Let P(t) be the population (in millions) of a certain city t years after 1990, a

Question

Let P(t) be the population (in millions) of a certain city t years after 1990, and suppose that P(t) satisfies the differential equation P' = 02P(t), P(0) = 2. The initial population was 2 million. (c) What is the growth constant? The growth constant is 02. (d) What was the population in 2000? The population in 2000 was 2.44 million. (e) Use the differential equation to determine how fast the population is growing when it reaches 5 million people. The growth rate is 100,000 people per year. (f) How large is the population when it is growing at the rate of 80,000 people per year? The population is million.

Explanation / Answer

intital population P(0)=2 million

c) growth constant =0.02

d) here P'(t)=0.02P(t)

(1/P(t))dP =0.02dt

ln(P(t)) =0.02t+C where C is constant

P(t) =e0.02t+C

at t=0 ; P(0) =2 putting it in above:

2 =e0.02*0+C

eC =2

therefore P(t) =2*e0.02t  million

hence at t=10; P(2000) =2*e0.02*10 =2.44 million

e)as P(t)=5 million

therefore P'(t) =0.02*5 million =100,000 people

f) here P'(t)=80000

therefore P(t)=P'(t)/0.02 =(80000/0.02) =4000000 people =4 million people

please revert for any clarification

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