MHT CET · Maths · Differential Equations
The population of a city increases at a rate proportional to the population at that time. If the population of the city increase from 20 lakhs to 40 lakhs in 30 years, then after another 15 years the population is
- A \(10 \sqrt{2}\) lakhs
- B \(40 \sqrt{2}\) lakh
- C \(30 \sqrt{2}\) lakhs
- D None of these
Answer & Solution
Correct Answer
(B) \(40 \sqrt{2}\) lakh
Step-by-step Solution
Detailed explanation
We have \(\frac{\mathrm{dP}}{\mathrm{dt}} \propto \mathrm{P} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\mathrm{kP}\)
\(
\therefore \int \frac{\mathrm{dP}}{\mathrm{dt}}=\int \mathrm{k} d t \Rightarrow \log \mathrm{P}=\mathrm{kt}+\mathrm{c}
\)
From given data, we write
\(\log 20=\mathrm{k}(0)+\mathrm{c} \Rightarrow \mathrm{c}=\log 20 \)
\( \therefore \log \mathrm{P}=\mathrm{kt}+\log 20 \)
\( \text { Also } \log 40=30 \mathrm{k}+\log 20 \)
\( \therefore \log 40-\log 20=30 \mathrm{k} \Rightarrow \mathrm{k}=\frac{1}{30} \log 2 \)
\( \therefore \log \mathrm{P}=\left(\frac{\log 2}{30}\right) \mathrm{t}+\log 20 \)
\( \text { When } \mathrm{t}=30+15=45 \)
\( \therefore \log \mathrm{P}=\left(\frac{\log 2}{30}\right)(45)+\log 20=(\log 2)\left(\frac{3}{2}\right)\) \(+\log 20 \)
\( =\log (2)^{\frac{3}{2}}+\log 20=\log (2 \sqrt{2} \times 20) \)
\( \therefore \mathrm{P}=40 \sqrt{2} \text { lakhs}\)
\(
\therefore \int \frac{\mathrm{dP}}{\mathrm{dt}}=\int \mathrm{k} d t \Rightarrow \log \mathrm{P}=\mathrm{kt}+\mathrm{c}
\)
From given data, we write
\(\log 20=\mathrm{k}(0)+\mathrm{c} \Rightarrow \mathrm{c}=\log 20 \)
\( \therefore \log \mathrm{P}=\mathrm{kt}+\log 20 \)
\( \text { Also } \log 40=30 \mathrm{k}+\log 20 \)
\( \therefore \log 40-\log 20=30 \mathrm{k} \Rightarrow \mathrm{k}=\frac{1}{30} \log 2 \)
\( \therefore \log \mathrm{P}=\left(\frac{\log 2}{30}\right) \mathrm{t}+\log 20 \)
\( \text { When } \mathrm{t}=30+15=45 \)
\( \therefore \log \mathrm{P}=\left(\frac{\log 2}{30}\right)(45)+\log 20=(\log 2)\left(\frac{3}{2}\right)\) \(+\log 20 \)
\( =\log (2)^{\frac{3}{2}}+\log 20=\log (2 \sqrt{2} \times 20) \)
\( \therefore \mathrm{P}=40 \sqrt{2} \text { lakhs}\)
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