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प्रश्न
The rate of growth of bacteria is proportional to the number present. If initially, there were 1000 bacteria and the number doubles in 1 hour, find the number of bacteria after `2 1/2` hours.
[Take `sqrt2 = 1.414`]
उत्तर
Let x be the number of bacteria at time t.
Then the rate of increase is `"dx"/"dt"` which is proportional to x.
∴ `"dx"/"dt" prop "x"`
∴ `"dx"/"dt"` = kx, where k is a constant
∴ `"dx"/"x" = "k dt"`
On integrating, we get
`int "dx"/"x" = "k" int "dt" + "c"`
∴ log x = kt + c
Initially, i.e. when t = 0, x = 1000
∴ log 1000 = k × 0 + c ∴ c = log 1000
∴ log x = ly + log 1000
∴ log x - log 1000 = kt
∴ log`("x"/1000) = "kt"` ...(1)
Now, when t = 1, x = 2 × 1000 = 2000
∴ `log (2000/1000)` = k ∴ k = log 2
∴ (1) becomes, log`("x"/1000) = t log 2
If t = `2 1/2 = 5/2,` then
log`("x"/1000) = 5/2 log 2 = log (2)^(5/2)`
∴ `("x"/1000) = (2)^(5/2) = 4sqrt2 = 4 xx 1.414 = 5.656`
∴ x = 5.656 × 1000 = 5656
∴ number of bacteria after `2 1/2`hours = 5656.
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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.
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?
