{"id":1991,"date":"2023-04-18T17:20:49","date_gmt":"2023-04-18T17:20:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&#038;p=1991"},"modified":"2025-08-28T19:15:22","modified_gmt":"2025-08-28T19:15:22","slug":"linear-and-geometric-growth-learn-it-3","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/linear-and-geometric-growth-learn-it-3\/","title":{"raw":"Linear and Geometric Growth: Learn It 3","rendered":"Linear and Geometric Growth: Learn It 3"},"content":{"raw":"<h2>Exponential (Population) Growth<\/h2>\r\n<p>Suppose that every year, only [latex]10\\%[\/latex] of the fish in a lake have surviving offspring. If there were [latex]100[\/latex] fish in the lake last year, there would now be [latex]110[\/latex] fish. If there were [latex]1000[\/latex] fish in the lake last year, there would now be [latex]1100[\/latex] fish. Absent any inhibiting factors, populations of people and animals tend to grow by a percent of the existing population each year.<\/p>\r\n<p>Suppose our lake began with [latex]1000[\/latex] fish, and [latex]10\\%[\/latex] of the fish have surviving offspring each year. Since we start with [latex]1000[\/latex] fish, [latex]P_0 = 1000[\/latex]. How do we calculate [latex]P_1[\/latex]? The new population will be the old population, plus an additional [latex]10\\%[\/latex]. Symbolically:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_{1}=P_{0}+0.10P_{0}[\/latex]<\/p>\r\n<section class=\"textbox recall\"><strong>representing percent as a decimal<\/strong>\r\n<p>The above statement can be read as \"the number of fish in the pond after the first year is equivalent to the initial number of fish plus [latex]10\\%[\/latex],\" where [latex]10\\%[\/latex] has been expressed in decimal form at [latex]0.10[\/latex].<\/p>\r\n<p>To rewrite a percent as a decimal, drop the [latex]\\%[\/latex] symbol and move the decimal point two places to the left.<\/p>\r\n<\/section>\r\n<p>Notice this could be condensed to a shorter form by factoring:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_1 = P_0 + 0.10P_0 = 1P_{0} + 0.10P_0 = (1+ 0.10)P_0 = 1.10P_0[\/latex]<\/p>\r\n<p>While [latex]10%[\/latex] is the <strong>growth rate<\/strong>, [latex]1.10[\/latex] is the <strong>growth multiplier<\/strong>. Notice that [latex]1.10[\/latex] can be thought of as \u201cthe original [latex]100\\%[\/latex] plus an additional [latex]10\\%[\/latex].\u201d<\/p>\r\n<p>For our fish population,<\/p>\r\n<p style=\"text-align: center;\">[latex]P_1 = 1.10(1000) = 1100[\/latex]<\/p>\r\n<p>We could then calculate the population in later years:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_2 = 1.10P_1 = 1.10(1100) = 1210[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_3 = 1.10P_2 = 1.10(1210) = 1331[\/latex]<\/p>\r\n<p>Notice that in the first year, the population grew by [latex]100[\/latex] fish; in the second year, the population grew by [latex]110[\/latex] fish; and in the third year the population grew by [latex]121[\/latex] fish.<\/p>\r\n<p>While there is a constant <em>percentage<\/em> growth, the actual increase in number of fish is increasing each year.<\/p>\r\n<p>Graphing these values we see that this growth doesn\u2019t quite appear linear.<\/p>\r\n<center>\r\n[caption id=\"attachment_6214\" align=\"aligncenter\" width=\"500\"]<img class=\"wp-image-6214\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2-300x181.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 200, from 800 to 1800. Measured horizontally: Years from now, measured in units of 1, from 0 to 5. Year 0 is at 1000; Year 1 is at roughly 1100; and subsequent years move in an increasing rate so that the overall line is curved to the upper right.\" width=\"500\" height=\"301\" \/> Figure 1. A graph showing the exponential growth of the fish population in 5 years[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p>A walk-through of this fish scenario can be viewed here:<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/3BiU7Ihxvxg\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Exponential+Growth+Model+Part+1.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth Model Part 1\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>\r\n<p>To get a better picture of how this percentage-based growth affects things, we need an explicit form, so we can quickly calculate values further out in the future.