Question

A population grows exponentially according to the differential equation dP, dt equals k times P , where P is the population, t is time, and k is a positive constant. If P(0) = A, what is the time for the population to quadruple its initial value?

Answer 1

`(\ln 4)/(k)`

Explanation:

Given equation that shows the change in population with respect to time,

`(dP)/(dt)=kP`

`(dP)/(P)=kdt`

On integrating,

`\ln P = kt + C`

`\implies P = e^(kt+C)`

`P=e^C e^(kt)`

Put
`e^C=P_0`

`P = P_0 e^(kt)`

According to the question,

If P = A if t = 0,

`A = P_0 e^0\implies A = P_0`

Thus, the required function,

`P = A e^(kt)`

If the final population = 4A,

`4A = A e^(kt)`

`4 = e^(kt)`

Taking natural log both sides,

`\ln 4 = \ln e^(kt)`

`\ln 4 = kt\ln e`

`\ln 4 = kt`

`\implies t = (\ln 4)/(k)`

Hence, after
`(\ln 4)/(k)` time, the population will be quadruple its initial value.