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 \(=1.60%\)

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 \(=1.60%\)