math problem_1－lgzrelv的部落格

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math problem

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math problem http://hk.myblog.yahoo.com/jw!Nfr4uoaTAkaBuxVryujr4w-- part 1 answer should be dP/dt=1/265 P(1-P/100) part 2 . 5.49 billion part 3. in billion:7.81, 27.72 part 4. in billion: 5.48, 7.61, 22.41

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The population of the world was about 5.3 billion in 1990. birth rates in the 1990s ranged from 35 to 40 million per year and death rates ranged from 15 to 20 million per year. Let’s assume that the carrying capacity for world population 100 billion. write the logistic differential equation for these data. (b/c the initial population is small compared to the carrying capacity, you can take k to be an estimate of the initial relative growth rate.) use the logistic model to estimate the world population in the year 2000 & compare with the actual population of 6.1 billion use the logistic model to predict the world population in the years 2100 and 2500 what are your predictions if the carrying capacity is 50 billion? ANSWER 1 dP/dt=rP(K-P)/K where K=100, r=(35-15)/5300=1/265 So the logistic differential equation is dP/dt=1/265 P(1-P/100) 2 Solve the logistic differential equation , we get P(t)=KP0/[(K-P0)e^(-rt)+P0] substitute t=10 P(10) =100*5.3/[(100-5.3)e^(-0.0037*10)+5.3] =5.49 billion less than the actual population of 6.1 billion P(110) =100*5.3/[(100-5.3)e^(-0.0037*110)+5.3] =7.81 billion P(510) =100*5.3/[(100-5.3)e^(-0.0037*510)+5.3] =27.72 billion using the logistic model , the world population in the years 2100 and 2500 are 7.81 and 27.72 billion respectively 4 The formula is P(t) =50(5.3)/[(50-5.3)e^(-0.0037t)+5.3] =265/[(44.7)e^(-0.0037t)+5.3] P(10)=265/[(44.7)e^(-0.0037*10)+5.3]=5.48 billion P(110)=265/[(44.7)e^(-0.0037*110)+5.3]=7.61 billion P(510)=265/[(44.7)e^(-0.0037*510)+5.3]=22.41 billion using the logistic model , the world population in the years 2000 2100 and 2500 are 5.48 7.61 and 22.41 billion respectively