{"id":295,"date":"2023-09-20T22:48:53","date_gmt":"2023-09-20T22:48:53","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/related-rates-problem-solving\/"},"modified":"2024-08-05T02:10:24","modified_gmt":"2024-08-05T02:10:24","slug":"related-rates-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/related-rates-learn-it-1\/","title":{"raw":"Related Rates: Learn It 1","rendered":"Related Rates: Learn It 1"},"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<p>We have seen that for quantities that are changing over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities.<\/p>\r\n<h3>Setting up Related-Rates Problems<\/h3>\r\n<p id=\"fs-id1165043085174\">In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, [latex]V[\/latex], is related to the rate of change in the radius, [latex]r[\/latex]. In this case, we say that [latex]\\frac{dV}{dt}[\/latex] and [latex]\\frac{dr}{dt}[\/latex] are <strong>related rates<\/strong> because [latex]V[\/latex] is related to [latex]r[\/latex]. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate of change.<\/p>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1165043084830\">A spherical balloon is being filled with air at the constant rate of [latex]2 \\, \\frac{\\text{cm}^3}{\\text{sec}}[\/latex] (Figure 1). How fast is the radius increasing when the radius is [latex]3\\, \\text{cm}[\/latex]?<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"900\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11210704\/CNX_Calc_Figure_04_01_001.jpg\" alt=\"Three balloons are shown at Times 1, 2, and 3. These balloons increase in volume and radius as time increases.\" width=\"900\" height=\"297\" \/> Figure 1. As the balloon is being filled with air, both the radius and the volume are increasing with respect to time.[\/caption]\r\n\r\n<p id=\"fs-id1165043021803\">The volume of a sphere of radius [latex]r[\/latex] centimeters is<\/p>\r\n<div id=\"fs-id1165042990072\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=\\frac{4}{3}\\pi r^3 \\, \\text{cm}^3[\/latex]<\/div>\r\n<p id=\"fs-id1165042881658\">Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, [latex]t[\/latex] seconds after beginning to fill the balloon with air, the volume of air in the balloon is<\/p>\r\n<div id=\"fs-id1165043013475\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(t)=\\frac{4}{3}\\pi [r(t)]^3 \\, \\text{cm}^3[\/latex]<\/div>\r\n<p id=\"fs-id1165042941218\">Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation<\/p>\r\n<div id=\"fs-id1165043119522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V^{\\prime}(t)=4\\pi [r(t)]^2 \\cdot r^{\\prime}(t)[\/latex]<\/div>\r\n<p id=\"fs-id1165043111378\">The balloon is being filled with air at the constant rate of 2 cm<sup>3<\/sup>\/sec, so [latex]V^{\\prime}(t)=2 \\, \\text{cm}^3 \/ \\sec[\/latex]. Therefore,<\/p>\r\n<div id=\"fs-id1165043009965\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]2 \\, \\text{cm}^3 \/ \\sec =(4\\pi [r(t)]^2 \\, \\text{cm}^2) \\cdot (r^{\\prime}(t) \\, \\text{cm\/sec})[\/latex],<\/div>\r\n<p id=\"fs-id1165043051118\">which implies<\/p>\r\n<div id=\"fs-id1165043060683\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]r^{\\prime}(t)=\\dfrac{1}{2\\pi [r(t)]^2} \\, \\text{cm\/sec}[\/latex]<\/div>\r\n<p id=\"fs-id1165043105262\">When the radius [latex]r=3 \\, \\text{cm}[\/latex],<\/p>\r\n<div id=\"fs-id1165043035926\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]r^{\\prime}(t)=\\dfrac{1}{18\\pi} \\, \\text{cm\/sec}[\/latex]<\/div>\r\n<\/section>\r\n<section class=\"textbox watchIt\">\r\n<p>Watch the following video to see the worked solution to the example above.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7SWaBGupLT0?controls=0&amp;start=32&amp;end=212&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\n<p>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]<br \/>\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.<\/p>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/4.1RelatedRates32to212_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"4.1 Related Rates\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p>What is the instantaneous rate of change of the radius when [latex]r=6 \\, \\text{cm}[\/latex]?<\/p>\r\n<p>[reveal-answer q=\"44780133\"]Hint[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"44780133\"]<\/p>\r\n<p id=\"fs-id1165042954639\">[latex]\\frac{dr}{dt}=\\dfrac{1}{2\\pi r^2}[\/latex]<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<p>[reveal-answer q=\"fs-id1165043018863\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1165043018863\"]<\/p>\r\n<p id=\"fs-id1165043018863\">[latex]\\dfrac{1}{72\\pi} \\, \\text{cm\/sec}[\/latex], or approximately 0.0044 cm\/sec<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<p id=\"fs-id1165042989383\">Before looking at other examples, let\u2019s outline the problem-solving strategy we will be using to solve related-rates problems.