{"id":1972,"date":"2023-04-18T13:49:49","date_gmt":"2023-04-18T13:49:49","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=1972"},"modified":"2024-10-18T20:57:15","modified_gmt":"2024-10-18T20:57:15","slug":"linear-and-geometric-growth-learn-it-2","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/linear-and-geometric-growth-learn-it-2\/","title":{"raw":"Linear and Geometric Growth: Learn It 2","rendered":"Linear and Geometric Growth: Learn It 2"},"content":{"raw":"

Linear Growth<\/h2>\r\n

In the previous example, Marco\u2019s collection grew by the same number 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>\r\n

\r\n
\r\n

linear growth<\/h3>\r\n

If a quantity starts at size [latex]P_0[\/latex]\u00a0and grows by [latex]d[\/latex] every time period, then the quantity after [latex]n[\/latex] time periods can be determined using either of these relations:<\/p>\r\n

Recursive form<\/h4>\r\n

[latex]P_n = P_{n-1} + d[\/latex]<\/p>\r\n

Explicit form<\/h4>\r\n

[latex]P_n = P_0 + d n[\/latex]<\/p>\r\n

 <\/p>\r\n

In this equation, [latex]d[\/latex]\u00a0represents the common difference<\/strong> \u2013 the amount that the population changes each time [latex]n[\/latex] increases by [latex]1[\/latex].<\/p>\r\n<\/div>\r\n<\/section>\r\n

Connection to Prior Learning: Slope and Intercept<\/strong>\r\n

You may recognize the common difference, [latex]d[\/latex], in our linear equation as slope. In fact, the entire explicit equation should look familiar \u2013 it is the same linear equation you learned in algebra, probably stated as [latex]y = mx + b[\/latex].<\/p>\r\n

In the standard algebraic equation [latex]y = mx + b[\/latex], [latex]b[\/latex] was the y-intercept, or the [latex]y[\/latex] value when [latex]x[\/latex] was zero. In the form of the equation we\u2019re using, we are using [latex]P_0[\/latex]\u00ad to represent that initial amount.<\/p>\r\n

In the [latex]y = mx + b[\/latex] equation, recall that [latex]m[\/latex] was the slope. You might remember this as \u201crise over run,\u201d or the change in [latex]y[\/latex] divided by the change in [latex]x[\/latex]. Either way, it represents the same thing as the common difference, [latex]d[\/latex], we are using \u2013 the amount the output [latex]P_n[\/latex] changes when the input [latex]n[\/latex] increases by [latex]1[\/latex].<\/p>\r\n

The equations [latex]y = mx + b[\/latex] and [latex]P_n = P_0 + d n[\/latex] mean the same thing and can be used the same ways.\u00a0We\u2019re just writing it somewhat differently.<\/p>\r\n<\/section>\r\n

The population of elk in a national forest was measured to be [latex]12,000[\/latex] in 2003, and was measured again to be [latex]15,000[\/latex] 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\"]To begin, we need to define how we\u2019re going to measure [latex]n[\/latex]. Remember that [latex]P_0[\/latex] is the population when [latex]n = 0[\/latex], so we probably don\u2019t want to literally use the year [latex]0[\/latex]. Since we already know the population in 2003, let us define [latex]n = 0[\/latex] to be the year 2003. Then [latex]P_0 = 12,000[\/latex].
\r\n
\r\nNext we need to find [latex]d[\/latex]. Remember [latex]d[\/latex] is the growth per time period, in this case growth per year. Between the two measurements, the population grew by [latex]15,000-12,000 = 3,000[\/latex], but it took [latex]2007-2003 = 4[\/latex] years to grow that much. To find the growth per year, we can divide: [latex]\\frac{3000 \\text{ elk}}{4}[\/latex] years =[latex]750[\/latex] elk in [latex]1[\/latex] year. 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.\r\n\r\n\r\n

[latex]d=\\text{slope}=\\frac{\\text{change in output}}{\\text{change in input}}=\\frac{15,000-12,000}{2007-2003}=\\frac{3,000}{4}=750[\/latex]<\/p>\r\n\r\n\r\nWe can now write our equation in whichever form is preferred.\r\n\r\n\r\n

Recursive form<\/h4>\r\n
[latex]P\u00ad_0 = 12,000[\/latex]<\/center>\r\n

 <\/p>\r\n

[latex]P_n = P_{n-1} + 750[\/latex]<\/center>\r\n

Explicit form<\/h4>\r\n

[latex]P_n = 12,000 + 750 n[\/latex]<\/p>\r\n

To answer the question, we need to first note that the year 2014 will be [latex]n = 11[\/latex], since 2014 is [latex]11[\/latex] years after 2003. The explicit form will be easier to use for this calculation:<\/p>\r\n

