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प्रश्न
In a certain culture of bacteria, the rate of increase is proportional to the number present. If it is found that the number doubles in 4 hours, complete the following activity to find the number of times the bacteria are increased in 12 hours.
उत्तर
Let N be the number of bacteria present at time t.
∵ The rate of increase is proportional to the number present.
`(dN)/(dt) ∝ N`
∴ `(dN)/(dt)` = k. N
Where k is the constant of proportionality.
∴ `(dN)/N` = k. dt
Integrating both sides,
`int 1/N. dN = k int dt`
log N = k. t + C ......(1)
Now at, t = 0, N = N0
∴ From (1),
log N0 = C ......(2)
At t = 4, N = 2N0
∴ From (1),
log 2N0 = 4K + log N0 ......[∵ C = log N0]
⇒ log 2N0 – log N0 = 4k
⇒ `log((2N_0)/N_0)` = 4k
⇒ log 2 = 4k
⇒ k = `1/4` log 2
When t = 12,
log N = `1/4 log 2 xx 12 + log N_0`
⇒ `log (N/N_0)` = 3 log 2 = log23
⇒ `log(N/N_0)` = log 8
⇒ N = 8N0.
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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`
∴ `(dN)/dt` = KN, where K is constant of proportionality
∴ `(dN)/N` = k . dt
∴ `int 1/N dN = K int 1 . dt`
∴ log N = `square` + C ...(1)
When t = 0, N = N0 where N0 is initial number of bacteria.
∴ log N0 = K × 0 + C
∴ C = log N0
Also when t = 4, N = 2N0
∴ log (2 N0) = K . 4 + `square` ...[From (1)]
∴ `log((2N_0)/N_0)` = 4K,
∴ log 2 = 4K
∴ K = `square` ...(2)
Now N = ? when t = 12
From (1) and (2)
log N = `1/4 log 2 . (12) + log N_0`
log N – log N0 = 3 log 2
∴ `log(N_0/N_0)` = `square`
∴ N = 8 N0
∴ Bacteria are increased 8 times in 12 hours.
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 1,00,000, when will the city have population 4,00,000?
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
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∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
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∴ log`("p"/(1,00,000)) = (1/25log2).t`
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∴ t = `square ` years
