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Question
If the population of a town increases at a rate proportional to the population at that time. If the population increases from 40 thousand to 60 thousand in 40 years, what will be the population in another 20 years? `("Given" sqrt(3/2) = 1.2247)`
Solution
Let 'x' be the population at time 't'.
∴ `("d"x)/("d"t) ∞ x`
∴ `("d"x)/("d"t)` = kx,
where k is the constant of proportionality.
`("d"x)/"dt"` = k dt
Integrating on both sides, we get
`int ("d"x)/x = "k" int 1 "dt"`
∴ log x = kt + c
∴ x = a.ekt, ......(i)
where a = ec
When t = 0, x = 40,000
∴ 40000 = a.e0
∴ a = 40000
∴ x = 40000.ekt ......(ii) ......[From (i)]
When t = 40, P = 60000
∴ 60000 = 40000.e40k
∴ e40k = `60000/40000 = 3/2` ......(iii)
Now, we have to find x when t = 40 + 20
= 60 years
∴ P = 40000.e60k .....[From (iii)]
= `40000 ("e"^(40"k"))^(3/2)`
= `40000 (3/2)^(3/2)` ......[From (iii)]
= `40000 (3/2) sqrt(3/2)`
= 60000(1.2247)
= 73482
∴ The required population will be 73482.
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Solution:
Let N be the number of bacteria present at time ‘t’.
Since the rate of increase of N is proportional to N, the differential equation can be written as –
`(dN)/dt αN`
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∴ `int 1/N dN = K int 1 . dt`
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∴ log N0 = K × 0 + C
∴ C = log N0
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∴ log (2 N0) = K . 4 + `square` ...[From (1)]
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∴ K = `square` ...(2)
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Bacteria increase at the rate proportional to the number of bacteria present. If the original number N doubles in 3 hours, find in how many hours the number of bacteria will be 4N?
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Let ‘p’ be the population at time ‘t’ years.
∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
(i) Where t = 0, p = 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
∴ from (ii) log2 = 25k
∴ k = `square`
∴ log`("p"/(1,00,000)) = (1/25log2).t`
(iii) ∴ when p = 4,00,000
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∴ t = `square ` years
