therefore, when the water is [latex]4 \\text{ft}[\/latex] deep, it's rising at a rate of approximately [latex]0.796 \\text{ft}\/\\text{min}[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>A street lamp is mounted at the top of a [latex]15[\/latex] foot tall pole. A man [latex]6 [\/latex] feet tall walks away from the pole with a speed of [latex]5 \\text{ft}\/\\text{sec}[\/latex] along a straight path. At what rate is the tip of his shadow moving when he is [latex]40 \\text{feet}[\/latex] from the pole?<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"442835\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"442835\"]<\/p>\r\n<p>Assign variables:<\/p>\r\n<p>[latex]x[\/latex] = distance of man from pole<br \/>\r\n[latex]s[\/latex] = length of shadow<br \/>\r\n[latex]h[\/latex] = height of pole ([latex]15 , \\text{ft}[\/latex])<br \/>\r\n[latex]H[\/latex] = height of man ([latex]6 , \\text{ft}[\/latex])<\/p>\r\n<p>Given information:<\/p>\r\n<p>[latex]\\frac{dx}{dt} = 5 , \\text{ft}\/\\text{sec}[\/latex]<br \/>\r\n[latex]x = 40 , \\text{ft}[\/latex]<br \/>\r\nNeed to find [latex]\\frac{ds}{dt}[\/latex]<\/p>\r\n<p>Develop equation:<\/p>\r\n<p>Similar triangles: [latex]\\frac{s}{x+s} = \\frac{h}{H}[\/latex]<br \/>\r\n[latex]\\frac{15}{6} = \\frac{x+s}{x}[\/latex]<br \/>\r\n[latex]5x = 2(x+s)[\/latex]<br \/>\r\n[latex]s = \\frac{3x}{2} - x = \\frac{x}{2}[\/latex]<\/p>\r\n<p>Differentiate with respect to time:<\/p>\r\n<p>[latex]\\frac{ds}{dt} = \\frac{1}{2}\\frac{dx}{dt}[\/latex]<\/p>\r\n<p>Substitute and solve:<\/p>\r\n<p>[latex]\\frac{ds}{dt} = \\frac{1}{2}(5) = 2.5 , \\text{ft}\/\\text{sec}[\/latex]<\/p>\r\n<p>Therefore, when the man is [latex]40 \\text{feet}[\/latex] from the pole, the tip of his shadow is moving at [latex]2.5 \\text{ft}\/\\text{sec}[\/latex].<\/p>\r\n<\/section>\r\n<section class=\"textbox example\" aria-label=\"Example\">\r\n<p>A circular oil slick is expanding at a constant rate of [latex]2[\/latex] m<sup>2 <\/sup>\/sec. How fast is the radius of the slick increasing when the radius is [latex]10[\/latex] meters?<\/p>\r\n<p><br \/>\r\n[reveal-answer q=\"110771\"]Show Answer[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"110771\"]<\/p>\r\n<p>Assign variables:<\/p>\r\n<p>[latex]A[\/latex] = area of oil slick<br \/>\r\n[latex]r[\/latex] = radius of oil slick<\/p>\r\n<p><br \/>\r\nGiven information:<\/p>\r\n<p>[latex]\\frac{dA}{dt} = 2 , \\text{m}^2\/\\text{sec}[\/latex]<br \/>\r\n[latex]r = 10 , \\text{m}[\/latex]<br \/>\r\nNeed to find [latex]\\frac{dr}{dt}[\/latex]<\/p>\r\n<p>Develop equation:<\/p>\r\n<p>Area of a circle: [latex]A = \\pi r^2[\/latex]<\/p>\r\n<p>Differentiate with respect to time:<\/p>\r\n<p>[latex]\\frac{dA}{dt} = 2\\pi r\\frac{dr}{dt}[\/latex]<\/p>\r\n<p>Substitute and solve:<\/p>\r\n<p>[latex]2 = 2\\pi(10)\\frac{dr}{dt}[\/latex]<br \/>\r\n[latex]\\frac{dr}{dt} = \\frac{2}{20\\pi} = \\frac{1}{10\\pi} \\approx 0.0318 , \\text{m}\/\\text{sec}[\/latex]<\/p>\r\n<p>Therefore, when the radius of the oil slick is [latex]10 \\text{meters}[\/latex], it's expanding at a rate of approximately [latex]0.0318 \\text{m}\/\\text{sec}[\/latex] or [latex]3.18 \\text{cm}\/\\text{sec}[\/latex].<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Show how quantities change using derivatives and explore how these changes are connected<\/li>\n<li>Apply the chain rule to calculate how one changing quantity affects another<\/li>\n<\/ul>\n<\/section>\n<h2>Related-Rates Problem-Solving<\/h2>\n<div class=\"textbox shaded\">\n<p><strong>The Main Idea\u00a0<\/strong><\/p>\n<ul>\n<li class=\"whitespace-normal break-words\">Related rates problems involve finding the rate of change of one quantity when it&#8217;s related to another quantity whose rate of change is known.