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प्रश्न
Find the population of a city at any time t, given that the rate of increase of population is proportional to the population at that instant and that in a period of 40 years the population increased from 3,00,000 to 4,00,000
उत्तर
Let P denote the population of a city
Given `"dP"/"dt"` = kP
This equation can be written as
`"dP"/"P"` = kdt
Taking integration on both sides, we get
`int "dP"/"P" = "k" int "dt"`
log P = kt + log c
log P – log c = kt
`log ("P"/"c")` = kt
`"P"/"c"` = ekt
P = cekt ........(1)
Initial condition:
Given when t = 0, P = 3,00,000
Equation (1) becomes,
3.0. 000 = cek(0) = ce°
3,00,000 = c .......[∵ e° = 1]
∴ (1) ⇒ P = 3,00,000 ekt ........(2)
Again when t = 40, P = 4,00,000
Equation (2) becomes,
4,00,000 = 3,00,000 e40k
`4,00,000/3,00,000` = e40k
`4/3` = e40k
Taking log,
`log (4/3)` 40k
k = `1/40 log 4/3`
k = `log (4/3)^(1/40)` ........[∵ n log m = log mn]
Substituting k values in equation (2), we get
p = 3,00,000 `"e"^(log(4/3)^(1/40))`
P = 3,00,000 `(4/3)^(1/40)` ........`[∵ "e"^(log"a"^x) = "a"^x]`
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