{"id":1229,"date":"2017-01-31T23:27:25","date_gmt":"2017-01-31T23:27:25","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/waymakermath4libarts\/?post_type=chapter&#038;p=1229"},"modified":"2020-12-18T22:57:36","modified_gmt":"2020-12-18T22:57:36","slug":"introduction-linear-and-geometric-growth","status":"publish","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/wmopen-mathforliberalarts\/chapter\/introduction-linear-and-geometric-growth\/","title":{"raw":"Linear and Geometric Growth","rendered":"Linear and Geometric Growth"},"content":{"raw":"<div class=\"textbox learning-objectives\">\r\n<h3>Learning Outcomes<\/h3>\r\n<ul>\r\n \t<li>Determine whether data or a scenario describe linear or geometric growth<\/li>\r\n \t<li>Identify growth rates, initial values, or point values expressed verbally, graphically, or numerically, and translate them into a format usable in calculation<\/li>\r\n \t<li>Calculate recursive and explicit equations for linear and geometric growth given sufficient information, and use those equations to make predictions<\/li>\r\n<\/ul>\r\n<\/div>\r\nHaving a constant rate of change is the defining characteristic of linear growth. Plotting coordinate pairs\u00a0associated with constant change will result in\u00a0a straight line, the shape of linear growth. In this section, we will formalize a way to describe linear growth using mathematical terms and concepts. By the end of this section, you will be able to write both a recursive and explicit equations for linear growth given starting conditions, or a constant of change. \u00a0You will also be able to recognize the difference between linear and geometric growth given a graph or an equation.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01232841\/growth-1140534_1280.jpg\"><img class=\" wp-image-1273 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01232841\/growth-1140534_1280-300x219.jpg\" alt=\"\" width=\"516\" height=\"377\" \/><\/a>\r\n<h2>Linear (Algebraic) Growth<\/h2>\r\n<h3>Predicting Growth<\/h3>\r\nMarco is a collector of antique soda bottles. His collection currently contains 437 bottles. Every year, he budgets enough money to buy 32 new bottles. Can we determine how many bottles he will have in 5 years, and how long it will take for his collection to reach 1000 bottles?\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/15183507\/15186624687_fc02ef5925_z.jpg\"><img class=\"aligncenter size-full wp-image-806\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/15183507\/15186624687_fc02ef5925_z.jpg\" alt=\"collection of empty glass soda bottles\" width=\"640\" height=\"426\" \/><\/a>\r\n\r\nWhile you could probably solve both of these questions without an equation or formal mathematics, we are going to formalize our approach to this problem to provide a means to answer more complicated questions.\r\n\r\nSuppose that <em>P\u00ad<sub>n <\/sub><\/em>represents the number, or population, of bottles Marco has after <em>n<\/em> years. So <em>P\u00ad<sub>0<\/sub><\/em> would represent the number of bottles now, <em>P\u00ad<sub>1<\/sub><\/em> would represent the number of bottles after 1 year, <em>P\u00ad<sub>2<\/sub><\/em> would represent the number of bottles after 2 years, and so on. We could describe how Marco\u2019s bottle collection is changing using:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 437<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P<sub>\u00adn-1<\/sub><\/em> + 32<\/p>\r\nThis is called a <strong>recursive relationship<\/strong>. A recursive relationship is a formula which relates the next value in a sequence to the previous values. Here, the number of bottles in year <em>n<\/em> can be found by adding 32 to the number of bottles in the previous year, <em>P\u00ad<sub>n-1<\/sub><\/em>. Using this relationship, we could calculate:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub> = P\u00ad<sub>0<\/sub><\/em> + 32 = 437 + 32 = 469<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub> = P\u00ad<sub>1<\/sub><\/em> + 32 = 469 + 32 = 501<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>3<\/sub> = P\u00ad<sub>2<\/sub><\/em> + 32 = 501 + 32 = 533<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub> = P\u00ad<sub>3<\/sub><\/em> + 32 = 533 + 32 = 565<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>5<\/sub> = P\u00ad<sub>4<\/sub><\/em> + 32 = 565 + 32 = 597<\/p>\r\nWe have answered the question of how many bottles Marco will have in 5 years.\u00a0<img class=\"aligncenter wp-image-358 \" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11200627\/bottles.png\" alt=\"Line chart. Vertical measures Bottles, with increments of 100 from 0 to 700. Horizontal measures Years From Now, with increments of 1 from 0 to 5. The line moves in a slow rise from left to right, from a little over 400 at year 0 to 600 at year 5.\" width=\"402\" height=\"325\" \/>\r\n\r\nHowever, solving how long it will take for his collection to reach 1000 bottles would require a lot more calculations.\r\n\r\nWhile recursive relationships are excellent for describing simply and cleanly <em>how<\/em> a quantity is changing, they are not convenient for making predictions or solving problems that stretch far into the future. For that, a closed or explicit form for the relationship is preferred. An <strong>explicit equation <\/strong>allows us to calculate <em>P\u00ad<sub>n <\/sub><\/em>directly, without needing to know <em>P<sub>\u00adn-1<\/sub><\/em>. While you may already be able to guess the explicit equation, let us derive it from the recursive formula. We can do so by selectively not simplifying as we go:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub> = <\/em>437 + 32 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 437 + 1(32)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub> = P\u00ad<sub>1<\/sub><\/em> + 32 = 437 + 32 + 32 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 437 + 2(32)<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>3<\/sub> = P\u00ad<sub>2<\/sub><\/em> + 32 = (437 + 2(32)) + 32 \u00a0\u00a0\u00a0 = 437 + 3(32)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub> = P<sub>\u00ad3<\/sub><\/em> + 32 = (437 + 3(32)) + 32 \u00a0\u00a0\u00a0 = 437 + 4(32)<\/p>\r\n<em>\u00a0<\/em>\r\n\r\nYou can probably see the pattern now, and generalize that\r\n<p style=\"text-align: center;\"><em>P<sub>\u00adn<\/sub> = <\/em>437 + <em>n<\/em>(32) = 437 + 32<em>n<\/em><\/p>\r\nUsing this equation, we can calculate how many bottles he\u2019ll have after 5 years:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>5<\/sub> =<\/em> 437 + 32(5) = 437 + 160 = 597<\/p>\r\nWe can now also solve for when the collection will reach 1000 bottles by substituting in 1000 for <em>P\u00ad<sub>n<\/sub><\/em> and solving for <em>n<\/em>\r\n<p style=\"text-align: center;\">1000 = 437 + 32<sub><em>n<\/em><\/sub><\/p>\r\n<p style=\"text-align: center;\">563 = 32<sub><em>n<\/em><\/sub><\/p>\r\n<p style=\"text-align: center;\"><em>n<\/em> = 563\/32 = 17.59<\/p>\r\nSo Marco will reach 1000 bottles in 18 years.\r\n\r\nThe steps of determining the formula and solving the problem of Marco's bottle collection are explained in detail in the following videos.\r\n\r\nhttps:\/\/youtu.be\/SJcAjN-HL_I\r\n\r\nhttps:\/\/youtu.be\/4Two_oduhrA\r\n\r\nhttps:\/\/youtu.be\/pZ4u3j8Vmzo\r\n\r\nIn this\u00a0example, Marco\u2019s collection grew by the <em>same numbe<\/em>r of bottles every year. This constant change is the defining characteristic of linear growth. Plotting the values we calculated for Marco\u2019s collection, we can see the values form a straight line, the shape of linear growth.\r\n<div class=\"textbox\">\r\n<h3>Linear Growth<\/h3>\r\nIf a quantity starts at size <em>P\u00ad<sub>0<\/sub><\/em> and grows by <em>d<\/em> every time period, then the quantity after <em>n<\/em> time periods can be determined using either of these relations:\r\n<h4>Recursive form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>n-1<\/sub><\/em> + <em>d<\/em><\/p>\r\n\r\n<h4>Explicit form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>0 <\/sub><\/em>+ <em>d n<\/em><\/p>\r\nIn this equation, <em>d<\/em> represents the <strong>common difference<\/strong> \u2013 the amount that the population changes each time <em>n<\/em> increases by 1.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Connection to Prior Learning: Slope and Intercept<\/h3>\r\nYou may recognize the common difference, <em>d<\/em>, in our linear equation as <em>slope<\/em>. In fact, the entire explicit equation should look familiar \u2013 it is the same linear equation you learned in algebra, probably stated as <em>y = mx + b<\/em>.\r\n\r\nIn the standard algebraic equation <em>y = mx + b<\/em>, <em>b<\/em> was the <em>y<\/em>-intercept, or the <em>y<\/em> value when <em>x<\/em> was zero. In the form of the equation we\u2019re using, we are using <em>P\u00ad0\u00ad<\/em> to represent that initial amount.\r\n\r\nIn the <em>y = mx + b<\/em> equation, recall that <em>m<\/em> was the slope. You might remember this as \u201crise over run,\u201d or the change in <em>y<\/em> divided by the change in <em>x<\/em>. Either way, it represents the same thing as the common difference, <em>d<\/em>, we are using \u2013 the amount the output <em>P<sub>n<\/sub><\/em> changes when the input <em>n<\/em> increases by 1.\r\n\r\nThe equations <em>y = mx + b<\/em> and <em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>0 <\/sub><\/em>+ <em>d n <\/em>mean the same thing and can be used the same ways.\u00a0We\u2019re just writing it somewhat differently.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nThe population of elk in a national forest was measured to be 12,000 in 2003, and was measured again to be 15,000 in 2007. If the population continues to grow linearly at this rate, what will the elk population be in 2014?\r\n[reveal-answer q=\"60252\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"60252\"]\r\n\r\nTo begin, we need to define how we\u2019re going to measure <em>n<\/em>.\u00a0\u00a0 Remember that <em>P\u00ad<sub>0<\/sub><\/em> is the population when <em>n<\/em> = 0, so we probably don\u2019t want to literally use the year 0. Since we already know the population in 2003, let us define <em>n<\/em> = 0 to be the year 2003.\r\n\r\nThen\u00a0<em>P\u00ad<sub>0<\/sub><\/em> = 12,000.\r\n\r\nNext we need to find <em>d<\/em>. Remember <em>d<\/em> is the growth per time period, in this case growth per year. Between the two measurements, the population grew by 15,000-12,000 = 3,000, but it took 2007-2003 = 4 years to grow that much. To find the growth per year, we can divide: 3000 elk \/ 4 years = 750 elk in 1 year.