[latex]P_{11}= 12,000 + 750(11) = 20,250[\/latex] elk<\/p>\r\n

View more about this example here.<\/p>\r\n

[embed]https:\/\/youtu.be\/J1XqqlKzYGs[\/embed]<\/p>\r\n

You can view the transcript for \u201cLinear Growth - Elk\u201d here (opens in new window).<\/a><\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n

Gasoline consumption in the US has been increasing steadily. Consumption data from 1992 to 2004 is shown below.[footnote]\"https:\/\/www.bts.gov\/archive\/publications\/national_transportation_statistics\/2005\/table_04_10\".[\/footnote] Find a model for this data, and use it to predict consumption in 2016. If the trend continues, when will consumption reach [latex]200[\/latex] billion gallons?\r\n\r\n\r\n\r\n\r\n\r\n\r\n
Year<\/td>\r\n'92<\/td>\r\n'93<\/td>\r\n'94<\/td>\r\n'95<\/td>\r\n'96<\/td>\r\n'97<\/td>\r\n'98<\/td>\r\n'99<\/td>\r\n'00<\/td>\r\n'01<\/td>\r\n'02<\/td>\r\n'03<\/td>\r\n'04<\/td>\r\n<\/tr>\r\n
Consumption (billion of gallons)<\/td>\r\n[latex]110[\/latex]<\/td>\r\n[latex]111[\/latex]<\/td>\r\n[latex]113[\/latex]<\/td>\r\n[latex]116[\/latex]<\/td>\r\n[latex]118[\/latex]<\/td>\r\n[latex]119[\/latex]<\/td>\r\n[latex]123[\/latex]<\/td>\r\n[latex]125[\/latex]<\/td>\r\n[latex]126[\/latex]<\/td>\r\n[latex]128[\/latex]<\/td>\r\n[latex]131[\/latex]<\/td>\r\n[latex]133[\/latex]<\/td>\r\n[latex]136[\/latex]<\/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\"]<\/p>\r\n

Plotting this data, it appears to have an approximately linear relationship:<\/p>\r\n

\"Graph.<\/center>\r\n

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>\r\n

Letting [latex]n = 0[\/latex] correspond with 1993 would give[latex]P_0 = 111[\/latex] billion gallons.<\/p>\r\n

To find [latex]d[\/latex], we need to know how much the gas consumption increased each year, on average. From 1993 to 2003 the gas consumption increased from [latex]111[\/latex] billion gallons to [latex]133[\/latex] billion gallons, a total change of [latex]133 \u2013 111 = 22[\/latex] billion gallons, over [latex]10[\/latex] years. This gives us an average change of [latex]\\frac{22 \\text{ billion gallons}}{10} \\text{ year} = 2.2[\/latex] billion gallons per year.<\/p>\r\n

Equivalently,<\/p>\r\n

[latex]d=\\text{slope}=\\frac{\\text{change in output}}{\\text{change in input}}=\\frac{133-111}{10-0}=\\frac{22}{10}=2.2[\/latex] billion gallons per year<\/p>\r\n

We can now write our equation in whichever form is preferred.<\/p>\r\n

Recursive form<\/h4>\r\n

[latex]P_0 = 111[\/latex]<\/p>\r\n

[latex]P_n = P_{n-1} + 2.2[\/latex]<\/p>\r\n

Explicit form<\/h4>\r\n

[latex]P_n = 111 + 2.2 n[\/latex]<\/p>\r\n

\"Graph.<\/center>\r\n

Calculating values using the explicit form and plotting them with the original data shows how well our model fits the data.<\/p>\r\n

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>\r\n

[latex]n = 23[\/latex] (2016 \u2013 1993 = 23 years later)<\/p>\r\n

[latex]P_23 = 111 + 2.2(23) = 161.6[\/latex]<\/p>\r\n

Our model predicts that the US will consume [latex]161.6[\/latex] billion gallons of gasoline in 2016 if the current trend continues.<\/p>\r\n

To find when the consumption will reach [latex]200[\/latex] billion gallons, we would set [latex]P_n = 200[\/latex], and solve for n:<\/p>\r\n

[latex]P_n = 200[\/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\u00a0\u00a0 Replace [latex]P_n[\/latex] with our model<\/p>\r\n

[latex]111 + 2.2 n = 200[\/latex]\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0\u00a0 Subtract [latex]111[\/latex] from both sides<\/p>\r\n

[latex]2.2 n = 89[\/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 Divide both sides by [latex]2.2[\/latex]<\/p>\r\n

