Answer:
Part A)
About 0.51% per year.
Part B)
About 0.30% per year.
Part C)
About 28.26%.
Step-by-step explanation:
We are given that the population of Americans age 55 and older as a percentange of the total population is approximated by the function:
![f(t) = 10.72(0.9t+10)^{0.3}\text{ where } 0 \leq t \leq 20](https://tex.z-dn.net/?f=f%28t%29%20%3D%2010.72%280.9t%2B10%29%5E%7B0.3%7D%5Ctext%7B%20where%20%7D%200%20%5Cleq%20t%20%5Cleq%2020)
Where <em>t</em> is measured in years with <em>t</em> = 0 being the year 2000.
Part A)
Recall that the rate of change of a function at a point is given by its derivative. Thus, find the derivative of our function:
![\displaystyle f'(t) = \frac{d}{dt} \left[ 10.72\left(0.9t+10\right)^{0.3}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28t%29%20%20%3D%20%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5B%2010.72%5Cleft%280.9t%2B10%5Cright%29%5E%7B0.3%7D%5Cright%5D)
Rewrite:
![\displaystyle f'(t) = 10.72\frac{d}{dt} \left[(0.9t+10)^{0.3}\right]](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28t%29%20%3D%2010.72%5Cfrac%7Bd%7D%7Bdt%7D%20%5Cleft%5B%280.9t%2B10%29%5E%7B0.3%7D%5Cright%5D)
We can use the chain rule. Recall that:
![\displaystyle \frac{d}{dx} [u(v(x))] = u'(v(x)) \cdot v'(x)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdx%7D%20%5Bu%28v%28x%29%29%5D%20%3D%20u%27%28v%28x%29%29%20%5Ccdot%20v%27%28x%29)
Let:
![\displaystyle u(t) = t^{0.3}\text{ and } v(t) = 0.9t+10 \text{ (so } u(v(t)) = (0.9t+10)^{0.3}\text{)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20u%28t%29%20%3D%20t%5E%7B0.3%7D%5Ctext%7B%20and%20%7D%20v%28t%29%20%3D%200.9t%2B10%20%5Ctext%7B%20%20%20%28so%20%7D%20u%28v%28t%29%29%20%3D%20%280.9t%2B10%29%5E%7B0.3%7D%5Ctext%7B%29%7D)
Then from the Power Rule:
![\displaystyle u'(t) = 0.3t^{-0.7}\text{ and } v'(t) = 0.9](https://tex.z-dn.net/?f=%5Cdisplaystyle%20u%27%28t%29%20%3D%200.3t%5E%7B-0.7%7D%5Ctext%7B%20and%20%7D%20v%27%28t%29%20%3D%200.9)
Thus:
![\displaystyle \frac{d}{dt}\left[(0.9t+10)^{0.3}\right]= 0.3(0.9t+10)^{-0.7}\cdot 0.9](https://tex.z-dn.net/?f=%5Cdisplaystyle%20%5Cfrac%7Bd%7D%7Bdt%7D%5Cleft%5B%280.9t%2B10%29%5E%7B0.3%7D%5Cright%5D%3D%200.3%280.9t%2B10%29%5E%7B-0.7%7D%5Ccdot%200.9)
Substitute:
![\displaystyle f'(t) = 10.72\left( 0.3(0.9t+10)^{-0.7}\cdot 0.9 \right)](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28t%29%20%3D%2010.72%5Cleft%28%20%20%20%200.3%280.9t%2B10%29%5E%7B-0.7%7D%5Ccdot%200.9%20%20%20%5Cright%29)
And simplify:
![\displaystyle f'(t) = 2.8944(0.9t+10)^{-0.7}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%28t%29%20%3D%202.8944%280.9t%2B10%29%5E%7B-0.7%7D)
For 2002, <em>t</em> = 2. Then the rate at which the percentage is changing will be:
![\displaystyle f'(2) = 2.8944(0.9(2)+10)^{-0.7} = 0.5143...\approx 0.51](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%282%29%20%3D%202.8944%280.9%282%29%2B10%29%5E%7B-0.7%7D%20%3D%200.5143...%5Capprox%200.51)
Contextually, this means the percentage is increasing by about 0.51% per year.
Part B)
Evaluate f'(t) when <em>t</em> = 17. This yields:
![\displaystyle f'(17) = 2.8944(0.9(17)+10)^{-0.7} =0.3015...\approx 0.30](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%27%2817%29%20%3D%202.8944%280.9%2817%29%2B10%29%5E%7B-0.7%7D%20%3D0.3015...%5Capprox%200.30)
Contextually, this means the percetange is increasing by about 0.30% per year.
Part C)
For this question, we will simply use the original function since it outputs the percentage of the American population 55 and older. Thus, evaluate f(t) when <em>t</em> = 17:
![\displaystyle f(17) = 10.72(0.9(17)+10)^{0.3}=28.2573...\approx 28.26](https://tex.z-dn.net/?f=%5Cdisplaystyle%20f%2817%29%20%3D%2010.72%280.9%2817%29%2B10%29%5E%7B0.3%7D%3D28.2573...%5Capprox%2028.26)
So, about 28.26% of the American population in 2017 are age 55 and older.