\r\n\r\nAlternatively, you can use the slope formula from algebra to determine the common difference, noting that the population is the output of the formula, and time is the input.\r\n\r\n[latex]d=slope=\\frac{\\text{changeinoutput}}{\\text{changeininput}}=\\frac{15,000-12,000}{2007-2003}=\\frac{3000}{4}=750[\/latex]\r\n\r\nWe can now write our equation in whichever form is preferred.\r\n<h4>Recursive form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 12,000<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00adn<\/sub> = P<sub>\u00adn-1<\/sub><\/em> + 750<\/p>\r\n\r\n<h4>Explicit form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 12,000 + 750(n)<\/p>\r\nTo answer the question, we need to first note that the year 2014 will be <em>n<\/em> = 11, since 2014 is 11 years after 2003. The explicit form will be easier to use for this calculation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>11<\/sub><\/em> = 12,000 + 750(11) = 20,250 elk<\/p>\r\n[\/hidden-answer]\r\n\r\nView more about this example here.\r\n\r\nhttps:\/\/youtu.be\/J1XqqlKzYGs\r\n\r\n<hr \/>\r\n\r\nGasoline consumption in the US has been increasing steadily. Consumption data from 1992 to 2004 is shown below.[footnote]<a href=\"http:\/\/www.bts.gov\/publications\/national_transportation_statistics\/2005\/html\/table_04_10.html\" target=\"_blank\" rel=\"noopener\">http:\/\/www.bts.gov\/publications\/national_transportation_statistics\/2005\/html\/table_04_10.html<\/a>[\/footnote] Find a model for this data, and use it to predict consumption in 2016. If the trend continues, when will consumption reach 200 billion gallons?\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Year<\/td>\r\n<td>'92<\/td>\r\n<td>'93<\/td>\r\n<td>'94<\/td>\r\n<td>'95<\/td>\r\n<td>'96<\/td>\r\n<td>'97<\/td>\r\n<td>'98<\/td>\r\n<td>'99<\/td>\r\n<td>'00<\/td>\r\n<td>'01<\/td>\r\n<td>'02<\/td>\r\n<td>'03<\/td>\r\n<td>'04<\/td>\r\n<\/tr>\r\n<tr>\r\n<td>Consumption (billion of gallons)<\/td>\r\n<td>110<\/td>\r\n<td>111<\/td>\r\n<td>113<\/td>\r\n<td>116<\/td>\r\n<td>118<\/td>\r\n<td>119<\/td>\r\n<td>123<\/td>\r\n<td>125<\/td>\r\n<td>126<\/td>\r\n<td>128<\/td>\r\n<td>131<\/td>\r\n<td>133<\/td>\r\n<td>136<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"307147\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"307147\"]\r\n\r\nPlotting this data, it appears to have an approximately linear relationship:\r\n\r\n<img class=\"aligncenter wp-image-359\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201004\/gasconsumption.png\" alt=\"Graph. Vertical measures Gas Consumption in increments of 10, from 100 to 140. Horizontal measures Year in increments of 4, from 1992 to 2004. Points identified in a generally upward trend, left to right, from 110 in 1992 to near 140 in 2004.\" width=\"350\" height=\"235\" \/>\r\n\r\nWhile there are more advanced statistical techniques that can be used to find an equation to model the data, to get an idea of what is happening, we can find an equation by using two pieces of the data \u2013 perhaps the data from 1993 and 2003.\r\n\r\nLetting <em>n<\/em> = 0 correspond with 1993 would give <em>P\u00ad<sub>0<\/sub><\/em> = 111 billion gallons.\r\n\r\nTo find <em>d<\/em>, we need to know how much the gas consumption increased each year, on average. From 1993 to 2003 the gas consumption increased from 111 billion gallons to 133 billion gallons, a total change of 133 \u2013 111 = 22 billion gallons, over 10 years. This gives us an average change of 22 billion gallons \/ 10 year = 2.2 billion gallons per year.\r\n\r\nEquivalently,\r\n\r\n[latex]d=slope=\\frac{\\text{changeinoutput}}{\\text{changeininput}}=\\frac{133-111}{10-0}=\\frac{22}{10}=2.2[\/latex]billion gallons per year\r\n\r\nWe can now write our equation in whichever form is preferred.\r\n<h4>Recursive form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 111<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>n-1<\/sub><\/em> + 2.2<\/p>\r\n\r\n<h4>Explicit form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 111 + 2.2<sub><em>n<\/em><\/sub><\/p>\r\n<em>\u00a0<\/em><em><img class=\"aligncenter wp-image-360\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201052\/gasconsumption2.png\" alt=\"Graph. Vertical measures Gas Consumption in increments of 10, from 100 to 140. Horizontal measures Year in increments of 4, from 1992 to 2004. Points identified in a generally upward trend, left to right, from 110 in 1992 to near 140 in 2004. This line connects dots with a solid line.\" width=\"350\" height=\"283\" \/><\/em>\r\n\r\nCalculating values using the explicit form and plotting them with the original data shows how well our model fits the data.\r\n\r\nWe can now use our model to make predictions about the future, assuming that the previous trend continues unchanged. To predict the gasoline consumption in 2016:\r\n<p style=\"text-align: center;\"><em>n<\/em> = 23 (2016 \u2013 1993 = 23 years later)<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad23<\/sub><\/em> = 111 + 2.2(23) = 161.6<\/p>\r\nOur model predicts that the US will consume 161.6 billion gallons of gasoline in 2016 if the current trend continues.\r\n\r\nTo find when the consumption will reach 200 billion gallons, we would set <em>P\u00ad<sub>n <\/sub><\/em>= 200, and solve for <em>n<\/em>:\r\n<p style=\"padding-left: 210px;\"><em>P<sub>n<\/sub><\/em> = 200\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Replace <em>P<sub>n<\/sub><\/em> with our model<\/p>\r\n<p style=\"padding-left: 210px;\">111 + 2.2<sub><em>n<\/em><\/sub> = 200\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract 111 from both sides<\/p>\r\n<p style=\"padding-left: 210px;\">2.2<sub><em>n<\/em><\/sub> = 89\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 2.2<\/p>\r\n<p style=\"padding-left: 210px;\"><em>n<\/em> = 40.4545<\/p>\r\nThis tells us that consumption will reach 200 billion about 40 years after 1993, which would be in the year 2033.\r\n\r\n[\/hidden-answer]\r\n\r\nThe steps for reaching this answer are detailed in the following video.\r\n\r\nhttps:\/\/youtu.be\/ApFxDWd6IbE\r\n\r\n<hr \/>\r\n\r\nThe cost, in dollars, of a gym membership for <em>n<\/em> months can be described by the explicit equation <em>P\u00ad<sub>n<\/sub><\/em> = 70 + 30<sub><em>n<\/em><\/sub>. What does this equation tell us?\r\n\r\n[reveal-answer q=\"438458\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"438458\"]\r\n\r\nThe value for <em>P\u00ad<sub>0<\/sub><\/em> in this equation is 70, so the initial starting cost is $70. This tells us that there must be an initiation or start-up fee of $70 to join the gym.\r\n\r\nThe value for <em>d<\/em> in the equation is 30, so the cost increases by $30 each month. This tells us that the monthly membership fee for the gym is $30 a month.\r\n\r\n[\/hidden-answer]\r\n\r\nThe explanation for this example is detailed below.\r\n\r\nhttps:\/\/youtu.be\/0Uwz5dmLTtk\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe number of stay-at-home fathers in Canada has been growing steadily[footnote]<a href=\"http:\/\/www.fira.ca\/article.php?id=140\" target=\"_blank\" rel=\"noopener\">http:\/\/www.fira.ca\/article.php?id=140<\/a>[\/footnote]. While the trend is not perfectly linear, it is fairly linear. Use the data from 1976 and 2010 to find an explicit formula for the number of stay-at-home fathers, then use it to predict the number in\u00a02020.\r\n<table>\r\n<tbody>\r\n<tr>\r\n<td>Year<\/td>\r\n<td>1976<\/td>\r\n<td>1984<\/td>\r\n<td>1991<\/td>\r\n<td>2000<\/td>\r\n<td>2010<\/td>\r\n<\/tr>\r\n<tr>\r\n<td># of Stay -at-home fathers<\/td>\r\n<td>20610<\/td>\r\n<td>28725<\/td>\r\n<td>43530<\/td>\r\n<td>47665<\/td>\r\n<td>53555<\/td>\r\n<\/tr>\r\n<\/tbody>\r\n<\/table>\r\n[reveal-answer q=\"688205\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"688205\"]\r\n<div>Letting n= 0 correspond with 1976, then [latex]P_0= 20,610[\/latex].<\/div>\r\n<div>From 1976 to 2010 the number of stay-at-home fathers increased by 53,555 \u2013 20,610 = 32,945<\/div>\r\n<div>This happened over 34 years, giving a common different <em>d\u00a0<\/em>of 32,945 \/ 34 = 969.<\/div>\r\n<div>[latex]P_n= 20,610 + 969n[\/latex]<\/div>\r\n<div>Predicting for 2020, we use n = 44, P(44) = 20,610 + 969(44) = 63,246 stay-at-home fathers in 2020.<\/div>\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6594&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>\r\n<h3>When Good Models Go Bad<\/h3>\r\nWhen using mathematical models to predict future behavior, it is important to keep in mind that very few trends will continue indefinitely.\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nSuppose a four year old boy is currently 39 inches tall, and you are told to expect him to grow 2.5 inches a year.\r\n\r\nWe can set up a growth model, with <em>n<\/em> = 0 corresponding to 4 years old.\r\n<h4>Recursive form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 39<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00adn<\/sub> = P\u00ad<sub>n-1<\/sub><\/em> + 2.5<\/p>\r\n\r\n<h4>Explicit form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 39 + 2.5(n)<\/p>\r\nSo at 6 years old, we would expect him to be\r\n<p style=\"text-align: center;\"><em>P\u00ad<\/em><sub>2<\/sub> = 39 + 2.5(2) = 44 inches tall<\/p>\r\nAny mathematical model will break down eventually. Certainly, we shouldn\u2019t expect this boy to continue to grow at the same rate all his life. If he did, at age 50 he would be\r\n<p style=\"text-align: center;\"><em>P<\/em><sub>\u00ad46<\/sub> = 39 + 2.5(46) = 154 inches tall = 12.8 feet tall!<\/p>\r\nWhen using any mathematical model, we have to consider which inputs are reasonable to use. Whenever we <strong>extrapolate<\/strong>, or make predictions into the future, we are assuming the model will continue to be valid.\r\n\r\nView a video explanation of this breakdown of the linear growth model here.\r\n\r\nhttps:\/\/youtu.be\/6zfXCsmcDzI\r\n\r\n<\/div>\r\n&nbsp;\r\n<h2>Exponential (Geometric ) Growth<\/h2>\r\n<h3>Population Growth<\/h3>\r\nSuppose that every year, only 10% of the fish in a lake have surviving offspring. If there were 100 fish in the lake last year, there would now be 110 fish. If there were 1000 fish in the lake last year, there would now be 1100 fish. Absent any inhibiting factors, populations of people and animals tend to grow by a percent of the existing population each year.\r\n\r\n<a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280.jpg\"><img class=\"aligncenter size-large wp-image-906\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280-1024x714.jpg\" alt=\"Two orange koi at the top of a pond, mouths open. Others are swimming beneath them.