The exponential models the population of the indicated country, A, in millions, t years after 20006. Which country has the greatest growth rate? By what percentage is the population of that country increasing each year? Country 1: A=26.3e^{0.029t} Country 2: A=127.7e^{0.007t} Country 3: A=148.5e^{-0.0091t} Country 4: A=1094.2e^{0.016t}

Question
Exponential models
asked 2021-02-06
The exponential models the population of the indicated country, A, in millions, t years after 20006. Which country has the greatest growth rate? By what percentage is the population of that country increasing each year?
Country 1: \(A=26.3e^{0.029t}\)
Country 2: \(A=127.7e^{0.007t}\)
Country 3: \(A=148.5e^{-0.0091t}\)
Country 4: \(A=1094.2e^{0.016t}\)

Answers (1)

2021-02-07
Given:
The exponential model that describe the population of different countries:
Country 1: \(A=26.3e^{0.029t}\)
Country 2: \(A=127.7e^{0.007t}\)
Country 3: \(A=148.5e^{-0.0091t}\)
Country 4: \(A=1094.2e^{0.016t}\)
Population growth is equal to differentiation of exponential model with respect to ‘t’,
So now differentiating the given function one by one:
\(\frac{d}{dt}(A_1)=\frac{d}{dt}(26.3e^{0.029t})\)
\(\frac{dA_1}{dt}=(26.3\times0.029e^{0.029t})\)
\(\frac{dA_1}{dt}=(0.7627e^{0.029t}\)
\(\frac{d}{dt}(A_2)=\frac{d}{dt}(127.7e^{0.007t})\)
\(\frac{dA_1}{dt}=(127.7\times0.007e^{0.007t}\)
\(\frac{dA_1}{dt}=(0.8939e^{0.007t})\)
\(\frac{d}{dt}(A_3)=\frac{d}{dt}(148.5e^{-0.009t})\)
\(\frac{dA_3}{dt}=\frac{d}{dt}(148.5\times(-0.009)e^{-0.009t})\)
\(\frac{dA_3}{dt}=-(1.3365e^{-0.009t})\)
\(\frac{d}{dt}(A_4)=\frac{d}{dt}(1094.2e^{0.016t}\)
\(\frac{dA_4}{dt}=\frac{d}{dt}(1094.2\times0.016e^{0.016t})\)
\(\frac{dA_4}{dt}=(17.5072e^{0.016t})\)
To check which one is larger, by comparing the quantities of growth rate it can be concluded that:
Country 4: \(A_4=1094.2e^{0.016t}\)
Has the greates growth rate.
To find the percentage change in population:
Country 4: \(A_4=1092.2e^{0.016t}\)
Now put t=0,
\(A_4=1094.2e^{0.016\times0}\)
\(A_4=1094.2e^0\)
\(A_4=1094.2\)
Now put \(t=1,\)
\(A_4=1094.2e^{0.016\times1}\)
\(A_4=1094.2e^{0.016}\)
\(A_4=1111.8\)
Now,
Percentage change \(=\frac{1111.8-1094.2}{1094.2}\times100\)
Percentage change \(=1.60%\)
Answer: The final answer is given below:
Country 4: \(A=1094.2e^{0.016t}\)
Has the greates growth rate.
Percentage change in population each year \(=1.60%\)
0

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