Answer to Question #247662 in Calculus for JaytheCreator

Question

The population 𝑃 of a town decreases at a rate proportional to the number by which the population exceeds a 1000. (i) Form the differential equation involving 𝑃, (ii) Show that 𝑃 = 1000 + 𝐴𝑒 −𝑘𝑡, where 𝐴 and 𝑘 are constants and 𝑃 is the solution of the differential equation, (iii) Initially, the population of the town was 2500. Ten years later, it had fallen to 1,900. In how many years will the population be 1500? (iv) What does the mathematical model predict about the population of the town in the long term?

Answer 1

Let $P(t)=$ the size of population at time $t.$

(i) Form the differential equation involving 𝑃

$$\dfrac{dP}{dt}=-k(P-1000), k>0$$

(ii)

$$\dfrac{dP}{P-1000}=-kdt$$

Integrate

$$\int \dfrac{dP}{P-1000}=-\int kdt$$

$$\ln(|P-1000|)=-kt+\ln A$$

$$P-1000=Ae^{-kt}$$

$$P(t)=1000+Ae^{-kt}$$

(iii)

Given $P(0)=2500, P(10)=1900$

$$2500=1000+Ae^{-k(0)}=>A=1500$$

$$P(t)=1000+1500e^{-kt}$$

$$1900=1000+1500e^{-k(10)}$$

$$e^{10k}=\dfrac{1500}{900}$$

$$10k=\ln (\dfrac{5}{3})$$

$$k=0.1\ln (\dfrac{5}{3})$$

$$P(t)=1000+1500(\dfrac{3}{5})^{0.1t}$$

In how many years will the population be 1500?

$$P(t_1)=1000+1500(\dfrac{3}{5})^{0.1t_1}=1500$$

$$(\dfrac{3}{5})^{0.1t_1}=\dfrac{1}{3}$$

$$0.1t_1=\dfrac{\ln3}{\ln(5/3)}$$

$$t_1=\dfrac{10\ln3}{\ln(5/3)}$$

$$t_1=21.5\ years$$

(iv)

$$\lim\limits_{t\to\infin}P(t)=\lim\limits_{t\to\infin}(1000+1500(\dfrac{3}{5})^{0.1t})$$

$$=1000+1500\lim\limits_{t\to\infin}((\dfrac{3}{5})^{0.1t})=1000+1500(0)$$

$$=1000$$

The size of population approaches to $1000$ in the long term.