\" width=\"1024\" height=\"714\" \/><\/a>\r\nSuppose our lake began with 1000 fish, and 10% of the fish have surviving offspring each year. Since we start with 1000 fish, <em>P\u00ad<sub>0<\/sub><\/em> = 1000. How do we calculate <em>P\u00ad<sub>1<\/sub><\/em>? The new population will be the old population, plus an additional 10%. Symbolically:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0<\/sub><\/em><\/p>\r\nNotice this could be condensed to a shorter form by factoring:\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P\u00ad<sub>0<\/sub><\/em> = 1<em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0 <\/sub><\/em>= (1+ 0.10)<em>P\u00ad<sub>0 <\/sub><\/em>= 1.10<em>P\u00ad<sub>0<\/sub><\/em><\/p>\r\nWhile 10% is the <strong>growth rate<\/strong>, 1.10 is the <strong>growth multiplier<\/strong>. Notice that 1.10 can be thought of as \u201cthe original 100% plus an additional 10%.\u201d\r\n\r\nFor our fish population,\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1000) = 1100<\/p>\r\nWe could then calculate the population in later years:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10<em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1100) = 1210<\/p>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad3<\/sub><\/em> = 1.10<em>P<sub>2<\/sub><\/em> = 1.10(1210) = 1331<\/p>\r\nNotice that in the first year, the population grew by 100 fish; in the second year, the population grew by 110 fish; and in the third year the population grew by 121 fish.\r\n\r\nWhile there is a constant <em>percentage<\/em> growth, the actual increase in number of fish is increasing each year.\r\n\r\nGraphing these values we see that this growth doesn\u2019t quite appear linear.\r\n\r\n<img class=\"aligncenter wp-image-363\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201507\/yearsfromnow1.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=\"350\" height=\"283\" \/>\r\n\r\nA walkthrough of this fish scenario can be viewed here:\r\n\r\nhttps:\/\/youtu.be\/3BiU7Ihxvxg\r\n\r\nTo 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.\r\n\r\nLike we did for the linear model, we will start building from the recursive equation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10(<em>P\u00ad<sub>0<\/sub><\/em>\u00a0)= 1.10(1000)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10(<em>P\u00ad<sub>1<\/sub><\/em>\u00a0)= 1.10(1.10(1000)) = 1.102(1000)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>3<\/sub><\/em> = 1.10(<em>P\u00ad<sub>2<\/sub><\/em>\u00a0)= 1.10(1.10<sup>2<\/sup>(1000)) = 1.103(1000)<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub><\/em> = 1.10(<em>P\u00ad<sub>3<\/sub><\/em>\u00a0)= 1.10(1.10<sup>3<\/sup>(1000)) = 1.104(1000)<\/p>\r\nObserving a pattern, we can generalize the explicit form to be:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 1.10<sup><em>n<\/em><\/sup>(1000), or equivalently, <em>P\u00ad<sub>n<\/sub><\/em> = 1000(1.10<em><sup>n<\/sup><\/em>)<\/p>\r\nFrom this, we can quickly calculate the number of fish in 10, 20, or 30 years:\r\n\r\n<img class=\"aligncenter wp-image-364\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201611\/yearsfromnow2.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=\"350\" height=\"281\" \/>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 1.10<sup><em>10<\/em><\/sup>(1000) = 2594<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>20<\/sub><\/em> = 1.10<sup><em>20<\/em><\/sup>(1000) = 6727<\/p>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>30<\/sub><\/em> = 1.10<sup><em>30<\/em><\/sup>(1000) = 17449<\/p>\r\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">Adding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly, by 100 fish each year, the population would have only reached 4000 in 30 years, compared to almost 18,000 with this percent-based growth, called <\/span><strong style=\"font-size: 1rem; text-align: initial;\">exponential growth.<\/strong><\/p>\r\nA video demonstrating the explicit model of this fish story can be viewed here:\r\n\r\nhttps:\/\/youtu.be\/tg2ysaZ8agY\r\n\r\nIn exponential growth, the population grows proportional to the size of the population, so as the population gets larger, the same percent growth will yield a larger numeric growth.\r\n\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Exponential Growth<\/h3>\r\nIf a quantity starts at size <em>P\u00ad<sub>0<\/sub><\/em> and grows by <em>R%<\/em> (written as a decimal, <em>r<\/em>) every time period, then the quantity after <em>n<\/em> time periods can be determined using either of these relations:\r\n<h4>Recursive form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>) <em>P\u00ad<sub>n-1<\/sub><\/em><\/p>\r\n\r\n<h4>Explicit form<\/h4>\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>)<em><sup>n<\/sup><\/em>\u00a0<em>P\u00ad<sub>0<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>or equivalently, <em>P<sub>\u00adn<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> (1+<em>r<\/em>)<em><sup>n<\/sup><\/em><\/p>\r\nWe call <em>r<\/em> the <strong>growth rate<\/strong>.\r\n\r\nThe term (1+<em>r<\/em>) is called the <strong>growth multiplier<\/strong>, or common ratio.\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nBetween 2007 and 2008, Olympia, WA grew almost 3% to a population of 245 thousand people. If this growth rate was to continue, what would the population of Olympia be in 2014?\r\n[reveal-answer q=\"54756\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"54756\"]\r\n\r\nAs we did before, we first need to define what year will correspond to <em>n<\/em> = 0. Since we know the population in 2008, it would make sense to have 2008 correspond to <em>n <\/em>= 0, so <em>P\u00ad<sub>0<\/sub><\/em> = 245,000.\u00a0\u00a0 The year 2014 would then be <em>n<\/em> = 6.\r\n\r\nWe know the growth rate is 3%, giving <em>r <\/em>= 0.03.\r\n\r\nUsing the explicit form:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>6<\/sub><\/em> = (1+0.03)<sup>6<\/sup> (245,000) = 1.19405(245,000) = 292,542.25<\/p>\r\nThe model predicts that in 2014, Olympia would have a population of about 293 thousand people.\r\n\r\n[\/hidden-answer]\r\n\r\nThe following video explains this example in detail.\r\n\r\nhttps:\/\/youtu.be\/CDI4xS65rxY\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox\">\r\n<h3>Evaluating exponents on the calculator<\/h3>\r\nTo evaluate expressions like (1.03)<sup>6<\/sup>, it will be easier to use a calculator than multiply 1.03 by itself six times. Most scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:\r\n<p style=\"text-align: center;\">^ ,\u00a0\u00a0 y<sup>x<\/sup> ,\u00a0\u00a0 or x<sup>y<\/sup> .<\/p>\r\nTo evaluate 1.03<sup>6<\/sup>\u00a0we\u2019d type 1.03 ^ 6, or 1.03 y<sup>x<\/sup> 6.\u00a0 Try it out - you should get an answer around 1.1940523.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nIndia is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, what will India\u2019s population grow to by 2020?\r\n[reveal-answer q=\"265475\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"265475\"]\r\n<div>Using n = 0 corresponding with 2008, [latex]P_12= (1+0.0134)12(1.14)[\/latex] = about 1.337 billion people in 2020<\/div>\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6673&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>\r\n&nbsp;\r\n<div class=\"textbox exercises\">\r\n<h3>Examples<\/h3>\r\nA friend is using the equation <em>P\u00ad<sub>n<\/sub><\/em> = 4600(1.072)<sup><em>n <\/em><\/sup>to predict the annual tuition at a local college. She says the formula is based on years after 2010. What does this equation tell us?\r\n\r\n[reveal-answer q=\"224261\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"224261\"]\r\n\r\nIn the equation, <em>P<sub>\u00ad0<\/sub><\/em> = 4600, which is the starting value of the tuition when <em>n<\/em> = 0. This tells us that the tuition in 2010 was $4,600.\r\n\r\nThe growth multiplier is 1.072, so the growth rate is 0.072, or 7.2%. This tells us that the tuition is expected to grow by 7.2% each year.\r\n\r\nPutting this together, we could say that the tuition in 2010 was $4,600, and is expected to grow by 7.2% each year.\r\n\r\n[\/hidden-answer]\r\n\r\nView the following to see this example worked out.\r\n\r\nhttps:\/\/youtu.be\/T8Yz94De5UM\r\n\r\n<hr \/>\r\n\r\nIn 1990, residential energy use in the US was responsible for 962 million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to 1182 million metric tons[footnote]<a href=\"http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html\" target=\"_blank\" rel=\"noopener\">http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html<\/a>[\/footnote]. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?\r\n[reveal-answer q=\"755963\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"755963\"]\r\n\r\nSimilar to before, we will correspond <em>n<\/em> = 0 with 1990, as that is the year for the first piece of data we have. That will make <em>P\u00ad<sub>0<\/sub><\/em> = 962 (million metric tons of CO<sub>2<\/sub>). In this problem, we are not given the growth rate, but instead are given that <em>P\u00ad<sub>10<\/sub><\/em> = 1182.\r\n\r\nWhen <em>n<\/em> = 10, the explicit equation looks like:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> <em>P\u00ad<sub>0<\/sub><\/em><\/p>\r\nWe know the value for <em>P\u00ad<sub>0<\/sub><\/em>, so we can put that into the equation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\r\nWe also know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, so substituting that in, we get\r\n<p style=\"text-align: center;\">1182 = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\r\nWe can now solve this equation for the growth rate, <em>r<\/em>. Start by dividing by 962.\r\n<p style=\"padding-left: 120px;\">[latex]\\frac{1182}{962}={{(1+r)}^{10}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Take the 10th root of both sides<\/p>\r\n<p style=\"padding-left: 120px;\">[latex]\\sqrt[10]{\\frac{1182}{962}}=1+r[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract 1 from both sides<\/p>\r\n<p style=\"padding-left: 120px;\">[latex]r=\\sqrt[10]{\\frac{1182}{962}}-1=0.0208[\/latex] = 2.08%<\/p>\r\nSo if the emissions are growing exponentially, they are growing by about 2.08% per year. We can now predict the emissions in 2050 by finding <em>P<sub>\u00ad60<\/sub><\/em>\r\n<p style=\"text-align: center;\"><em>P<sub>\u00ad60<\/sub><\/em> = (1+0.0208)<sup>60<\/sup> 962 = 3308.4 million metric tons of CO<sub>2<\/sub> in 2050<\/p>\r\n[\/hidden-answer]\r\n\r\nView more about this example here.