<\/p>\r\n<p>Like we did for the linear model, we will start building from the recursive equation:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_1 = 1.10P_0 = 1.10(1000)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P\u00ad_2 = 1.10P_1 = 1.10(1.10(1000)) = 1.10^{2}(1000)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_3 = 1.10P_2 = 1.10(1.102(1000)) = 1.10^{3}(1000)[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_4 = 1.10P_3 = 1.10(1.103(1000)) = 1.10^{4}(1000)[\/latex]<\/p>\r\n<p>Observing a pattern, we can generalize the explicit form to be:<\/p>\r\n<p style=\"text-align: center;\">[latex]P_n= 1.10^{n}(1000)[\/latex], or equivalently, [latex]P_n = 1000(1.10^{n})[\/latex]<\/p>\r\n<p>From this, we can quickly calculate the number of fish in [latex]10[\/latex], [latex]20[\/latex], or [latex]30[\/latex] years:<\/p>\r\n<center>\r\n[caption id=\"attachment_6216\" align=\"aligncenter\" width=\"500\"]<img class=\"wp-image-6216\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3-300x181.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 3000, from 0 to 18,000. Measured horizontally: Years from now, measured in units of 5, from 0 to 30. Year 0 is at 1000, with a tight cluster of dots in the lower left quadrant. The exponential increase becomes more dramatic as time advances, so that year 30 is at 18,000 fish with more spacing between dots.\" width=\"500\" height=\"301\" \/> Figure 2. A graph showing the exponential growth of the fish population in 30 years[\/caption]\r\n<\/center>\r\n<p>&nbsp;<\/p>\r\n<p style=\"text-align: center;\">[latex]P_{10}= 1.10^{10}(1000) = 2594[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_{20} = 1.10^{20}(1000) = 6727[\/latex]<\/p>\r\n<p style=\"text-align: center;\">[latex]P_{30} = 1.10^{30}(1000) = 17449[\/latex]<\/p>\r\n<p>Adding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly, by [latex]100[\/latex] fish each year, the population would have only reached [latex]4000[\/latex] in 30 years, compared to almost [latex]18,000[\/latex] with this percent-based growth, called <strong>exponential growth.<\/strong><\/p>\r\n<p>A video demonstrating the explicit model of this fish story can be viewed here:<\/p>\r\n<section class=\"textbox watchIt\"><iframe title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tg2ysaZ8agY\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe>\r\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Exponential+Growth+Model+Part+2.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth Model Part 2\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>","rendered":"<h2>Exponential (Population) Growth<\/h2>\n<p>Suppose that every year, only [latex]10\\%[\/latex] of the fish in a lake have surviving offspring. If there were [latex]100[\/latex] fish in the lake last year, there would now be [latex]110[\/latex] fish. If there were [latex]1000[\/latex] fish in the lake last year, there would now be [latex]1100[\/latex] fish. Absent any inhibiting factors, populations of people and animals tend to grow by a percent of the existing population each year.<\/p>\n<p>Suppose our lake began with [latex]1000[\/latex] fish, and [latex]10\\%[\/latex] of the fish have surviving offspring each year. Since we start with [latex]1000[\/latex] fish, [latex]P_0 = 1000[\/latex]. How do we calculate [latex]P_1[\/latex]? The new population will be the old population, plus an additional [latex]10\\%[\/latex]. Symbolically:<\/p>\n<p style=\"text-align: center;\">[latex]P_{1}=P_{0}+0.10P_{0}[\/latex]<\/p>\n<section class=\"textbox recall\"><strong>representing percent as a decimal<\/strong><\/p>\n<p>The above statement can be read as &#8220;the number of fish in the pond after the first year is equivalent to the initial number of fish plus [latex]10\\%[\/latex],&#8221; where [latex]10\\%[\/latex] has been expressed in decimal form at [latex]0.10[\/latex].<\/p>\n<p>To rewrite a percent as a decimal, drop the [latex]\\%[\/latex] symbol and move the decimal point two places to the left.