Answer
(a) $ 500$ insects.
(b) $0.02$ or $2\%$.
(c) $ 611$ insects.
(d) $ 23.5$ days.
(e) $ 34.7$ days.
Work Step by Step
Given $P(t)=500e^{0.02t}$, we have:
(a) $P(0)=500e^{0}=500$ insects.
(b) The growth rate of the insect population is $0.02$ or $2\%$.
(c) $P(10)=500e^{0.02(10)}\approx611$ insects.
(d) $P(t)=500e^{0.02t}=800\Longrightarrow t=\frac{ln(8/5)}{0.02}\approx23.5$ days.
(e) $P(t)=500e^{0.02t}=2(500) \Longrightarrow t=\frac{ln(2)}{0.02}\approx34.7$ days.
