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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? - Mathematics and Statistics

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Question

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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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 = log 100000

∴ 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 = `1/25 log 2`

∴ equation (i) becomes, `log("p"/100000)` = `"t"/25 log 2`

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 = `25 (2 log 2)/(log 2)`

∴ t = 50 years

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Application of Differential Equations
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Chapter 1.8: Differential Equation and Applications - Q.6

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SCERT Maharashtra Mathematics and Statistics (Commerce) [English] 12 Standard HSC
Chapter 1.8 Differential Equation and Applications
Q.6 | Q 3

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