\r\n\r\nhttps:\/\/youtu.be\/9Zu2uONfLkQ\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Rounding<\/h3>\r\nAs a note on rounding, notice that if we had rounded the growth rate to 2.1%, our calculation for the emissions in 2050 would have been 3347.\u00a0\u00a0 Rounding to 2% would have changed our result to 3156. A very small difference in the growth rates gets magnified greatly in exponential growth. For this reason, it is recommended to round the growth rate as little as possible.\r\n\r\nIf you need to round, <strong>keep at least three significant digits<\/strong> - numbers after any leading zeros.\u00a0\u00a0 So 0.4162 could be reasonably rounded to 0.416. A growth rate of 0.001027 could be reasonably rounded to 0.00103.\r\n\r\n<\/div>\r\n<div class=\"textbox\">\r\n<h3>Evaluating roots on the calculator<\/h3>\r\nIn the previous example, we had to calculate the 10th root of a number. This is different than taking the basic square root, \u221a. Many scientific calculators have a button for general roots.\u00a0 It is typically labeled like:\r\n<p style=\"text-align: center;\">[latex]\\sqrt[y]{x}[\/latex]<\/p>\r\nTo evaluate the 3rd root of 8, for example, we\u2019d either type 3 [latex]\\sqrt[x]{{}}[\/latex] 8, or 8 [latex]\\sqrt[x]{{}}[\/latex] 3, depending on the calculator. Try it on yours to see which to use \u2013 you should get an answer of 2.\r\n\r\nIf your calculator does not have a general root button, all is not lost. You can instead use the property of exponents which states that:\r\n<p style=\"text-align: center;\"><span style=\"color: #000000;\">[latex]\\sqrt[n]{a}={a}^{\\frac{1}{2}}[\/latex]<\/span>.<\/p>\r\nSo, to compute the 3rd root of 8, you could use your calculator\u2019s exponent key to evaluate 8<sup>1\/3<\/sup>. To do this, type:\r\n<p style=\"text-align: center;\">8 y<sup>x<\/sup> ( 1 \u00f7 3 )<\/p>\r\nThe parentheses tell the calculator to divide 1\/3 before doing the exponent.\r\n\r\n<\/div>\r\n<div class=\"textbox key-takeaways\">\r\n<h3>Try It<\/h3>\r\nThe number of users on a social networking site was 45 thousand in February when they officially went public, and grew to 60 thousand by October. If the site is growing exponentially, and growth continues at the same rate, how many users should they expect two years after they went public?\r\n\r\n[reveal-answer q=\"774576\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"774576\"]\r\n\r\nHere we will measure n in months rather than years, with n = 0 corresponding to the February when they went public. This gives [latex]P_0= 45[\/latex] thousand. October is 8 months later, so [latex]P_8= 60[\/latex].\r\n[latex]P_8=(1+r)^{8}P_0[\/latex]\r\n[latex]60=(1+r)^{8}45[\/latex]\r\n[latex]\\frac{60}{45}=(1+r)^8[\/latex]\r\n[latex]\\sqrt[8]{\\frac{60}{45}}=1+r[\/latex]\r\n[latex]r=\\sqrt[8]{\\frac{60}{45}}-1=0.0366\\text{ or }3.66%[\/latex]\r\nThe general explicit equation is [latex]P_n =(1.0366)^{n}45[\/latex]. Predicting 24 months after they went public gives [latex]P_{24}=(1.0366)^{24}45=106.63[\/latex] thousand users.\r\n[\/hidden-answer]\r\n\r\n<iframe id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6598&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"100%\"><\/iframe>\r\n\r\n<\/div>\r\n<div class=\"textbox exercises\">\r\n<h3>Example<\/h3>\r\nLooking back at the last example, for the sake of comparison, what would the carbon emissions be in 2050 if emissions grow linearly at the same rate?\r\n[reveal-answer q=\"355767\"]Show Solution[\/reveal-answer]\r\n[hidden-answer a=\"355767\"]\r\n\r\nAgain we will get <em>n<\/em> = 0 correspond with 1990, giving <em>P<sub>\u00ad0<\/sub><\/em> = 962. To find <em>d<\/em>, we could take the same approach as earlier, noting that the emissions increased by 220 million metric tons in 10 years, giving a common difference of 22 million metric tons each year.\r\n\r\nAlternatively, we could use an approach similar to that which we used to find the exponential equation. When <em>n<\/em> = 10, the explicit linear equation looks like:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = <em>P<sub>\u00ad0<\/sub><\/em> + 10<em>d<\/em><em>\u00a0<\/em><\/p>\r\nWe know the value for <em>P\u00ad0<\/em>, so we can put that into the equation:\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 962 + 10<em>d<\/em><\/p>\r\nSince we know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, substituting that in we get\r\n<p style=\"text-align: center;\">1182 = 962 + 10<em>d<\/em><\/p>\r\nWe can now solve this equation for the common difference, <em>d<\/em>.\r\n<p style=\"text-align: center;\">1182 \u2013 962 = 10<em>d<\/em><\/p>\r\n<p style=\"text-align: center;\">220 = 10<em>d<\/em><\/p>\r\n<p style=\"text-align: center;\"><em>d<\/em> = 22<\/p>\r\nThis tells us that if the emissions are changing linearly, they are growing by 22 million metric tons each year. Predicting the emissions in 2050,\r\n<p style=\"text-align: center;\"><em>P\u00ad<sub>60<\/sub><\/em> = 962 + 22(60) = 2282 million metric tons.<\/p>\r\nYou will notice that this number is substantially smaller than the prediction from the exponential growth model. Calculating and plotting more values helps illustrate the differences.\r\n\r\n<img class=\"aligncenter wp-image-365\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201942\/yearsafter1990.png\" alt=\"Line graph. Vertical measures Emissions (Millions Metric Tons) in increments of 500, from 0 to 3500. Horizontal measures Years after 1990, in increments of 10, from 0 to 60. A blue line shows a linear growth from 1000 in year 0 to over 2000 in year 60. A pink line shows an exponential growth from 1000 in year 0 to 3500 in year 60.\" width=\"350\" height=\"283\" \/>\r\n\r\n[\/hidden-answer]\r\n\r\nA demonstration of this example can be seen in the following video.\r\n\r\nhttps:\/\/youtu.be\/yiuZoiRMtYM\r\n\r\n<\/div>\r\nSo how do we know which growth model to use when working with data? There are two approaches which should be used together whenever possible:\r\n<ol>\r\n \t<li>Find more than two pieces of data. Plot the values, and look for a trend. Does the data appear to be changing like a line, or do the values appear to be curving upwards?<\/li>\r\n \t<li>Consider the factors contributing to the data. Are they things you would expect to change linearly or exponentially? For example, in the case of carbon emissions, we could expect that, absent other factors, they would be tied closely to population values, which tend to change exponentially.<\/li>\r\n<\/ol>","rendered":"<div class=\"textbox learning-objectives\">\n<h3>Learning Outcomes<\/h3>\n<ul>\n<li>Determine whether data or a scenario describe linear or geometric growth<\/li>\n<li>Identify growth rates, initial values, or point values expressed verbally, graphically, or numerically, and translate them into a format usable in calculation<\/li>\n<li>Calculate recursive and explicit equations for linear and geometric growth given sufficient information, and use those equations to make predictions<\/li>\n<\/ul>\n<\/div>\n<p>Having a constant rate of change is the defining characteristic of linear growth. Plotting coordinate pairs\u00a0associated with constant change will result in\u00a0a straight line, the shape of linear growth. In this section, we will formalize a way to describe linear growth using mathematical terms and concepts. By the end of this section, you will be able to write both a recursive and explicit equations for linear growth given starting conditions, or a constant of change. \u00a0You will also be able to recognize the difference between linear and geometric growth given a graph or an equation.<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01232841\/growth-1140534_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"wp-image-1273 aligncenter\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2017\/02\/01232841\/growth-1140534_1280-300x219.jpg\" alt=\"\" width=\"516\" height=\"377\" \/><\/a><\/p>\n<h2>Linear (Algebraic) Growth<\/h2>\n<h3>Predicting Growth<\/h3>\n<p>Marco is a collector of antique soda bottles. His collection currently contains 437 bottles. Every year, he budgets enough money to buy 32 new bottles. Can we determine how many bottles he will have in 5 years, and how long it will take for his collection to reach 1000 bottles?<\/p>\n<p><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/15183507\/15186624687_fc02ef5925_z.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-full wp-image-806\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/15183507\/15186624687_fc02ef5925_z.jpg\" alt=\"collection of empty glass soda bottles\" width=\"640\" height=\"426\" \/><\/a><\/p>\n<p>While you could probably solve both of these questions without an equation or formal mathematics, we are going to formalize our approach to this problem to provide a means to answer more complicated questions.<\/p>\n<p>Suppose that <em>P\u00ad<sub>n <\/sub><\/em>represents the number, or population, of bottles Marco has after <em>n<\/em> years. So <em>P\u00ad<sub>0<\/sub><\/em> would represent the number of bottles now, <em>P\u00ad<sub>1<\/sub><\/em> would represent the number of bottles after 1 year, <em>P\u00ad<sub>2<\/sub><\/em> would represent the number of bottles after 2 years, and so on. We could describe how Marco\u2019s bottle collection is changing using:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 437<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P<sub>\u00adn-1<\/sub><\/em> + 32<\/p>\n<p>This is called a <strong>recursive relationship<\/strong>. A recursive relationship is a formula which relates the next value in a sequence to the previous values. Here, the number of bottles in year <em>n<\/em> can be found by adding 32 to the number of bottles in the previous year, <em>P\u00ad<sub>n-1<\/sub><\/em>. Using this relationship, we could calculate:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub> = P\u00ad<sub>0<\/sub><\/em> + 32 = 437 + 32 = 469<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub> = P\u00ad<sub>1<\/sub><\/em> + 32 = 469 + 32 = 501<\/p>\n<p style=\"text-align: center;\"><em>P<sub>3<\/sub> = P\u00ad<sub>2<\/sub><\/em> + 32 = 501 + 32 = 533<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub> = P\u00ad<sub>3<\/sub><\/em> + 32 = 533 + 32 = 565<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>5<\/sub> = P\u00ad<sub>4<\/sub><\/em> + 32 = 565 + 32 = 597<\/p>\n<p>We have answered the question of how many bottles Marco will have in 5 years.