{"id":357,"date":"2016-10-11T20:12:40","date_gmt":"2016-10-11T20:12:40","guid":{"rendered":"https:\/\/courses.lumenlearning.com\/math4libarts\/?post_type=chapter&#038;p=357"},"modified":"2021-02-05T23:55:55","modified_gmt":"2021-02-05T23:55:55","slug":"linear-algebraic-growth","status":"web-only","type":"chapter","link":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/chapter\/linear-algebraic-growth\/","title":{"raw":"Linear (Algebraic) Growth","rendered":"Linear (Algebraic) 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 growth and use those equations to make predictions<\/li>\r\n<\/ul>\r\n<\/div>\r\n<h2>Predicting Growth<\/h2>\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<div class=\"textbox examples\">\r\n<h3>Recall recursive relationships<\/h3>\r\nYou saw recursive relationships before when studying fractals, objects created using self-similarity, in which each step of the process relates to the previous step.\r\n\r\nThe subscripts on the variables below, [latex]P_n[\/latex], represent the order of the steps.\r\n\r\n<\/div>\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<div class=\"textbox examples\">\r\n<h3>recall translating between words and mathematical operations<\/h3>\r\nIn the description below, the desired equation is built up by describing and translating between math notation and words. If you read the lines in words as you go, it may help reveal what's being done.\r\n\r\nEx. [latex]P_1=437+32[\/latex] may be read, \"increasing the initial number of 437 bottles by 32 in the first year\". Furthermore, that tells us that at the end of the first year, Marco will have 437 bottles plus 1 set of 32 added during the year.\r\n\r\nEx. [latex]P_2=437 + 2(32)[\/latex] may be read, \"at the end of the 2nd year, the initial collection of 437 bottles has been increased by 2 sets of 32, since 32 bottles were added during each of the years.\"\r\n\r\nTranslating between the real-world situation and the mathematical notation as you work through an example can help you to understand the process being used to create the explicit equation that describes the situation.\r\n\r\n<\/div>\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>You 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<sub><em>n<\/em><\/sub><\/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[embed]https:\/\/www.myopenmath.com\/multiembedq.php?id=6594&amp;theme=oea&amp;iframe_resize_id=mom3[\/embed]\r\n\r\n<\/div>\r\n<h2>When Good Models Go Bad<\/h2>\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<sub><em>n<\/em><\/sub><\/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;","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 growth and use those equations to make predictions<\/li>\n<\/ul>\n<\/div>\n<h2>Predicting Growth<\/h2>\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<div class=\"textbox examples\">\n<h3>Recall recursive relationships<\/h3>\n<p>You saw recursive relationships before when studying fractals, objects created using self-similarity, in which each step of the process relates to the previous step.<\/p>\n<p>The subscripts on the variables below, [latex]P_n[\/latex], represent the order of the steps.<\/p>\n<\/div>\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<div class=\"textbox examples\">\n<h3>recall translating between words and mathematical operations<\/h3>\n<p>In the description below, the desired equation is built up by describing and translating between math notation and words. If you read the lines in words as you go, it may help reveal what&#8217;s being done.<\/p>\n<p>Ex. [latex]P_1=437+32[\/latex] may be read, &#8220;increasing the initial number of 437 bottles by 32 in the first year&#8221;. Furthermore, that tells us that at the end of the first year, Marco will have 437 bottles plus 1 set of 32 added during the year.<\/p>\n<p>Ex. [latex]P_2=437 + 2(32)[\/latex] may be read, &#8220;at the end of the 2nd year, the initial collection of 437 bottles has been increased by 2 sets of 32, since 32 bottles were added during each of the years.&#8221;<\/p>\n<p>Translating between the real-world situation and the mathematical notation as you work through an example can help you to understand the process being used to create the explicit equation that describes the situation.<\/p>\n<\/div>\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>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<sub><em>n<\/em><\/sub><\/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-357-1\" href=\"#footnote-357-1\" 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-357-2\" href=\"#footnote-357-2\" 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><a href=\"https:\/\/www.myopenmath.com\/multiembedq.php?id=6594&#38;theme=oea&#38;iframe_resize_id=mom3\">https:\/\/www.myopenmath.com\/multiembedq.php?id=6594&amp;theme=oea&amp;iframe_resize_id=mom3<\/a><\/p>\n<\/div>\n<h2>When Good Models Go Bad<\/h2>\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<sub><em>n<\/em><\/sub><\/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\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-357\">\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>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>Linear (Algebraic) Growth. <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>Project<\/strong>: Math in Society. <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><\/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-357-1\"><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-357-1\" class=\"return-footnote\" aria-label=\"Return to footnote 1\">&crarr;<\/a><\/li><li id=\"footnote-357-2\"><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-357-2\" class=\"return-footnote\" aria-label=\"Return to footnote 2\">&crarr;<\/a><\/li><\/ol><\/div>","protected":false},"author":20,"menu_order":17,"template":"","meta":{"_candela_citation":"[{\"type\":\"cc\",\"description\":\"Linear (Algebraic) Growth\",\"author\":\"David Lippman\",\"organization\":\"\",\"url\":\"http:\/\/www.opentextbookstore.com\/mathinsociety\/\",\"project\":\"Math in Society\",\"license\":\"cc-by-sa\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Feira Tom Jobim - BH\",\"author\":\"Antonio Thomas Koenigkam Oliveira\",\"organization\":\"\",\"url\":\"https:\/\/flic.kr\/p\/p8ZqqF\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Linear Growth Part 1\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/SJcAjN-HL_I\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Linear Growth Part 2\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/4Two_oduhrA\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Linear Growth Part 3\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/pZ4u3j8Vmzo\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Linear Growth - Elk\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/J1XqqlKzYGs\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Finding linear model for gas consumption\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/ApFxDWd6IbE\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Interpreting a linear model\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/0Uwz5dmLTtk\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Linear model breakdown\",\"author\":\"OCLPhase2\\'s channel\",\"organization\":\"\",\"url\":\"https:\/\/youtu.be\/6zfXCsmcDzI\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"original\",\"description\":\"Revision and Adaptation\",\"author\":\"\",\"organization\":\"Lumen Learning\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"\"},{\"type\":\"cc\",\"description\":\"Question ID 6594\",\"author\":\"Lippman,David\",\"organization\":\"\",\"url\":\"\",\"project\":\"\",\"license\":\"cc-by\",\"license_terms\":\"IMathAS Community License CC-BY + GPL\"}]","CANDELA_OUTCOMES_GUID":"6c864afc-ae5e-465d-8351-c6fb40593821","pb_show_title":"on","pb_short_title":"","pb_subtitle":"","pb_authors":[],"pb_section_license":""},"chapter-type":[],"contributor":[],"license":[],"class_list":["post-357","chapter","type-chapter","status-web-only","hentry"],"part":356,"_links":{"self":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/357","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters"}],"about":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/types\/chapter"}],"author":[{"embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/users\/20"}],"version-history":[{"count":19,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/357\/revisions"}],"predecessor-version":[{"id":3997,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/357\/revisions\/3997"}],"part":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/parts\/356"}],"metadata":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapters\/357\/metadata\/"}],"wp:attachment":[{"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/media?parent=357"}],"wp:term":[{"taxonomy":"chapter-type","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/pressbooks\/v2\/chapter-type?post=357"},{"taxonomy":"contributor","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/contributor?post=357"},{"taxonomy":"license","embeddable":true,"href":"https:\/\/courses.lumenlearning.com\/slcc-mathforliberalartscorequisite\/wp-json\/wp\/v2\/license?post=357"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}