{"id":244,"date":"2023-09-20T22:48:32","date_gmt":"2023-09-20T22:48:32","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/amount-of-change-formula\/"},"modified":"2024-08-05T12:42:17","modified_gmt":"2024-08-05T12:42:17","slug":"derivatives-as-rates-of-change-learn-it-1","status":"publish","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/calculus1\/chapter\/derivatives-as-rates-of-change-learn-it-1\/","title":{"raw":"Derivatives as Rates of Change: Learn It 1","rendered":"Derivatives as Rates of Change: Learn It 1"},"content":{"raw":"<section class=\"textbox learningGoals\">\r\n<ul>\r\n\t<li>Calculate how quantities change on average over time<\/li>\r\n\t<li>Use rates of change to figure out how an object\u2019s position, speed, and acceleration are changing over time<\/li>\r\n\t<li>Estimate future population sizes using current data and how fast the population is growing<\/li>\r\n\t<li>Use derivatives to determine the cost and revenue of producing one more unit in a business<\/li>\r\n<\/ul>\r\n<\/section>\r\n<h2>Amount of Change Formula<\/h2>\r\n<p id=\"fs-id1169739001719\">One application of derivatives is to estimate an unknown value of a function at a point by using a known value of the function at some given point together with its rate of change at that given point.<\/p>\r\n<p>If [latex]f(x)[\/latex] is a function defined on an interval [latex][a,a+h][\/latex], then the <strong>amount of change<\/strong> of [latex]f(x)[\/latex] over the interval is the change in the [latex]y[\/latex] values of the function over that interval and is given by:<\/p>\r\n<div id=\"fs-id1169738947436\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+h)-f(a)[\/latex]<\/div>\r\n<p id=\"fs-id1169739019267\">The <strong>average rate of change<\/strong> of the function [latex]f[\/latex] over that same interval is the ratio of the amount of change over that interval to the corresponding change in the [latex]x[\/latex] values. It is given by:<\/p>\r\n<div id=\"fs-id1169739005180\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(a+h)-f(a)}{h}[\/latex]<\/div>\r\n<p id=\"fs-id1169738865347\">As we already know, the <strong>instantaneous rate of change<\/strong> of [latex]f(x)[\/latex] at [latex]a[\/latex] is its derivative,<\/p>\r\n<div id=\"fs-id1169738975674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(a)=\\underset{h\\to 0}{\\lim}\\dfrac{f(a+h)-f(a)}{h}[\/latex]<\/div>\r\n<p id=\"fs-id1169739038407\">For small enough values of [latex]h, \\, f^{\\prime}(a)\\approx \\frac{f(a+h)-f(a)}{h}[\/latex]. We can then solve for [latex]f(a+h)[\/latex] to get the amount of change formula:<\/p>\r\n<div id=\"fs-id1169739034265\" class=\"equation\" style=\"text-align: center;\">[latex]f(a+h)\\approx f(a)+f^{\\prime}(a)h[\/latex]<\/div>\r\n<p id=\"fs-id1169738850663\">We can use this formula if we know only [latex]f(a)[\/latex] and [latex]f^{\\prime}(a)[\/latex] and wish to estimate the value of [latex]f(a+h)[\/latex]. For example, we may use the current population of a city and the rate at which it is growing to estimate its population in the near future. <br \/>\r\n<br \/>\r\nAs we can see in Figure 1, we are approximating [latex]f(a+h)[\/latex] by the [latex]y[\/latex] coordinate at [latex]a+h[\/latex] on the line tangent to [latex]f(x)[\/latex] at [latex]x=a[\/latex]. Observe that the accuracy of this estimate depends on the value of [latex]h[\/latex] as well as the value of [latex]f^{\\prime}(a)[\/latex].<\/p>\r\n\r\n[caption id=\"\" align=\"aligncenter\" width=\"478\"]<img src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/2332\/2018\/01\/11205357\/CNX_Calc_Figure_03_04_001.jpg\" alt=\"On the Cartesian coordinate plane with a and a + h marked on the x axis, the function f is graphed. It passes through (a, f(a)) and (a + h, f(a + h)). A straight line is drawn through (a, f(a)) with its slope being the derivative at that point. This straight line passes through (a + h, f(a) + f\u2019(a)h). There is a line segment connecting (a + h, f(a + h)) and (a + h, f(a) + f\u2019(a)h), and it is marked that this is the error in using f(a) + f\u2019(a)h to estimate f(a + h).\" width=\"478\" height=\"347\" \/> Figure 1. The new value of a changed quantity equals the original value plus the rate of change times the interval of change: [latex]f(a+h)\\approx f(a)+f^{\\prime}(a)h[\/latex].