{"id":1994,"date":"2023-04-18T17:25:18","date_gmt":"2023-04-18T17:25:18","guid":{"rendered":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/?post_type=chapter&p=1994"},"modified":"2024-10-18T20:57:15","modified_gmt":"2024-10-18T20:57:15","slug":"linear-and-geometric-growth-learn-it-4","status":"web-only","type":"chapter","link":"https:\/\/content.one.lumenlearning.com\/quantitativereasoning\/chapter\/linear-and-geometric-growth-learn-it-4\/","title":{"raw":"Linear and Geometric Growth: Learn It 4","rendered":"Linear and Geometric Growth: Learn It 4"},"content":{"raw":"

Exponential (Population) Growth Cont.<\/h2>\r\n
\r\n
\r\n

exponential growth<\/h3>\r\n

If a quantity starts at size [latex]P_0[\/latex] and grows by [latex]R\\%[\/latex] (written as a decimal, [latex]r[\/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 = (1+r)P_{n-1}[\/latex]<\/p>\r\n

Explicit form<\/h4>\r\n

[latex]P_n = (1+r)^{n}P_0[\/latex] or equivalently, [latex]P_n= P_0(1+r)^{n}[\/latex]<\/p>\r\n

 <\/p>\r\n

We call\u00a0[latex]r[\/latex] the growth rate<\/strong>.<\/p>\r\n

 <\/p>\r\n

The term\u00a0[latex](1+r)[\/latex] is called the growth multiplier<\/strong>, or common ratio.<\/p>\r\n<\/div>\r\n<\/section>\r\n

In 1990, residential energy use in the US was responsible for [latex]962[\/latex] million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to [latex]1182[\/latex] million metric tons[footnote]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\"]Similar to before, we will correspond [latex]n = 0[\/latex] with 1990, as that is the year for the first piece of data we have. That will make [latex]P_0 = 962[\/latex] (million metric tons of CO2<\/sub>). In this problem, we are not given the growth rate, but instead are given that [latex]P\u00ad_{10} = 1182[\/latex]. When [latex]n = 10[\/latex], the explicit equation looks like:\r\n\r\n\r\n

[latex]P_{10} =P\u00ad_0 (1+r)^{10}[\/latex]<\/p>\r\n

We know the value for [latex]P_0[\/latex]\u00a0so we can put that into the equation:<\/p>\r\n

[latex]P\u00ad_{10} =962 (1+r)^{10}[\/latex]<\/p>\r\n

We also know that [latex]P\u00ad_{10}= 1182[\/latex], so substituting that in, we get<\/p>\r\n

[latex]1182 =962 (1+r)^{10}[\/latex]<\/p>\r\n

We can now solve this equation for the growth rate, [latex]r[\/latex]. Start by dividing by [latex]962[\/latex].<\/p>\r\n

[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 [latex]10th[\/latex] root of both sides<\/p>\r\n

[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 [latex]1[\/latex] from both sides<\/p>\r\n

[latex]r=\\sqrt[10]{\\frac{1182}{962}}-1=0.0208[\/latex] [latex]= 2.08\\%[\/latex]<\/p>\r\n

So if the emissions are growing exponentially, they are growing by about [latex]2.08\\%[\/latex] per year. We can now predict the emissions in 2050 by finding [latex]P_60[\/latex]<\/p>\r\n

[latex]P_{60} = 962 (1+0.0208)^{60}= 3308.4[\/latex] million metric tons of CO2<\/sub> in 2050<\/p>\r\n

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

[embed]https:\/\/youtu.be\/9Zu2uONfLkQ[\/embed]<\/p>\r\n

You can view the transcript for \u201cFinding an Exponential Model\u201d here (opens in new window).<\/a><\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n

\r\nEvaluating exponents on the calculator<\/strong>\r\n

To evaluate expressions like [latex](1.03)^{6}[\/latex], it will be easier to use a calculator than multiply [latex]1.03[\/latex] by itself six times. Most scientific calculators have a button for exponents.\u00a0 It is typically either labeled like:<\/p>\r\n

^,\u00a0\u00a0[latex]y^{x}[\/latex]\u00a0,\u00a0\u00a0 or [latex]x^{y}[\/latex].<\/p>\r\n

To evaluate [latex]1.03^{6}[\/latex]\u00a0we\u2019d type [latex]1.03 ^ 6[\/latex], or [latex]1.03 y^{x} 6[\/latex].\u00a0 Try it out - you should get an answer around [latex]1.1940523[\/latex].<\/p>\r\n<\/section>\r\n

