Question

topic 11: population growth

1. the population of duck county for a few recent years is shown in the table below. variable t represents the number of years since 2010 and p(t) represents the population in millions.

a. calculate the 4 common ratio calculations to see if it is growing exponentially and the 4 common difference calculations to see if it is growing linearly.
b. write a function that describes the data.
c. what is the population in 2020?
d. what is the population in 2030?

year	year since 2010	population (millions)
2011	1	1.756
2012	2	1.789
2013	3	1.822
2014	4	1.856

Explanation:

Part a: Check for Exponential (Common Ratio) and Linear (Common Difference) Growth

Exponential Growth (Common Ratio)

To check for exponential growth, we calculate the ratio of consecutive population values. The formula for the common ratio \( r \) between \( P(t) \) and \( P(t - 1) \) is \( r=\frac{P(t)}{P(t - 1)} \).

• For \( t = 1 \) (2011) and \( t = 0 \) (2010):

\( r_1=\frac{1.756}{1.723}\approx1.019 \)

• For \( t = 2 \) (2012) and \( t = 1 \) (2011):

\( r_2=\frac{1.789}{1.756}\approx1.019 \)

• For \( t = 3 \) (2013) and \( t = 2 \) (2012):

\( r_3=\frac{1.822}{1.789}\approx1.019 \)

• For \( t = 4 \) (2014) and \( t = 3 \) (2013):

\( r_4=\frac{1.856}{1.822}\approx1.019 \)

Linear Growth (Common Difference)

To check for linear growth, we calculate the difference between consecutive population values. The formula for the common difference \( d \) between \( P(t) \) and \( P(t - 1) \) is \( d = P(t)-P(t - 1) \).

• For \( t = 1 \) (2011) and \( t = 0 \) (2010):

\( d_1=1.756 - 1.723 = 0.033 \)

• For \( t = 2 \) (2012) and \( t = 1 \) (2011):

\( d_2=1.789 - 1.756 = 0.033 \)

• For \( t = 3 \) (2013) and \( t = 2 \) (2012):

\( d_3=1.822 - 1.789 = 0.033 \)

• For \( t = 4 \) (2014) and \( t = 3 \) (2013):

\( d_4=1.856 - 1.822 = 0.034 \) (slight rounding difference, but approximately constant)

Part b: Write the Function

Exponential Function (Preferred, since ratios are constant)

The general form of an exponential function is \( P(t)=P_0\cdot r^t \), where \( P_0 \) is the initial population, and \( r \) is the common ratio.

• \( P_0 = 1.723 \) (population in 2010, when \( t = 0 \))
• \( r\approx1.019 \)

Thus, the exponential function is:
\( P(t)=1.723\cdot(1.019)^t \)

Linear Function (Alternative, since differences are approximately constant)

The general form of a linear function is \( P(t)=P_0 + d\cdot t \), where \( P_0 \) is the initial population, and \( d \) is the common difference.

• \( P_0 = 1.723 \)
• \( d\approx0.033 \)

Thus, the linear function is:
\( P(t)=1.723 + 0.033t \)

Part c: Population in 2020

2020 is 10 years since 2010, so \( t = 10 \).

Using Exponential Function:

\( P(10)=1.723\cdot(1.019)^{10} \)
Calculate \( (1.019)^{10}\approx1.207 \) (using a calculator or logarithm properties).
\( P(10)\approx1.723\cdot1.207\approx2.080 \) million.

Using Linear Function:

\( P(10)=1.723 + 0.033\cdot10 \)
\( P(10)=1.723 + 0.33 = 2.053 \) million.

Part d: Population in 2030

2030 is 20 years since 2010, so \( t = 20 \).

Using Exponential Function:

\( P(20)=1.723\cdot(1.019)^{20} \)
Calculate \( (1.019)^{20}\approx(1.019^{10})^2\approx(1.207)^2\approx1.457 \)
\( P(20)\approx1.723\cdot1.457\approx2.511 \) million.

Using Linear Function:

\( P(20)=1.723 + 0.033\cdot20 \)
\( P(20)=1.723 + 0.66 = 2.383 \) million.

Final Answers (Using Exponential Function, more accurate for population growth)

• Part a: Common ratios are approximately \( 1.019 \) (exponential growth), common differences are approximately \( 0.033 \) (linear growth).
• Part b: Exponential: \( \boldsymbol{P(t)=1.723\cdot(1.019)^t} \); Linear: \( \boldsymbol{P(t)=1.723 + 0.033t} \)
• Part c: Population in 2020: \( \boldsymbol{\approx2.08} \) million (exponential) or \( \boldsymbol{\approx2.05} \) million (linear).
• Part d: Population in 2030: \( \boldsymbol{\approx2.51} \) million (exponential) or \( \boldsymbol{\approx2.38} \) million (linear).