[latex]n = 40.4545[\/latex]<\/p>\r\n

This tells us that consumption will reach [latex]200[latex] billion about 40 years after 1993, which would be in the year 2033.<\/p>\r\n

The steps for reaching this answer are detailed in the following video.<\/p>\r\n

[embed]https:\/\/youtu.be\/ApFxDWd6IbE[\/embed]<\/p>\r\n

You can view the transcript for \u201cFinding linear model for gas consumption\u201d here (opens in new window).<\/a><\/p>\r\n

[\/hidden-answer]<\/p>\r\n<\/section>\r\n

[ohm2_question hide_question_numbers=1]6916[\/ohm2_question]<\/section>\r\n

When Good Models Go Bad<\/h2>\r\n
Suppose a four year old boy is currently [latex]39[\/latex] inches tall, and you are told to expect him to grow [latex]2.5[\/latex] inches a year. We can set up a growth model, with [latex]n = 0[\/latex] corresponding to [latex]4[\/latex] years old.\r\n\r\n\r\n

Recursive form<\/h4>\r\n

[latex]P_0 = 39[\/latex]<\/p>\r\n

[latex]P_n = P_{n-1} + 2.5[\/latex]<\/p>\r\n

Explicit form<\/h4>\r\n

[latex]P_n = 39 + 2.5 n[\/latex]<\/p>\r\n

So at 6 years old, we would expect him to be<\/p>\r\n

[latex]P_2 = 39 + 2.5(2) = 44[\/latex] inches tall<\/p>\r\n

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>\r\n

[latex]P_{46} = 39 + 2.5(46) = 154[\/latex] inches tall [latex]= 12.8[\/latex] feet tall!<\/p>\r\n

When using any mathematical model, we have to consider which inputs are reasonable to use. Whenever we extrapolate<\/strong>, or make predictions into the future, we are assuming the model will continue to be valid.<\/p>\r\n

View a video explanation of this breakdown of the linear growth model here.<\/p>\r\n

[embed]https:\/\/youtu.be\/6zfXCsmcDzI[\/embed]<\/p>\r\n

You can view the transcript for \u201cLinear model breakdown\u201d here (opens in new window).<\/a><\/p>\r\n<\/section>","rendered":"

Linear Growth<\/h2>\n

In the previous example, Marco\u2019s collection grew by the same number 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

\n
\n

linear growth<\/h3>\n

If a quantity starts at size [latex]P_0[\/latex]\u00a0and grows by [latex]d[\/latex] every time period, then the quantity after [latex]n[\/latex] time periods can be determined using either of these relations:<\/p>\n

Recursive form<\/h4>\n

[latex]P_n = P_{n-1} + d[\/latex]<\/p>\n

Explicit form<\/h4>\n

[latex]P_n = P_0 + d n[\/latex]<\/p>\n

 <\/p>\n

In this equation, [latex]d[\/latex]\u00a0represents the common difference<\/strong> \u2013 the amount that the population changes each time [latex]n[\/latex] increases by [latex]1[\/latex].<\/p>\n<\/div>\n<\/section>\n

Connection to Prior Learning: Slope and Intercept<\/strong><\/p>\n

You may recognize the common difference, [latex]d[\/latex], in our linear equation as slope. In fact, the entire explicit equation should look familiar \u2013 it is the same linear equation you learned in algebra, probably stated as [latex]y = mx + b[\/latex].<\/p>\n

In the standard algebraic equation [latex]y = mx + b[\/latex], [latex]b[\/latex] was the y-intercept, or the [latex]y[\/latex] value when [latex]x[\/latex] was zero. In the form of the equation we\u2019re using, we are using [latex]P_0[\/latex]\u00ad to represent that initial amount.<\/p>\n

In the [latex]y = mx + b[\/latex] equation, recall that [latex]m[\/latex] was the slope. You might remember this as \u201crise over run,\u201d or the change in [latex]y[\/latex] divided by the change in [latex]x[\/latex]. Either way, it represents the same thing as the common difference, [latex]d[\/latex], we are using \u2013 the amount the output [latex]P_n[\/latex] changes when the input [latex]n[\/latex] increases by [latex]1[\/latex].<\/p>\n

The equations [latex]y = mx + b[\/latex] and [latex]P_n = P_0 + d n[\/latex] mean the same thing and can be used the same ways.\u00a0We\u2019re just writing it somewhat differently.<\/p>\n<\/section>\n

The population of elk in a national forest was measured to be [latex]12,000[\/latex] in 2003, and was measured again to be [latex]15,000[\/latex] in 2007. If the population continues to grow linearly at this rate, what will the elk population be in 2014?<\/p>\n