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?
Accepted Solution
A:
Answer:[tex]\frac{\ln 4}{k}[/tex]Step-by-step explanation:Given equation that shows the change in population with respect to time,[tex]\frac{dP}{dt}=kP[/tex][tex]\frac{dP}{P}=kdt[/tex]On integrating,[tex]\ln P = kt + C[/tex][tex]\implies P = e^{kt+C}[/tex][tex]P=e^C e^{kt}[/tex]Put [tex]e^C=P_0[/tex][tex]P = P_0 e^{kt}[/tex]According to the question,If P = A if t = 0,[tex]A = P_0 e^0\implies A = P_0[/tex]Thus, the required function,[tex]P = A e^{kt}[/tex]If the final population = 4A,[tex]4A = A e^{kt}[/tex][tex]4 = e^{kt}[/tex]Taking natural log both sides,[tex]\ln 4 = \ln e^{kt}[/tex][tex]\ln 4 = kt\ln e[/tex][tex]\ln 4 = kt[/tex][tex]\implies t = \frac{\ln 4}{k}[/tex]Hence, after [tex]\frac{\ln 4}{k}[/tex] time, the population will be quadruple its initial value.