The population of a community is known to increase at a rate proportional to the number of
people present at time t. The initial population
P0
has doubled in 5 years.
Suppose it is known that the population is 11,000 after 3 years. What was the initial population
P0? (Round your answer to one decimal place.)
P0 = 1
What will be the population in 10 years? (Round your answer to the nearest person.)
2 persons
How fast is the population growing at
t = 10?
(Round your answer to the nearest person.)
3 persons/year
Solution
dP/dt = P(t)
P = ? P(t) dt
P = e^(kt)
So
P(t) = P0*e^(kt)
If P(5) = 2P0:
P(5) = P0*e^(k5)
2P0 = P0*e^(5k)
2 = e^(5k)
ln(2) = 5k
k = ln(2)/5
P(t) = P0*e^(ln(2)t/5)
P(t) = P0*2^(t/5)
2) P(t) = 11,000 when t = 3
P(3) = 11,000
P0*2^(3/5) = 11,000
P0 = 11000/(2^(3/5))
P0 = 7257.29, or rounded to 7257
P(10) = 7257*2^(10/5)
P(10) = 7257*2.
Historical philosophical, theoretical, and legal foundations of special and i...
The population of a community is known to increase at a rate proport.pdf
1. The population of a community is known to increase at a rate proportional to the number of
people present at time t. The initial population
P0
has doubled in 5 years.
Suppose it is known that the population is 11,000 after 3 years. What was the initial population
P0? (Round your answer to one decimal place.)
P0 = 1
What will be the population in 10 years? (Round your answer to the nearest person.)
2 persons
How fast is the population growing at
t = 10?
(Round your answer to the nearest person.)
3 persons/year
Solution
dP/dt = P(t)
P = ? P(t) dt
P = e^(kt)
So
P(t) = P0*e^(kt)
If P(5) = 2P0:
P(5) = P0*e^(k5)
2P0 = P0*e^(5k)
2 = e^(5k)
ln(2) = 5k
k = ln(2)/5
P(t) = P0*e^(ln(2)t/5)
P(t) = P0*2^(t/5)
2) P(t) = 11,000 when t = 3
P(3) = 11,000
2. P0*2^(3/5) = 11,000
P0 = 11000/(2^(3/5))
P0 = 7257.29, or rounded to 7257
P(10) = 7257*2^(10/5)
P(10) = 7257*2
