The rate of decay of certain substances is directly proportional to the amount present at that instant. Initially, there is 25 gm of certain substance and two hours later it is found that 9 gm are left. Find the amount left after one more hour.
Solution
Let x gm be the amount of the substance left at time t.
Then the rate of decay is `"dx"/"dt"`, which is proportional to x.
∴ `"dx"/"dt" prop "x"`
∴ `"dx"/"dt"` = - kx, where k > 0
∴ `1/"x" "dx"` = - k dt
On integrating, we get
`int1/"x" "dx" = - "k" int "dt" + "c"`
∴ log x = - kt + c
Initially i.e. when t = 0, x = 25
∴ log 25 = - k ×0 + c ∴ c = log 25
∴ log x = - kt + log 25
∴ log x - log 25 = - kt
∴ log`("x"/25) = - "kt"` ....(1)
Now, when t = 2, x = 9
∴ `log(9/25) = - 2"k"`
∴ `- 2"k" = log (3/5)^2 = 2 log(3/5)`
∴ k = - log`(3/5)`
∴ (1) becomes, `log ("x"/25) = "t" log(3/5)`
When t = 3, then
`log ("x"/25) = 3 log (3/5) = log (3/5)^3`
∴ `"x"/25 = 27/125` ∴ x = `27/5`
∴ the amount left after 3 hours = `27/5` gm.