\u00a0<img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-358\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11200627\/bottles.png\" alt=\"Line chart. Vertical measures Bottles, with increments of 100 from 0 to 700. Horizontal measures Years From Now, with increments of 1 from 0 to 5. The line moves in a slow rise from left to right, from a little over 400 at year 0 to 600 at year 5.\" width=\"402\" height=\"325\" \/><\/p>\n<p>However, solving how long it will take for his collection to reach 1000 bottles would require a lot more calculations.<\/p>\n<p>While recursive relationships are excellent for describing simply and cleanly <em>how<\/em> a quantity is changing, they are not convenient for making predictions or solving problems that stretch far into the future. For that, a closed or explicit form for the relationship is preferred. An <strong>explicit equation <\/strong>allows us to calculate <em>P\u00ad<sub>n <\/sub><\/em>directly, without needing to know <em>P<sub>\u00adn-1<\/sub><\/em>. While you may already be able to guess the explicit equation, let us derive it from the recursive formula. We can do so by selectively not simplifying as we go:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub> = <\/em>437 + 32 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0 \u00a0= 437 + 1(32)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub> = P\u00ad<sub>1<\/sub><\/em> + 32 = 437 + 32 + 32 \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 = 437 + 2(32)<\/p>\n<p style=\"text-align: center;\"><em>P<sub>3<\/sub> = P\u00ad<sub>2<\/sub><\/em> + 32 = (437 + 2(32)) + 32 \u00a0\u00a0\u00a0 = 437 + 3(32)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub> = P<sub>\u00ad3<\/sub><\/em> + 32 = (437 + 3(32)) + 32 \u00a0\u00a0\u00a0 = 437 + 4(32)<\/p>\n<p><em>\u00a0<\/em><\/p>\n<p>You can probably see the pattern now, and generalize that<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00adn<\/sub> = <\/em>437 + <em>n<\/em>(32) = 437 + 32<em>n<\/em><\/p>\n<p>Using this equation, we can calculate how many bottles he\u2019ll have after 5 years:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>5<\/sub> =<\/em> 437 + 32(5) = 437 + 160 = 597<\/p>\n<p>We can now also solve for when the collection will reach 1000 bottles by substituting in 1000 for <em>P\u00ad<sub>n<\/sub><\/em> and solving for <em>n<\/em><\/p>\n<p style=\"text-align: center;\">1000 = 437 + 32<sub><em>n<\/em><\/sub><\/p>\n<p style=\"text-align: center;\">563 = 32<sub><em>n<\/em><\/sub><\/p>\n<p style=\"text-align: center;\"><em>n<\/em> = 563\/32 = 17.59<\/p>\n<p>So Marco will reach 1000 bottles in 18 years.<\/p>\n<p>The steps of determining the formula and solving the problem of Marco&#8217;s bottle collection are explained in detail in the following videos.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-1\" title=\"Linear Growth Part 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/SJcAjN-HL_I?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-2\" title=\"Linear Growth Part 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/4Two_oduhrA?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-3\" title=\"Linear Growth Part 3\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/pZ4u3j8Vmzo?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In this\u00a0example, Marco\u2019s collection grew by the <em>same numbe<\/em>r of bottles every year. This constant change is the defining characteristic of linear growth. Plotting the values we calculated for Marco\u2019s collection, we can see the values form a straight line, the shape of linear growth.<\/p>\n<div class=\"textbox\">\n<h3>Linear Growth<\/h3>\n<p>If a quantity starts at size <em>P\u00ad<sub>0<\/sub><\/em> and grows by <em>d<\/em> every time period, then the quantity after <em>n<\/em> time periods can be determined using either of these relations:<\/p>\n<h4>Recursive form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>n-1<\/sub><\/em> + <em>d<\/em><\/p>\n<h4>Explicit form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>0 <\/sub><\/em>+ <em>d n<\/em><\/p>\n<p>In this equation, <em>d<\/em> represents the <strong>common difference<\/strong> \u2013 the amount that the population changes each time <em>n<\/em> increases by 1.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Connection to Prior Learning: Slope and Intercept<\/h3>\n<p>You may recognize the common difference, <em>d<\/em>, in our linear equation as <em>slope<\/em>. In fact, the entire explicit equation should look familiar \u2013 it is the same linear equation you learned in algebra, probably stated as <em>y = mx + b<\/em>.<\/p>\n<p>In the standard algebraic equation <em>y = mx + b<\/em>, <em>b<\/em> was the <em>y<\/em>-intercept, or the <em>y<\/em> value when <em>x<\/em> was zero. In the form of the equation we\u2019re using, we are using <em>P\u00ad0\u00ad<\/em> to represent that initial amount.<\/p>\n<p>In the <em>y = mx + b<\/em> equation, recall that <em>m<\/em> was the slope. You might remember this as \u201crise over run,\u201d or the change in <em>y<\/em> divided by the change in <em>x<\/em>. Either way, it represents the same thing as the common difference, <em>d<\/em>, we are using \u2013 the amount the output <em>P<sub>n<\/sub><\/em> changes when the input <em>n<\/em> increases by 1.<\/p>\n<p>The equations <em>y = mx + b<\/em> and <em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>0 <\/sub><\/em>+ <em>d n <\/em>mean the same thing and can be used the same ways.\u00a0We\u2019re just writing it somewhat differently.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>The population of elk in a national forest was measured to be 12,000 in 2003, and was measured again to be 15,000 in 2007. If the population continues to grow linearly at this rate, what will the elk population be in 2014?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q60252\">Show Solution<\/span><\/p>\n<div id=\"q60252\" class=\"hidden-answer\" style=\"display: none\">\n<p>To begin, we need to define how we\u2019re going to measure <em>n<\/em>.\u00a0\u00a0 Remember that <em>P\u00ad<sub>0<\/sub><\/em> is the population when <em>n<\/em> = 0, so we probably don\u2019t want to literally use the year 0. Since we already know the population in 2003, let us define <em>n<\/em> = 0 to be the year 2003.<\/p>\n<p>Then\u00a0<em>P\u00ad<sub>0<\/sub><\/em> = 12,000.<\/p>\n<p>Next we need to find <em>d<\/em>. Remember <em>d<\/em> is the growth per time period, in this case growth per year. Between the two measurements, the population grew by 15,000-12,000 = 3,000, but it took 2007-2003 = 4 years to grow that much. To find the growth per year, we can divide: 3000 elk \/ 4 years = 750 elk in 1 year.<\/p>\n<p>Alternatively, you can use the slope formula from algebra to determine the common difference, noting that the population is the output of the formula, and time is the input.<\/p>\n<p>[latex]d=slope=\\frac{\\text{changeinoutput}}{\\text{changeininput}}=\\frac{15,000-12,000}{2007-2003}=\\frac{3000}{4}=750[\/latex]<\/p>\n<p>We can now write our equation in whichever form is preferred.<\/p>\n<h4>Recursive form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 12,000<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00adn<\/sub> = P<sub>\u00adn-1<\/sub><\/em> + 750<\/p>\n<h4>Explicit form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 12,000 + 750(n)<\/p>\n<p>To answer the question, we need to first note that the year 2014 will be <em>n<\/em> = 11, since 2014 is 11 years after 2003. The explicit form will be easier to use for this calculation:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>11<\/sub><\/em> = 12,000 + 750(11) = 20,250 elk<\/p>\n<\/div>\n<\/div>\n<p>View more about this example here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-4\" title=\"Linear Growth - Elk\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/J1XqqlKzYGs?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<hr \/>\n<p>Gasoline consumption in the US has been increasing steadily. Consumption data from 1992 to 2004 is shown below.<a class=\"footnote\" title=\"http:\/\/www.bts.gov\/publications\/national_transportation_statistics\/2005\/html\/table_04_10.html\" id=\"return-footnote-1229-1\" href=\"#footnote-1229-1\" aria-label=\"Footnote 1\"><sup class=\"footnote\">[1]<\/sup><\/a> Find a model for this data, and use it to predict consumption in 2016. If the trend continues, when will consumption reach 200 billion gallons?<\/p>\n<table>\n<tbody>\n<tr>\n<td>Year<\/td>\n<td>&#8217;92<\/td>\n<td>&#8217;93<\/td>\n<td>&#8217;94<\/td>\n<td>&#8217;95<\/td>\n<td>&#8217;96<\/td>\n<td>&#8217;97<\/td>\n<td>&#8217;98<\/td>\n<td>&#8217;99<\/td>\n<td>&#8217;00<\/td>\n<td>&#8217;01<\/td>\n<td>&#8217;02<\/td>\n<td>&#8217;03<\/td>\n<td>&#8217;04<\/td>\n<\/tr>\n<tr>\n<td>Consumption (billion of gallons)<\/td>\n<td>110<\/td>\n<td>111<\/td>\n<td>113<\/td>\n<td>116<\/td>\n<td>118<\/td>\n<td>119<\/td>\n<td>123<\/td>\n<td>125<\/td>\n<td>126<\/td>\n<td>128<\/td>\n<td>131<\/td>\n<td>133<\/td>\n<td>136<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q307147\">Show Solution<\/span><\/p>\n<div id=\"q307147\" class=\"hidden-answer\" style=\"display: none\">\n<p>Plotting this data, it appears to have an approximately linear relationship:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-359\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201004\/gasconsumption.png\" alt=\"Graph. Vertical measures Gas Consumption in increments of 10, from 100 to 140. Horizontal measures Year in increments of 4, from 1992 to 2004. Points identified in a generally upward trend, left to right, from 110 in 1992 to near 140 in 2004.\" width=\"350\" height=\"235\" \/><\/p>\n<p>While there are more advanced statistical techniques that can be used to find an equation to model the data, to get an idea of what is happening, we can find an equation by using two pieces of the data \u2013 perhaps the data from 1993 and 2003.<\/p>\n<p>Letting <em>n<\/em> = 0 correspond with 1993 would give <em>P\u00ad<sub>0<\/sub><\/em> = 111 billion gallons.<\/p>\n<p>To find <em>d<\/em>, we need to know how much the gas consumption increased each year, on average. From 1993 to 2003 the gas consumption increased from 111 billion gallons to 133 billion gallons, a total change of 133 \u2013 111 = 22 billion gallons, over 10 years. This gives us an average change of 22 billion gallons \/ 10 year = 2.2 billion gallons per year.<\/p>\n<p>Equivalently,<\/p>\n<p>[latex]d=slope=\\frac{\\text{changeinoutput}}{\\text{changeininput}}=\\frac{133-111}{10-0}=\\frac{22}{10}=2.2[\/latex]billion gallons per year<\/p>\n<p>We can now write our equation in whichever form is preferred.