<\/p>\n<\/section>\n<p>Notice this could be condensed to a shorter form by factoring:<\/p>\n<p style=\"text-align: center;\">[latex]P_1 = P_0 + 0.10P_0 = 1P_{0} + 0.10P_0 = (1+ 0.10)P_0 = 1.10P_0[\/latex]<\/p>\n<p>While [latex]10%[\/latex] is the <strong>growth rate<\/strong>, [latex]1.10[\/latex] is the <strong>growth multiplier<\/strong>. Notice that [latex]1.10[\/latex] can be thought of as \u201cthe original [latex]100\\%[\/latex] plus an additional [latex]10\\%[\/latex].\u201d<\/p>\n<p>For our fish population,<\/p>\n<p style=\"text-align: center;\">[latex]P_1 = 1.10(1000) = 1100[\/latex]<\/p>\n<p>We could then calculate the population in later years:<\/p>\n<p style=\"text-align: center;\">[latex]P_2 = 1.10P_1 = 1.10(1100) = 1210[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_3 = 1.10P_2 = 1.10(1210) = 1331[\/latex]<\/p>\n<p>Notice that in the first year, the population grew by [latex]100[\/latex] fish; in the second year, the population grew by [latex]110[\/latex] fish; and in the third year the population grew by [latex]121[\/latex] fish.<\/p>\n<p>While there is a constant <em>percentage<\/em> growth, the actual increase in number of fish is increasing each year.<\/p>\n<p>Graphing these values we see that this growth doesn\u2019t quite appear linear.<\/p>\n<div style=\"text-align: center;\">\n<figure id=\"attachment_6214\" aria-describedby=\"caption-attachment-6214\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-6214\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2-300x181.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 200, from 800 to 1800. Measured horizontally: Years from now, measured in units of 1, from 0 to 5. Year 0 is at 1000; Year 1 is at roughly 1100; and subsequent years move in an increasing rate so that the overall line is curved to the upper right.\" width=\"500\" height=\"301\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2-300x181.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2-65x39.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2-225x136.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2-350x211.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07163409\/Linegraph2.png 715w\" sizes=\"(max-width: 500px) 100vw, 500px\" \/><figcaption id=\"caption-attachment-6214\" class=\"wp-caption-text\">Figure 1. A graph showing the exponential growth of the fish population in 5 years<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p>A walk-through of this fish scenario can be viewed here:<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/3BiU7Ihxvxg\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Exponential+Growth+Model+Part+1.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth Model Part 1\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n<p>To get a better picture of how this percentage-based growth affects things, we need an explicit form, so we can quickly calculate values further out in the future.<\/p>\n<p>Like we did for the linear model, we will start building from the recursive equation:<\/p>\n<p style=\"text-align: center;\">[latex]P_1 = 1.10P_0 = 1.10(1000)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P\u00ad_2 = 1.10P_1 = 1.10(1.10(1000)) = 1.10^{2}(1000)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_3 = 1.10P_2 = 1.10(1.102(1000)) = 1.10^{3}(1000)[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_4 = 1.10P_3 = 1.10(1.103(1000)) = 1.10^{4}(1000)[\/latex]<\/p>\n<p>Observing a pattern, we can generalize the explicit form to be:<\/p>\n<p style=\"text-align: center;\">[latex]P_n= 1.10^{n}(1000)[\/latex], or equivalently, [latex]P_n = 1000(1.10^{n})[\/latex]<\/p>\n<p>From this, we can quickly calculate the number of fish in [latex]10[\/latex], [latex]20[\/latex], or [latex]30[\/latex] years:<\/p>\n<div style=\"text-align: center;\">\n<figure id=\"attachment_6216\" aria-describedby=\"caption-attachment-6216\" style=\"width: 500px\" class=\"wp-caption aligncenter\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-6216\" src=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3-300x181.png\" alt=\"Line graph. Measured vertically: Fish, in increments of 3000, from 0 to 18,000. Measured horizontally: Years from now, measured in units of 5, from 0 to 30. Year 0 is at 1000, with a tight cluster of dots in the lower left quadrant. The exponential increase becomes more dramatic as time advances, so that year 30 is at 18,000 fish with more spacing between dots.