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Exercise 10.8 · Q2

Q.Find the population of a city at any time tt, given that the rate of increase of population is proportional to the population at that instant and that in a period of 4040 years the population increased from 3,00,0003,00,000 to 4,00,0004,00,000.

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Take t=0t=0 at the first census reading, fix kk from the second reading 4040 years later, and write the population formula for general tt.

Step 1. Set up the model. dPdt=kP ⟹ P(t)=P0ekt\dfrac{dP}{dt}=kP\ \Longrightarrow\ P(t)=P_0e^{kt}, with P0=3,00,000P_0=3{,}00{,}000 at t=0t=0.

Step 2. Use the 4040-year data point. P(40)=4,00,000 ⟹ 3,00,000 e40k=4,00,000 ⟹ e40k=43 ⟹ k=140ln⁡43P(40)=4{,}00{,}000\ \Longrightarrow\ 3{,}00{,}000\,e^{40k}=4{,}00{,}000\ \Longrightarrow\ e^{40k}=\dfrac43\ \Longrightarrow\ k=\dfrac1{40}\ln\dfrac43.

Step 3. Write P(t)P(t) using this kk. P(t)=3,00,000 et40ln⁡(4/3)=3,00,000(43)t/40P(t)=3{,}00{,}000\,e^{\frac{t}{40}\ln(4/3)}=3{,}00{,}000\left(\dfrac43\right)^{t/40}.

✓Final answer

P(t)=3,00,000(43)t/40P(t)=3{,}00{,}000\left(\dfrac43\right)^{t/40}

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