<\/p>\r\n<section class=\"textbox questionHelp\">\r\n<p><strong>How to: Solve a Related-Rates Problem<\/strong><\/p>\r\n<ol id=\"fs-id1165042985160\">\r\n\t<li><strong>Variable Assignment:<\/strong> Assign symbols to all variables involved in the problem. If it helps, draw a diagram to visualize the scenario.<\/li>\r\n\t<li><strong>Information Setup:<\/strong> Clearly state the given information and what needs to be determined, using the assigned variables.<\/li>\r\n\t<li><strong>Equation Development:<\/strong> Formulate an equation that connects the variables. This relationship should encapsulate the dynamics of the problem.<\/li>\r\n\t<li><strong>Differentiation:<\/strong> Apply the chain rule to differentiate the equation with respect to time. This step transforms the static equation into one that describes rates of change.<\/li>\r\n\t<li><strong>Substitution and Solution:<\/strong> Plug in all known values of rates and other variables into the differentiated equation and solve for the unknown rate.<\/li>\r\n<\/ol>\r\n<\/section>\r\n<p>We are able to solve related-rates problems using a similar approach to implicit differentiation.\u00a0<\/p>\r\n<section class=\"textbox proTip\">\r\n<p>Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this potential error in the first example on the next page.<\/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<p>We have seen that for quantities that are changing over time, the rates at which these quantities change are given by derivatives. If two related quantities are changing over time, the rates at which the quantities change are related. For example, if a balloon is being filled with air, both the radius of the balloon and the volume of the balloon are increasing. In this section, we consider several problems in which two or more related quantities are changing and we study how to determine the relationship between the rates of change of these quantities.<\/p>\n<h3>Setting up Related-Rates Problems<\/h3>\n<p id=\"fs-id1165043085174\">In many real-world applications, related quantities are changing with respect to time. For example, if we consider the balloon example again, we can say that the rate of change in the volume, [latex]V[\/latex], is related to the rate of change in the radius, [latex]r[\/latex]. In this case, we say that [latex]\\frac{dV}{dt}[\/latex] and [latex]\\frac{dr}{dt}[\/latex] are <strong>related rates<\/strong> because [latex]V[\/latex] is related to [latex]r[\/latex]. Here we study several examples of related quantities that are changing with respect to time and we look at how to calculate one rate of change given another rate of change.<\/p>\n<section class=\"textbox example\">\n<p id=\"fs-id1165043084830\">A spherical balloon is being filled with air at the constant rate of [latex]2 \\, \\frac{\\text{cm}^3}{\\text{sec}}[\/latex] (Figure 1). How fast is the radius increasing when the radius is [latex]3\\, \\text{cm}[\/latex]?<\/p>\n<figure style=\"width: 900px\" 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\/11210704\/CNX_Calc_Figure_04_01_001.jpg\" alt=\"Three balloons are shown at Times 1, 2, and 3. These balloons increase in volume and radius as time increases.\" width=\"900\" height=\"297\" \/><figcaption class=\"wp-caption-text\">Figure 1. As the balloon is being filled with air, both the radius and the volume are increasing with respect to time.<\/figcaption><\/figure>\n<p id=\"fs-id1165043021803\">The volume of a sphere of radius [latex]r[\/latex] centimeters is<\/p>\n<div id=\"fs-id1165042990072\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V=\\frac{4}{3}\\pi r^3 \\, \\text{cm}^3[\/latex]<\/div>\n<p id=\"fs-id1165042881658\">Since the balloon is being filled with air, both the volume and the radius are functions of time. Therefore, [latex]t[\/latex] seconds after beginning to fill the balloon with air, the volume of air in the balloon is<\/p>\n<div id=\"fs-id1165043013475\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V(t)=\\frac{4}{3}\\pi [r(t)]^3 \\, \\text{cm}^3[\/latex]<\/div>\n<p id=\"fs-id1165042941218\">Differentiating both sides of this equation with respect to time and applying the chain rule, we see that the rate of change in the volume is related to the rate of change in the radius by the equation<\/p>\n<div id=\"fs-id1165043119522\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]V^{\\prime}(t)=4\\pi [r(t)]^2 \\cdot r^{\\prime}(t)[\/latex]<\/div>\n<p id=\"fs-id1165043111378\">The balloon is being filled with air at the constant rate of 2 cm<sup>3<\/sup>\/sec, so [latex]V^{\\prime}(t)=2 \\, \\text{cm}^3 \/ \\sec[\/latex]. Therefore,<\/p>\n<div id=\"fs-id1165043009965\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]2 \\, \\text{cm}^3 \/ \\sec =(4\\pi [r(t)]^2 \\, \\text{cm}^2) \\cdot (r^{\\prime}(t) \\, \\text{cm\/sec})[\/latex],<\/div>\n<p id=\"fs-id1165043051118\">which implies<\/p>\n<div id=\"fs-id1165043060683\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]r^{\\prime}(t)=\\dfrac{1}{2\\pi [r(t)]^2} \\, \\text{cm\/sec}[\/latex]<\/div>\n<p id=\"fs-id1165043105262\">When the radius [latex]r=3 \\, \\text{cm}[\/latex],<\/p>\n<div id=\"fs-id1165043035926\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]r^{\\prime}(t)=\\dfrac{1}{18\\pi} \\, \\text{cm\/sec}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox watchIt\">\n<p>Watch the following video to see the worked solution to the example above.