Question 1087535
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.