Respuesta :
[tex]\begin{gathered} \text{ (a) }Given\text{ that p(t) = }800,000e^{0.05t} \\ \text{ In the year 2014, t=0} \\ \text{substitute t=0 into p(t) above} \\ p(0)\text{ = 800000}e^{0.05(0)} \\ p(0)\text{ = 800000}e^0 \\ p(0)\text{ = 800000}(1) \\ p(0)\text{ = 800000} \\ \text{The population at 2014 is 800,000} \end{gathered}[/tex]
(b) To determine the year the population will reach 1,200,000, substitute p(t) = 1,200,000 into p(t) above and calculate the value of t
[tex]\begin{gathered} \text{ p(t) = 1,200,000} \\ 1,200,000=800,000e^{0.05t} \\ \text{divide both sides by 800,000} \\ \frac{1200000}{800000}\text{ =}\frac{\text{ 800000}e^{0.05t}}{800000} \\ \\ 1.5=e^{0.05t} \\ \text{take }\ln \text{ of both sides} \\ \ln 1.5\text{ = }\ln e^{0.05t} \\ \ln 1.5=0.05t\ln e \\ \ln 1.5=0.05t(1) \\ \ln 1.5\text{ = 0.05t} \\ \text{divide both sides by 0.05} \\ \frac{\ln 1.5}{0.05}\text{ = t} \\ t=8 \end{gathered}[/tex]The year when t= 8, given that t=0 in 2014 is the year 2022
(c) the rate of change of the population in the year 2019
Firstly, differentiate p(t) with respect to t
[tex]\begin{gathered} p(t)\text{ = }800,000e^{0.05t} \\ \frac{\text{ dp}}{\text{ dt}}\text{ = 0.05(800,000)}e^{0.05t} \\ \frac{dp}{\mathrm{d}t}\text{ = 40,000}e^{0.05t} \\ \text{ This means that the rate of change of the function at any time t is 40,000}e^{0.05t} \\ \end{gathered}[/tex]In the year 2019, t = 5
Substitute t=5 into the rate of change above
[tex]\begin{gathered} \text{ at t= 5, } \\ \text{rate of change = 40,000}e^{0.05(5)} \\ =\text{ 40,000}e^{0.25} \\ =40,000(1.284025) \\ =51361 \end{gathered}[/tex]
