# The rate of decay of a certain substances is directly proportional to the amount present at that instant. - Mathematics and Statistics

Sum

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.

Concept: Application of Differential Equations
Is there an error in this question or solution?
Chapter 6: Differential Equations - Exercise 6.6 [Page 213]

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