[\/caption]\r\n\r\n<section class=\"textbox keyTakeaway\">\r\n<h3>average rate of change<\/h3>\r\n<p>The average rate of change of a function [latex]f[\/latex] over the interval [latex][a,a+h][\/latex] is the ratio of the amount of change in [latex]f[\/latex] over that interval to the corresponding change in [latex]x[\/latex]\u00a0values.<br \/>\r\n<br \/>\r\nThe average rate of change is given by:<br \/>\r\n<br \/>\r\n<\/p>\r\n<center>[latex]\\dfrac{f(a+h)-f(a)}{h}[\/latex]<\/center><\/section>\r\n<section class=\"textbox interact\">\r\n<p><a href=\"https:\/\/demonstrations.wolfram.com\/ASnowballsRateOfChange\/\" target=\"_blank\" rel=\"noopener\">Here is an interesting demonstration of rate of change.<\/a><\/p>\r\n<\/section>\r\n<section class=\"textbox example\">\r\n<p id=\"fs-id1169738865923\">If [latex]f(3)=2[\/latex] and [latex]f^{\\prime}(3)=5[\/latex], estimate [latex]f(3.2)[\/latex].<\/p>\r\n<p>[reveal-answer q=\"fs-id1169738857080\"]Show Solution[\/reveal-answer]<br \/>\r\n[hidden-answer a=\"fs-id1169738857080\"]<\/p>\r\n<p id=\"fs-id1169738857080\">Begin by finding [latex]h[\/latex]. We have [latex]h=3.2-3=0.2[\/latex]. Thus,<\/p>\r\n<div class=\"equation unnumbered\">[latex]f(3.2)=f(3+0.2)\\approx f(3)+(0.2)f^{\\prime}(3)=2+0.2(5)=3[\/latex].<\/div>\r\n<p>&nbsp;<\/p>\r\n<p>Watch the following video to see the worked solution to this example.<\/p>\r\n<center><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/XOgb95xhWF0?controls=0&amp;start=110&amp;end=190&amp;autoplay=0\" width=\"750\" height=\"450\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\" data-mce-fragment=\"1\"><\/iframe><\/center>\r\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>\r\n<p>You can view the <a href=\"https:\/\/oerfiles.s3.us-west-2.amazonaws.com\/Calculus\/Calculus1+Videos\/3.4DerivativesAsRatesOfChange110to190_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of \"3.4 Derivatives as Rates of Change\" here (opens in new window)<\/a>.<\/p>\r\n<p>[\/hidden-answer]<\/p>\r\n<\/section>\r\n<section class=\"textbox tryIt\">\r\n<p>[ohm_question hide_question_numbers=1]288384[\/ohm_question]<\/p>\r\n<\/section>","rendered":"<section class=\"textbox learningGoals\">\n<ul>\n<li>Calculate how quantities change on average over time<\/li>\n<li>Use rates of change to figure out how an object\u2019s position, speed, and acceleration are changing over time<\/li>\n<li>Estimate future population sizes using current data and how fast the population is growing<\/li>\n<li>Use derivatives to determine the cost and revenue of producing one more unit in a business<\/li>\n<\/ul>\n<\/section>\n<h2>Amount of Change Formula<\/h2>\n<p id=\"fs-id1169739001719\">One application of derivatives is to estimate an unknown value of a function at a point by using a known value of the function at some given point together with its rate of change at that given point.<\/p>\n<p>If [latex]f(x)[\/latex] is a function defined on an interval [latex][a,a+h][\/latex], then the <strong>amount of change<\/strong> of [latex]f(x)[\/latex] over the interval is the change in the [latex]y[\/latex] values of the function over that interval and is given by:<\/p>\n<div id=\"fs-id1169738947436\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+h)-f(a)[\/latex]<\/div>\n<p id=\"fs-id1169739019267\">The <strong>average rate of change<\/strong> of the function [latex]f[\/latex] over that same interval is the ratio of the amount of change over that interval to the corresponding change in the [latex]x[\/latex] values. It is given by:<\/p>\n<div id=\"fs-id1169739005180\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]\\dfrac{f(a+h)-f(a)}{h}[\/latex]<\/div>\n<p id=\"fs-id1169738865347\">As we already know, the <strong>instantaneous rate of change<\/strong> of [latex]f(x)[\/latex] at [latex]a[\/latex] is its derivative,<\/p>\n<div id=\"fs-id1169738975674\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f^{\\prime}(a)=\\underset{h\\to 0}{\\lim}\\dfrac{f(a+h)-f(a)}{h}[\/latex]<\/div>\n<p id=\"fs-id1169739038407\">For small enough values of [latex]h, \\, f^{\\prime}(a)\\approx \\frac{f(a+h)-f(a)}{h}[\/latex]. We can then solve for [latex]f(a+h)[\/latex] to get the amount of change formula:<\/p>\n<div id=\"fs-id1169739034265\" class=\"equation\" style=\"text-align: center;\">[latex]f(a+h)\\approx f(a)+f^{\\prime}(a)h[\/latex]<\/div>\n<p id=\"fs-id1169738850663\">We can use this formula if we know only [latex]f(a)[\/latex] and [latex]f^{\\prime}(a)[\/latex] and wish to estimate the value of [latex]f(a+h)[\/latex]. For example, we may use the current population of a city and the rate at which it is growing to estimate its population in the near future. <\/p>\n<p>As we can see in Figure 1, we are approximating [latex]f(a+h)[\/latex] by the [latex]y[\/latex] coordinate at [latex]a+h[\/latex] on the line tangent to [latex]f(x)[\/latex] at [latex]x=a[\/latex]. Observe that the accuracy of this estimate depends on the value of [latex]h[\/latex] as well as the value of [latex]f^{\\prime}(a)[\/latex].