\" width=\"500\" height=\"301\" srcset=\"https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3-300x181.png 300w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3-65x39.png 65w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3-225x136.png 225w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3-350x211.png 350w, https:\/\/content-cdn.one.lumenlearning.com\/wp-content\/uploads\/sites\/18\/2023\/04\/07164043\/Linegraph3.png 715w\" sizes=\"(max-width: 500px) 100vw, 500px\" \/><figcaption id=\"caption-attachment-6216\" class=\"wp-caption-text\">Figure 2. A graph showing the exponential growth of the fish population in 30 years<\/figcaption><\/figure>\n<\/div>\n<p>&nbsp;<\/p>\n<p style=\"text-align: center;\">[latex]P_{10}= 1.10^{10}(1000) = 2594[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_{20} = 1.10^{20}(1000) = 6727[\/latex]<\/p>\n<p style=\"text-align: center;\">[latex]P_{30} = 1.10^{30}(1000) = 17449[\/latex]<\/p>\n<p>Adding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly, by [latex]100[\/latex] fish each year, the population would have only reached [latex]4000[\/latex] in 30 years, compared to almost [latex]18,000[\/latex] with this percent-based growth, called <strong>exponential growth.<\/strong><\/p>\n<p>A video demonstrating the explicit model of this fish story can be viewed here:<\/p>\n<section class=\"textbox watchIt\"><iframe loading=\"lazy\" title=\"YouTube video player\" src=\"https:\/\/www.youtube.com\/embed\/tg2ysaZ8agY\" width=\"560\" height=\"315\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>You can view the <a href=\"https:\/\/course-building.s3.us-west-2.amazonaws.com\/Quantitative+Reasoning+-+2023+Build\/Transcriptions\/Exponential+Growth+Model+Part+2.txt\" target=\"_blank\" rel=\"noopener\">transcript for \u201cExponential Growth Model Part 2\u201d here (opens in new window).<\/a><\/p>\n<\/section>\n","protected":false},"author":15,"menu_order":7,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Math in Society\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc-attribution\",\"description\":\"Math in Society (Lippman)\",\"author\":\"David Lippman\",\"organization\":\"LibreTexts Mathematics\",\"url\":\"https:\/\/math.libretexts.org\/Bookshelves\/Applied_Mathematics\/Math_in_Society_(Lippman)\/08%3A_Growth_Models\/8.04%3A_Exponential_(Geometric)_Growth\",\"project\":\"8.4: Exponential (Geometric) Growth\",\"license\":\"cc-by-sa\",\"license_terms\":\"Access for free at https:\/\/math.libretexts.org\/Bookshelves\/Applied_Mathematics\/Math_in_Society_(Lippman)\"}]","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"part":87,"module-header":"learn_it","content_attributions":[{"type":"cc","description":"Math in Society","author":"David Lippman","organization":"","url":"http:\/\/www.opentextbookstore.com\/mathinsociety\/","project":"","license":"cc-by","license_terms":""},{"type":"cc-attribution","description":"Math in Society (Lippman)","author":"David Lippman","organization":"LibreTexts Mathematics","url":"https:\/\/math.libretexts.org\/Bookshelves\/Applied_Mathematics\/Math_in_Society_(Lippman)\/08%3A_Growth_Models\/8.04%3A_Exponential_(Geometric)_Growth","project":"8.4: Exponential (Geometric) Growth","license":"cc-by-sa","license_terms":"Access for free at https:\/\/math.libretexts.org\/Bookshelves\/Applied_Mathematics\/Math_in_Society_(Lippman)"}],"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\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/1991"}],"collection":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/users\/15"}],"version-history":[{"count":25,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/1991\/revisions"}],"predecessor-version":[{"id":15867,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/1991\/revisions\/15867"}],"part":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/parts\/87"}],"metadata":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapters\/1991\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/media?parent=1991"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/pressbooks\/v2\/chapter-type?post=1991"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/contributor?post=1991"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/wp-json\/wp\/v2\/license?post=1991"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}