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 = 0 corresponds to 2000 and 0 ≤ t ≤ 50.
a(t)=11.16(1.063)^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.
(a) what is the expression for the average rate of change? select the correct answer below and fill in the answer boxes to complete your choice
(type whole numbers. use descending order.)
a. lim(h→0) (f(□ + h)-f(□))/h
b. (f(□)-f(□))/(□ - □)

Explanation:

Step1: Identify the time - values

For 2000, \(t_1 = 0\); for 2014, \(t_2=2014 - 2000=14\). The average rate of change of a function \(y = A(t)\) over the interval \([t_1,t_2]\) is given by \(\frac{A(t_2)-A(t_1)}{t_2 - t_1}\). So the correct answer for part (a) of the multiple - choice is B, and the expression is \(\frac{A(14)-A(0)}{14 - 0}\).
First, find \(A(0)\):
Substitute \(t = 0\) into \(A(t)=11.16(1.063)^{t}\), we get \(A(0)=11.16(1.063)^{0}=11.16\) (since any non - zero number to the power of 0 is 1).

Step2: Find \(A(14)\)

Substitute \(t = 14\) into \(A(t)=11.16(1.063)^{t}\), we have \(A(14)=11.16\times(1.063)^{14}\).
Using a calculator, \((1.063)^{14}\approx2.377\), so \(A(14)=11.16\times2.377 = 11.16\times2.377=26.53732\).

Step3: Calculate the average rate of change

The average rate of change \(\frac{A(14)-A(0)}{14 - 0}=\frac{26.53732 - 11.16}{14}=\frac{15.37732}{14}\approx1.098\) million per year.

Step4: Find the derivative of \(A(t)\) for part (b)

The derivative of \(y = a\cdot b^{t}\) (where \(a = 11.16\) and \(b = 1.063\)) using the formula \(\frac{d}{dt}(a\cdot b^{t})=a\cdot b^{t}\ln(b)\) is \(A^\prime(t)=11.16\times1.063^{t}\ln(1.063)\).

Step5: Evaluate the derivative at \(t = 14\)

Substitute \(t = 14\) into \(A^\prime(t)\):
\(A^\prime(14)=11.16\times(1.063)^{14}\ln(1.063)\).
Since \((1.063)^{14}\approx2.377\) and \(\ln(1.063)\approx0.0611\), then \(A^\prime(14)=11.16\times2.377\times0.0611\).
\(A^\prime(14)=11.16\times2.377\times0.0611 = 1.622\) million per year.

Answer:

(a) The average rate of change is approximately \(1.098\) million per year.
(b) The instantaneous rate of change in 2014 is approximately \(1.622\) million per year.