<\/p>\n<div style=\"text-align: center;\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/7SWaBGupLT0?controls=0&amp;start=32&amp;end=212&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/button><\/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\/4.1RelatedRates32to212_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;4.1 Related Rates&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox example\">\n<p>What is the instantaneous rate of change of the radius when [latex]r=6 \\, \\text{cm}[\/latex]?<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"q44780133\">Hint<\/button><\/p>\n<div id=\"q44780133\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165042954639\">[latex]\\frac{dr}{dt}=\\dfrac{1}{2\\pi r^2}[\/latex]<\/p>\n<\/div>\n<\/div>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1165043018863\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1165043018863\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1165043018863\">[latex]\\dfrac{1}{72\\pi} \\, \\text{cm\/sec}[\/latex], or approximately 0.0044 cm\/sec<\/p>\n<\/div>\n<\/div>\n<\/section>\n<p id=\"fs-id1165042989383\">Before looking at other examples, let\u2019s outline the problem-solving strategy we will be using to solve related-rates problems.<\/p>\n<section class=\"textbox questionHelp\">\n<p><strong>How to: Solve a Related-Rates Problem<\/strong><\/p>\n<ol id=\"fs-id1165042985160\">\n<li><strong>Variable Assignment:<\/strong> Assign symbols to all variables involved in the problem. If it helps, draw a diagram to visualize the scenario.<\/li>\n<li><strong>Information Setup:<\/strong> Clearly state the given information and what needs to be determined, using the assigned variables.<\/li>\n<li><strong>Equation Development:<\/strong> Formulate an equation that connects the variables. This relationship should encapsulate the dynamics of the problem.<\/li>\n<li><strong>Differentiation:<\/strong> Apply the chain rule to differentiate the equation with respect to time. This step transforms the static equation into one that describes rates of change.<\/li>\n<li><strong>Substitution and Solution:<\/strong> Plug in all known values of rates and other variables into the differentiated equation and solve for the unknown rate.<\/li>\n<\/ol>\n<\/section>\n<p>We are able to solve related-rates problems using a similar approach to implicit differentiation.\u00a0<\/p>\n<section class=\"textbox proTip\">\n<p>Note that when solving a related-rates problem, it is crucial not to substitute known values too soon. For example, if the value for a changing quantity is substituted into an equation before both sides of the equation are differentiated, then that quantity will behave as a constant and its derivative will not appear in the new equation found in step 4. We examine this potential error in the first example on the next page.<\/p>\n<\/section>\n","protected":false},"author":6,"menu_order":6,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Calculus Volume 1\",\"author\":\"Gilbert Strang, Edwin (Jed) Herman\",\"organization\":\"OpenStax\",\"url\":\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\",\"project\":\"\",\"license\":\"cc-by-nc-sa\",\"license_terms\":\"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction\"},{\"type\":\"original\",\"description\":\"4.1 Related Rates\",\"author\":\"Ryan Melton\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":652,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Calculus Volume 1","author":"Gilbert Strang, Edwin (Jed) Herman","organization":"OpenStax","url":"https:\/\/openstax.org\/details\/books\/calculus-volume-1","project":"","license":"cc-by-nc-sa","license_terms":"Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction"},{"type":"original","description":"4.1 Related Rates","author":"Ryan Melton","organization":"","url":"","project":"","license":"cc-by","license_terms":""}],"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\/295"}],"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\/6"}],"version-history":[{"count":7,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/295\/revisions"}],"predecessor-version":[{"id":3207,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/295\/revisions\/3207"}],"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\/295\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=295"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=295"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=295"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=295"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}