<\/li>\n<li class=\"whitespace-normal break-words\">Use the chain rule to relate the rates of change of different quantities.<\/li>\n<li class=\"whitespace-normal break-words\">Common Applications:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Geometric problems (e.g., expanding circles, changing triangles)<\/li>\n<li class=\"whitespace-normal break-words\">Physical scenarios (e.g., water flowing, objects moving)<\/li>\n<\/ul>\n<\/li>\n<li class=\"whitespace-normal break-words\">Important Considerations:\n<ul class=\"-mt-1 list-disc space-y-2 pl-8\">\n<li class=\"whitespace-normal break-words\">Draw a diagram if possible<\/li>\n<li class=\"whitespace-normal break-words\">Don&#8217;t substitute values too early (before differentiation)<\/li>\n<li class=\"whitespace-normal break-words\">Pay attention to units and sign conventions<\/li>\n<\/ul>\n<\/li>\n<\/ul>\n<p><strong>Problem-Solving Strategy<\/strong><\/p>\n<ol>\n<li>Assign variables to all relevant quantities<\/li>\n<li>Establish the given information and what needs to be found<\/li>\n<li>Develop an equation relating the variables<\/li>\n<li>Differentiate the equation with respect to time<\/li>\n<li>Substitute known values and solve for the unknown rate<\/li>\n<\/ol>\n<\/div>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A conical water tank is being filled at a rate of [latex]10 \\text{ft}^3\/\\text{min}[\/latex]. The tank has a height of [latex]12 \\text{ft}[\/latex] and a base radius of [latex]6 \\text{ft}[\/latex]. At what rate is the water level rising when the water is [latex]4 \\text{ft}[\/latex] deep?<\/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=\"q16696\">Show Answer<\/button><\/p>\n<div id=\"q16696\" class=\"hidden-answer\" style=\"display: none\">\n<p>Assign variables:<\/p>\n<p>[latex]r[\/latex] = radius of water surface<br \/>\n[latex]h[\/latex] = height of water<br \/>\n[latex]V[\/latex] = volume of water<\/p>\n<p>\nGiven information:<\/p>\n<p>[latex]\\frac{dV}{dt} = 10[\/latex] ft\u00b3\/min<br \/>\nTank height = [latex]12[\/latex] ft, base radius = [latex]6[\/latex] ft<br \/>\nNeed to find [latex]\\frac{dh}{dt}[\/latex] when [latex]h = 4[\/latex] ft<\/p>\n<p>\nDevelop equation:<\/p>\n<p>Volume of a cone: [latex]V = \\frac{1}{3}\\pi r^2h[\/latex]<br \/>\nSimilar triangles: [latex]\\frac{r}{h} = \\frac{6}{12} = \\frac{1}{2}[\/latex], so [latex]r = \\frac{h}{2}[\/latex]<br \/>\n[latex]V = \\frac{1}{3}\\pi(\\frac{h}{2})^2h = \\frac{\\pi}{12}h^3[\/latex]<\/p>\n<p>Differentiate with respect to time:<\/p>\n<p>[latex]\\frac{dV}{dt} = \\frac{\\pi}{4}h^2\\frac{dh}{dt}[\/latex]<\/p>\n<p>Substitute and solve:<\/p>\n<p>[latex]10 = \\frac{\\pi}{4}(4^2)\\frac{dh}{dt}[\/latex]<br \/>\n[latex]\\frac{dh}{dt} = \\frac{10}{4\\pi} \\approx 0.796 , \\text{ft}\/\\text{min}[\/latex]<\/p>\n<p>Therefore, when the water is [latex]4 \\text{ft}[\/latex] deep, it&#8217;s rising at a rate of approximately [latex]0.796 \\text{ft}\/\\text{min}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A street lamp is mounted at the top of a [latex]15[\/latex] foot tall pole. A man [latex]6[\/latex] feet tall walks away from the pole with a speed of [latex]5 \\text{ft}\/\\text{sec}[\/latex] along a straight path. At what rate is the tip of his shadow moving when he is [latex]40 \\text{feet}[\/latex] from the pole?