<\/p>\n<h4>Recursive form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 111<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub> = P\u00ad<sub>n-1<\/sub><\/em> + 2.2<\/p>\n<h4>Explicit form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 111 + 2.2<sub><em>n<\/em><\/sub><\/p>\n<p><em>\u00a0<\/em><em><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-360\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201052\/gasconsumption2.png\" alt=\"Graph. Vertical measures Gas Consumption in increments of 10, from 100 to 140. Horizontal measures Year in increments of 4, from 1992 to 2004. Points identified in a generally upward trend, left to right, from 110 in 1992 to near 140 in 2004. This line connects dots with a solid line.\" width=\"350\" height=\"283\" \/><\/em><\/p>\n<p>Calculating values using the explicit form and plotting them with the original data shows how well our model fits the data.<\/p>\n<p>We can now use our model to make predictions about the future, assuming that the previous trend continues unchanged. To predict the gasoline consumption in 2016:<\/p>\n<p style=\"text-align: center;\"><em>n<\/em> = 23 (2016 \u2013 1993 = 23 years later)<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad23<\/sub><\/em> = 111 + 2.2(23) = 161.6<\/p>\n<p>Our model predicts that the US will consume 161.6 billion gallons of gasoline in 2016 if the current trend continues.<\/p>\n<p>To find when the consumption will reach 200 billion gallons, we would set <em>P\u00ad<sub>n <\/sub><\/em>= 200, and solve for <em>n<\/em>:<\/p>\n<p style=\"padding-left: 210px;\"><em>P<sub>n<\/sub><\/em> = 200\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Replace <em>P<sub>n<\/sub><\/em> with our model<\/p>\n<p style=\"padding-left: 210px;\">111 + 2.2<sub><em>n<\/em><\/sub> = 200\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract 111 from both sides<\/p>\n<p style=\"padding-left: 210px;\">2.2<sub><em>n<\/em><\/sub> = 89\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Divide both sides by 2.2<\/p>\n<p style=\"padding-left: 210px;\"><em>n<\/em> = 40.4545<\/p>\n<p>This tells us that consumption will reach 200 billion about 40 years after 1993, which would be in the year 2033.<\/p>\n<\/div>\n<\/div>\n<p>The steps for reaching this answer are detailed in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-5\" title=\"Finding linear model for gas consumption\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/ApFxDWd6IbE?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<hr \/>\n<p>The cost, in dollars, of a gym membership for <em>n<\/em> months can be described by the explicit equation <em>P\u00ad<sub>n<\/sub><\/em> = 70 + 30<sub><em>n<\/em><\/sub>. What does this equation tell us?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q438458\">Show Solution<\/span><\/p>\n<div id=\"q438458\" class=\"hidden-answer\" style=\"display: none\">\n<p>The value for <em>P\u00ad<sub>0<\/sub><\/em> in this equation is 70, so the initial starting cost is $70. This tells us that there must be an initiation or start-up fee of $70 to join the gym.<\/p>\n<p>The value for <em>d<\/em> in the equation is 30, so the cost increases by $30 each month. This tells us that the monthly membership fee for the gym is $30 a month.<\/p>\n<\/div>\n<\/div>\n<p>The explanation for this example is detailed below.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-6\" title=\"Interpreting a linear model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/0Uwz5dmLTtk?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The number of stay-at-home fathers in Canada has been growing steadily<a class=\"footnote\" title=\"http:\/\/www.fira.ca\/article.php?id=140\" id=\"return-footnote-1229-2\" href=\"#footnote-1229-2\" aria-label=\"Footnote 2\"><sup class=\"footnote\">[2]<\/sup><\/a>. While the trend is not perfectly linear, it is fairly linear. Use the data from 1976 and 2010 to find an explicit formula for the number of stay-at-home fathers, then use it to predict the number in\u00a02020.<\/p>\n<table>\n<tbody>\n<tr>\n<td>Year<\/td>\n<td>1976<\/td>\n<td>1984<\/td>\n<td>1991<\/td>\n<td>2000<\/td>\n<td>2010<\/td>\n<\/tr>\n<tr>\n<td># of Stay -at-home fathers<\/td>\n<td>20610<\/td>\n<td>28725<\/td>\n<td>43530<\/td>\n<td>47665<\/td>\n<td>53555<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q688205\">Show Solution<\/span><\/p>\n<div id=\"q688205\" class=\"hidden-answer\" style=\"display: none\">\n<div>Letting n= 0 correspond with 1976, then [latex]P_0= 20,610[\/latex].<\/div>\n<div>From 1976 to 2010 the number of stay-at-home fathers increased by 53,555 \u2013 20,610 = 32,945<\/div>\n<div>This happened over 34 years, giving a common different <em>d\u00a0<\/em>of 32,945 \/ 34 = 969.<\/div>\n<div>[latex]P_n= 20,610 + 969n[\/latex]<\/div>\n<div>Predicting for 2020, we use n = 44, P(44) = 20,610 + 969(44) = 63,246 stay-at-home fathers in 2020.<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6594&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n<h3>When Good Models Go Bad<\/h3>\n<p>When using mathematical models to predict future behavior, it is important to keep in mind that very few trends will continue indefinitely.<\/p>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Suppose a four year old boy is currently 39 inches tall, and you are told to expect him to grow 2.5 inches a year.<\/p>\n<p>We can set up a growth model, with <em>n<\/em> = 0 corresponding to 4 years old.<\/p>\n<h4>Recursive form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>0<\/sub><\/em> = 39<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00adn<\/sub> = P\u00ad<sub>n-1<\/sub><\/em> + 2.5<\/p>\n<h4>Explicit form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 39 + 2.5(n)<\/p>\n<p>So at 6 years old, we would expect him to be<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<\/em><sub>2<\/sub> = 39 + 2.5(2) = 44 inches tall<\/p>\n<p>Any mathematical model will break down eventually. Certainly, we shouldn\u2019t expect this boy to continue to grow at the same rate all his life. If he did, at age 50 he would be<\/p>\n<p style=\"text-align: center;\"><em>P<\/em><sub>\u00ad46<\/sub> = 39 + 2.5(46) = 154 inches tall = 12.8 feet tall!<\/p>\n<p>When using any mathematical model, we have to consider which inputs are reasonable to use. Whenever we <strong>extrapolate<\/strong>, or make predictions into the future, we are assuming the model will continue to be valid.<\/p>\n<p>View a video explanation of this breakdown of the linear growth model here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-7\" title=\"Linear model breakdown\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/6zfXCsmcDzI?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<h2>Exponential (Geometric ) Growth<\/h2>\n<h3>Population Growth<\/h3>\n<p>Suppose that every year, only 10% of the fish in a lake have surviving offspring. If there were 100 fish in the lake last year, there would now be 110 fish. If there were 1000 fish in the lake last year, there would now be 1100 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><a href=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280.jpg\"><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter size-large wp-image-906\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/1141\/2016\/12\/21172235\/calf-1869984_1280-1024x714.jpg\" alt=\"Two orange koi at the top of a pond, mouths open. Others are swimming beneath them.\" width=\"1024\" height=\"714\" \/><\/a><br \/>\nSuppose our lake began with 1000 fish, and 10% of the fish have surviving offspring each year. Since we start with 1000 fish, <em>P\u00ad<sub>0<\/sub><\/em> = 1000. How do we calculate <em>P\u00ad<sub>1<\/sub><\/em>? The new population will be the old population, plus an additional 10%. Symbolically:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0<\/sub><\/em><\/p>\n<p>Notice this could be condensed to a shorter form by factoring:<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad1<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P\u00ad<sub>0<\/sub><\/em> = 1<em>P\u00ad<sub>0<\/sub><\/em> + 0.10<em>P<sub>\u00ad0 <\/sub><\/em>= (1+ 0.10)<em>P\u00ad<sub>0 <\/sub><\/em>= 1.10<em>P\u00ad<sub>0<\/sub><\/em><\/p>\n<p>While 10% is the <strong>growth rate<\/strong>, 1.10 is the <strong>growth multiplier<\/strong>. Notice that 1.10 can be thought of as \u201cthe original 100% plus an additional 10%.\u201d<\/p>\n<p>For our fish population,<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1000) = 1100<\/p>\n<p>We could then calculate the population in later years:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10<em>P\u00ad<sub>1<\/sub><\/em> = 1.10(1100) = 1210<\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad3<\/sub><\/em> = 1.10<em>P<sub>2<\/sub><\/em> = 1.10(1210) = 1331<\/p>\n<p>Notice that in the first year, the population grew by 100 fish; in the second year, the population grew by 110 fish; and in the third year the population grew by 121 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<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-363\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201507\/yearsfromnow1.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=\"350\" height=\"283\" \/><\/p>\n<p>A walkthrough of this fish scenario can be viewed here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-8\" title=\"Exponential Growth Model Part 1\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/3BiU7Ihxvxg?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\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;\"><em>P\u00ad<sub>1<\/sub><\/em> = 1.10(<em>P\u00ad<sub>0<\/sub><\/em>\u00a0)= 1.10(1000)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>2<\/sub><\/em> = 1.10(<em>P\u00ad<sub>1<\/sub><\/em>\u00a0)= 1.10(1.10(1000)) = 1.102(1000)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>3<\/sub><\/em> = 1.10(<em>P\u00ad<sub>2<\/sub><\/em>\u00a0)= 1.10(1.10<sup>2<\/sup>(1000)) = 1.103(1000)<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>4<\/sub><\/em> = 1.10(<em>P\u00ad<sub>3<\/sub><\/em>\u00a0)= 1.10(1.10<sup>3<\/sup>(1000)) = 1.104(1000)<\/p>\n<p>Observing a pattern, we can generalize the explicit form to be:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = 1.10<sup><em>n<\/em><\/sup>(1000), or equivalently, <em>P\u00ad<sub>n<\/sub><\/em> = 1000(1.10<em><sup>n<\/sup><\/em>)<\/p>\n<p>From this, we can quickly calculate the number of fish in 10, 20, or 30 years:<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-364\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201611\/yearsfromnow2.