Question

Based on a study of population projections for 2000 to​ 2050, the projected population of a group of people​ (in millions) can be modeled by the following exponential​ function, where t equals 0 corresponds to 2000 and less than or equalsxless than or equals50. ​A(t)equals11.19 (1.009 )Superscript t ​
(a) Use A to estimate the average rate of change in the population from 2000 to 2014.
​(b) Estimate the instantaneous rate of change in the population in 2014.

Answer 1

Answer:

(a) 0.107 million per year

(b) 0.114 million per year

Step-by-step explanation:

$A(t) = 11.19(1.009)^t$

(a) The average rate of change between 2000 and 2014 is determined by dividing the difference in the populations in the two years by the number of years. In the year 2000, $t=0$ and in 2014, $t=14$. Mathematically,

$\text{Rate}=\dfrac{A(2014) - A(2000)}{2014-2000}=\dfrac{11.19(1.009)^14-11.19(1.009)^0} {14}$

$A(0)=\dfrac{12.69-11.19}{14}=\dfrac{1.5}{14}=0.107$

(b) The instantaneous rate of change is determined by finding the differential derivative at that year.

The result of differentiating functions of the firm $y=a^x$ (where $a$ is a constant) is $\dfrac{dy}{dx}=a^x\ln a$. Let's use in this in finding the derivative of $A(t)$.

$A\prime(t) = \dfrac{A}{t}=11.19\cdot1.009^t\ln1.009$

In the year 2014, $t=14$.

$A\prime(14) =11.19\cdot1.009^14\ln1.009=0.114$