SOLUTION: The size P of a certain insect population at time t (in days) obeys the function p(t)=700e^0.05t
a)Determine the number of insects at t=0 days.
b)What is the growth rate of t
Algebra ->
Finance
-> SOLUTION: The size P of a certain insect population at time t (in days) obeys the function p(t)=700e^0.05t
a)Determine the number of insects at t=0 days.
b)What is the growth rate of t
Log On
Question 1087535: The size P of a certain insect population at time t (in days) obeys the function p(t)=700e^0.05t
a)Determine the number of insects at t=0 days.
b)What is the growth rate of the insect population?
c)What is the population after 10 days?
d)When will the insect population reach 1050?
e)When will the insect population double? Answer by mathmate(429) (Show Source):
You can put this solution on YOUR website! Question:
The size P of a certain insect population at time t (in days) obeys the function p(t)=700e^0.05t
a)Determine the number of insects at t=0 days.
b)What is the growth rate of the insect population?
c)What is the population after 10 days?
d)When will the insect population reach 1050?
e)When will the insect population double?
Solution:
Given : P(t)=700e^(0.5t), t in days
(a) number of insects at t=0 days
number of insects = P(0)=700(1)=700.
(b) the growth rate is the percentage increase between days
=P(t+1)/P(t)=700e^(0.5(t+1))/700e^(0.5t)=e^(0.5(1))=e^0.5=1.64872
therefore the growth rate is 64.872% per day.
(c) population after 10 days
P(10)=700e^(0.5*10)=700(148.41)=103889
(d) the population will reach 1050 when P(t)=1050, or
700e^0.5t=1050 =>
e^(0.5t)=1050/700=1.5
take log
0.5t=log(3/2) =>
t=2log(3/2)=0.8109 days
=approximately 19.5 hours
(e) population will double when P(t)=2*700=1400
700e^(0.5t)=1400 =>
e^(0.5t)=1400/700=2
take log
0.5t=log(2)
t=2log(2)=1.3863
=33.27 hours.
Note: log(x) is log to the base e, or natural log. Log10 is mostly used in high-school.
