Question

The rate of growth of the population of a city at any time t is proportional to the size of the population. For a certain city it is found that the constant of proportionality is 0.04. Find the population of the city after 25 years if the initial population is 10,000.[Take e = 2.7182]

Answer 1

I think it is 250000

Answer 2

Answer:

27182.

Step-by-step explanation:

Let the population of the city at any time t be $\rm N(t)$.

Given that, the rate of growth of the population of a city at any time t is proportional to the size of the population.

Therefore,

$\rm \dfrac{d N(t)}{dt}\propto N(t).\\\dfrac{d N(t)}{dt}=k N(t).$

k is the constant of proportionality.

The population of the city at any time t can be found as

$\rm \dfrac{d N(t)}{dt}=k N(t).\\\dfrac{dN(t)}{N(t)} = k\ dt\\\\\text{Integrating both the sides,}\\\int \dfrac{dN(t)}{N(t)} = \int k\ dt\\\ln(N(t))=kt+C$

C is the constant of integration.

We know that at time t = 0 s, the population of the city is initial population, $\rm N_o$.

Therefore,

$\rm \ln(N(t))_{t=0}=k\cdot 0+C = \ln(N_o)\\\Rightarrow C=\ln(N_o).$

Putting this value of C,

$\rm \ln(N(t))=kt+\ln(N_o)\\\ln(N(t))-\ln(N_o)=kt\\\ln \left (\dfrac{N(t)}{N_o} \right ) = kt\\\dfrac{N(t)}{N_o}=e^{kt}\\N(t) = N_oe^{kt}$

Given values are:

• Proportionality constant, $\rm k=0.04 year^{-1}.$
• Initial population, $\rm N_o = 10,000$

NOTE: Here I have taken the units of proportionality constant to be $\rm year^{-1}$ as it is not mentioned in the question. If the unit of the constant is different from what I have taken then the answer would be different.

The population of the city is asked after 25 years, such that, $\rm t=25\ years.$ is given by

$\rm N(t=25\ years) = 10000\times e^{0.04\times 25} \\= 10000\times e^1\\=10000\times 2.7182\\=27182.$