Question 19, 2.6.41-GI Part 3 of 5 The population of Nilam doubles in size every $6 \mathrm{yr}$. In 1992, its population was 10,000. a) Find an exponential function of the form $P(t)=P_{0} n^{\frac{t}{T}}$ that models Nilam's population after $t$ years. b) Find the equivalent exponential model of the form $P(t)=P_{0} e^{r t}$. c) What is Nilam's yearly percentage growth rate? d) Without using a calculator, find Nilam's population in 2010. e) How fast was Nilam's population changing in 2002 ? a) Find an exponential function of the form $P(t)=P_{0} n^{\frac{t}{T}}$ that models the situation. The exponential function is $P(t)=10000 \times 2^{\frac{t}{6}}$. (Use integers or fractions for any numbers in the expression.) b) Find the equivalent exponential model of the form $P(t)=P_{0} e^{r t}$. The exponential model is $\mathrm{P}(\mathrm{t})=10000 e^{0.1155 \mathrm{t}}$. (Round to four decimal places as needed.) c) Nilam's yearly percentage growth rate is $\%$. (Round to two decimal places as needed.) Textbook
Solution
Step 1 ：Given that the population of Nilam doubles every 6 years, we can model this growth with an exponential function of the form \(P(t)=P_{0} n^{\frac{t}{T}}\), where \(P_{0}\) is the initial population, \(n\) is the growth factor, \(t\) is the time in years, and \(T\) is the time it takes for the population to double.
Step 2 ：Substituting the given values into the equation, we get \(P(t)=10000 \times 2^{\frac{t}{6}}\).
Step 3 ：We can also express this exponential growth in the form \(P(t)=P_{0} e^{r t}\), where \(r\) is the growth rate. To find \(r\), we can use the formula \(r=\frac{\ln(n)}{T}\).
Step 4 ：Substituting the given values into the equation, we get \(r=\frac{\ln(2)}{6}\approx 0.1155\). So, the equivalent exponential model is \(P(t)=10000 e^{0.1155 t}\).
Step 5 ：The yearly percentage growth rate can be found by multiplying the growth rate \(r\) by 100. So, Nilam's yearly percentage growth rate is \(0.1155 \times 100\% \approx 11.55\%\).
Step 6 ：\(\boxed{P(t)=10000 \times 2^{\frac{t}{6}}, P(t)=10000 e^{0.1155 t}, 11.55\%}\)
