{"id":111898,"date":"2022-02-03T13:48:10","date_gmt":"2022-02-03T19:48:10","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=111898"},"modified":"2026-03-28T11:49:23","modified_gmt":"2026-03-28T16:49:23","slug":"related-rates","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/related-rates\/","title":{"rendered":"Related Rates"},"content":{"rendered":"<h1>Related Rates<\/h1>\n\n\t\t\t<div id=\"mmDeferVideoEncompass_wXXsPC2BfeY\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_wXXsPC2BfeY\" data-source-videoID=\"wXXsPC2BfeY\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Related Rates Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Related Rates\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_wXXsPC2BfeY:hover {cursor:pointer;} img#videoThumbnailImage_wXXsPC2BfeY {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1761-thumb-final-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_wXXsPC2BfeY\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_wXXsPC2BfeY\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Related Rates\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_wXXsPC2BfeY\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_wXXsPC2BfeY\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_wXXsPC2BfeY\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Consider the following problem: one day it snows, and you and a friend decide to form a giant snowball. The next day, however, temperatures are back up above freezing, and the snowball starts to melt. If the snowball\u2019s radius decreases at a rate of one centimeter per hour, can we determine the rate at which the snowball\u2019s volume is decreasing as well? We can! Let\u2019s discuss how.<\/p>\n<p>This snowball problem belongs to a type of calculus problem we like to call \u201c<strong>related rates<\/strong>.\u201d These kinds of problems have that name because we are given information about one rate of change and are asked about <em>another<\/em> rate of change, where the two are related to each other in some way. For example, we know from geometry that the radius and volume of a sphere are related by the equation \\(V=\\frac{4}{3}\\pi r^{3}\\).<\/p>\n<p>Let\u2019s work through how to solve the problem now. We want to know the rate at which the snowball\u2019s volume is decreasing.<\/p>\n<p>To start, let\u2019s ask: \u201cwhat do we already know?\u201d We know that the radius of the snowball decreases at a rate of one centimeter per hour, so we can write this as \\(r'(t)=\\frac{dr}{dt}=-1\\frac{cm}{hr}\\). I chose the letter \\(r\\) because we are using this equation to represent the radius. \\(r\\) is prime because we are describing the radius\u2019s rate of change. This rate is described in units \u201ccentimeters per hour,\u201d so mathematically we can express it with \\(\\frac{dr}{dt}\\), the derivative of \\(r\\) with respect to the time \\(t\\).<\/p>\n<p>What else do we know? As I mentioned earlier, we know that the volume of a sphere is:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(V=\\frac{4}{3}\\pi r^{3}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s use this information to write a function describing the snowball\u2019s volume at time \\(t\\).<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(V(t)=\\frac{4}{3}\\pi \\cdot r(t)^{3}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Here, instead of just writing \\(r^{3}\\), we need to write \\(r(t)^{3}\\) to remind ourselves that the radius is changing with time just like the volume is. Since we are trying to find the rate at which the snowball\u2019s volume is decreasing, all we have to do from here is calculate \\(V'(t)\\).<\/p>\n<p>Notice that because \\(r\\) is a function of \\(t\\), <em>within the function V<\/em>, we need to use the <a href=\"https:\/\/www.mometrix.com\/academy\/chain-rule\/\"><strong>chain rule<\/strong><\/a> to compute this derivative.<\/p>\n<p>The derivative of \\(V(t)\\) is then:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(V'(t)=\\frac{d}{dt}\\left [ \\frac{4}{3}\\pi r(t)^{3} \\right ]\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Taking the derivative with the respect to \\(t\\), the <a href=\"https:\/\/www.mometrix.com\/academy\/definition-of-the-derivative\/\"><strong>power rule<\/strong><\/a> states that we multiply by 3 and drop the exponent to 2. This will gives us:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(V'(t)=\\frac{4}{3}\\pi \\cdot 3r(t)^{2}\\cdot \\frac{dr}{dt}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>The \\(dr\\) at the end is to help us to remember to use the chain rule. We now need to multiply by the derivative of \\(r\\), \\(r'(t)\\), which is -1.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(V'(t)=\\frac{4}{3}\\pi \\cdot 3[r(t)]^{2}\\cdot (-1)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Then, all this will simplify to:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(V'(t)=-4\\pi [r(t)]^{2}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>This means that the rate of change in volume at \\(t\\) hours is equal to \\(-4\\pi\\) times the radius at that time squared.<\/p>\n<p>Let\u2019s look at another example.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:85%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">Elizabeth throws a stone into a lake and observes that the ripples grow in diameter at the rate of 2 feet per second. By what speed does the area inside the ripples grow when they are 8 feet across?<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Let\u2019s start by asking, \u201cwhat do we want to know?\u201d We want to know the rate of change in area at a particular time. Let\u2019s call area at time \\(t\\), \\(A(t)\\), and let\u2019s call the rate of change in area \\(A'(t)\\). Once we find \\(A'(t)\\), we will need to plug in the time when diameter equals 8 feet to get the final solution.<\/p>\n<p>Now, what do we already know? So, we know that the ripples grow in diameter by 2 feet per second. Let\u2019s call this rate of change \\(d'(t)=2\\). We will let \\(d(t)\\) represent the diameter at time \\(t\\).<\/p>\n<p>We also know that a circle\u2019s diameter is related to its area because \\(A=\\pi r^{2}=\\pi (\\frac{d}{2})^{2}\\).<\/p>\n<p>This is everything we need to solve the problem. Remember, since we want to know \\(A'(t)\\), we first need to write \\(A(t)\\). Then we\u2019ll take its derivative.<\/p>\n<p>\\(A(t)\\), or area at time \\(t\\), is equal to \\(\\pi\\) times half the diameter at time \\(t\\), squared.