The population P(t) of mosquitoes in a certain area increases at a rate proporti
ID: 1719966 • Letter: T
Question
The population P(t) of mosquitoes in a certain area increases at a rate proportional to the current population where the growth rate r = ln(2)/7. If there are initially 500 (thousand) mosquitoes in the area and predators (bats, birds, frogs, etc.) eat 80 (thousand) per day, determine an expression for the population of mosquitoes in the area at any time t.
a. P(t) = 80 e^(t ln(2)/7) - 560/ln(2) (e^(t ln(2)/7) - 1)
b. P(t) = 80 e^(7 t) - 560 (e^(7 t ) - 1)
c. P(t) = 500 e^(t ln(2)/7) - 560/ln(2) (e^(t ln(2)/7) - 1)
d. P(t) = 500 e^(t ln(2)) - 560/ln(2) (e^(t ln(2)) - 1)
e. P(t) = 500 e^(t ln(2)/7) - 80/ln(2) (e^(t ln(2)/7) - 1)
Explanation / Answer
P' = (ln2)/7 P -80
=>
P' = [(ln2)/7]*[P - 560/ln2]
=>
ln( P-560/ln2) = [(ln2)/7]t +c
P(0) = 500
=>
c = ln(500-560/ln2)
=>
ln( P-560/ln2) = [(ln2)/7]t + ln(500-560/ln2)
=>
P -560/ln2 = e^[(ln2)/7 *t]* [ 500 -560/ln2]
=>
P = 500e^[(ln2)/7 *t] -560/ln2 [e^((ln2)/7 *t) -1]
option ;C