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महाराष्ट्र राज्य शिक्षण मंडळएचएससी वाणिज्य (इंग्रजी माध्यम) इयत्ता १२ वी

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 4,00,000? - Mathematics and Statistics

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प्रश्न

The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and the present population is 1 lac, when will the city have population 4,00,000?

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उत्तर

Let ‘x’ be the population at time ‘t’ years.

∴ `dx/dt prop x`

∴ `dx/dt = kx`, where k is the constant of proportionality.

∴ `dx/x = kdt`

Integrating on both sides, we get

`int dx/x = int kdt`

∴ logx = kt + c  …(i)

When t = 0, x = 50,000

∴ log(50,000) = k(0) + c

∴ c = log(50,000)

∴ logx = kt + log(50,000) ...(ii) [From (i)]

When t = 25, x = 1,00,000, we have

log(1,00,000) = 25k + log(50000)

∴ log2 = 25k

∴ k = `1/25 log 2` …(iii)

When x = 4,00,000, we get

`log(4,00,000) = [1/25 log (2)]t + log(50,000)` ...[From (ii) and (iii)]

∴ `log[400000/50000] = t/25 log2`

∴ log8 = `t/25 log2`

∴ 3log2 = `t/25 log2`

∴ 3 = `t/25`

∴ t = 75 years.

Thus, the population will be 4,00,000 after 75 – 25 = 50 years from present date.

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पाठ 8: Differential Equation and Applications - Miscellaneous Exercise 8 [पृष्ठ १७३]

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बालभारती Mathematics and Statistics 1 (Commerce) [English] Standard 12 Maharashtra State Board
पाठ 8 Differential Equation and Applications
Miscellaneous Exercise 8 | Q 4.09 | पृष्ठ १७३
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