{"id":340,"date":"2021-02-04T01:15:40","date_gmt":"2021-02-04T01:15:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/calculus1\/?post_type=chapter&#038;p=340"},"modified":"2022-03-16T05:32:07","modified_gmt":"2022-03-16T05:32:07","slug":"amount-of-change-formula","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/calculus1\/chapter\/amount-of-change-formula\/","title":{"raw":"Amount of Change Formula","rendered":"Amount of Change Formula"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine a new value of a quantity from the old value and the amount of change<\/li>\r\n \t<li>Calculate the average rate of change and explain how it differs from the instantaneous rate of change<\/li>\r\n<\/ul>\r\n<\/div>\r\n<div id=\"fs-id1169739031094\" class=\"bc-section section\">\r\n<p id=\"fs-id1169739001719\">One application for derivatives is to estimate an unknown value of a function at a point by using a known value of a function at some given point together with its rate of change at the given point. 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\r\n<div id=\"fs-id1169738947436\" class=\"equation unnumbered\" style=\"text-align: center;\">[latex]f(a+h)-f(a)[\/latex]<\/div>\r\n&nbsp;\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\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&nbsp;\r\n<p id=\"fs-id1169738865347\">As we already know, the instantaneous rate of change of [latex]f(x)[\/latex] at [latex]a[\/latex] is its derivative<\/p>\r\n\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&nbsp;\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\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&nbsp;\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. 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>\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<div class=\"wp-caption-text\"><\/div>\r\n<div id=\"fs-id1169739182517\" class=\"textbox tryit\">\r\n<h3>Interactive<\/h3>\r\n<a href=\"https:\/\/demonstrations.wolfram.com\/ASnowballsRateOfChange\/\" target=\"_blank\" rel=\"noopener\">Here is an interesting demonstration of rate of change.<\/a>\r\n\r\n<\/div>\r\n<div id=\"fs-id1169738907373\" class=\"textbook exercises\">\r\n<h3>Example: Estimating the Value of a Function<\/h3>\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[reveal-answer q=\"fs-id1169738857080\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169738857080\"]\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\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[\/hidden-answer]\r\n\r\n<\/div>\r\nWatch the following video to see the worked solution to Example: Estimating the Value of a Function.\r\n\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\"><\/iframe><\/center>[reveal-answer q=\"266834\"]Closed Captioning and Transcript Information for Video[\/reveal-answer]\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.\r\n\r\nYou 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>.[\/hidden-answer]\r\n<div id=\"fs-id1169739230273\" class=\"textbook key-takeaways\">\r\n<h3>Try It<\/h3>\r\n<p id=\"fs-id1169738905166\">Given [latex]f(10)=-5[\/latex] and [latex]f^{\\prime}(10)=6[\/latex], estimate [latex]f(10.1)[\/latex].<\/p>\r\n[reveal-answer q=\"762225\"]Hint[\/reveal-answer]\r\n[hidden-answer a=\"762225\"]\r\n<p id=\"fs-id1169738908581\">Use the same process as in the preceding example.<\/p>\r\n[\/hidden-answer]\r\n\r\n[reveal-answer q=\"fs-id1169739097606\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"fs-id1169739097606\"]\r\n<p id=\"fs-id1169739097606\">-4.4<\/p>\r\n[\/hidden-answer]\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\n[ohm_question]16098[\/ohm_question]\r\n\r\n<\/div>\r\n<\/div>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine a new value of a quantity from the old value and the amount of change<\/li>\n<li>Calculate the average rate of change and explain how it differs from the instantaneous rate of change<\/li>\n<\/ul>\n<\/div>\n<div id=\"fs-id1169739031094\" class=\"bc-section section\">\n<p id=\"fs-id1169739001719\">One application for derivatives is to estimate an unknown value of a function at a point by using a known value of a function at some given point together with its rate of change at the given point. 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>&nbsp;<\/p>\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>&nbsp;<\/p>\n<p id=\"fs-id1169738865347\">As we already know, the instantaneous rate of change 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>&nbsp;<\/p>\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>&nbsp;<\/p>\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. 