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=\"350\" height=\"281\" \/><\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 1.10<sup><em>10<\/em><\/sup>(1000) = 2594<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>20<\/sub><\/em> = 1.10<sup><em>20<\/em><\/sup>(1000) = 6727<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>30<\/sub><\/em> = 1.10<sup><em>30<\/em><\/sup>(1000) = 17449<\/p>\n<p style=\"text-align: center;\"><span style=\"font-size: 1rem; text-align: initial;\">Adding these values to our graph reveals a shape that is definitely not linear. If our fish population had been growing linearly, by 100 fish each year, the population would have only reached 4000 in 30 years, compared to almost 18,000 with this percent-based growth, called <\/span><strong style=\"font-size: 1rem; text-align: initial;\">exponential growth.<\/strong><\/p>\n<p>A video demonstrating the explicit model of this fish story can be viewed here:<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-9\" title=\"Exponential Growth Model Part 2\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/tg2ysaZ8agY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<p>In exponential growth, the population grows proportional to the size of the population, so as the population gets larger, the same percent growth will yield a larger numeric growth.<\/p>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Exponential Growth<\/h3>\n<p>If a quantity starts at size <em>P\u00ad<sub>0<\/sub><\/em> and grows by <em>R%<\/em> (written as a decimal, <em>r<\/em>) every time period, then the quantity after <em>n<\/em> time periods can be determined using either of these relations:<\/p>\n<h4>Recursive form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>) <em>P\u00ad<sub>n-1<\/sub><\/em><\/p>\n<h4>Explicit form<\/h4>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>n<\/sub><\/em> = (1+<em>r<\/em>)<em><sup>n<\/sup><\/em>\u00a0<em>P\u00ad<sub>0<\/sub> \u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 <\/em>or equivalently, <em>P<sub>\u00adn<\/sub><\/em> = <em>P\u00ad<sub>0<\/sub><\/em> (1+<em>r<\/em>)<em><sup>n<\/sup><\/em><\/p>\n<p>We call <em>r<\/em> the <strong>growth rate<\/strong>.<\/p>\n<p>The term (1+<em>r<\/em>) is called the <strong>growth multiplier<\/strong>, or common ratio.<\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Between 2007 and 2008, Olympia, WA grew almost 3% to a population of 245 thousand people. If this growth rate was to continue, what would the population of Olympia be in 2014?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q54756\">Show Solution<\/span><\/p>\n<div id=\"q54756\" class=\"hidden-answer\" style=\"display: none\">\n<p>As we did before, we first need to define what year will correspond to <em>n<\/em> = 0. Since we know the population in 2008, it would make sense to have 2008 correspond to <em>n <\/em>= 0, so <em>P\u00ad<sub>0<\/sub><\/em> = 245,000.\u00a0\u00a0 The year 2014 would then be <em>n<\/em> = 6.<\/p>\n<p>We know the growth rate is 3%, giving <em>r <\/em>= 0.03.<\/p>\n<p>Using the explicit form:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>6<\/sub><\/em> = (1+0.03)<sup>6<\/sup> (245,000) = 1.19405(245,000) = 292,542.25<\/p>\n<p>The model predicts that in 2014, Olympia would have a population of about 293 thousand people.<\/p>\n<\/div>\n<\/div>\n<p>The following video explains this example in detail.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-10\" title=\"Predicting future population using an exponential model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/CDI4xS65rxY?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox\">\n<h3>Evaluating exponents on the calculator<\/h3>\n<p>To evaluate expressions like (1.03)<sup>6<\/sup>, it will be easier to use a calculator than multiply 1.03 by itself six times. Most scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:<\/p>\n<p style=\"text-align: center;\">^ ,\u00a0\u00a0 y<sup>x<\/sup> ,\u00a0\u00a0 or x<sup>y<\/sup> .<\/p>\n<p>To evaluate 1.03<sup>6<\/sup>\u00a0we\u2019d type 1.03 ^ 6, or 1.03 y<sup>x<\/sup> 6.\u00a0 Try it out &#8211; you should get an answer around 1.1940523.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>India is the second most populous country in the world, with a population in 2008 of about 1.14 billion people. The population is growing by about 1.34% each year. If this trend continues, what will India\u2019s population grow to by 2020?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q265475\">Show Solution<\/span><\/p>\n<div id=\"q265475\" class=\"hidden-answer\" style=\"display: none\">\n<div>Using n = 0 corresponding with 2008, [latex]P_12= (1+0.0134)12(1.14)[\/latex] = about 1.337 billion people in 2020<\/div>\n<\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom3\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6673&amp;theme=oea&amp;iframe_resize_id=mom3\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n<p>&nbsp;<\/p>\n<div class=\"textbox exercises\">\n<h3>Examples<\/h3>\n<p>A friend is using the equation <em>P\u00ad<sub>n<\/sub><\/em> = 4600(1.072)<sup><em>n <\/em><\/sup>to predict the annual tuition at a local college. She says the formula is based on years after 2010. What does this equation tell us?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q224261\">Show Solution<\/span><\/p>\n<div id=\"q224261\" class=\"hidden-answer\" style=\"display: none\">\n<p>In the equation, <em>P<sub>\u00ad0<\/sub><\/em> = 4600, which is the starting value of the tuition when <em>n<\/em> = 0. This tells us that the tuition in 2010 was $4,600.<\/p>\n<p>The growth multiplier is 1.072, so the growth rate is 0.072, or 7.2%. This tells us that the tuition is expected to grow by 7.2% each year.<\/p>\n<p>Putting this together, we could say that the tuition in 2010 was $4,600, and is expected to grow by 7.2% each year.<\/p>\n<\/div>\n<\/div>\n<p>View the following to see this example worked out.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-11\" title=\"Interpreting an Exponential Equation\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/T8Yz94De5UM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<hr \/>\n<p>In 1990, residential energy use in the US was responsible for 962 million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to 1182 million metric tons<a class=\"footnote\" title=\"http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html\" id=\"return-footnote-1229-3\" href=\"#footnote-1229-3\" aria-label=\"Footnote 3\"><sup class=\"footnote\">[3]<\/sup><\/a>. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q755963\">Show Solution<\/span><\/p>\n<div id=\"q755963\" class=\"hidden-answer\" style=\"display: none\">\n<p>Similar to before, we will correspond <em>n<\/em> = 0 with 1990, as that is the year for the first piece of data we have. That will make <em>P\u00ad<sub>0<\/sub><\/em> = 962 (million metric tons of CO<sub>2<\/sub>). In this problem, we are not given the growth rate, but instead are given that <em>P\u00ad<sub>10<\/sub><\/em> = 1182.<\/p>\n<p>When <em>n<\/em> = 10, the explicit equation looks like:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> <em>P\u00ad<sub>0<\/sub><\/em><\/p>\n<p>We know the value for <em>P\u00ad<sub>0<\/sub><\/em>, so we can put that into the equation:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\n<p>We also know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, so substituting that in, we get<\/p>\n<p style=\"text-align: center;\">1182 = (1+<em>r<\/em>)<sup>10<\/sup> 962<\/p>\n<p>We can now solve this equation for the growth rate, <em>r<\/em>. Start by dividing by 962.<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\frac{1182}{962}={{(1+r)}^{10}}[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Take the 10th root of both sides<\/p>\n<p style=\"padding-left: 120px;\">[latex]\\sqrt[10]{\\frac{1182}{962}}=1+r[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract 1 from both sides<\/p>\n<p style=\"padding-left: 120px;\">[latex]r=\\sqrt[10]{\\frac{1182}{962}}-1=0.0208[\/latex] = 2.08%<\/p>\n<p>So if the emissions are growing exponentially, they are growing by about 2.08% per year. We can now predict the emissions in 2050 by finding <em>P<sub>\u00ad60<\/sub><\/em><\/p>\n<p style=\"text-align: center;\"><em>P<sub>\u00ad60<\/sub><\/em> = (1+0.0208)<sup>60<\/sup> 962 = 3308.4 million metric tons of CO<sub>2<\/sub> in 2050<\/p>\n<\/div>\n<\/div>\n<p>View more about this example here.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-12\" title=\"Finding an Exponential Model\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/9Zu2uONfLkQ?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Rounding<\/h3>\n<p>As a note on rounding, notice that if we had rounded the growth rate to 2.1%, our calculation for the emissions in 2050 would have been 3347.\u00a0\u00a0 Rounding to 2% would have changed our result to 3156. A very small difference in the growth rates gets magnified greatly in exponential growth. For this reason, it is recommended to round the growth rate as little as possible.<\/p>\n<p>If you need to round, <strong>keep at least three significant digits<\/strong> &#8211; numbers after any leading zeros.\u00a0\u00a0 So 0.4162 could be reasonably rounded to 0.416. A growth rate of 0.001027 could be reasonably rounded to 0.00103.<\/p>\n<\/div>\n<div class=\"textbox\">\n<h3>Evaluating roots on the calculator<\/h3>\n<p>In the previous example, we had to calculate the 10th root of a number. This is different than taking the basic square root, \u221a. Many scientific calculators have a button for general roots.\u00a0 It is typically labeled like:<\/p>\n<p style=\"text-align: center;\">[latex]\\sqrt[y]{x}[\/latex]<\/p>\n<p>To evaluate the 3rd root of 8, for example, we\u2019d either type 3 [latex]\\sqrt[x]{{}}[\/latex] 8, or 8 [latex]\\sqrt[x]{{}}[\/latex] 3, depending on the calculator. Try it on yours to see which to use \u2013 you should get an answer of 2.<\/p>\n<p>If your calculator does not have a general root button, all is not lost. You can instead use the property of exponents which states that:<\/p>\n<p style=\"text-align: center;\"><span style=\"color: #000000;\">[latex]\\sqrt[n]{a}={a}^{\\frac{1}{2}}[\/latex]<\/span>.<\/p>\n<p>So, to compute the 3rd root of 8, you could use your calculator\u2019s exponent key to evaluate 8<sup>1\/3<\/sup>. To do this, type:<\/p>\n<p style=\"text-align: center;\">8 y<sup>x<\/sup> ( 1 \u00f7 3 )<\/p>\n<p>The parentheses tell the calculator to divide 1\/3 before doing the exponent.<\/p>\n<\/div>\n<div class=\"textbox key-takeaways\">\n<h3>Try It<\/h3>\n<p>The number of users on a social networking site was 45 thousand in February when they officially went public, and grew to 60 thousand by October. If the site is growing exponentially, and growth continues at the same rate, how many users should they expect two years after they went public?