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Question
The population of city doubles in 80 years, in how many years will it be triple when the rate of increase is proportional to the number of inhabitants. `("Given" log3/log2 = 1.5894)`
Solution: Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
∴ `"dp"/"p"` = kdt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e., when t = 0, let p = N
∴ log N = k × 0 + c
∴ c = `square`
When t = 80, p = 2N
∴ log 2N = 80k + log N
∴ log 2N – log N = 80k
∴ `log ((2"N")/"N")` = 80k
∴ log (2) = 80k
∴ k = `square`
∴ p = 3N, then t = ?
∴ log p = `log2/80 "t" + log "N"`
∴ log 3N – log N = `square`
∴ t = `square` = `square` years
Solution
Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" ∝ "p"`
∴ `"dp"/"dt"` = kp, where k is a constant
∴ `"dp"/"p"` = kdt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e., when t = 0, let p = N
∴ log N = k × 0 + c
∴ c = log N
When t = 80, p = 2N
∴ log 2N = 80k + log N
∴ log 2N – log N = 80k
∴ `log ((2"N")/"N")` = 80k
∴ log (2) = 80k
∴ k = `1/80 log 2`
∴ p = 3N, then t = ?
∴ log p = `log2/80 "t" + log "N"`
∴ log 3N – log N = `"t"/80 log 2`
∴ `log ((3"N")/"N") = "t"/80 log 2`
∴ log 3 = `"t"/80 log 2`
∴ t = `80 (log3)/(log 2)`
= 80(1.5894)
= 127.15 years
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On integrating, we get
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∴ `log (200000/100000)` = 25k
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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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Solution: Let p be the population at time t.
Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
∴ `"dp"/"dt" prop "p"`
∴ `"dp"/"dt"` = kp, where k is a constant.
∴ `"dp"/"p"` = k dt
On integrating, we get
`int "dp"/"p" = "k" int "dt"`
∴ log p = kt + c
Initially, i.e. when t = 0, let p = 30000
∴ log 30000 = k × 0 + c
∴ c = `square`
∴ log p = kt + log 30000
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∴ `log (40000/30000) = 40"k"`
∴ k = `square`
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∴ p = `square`
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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 `("d"x)/"dt"` which is proportional to x.
∴ `("d"x)/"dt" ∝ x`
∴ `("d"x)/"dt"` = kx, where k is a constant
∴ `("d"x)/x` = kdt
On integrating, we get
`int ("d"x)/x = "k" int "dt"`
∴ log x = kt + c .....(1)
∴ x = aekt where a = ec
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∴ N = aek(0)
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∴ a = N, x = Nekt ......(2)
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From equation (2), 2N = Ne4k
∴ e4k = 2
∴ ek = `square`
Now we have to find out t, when x = 16N
From equation (2),
16N = Nekt
∴ 16 = ekt
∴ `"t"/4 = square` hours
Hence, number of bacteria will be 16N in `square` hours
If the population grows at the rate of 8% per year, then the time taken for the population to be doubled, is (Given log 2 = 0.6912).
The rate of increase of bacteria in a certain culture is proportional to the number present. If it doubles in 7 hours, then in 35 hours its number would be ______.
The rate of disintegration of a radioactive element at time t is proportional to its mass at that time. The original mass of 800 gm will disintegrate into its mass of 400 gm after 5 days. Find the mass remaining after 30 days.
Solution: If x is the amount of material present at time t then `dx/dt = square`, where k is constant of proportionality.
`int dx/x = square + c`
∴ logx = `square`
x = `square` = `square`.ec
∴ x = `square`.a where a = ec
At t = 0, x = 800
∴ a = `square`
At t = 5, x = 400
∴ e–5k = `square`
Now when t = 30
x = `square` × `square` = 800 × (e–5k)6 = 800 × `square` = `square`.
The mass remaining after 30 days will be `square` mg.
