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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?
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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The rate of growth of population is proportional to the number present. If the population doubled in the last 25 years and present population is 1 lac., when will the city have population 4,00,000?
Solution: Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
∴ `"dp"/"p"` = kdt
On integrating, we get
`int "dp"/"p" = "k"int "dt"`
∴ log p = kt + c
Initially, i.e., when t = 0, let p = 100000
∴ log 100000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 100000
∴ log p – log 100000 = kt
∴ `log ("P"/100000)` = kt ......(i)
Since the number doubled in 25 years, i.e., when t = 25, p = 200000
∴ `log (200000/100000)` = 25k
∴ k = `square`
∴ equation (i) becomes, `log("p"/100000) = square`
When p = 400000, then find t.
∴ `log(400000/100000) = "t"/25 log 2`
∴ `log 4 = "t"/25 log 2`
∴ t = `25 (log 4)/(log 2)`
∴ t = `square` years
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∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
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∴ logx = `square`
x = `square` = `square`.ec
∴ x = `square`.a where a = ec
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