<\/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=\"q442835\">Show Answer<\/button><\/p>\n<div id=\"q442835\" class=\"hidden-answer\" style=\"display: none\">\n<p>Assign variables:<\/p>\n<p>[latex]x[\/latex] = distance of man from pole<br \/>\n[latex]s[\/latex] = length of shadow<br \/>\n[latex]h[\/latex] = height of pole ([latex]15 , \\text{ft}[\/latex])<br \/>\n[latex]H[\/latex] = height of man ([latex]6 , \\text{ft}[\/latex])<\/p>\n<p>Given information:<\/p>\n<p>[latex]\\frac{dx}{dt} = 5 , \\text{ft}\/\\text{sec}[\/latex]<br \/>\n[latex]x = 40 , \\text{ft}[\/latex]<br \/>\nNeed to find [latex]\\frac{ds}{dt}[\/latex]<\/p>\n<p>Develop equation:<\/p>\n<p>Similar triangles: [latex]\\frac{s}{x+s} = \\frac{h}{H}[\/latex]<br \/>\n[latex]\\frac{15}{6} = \\frac{x+s}{x}[\/latex]<br \/>\n[latex]5x = 2(x+s)[\/latex]<br \/>\n[latex]s = \\frac{3x}{2} - x = \\frac{x}{2}[\/latex]<\/p>\n<p>Differentiate with respect to time:<\/p>\n<p>[latex]\\frac{ds}{dt} = \\frac{1}{2}\\frac{dx}{dt}[\/latex]<\/p>\n<p>Substitute and solve:<\/p>\n<p>[latex]\\frac{ds}{dt} = \\frac{1}{2}(5) = 2.5 , \\text{ft}\/\\text{sec}[\/latex]<\/p>\n<p>Therefore, when the man is [latex]40 \\text{feet}[\/latex] from the pole, the tip of his shadow is moving at [latex]2.5 \\text{ft}\/\\text{sec}[\/latex].<\/p>\n<\/section>\n<section class=\"textbox example\" aria-label=\"Example\">\n<p>A circular oil slick is expanding at a constant rate of [latex]2[\/latex] m<sup>2 <\/sup>\/sec. How fast is the radius of the slick increasing when the radius is [latex]10[\/latex] meters?<\/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=\"q110771\">Show Answer<\/button><\/p>\n<div id=\"q110771\" class=\"hidden-answer\" style=\"display: none\"><\/div>\n<\/div>\n<p>Assign variables:<\/p>\n<p>[latex]A[\/latex] = area of oil slick<br \/>\n[latex]r[\/latex] = radius of oil slick<\/p>\n<p>\nGiven information:<\/p>\n<p>[latex]\\frac{dA}{dt} = 2 , \\text{m}^2\/\\text{sec}[\/latex]<br \/>\n[latex]r = 10 , \\text{m}[\/latex]<br \/>\nNeed to find [latex]\\frac{dr}{dt}[\/latex]<\/p>\n<p>Develop equation:<\/p>\n<p>Area of a circle: [latex]A = \\pi r^2[\/latex]<\/p>\n<p>Differentiate with respect to time:<\/p>\n<p>[latex]\\frac{dA}{dt} = 2\\pi r\\frac{dr}{dt}[\/latex]<\/p>\n<p>Substitute and solve:<\/p>\n<p>[latex]2 = 2\\pi(10)\\frac{dr}{dt}[\/latex]<br \/>\n[latex]\\frac{dr}{dt} = \\frac{2}{20\\pi} = \\frac{1}{10\\pi} \\approx 0.0318 , \\text{m}\/\\text{sec}[\/latex]<\/p>\n<p>Therefore, when the radius of the oil slick is [latex]10 \\text{meters}[\/latex], it&#8217;s expanding at a rate of approximately [latex]0.0318 \\text{m}\/\\text{sec}[\/latex] or [latex]3.18 \\text{cm}\/\\text{sec}[\/latex].<\/p>\n<\/div>\n<\/div>\n<\/section>\n","protected":false},"author":15,"menu_order":9,"template":"","meta":{"_candela_citation":"[]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":652,"module-header":"fresh_take","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\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3200"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":5,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3200\/revisions"}],"predecessor-version":[{"id":3830,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3200\/revisions\/3830"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/652"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/3200\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=3200"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=3200"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=3200"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=3200"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}