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<div style=\"width: 488px\" 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\" \/><\/p>\n<p 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].<\/p>\n<\/div>\n<div class=\"wp-caption-text\"><\/div>\n<div id=\"fs-id1169739182517\" class=\"textbox tryit\">\n<h3>Interactive<\/h3>\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<\/div>\n<div id=\"fs-id1169738907373\" class=\"textbook exercises\">\n<h3>Example: Estimating the Value of a Function<\/h3>\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<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169738857080\">Show Solution<\/span><\/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<\/div>\n<\/div>\n<\/div>\n<p>Watch the following video to see the worked solution to Example: Estimating the Value of a Function.<\/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\"><\/iframe><\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q266834\">Closed Captioning and Transcript Information for Video<\/span><\/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\/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>.<\/div>\n<\/div>\n<div id=\"fs-id1169739230273\" class=\"textbook key-takeaways\">\n<h3>Try It<\/h3>\n<p id=\"fs-id1169738905166\">Given [latex]f(10)=-5[\/latex] and [latex]f^{\\prime}(10)=6[\/latex], estimate [latex]f(10.1)[\/latex].<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q762225\">Hint<\/span><\/p>\n<div id=\"q762225\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169738908581\">Use the same process as in the preceding example.<\/p>\n<\/div>\n<\/div>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"qfs-id1169739097606\">Show Solution<\/span><\/p>\n<div id=\"qfs-id1169739097606\" class=\"hidden-answer\" style=\"display: none\">\n<p id=\"fs-id1169739097606\">-4.4<\/p>\n<\/div>\n<\/div>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p><iframe loading=\"lazy\" id=\"ohm16098\" class=\"resizable\" src=\"https:\/\/ohm.lumenlearning.com\/multiembedq.php?id=16098&theme=oea&iframe_resize_id=ohm16098&show_question_numbers\" width=\"100%\" height=\"150\"><\/iframe><\/p>\n<\/div>\n<\/div>\n\n\t\t\t <section class=\"citations-section\" role=\"contentinfo\">\n\t\t\t <h3>Candela Citations<\/h3>\n\t\t\t\t\t <div>\n\t\t\t\t\t\t <div id=\"citation-list-340\">\n\t\t\t\t\t\t\t <div class=\"licensing\"><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Original<\/div><ul class=\"citation-list\"><li>3.4 Derivatives as Rates of Change. <strong>Authored by<\/strong>: Ryan Melton. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">CC licensed content, Shared previously<\/div><ul class=\"citation-list\"><li>Calculus Volume 1. <strong>Authored by<\/strong>: Gilbert Strang, Edwin (Jed) Herman. <strong>Provided by<\/strong>: OpenStax. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/openstax.org\/details\/books\/calculus-volume-1\">https:\/\/openstax.org\/details\/books\/calculus-volume-1<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-nc-sa\/4.0\/\">CC BY-NC-SA: Attribution-NonCommercial-ShareAlike<\/a><\/em>. <strong>License Terms<\/strong>: Access for free at https:\/\/openstax.org\/books\/calculus-volume-1\/pages\/1-introduction<\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section>","protected":false},"author":17533,"menu_order":17,"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\":\"\"}]","CANDELA_OUTCOMES_GUID":"","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-340","chapter","type-chapter","status-publish","hentry"],"part":35,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/340","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/users\/17533"}],"version-history":[{"count":15,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/340\/revisions"}],"predecessor-version":[{"id":4809,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/340\/revisions\/4809"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/parts\/35"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapters\/340\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/media?parent=340"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/pressbooks\/v2\/chapter-type?post=340"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/contributor?post=340"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/calculus1\/wp-json\/wp\/v2\/license?post=340"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}