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Question
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.
Solution
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 x be the number of bacteria in the culture at time t.
Then the rate of increase of x is `"dx"/"dt"` which is proportional to x.
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∴ `square`
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