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Question
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`]
Solution
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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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, find the number of times the bacteria are increased in 12 hours.
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∴ `(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.
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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 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"`
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where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
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∴ 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`
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