[ohm2_question hide_question_numbers=1]6919[\/ohm2_question]<\/section>\r\n
A friend is using the equation [latex]P_n = 4600(1.072)^{n}[\/latex] 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?[reveal-answer q=\"224261\"]Show Solution[\/reveal-answer]
\r\n[hidden-answer a=\"224261\"]In the equation, [latex]P_0 = 4600[\/latex], which is the starting value of the tuition when [latex]n = 0[\/latex]. This tells us that the tuition in 2010 was [latex]$4,600[\/latex].The growth multiplier is [latex]1.072[\/latex], so the growth rate is [latex]0.072[\/latex], or [latex]7.2\\%[\/latex]. This tells us that the tuition is expected to grow by [latex]7.2\\%[\/latex] each year.Putting this together, we could say that the tuition in 2010 was [latex]$4,600[\/latex], and is expected to grow by [latex]7.2\\%[\/latex] each year.\r\n\r\n\r\n

View the following to see this example worked out.<\/p>\r\n

[embed]https:\/\/youtu.be\/T8Yz94De5UM[\/embed]<\/p>\r\n

You can view the transcript for \u201cInterpreting an Exponential Equation\u201d here (opens in new window).<\/a><\/p>\r\n\r\n[\/hidden-answer]<\/section>\r\n

\r\nRounding<\/strong>\r\n

As a note on rounding, notice that if we had rounded the growth rate to [latex]2.1\\%[\/latex], our calculation for the emissions in 2050 would have been [latex]3347[\/latex].\u00a0\u00a0 Rounding to [latex]2\\%[\/latex] would have changed our result to [latex]3156[\/latex]. 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>\r\n

If you need to round, keep at least three significant digits<\/strong> - numbers after any leading zeros.\u00a0\u00a0 So [latex]0.4162[\/latex] could be reasonably rounded to [latex]0.416[\/latex]. A growth rate of [latex]0.001027[\/latex] could be reasonably rounded to [latex]0.00103[\/latex].<\/p>\r\n<\/section>\r\n

\r\nEvaluating roots on the calculator<\/strong>\r\n

In the previous example, we had to calculate the [latex]10th[\/latex] 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>\r\n

[latex]\\sqrt[y]{x}[\/latex]<\/p>\r\n

To evaluate the [latex]3rd[\/latex] root of [latex]8[\/latex], for example, we\u2019d either type [latex]3[\/latex] [latex]\\sqrt[x]{{}}[\/latex][latex]8[\/latex], or [latex]8[\/latex][latex]\\sqrt[x]{{}}[\/latex][latex]3[\/latex], depending on the calculator. Try it on yours to see which to use \u2013 you should get an answer of [latex]2[\/latex].<\/p>\r\n

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>\r\n

[latex]\\sqrt[n]{a}={a}^{\\frac{1}{2}}[\/latex]<\/span>.<\/p>\r\n

So, to compute the [latex]3rd[\/latex] root of [latex]8[\/latex], you could use your calculator\u2019s exponent key to evaluate [latex]8^{\\frac{1}{3}}[\/latex]. To do this, type:<\/p>\r\n

[latex]8 y^{x} ( 1 \u00f7 3 )[\/latex]<\/p>\r\n

The parentheses tell the calculator to divide [latex]\\frac{1}{3}[\/latex] before doing the exponent.<\/p>\r\n<\/section>\r\n

[ohm2_question hide_question_numbers=1]6922[\/ohm2_question]<\/section>\r\n
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:\r\n\r\n\r\n
    \r\n\t
  1. 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
  2. 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>\r\n<\/section>","rendered":"

    Exponential (Population) Growth Cont.<\/h2>\n
    \n
    \n

    exponential growth<\/h3>\n

    If a quantity starts at size [latex]P_0[\/latex] and grows by [latex]R\\%[\/latex] (written as a decimal, [latex]r[\/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 = (1+r)P_{n-1}[\/latex]<\/p>\n

    Explicit form<\/h4>\n

    [latex]P_n = (1+r)^{n}P_0[\/latex] or equivalently, [latex]P_n= P_0(1+r)^{n}[\/latex]<\/p>\n

     <\/p>\n

    We call\u00a0[latex]r[\/latex] the growth rate<\/strong>.<\/p>\n

     <\/p>\n

    The term\u00a0[latex](1+r)[\/latex] is called the growth multiplier<\/strong>, or common ratio.<\/p>\n<\/div>\n<\/section>\n

    In 1990, residential energy use in the US was responsible for [latex]962[\/latex] million metric tons of carbon dioxide emissions. By the year 2000, that number had risen to [latex]1182[\/latex] million metric tons[1]<\/sup><\/a>. If the emissions grow exponentially and continue at the same rate, what will the emissions grow to by 2050?<\/p>\n