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(A(t)=\\pi \\left [ \\frac{d(t)}{2} \\right ]^{2}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Notice that we can go ahead and square this out. That will give us:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(A(t)=\\pi \\frac{[d(t)]^{2}}{4}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Then we can rewrite this as:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(A(t)=\\frac{\\pi}{4}[d(t)]^{2}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Now let\u2019s take the derivative. Since \\(\\frac{\\pi}{4}\\) is constant, its derivative is 0, so we can focus on taking the derivative of \\([d(t)]^{2}\\). We need to use the chain rule here because \\(d(t)\\) is of course a function. So \\(A'(t)\\) will equal:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(A'(t)=\\frac{\\pi}{4}\\cdot 2\\cdot d(t)\\cdot d'(t)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Remember, we were told in the problem statement that \\(d'(t)=2\\), so we can plug this in.<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(A'(t)=\\frac{\\pi}{4}\\cdot 2\\cdot d(t)\\cdot (2)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Then, we can simplify this out and get:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(A'(t)=\\pi d(t)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>At this point, we have found the rate of change in area at time \\(t\\). The last step is to find the rate of change in area when the ripple is 8 feet across. In other words, when \\(d(t)=8\\). So, our final answer is that the rate of change when \\(d\\) of \\(t\\) is equal to 8 is:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(\\pi d(t)=\\pi (8)=8\\pi ft^{2}\/sec\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>I have one more problem to go over with you in this video, but for this one, let\u2019s see if you can work it out on your own!<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:85%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">The sides of a cube are growing at the rate of 4 inches per minute. By what speed will the cube\u2019s surface area be growing when the sides of the cube are 24 inches long?<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Pause the video now and work it out on your own. Then we will compare our work.<\/p>\n<p>Let\u2019s jump right in. For this problem, we want to know the rate of change in the cube\u2019s surface area at a specific time. This means we need to write a formula for surface area, then take its derivative to get that rate of change. <\/p>\n<p>We are told that the sides of a cube are growing in length at the rate of 4 inches per minute. I\u2019m going to write this fact as \\(s'(m)=4\\), where \\(s\\) represents side length at time \\(m\\). <\/p>\n<p>Now, because cubes have 6 sides of equal size, we can write the surface area with respect to time \\(m\\), in minutes, as:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(SA(m)=6[s(m)]^{2}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Time to take the derivative, once again using the chain rule. The surface area has rate of change:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:85%; text-align:center;\">\\(SA'(m)=6\\cdot 2\\cdot s(m )\\cdot s'(m)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Since \\(s'(m)=4\\), we can rewrite this so that it\u2019s:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(SA'(m)=6\\cdot 2\\cdot s(m)\\cdot (4)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Then, this will simplify down to:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:90%; text-align:center;\">\\(SA'(m)=48s(m)\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>So the rate of change in surface area at a particular time equals 48 times the side length at that time.<\/p>\n<p>We are interested in the rate of change in surface area when sides have length 24 inches, so substitute in 24 where you see \\(s(m)\\). The solution is then:<\/p>\n<blockquote style=\"border: 0px; padding: 30px; background-color: #eee; box-shadow: 3px 3px 10px grey; width:80%; margin: auto;\">\n<div style=\"font-style:normal; font-size:85%; text-align:center;\">\\(48s(m)=48(24)\\)\\(<br \/>\n=1,152\\text{ in}^{2}\/\\text{min}\\)<\/div>\n<\/blockquote>\n<p>\n&nbsp;<\/p>\n<p>Many students and teachers acknowledge that related rates is typically the most difficult section in Calculus 1. Even so, these problems are certainly doable if you keep these main steps in mind:<\/p>\n<ul>\n<li>Ask yourself, \u201cWhat is the problem asking for? What do we want to know?\u201d Typically, you want to find some rate of change. Write down exactly what you\u2019re looking for.<\/li>\n<li>Next, ask, \u201cWhat do I already know from the problem?\u201d Write down the information that you\u2019re given in the problem statement, which should include some rate of change. Also make sure to write down the relationship between the two rates of interest (for example, \\(V=\\frac{4}{3}\\pi r^{3}\\)).<\/li>\n<li>Use what you know to write a formula with the respect to time.<\/li>\n<li>Then take the derivative to acquire the rate of change you\u2019re interested in. Remember that this step almost always incorporates the chain rule.<\/li>\n<li>Finally, plug in the appropriate condition to get your solution.<\/li>\n<\/ul>\n<p>Even though that seems like a lot, you\u2019ll get the hang of it with practice.<\/p>\n<p>Thanks for watching, and happy studying!<\/p>\n<\/div>\n<\/div>\n\n<p>&nbsp;<\/p>\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/calculus\/\"><strong>Return to Calculus Videos<\/strong><\/a><\/p>\n","protected":false},"excerpt":{"rendered":"<p>Related Rates &nbsp; Return to Calculus Videos<\/p>\n","protected":false},"author":22,"featured_media":111901,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-111898","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-calculus-videos","7":"page_category-video-pages-for-study-course-sidebar-ad","8":"page_type-video","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111898","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/22"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=111898"}],"version-history":[{"count":4,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111898\/revisions"}],"predecessor-version":[{"id":168074,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/111898\/revisions\/168074"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/111901"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=111898"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}