Answer to Question #122450 in Differential Equations for JSE

Question #122450
The population of a town grows at a rate proportional to the population present at time t. The initial population of 500 increases by 15% in 10 years. What will be the population in 50 years? (Round your answer to the nearest person.)

____ persons

How fast is the population growing at t = 50? (Round your answer to two decimal places.)

______ persons / year
1
Expert's answer
2020-06-22T15:35:01-0400

According to the question,

dPdt=kP\frac{dP}{dt} = kP where P is population at any time


solving equation,

dPP=kdt\int \frac{dP}{P} = \int kdt


lnP=kt+ClnP = kt + C (1)

Applying the conditions,

at t=0, P = 500


ln500=Cln500 = C (2)


at t=10, P= 575


ln(575)=k(10)+ln(500)ln(575) = k(10) + ln(500)

k=110ln(575500)=0.0139762k = \frac{1}{10} ln (\frac{575}{500}) = 0.0139762


So Equation of the Population growth is given by


P=500e0.0139762tP = 500 e^{0.0139762t} (3)



Population in 50 years,

P=500e0.013976250=1005.61006P = 500 e^{0.0139762*50} = 1005.6 \approx 1006 people



Rate of population growth is

dPdt=d(500e0.0139762t)dt=(5000.0139762)e0.0139762t\frac {dP}{dt} = \frac {d (500 e^{0.0139762t})}{dt} = (500*0.0139762)e^{0.0139762t}

putting t = 50, we get


dPdt=(5000.0139762)e0.013976250=14\frac {dP}{dt} =(500*0.0139762) e^{0.0139762*50} = 14 people/year (approx)


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Comments

Assignment Expert
28.02.21, 14:07

Dear Kabilen, please use the panel for submitting new questions.

Kabilen
27.02.21, 15:42

The population of a town grows at a rate proportional to the population present at time t. The initial population of 5000 increases by 50% in 10 years. What will be the population in 30 years?

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