Question

The population P of a town decreases at a rate proportional to the number by which the population exceeds-10 30, proportionality constant being k > O, Find Population at any time t, given initial population of the town being 2500., then $P=1000+1500e^{-kt}$ then k= ?

• A. $k=\frac{1}{10}\ln \left(\frac{5}{3}\right)$
• B. $k=\frac{-1}{10}\ln \left(\frac{5}{3}\right)$
• C. $k=\ln \left(\frac{5}{3}\right)$
• D. $k=\frac{1}{10}\ln \left(\frac{3}{5}\right)$

Answer 1

The correct option is A $k=\frac{1}{10}\ln \left(\frac{5}{3}\right)$

$\frac{dP}{dt}\alpha (P-100)\Rightarrow \frac{-dP}{dt}=K(P-1000)\Rightarrow \int -dPP-1000=\int kt\Rightarrow ln(P-1000)=-kt+c$

At t=0

$P=2500C=ln1500P=1000+1500e^{-kt}$

At t=10

$P=19001900=1000+1500e^{-10k}\Rightarrow e^{-10k}=\frac{900}{1500}=\frac{3}{5}\Rightarrow k=\frac{1}{10}ln\left(\frac{5}{3}\right)$