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q774576\">Show Solution<\/span><\/p>\n<div id=\"q774576\" class=\"hidden-answer\" style=\"display: none\">\n<p>Here we will measure n in months rather than years, with n = 0 corresponding to the February when they went public. This gives [latex]P_0= 45[\/latex] thousand. October is 8 months later, so [latex]P_8= 60[\/latex].<br \/>\n[latex]P_8=(1+r)^{8}P_0[\/latex]<br \/>\n[latex]60=(1+r)^{8}45[\/latex]<br \/>\n[latex]\\frac{60}{45}=(1+r)^8[\/latex]<br \/>\n[latex]\\sqrt[8]{\\frac{60}{45}}=1+r[\/latex]<br \/>\n[latex]r=\\sqrt[8]{\\frac{60}{45}}-1=0.0366\\text{ or }3.66%[\/latex]<br \/>\nThe general explicit equation is [latex]P_n =(1.0366)^{n}45[\/latex]. Predicting 24 months after they went public gives [latex]P_{24}=(1.0366)^{24}45=106.63[\/latex] thousand users.\n<\/p><\/div>\n<\/div>\n<p><iframe loading=\"lazy\" id=\"mom4\" class=\"resizable\" src=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6598&amp;theme=oea&amp;iframe_resize_id=mom4\" width=\"100%\" height=\"100%\"><\/iframe><\/p>\n<\/div>\n<div class=\"textbox exercises\">\n<h3>Example<\/h3>\n<p>Looking back at the last example, for the sake of comparison, what would the carbon emissions be in 2050 if emissions grow linearly at the same rate?<\/p>\n<div class=\"qa-wrapper\" style=\"display: block\"><span class=\"show-answer collapsed\" style=\"cursor: pointer\" data-target=\"q355767\">Show Solution<\/span><\/p>\n<div id=\"q355767\" class=\"hidden-answer\" style=\"display: none\">\n<p>Again we will get <em>n<\/em> = 0 correspond with 1990, giving <em>P<sub>\u00ad0<\/sub><\/em> = 962. To find <em>d<\/em>, we could take the same approach as earlier, noting that the emissions increased by 220 million metric tons in 10 years, giving a common difference of 22 million metric tons each year.<\/p>\n<p>Alternatively, we could use an approach similar to that which we used to find the exponential equation. When <em>n<\/em> = 10, the explicit linear equation looks like:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = <em>P<sub>\u00ad0<\/sub><\/em> + 10<em>d<\/em><em>\u00a0<\/em><\/p>\n<p>We know the value for <em>P\u00ad0<\/em>, so we can put that into the equation:<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>10<\/sub><\/em> = 962 + 10<em>d<\/em><\/p>\n<p>Since we know that <em>P\u00ad<sub>10<\/sub><\/em> = 1182, substituting that in we get<\/p>\n<p style=\"text-align: center;\">1182 = 962 + 10<em>d<\/em><\/p>\n<p>We can now solve this equation for the common difference, <em>d<\/em>.<\/p>\n<p style=\"text-align: center;\">1182 \u2013 962 = 10<em>d<\/em><\/p>\n<p style=\"text-align: center;\">220 = 10<em>d<\/em><\/p>\n<p style=\"text-align: center;\"><em>d<\/em> = 22<\/p>\n<p>This tells us that if the emissions are changing linearly, they are growing by 22 million metric tons each year. Predicting the emissions in 2050,<\/p>\n<p style=\"text-align: center;\"><em>P\u00ad<sub>60<\/sub><\/em> = 962 + 22(60) = 2282 million metric tons.<\/p>\n<p>You will notice that this number is substantially smaller than the prediction from the exponential growth model. Calculating and plotting more values helps illustrate the differences.<\/p>\n<p><img loading=\"lazy\" decoding=\"async\" class=\"aligncenter wp-image-365\" src=\"https:\/\/s3-us-west-2.amazonaws.com\/courses-images\/wp-content\/uploads\/sites\/276\/2016\/10\/11201942\/yearsafter1990.png\" alt=\"Line graph. Vertical measures Emissions (Millions Metric Tons) in increments of 500, from 0 to 3500. Horizontal measures Years after 1990, in increments of 10, from 0 to 60. A blue line shows a linear growth from 1000 in year 0 to over 2000 in year 60. A pink line shows an exponential growth from 1000 in year 0 to 3500 in year 60.\" width=\"350\" height=\"283\" \/><\/p>\n<\/div>\n<\/div>\n<p>A demonstration of this example can be seen in the following video.<\/p>\n<p><iframe loading=\"lazy\" id=\"oembed-13\" title=\"Comparing exponential to linear growth\" width=\"500\" height=\"281\" src=\"https:\/\/www.youtube.com\/embed\/yiuZoiRMtYM?feature=oembed&#38;rel=0\" frameborder=\"0\" allowfullscreen=\"allowfullscreen\"><\/iframe><\/p>\n<\/div>\n<p>So how do we know which growth model to use when working with data? There are two approaches which should be used together whenever possible:<\/p>\n<ol>\n<li>Find more than two pieces of data. Plot the values, and look for a trend. Does the data appear to be changing like a line, or do the values appear to be curving upwards?<\/li>\n<li>Consider the factors contributing to the data. Are they things you would expect to change linearly or exponentially? For example, in the case of carbon emissions, we could expect that, absent other factors, they would be tied closely to population values, which tend to change exponentially.<\/li>\n<\/ol>\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-1229\">\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>Introduction and Learning Outcomes. <strong>Provided by<\/strong>: Lumen Learning. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Revision and Adaptation. <strong>Provided by<\/strong>: Lumen Learning. <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>Math in Society. <strong>Authored by<\/strong>: David Lippman. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\">http:\/\/www.opentextbookstore.com\/mathinsociety\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by-sa\/4.0\/\">CC BY-SA: Attribution-ShareAlike<\/a><\/em><\/li><li>Feira Tom Jobim - BH. <strong>Authored by<\/strong>: Antonio Thomas Koenigkam Oliveira. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/flic.kr\/p\/p8ZqqF\">https:\/\/flic.kr\/p\/p8ZqqF<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Linear Growth Part 1. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/SJcAjN-HL_I\">https:\/\/youtu.be\/SJcAjN-HL_I<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Linear Growth Part 2. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/4Two_oduhrA\">https:\/\/youtu.be\/4Two_oduhrA<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Linear Growth Part 3. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/pZ4u3j8Vmzo\">https:\/\/youtu.be\/pZ4u3j8Vmzo<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Linear Growth - Elk. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/J1XqqlKzYGs\">https:\/\/youtu.be\/J1XqqlKzYGs<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Finding linear model for gas consumption. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/ApFxDWd6IbE\">https:\/\/youtu.be\/ApFxDWd6IbE<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Interpreting a linear model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/0Uwz5dmLTtk\">https:\/\/youtu.be\/0Uwz5dmLTtk<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Linear model breakdown. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/6zfXCsmcDzI\">https:\/\/youtu.be\/6zfXCsmcDzI<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6594. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><li>calf-fish-lake-pond-river. <strong>Authored by<\/strong>: Pexels. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/pixabay.com\/en\/calf-fish-lake-pond-river-1869984\/\">https:\/\/pixabay.com\/en\/calf-fish-lake-pond-river-1869984\/<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/cc0\">CC0: No Rights Reserved<\/a><\/em><\/li><li>Exponential Growth Model Part 1. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/3BiU7Ihxvxg\">https:\/\/youtu.be\/3BiU7Ihxvxg<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Exponential Growth Model Part 2. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/tg2ysaZ8agY\">https:\/\/youtu.be\/tg2ysaZ8agY<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Predicting future population using an exponential model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/CDI4xS65rxY\">https:\/\/youtu.be\/CDI4xS65rxY<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Interpreting an Exponential Equation. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/T8Yz94De5UM\">https:\/\/youtu.be\/T8Yz94De5UM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Finding an Exponential Model. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/9Zu2uONfLkQ\">https:\/\/youtu.be\/9Zu2uONfLkQ<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Comparing exponential to linear growth. <strong>Authored by<\/strong>: OCLPhase2&#039;s channel. <strong>Located at<\/strong>: <a target=\"_blank\" href=\"https:\/\/youtu.be\/yiuZoiRMtYM\">https:\/\/youtu.be\/yiuZoiRMtYM<\/a>. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em><\/li><li>Question ID 6598. <strong>Authored by<\/strong>: Lippman, David. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/licenses\/by\/4.0\/\">CC BY: Attribution<\/a><\/em>. <strong>License Terms<\/strong>: IMathAS Community License CC-BY + GPL<\/li><\/ul><div class=\"license-attribution-dropdown-subheading\">Public domain content<\/div><ul class=\"citation-list\"><li>Growth. <strong>Authored by<\/strong>: bykst. <strong>License<\/strong>: <em><a target=\"_blank\" rel=\"license\" href=\"https:\/\/creativecommons.org\/about\/pdm\">Public Domain: No Known Copyright<\/a><\/em><\/li><\/ul><\/div>\n\t\t\t\t\t\t <\/div>\n\t\t\t\t\t <\/div>\n\t\t\t <\/section><hr class=\"before-footnotes clear\" \/><div class=\"footnotes\"><ol><li id=\"footnote-1229-1\"><a href=\"http:\/\/www.bts.gov\/publications\/national_transportation_statistics\/2005\/html\/table_04_10.html\" target=\"_blank\" rel=\"noopener\">http:\/\/www.bts.gov\/publications\/national_transportation_statistics\/2005\/html\/table_04_10.html<\/a> <a href=\"#return-footnote-1229-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-1229-2\"><a href=\"http:\/\/www.fira.ca\/article.php?id=140\" target=\"_blank\" rel=\"noopener\">http:\/\/www.fira.ca\/article.php?id=140<\/a> <a href=\"#return-footnote-1229-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><li id=\"footnote-1229-3\"><a href=\"http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html\" target=\"_blank\" rel=\"noopener\">http:\/\/www.eia.doe.gov\/oiaf\/1605\/ggrpt\/carbon.html<\/a> <a href=\"#return-footnote-1229-3\" class=\"return-footnote\" aria-label=\"Return to footnote 3\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":21,"menu_order":2,"template":"","meta":{"_candela_citation":"[{\"type\":\"original\",\"description\":\"Introduction and Learning Outcomes\",\"author\":\"\",\"organization\":\"Lumen 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