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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 `5/2` hours `("Given" sqrt(2) = 1.414)`
Solution
Let ‘x’ be the number of bacteria present at time ‘t’.
∴ `("d"x)/"dt" ∞ x`
∴ `("d"x)/"dt"` = kx,
where k is the constant of proportionality.
∴ `("d"x)/x` = kdt
Integrating on both sides, we get
`int ("d"x)/x = "k" int "dt"`
∴ log x = kt + c .....(i)
When t = 0, x = 1000
∴ log (1000) = k(0) + c
∴ c = log (1000)
∴ log x = kt + log (1000) ....(ii) ....[From (i)]
When t = 1, x = 2000
∴ log (2000) = k(1) + log (1000)
∴ log (2000) − log (1000) = k
∴ k = `log(2000/1000)`
= log 2 .....(iii)
When t =`5/2`, we get
log x = `5/2 "k" + log(1000)` .......[From (ii)]
∴ log x = `(5/2) log 2 + log(1000)` .....[From (iii)]
= `log (2^(5/2)) + log(1000)`
= `log (4sqrt(2)) + log(1000)`
= `log (4000 sqrt(2))`
= log (4000 × 1.414)
∴ log x = log (5656)
∴ x = 5656
Thus, there will be 5656 bacteria after `5/2` hours.
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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.
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)
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∴ 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`
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∴ 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?
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
∴ 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
`log ((4,00,000)/(1,00,000)) = (1/25log2).t`
∴ `log 4 = (1/25 log2).t`
∴ t = `square ` years