<\/p>\n<figure style=\"width: 478px\" 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\/11205357\/CNX_Calc_Figure_03_04_001.jpg\" alt=\"On the Cartesian coordinate plane with a and a + h marked on the x axis, the function f is graphed. It passes through (a, f(a)) and (a + h, f(a + h)). A straight line is drawn through (a, f(a)) with its slope being the derivative at that point. This straight line passes through (a + h, f(a) + f\u2019(a)h). There is a line segment connecting (a + h, f(a + h)) and (a + h, f(a) + f\u2019(a)h), and it is marked that this is the error in using f(a) + f\u2019(a)h to estimate f(a + h).\" width=\"478\" height=\"347\" \/><figcaption class=\"wp-caption-text\">Figure 1. The new value of a changed quantity equals the original value plus the rate of change times the interval of change: [latex]f(a+h)\\approx f(a)+f^{\\prime}(a)h[\/latex].<\/figcaption><\/figure>\n<section class=\"textbox keyTakeaway\">\n<h3>average rate of change<\/h3>\n<p>The average rate of change of a function [latex]f[\/latex] over the interval [latex][a,a+h][\/latex] is the ratio of the amount of change in [latex]f[\/latex] over that interval to the corresponding change in [latex]x[\/latex]\u00a0values.<\/p>\n<p>The average rate of change is given by:<\/p>\n<div style=\"text-align: center;\">[latex]\\dfrac{f(a+h)-f(a)}{h}[\/latex]<\/div>\n<\/section>\n<section class=\"textbox interact\">\n<p><a href=\"https:\/\/demonstrations.wolfram.com\/ASnowballsRateOfChange\/\" target=\"_blank\" rel=\"noopener\">Here is an interesting demonstration of rate of change.<\/a><\/p>\n<\/section>\n<section class=\"textbox example\">\n<p id=\"fs-id1169738865923\">If [latex]f(3)=2[\/latex] and [latex]f^{\\prime}(3)=5[\/latex], estimate [latex]f(3.2)[\/latex].<\/p>\n<p><div class=\"qa-wrapper\" style=\"display: block\"><button class=\"show-answer show-answer-button collapsed\" data-target=\"qfs-id1169738857080\">Show Solution<\/button><\/p>\n<div id=\"qfs-id1169738857080\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738857080\">Begin by finding [latex]h[\/latex]. We have [latex]h=3.2-3=0.2[\/latex]. Thus,<\/p>\n<div class=\"equation unnumbered\">[latex]f(3.2)=f(3+0.2)\\approx f(3)+(0.2)f^{\\prime}(3)=2+0.2(5)=3[\/latex].<\/div>\n<p>&nbsp;<\/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\/XOgb95xhWF0?controls=0&amp;start=110&amp;end=190&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\/3.4DerivativesAsRatesOfChange110to190_transcript.txt\" target=\"_blank\" rel=\"noopener\">transcript for this segmented clip of &#8220;3.4 Derivatives as Rates of Change&#8221; here (opens in new window)<\/a>.<\/p>\n<\/div>\n<\/div>\n<\/section>\n<section class=\"textbox tryIt\">\n<iframe loading=\"lazy\" id=\"ohm288384\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=288384&theme=lumen&iframe_resize_id=ohm288384&source=tnh\" width=\"100%\" height=\"150\"><\/iframe><br \/>\n<\/section>\n","protected":false},"author":6,"menu_order":23,"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\":\"3.4 Derivatives as Rates of Change\",\"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":503,"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":"3.4 Derivatives as Rates of Change","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\/244"}],"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":11,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/244\/revisions"}],"predecessor-version":[{"id":4657,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/244\/revisions\/4657"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/503"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/244\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=244"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=244"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=244"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=244"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}