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Question 102 of 126

Q.The rate at which the population of a city increases at any time is proportional to the population at that time. If there were 1,30,000 people in the city in 1960 and 1,60,000 in 1990, what approximate population may be anticipated in 2020? [log⁡e(1613)=0.2070, e0.42=1.52]\left[\log_e\left(\dfrac{16}{13}\right) = 0.2070,\ e^{0.42} = 1.52\right]

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2018Subjective· 10mImportance★★★★★
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Modelling the population as P=P0ektP=P_0e^{kt}, the growth rate kk is found from the 1960–1990 data using the given log⁡e(16/13)=0.2070\log_e(16/13)=0.2070, and then projected to 2020 using the given e0.42=1.52e^{0.42}=1.52, giving about 1,97,6001{,}97{,}600 people.

  1. Since the rate of increase of population is proportional to the population at that time, dPdt=kP\dfrac{dP}{dt}=kP for a constant kk, whose solution is P(t)=P0ektP(t)=P_0e^{kt}, where P0P_0 is the population at t=0t=0.
  2. Take t=0t=0 at the year 19601960, so P0=1,30,000P_0 = 1{,}30{,}000.
  3. In 19901990, t=30t=30 years, and P(30)=1,60,000P(30)=1{,}60{,}000:   1,60,000=1,30,000 e30k\;1{,}60{,}000 = 1{,}30{,}000\,e^{30k}.
  4. So e30k=1,60,0001,30,000=1613e^{30k} = \dfrac{1{,}60{,}000}{1{,}30{,}000} = \dfrac{16}{13}.
  5. Taking log⁡e\log_e of both sides: 30k=log⁡e ⁣(1613)=0.207030k = \log_e\!\left(\dfrac{16}{13}\right) = 0.2070 (given).
  6. So k=0.207030=0.0069k = \dfrac{0.2070}{30} = 0.0069 per year. …

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