Answer:

Part a: Check for Exponential (Common Ratio) and Linear (Common Difference) Growth

Exponential Growth (Common Ratio)

To check for exponential growth, we calculate the ratio of consecutive population values. The formula for the common ratio \( r \) between \( P(t) \) and \( P(t - 1) \) is \( r=\frac{P(t)}{P(t - 1)} \).

• For \( t = 1 \) (2011) and \( t = 0 \) (2010):

\( r_1=\frac{1.756}{1.723}\approx1.019 \)

• For \( t = 2 \) (2012) and \( t = 1 \) (2011):

\( r_2=\frac{1.789}{1.756}\approx1.019 \)

• For \( t = 3 \) (2013) and \( t = 2 \) (2012):

\( r_3=\frac{1.822}{1.789}\approx1.019 \)

• For \( t = 4 \) (2014) and \( t = 3 \) (2013):

\( r_4=\frac{1.856}{1.822}\approx1.019 \)

Linear Growth (Common Difference)

To check for linear growth, we calculate the difference between consecutive population values. The formula for the common difference \( d \) between \( P(t) \) and \( P(t - 1) \) is \( d = P(t)-P(t - 1) \).

• For \( t = 1 \) (2011) and \( t = 0 \) (2010):

\( d_1=1.756 - 1.723 = 0.033 \)

• For \( t = 2 \) (2012) and \( t = 1 \) (2011):

\( d_2=1.789 - 1.756 = 0.033 \)

• For \( t = 3 \) (2013) and \( t = 2 \) (2012):

\( d_3=1.822 - 1.789 = 0.033 \)

• For \( t = 4 \) (2014) and \( t = 3 \) (2013):

\( d_4=1.856 - 1.822 = 0.034 \) (slight rounding difference, but approximately constant)

Part b: Write the Function

Exponential Function (Preferred, since ratios are constant)

The general form of an exponential function is \( P(t)=P_0\cdot r^t \), where \( P_0 \) is the initial population, and \( r \) is the common ratio.

• \( P_0 = 1.723 \) (population in 2010, when \( t = 0 \))
• \( r\approx1.019 \)

Thus, the exponential function is:
\( P(t)=1.723\cdot(1.019)^t \)

Linear Function (Alternative, since differences are approximately constant)

The general form of a linear function is \( P(t)=P_0 + d\cdot t \), where \( P_0 \) is the initial population, and \( d \) is the common difference.

• \( P_0 = 1.723 \)
• \( d\approx0.033 \)

Thus, the linear function is:
\( P(t)=1.723 + 0.033t \)

Part c: Population in 2020

2020 is 10 years since 2010, so \( t = 10 \).

Using Exponential Function:

\( P(10)=1.723\cdot(1.019)^{10} \)
Calculate \( (1.019)^{10}\approx1.207 \) (using a calculator or logarithm properties).
\( P(10)\approx1.723\cdot1.207\approx2.080 \) million.

Using Linear Function:

\( P(10)=1.723 + 0.033\cdot10 \)
\( P(10)=1.723 + 0.33 = 2.053 \) million.

Part d: Population in 2030

2030 is 20 years since 2010, so \( t = 20 \).

Using Exponential Function:

\( P(20)=1.723\cdot(1.019)^{20} \)
Calculate \( (1.019)^{20}\approx(1.019^{10})^2\approx(1.207)^2\approx1.457 \)
\( P(20)\approx1.723\cdot1.457\approx2.511 \) million.

Using Linear Function:

\( P(20)=1.723 + 0.033\cdot20 \)
\( P(20)=1.723 + 0.66 = 2.383 \) million.

Final Answers (Using Exponential Function, more accurate for population growth)

• Part a: Common ratios are approximately \( 1.019 \) (exponential growth), common differences are approximately \( 0.033 \) (linear growth).
• Part b: Exponential: \( \boldsymbol{P(t)=1.723\cdot(1.019)^t} \); Linear: \( \boldsymbol{P(t)=1.723 + 0.033t} \)
• Part c: Population in 2020: \( \boldsymbol{\approx2.08} \) million (exponential) or \( \boldsymbol{\approx2.05} \) million (linear).
• Part d: Population in 2030: \( \boldsymbol{\approx2.51} \) million (exponential) or \( \boldsymbol{\approx2.38} \) million (linear).