From $1945$ to $2008$ , one country's per capita gross domestic product (GDPPC) increased about $6\%$ per year, ending at $19,600$ . For the next few years, the growth slowed to an increase of about $1,300$ per year. To the nearest dollar, how much less did the GDPPC grow during the next $6$ years than it would have if it had maintained its earlier growth rate?

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Exponential growth and decay: word problems

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Q. From

1945

2008

, one country's per capita gross domestic product (GDPPC) increased about

6\%

per year, ending at

19,600

. For the next few years, the growth slowed to an increase of about

1,300

per year. To the nearest dollar, how much less did the GDPPC grow during the next

6

years than it would have if it had maintained its earlier growth rate?

Calculate GDPPC Growth: First, we need to calculate the GDPPC growth over $6$ years at the original growth rate of $6\%$ per year. $\newline$ We use the formula for compound interest: $A = P(1 + r/n)^{nt}$ , where: $\newline$ - $A$ is the amount of money accumulated after $n$ years, including interest. $\newline$ - $P$ is the principal amount (the initial amount of money). $\newline$ - $r$ is the annual interest rate (decimal). $\newline$ - $n$ is the number of times that interest is compounded per year. $\newline$ - $t$ is the time the money is invested for, in years. $\newline$ In this case, $P = 19,600$ , $6\%$ $0$ ( $6\%$ ), $6\%$ $2$ (compounded annually), and $6\%$ $3$ years.
Calculate GDPPC After $6$ Years: We calculate the GDPPC after $6$ years with the original growth rate: $\newline$ $A = 19,600(1 + 0.06/1)^{(1*6)}$ $\newline$ $A = 19,600(1 + 0.06)^6$ $\newline$ $A = 19,600(1.06)^6$
Calculate Linear Growth: Now we perform the actual calculation: $\newline$ $A = 19,600 \times 1.06^6$ $\newline$ $A \approx 19,600 \times 1.4185$ $\newline$ $A \approx 27,802.36$ $\newline$ This is the GDPPC after $6$ years with a $6\%$ growth rate per year.
Calculate GDPPC After Slowed Growth: Next, we calculate the GDPPC growth over $6$ years at the slowed growth rate of $1,300$ per year. $\newline$ This is a linear growth, so we simply add $1,300$ for each year. $\newline$ Growth over $6$ years = $1,300 \times 6$
Find Difference: We perform the calculation for the linear growth: $\newline$ Growth over $6$ years = $1,300 \times 6$ $\newline$ Growth over $6$ years = $7,800$
Find Difference: We perform the calculation for the linear growth: $\newline$ Growth over $6$ years = $1,300 \times 6$ $\newline$ Growth over $6$ years = $7,800$ Now we calculate the GDPPC after $6$ years with the slowed growth rate: $\newline$ GDPPC after $6$ years = Initial GDPPC + Growth over $6$ years $\newline$ GDPPC after $6$ years = $19,600 + 7,800$ $\newline$ GDPPC after $6$ years = $27,400$
Find Difference: We perform the calculation for the linear growth: $\newline$ Growth over $6$ years = $1,300 \times 6$ $\newline$ Growth over $6$ years = $7,800$ Now we calculate the GDPPC after $6$ years with the slowed growth rate: $\newline$ GDPPC after $6$ years = Initial GDPPC + Growth over $6$ years $\newline$ GDPPC after $6$ years = $19,600 + 7,800$ $\newline$ GDPPC after $6$ years = $27,400$ Finally, we find the difference between the GDPPC growth at the original rate and the slowed rate: $\newline$ Difference = GDPPC with original growth rate - GDPPC with slowed growth rate $\newline$ Difference = $27,802.36 - 27,400$ $\newline$ Difference $\approx 402.36$