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Question
Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3 mm and 1 hour later has been reduced to 2 mm, find an expression for the radius of the raindrop at any time t.
Solution
Let r be the radius, V be the volume and S be the surface area of the spherical raindrop at time t.
Then V = `4/3 pi"r"^3` and S = 4πr2
The rate at which the raindrop evaporates is `"dV"/"dt"` which is proportional to the surface area.
∴ `"dV"/"dt" prop "S"`
∴ `"dV"/"dt"` = - kS, where k > 0 ...(1)
Now, V = `4/3pi"r"^3` and S = 4πr2
∴ `"dV"/"dt" = (4pi)/3 xx 3"r"^2 "dr"/"dt" = 4 pi "r"^2 "dr"/"dt"`
∴ (1) becomes, `4 pi "r"^2 "dr"/"dt" = - "k"(4 pi "r"^2)`
∴ `"dr"/"dt"` = - k
∴ dr = - k dt
On integrating, we get
`int "dr" = - "k" int "dt" + "c"`
∴ r = - kt + c
Initially, i.e. when t = 0, r = 3
∴ 3 = - k × 0 + c ∴ c = 3
∴ r = - kt + 3
When t = 1, r = 2
∴ 2 = - k × 1 + 3
∴ k = 1
∴ r = - t + 3
∴ r = 3 - t, where 0 ≤ t ≤ 3.
This is the required expression for the radius of the raindrop at any time t.
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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"`
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`square` = kt + c ...(i)
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