Question 19, 2.6.41-GI
Part 3 of 5
The population of Nilam doubles in size every $6 yr$ . 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 $P (t) = 10000 e^{0.1155 t}$ .
(Round to four decimal places as needed.)
c) Nilam's yearly percentage growth rate is $%$ . (Round to two decimal places as needed.)
Textbook

Answer

Expert–verified

Hide Steps

Answer

$P (t) = 10000 \times 2^{\frac{t}{6}}, P (t) = 10000 e^{0.1155 t}, 11.55 %$

Steps

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 ： $P (t) = 10000 \times 2^{\frac{t}{6}}, P (t) = 10000 e^{0.1155 t}, 11.55 %$

Answer

Popular Problems